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Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers, and let
$$
S=x_{1}+x_{2}+\cdots+x_{n}
$$
Prove that
$$
\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \leq 1+S+\frac{S^{2}}{2!}+\frac{S^{3}}{3!}+\cdots+\frac{S^{n}}{n!}
$$
|
Let $\sigma_{k}$ be the $k$ th symmetric polynomial, namely
$$
\sigma_{k}=\sum_{\substack{|S|=k \\ S \subseteq\{1,2, \ldots, n\}}} \prod_{i \in S} x_{i},
$$
and more explicitly
$$
\sigma_{1}=S, \quad \sigma_{2}=x_{1} x_{2}+x_{1} x_{3}+\cdots+x_{n-1} x_{n}, \quad \text { and so on. }
$$
Then
$$
\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right)=1+\sigma_{1}+\sigma_{2}+\cdots+\sigma_{n}
$$
The expansion of
$$
S^{k}=\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{k}=\underbrace{\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(x_{1}+x_{2}+\cdots+x_{n}\right) \cdots\left(x_{1}+x_{2}+\cdots+x_{n}\right)}_{k \text { times }}
$$
has at least $k$ ! occurrences of $\prod_{i \in S} x_{i}$ for each subset $S$ with $k$ indices from $\{1,2, \ldots, n\}$. In fact, if $\pi$ is a permutation of $S$, we can choose each $x_{\pi(i)}$ from the $i$ th factor of $\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{k}$. Then each term appears at least $k$ ! times, and
$$
S^{k} \geq k!\sigma_{k} \Longleftrightarrow \sigma_{k} \leq \frac{S^{k}}{k!}
$$
Summing the obtained inequalities for $k=1,2, \ldots, n$ yields the result.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers, and let
$$
S=x_{1}+x_{2}+\cdots+x_{n}
$$
Prove that
$$
\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \leq 1+S+\frac{S^{2}}{2!}+\frac{S^{3}}{3!}+\cdots+\frac{S^{n}}{n!}
$$
|
Let $\sigma_{k}$ be the $k$ th symmetric polynomial, namely
$$
\sigma_{k}=\sum_{\substack{|S|=k \\ S \subseteq\{1,2, \ldots, n\}}} \prod_{i \in S} x_{i},
$$
and more explicitly
$$
\sigma_{1}=S, \quad \sigma_{2}=x_{1} x_{2}+x_{1} x_{3}+\cdots+x_{n-1} x_{n}, \quad \text { and so on. }
$$
Then
$$
\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right)=1+\sigma_{1}+\sigma_{2}+\cdots+\sigma_{n}
$$
The expansion of
$$
S^{k}=\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{k}=\underbrace{\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(x_{1}+x_{2}+\cdots+x_{n}\right) \cdots\left(x_{1}+x_{2}+\cdots+x_{n}\right)}_{k \text { times }}
$$
has at least $k$ ! occurrences of $\prod_{i \in S} x_{i}$ for each subset $S$ with $k$ indices from $\{1,2, \ldots, n\}$. In fact, if $\pi$ is a permutation of $S$, we can choose each $x_{\pi(i)}$ from the $i$ th factor of $\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{k}$. Then each term appears at least $k$ ! times, and
$$
S^{k} \geq k!\sigma_{k} \Longleftrightarrow \sigma_{k} \leq \frac{S^{k}}{k!}
$$
Summing the obtained inequalities for $k=1,2, \ldots, n$ yields the result.
|
{
"resource_path": "APMO/segmented/en-apmo1989_sol.jsonl",
"problem_match": "# Problem 1",
"solution_match": "# Solution 1"
}
|
b68eb2f6-098f-5fca-a42e-60c8dd41aef2
| 604,152
|
Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers, and let
$$
S=x_{1}+x_{2}+\cdots+x_{n}
$$
Prove that
$$
\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \leq 1+S+\frac{S^{2}}{2!}+\frac{S^{3}}{3!}+\cdots+\frac{S^{n}}{n!}
$$
|
By AM-GM,
$$
\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \leq\left(\frac{\left(1+x_{1}\right)+\left(1+x_{2}\right)+\cdots+\left(1+x_{n}\right)}{n}\right)^{n}=\left(1+\frac{S}{n}\right)^{n}
$$
By the binomial theorem,
$$
\left(1+\frac{S}{n}\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}\left(\frac{S}{n}\right)^{k}=\sum_{k=0}^{n} \frac{1}{k!} \frac{n(n-1) \ldots(n-k+1)}{n^{k}} S^{k} \leq \sum_{k=0}^{n} \frac{S^{k}}{k!}
$$
and the result follows.
Comment: Maclaurin's inequality states that
$$
\frac{\sigma_{1}}{n} \geq \sqrt{\frac{\sigma_{2}}{\binom{n}{2}}} \geq \cdots \geq \sqrt[k]{\frac{\sigma_{k}}{\binom{n}{k}}} \geq \cdots \geq \sqrt[n]{\frac{\sigma_{n}}{\binom{n}{n}}}
$$
Then $\sigma_{k} \leq\binom{ n}{k} \frac{S^{k}}{n^{k}}=\frac{1}{k!} \frac{n(n-1) \ldots(n-k+1)}{n^{k}} S^{k} \leq \frac{S^{k}}{k!}$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers, and let
$$
S=x_{1}+x_{2}+\cdots+x_{n}
$$
Prove that
$$
\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \leq 1+S+\frac{S^{2}}{2!}+\frac{S^{3}}{3!}+\cdots+\frac{S^{n}}{n!}
$$
|
By AM-GM,
$$
\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \leq\left(\frac{\left(1+x_{1}\right)+\left(1+x_{2}\right)+\cdots+\left(1+x_{n}\right)}{n}\right)^{n}=\left(1+\frac{S}{n}\right)^{n}
$$
By the binomial theorem,
$$
\left(1+\frac{S}{n}\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}\left(\frac{S}{n}\right)^{k}=\sum_{k=0}^{n} \frac{1}{k!} \frac{n(n-1) \ldots(n-k+1)}{n^{k}} S^{k} \leq \sum_{k=0}^{n} \frac{S^{k}}{k!}
$$
and the result follows.
Comment: Maclaurin's inequality states that
$$
\frac{\sigma_{1}}{n} \geq \sqrt{\frac{\sigma_{2}}{\binom{n}{2}}} \geq \cdots \geq \sqrt[k]{\frac{\sigma_{k}}{\binom{n}{k}}} \geq \cdots \geq \sqrt[n]{\frac{\sigma_{n}}{\binom{n}{n}}}
$$
Then $\sigma_{k} \leq\binom{ n}{k} \frac{S^{k}}{n^{k}}=\frac{1}{k!} \frac{n(n-1) \ldots(n-k+1)}{n^{k}} S^{k} \leq \frac{S^{k}}{k!}$.
|
{
"resource_path": "APMO/segmented/en-apmo1989_sol.jsonl",
"problem_match": "# Problem 1",
"solution_match": "# Solution 2"
}
|
b68eb2f6-098f-5fca-a42e-60c8dd41aef2
| 604,152
|
Prove that the equation
$$
6\left(6 a^{2}+3 b^{2}+c^{2}\right)=5 n^{2}
$$
has no solutions in integers except $a=b=c=n=0$.
|
We can suppose without loss of generality that $a, b, c, n \geq 0$. Let $(a, b, c, n)$ be a solution with minimum sum $a+b+c+n$. Suppose, for the sake of contradiction, that $a+b+c+n>0$. Since 6 divides $5 n^{2}, n$ is a multiple of 6 . Let $n=6 n_{0}$. Then the equation reduces to
$$
6 a^{2}+3 b^{2}+c^{2}=30 n_{0}^{2}
$$
The number $c$ is a multiple of 3 , so let $c=3 c_{0}$. The equation now reduces to
$$
2 a^{2}+b^{2}+3 c_{0}^{2}=10 n_{0}^{2}
$$
Now look at the equation modulo 8:
$$
b^{2}+3 c_{0}^{2} \equiv 2\left(n_{0}^{2}-a^{2}\right) \quad(\bmod 8)
$$
Integers $b$ and $c_{0}$ have the same parity. Either way, since $x^{2}$ is congruent to 0 or 1 modulo 4 , $b^{2}+3 c_{0}^{2}$ is a multiple of 4 , so $n_{0}^{2}-a^{2}=\left(n_{0}-a\right)\left(n_{0}+a\right)$ is even, and therefore also a multiple of 4 , since $n_{0}-a$ and $n_{0}+a$ have the same parity. Hence $2\left(n_{0}^{2}-a^{2}\right)$ is a multiple of 8 , and
$$
b^{2}+3 c_{0}^{2} \equiv 0 \quad(\bmod 8)
$$
If $b$ and $c_{0}$ are both odd, $b^{2}+3 c_{0}^{2} \equiv 4(\bmod 8)$, which is impossible. Then $b$ and $c_{0}$ are both even. Let $b=2 b_{0}$ and $c_{0}=2 c_{1}$, and we find
$$
a^{2}+2 b_{0}^{2}+6 c_{1}^{2}=5 n_{0}^{2}
$$
Look at the last equation modulo 8:
$$
a^{2}+3 n_{0}^{2} \equiv 2\left(c_{1}^{2}-b_{0}^{2}\right) \quad(\bmod 8)
$$
A similar argument shows that $a$ and $n_{0}$ are both even.
We have proven that $a, b, c, n$ are all even. Then, dividing the original equation by 4 we find
$$
6\left(6(a / 2)^{2}+3(b / 2)^{2}+(c / 2)^{2}\right)=5(n / 2)^{2}
$$
and we find that $(a / 2, b / 2, c / 2, n / 2)$ is a new solution with smaller sum. This is a contradiction, and the only solution is $(a, b, c, n)=(0,0,0,0)$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Prove that the equation
$$
6\left(6 a^{2}+3 b^{2}+c^{2}\right)=5 n^{2}
$$
has no solutions in integers except $a=b=c=n=0$.
|
We can suppose without loss of generality that $a, b, c, n \geq 0$. Let $(a, b, c, n)$ be a solution with minimum sum $a+b+c+n$. Suppose, for the sake of contradiction, that $a+b+c+n>0$. Since 6 divides $5 n^{2}, n$ is a multiple of 6 . Let $n=6 n_{0}$. Then the equation reduces to
$$
6 a^{2}+3 b^{2}+c^{2}=30 n_{0}^{2}
$$
The number $c$ is a multiple of 3 , so let $c=3 c_{0}$. The equation now reduces to
$$
2 a^{2}+b^{2}+3 c_{0}^{2}=10 n_{0}^{2}
$$
Now look at the equation modulo 8:
$$
b^{2}+3 c_{0}^{2} \equiv 2\left(n_{0}^{2}-a^{2}\right) \quad(\bmod 8)
$$
Integers $b$ and $c_{0}$ have the same parity. Either way, since $x^{2}$ is congruent to 0 or 1 modulo 4 , $b^{2}+3 c_{0}^{2}$ is a multiple of 4 , so $n_{0}^{2}-a^{2}=\left(n_{0}-a\right)\left(n_{0}+a\right)$ is even, and therefore also a multiple of 4 , since $n_{0}-a$ and $n_{0}+a$ have the same parity. Hence $2\left(n_{0}^{2}-a^{2}\right)$ is a multiple of 8 , and
$$
b^{2}+3 c_{0}^{2} \equiv 0 \quad(\bmod 8)
$$
If $b$ and $c_{0}$ are both odd, $b^{2}+3 c_{0}^{2} \equiv 4(\bmod 8)$, which is impossible. Then $b$ and $c_{0}$ are both even. Let $b=2 b_{0}$ and $c_{0}=2 c_{1}$, and we find
$$
a^{2}+2 b_{0}^{2}+6 c_{1}^{2}=5 n_{0}^{2}
$$
Look at the last equation modulo 8:
$$
a^{2}+3 n_{0}^{2} \equiv 2\left(c_{1}^{2}-b_{0}^{2}\right) \quad(\bmod 8)
$$
A similar argument shows that $a$ and $n_{0}$ are both even.
We have proven that $a, b, c, n$ are all even. Then, dividing the original equation by 4 we find
$$
6\left(6(a / 2)^{2}+3(b / 2)^{2}+(c / 2)^{2}\right)=5(n / 2)^{2}
$$
and we find that $(a / 2, b / 2, c / 2, n / 2)$ is a new solution with smaller sum. This is a contradiction, and the only solution is $(a, b, c, n)=(0,0,0,0)$.
|
{
"resource_path": "APMO/segmented/en-apmo1989_sol.jsonl",
"problem_match": "# Problem 2",
"solution_match": "# Solution\n\n"
}
|
f926320b-a7a4-5a7e-adab-23b4749bfd74
| 604,164
|
Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers, and let $S_{k}$ be the sum of products of $a_{1}, a_{2}, \ldots, a_{n}$ taken $k$ at a time.
Show that
$$
S_{k} S_{n-k} \geq\binom{ n}{k}^{2} a_{1} a_{2} \ldots a_{n}, \quad \text { for } \quad k=1,2, \ldots, n-1
$$
|
$$
\binom{n}{k} a_{1} a_{2} \ldots a_{n}
$$
$2=\sum_{1 \leq i_{1}<i_{2}<\ldots<i_{k} \leq n} a_{i_{1}} a_{i_{2}} \ldots a_{i_{k}} \cdot a_{1} a_{2} \ldots a_{n} / a_{i_{1}} a_{i_{2}} \ldots a_{i_{k}}$
(and using the Cauchy-Schwarz inequality)
$$
\begin{aligned}
& \leq\left(\sum_{1 \leq i_{1}<i_{2}<\ldots<i_{k} \leq n} a_{i_{1}} a_{i_{2}} \ldots a_{i_{k}}\right)^{\frac{1}{2}} \cdot\left(\sum_{1 \leq i_{1}<i_{2}<\ldots<i_{k} \leq n} a_{1} a_{2} \ldots a_{n} / a_{i_{1}} a_{i_{2}} \ldots a_{i_{k}}\right)^{\frac{1}{2}} \\
& =S_{k}^{\frac{1}{2}} \cdot S_{n-k}^{\frac{1}{2}}
\end{aligned}
$$
Therefore
$$
\binom{n}{k}^{2} a_{1} a_{2} \ldots a_{n} \leq S_{k} S_{n-k}
$$
q.e.d.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers, and let $S_{k}$ be the sum of products of $a_{1}, a_{2}, \ldots, a_{n}$ taken $k$ at a time.
Show that
$$
S_{k} S_{n-k} \geq\binom{ n}{k}^{2} a_{1} a_{2} \ldots a_{n}, \quad \text { for } \quad k=1,2, \ldots, n-1
$$
|
$$
\binom{n}{k} a_{1} a_{2} \ldots a_{n}
$$
$2=\sum_{1 \leq i_{1}<i_{2}<\ldots<i_{k} \leq n} a_{i_{1}} a_{i_{2}} \ldots a_{i_{k}} \cdot a_{1} a_{2} \ldots a_{n} / a_{i_{1}} a_{i_{2}} \ldots a_{i_{k}}$
(and using the Cauchy-Schwarz inequality)
$$
\begin{aligned}
& \leq\left(\sum_{1 \leq i_{1}<i_{2}<\ldots<i_{k} \leq n} a_{i_{1}} a_{i_{2}} \ldots a_{i_{k}}\right)^{\frac{1}{2}} \cdot\left(\sum_{1 \leq i_{1}<i_{2}<\ldots<i_{k} \leq n} a_{1} a_{2} \ldots a_{n} / a_{i_{1}} a_{i_{2}} \ldots a_{i_{k}}\right)^{\frac{1}{2}} \\
& =S_{k}^{\frac{1}{2}} \cdot S_{n-k}^{\frac{1}{2}}
\end{aligned}
$$
Therefore
$$
\binom{n}{k}^{2} a_{1} a_{2} \ldots a_{n} \leq S_{k} S_{n-k}
$$
q.e.d.
|
{
"resource_path": "APMO/segmented/en-apmo1990_sol.jsonl",
"problem_match": "# Question 2",
"solution_match": "# FIRST SOLUTION\n\n"
}
|
92756af4-1c9d-52cf-ab66-3f4f0bc5e602
| 604,264
|
Show that for every integer $n \geq 6$, there exists a convex hexagon which can be dissected into exactly $n$ congruent triangles.
|
(provided by the Canadian Problems Committee).
The basic building blocks will be right angled triangles with sides $p, q$ (which are positive integers) adjacent to the right angle.
In the first instance, we take $p=q=1$ and construct five basic building blocks: $L_{1}, L_{2}, M, R_{1}$ and $R_{2}$ 。
## [3)
$L_{1}$
$L_{2}$
M
$R_{1}$
$R_{2}$
We shall now build convex hexagons by taking, on the left, one of the blocks $L_{i}$, attaching $n$ copies of the block $M$, and finally attaching one of the blocks $R_{j}$. We must therefore exclude the case when $(i, j)=(2,1)$ for this does not generate a hexagon. Further, for $(i, j)=(1,1)$ or $(i, j)=(1,2)$, we require that $n \geq 1$, whereas for $(i, j)=$ $(2,2)$, we only need require that $n \geq 0$.
Thus, with the obvious interpretation:
$L_{1}+n M+R_{1}$ gives a convex hexagon containing $2+4 n+2=4 n+4 \quad(n \geq 1)$ congruent triangles;
$L_{1}+n M+R_{2}$ gives a convex hexagon containing $2+4 n+3=4 n+5(n \geq 1)$ congruent triangles; and
$L_{2}+n M+R_{2}$ gives a convex hexagon containing $3+4 n+3=4 n+6 \quad(n \geq 0)$ congruent triangles, or $4 n+2(n \geq 1)$ congruent triangles.
We shall now modify the lengths of the sides of the right triangle to obtain the case of $4 n+3 \quad(n \geq 1)$ congruent triangles.
So we have $2 n+1$ triangles in the top part and $2 n+2$ triangles in the bottom part. In order to match, we need
$$
(n+1) p=(n+2) q
$$
so we take
$$
q=n+1 \quad \text { and } \quad p=n+2
$$
This completes the solution.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Show that for every integer $n \geq 6$, there exists a convex hexagon which can be dissected into exactly $n$ congruent triangles.
|
(provided by the Canadian Problems Committee).
The basic building blocks will be right angled triangles with sides $p, q$ (which are positive integers) adjacent to the right angle.
In the first instance, we take $p=q=1$ and construct five basic building blocks: $L_{1}, L_{2}, M, R_{1}$ and $R_{2}$ 。
## [3)
$L_{1}$
$L_{2}$
M
$R_{1}$
$R_{2}$
We shall now build convex hexagons by taking, on the left, one of the blocks $L_{i}$, attaching $n$ copies of the block $M$, and finally attaching one of the blocks $R_{j}$. We must therefore exclude the case when $(i, j)=(2,1)$ for this does not generate a hexagon. Further, for $(i, j)=(1,1)$ or $(i, j)=(1,2)$, we require that $n \geq 1$, whereas for $(i, j)=$ $(2,2)$, we only need require that $n \geq 0$.
Thus, with the obvious interpretation:
$L_{1}+n M+R_{1}$ gives a convex hexagon containing $2+4 n+2=4 n+4 \quad(n \geq 1)$ congruent triangles;
$L_{1}+n M+R_{2}$ gives a convex hexagon containing $2+4 n+3=4 n+5(n \geq 1)$ congruent triangles; and
$L_{2}+n M+R_{2}$ gives a convex hexagon containing $3+4 n+3=4 n+6 \quad(n \geq 0)$ congruent triangles, or $4 n+2(n \geq 1)$ congruent triangles.
We shall now modify the lengths of the sides of the right triangle to obtain the case of $4 n+3 \quad(n \geq 1)$ congruent triangles.
So we have $2 n+1$ triangles in the top part and $2 n+2$ triangles in the bottom part. In order to match, we need
$$
(n+1) p=(n+2) q
$$
so we take
$$
q=n+1 \quad \text { and } \quad p=n+2
$$
This completes the solution.
|
{
"resource_path": "APMO/segmented/en-apmo1990_sol.jsonl",
"problem_match": "# Question 5",
"solution_match": "\nFIRST SOLUTION "
}
|
73c9ec95-e9ca-5aef-bf23-74672775d15a
| 54,403
|
Show that for every integer $n \geq 6$, there exists a convex hexagon which can be dissected into exactly $n$ congruent triangles.
|
(provided by the Canadian Problems Committee):
The basic building blocks will be right angled triangles with sides $m, n$ (which are positive integers) adjacent to the right angle.
We construct an "UPPER CONFIGURATION", being a rectangle consisting of $m$ building block units of pairs of triangles with the side of length $n$ as base. This gives a base length of $n m$ across the configuration.
We further construct a "LOWER CONFIGURATION", being a triangle with base up, consisting along the base of $n$ building block units. Again, we have a base length of $m n$ across the configuration.
Two triangles in the upper configuration are shaded horizontally. One triangle in the lower configuration is also shaded horizontally. Another triangle in the lower configuration is shaded vertically.
Now consider the figure obtained by joining the two configurations along the base line of common length nm. To create the classes of hexagons defined below, it is necessary that both $n \geq 3$ and $m \geq 3$.
We create a class of convex hexagons (class 1 ) by omitting the three triangles that are shaded horizontally. The other class of convex hexagons (class 2) is obtained by omitting all shaded triangles.
Now count the total number of triangles in the full configuration.
The upper configuration gives $2 m$ triangles. The lower configuration gives
$$
\sum_{k=1}^{n}(2 k-1)=n^{2} \quad \text { triangles. }
$$
Thus the total number of triangles in a hexagon in class 1 is
$$
2 m-2+n^{2}-1
$$
and the total number of triangle in a hexagon in class 2 is
$$
2 m-2+n^{2}-2
$$
These, together with the restrictions on $n$ and $m$, generate all positive integers greater than or equal to 11.
For the integers $6,7,8,9$ and 10 , we give specific examples:
6
7
8
9
10
This completes the solution.
There are $\binom{n}{k}$ products of the $a_{i}$ taken $k$ at a time. Amongst these products any given $a_{i}$ will appear $\binom{n-1}{k-1}$ times, since $\binom{n-1}{k-1}$ is the number of ways of choosing the other factors of the product. So the $\mathrm{AM} / \mathrm{GM}$ inequality gives
## ④
$$
\frac{S_{k}}{\binom{n}{k}} \geq\left[\prod_{i=1}^{n} a_{i}^{\binom{n-1}{k-1}}\right]^{\frac{1}{n}\binom{n}{n}}
$$
But $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$, leading to
6 S $\quad S_{k} \geq\binom{ n}{k}\left(\prod_{i=1}^{n} a_{i}\right)^{\frac{k}{n}}$.
Hence
田
$$
S_{k} S_{n-k} \geq\binom{ n}{k}\left(\prod_{i=1}^{n} a_{i}\right)^{\frac{k}{n}}\binom{n}{n-k}\left(\prod_{i=1}^{n} a_{i}\right)^{\frac{n-k}{n}}=\binom{n}{k}^{2}\left(\prod_{1}^{n} a_{i}\right) .
$$
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Show that for every integer $n \geq 6$, there exists a convex hexagon which can be dissected into exactly $n$ congruent triangles.
|
(provided by the Canadian Problems Committee):
The basic building blocks will be right angled triangles with sides $m, n$ (which are positive integers) adjacent to the right angle.
We construct an "UPPER CONFIGURATION", being a rectangle consisting of $m$ building block units of pairs of triangles with the side of length $n$ as base. This gives a base length of $n m$ across the configuration.
We further construct a "LOWER CONFIGURATION", being a triangle with base up, consisting along the base of $n$ building block units. Again, we have a base length of $m n$ across the configuration.
Two triangles in the upper configuration are shaded horizontally. One triangle in the lower configuration is also shaded horizontally. Another triangle in the lower configuration is shaded vertically.
Now consider the figure obtained by joining the two configurations along the base line of common length nm. To create the classes of hexagons defined below, it is necessary that both $n \geq 3$ and $m \geq 3$.
We create a class of convex hexagons (class 1 ) by omitting the three triangles that are shaded horizontally. The other class of convex hexagons (class 2) is obtained by omitting all shaded triangles.
Now count the total number of triangles in the full configuration.
The upper configuration gives $2 m$ triangles. The lower configuration gives
$$
\sum_{k=1}^{n}(2 k-1)=n^{2} \quad \text { triangles. }
$$
Thus the total number of triangles in a hexagon in class 1 is
$$
2 m-2+n^{2}-1
$$
and the total number of triangle in a hexagon in class 2 is
$$
2 m-2+n^{2}-2
$$
These, together with the restrictions on $n$ and $m$, generate all positive integers greater than or equal to 11.
For the integers $6,7,8,9$ and 10 , we give specific examples:
6
7
8
9
10
This completes the solution.
There are $\binom{n}{k}$ products of the $a_{i}$ taken $k$ at a time. Amongst these products any given $a_{i}$ will appear $\binom{n-1}{k-1}$ times, since $\binom{n-1}{k-1}$ is the number of ways of choosing the other factors of the product. So the $\mathrm{AM} / \mathrm{GM}$ inequality gives
## ④
$$
\frac{S_{k}}{\binom{n}{k}} \geq\left[\prod_{i=1}^{n} a_{i}^{\binom{n-1}{k-1}}\right]^{\frac{1}{n}\binom{n}{n}}
$$
But $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$, leading to
6 S $\quad S_{k} \geq\binom{ n}{k}\left(\prod_{i=1}^{n} a_{i}\right)^{\frac{k}{n}}$.
Hence
田
$$
S_{k} S_{n-k} \geq\binom{ n}{k}\left(\prod_{i=1}^{n} a_{i}\right)^{\frac{k}{n}}\binom{n}{n-k}\left(\prod_{i=1}^{n} a_{i}\right)^{\frac{n-k}{n}}=\binom{n}{k}^{2}\left(\prod_{1}^{n} a_{i}\right) .
$$
|
{
"resource_path": "APMO/segmented/en-apmo1990_sol.jsonl",
"problem_match": "# Question 5",
"solution_match": "\nSECOND SOLUTION "
}
|
73c9ec95-e9ca-5aef-bf23-74672775d15a
| 54,403
|
Let $G$ be the centroid of triangle $A B C$ and $M$ be the midpoint of $B C$. Let $X$ be on $A B$ and $Y$ on $A C$ such that the points $X, Y$, and $G$ are collinear and $X Y$ and $B C$ are parallel. Suppose that $X C$ and $G B$ intersect at $Q$ and $Y B$ and $G C$ intersect at $P$. Show that triangle $M P Q$ is similar to triangle $A B C$.
|
Let $R$ be the midpoint of $A C$; so $B R$ is a median and contains the centroid $G$.
It is well known that $\frac{A G}{A M}=\frac{2}{3}$; thus the ratio of the similarity between $A X Y$ and $A B C$ is $\frac{2}{3}$. Hence $G X=\frac{1}{2} X Y=\frac{1}{3} B C$.
Now look at the similarity between triangles $Q B C$ and $Q G X$ :
$$
\frac{Q G}{Q B}=\frac{G X}{B C}=\frac{1}{3} \Longrightarrow Q B=3 Q G \Longrightarrow Q B=\frac{3}{4} B G=\frac{3}{4} \cdot \frac{2}{3} B R=\frac{1}{2} B R .
$$
Finally, since $\frac{B M}{B C}=\frac{B Q}{B R}, M Q$ is a midline in $B C R$. Therefore $M Q=\frac{1}{2} C R=\frac{1}{4} A C$ and $M Q \| A C$. Similarly, $M P=\frac{1}{4} A B$ and $M P \| A B$. This is sufficient to establish that $M P Q$ and $A B C$ are similar (with similarity ratio $\frac{1}{4}$ ).
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $G$ be the centroid of triangle $A B C$ and $M$ be the midpoint of $B C$. Let $X$ be on $A B$ and $Y$ on $A C$ such that the points $X, Y$, and $G$ are collinear and $X Y$ and $B C$ are parallel. Suppose that $X C$ and $G B$ intersect at $Q$ and $Y B$ and $G C$ intersect at $P$. Show that triangle $M P Q$ is similar to triangle $A B C$.
|
Let $R$ be the midpoint of $A C$; so $B R$ is a median and contains the centroid $G$.
It is well known that $\frac{A G}{A M}=\frac{2}{3}$; thus the ratio of the similarity between $A X Y$ and $A B C$ is $\frac{2}{3}$. Hence $G X=\frac{1}{2} X Y=\frac{1}{3} B C$.
Now look at the similarity between triangles $Q B C$ and $Q G X$ :
$$
\frac{Q G}{Q B}=\frac{G X}{B C}=\frac{1}{3} \Longrightarrow Q B=3 Q G \Longrightarrow Q B=\frac{3}{4} B G=\frac{3}{4} \cdot \frac{2}{3} B R=\frac{1}{2} B R .
$$
Finally, since $\frac{B M}{B C}=\frac{B Q}{B R}, M Q$ is a midline in $B C R$. Therefore $M Q=\frac{1}{2} C R=\frac{1}{4} A C$ and $M Q \| A C$. Similarly, $M P=\frac{1}{4} A B$ and $M P \| A B$. This is sufficient to establish that $M P Q$ and $A B C$ are similar (with similarity ratio $\frac{1}{4}$ ).
|
{
"resource_path": "APMO/segmented/en-apmo1991_sol.jsonl",
"problem_match": "# Problem 1",
"solution_match": "# Solution 1"
}
|
bd1d8513-fe5c-50c2-a9d8-0c6ddc4272c9
| 604,343
|
Let $G$ be the centroid of triangle $A B C$ and $M$ be the midpoint of $B C$. Let $X$ be on $A B$ and $Y$ on $A C$ such that the points $X, Y$, and $G$ are collinear and $X Y$ and $B C$ are parallel. Suppose that $X C$ and $G B$ intersect at $Q$ and $Y B$ and $G C$ intersect at $P$. Show that triangle $M P Q$ is similar to triangle $A B C$.
|
Let $S$ and $R$ be the midpoints of $A B$ and $A C$, respectively. Since $G$ is the centroid, it lies in the medians $B R$ and $C S$.
Due to the similarity between triangles $Q B C$ and $Q G X$ (which is true because $G X \| B C$ ), there is an inverse homothety with center $Q$ and ratio $-\frac{X G}{B C}=\frac{X Y}{2 B C}$ that takes $B$ to $G$ and $C$ to $X$. This homothety takes the midpoint $M$ of $B C$ to the midpoint $K$ of $G X$.
Now consider the homothety that takes $B$ to $X$ and $C$ to $G$. This new homothety, with ratio $\frac{X Y}{2 B C}$, also takes $M$ to $K$. Hence lines $B X$ (which contains side $A B$ ), $C G$ (which contains the median $C S$ ), and $M K$ have a common point, which is $S$. Thus $Q$ lies on midline $M S$.
The same reasoning proves that $P$ lies on midline $M R$. Since all homothety ratios are the same, $\frac{M Q}{M S}=\frac{M P}{M R}$, which shows that $M P Q$ is similar to $M R S$, which in turn is similar to $A B C$, and we are done.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $G$ be the centroid of triangle $A B C$ and $M$ be the midpoint of $B C$. Let $X$ be on $A B$ and $Y$ on $A C$ such that the points $X, Y$, and $G$ are collinear and $X Y$ and $B C$ are parallel. Suppose that $X C$ and $G B$ intersect at $Q$ and $Y B$ and $G C$ intersect at $P$. Show that triangle $M P Q$ is similar to triangle $A B C$.
|
Let $S$ and $R$ be the midpoints of $A B$ and $A C$, respectively. Since $G$ is the centroid, it lies in the medians $B R$ and $C S$.
Due to the similarity between triangles $Q B C$ and $Q G X$ (which is true because $G X \| B C$ ), there is an inverse homothety with center $Q$ and ratio $-\frac{X G}{B C}=\frac{X Y}{2 B C}$ that takes $B$ to $G$ and $C$ to $X$. This homothety takes the midpoint $M$ of $B C$ to the midpoint $K$ of $G X$.
Now consider the homothety that takes $B$ to $X$ and $C$ to $G$. This new homothety, with ratio $\frac{X Y}{2 B C}$, also takes $M$ to $K$. Hence lines $B X$ (which contains side $A B$ ), $C G$ (which contains the median $C S$ ), and $M K$ have a common point, which is $S$. Thus $Q$ lies on midline $M S$.
The same reasoning proves that $P$ lies on midline $M R$. Since all homothety ratios are the same, $\frac{M Q}{M S}=\frac{M P}{M R}$, which shows that $M P Q$ is similar to $M R S$, which in turn is similar to $A B C$, and we are done.
|
{
"resource_path": "APMO/segmented/en-apmo1991_sol.jsonl",
"problem_match": "# Problem 1",
"solution_match": "# Solution 2"
}
|
bd1d8513-fe5c-50c2-a9d8-0c6ddc4272c9
| 604,343
|
In a circle $C$ with centre $O$ and radius $r$, let $C_{1}, C_{2}$ be two circles with centres $O_{1}, O_{2}$ and radii $r_{1}, r_{2}$ respectively, so that each circle $C_{i}$ is internally tangent to $C$ at $A_{i}$ and so that $C_{1}, C_{2}$ are externally tangent to each other at $A$.
Prove that the three lines $O A, O_{1} A_{2}$, and $O_{2} A_{1}$ are concurrent.
|
Because of the tangencies, the following triples of points (two centers and a tangency point) are collinear:
$$
O_{1} ; O_{2} ; A, \quad O ; O_{1} ; A_{1}, \quad O ; O_{2} ; A_{2}
$$
Because of that we can ignore the circles and only draw their centers and tangency points.
Now the problem is immediate from Ceva's theorem in triangle $O O_{1} O_{2}$, because
$$
\frac{O A_{1}}{A_{1} O_{1}} \cdot \frac{O_{1} A}{A O_{2}} \cdot \frac{O_{2} A_{2}}{A_{2} O}=\frac{r}{r_{1}} \cdot \frac{r_{1}}{r_{2}} \cdot \frac{r_{2}}{r}=1
$$
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In a circle $C$ with centre $O$ and radius $r$, let $C_{1}, C_{2}$ be two circles with centres $O_{1}, O_{2}$ and radii $r_{1}, r_{2}$ respectively, so that each circle $C_{i}$ is internally tangent to $C$ at $A_{i}$ and so that $C_{1}, C_{2}$ are externally tangent to each other at $A$.
Prove that the three lines $O A, O_{1} A_{2}$, and $O_{2} A_{1}$ are concurrent.
|
Because of the tangencies, the following triples of points (two centers and a tangency point) are collinear:
$$
O_{1} ; O_{2} ; A, \quad O ; O_{1} ; A_{1}, \quad O ; O_{2} ; A_{2}
$$
Because of that we can ignore the circles and only draw their centers and tangency points.
Now the problem is immediate from Ceva's theorem in triangle $O O_{1} O_{2}$, because
$$
\frac{O A_{1}}{A_{1} O_{1}} \cdot \frac{O_{1} A}{A O_{2}} \cdot \frac{O_{2} A_{2}}{A_{2} O}=\frac{r}{r_{1}} \cdot \frac{r_{1}}{r_{2}} \cdot \frac{r_{2}}{r}=1
$$
|
{
"resource_path": "APMO/segmented/en-apmo1992_sol.jsonl",
"problem_match": "# Problem 2",
"solution_match": "# Solution\n\n"
}
|
ffb81a07-18b5-55d8-b41e-205d7c2461c9
| 604,440
|
Let $n$ be an integer such that $n>3$. Suppose that we choose three numbers from the set $\{1,2, \ldots, n\}$. Using each of these three numbers only once and using addition, multiplication, and parenthesis, let us form all possible combinations.
(a) Show that if we choose all three numbers greater than $n / 2$, then the values of these combinations are all distinct.
(b) Let $p$ be a prime number such that $p \leq \sqrt{n}$. Show that the number of ways of choosing three numbers so that the smallest one is $p$ and the values of the combinations are not all distinct is precisely the number of positive divisors of $p-1$.
|
In both items, the smallest chosen number is at least 2: in part (a), $n / 2>1$ and in part (b), $p$ is a prime. So let $1<x<y<z$ be the chosen numbers. Then all possible combinations are
$$
x+y+z, \quad x+y z, \quad x y+z, \quad y+z x, \quad(x+y) z, \quad(z+x) y, \quad(x+y) z, \quad x y z
$$
Since, for $1<m<n$ and $t>1,(m-1)(n-1) \geq 1 \cdot 2 \Longrightarrow m n>m+n, t n+m-(t m+n)=$ $(t-1)(n-m)>0 \Longrightarrow t n+m>t m+n$, and $(t+m) n-(t+n) m=t(n-m)>0$,
$$
x+y+z<z+x y<y+z x<x+y z
$$
and
$$
(y+z) x<(x+z) y<(x+y) z<x y z .
$$
Also, $(y+z) x-(y+z x)=(x-1) y>0 \Longrightarrow(y+z) x>y+z x$ and $(x+z) y-(x+y z)=$ $(y-1) x>0 \Longrightarrow(x+z) y>x+y z$. Therefore the only numbers that can be equal are $x+y z$ and $(y+z) x$. In this case,
$$
x+y z=(y+z) x \Longleftrightarrow(y-x)(z-x)=x(x-1)
$$
Now we can solve the items.
(a) if $n / 2<x<y<z$ then $z-x<n / 2$, and since $y-x<z-x, y-x<n / 2-1$; then
$$
(y-x)(z-x)<\frac{n}{2}\left(\frac{n}{2}-1\right)<x(x-1)
$$
and therefore $x+y z<(y+z) x$.
(b) if $x=p$, then $(y-p)(z-p)=p(p-1)$. Since $y-p<z-p,(y-p)^{2}<(y-p)(z-p)=$ $p(p-1) \Longrightarrow y-p<p$, that is, $p$ does not divide $y-p$. Then $y-p$ is a divisor $d$ of $p-1$ and $z-p=\frac{p(p-1)}{d}$. Therefore,
$$
x=p, \quad, y=p+d, \quad z=p+\frac{p(p-1)}{d}
$$
which is a solution for every divisor $d$ of $p-1$ because
$$
x=p<y=p+d<2 p \leq p+p \cdot \frac{p-1}{d}=z .
$$
Comment: If $x=1$ was allowed, then any choice $1, y, z$ would have repeated numbers in the combination, as $1 \cdot y+z=y+1 \cdot z$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $n$ be an integer such that $n>3$. Suppose that we choose three numbers from the set $\{1,2, \ldots, n\}$. Using each of these three numbers only once and using addition, multiplication, and parenthesis, let us form all possible combinations.
(a) Show that if we choose all three numbers greater than $n / 2$, then the values of these combinations are all distinct.
(b) Let $p$ be a prime number such that $p \leq \sqrt{n}$. Show that the number of ways of choosing three numbers so that the smallest one is $p$ and the values of the combinations are not all distinct is precisely the number of positive divisors of $p-1$.
|
In both items, the smallest chosen number is at least 2: in part (a), $n / 2>1$ and in part (b), $p$ is a prime. So let $1<x<y<z$ be the chosen numbers. Then all possible combinations are
$$
x+y+z, \quad x+y z, \quad x y+z, \quad y+z x, \quad(x+y) z, \quad(z+x) y, \quad(x+y) z, \quad x y z
$$
Since, for $1<m<n$ and $t>1,(m-1)(n-1) \geq 1 \cdot 2 \Longrightarrow m n>m+n, t n+m-(t m+n)=$ $(t-1)(n-m)>0 \Longrightarrow t n+m>t m+n$, and $(t+m) n-(t+n) m=t(n-m)>0$,
$$
x+y+z<z+x y<y+z x<x+y z
$$
and
$$
(y+z) x<(x+z) y<(x+y) z<x y z .
$$
Also, $(y+z) x-(y+z x)=(x-1) y>0 \Longrightarrow(y+z) x>y+z x$ and $(x+z) y-(x+y z)=$ $(y-1) x>0 \Longrightarrow(x+z) y>x+y z$. Therefore the only numbers that can be equal are $x+y z$ and $(y+z) x$. In this case,
$$
x+y z=(y+z) x \Longleftrightarrow(y-x)(z-x)=x(x-1)
$$
Now we can solve the items.
(a) if $n / 2<x<y<z$ then $z-x<n / 2$, and since $y-x<z-x, y-x<n / 2-1$; then
$$
(y-x)(z-x)<\frac{n}{2}\left(\frac{n}{2}-1\right)<x(x-1)
$$
and therefore $x+y z<(y+z) x$.
(b) if $x=p$, then $(y-p)(z-p)=p(p-1)$. Since $y-p<z-p,(y-p)^{2}<(y-p)(z-p)=$ $p(p-1) \Longrightarrow y-p<p$, that is, $p$ does not divide $y-p$. Then $y-p$ is a divisor $d$ of $p-1$ and $z-p=\frac{p(p-1)}{d}$. Therefore,
$$
x=p, \quad, y=p+d, \quad z=p+\frac{p(p-1)}{d}
$$
which is a solution for every divisor $d$ of $p-1$ because
$$
x=p<y=p+d<2 p \leq p+p \cdot \frac{p-1}{d}=z .
$$
Comment: If $x=1$ was allowed, then any choice $1, y, z$ would have repeated numbers in the combination, as $1 \cdot y+z=y+1 \cdot z$.
|
{
"resource_path": "APMO/segmented/en-apmo1992_sol.jsonl",
"problem_match": "# Problem 3",
"solution_match": "# Solution\n\n"
}
|
b52e4aa1-0199-5e11-88b8-bb3a94eec199
| 604,457
|
Let $A B C D$ be a quadrilateral such that all sides have equal length and angle $\angle A B C$ is 60 degrees. Let $\ell$ be a line passing through $D$ and not intersecting the quadrilateral (except at $D)$. Let $E$ and $F$ be the points of intersection of $\ell$ with $A B$ and $B C$ respectively. Let $M$ be the point of intersection of $C E$ and $A F$.
Prove that $C A^{2}=C M \times C E$.
|
Triangles $A E D$ and $C D F$ are similar, because $A D \| C F$ and $A E \| C D$. Thus, since $A B C$ and $A C D$ are equilateral triangles,
$$
\frac{A E}{C D}=\frac{A D}{C F} \Longleftrightarrow \frac{A E}{A C}=\frac{A C}{C F}
$$
The last equality combined with
$$
\angle E A C=180^{\circ}-\angle B A C=120^{\circ}=\angle A C F
$$
shows that triangles $E A C$ and $A C F$ are also similar. Therefore $\angle C A M=\angle C A F=\angle A E C$, which implies that line $A C$ is tangent to the circumcircle of $A M E$. By the power of a point, $C A^{2}=C M \cdot C E$, and we are done.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a quadrilateral such that all sides have equal length and angle $\angle A B C$ is 60 degrees. Let $\ell$ be a line passing through $D$ and not intersecting the quadrilateral (except at $D)$. Let $E$ and $F$ be the points of intersection of $\ell$ with $A B$ and $B C$ respectively. Let $M$ be the point of intersection of $C E$ and $A F$.
Prove that $C A^{2}=C M \times C E$.
|
Triangles $A E D$ and $C D F$ are similar, because $A D \| C F$ and $A E \| C D$. Thus, since $A B C$ and $A C D$ are equilateral triangles,
$$
\frac{A E}{C D}=\frac{A D}{C F} \Longleftrightarrow \frac{A E}{A C}=\frac{A C}{C F}
$$
The last equality combined with
$$
\angle E A C=180^{\circ}-\angle B A C=120^{\circ}=\angle A C F
$$
shows that triangles $E A C$ and $A C F$ are also similar. Therefore $\angle C A M=\angle C A F=\angle A E C$, which implies that line $A C$ is tangent to the circumcircle of $A M E$. By the power of a point, $C A^{2}=C M \cdot C E$, and we are done.
|
{
"resource_path": "APMO/segmented/en-apmo1993_sol.jsonl",
"problem_match": "# Problem 1",
"solution_match": "# Solution\n\n"
}
|
93fdf523-4aa5-5c8a-9d13-256f9c9907cd
| 604,496
|
Let
$$
f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0} \quad \text { and } \quad g(x)=c_{n+1} x^{n+1}+c_{n} x^{n}+\cdots+c_{0}
$$
be non-zero polynomials with real coefficients such that $g(x)=(x+r) f(x)$ for some real number $r$. If $a=\max \left(\left|a_{n}\right|, \ldots,\left|a_{0}\right|\right)$ and $c=\max \left(\left|c_{n+1}\right|, \ldots,\left|c_{0}\right|\right)$, prove that $\frac{a}{c} \leq n+1$.
|
Expanding $(x+r) f(x)$, we find that $c_{n+1}=a_{n}, c_{k}=a_{k-1}+r a_{k}$ for $k=1,2, \ldots, n$, and $c_{0}=r a_{0}$. Consider three cases:
- $r=0$. Then $c_{0}=0$ and $c_{k}=a_{k-1}$ for $k=1,2, \ldots, n$, and $a=c \Longrightarrow \frac{a}{c}=1 \leq n+1$.
- $|r| \geq 1$. Then
$$
\begin{gathered}
\left|a_{0}\right|=\left|\frac{c_{0}}{r}\right| \leq c \\
\left|a_{1}\right|=\left|\frac{c_{1}-a_{0}}{r}\right| \leq\left|c_{1}\right|+\left|a_{0}\right| \leq 2 c
\end{gathered}
$$
and inductively if $\left|a_{k}\right| \leq(k+1) c$
$$
\left|a_{k+1}\right|=\left|\frac{c_{k+1}-a_{k}}{r}\right| \leq\left|c_{k+1}\right|+\left|a_{k}\right| \leq c+(k+1) c=(k+2) c
$$
Therefore, $\left|a_{k}\right| \leq(k+1) c \leq(n+1) c$ for all $k$, and $a \leq(n+1) c \Longleftrightarrow \frac{a}{c} \leq n+1$.
- $0<|r|<1$. Now work backwards: $\left|a_{n}\right|=\left|c_{n+1}\right| \leq c$,
$$
\left|a_{n-1}\right|=\left|c_{n}-r a_{n}\right| \leq\left|c_{n}\right|+\left|r a_{n}\right|<c+c=2 c,
$$
and inductively if $\left|a_{n-k}\right| \leq(k+1) c$
$$
\left|a_{n-k-1}\right|=\left|c_{n-k}-r a_{n-k}\right| \leq\left|c_{n-k}\right|+\left|r a_{n-k}\right|<c+(k+1) c=(k+2) c .
$$
Therefore, $\left|a_{n-k}\right| \leq(k+1) c \leq(n+1) c$ for all $k$, and $a \leq(n+1) c$ again.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let
$$
f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0} \quad \text { and } \quad g(x)=c_{n+1} x^{n+1}+c_{n} x^{n}+\cdots+c_{0}
$$
be non-zero polynomials with real coefficients such that $g(x)=(x+r) f(x)$ for some real number $r$. If $a=\max \left(\left|a_{n}\right|, \ldots,\left|a_{0}\right|\right)$ and $c=\max \left(\left|c_{n+1}\right|, \ldots,\left|c_{0}\right|\right)$, prove that $\frac{a}{c} \leq n+1$.
|
Expanding $(x+r) f(x)$, we find that $c_{n+1}=a_{n}, c_{k}=a_{k-1}+r a_{k}$ for $k=1,2, \ldots, n$, and $c_{0}=r a_{0}$. Consider three cases:
- $r=0$. Then $c_{0}=0$ and $c_{k}=a_{k-1}$ for $k=1,2, \ldots, n$, and $a=c \Longrightarrow \frac{a}{c}=1 \leq n+1$.
- $|r| \geq 1$. Then
$$
\begin{gathered}
\left|a_{0}\right|=\left|\frac{c_{0}}{r}\right| \leq c \\
\left|a_{1}\right|=\left|\frac{c_{1}-a_{0}}{r}\right| \leq\left|c_{1}\right|+\left|a_{0}\right| \leq 2 c
\end{gathered}
$$
and inductively if $\left|a_{k}\right| \leq(k+1) c$
$$
\left|a_{k+1}\right|=\left|\frac{c_{k+1}-a_{k}}{r}\right| \leq\left|c_{k+1}\right|+\left|a_{k}\right| \leq c+(k+1) c=(k+2) c
$$
Therefore, $\left|a_{k}\right| \leq(k+1) c \leq(n+1) c$ for all $k$, and $a \leq(n+1) c \Longleftrightarrow \frac{a}{c} \leq n+1$.
- $0<|r|<1$. Now work backwards: $\left|a_{n}\right|=\left|c_{n+1}\right| \leq c$,
$$
\left|a_{n-1}\right|=\left|c_{n}-r a_{n}\right| \leq\left|c_{n}\right|+\left|r a_{n}\right|<c+c=2 c,
$$
and inductively if $\left|a_{n-k}\right| \leq(k+1) c$
$$
\left|a_{n-k-1}\right|=\left|c_{n-k}-r a_{n-k}\right| \leq\left|c_{n-k}\right|+\left|r a_{n-k}\right|<c+(k+1) c=(k+2) c .
$$
Therefore, $\left|a_{n-k}\right| \leq(k+1) c \leq(n+1) c$ for all $k$, and $a \leq(n+1) c$ again.
|
{
"resource_path": "APMO/segmented/en-apmo1993_sol.jsonl",
"problem_match": "# Problem 3",
"solution_match": "# Solution\n\n"
}
|
620b5678-bb8e-5e62-b80e-3aa4b8fd846e
| 604,520
|
Let $P_{1}, P_{2}, \ldots, P_{1993}=P_{0}$ be distinct points in the $x y$-plane with the following properties:
(i) both coordinates of $P_{i}$ are integers, for $i=1,2, \ldots, 1993$;
(ii) there is no point other than $P_{i}$ and $P_{i+1}$ on the line segment joining $P_{i}$ with $P_{i+1}$ whose coordinates are both integers, for $i=0,1, \ldots, 1992$.
Prove that for some $i, 0 \leq i \leq 1992$, there exists a point $Q$ with coordinates $\left(q_{x}, q_{y}\right)$ on the line segment joining $P_{i}$ with $P_{i+1}$ such that both $2 q_{x}$ and $2 q_{y}$ are odd integers.
|
Call a point $(x, y) \in \mathbb{Z}^{2}$ even or odd according to the parity of $x+y$. Since there are an odd number of points, there are two points $P_{i}=(a, b)$ and $P_{i+1}=(c, d), 0 \leq i \leq 1992$ with the same parity. This implies that $a+b+c+d$ is even. We claim that the midpoint of $P_{i} P_{i+1}$ is the desired point $Q$.
In fact, since $a+b+c+d=(a+c)+(b+d)$ is even, $a$ and $c$ have the same parity if and only if $b$ and $d$ also have the same parity. If both happen then the midpoint of $P_{i} P_{i+1}, Q=\left(\frac{a+c}{2}, \frac{b+d}{2}\right)$, has integer coordinates, which violates condition (ii). Then $a$ and $c$, as well as $b$ and $d$, have different parities, and $2 q_{x}=a+c$ and $2 q_{y}=b+d$ are both odd integers.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $P_{1}, P_{2}, \ldots, P_{1993}=P_{0}$ be distinct points in the $x y$-plane with the following properties:
(i) both coordinates of $P_{i}$ are integers, for $i=1,2, \ldots, 1993$;
(ii) there is no point other than $P_{i}$ and $P_{i+1}$ on the line segment joining $P_{i}$ with $P_{i+1}$ whose coordinates are both integers, for $i=0,1, \ldots, 1992$.
Prove that for some $i, 0 \leq i \leq 1992$, there exists a point $Q$ with coordinates $\left(q_{x}, q_{y}\right)$ on the line segment joining $P_{i}$ with $P_{i+1}$ such that both $2 q_{x}$ and $2 q_{y}$ are odd integers.
|
Call a point $(x, y) \in \mathbb{Z}^{2}$ even or odd according to the parity of $x+y$. Since there are an odd number of points, there are two points $P_{i}=(a, b)$ and $P_{i+1}=(c, d), 0 \leq i \leq 1992$ with the same parity. This implies that $a+b+c+d$ is even. We claim that the midpoint of $P_{i} P_{i+1}$ is the desired point $Q$.
In fact, since $a+b+c+d=(a+c)+(b+d)$ is even, $a$ and $c$ have the same parity if and only if $b$ and $d$ also have the same parity. If both happen then the midpoint of $P_{i} P_{i+1}, Q=\left(\frac{a+c}{2}, \frac{b+d}{2}\right)$, has integer coordinates, which violates condition (ii). Then $a$ and $c$, as well as $b$ and $d$, have different parities, and $2 q_{x}=a+c$ and $2 q_{y}=b+d$ are both odd integers.
|
{
"resource_path": "APMO/segmented/en-apmo1993_sol.jsonl",
"problem_match": "# Problem 5",
"solution_match": "# Solution\n\n"
}
|
7aa3067d-2293-5b49-8533-29510f81af65
| 604,546
|
Given a nondegenerate triangle $A B C$, with circumcentre $O$, orthocentre $H$, and circumradius $R$, prove that $|O H|<3 R$.
|
Suppose with loss of generality that $\angle A<90^{\circ}$. Let $B D$ be an altitude. Then
$$
A H=\frac{A D}{\cos \left(90^{\circ}-C\right)}=\frac{A B \cos A}{\sin C}=2 R \cos A
$$
By the triangle inequality,
$$
O H<A O+A H=R+2 R \cos A<3 R
$$
Comment: With a bit more work, if $a, b, c$ are the sidelengths of $A B C$, one can show that
$$
O H^{2}=9 R^{2}-a^{2}-b^{2}-c^{2} .
$$
In fact, using vectors in a coordinate system with $O$ as origin, by the Euler line
$$
\overrightarrow{O H}=3 \overrightarrow{O G}=3 \cdot \frac{\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}}{3}=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}
$$
so
$$
O H^{2}=\overrightarrow{O H} \cdot \overrightarrow{O H}=(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \cdot(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C})
$$
Expanding and using the fact that $\overrightarrow{O X} \cdot \overrightarrow{O X}=O X^{2}=R^{2}$ for $X \in\{A, B, C\}$, as well as
$\overrightarrow{O A} \cdot \overrightarrow{O B}=O A \cdot O B \cdot \cos \angle A O B=R^{2} \cos 2 C=R^{2}\left(1-2 \sin ^{2} C\right)=R^{2}\left(1-2\left(\frac{c}{2 R}\right)^{2}\right)=R^{2}-\frac{c^{2}}{2}$, we find that
$$
\begin{aligned}
O H^{2} & =\overrightarrow{O A} \cdot \overrightarrow{O A}+\overrightarrow{O B} \cdot \overrightarrow{O B}+\overrightarrow{O C} \cdot \overrightarrow{O C}+2 \overrightarrow{O A} \cdot \overrightarrow{O B}+2 \overrightarrow{O A} \cdot \overrightarrow{O C}+2 \overrightarrow{O B} \cdot \overrightarrow{O C} \\
& =3 R^{2}+\left(2 R^{2}-c^{2}\right)+\left(2 R^{2}-b^{2}\right)+\left(2 R^{2}-a^{2}\right) \\
& =9 R^{2}-a^{2}-b^{2}-c^{2}
\end{aligned}
$$
as required.
This proves that $O H^{2}<9 R^{2} \Longrightarrow O H<3 R$, and since $a, b, c$ can be arbitrarily small (fix the circumcircle and choose $A, B, C$ arbitrarily close in this circle), the bound is sharp.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Given a nondegenerate triangle $A B C$, with circumcentre $O$, orthocentre $H$, and circumradius $R$, prove that $|O H|<3 R$.
|
Suppose with loss of generality that $\angle A<90^{\circ}$. Let $B D$ be an altitude. Then
$$
A H=\frac{A D}{\cos \left(90^{\circ}-C\right)}=\frac{A B \cos A}{\sin C}=2 R \cos A
$$
By the triangle inequality,
$$
O H<A O+A H=R+2 R \cos A<3 R
$$
Comment: With a bit more work, if $a, b, c$ are the sidelengths of $A B C$, one can show that
$$
O H^{2}=9 R^{2}-a^{2}-b^{2}-c^{2} .
$$
In fact, using vectors in a coordinate system with $O$ as origin, by the Euler line
$$
\overrightarrow{O H}=3 \overrightarrow{O G}=3 \cdot \frac{\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}}{3}=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}
$$
so
$$
O H^{2}=\overrightarrow{O H} \cdot \overrightarrow{O H}=(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \cdot(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C})
$$
Expanding and using the fact that $\overrightarrow{O X} \cdot \overrightarrow{O X}=O X^{2}=R^{2}$ for $X \in\{A, B, C\}$, as well as
$\overrightarrow{O A} \cdot \overrightarrow{O B}=O A \cdot O B \cdot \cos \angle A O B=R^{2} \cos 2 C=R^{2}\left(1-2 \sin ^{2} C\right)=R^{2}\left(1-2\left(\frac{c}{2 R}\right)^{2}\right)=R^{2}-\frac{c^{2}}{2}$, we find that
$$
\begin{aligned}
O H^{2} & =\overrightarrow{O A} \cdot \overrightarrow{O A}+\overrightarrow{O B} \cdot \overrightarrow{O B}+\overrightarrow{O C} \cdot \overrightarrow{O C}+2 \overrightarrow{O A} \cdot \overrightarrow{O B}+2 \overrightarrow{O A} \cdot \overrightarrow{O C}+2 \overrightarrow{O B} \cdot \overrightarrow{O C} \\
& =3 R^{2}+\left(2 R^{2}-c^{2}\right)+\left(2 R^{2}-b^{2}\right)+\left(2 R^{2}-a^{2}\right) \\
& =9 R^{2}-a^{2}-b^{2}-c^{2}
\end{aligned}
$$
as required.
This proves that $O H^{2}<9 R^{2} \Longrightarrow O H<3 R$, and since $a, b, c$ can be arbitrarily small (fix the circumcircle and choose $A, B, C$ arbitrarily close in this circle), the bound is sharp.
|
{
"resource_path": "APMO/segmented/en-apmo1994_sol.jsonl",
"problem_match": "# Problem 2",
"solution_match": "# Solution 2"
}
|
392d101e-6071-52b8-9672-760a0d3498f8
| 604,567
|
Is there an infinite set of points in the plane such that no three points are collinear, and the distance between any two points is rational?
Answer: Yes.
|
The answer is yes and we present the following construction: the idea is considering points in the unit circle of the form $P_{n}=(\cos (2 n \theta), \sin (2 n \theta))$ for an appropriate $\theta$. Then the distance $P_{m} P_{n}$ is the length of the chord with central angle $(2 m-2 n) \theta \bmod \pi$, that is, $2|\sin ((m-n) \theta)|$. Our task is then finding $\theta$ such that (i) $\sin (k \theta)$ is rational for all $k \in \mathbb{Z}$; (ii) points $P_{n}$ are all distinct. We claim that $\theta \in(0, \pi / 2)$ such that $\cos \theta=\frac{3}{5}$ and therefore $\sin \theta=\frac{4}{5}$ does the job. Proof of (i): We know that $\sin ((n+1) \theta)+\sin ((n-1) \theta)=2 \sin (n \theta) \cos \theta$, so if $\sin ((n-1) \theta$ and $\sin (n \theta)$ are both rational then $\sin ((n+1) \theta)$ also is. Since $\sin (0 \theta)=0$ and $\sin \theta$ are rational, an induction shows that $\sin (n \theta)$ is rational for $n \in \mathbb{Z}_{>0}$; the result is also true if $n$ is negative because $\sin$ is an odd function.
Proof of (ii): $P_{m}=P_{n} \Longleftrightarrow 2 n \theta=2 m \theta+2 k \pi$ for some $k \in \mathbb{Z}$, which implies $\sin ((n-m) \theta)=$ $\sin (k \pi)=0$. We show that $\sin (k \theta) \neq 0$ for all $k \neq 0$.
We prove a stronger result: let $\sin (k \theta)=\frac{a_{k}}{5^{k}}$. Then
$$
\begin{aligned}
\sin ((k+1) \theta)+\sin ((k-1) \theta)=2 \sin (k \theta) \cos \theta & \Longleftrightarrow \frac{a_{k+1}}{5^{k+1}}+\frac{a_{k-1}}{5^{k-1}}=2 \cdot \frac{a_{k}}{5^{k}} \cdot \frac{3}{5} \\
& \Longleftrightarrow a_{k+1}=6 a_{k}-25 a_{k-1}
\end{aligned}
$$
Since $a_{0}=0$ and $a_{1}=4, a_{k}$ is an integer for $k \geq 0$, and $a_{k+1} \equiv a_{k}(\bmod 5)$ for $k \geq 1$ (note that $a_{-1}=-\frac{4}{25}$ is not an integer!). Thus $a_{k} \equiv 4(\bmod 5)$ for all $k \geq 1$, and $\sin (k \theta)=\frac{a_{k}}{5^{k}}$ is an irreducible fraction with $5^{k}$ as denominator and $a_{k} \equiv 4(\bmod 5)$. This proves (ii) and we are done.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Is there an infinite set of points in the plane such that no three points are collinear, and the distance between any two points is rational?
Answer: Yes.
|
The answer is yes and we present the following construction: the idea is considering points in the unit circle of the form $P_{n}=(\cos (2 n \theta), \sin (2 n \theta))$ for an appropriate $\theta$. Then the distance $P_{m} P_{n}$ is the length of the chord with central angle $(2 m-2 n) \theta \bmod \pi$, that is, $2|\sin ((m-n) \theta)|$. Our task is then finding $\theta$ such that (i) $\sin (k \theta)$ is rational for all $k \in \mathbb{Z}$; (ii) points $P_{n}$ are all distinct. We claim that $\theta \in(0, \pi / 2)$ such that $\cos \theta=\frac{3}{5}$ and therefore $\sin \theta=\frac{4}{5}$ does the job. Proof of (i): We know that $\sin ((n+1) \theta)+\sin ((n-1) \theta)=2 \sin (n \theta) \cos \theta$, so if $\sin ((n-1) \theta$ and $\sin (n \theta)$ are both rational then $\sin ((n+1) \theta)$ also is. Since $\sin (0 \theta)=0$ and $\sin \theta$ are rational, an induction shows that $\sin (n \theta)$ is rational for $n \in \mathbb{Z}_{>0}$; the result is also true if $n$ is negative because $\sin$ is an odd function.
Proof of (ii): $P_{m}=P_{n} \Longleftrightarrow 2 n \theta=2 m \theta+2 k \pi$ for some $k \in \mathbb{Z}$, which implies $\sin ((n-m) \theta)=$ $\sin (k \pi)=0$. We show that $\sin (k \theta) \neq 0$ for all $k \neq 0$.
We prove a stronger result: let $\sin (k \theta)=\frac{a_{k}}{5^{k}}$. Then
$$
\begin{aligned}
\sin ((k+1) \theta)+\sin ((k-1) \theta)=2 \sin (k \theta) \cos \theta & \Longleftrightarrow \frac{a_{k+1}}{5^{k+1}}+\frac{a_{k-1}}{5^{k-1}}=2 \cdot \frac{a_{k}}{5^{k}} \cdot \frac{3}{5} \\
& \Longleftrightarrow a_{k+1}=6 a_{k}-25 a_{k-1}
\end{aligned}
$$
Since $a_{0}=0$ and $a_{1}=4, a_{k}$ is an integer for $k \geq 0$, and $a_{k+1} \equiv a_{k}(\bmod 5)$ for $k \geq 1$ (note that $a_{-1}=-\frac{4}{25}$ is not an integer!). Thus $a_{k} \equiv 4(\bmod 5)$ for all $k \geq 1$, and $\sin (k \theta)=\frac{a_{k}}{5^{k}}$ is an irreducible fraction with $5^{k}$ as denominator and $a_{k} \equiv 4(\bmod 5)$. This proves (ii) and we are done.
|
{
"resource_path": "APMO/segmented/en-apmo1994_sol.jsonl",
"problem_match": "# Problem 4",
"solution_match": "# Solution 1"
}
|
1b841da3-cad4-5ef6-90dc-95513070c396
| 604,608
|
Is there an infinite set of points in the plane such that no three points are collinear, and the distance between any two points is rational?
Answer: Yes.
|
We present a different construction. Consider the (collinear) points
$$
P_{k}=\left(1, \frac{x_{k}}{y_{k}}\right),
$$
such that the distance $O P_{k}$ from the origin $O$,
$$
O P_{k}=\frac{\sqrt{x_{k}^{2}+y_{k}^{2}}}{y_{k}}
$$
is rational, and $x_{k}$ and $y_{k}$ are integers. Clearly, $P_{i} P_{j}=\left|\frac{x_{i}}{y_{i}}-\frac{x_{j}}{y_{j}}\right|$ is rational.
Perform an inversion with center $O$ and unit radius. It maps the line $x=1$, which contains all points $P_{k}$, to a circle (minus the origin). Let $Q_{k}$ be the image of $P_{k}$ under this inversion. Then
$$
Q_{i} Q_{j}=\frac{1^{2} P_{i} P_{j}}{O P_{i} \cdot O P_{j}}
$$
is rational and we are done if we choose $x_{k}$ and $y_{k}$ accordingly. But this is not hard, as we can choose the legs of a Pythagorean triple, say
$$
x_{k}=k^{2}-1, \quad y_{k}=2 k
$$
This implies $O P_{k}=\frac{k^{2}+1}{2 k}$, and then
$$
Q_{i} Q_{j}=\frac{\left|\frac{i^{2}-1}{i}-\frac{j^{2}-1}{j}\right|}{\frac{i^{2}+1}{2 i} \cdot \frac{j^{2}+1}{2 j}}=\frac{|4(i-j)(i j+1)|}{\left(i^{2}+1\right)\left(j^{2}+1\right)}
$$
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Is there an infinite set of points in the plane such that no three points are collinear, and the distance between any two points is rational?
Answer: Yes.
|
We present a different construction. Consider the (collinear) points
$$
P_{k}=\left(1, \frac{x_{k}}{y_{k}}\right),
$$
such that the distance $O P_{k}$ from the origin $O$,
$$
O P_{k}=\frac{\sqrt{x_{k}^{2}+y_{k}^{2}}}{y_{k}}
$$
is rational, and $x_{k}$ and $y_{k}$ are integers. Clearly, $P_{i} P_{j}=\left|\frac{x_{i}}{y_{i}}-\frac{x_{j}}{y_{j}}\right|$ is rational.
Perform an inversion with center $O$ and unit radius. It maps the line $x=1$, which contains all points $P_{k}$, to a circle (minus the origin). Let $Q_{k}$ be the image of $P_{k}$ under this inversion. Then
$$
Q_{i} Q_{j}=\frac{1^{2} P_{i} P_{j}}{O P_{i} \cdot O P_{j}}
$$
is rational and we are done if we choose $x_{k}$ and $y_{k}$ accordingly. But this is not hard, as we can choose the legs of a Pythagorean triple, say
$$
x_{k}=k^{2}-1, \quad y_{k}=2 k
$$
This implies $O P_{k}=\frac{k^{2}+1}{2 k}$, and then
$$
Q_{i} Q_{j}=\frac{\left|\frac{i^{2}-1}{i}-\frac{j^{2}-1}{j}\right|}{\frac{i^{2}+1}{2 i} \cdot \frac{j^{2}+1}{2 j}}=\frac{|4(i-j)(i j+1)|}{\left(i^{2}+1\right)\left(j^{2}+1\right)}
$$
|
{
"resource_path": "APMO/segmented/en-apmo1994_sol.jsonl",
"problem_match": "# Problem 4",
"solution_match": "# Solution 2"
}
|
1b841da3-cad4-5ef6-90dc-95513070c396
| 604,608
|
You are given three lists $A, B$, and $C$. List $A$ contains the numbers of the form $10^{k}$ in base 10, with $k$ any integer greater than or equal to 1 . Lists $B$ and $C$ contain the same numbers translated into base 2 and 5 respectively:
| $A$ | $B$ | $C$ |
| :--- | :--- | :--- |
| 10 | 1010 | 20 |
| 100 | 1100100 | 400 |
| 1000 | 1111101000 | 13000 |
| $\vdots$ | $\vdots$ | $\vdots$ |
Prove that for every integer $n>1$, there is exactly one number in exactly one of the lists $B$ or $C$ that has exactly $n$ digits.
|
Let $b_{k}$ and $c_{k}$ be the number of digits in the $k$ th term in lists $B$ and $C$, respectively. Then
$$
2^{b_{k}-1} \leq 10^{k}<2^{b_{k}} \Longleftrightarrow \log _{2} 10^{k}<b_{k} \leq \log _{2} 10^{k}+1 \Longleftrightarrow b_{k}=\left\lfloor k \cdot \log _{2} 10\right\rfloor+1
$$
and, similarly
$$
c_{k}=\left\lfloor k \cdot \log _{5} 10\right\rfloor+1
$$
Beatty's theorem states that if $\alpha$ and $\beta$ are irrational positive numbers such that
$$
\frac{1}{\alpha}+\frac{1}{\beta}=1
$$
then the sequences $\lfloor k \alpha\rfloor$ and $\lfloor k \beta\rfloor, k=1,2, \ldots$, partition the positive integers.
Then, since
$$
\frac{1}{\log _{2} 10}+\frac{1}{\log _{5} 10}=\log _{10} 2+\log _{10} 5=\log _{10}(2 \cdot 5)=1
$$
the sequences $b_{k}-1$ and $c_{k}-1$ partition the positive integers, and therefore each integer greater than 1 appears in $b_{k}$ or $c_{k}$ exactly once. We are done.
Comment: For the sake of completeness, a proof of Beatty's theorem follows.
Let $x_{n}=\alpha n$ and $y_{n}=\beta n, n \geq 1$ integer. Note that, since $\alpha m=\beta n$ implies that $\frac{\alpha}{\beta}$ is rational but
$$
\frac{\alpha}{\beta}=\alpha \cdot \frac{1}{\beta}=\alpha\left(1-\frac{1}{\alpha}\right)=\alpha-1
$$
is irrational, the sequences have no common terms, and all terms in both sequences are irrational.
The theorem is equivalent to proving that exactly one term of either $x_{n}$ of $y_{n}$ lies in the interval $(N, N+1)$ for each $N$ positive integer. For that purpose we count the number of terms of the union of the two sequences in the interval $(0, N)$ : since $n \alpha<N \Longleftrightarrow n<\frac{N}{\alpha}$, there are $\left\lfloor\frac{N}{\alpha}\right\rfloor$ terms of $x_{n}$ in the interval and, similarly, $\left\lfloor\frac{N}{\beta}\right\rfloor$ terms of $y_{n}$ in the same interval. Since the sequences are disjoint, the total of numbers is
$$
T(N)=\left\lfloor\frac{N}{\alpha}\right\rfloor+\left\lfloor\frac{N}{\beta}\right\rfloor
$$
However, $x-1<\lfloor x\rfloor<x$ for nonintegers $x$, so
$$
\begin{aligned}
\frac{N}{\alpha}-1+\frac{N}{\beta}-1<T(N)<\frac{N}{\alpha}+\frac{N}{\beta} & \Longleftrightarrow N\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)-2<T(N)<N\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \\
& \Longleftrightarrow N-2<T(N)<N,
\end{aligned}
$$
that is, $T(N)=N-1$.
Therefore the number of terms in $(N, N+1)$ is $T(N+1)-T(N)=N-(N-1)=1$, and the result follows.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
You are given three lists $A, B$, and $C$. List $A$ contains the numbers of the form $10^{k}$ in base 10, with $k$ any integer greater than or equal to 1 . Lists $B$ and $C$ contain the same numbers translated into base 2 and 5 respectively:
| $A$ | $B$ | $C$ |
| :--- | :--- | :--- |
| 10 | 1010 | 20 |
| 100 | 1100100 | 400 |
| 1000 | 1111101000 | 13000 |
| $\vdots$ | $\vdots$ | $\vdots$ |
Prove that for every integer $n>1$, there is exactly one number in exactly one of the lists $B$ or $C$ that has exactly $n$ digits.
|
Let $b_{k}$ and $c_{k}$ be the number of digits in the $k$ th term in lists $B$ and $C$, respectively. Then
$$
2^{b_{k}-1} \leq 10^{k}<2^{b_{k}} \Longleftrightarrow \log _{2} 10^{k}<b_{k} \leq \log _{2} 10^{k}+1 \Longleftrightarrow b_{k}=\left\lfloor k \cdot \log _{2} 10\right\rfloor+1
$$
and, similarly
$$
c_{k}=\left\lfloor k \cdot \log _{5} 10\right\rfloor+1
$$
Beatty's theorem states that if $\alpha$ and $\beta$ are irrational positive numbers such that
$$
\frac{1}{\alpha}+\frac{1}{\beta}=1
$$
then the sequences $\lfloor k \alpha\rfloor$ and $\lfloor k \beta\rfloor, k=1,2, \ldots$, partition the positive integers.
Then, since
$$
\frac{1}{\log _{2} 10}+\frac{1}{\log _{5} 10}=\log _{10} 2+\log _{10} 5=\log _{10}(2 \cdot 5)=1
$$
the sequences $b_{k}-1$ and $c_{k}-1$ partition the positive integers, and therefore each integer greater than 1 appears in $b_{k}$ or $c_{k}$ exactly once. We are done.
Comment: For the sake of completeness, a proof of Beatty's theorem follows.
Let $x_{n}=\alpha n$ and $y_{n}=\beta n, n \geq 1$ integer. Note that, since $\alpha m=\beta n$ implies that $\frac{\alpha}{\beta}$ is rational but
$$
\frac{\alpha}{\beta}=\alpha \cdot \frac{1}{\beta}=\alpha\left(1-\frac{1}{\alpha}\right)=\alpha-1
$$
is irrational, the sequences have no common terms, and all terms in both sequences are irrational.
The theorem is equivalent to proving that exactly one term of either $x_{n}$ of $y_{n}$ lies in the interval $(N, N+1)$ for each $N$ positive integer. For that purpose we count the number of terms of the union of the two sequences in the interval $(0, N)$ : since $n \alpha<N \Longleftrightarrow n<\frac{N}{\alpha}$, there are $\left\lfloor\frac{N}{\alpha}\right\rfloor$ terms of $x_{n}$ in the interval and, similarly, $\left\lfloor\frac{N}{\beta}\right\rfloor$ terms of $y_{n}$ in the same interval. Since the sequences are disjoint, the total of numbers is
$$
T(N)=\left\lfloor\frac{N}{\alpha}\right\rfloor+\left\lfloor\frac{N}{\beta}\right\rfloor
$$
However, $x-1<\lfloor x\rfloor<x$ for nonintegers $x$, so
$$
\begin{aligned}
\frac{N}{\alpha}-1+\frac{N}{\beta}-1<T(N)<\frac{N}{\alpha}+\frac{N}{\beta} & \Longleftrightarrow N\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)-2<T(N)<N\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \\
& \Longleftrightarrow N-2<T(N)<N,
\end{aligned}
$$
that is, $T(N)=N-1$.
Therefore the number of terms in $(N, N+1)$ is $T(N+1)-T(N)=N-(N-1)=1$, and the result follows.
|
{
"resource_path": "APMO/segmented/en-apmo1994_sol.jsonl",
"problem_match": "# Problem 5",
"solution_match": "# Solution\n\n"
}
|
2c3fb340-39e9-52e3-a2a8-1aeebdcc1da4
| 604,631
|
Let $a_{1}, a_{2}, \cdots$ be a sequence of real numbers satisfying $a_{i+j} \leq a_{i}+a_{j}$ for all $i, j=1,2, \cdots$. Prove that
$$
a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n} \geq a_{n}
$$
for each positive integer $n$.
|
and Marking Scheme:
Letting $b_{i}=a_{i} / i,(i=1,2, \cdots)$, we prove that
$$
b_{1}+\cdots+b_{n} \geq a_{n} \quad(n=1,2, \cdots)
$$
by induction on $n$. For $n=1, b_{1}=a_{1} \geq a_{1}$, and the induction starts. Assume that
$$
b_{1}+\cdots+b_{k} \geq a_{k}
$$
for all $k=1,2, \cdots, n-1$. It suffices to prove that $b_{1}+\cdots+b_{n} \geq a_{n}$ or equivalently that
$$
\begin{aligned}
& n b_{1}+\cdots+n b_{n-1} \geq(n-1) a_{n} . \\
& 3 \text { POINTS for separating } a_{n} \text { from } b_{1}, \cdots, b_{n-1} \text {. } \\
& n b_{1}+\cdots+n b_{n-1}=(n-1) b_{1}+(n-2) b_{2}+\cdots+b_{n-1}+b_{1}+2 b_{2}+\cdots+(n-1) b_{n-1} \\
& =b_{1}+\left(b_{1}+b_{2}\right)+\cdots+\left(b_{1}+b_{2}+\cdots+b_{n-1}\right)+\left(a_{1}+a_{2}+\cdots+a_{n-1}\right) \\
& \geq 2\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)=\sum_{i=1}^{n-1}\left(a_{i}+a_{n-i}\right) \geq(n-1) a_{n} .
\end{aligned}
$$
3 POINTS for the first inequaliny and 1 POINT for the rest.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a_{1}, a_{2}, \cdots$ be a sequence of real numbers satisfying $a_{i+j} \leq a_{i}+a_{j}$ for all $i, j=1,2, \cdots$. Prove that
$$
a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n} \geq a_{n}
$$
for each positive integer $n$.
|
and Marking Scheme:
Letting $b_{i}=a_{i} / i,(i=1,2, \cdots)$, we prove that
$$
b_{1}+\cdots+b_{n} \geq a_{n} \quad(n=1,2, \cdots)
$$
by induction on $n$. For $n=1, b_{1}=a_{1} \geq a_{1}$, and the induction starts. Assume that
$$
b_{1}+\cdots+b_{k} \geq a_{k}
$$
for all $k=1,2, \cdots, n-1$. It suffices to prove that $b_{1}+\cdots+b_{n} \geq a_{n}$ or equivalently that
$$
\begin{aligned}
& n b_{1}+\cdots+n b_{n-1} \geq(n-1) a_{n} . \\
& 3 \text { POINTS for separating } a_{n} \text { from } b_{1}, \cdots, b_{n-1} \text {. } \\
& n b_{1}+\cdots+n b_{n-1}=(n-1) b_{1}+(n-2) b_{2}+\cdots+b_{n-1}+b_{1}+2 b_{2}+\cdots+(n-1) b_{n-1} \\
& =b_{1}+\left(b_{1}+b_{2}\right)+\cdots+\left(b_{1}+b_{2}+\cdots+b_{n-1}\right)+\left(a_{1}+a_{2}+\cdots+a_{n-1}\right) \\
& \geq 2\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)=\sum_{i=1}^{n-1}\left(a_{i}+a_{n-i}\right) \geq(n-1) a_{n} .
\end{aligned}
$$
3 POINTS for the first inequaliny and 1 POINT for the rest.
|
{
"resource_path": "APMO/segmented/en-apmo1999_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "# Solution "
}
|
c5defa5c-148f-53da-87e8-0326891c55f5
| 604,662
|
Let $\Gamma_{1}$ and $\Gamma_{2}$ be two circles intersecting at $P$ and $Q$. The common tangent, closer to $P$, of $\Gamma_{1}$ and $\Gamma_{2}$ touches $\Gamma_{1}$ at $A$ and $\Gamma_{2}$ at $B$. The tangent of $\Gamma_{1}$ at $P$ meets $\Gamma_{2}$ at $C$, which is different from $P$ and the extension of $A P$ meets $B C$ at $R$. Prove that the circumcircle of triangle $P Q R$ is tangent to $B P$ and $B R$.
|
and Marking Scheme:
Let $\alpha=\angle P A B, \beta=\angle A B P$ y $\gamma=\angle Q A P$. Then, since $P C$ is tangent to $\Gamma_{1}$, we have $\angle Q P C=$ $\angle Q B C=\gamma$. Thus $A, B, R, Q$ are concyclic.
3 POINTS for proving that $A, B, R, Q$ are concyclic.
Since $A B$ is a common tangent to $\Gamma_{1}$ and $\Gamma_{2}$ then $\angle A Q P=\alpha$ and $\angle P Q B=\angle P C B=\beta$. Therefore, since $A, B, R, Q$ are concyclic, $\angle A R B=\angle A Q B=\alpha+\beta$ and $\angle B Q R=\alpha$. Thus $\angle P Q R=\angle P Q B+$ $\angle B Q R=\alpha+\beta$.
$$
2 \text { POINTS for proving that } \angle P Q R=\angle P R B=\alpha+\beta
$$
Since $\angle B P R$ is an exterior angle of triangle $A B P, \angle B P R=\alpha+\beta$. We have
$$
\angle P Q R=\angle B P R=\angle B R P
$$
1 POINT for proving $\angle B P R=\alpha+\beta$.
So circumcircle of $P Q R$ is tangent to $B P$ and $B R$.
1 POINT for concluding.
Remark. 2POINTS can be given for proving that $\angle P R B=\angle R P B$ and 1 more POINT for attempting to prove (unsuccessfully) that $\angle P R B=\angle R P B=\angle P Q R$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $\Gamma_{1}$ and $\Gamma_{2}$ be two circles intersecting at $P$ and $Q$. The common tangent, closer to $P$, of $\Gamma_{1}$ and $\Gamma_{2}$ touches $\Gamma_{1}$ at $A$ and $\Gamma_{2}$ at $B$. The tangent of $\Gamma_{1}$ at $P$ meets $\Gamma_{2}$ at $C$, which is different from $P$ and the extension of $A P$ meets $B C$ at $R$. Prove that the circumcircle of triangle $P Q R$ is tangent to $B P$ and $B R$.
|
and Marking Scheme:
Let $\alpha=\angle P A B, \beta=\angle A B P$ y $\gamma=\angle Q A P$. Then, since $P C$ is tangent to $\Gamma_{1}$, we have $\angle Q P C=$ $\angle Q B C=\gamma$. Thus $A, B, R, Q$ are concyclic.
3 POINTS for proving that $A, B, R, Q$ are concyclic.
Since $A B$ is a common tangent to $\Gamma_{1}$ and $\Gamma_{2}$ then $\angle A Q P=\alpha$ and $\angle P Q B=\angle P C B=\beta$. Therefore, since $A, B, R, Q$ are concyclic, $\angle A R B=\angle A Q B=\alpha+\beta$ and $\angle B Q R=\alpha$. Thus $\angle P Q R=\angle P Q B+$ $\angle B Q R=\alpha+\beta$.
$$
2 \text { POINTS for proving that } \angle P Q R=\angle P R B=\alpha+\beta
$$
Since $\angle B P R$ is an exterior angle of triangle $A B P, \angle B P R=\alpha+\beta$. We have
$$
\angle P Q R=\angle B P R=\angle B R P
$$
1 POINT for proving $\angle B P R=\alpha+\beta$.
So circumcircle of $P Q R$ is tangent to $B P$ and $B R$.
1 POINT for concluding.
Remark. 2POINTS can be given for proving that $\angle P R B=\angle R P B$ and 1 more POINT for attempting to prove (unsuccessfully) that $\angle P R B=\angle R P B=\angle P Q R$.
|
{
"resource_path": "APMO/segmented/en-apmo1999_sol.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "# Solution "
}
|
5a5af135-15a1-5ad2-9952-8863669c59e7
| 604,674
|
Let $A B C$ be a triangle. Let $M$ and $N$ be the points in which the median and angle bisector, respectively, at $A$ meet the side $B C$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $N A$ meets $M A$ and $B A$, respectively, and $O$ be the point in which the perpendicular at $P$ to $B A$ meets $A N$ produced. Prove that $Q O$ is perpendicular to $B C$.
|
Let $A N$ meet the circumcircle of $A B C$ at point $K$, the midpoint of arc $B C$ that does not contain $A$.
The orthogonal projection of $K$ onto side $B C$ is $M$. Let $R$ and $S$ be the orthogonal projections of $K$ onto lines $A B$ and $A C$, respectively. Points $R, M$, and $S$ lie in the Simson line of $K$ with respect to $A B C$. Since $K$ is in the bisector of $\angle B A C, A R K S$ is a kite, and the Simson line $R M S$ is perpendicular to $A N$, and therefore parallel to $P Q$.
Now consider the homothety with center $A$ that takes $O$ to $K$. Since $O P \perp A B$ and $K R \perp A B$, $O P$ and $K R$ are parallel, which means that $P$ is taken to $R$. Finally, line $P Q$ is parallel to line $R S$, so line $P Q$ is taken to line $R S$ by the homothety. Then $Q$ is taken to $M$, and since $O$ is taken to $K$, line $O Q$ is taken to line $M K$. We are done now: this means that $O Q$ is parallel to $M K$, which is perpendicular to $B C$ (it is its perpendicular bisector, as $M B=M C$ and $K B=K C$.)
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle. Let $M$ and $N$ be the points in which the median and angle bisector, respectively, at $A$ meet the side $B C$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $N A$ meets $M A$ and $B A$, respectively, and $O$ be the point in which the perpendicular at $P$ to $B A$ meets $A N$ produced. Prove that $Q O$ is perpendicular to $B C$.
|
Let $A N$ meet the circumcircle of $A B C$ at point $K$, the midpoint of arc $B C$ that does not contain $A$.
The orthogonal projection of $K$ onto side $B C$ is $M$. Let $R$ and $S$ be the orthogonal projections of $K$ onto lines $A B$ and $A C$, respectively. Points $R, M$, and $S$ lie in the Simson line of $K$ with respect to $A B C$. Since $K$ is in the bisector of $\angle B A C, A R K S$ is a kite, and the Simson line $R M S$ is perpendicular to $A N$, and therefore parallel to $P Q$.
Now consider the homothety with center $A$ that takes $O$ to $K$. Since $O P \perp A B$ and $K R \perp A B$, $O P$ and $K R$ are parallel, which means that $P$ is taken to $R$. Finally, line $P Q$ is parallel to line $R S$, so line $P Q$ is taken to line $R S$ by the homothety. Then $Q$ is taken to $M$, and since $O$ is taken to $K$, line $O Q$ is taken to line $M K$. We are done now: this means that $O Q$ is parallel to $M K$, which is perpendicular to $B C$ (it is its perpendicular bisector, as $M B=M C$ and $K B=K C$.)
|
{
"resource_path": "APMO/segmented/en-apmo2000_sol.jsonl",
"problem_match": "# Problem 3",
"solution_match": "# Solution 1"
}
|
8fe249f3-03d2-50eb-80d7-503be84cddfd
| 295,099
|
Let $A B C$ be a triangle. Let $M$ and $N$ be the points in which the median and angle bisector, respectively, at $A$ meet the side $B C$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $N A$ meets $M A$ and $B A$, respectively, and $O$ be the point in which the perpendicular at $P$ to $B A$ meets $A N$ produced. Prove that $Q O$ is perpendicular to $B C$.
|
Consider a cartesian plane with $A=(0,0)$ as the origin and the bisector $A N$ as $x$-axis. Thus $A B$ has equation $y=m x$ and $A C$ has equation $y=-m x$. Let $B=(b, m b)$ and $C=(c,-m c)$. By symmetry, the problem is immediate if $A B=A C$, that is, if $b=c$. Suppose that $b \neq c$ from now on. Line $B C$ has slope $\frac{m b-(-m c)}{b-c}=\frac{m(b+c)}{b-c}$. Let $N=(n, 0)$.
Point $M$ is the midpoint $\left(\frac{b+c}{2}, \frac{m b-m c}{2}\right)$ of $B C$, so $A M$ has slope $\frac{m(b-c)}{b+c}$.
The line through $N$ that is perpendicular to the $x$-axis $A N$ is $x=n$. Therefore
$$
P=(n, m n) \quad \text { and } \quad Q=\left(n, \frac{m(b-c) n}{b+c}\right) .
$$
In the right triangle $A P O$, with altitude $A N, A N \cdot A O=A P^{2}$. Thus
$$
n \cdot A O=(0-n)^{2}+(0-m n)^{2} \Longleftrightarrow A O=n\left(m^{2}+1\right) \Longrightarrow O=\left(n\left(m^{2}+1\right), 0\right)
$$
Finally, the slope of $O Q$ is
$$
\frac{\frac{m(b-c) n}{b+c}-0}{n-n\left(m^{2}+1\right)}=-\frac{b-c}{(b+c) m}
$$
Since the product of the slopes of $O Q$ and $B C$ is
$$
-\frac{b-c}{(b+c) m} \cdot \frac{m(b+c)}{b-c}=-1
$$
$O Q$ and $B C$ are perpendicular, and we are done.
Comment: The second solution shows that $N$ can be any point in the bisector of $\angle A$. In fact, if we move $N$ in the bisector and construct $O, P$ and $Q$ accordingly, then all lines $O Q$ obtained are parallel: just consider a homothety with center $A$ and variable ratios.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle. Let $M$ and $N$ be the points in which the median and angle bisector, respectively, at $A$ meet the side $B C$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $N A$ meets $M A$ and $B A$, respectively, and $O$ be the point in which the perpendicular at $P$ to $B A$ meets $A N$ produced. Prove that $Q O$ is perpendicular to $B C$.
|
Consider a cartesian plane with $A=(0,0)$ as the origin and the bisector $A N$ as $x$-axis. Thus $A B$ has equation $y=m x$ and $A C$ has equation $y=-m x$. Let $B=(b, m b)$ and $C=(c,-m c)$. By symmetry, the problem is immediate if $A B=A C$, that is, if $b=c$. Suppose that $b \neq c$ from now on. Line $B C$ has slope $\frac{m b-(-m c)}{b-c}=\frac{m(b+c)}{b-c}$. Let $N=(n, 0)$.
Point $M$ is the midpoint $\left(\frac{b+c}{2}, \frac{m b-m c}{2}\right)$ of $B C$, so $A M$ has slope $\frac{m(b-c)}{b+c}$.
The line through $N$ that is perpendicular to the $x$-axis $A N$ is $x=n$. Therefore
$$
P=(n, m n) \quad \text { and } \quad Q=\left(n, \frac{m(b-c) n}{b+c}\right) .
$$
In the right triangle $A P O$, with altitude $A N, A N \cdot A O=A P^{2}$. Thus
$$
n \cdot A O=(0-n)^{2}+(0-m n)^{2} \Longleftrightarrow A O=n\left(m^{2}+1\right) \Longrightarrow O=\left(n\left(m^{2}+1\right), 0\right)
$$
Finally, the slope of $O Q$ is
$$
\frac{\frac{m(b-c) n}{b+c}-0}{n-n\left(m^{2}+1\right)}=-\frac{b-c}{(b+c) m}
$$
Since the product of the slopes of $O Q$ and $B C$ is
$$
-\frac{b-c}{(b+c) m} \cdot \frac{m(b+c)}{b-c}=-1
$$
$O Q$ and $B C$ are perpendicular, and we are done.
Comment: The second solution shows that $N$ can be any point in the bisector of $\angle A$. In fact, if we move $N$ in the bisector and construct $O, P$ and $Q$ accordingly, then all lines $O Q$ obtained are parallel: just consider a homothety with center $A$ and variable ratios.
|
{
"resource_path": "APMO/segmented/en-apmo2000_sol.jsonl",
"problem_match": "# Problem 3",
"solution_match": "# Solution 2"
}
|
8fe249f3-03d2-50eb-80d7-503be84cddfd
| 295,099
|
Let $n, k$ be given positive integers with $n>k$. Prove that
$$
\frac{1}{n+1} \cdot \frac{n^{n}}{k^{k}(n-k)^{n-k}}<\frac{n!}{k!(n-k)!}<\frac{n^{n}}{k^{k}(n-k)^{n-k}} .
$$
|
The inequality is equivalent to
$$
\frac{n^{n}}{n+1}<\binom{n}{k} k^{k}(n-k)^{n-k}<n^{n}
$$
which suggests investigating the binomial expansion of
$$
n^{n}=((n-k)+k)^{n}=\sum_{i=0}^{n}\binom{n}{i}(n-k)^{n-i} k^{i}
$$
The $(k+1)$ th term $T_{k+1}$ of the expansion is $\binom{n}{k} k^{k}(n-k)^{n-k}$, and all terms in the expansion are positive, which implies the right inequality.
Now, for $1 \leq i \leq n$,
$$
\frac{T_{i+1}}{T_{i}}=\frac{\binom{n}{i}(n-k)^{n-i} k^{i}}{\binom{n}{i-1}(n-k)^{n-i+1} k^{i-1}}=\frac{(n-i+1) k}{i(n-k)}
$$
and
$$
\frac{T_{i+1}}{T_{i}}>1 \Longleftrightarrow(n-i+1) k>i(n-k) \Longleftrightarrow i<k+\frac{k}{n} \Longleftrightarrow i \leq k
$$
This means that
$$
T_{1}<T_{2}<\cdots<T_{k+1}>T_{k+2}>\cdots>T_{n+1}
$$
that is, $T_{k+1}=\binom{n}{k} k^{k}(n-k)^{n-k}$ is the largest term in the expansion. The maximum term is greater that the average, which is the sum $n^{n}$ divided by the quantity $n+1$, therefore
$$
\binom{n}{k} k^{k}(n-k)^{n-k}>\frac{n^{n}}{n+1}
$$
as required.
Comment: If we divide further by $n^{n}$ one finds
$$
\frac{1}{n+1}<\binom{n}{k}\left(\frac{k}{n}\right)^{k}\left(1-\frac{k}{n}\right)^{n-k}<1
$$
The middle term is the probability $P(X=k)$ of $k$ successes in a binomial distribution with $n$ trials and success probability $p=\frac{k}{n}$. The right inequality is immediate from the fact that $P(X=k)$ is not the only possible event in this distribution, and the left inequality comes from the fact that the mode of the binomial distribution are given by $\lfloor(n+1) p\rfloor=\left\lfloor(n+1) \frac{k}{n}\right\rfloor=k$ and $\lceil(n+1) p-1\rceil=k$. However, the proof of this fact is identical to the above solution.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $n, k$ be given positive integers with $n>k$. Prove that
$$
\frac{1}{n+1} \cdot \frac{n^{n}}{k^{k}(n-k)^{n-k}}<\frac{n!}{k!(n-k)!}<\frac{n^{n}}{k^{k}(n-k)^{n-k}} .
$$
|
The inequality is equivalent to
$$
\frac{n^{n}}{n+1}<\binom{n}{k} k^{k}(n-k)^{n-k}<n^{n}
$$
which suggests investigating the binomial expansion of
$$
n^{n}=((n-k)+k)^{n}=\sum_{i=0}^{n}\binom{n}{i}(n-k)^{n-i} k^{i}
$$
The $(k+1)$ th term $T_{k+1}$ of the expansion is $\binom{n}{k} k^{k}(n-k)^{n-k}$, and all terms in the expansion are positive, which implies the right inequality.
Now, for $1 \leq i \leq n$,
$$
\frac{T_{i+1}}{T_{i}}=\frac{\binom{n}{i}(n-k)^{n-i} k^{i}}{\binom{n}{i-1}(n-k)^{n-i+1} k^{i-1}}=\frac{(n-i+1) k}{i(n-k)}
$$
and
$$
\frac{T_{i+1}}{T_{i}}>1 \Longleftrightarrow(n-i+1) k>i(n-k) \Longleftrightarrow i<k+\frac{k}{n} \Longleftrightarrow i \leq k
$$
This means that
$$
T_{1}<T_{2}<\cdots<T_{k+1}>T_{k+2}>\cdots>T_{n+1}
$$
that is, $T_{k+1}=\binom{n}{k} k^{k}(n-k)^{n-k}$ is the largest term in the expansion. The maximum term is greater that the average, which is the sum $n^{n}$ divided by the quantity $n+1$, therefore
$$
\binom{n}{k} k^{k}(n-k)^{n-k}>\frac{n^{n}}{n+1}
$$
as required.
Comment: If we divide further by $n^{n}$ one finds
$$
\frac{1}{n+1}<\binom{n}{k}\left(\frac{k}{n}\right)^{k}\left(1-\frac{k}{n}\right)^{n-k}<1
$$
The middle term is the probability $P(X=k)$ of $k$ successes in a binomial distribution with $n$ trials and success probability $p=\frac{k}{n}$. The right inequality is immediate from the fact that $P(X=k)$ is not the only possible event in this distribution, and the left inequality comes from the fact that the mode of the binomial distribution are given by $\lfloor(n+1) p\rfloor=\left\lfloor(n+1) \frac{k}{n}\right\rfloor=k$ and $\lceil(n+1) p-1\rceil=k$. However, the proof of this fact is identical to the above solution.
|
{
"resource_path": "APMO/segmented/en-apmo2000_sol.jsonl",
"problem_match": "# Problem 4",
"solution_match": "# Solution\n\n"
}
|
8f67957e-a757-5eb8-81fd-23175a1dd233
| 604,792
|
Let $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ be a sequence of non-negative integers, where $n$ is a positive integer.
Let
$$
A_{n}=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}
$$
Prove that
$$
a_{1}!a_{2}!\ldots a_{n}!\geq\left(\left\lfloor A_{n}\right\rfloor!\right)^{n}
$$
where $\left\lfloor A_{n}\right\rfloor$ is the greatest integer less than or equal to $A_{n}$, and $a!=1 \times 2 \times \cdots \times a$ for $a \geq 1$ (and $0!=1$ ). When does equality hold?
|
Assume without loss of generality that $a_{1} \geq a_{2} \geq \cdots \geq a_{n} \geq 0$, and let $s=\left\lfloor A_{n}\right\rfloor$. Let $k$ be any (fixed) index for which $a_{k} \geq s \geq a_{k+1}$.
Our inequality is equivalent to proving that
$$
\frac{a_{1}!}{s!} \cdot \frac{a_{2}!}{s!} \cdot \ldots \cdot \frac{a_{k}!}{s!} \geq \frac{s!}{a_{k+1}!} \cdot \frac{s!}{a_{k+2}!} \cdot \ldots \cdot \frac{s!}{a_{n}!}
$$
Now for $i=1,2, \ldots, k, a_{i}!/ s!$ is the product of $a_{i}-s$ factors. For example, 9!/5! $=9 \cdot 8 \cdot 7 \cdot 6$. The left side of inequality (1) therefore is the product of $A=a_{1}+a_{2}+\cdots+a_{k}-k s$ factors, all of which are greater than $s$. Similarly, the right side of (1) is the product of $B=(n-k) s-$ $\left(a_{k+1}+a_{k+2}+\cdots+a_{n}\right)$ factors, all of which are at most $s$. Since $\sum_{i=1}^{n} a_{i}=n A_{n} \geq n s, A \geq B$. This proves the inequality. [ 5 marks to here.]
Equality in (1) holds if and only if either:
(i) $A=B=0$, that is, both sides of (1) are the empty product, which occurs if and only if $a_{1}=a_{2}=\cdots=a_{n}$; or
(ii) $a_{1}=1$ and $s=0$, that is, the only factors on either side of (1) are 1 's, which occurs if and only if $a_{i} \in\{0,1\}$ for all $i$. [2 marks for both (i) and (ii), no marks for (i) only.]
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ be a sequence of non-negative integers, where $n$ is a positive integer.
Let
$$
A_{n}=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}
$$
Prove that
$$
a_{1}!a_{2}!\ldots a_{n}!\geq\left(\left\lfloor A_{n}\right\rfloor!\right)^{n}
$$
where $\left\lfloor A_{n}\right\rfloor$ is the greatest integer less than or equal to $A_{n}$, and $a!=1 \times 2 \times \cdots \times a$ for $a \geq 1$ (and $0!=1$ ). When does equality hold?
|
Assume without loss of generality that $a_{1} \geq a_{2} \geq \cdots \geq a_{n} \geq 0$, and let $s=\left\lfloor A_{n}\right\rfloor$. Let $k$ be any (fixed) index for which $a_{k} \geq s \geq a_{k+1}$.
Our inequality is equivalent to proving that
$$
\frac{a_{1}!}{s!} \cdot \frac{a_{2}!}{s!} \cdot \ldots \cdot \frac{a_{k}!}{s!} \geq \frac{s!}{a_{k+1}!} \cdot \frac{s!}{a_{k+2}!} \cdot \ldots \cdot \frac{s!}{a_{n}!}
$$
Now for $i=1,2, \ldots, k, a_{i}!/ s!$ is the product of $a_{i}-s$ factors. For example, 9!/5! $=9 \cdot 8 \cdot 7 \cdot 6$. The left side of inequality (1) therefore is the product of $A=a_{1}+a_{2}+\cdots+a_{k}-k s$ factors, all of which are greater than $s$. Similarly, the right side of (1) is the product of $B=(n-k) s-$ $\left(a_{k+1}+a_{k+2}+\cdots+a_{n}\right)$ factors, all of which are at most $s$. Since $\sum_{i=1}^{n} a_{i}=n A_{n} \geq n s, A \geq B$. This proves the inequality. [ 5 marks to here.]
Equality in (1) holds if and only if either:
(i) $A=B=0$, that is, both sides of (1) are the empty product, which occurs if and only if $a_{1}=a_{2}=\cdots=a_{n}$; or
(ii) $a_{1}=1$ and $s=0$, that is, the only factors on either side of (1) are 1 's, which occurs if and only if $a_{i} \in\{0,1\}$ for all $i$. [2 marks for both (i) and (ii), no marks for (i) only.]
|
{
"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl",
"problem_match": "\n1. ",
"solution_match": "# Solution 1."
}
|
7d52eb5a-0c52-5bc1-a0a3-cce54fdf8e25
| 607,114
|
Let $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ be a sequence of non-negative integers, where $n$ is a positive integer.
Let
$$
A_{n}=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}
$$
Prove that
$$
a_{1}!a_{2}!\ldots a_{n}!\geq\left(\left\lfloor A_{n}\right\rfloor!\right)^{n}
$$
where $\left\lfloor A_{n}\right\rfloor$ is the greatest integer less than or equal to $A_{n}$, and $a!=1 \times 2 \times \cdots \times a$ for $a \geq 1$ (and $0!=1$ ). When does equality hold?
|
Assume without loss of generality that $0 \leq a_{1} \leq a_{2} \leq \cdots \leq a_{n}$. Let $d=a_{n}-a_{1}$ and $m=\left|\left\{i: a_{i}=a_{1}\right\}\right|$. Our proof is by induction on $d$.
We first do the case $d=a_{n}-a_{1}=0$ or 1 separately. Then $a_{1}=a_{2}=\cdots=a_{m}=a$ and $a_{m+1}=\cdots=a_{n}=a+1$ for some $1 \leq m \leq n$ and $a \geq 0$. In this case we have $\left\lfloor A_{n}\right\rfloor=a$, so the inequality to be proven is just $a_{1}!a_{2}!\ldots a_{n}!\geq(a!)^{n}$, which is obvious. Equality holds if and only if either $m=n$, that is, $a_{1}=a_{2}=\cdots=a_{n}=a$; or if $a=0$, that is, $a_{1}=\cdots=a_{m}=0$ and $a_{m+1}=\cdots=a_{n}=1$. [ 2 marks to here.]
So assume that $d=a_{n}-a_{1} \geq 2$ and that the inequality holds for all sequences with smaller values of $d$, or with the same value of $d$ and smaller values of $m$. Then the sequence
$$
a_{1}+1, a_{2}, a_{3}, \ldots, a_{n-1}, a_{n}-1
$$
though not necessarily in non-decreasing order any more, does have either a smaller value of $d$, or the same value of $d$ and a smaller value of $m$, but in any case has the same value of $A_{n}$. Thus, by induction and since $a_{n}>a_{1}+1$,
$$
\begin{aligned}
a_{1}!a_{2}!\ldots a_{n}! & =\left(a_{1}+1\right)!a_{2}!\ldots a_{n-1}!\left(a_{n}-1\right)!\cdot \frac{a_{n}}{a_{1}+1} \\
& \geq\left(\left\lfloor A_{n}\right\rfloor!\right)^{n} \cdot \frac{a_{n}}{a_{1}+1} \\
& >\left(\left\lfloor A_{n}\right\rfloor!\right)^{n}
\end{aligned}
$$
which completes the proof. Equality cannot hold in this case.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ be a sequence of non-negative integers, where $n$ is a positive integer.
Let
$$
A_{n}=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}
$$
Prove that
$$
a_{1}!a_{2}!\ldots a_{n}!\geq\left(\left\lfloor A_{n}\right\rfloor!\right)^{n}
$$
where $\left\lfloor A_{n}\right\rfloor$ is the greatest integer less than or equal to $A_{n}$, and $a!=1 \times 2 \times \cdots \times a$ for $a \geq 1$ (and $0!=1$ ). When does equality hold?
|
Assume without loss of generality that $0 \leq a_{1} \leq a_{2} \leq \cdots \leq a_{n}$. Let $d=a_{n}-a_{1}$ and $m=\left|\left\{i: a_{i}=a_{1}\right\}\right|$. Our proof is by induction on $d$.
We first do the case $d=a_{n}-a_{1}=0$ or 1 separately. Then $a_{1}=a_{2}=\cdots=a_{m}=a$ and $a_{m+1}=\cdots=a_{n}=a+1$ for some $1 \leq m \leq n$ and $a \geq 0$. In this case we have $\left\lfloor A_{n}\right\rfloor=a$, so the inequality to be proven is just $a_{1}!a_{2}!\ldots a_{n}!\geq(a!)^{n}$, which is obvious. Equality holds if and only if either $m=n$, that is, $a_{1}=a_{2}=\cdots=a_{n}=a$; or if $a=0$, that is, $a_{1}=\cdots=a_{m}=0$ and $a_{m+1}=\cdots=a_{n}=1$. [ 2 marks to here.]
So assume that $d=a_{n}-a_{1} \geq 2$ and that the inequality holds for all sequences with smaller values of $d$, or with the same value of $d$ and smaller values of $m$. Then the sequence
$$
a_{1}+1, a_{2}, a_{3}, \ldots, a_{n-1}, a_{n}-1
$$
though not necessarily in non-decreasing order any more, does have either a smaller value of $d$, or the same value of $d$ and a smaller value of $m$, but in any case has the same value of $A_{n}$. Thus, by induction and since $a_{n}>a_{1}+1$,
$$
\begin{aligned}
a_{1}!a_{2}!\ldots a_{n}! & =\left(a_{1}+1\right)!a_{2}!\ldots a_{n-1}!\left(a_{n}-1\right)!\cdot \frac{a_{n}}{a_{1}+1} \\
& \geq\left(\left\lfloor A_{n}\right\rfloor!\right)^{n} \cdot \frac{a_{n}}{a_{1}+1} \\
& >\left(\left\lfloor A_{n}\right\rfloor!\right)^{n}
\end{aligned}
$$
which completes the proof. Equality cannot hold in this case.
|
{
"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nSolution 2."
}
|
7d52eb5a-0c52-5bc1-a0a3-cce54fdf8e25
| 607,114
|
Let $x, y, z$ be positive numbers such that
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1
$$
Show that
$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z}
$$
|
$$
\begin{aligned}
\sum_{\text {cyclic }} \sqrt{x+y z} & =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\frac{1}{x}+\frac{1}{y z}} \\
& =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\frac{1}{x}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\frac{1}{y z}} \quad[1 \text { mark. }] \\
& =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)} \quad[1 \text { mark.] }
\end{aligned}
$$
$$
\begin{aligned}
& =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\left(\frac{1}{x}+\frac{1}{\sqrt{y z}}\right)^{2}+\frac{(\sqrt{y}-\sqrt{z})^{2}}{x y z}} \quad[2 \text { marks. }] \\
& \geq \sqrt{x y z} \sum_{\text {cyclic }}\left(\frac{1}{x}+\frac{1}{\sqrt{y z}}\right) \quad[1 \text { mark. }] \\
& =\sqrt{x y z}\left(1+\sum_{\text {cyclic }} \frac{1}{\sqrt{y z}}\right) \quad[1 \text { mark. }] \\
& =\sqrt{x y z}+\sum_{\text {cyclic }} \sqrt{x} \quad[1 \text { mark. }]
\end{aligned}
$$
Note. It is easy to check that equality holds if and only if $x=y=z=3$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $x, y, z$ be positive numbers such that
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1
$$
Show that
$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z}
$$
|
$$
\begin{aligned}
\sum_{\text {cyclic }} \sqrt{x+y z} & =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\frac{1}{x}+\frac{1}{y z}} \\
& =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\frac{1}{x}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\frac{1}{y z}} \quad[1 \text { mark. }] \\
& =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)} \quad[1 \text { mark.] }
\end{aligned}
$$
$$
\begin{aligned}
& =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\left(\frac{1}{x}+\frac{1}{\sqrt{y z}}\right)^{2}+\frac{(\sqrt{y}-\sqrt{z})^{2}}{x y z}} \quad[2 \text { marks. }] \\
& \geq \sqrt{x y z} \sum_{\text {cyclic }}\left(\frac{1}{x}+\frac{1}{\sqrt{y z}}\right) \quad[1 \text { mark. }] \\
& =\sqrt{x y z}\left(1+\sum_{\text {cyclic }} \frac{1}{\sqrt{y z}}\right) \quad[1 \text { mark. }] \\
& =\sqrt{x y z}+\sum_{\text {cyclic }} \sqrt{x} \quad[1 \text { mark. }]
\end{aligned}
$$
Note. It is easy to check that equality holds if and only if $x=y=z=3$.
|
{
"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution 1."
}
|
2ce20b58-70b3-5744-a593-c218e443a610
| 607,117
|
Let $x, y, z$ be positive numbers such that
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1
$$
Show that
$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z}
$$
|
Squaring both sides of the given inequality, we obtain
$$
\begin{aligned}
& \sum_{\text {cyclic }} x+\sum_{\text {cyclic }} y z+2 \sum_{\text {cyclic }} \sqrt{x+y z} \sqrt{y+z x} \\
& \quad \geq x y z+2 \sqrt{x y z} \sum_{\text {cyclic }} \sqrt{x}+\sum_{\text {cyclic }} x+2 \sum_{\text {cyclic }} \sqrt{x y} \quad \text { [1 mark.] }
\end{aligned}
$$
It follows from the given condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ that $x y z=\sum_{\text {cyclic }} x y$. Therefore, the given inequality is equivalent to
$$
\sum_{\text {cyclic }} \sqrt{x+y z} \sqrt{y+z x} \geq \sqrt{x y z} \sum_{\text {cyclic }} \sqrt{x}+\sum_{\text {cyclic }} \sqrt{x y} . \quad[2 \text { marks.] }
$$
Using the Cauchy-Schwarz inequality [or just $x^{2}+y^{2} \geq 2 x y$ ], we see that
$$
(x+y z)(y+z x) \geq\left(\sqrt{x y}+\sqrt{x y z^{2}}\right)^{2}, \quad[1 \text { mark. }]
$$
or
$$
\sqrt{x+y z} \sqrt{y+z x} \geq \sqrt{x y}+\sqrt{z} \sqrt{x y z} . \quad[1 \text { mark. }]
$$
Taking the cyclic sum of this inequality over $x, y$ and $z$, we get the desired inequality. [2 marks.]
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $x, y, z$ be positive numbers such that
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1
$$
Show that
$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z}
$$
|
Squaring both sides of the given inequality, we obtain
$$
\begin{aligned}
& \sum_{\text {cyclic }} x+\sum_{\text {cyclic }} y z+2 \sum_{\text {cyclic }} \sqrt{x+y z} \sqrt{y+z x} \\
& \quad \geq x y z+2 \sqrt{x y z} \sum_{\text {cyclic }} \sqrt{x}+\sum_{\text {cyclic }} x+2 \sum_{\text {cyclic }} \sqrt{x y} \quad \text { [1 mark.] }
\end{aligned}
$$
It follows from the given condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ that $x y z=\sum_{\text {cyclic }} x y$. Therefore, the given inequality is equivalent to
$$
\sum_{\text {cyclic }} \sqrt{x+y z} \sqrt{y+z x} \geq \sqrt{x y z} \sum_{\text {cyclic }} \sqrt{x}+\sum_{\text {cyclic }} \sqrt{x y} . \quad[2 \text { marks.] }
$$
Using the Cauchy-Schwarz inequality [or just $x^{2}+y^{2} \geq 2 x y$ ], we see that
$$
(x+y z)(y+z x) \geq\left(\sqrt{x y}+\sqrt{x y z^{2}}\right)^{2}, \quad[1 \text { mark. }]
$$
or
$$
\sqrt{x+y z} \sqrt{y+z x} \geq \sqrt{x y}+\sqrt{z} \sqrt{x y z} . \quad[1 \text { mark. }]
$$
Taking the cyclic sum of this inequality over $x, y$ and $z$, we get the desired inequality. [2 marks.]
|
{
"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution 2."
}
|
2ce20b58-70b3-5744-a593-c218e443a610
| 607,117
|
Let $x, y, z$ be positive numbers such that
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1
$$
Show that
$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z}
$$
|
This is another way of presenting the idea in the first solution.
Using the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ and the AM-GM inequality, we have
$$
\begin{aligned}
x+y z-\left(\sqrt{\frac{y z}{x}}+\sqrt{x}\right)^{2} & =y z\left(1-\frac{1}{x}\right)-2 \sqrt{y z} \\
& =y z\left(\frac{1}{y}+\frac{1}{z}\right)-2 \sqrt{y z}=y+z-2 \sqrt{y z} \geq 0
\end{aligned}
$$
which gives
$$
\sqrt{x+y z} \geq \sqrt{\frac{y z}{x}}+\sqrt{x} . \quad[3 \text { marks. }]
$$
Similarly, we have
$$
\sqrt{y+z x} \geq \sqrt{\frac{z x}{y}}+\sqrt{y} \text { and } \sqrt{z+x y} \geq \sqrt{\frac{x y}{z}}+\sqrt{z}
$$
Addition yields
$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}+\sqrt{x}+\sqrt{y}+\sqrt{z}
$$
[2 marks.] Using the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ again, we have
$$
\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}=\sqrt{x y z}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\sqrt{x y z}, \quad[1 \text { mark. }]
$$
and thus
$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z} . \quad[1 \text { mark. }]
$$
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $x, y, z$ be positive numbers such that
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1
$$
Show that
$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z}
$$
|
This is another way of presenting the idea in the first solution.
Using the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ and the AM-GM inequality, we have
$$
\begin{aligned}
x+y z-\left(\sqrt{\frac{y z}{x}}+\sqrt{x}\right)^{2} & =y z\left(1-\frac{1}{x}\right)-2 \sqrt{y z} \\
& =y z\left(\frac{1}{y}+\frac{1}{z}\right)-2 \sqrt{y z}=y+z-2 \sqrt{y z} \geq 0
\end{aligned}
$$
which gives
$$
\sqrt{x+y z} \geq \sqrt{\frac{y z}{x}}+\sqrt{x} . \quad[3 \text { marks. }]
$$
Similarly, we have
$$
\sqrt{y+z x} \geq \sqrt{\frac{z x}{y}}+\sqrt{y} \text { and } \sqrt{z+x y} \geq \sqrt{\frac{x y}{z}}+\sqrt{z}
$$
Addition yields
$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}+\sqrt{x}+\sqrt{y}+\sqrt{z}
$$
[2 marks.] Using the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ again, we have
$$
\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}=\sqrt{x y z}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\sqrt{x y z}, \quad[1 \text { mark. }]
$$
and thus
$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z} . \quad[1 \text { mark. }]
$$
|
{
"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution 3."
}
|
2ce20b58-70b3-5744-a593-c218e443a610
| 607,117
|
Let $x, y, z$ be positive numbers such that
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1
$$
Show that
$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z}
$$
|
This is also another way of presenting the idea in the first solution.
We make the substitution $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$. Then it is enough to show that
$$
\sqrt{\frac{1}{a}+\frac{1}{b c}}+\sqrt{\frac{1}{b}+\frac{1}{c a}}+\sqrt{\frac{1}{c}+\frac{1}{a b}} \geq \sqrt{\frac{1}{a b c}}+\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}},
$$
where $a+b+c=1$. Multiplying this inequality by $\sqrt{a b c}$, we find that it can be written
$$
\sqrt{a+b c}+\sqrt{b+c a}+\sqrt{c+a b} \geq 1+\sqrt{b c}+\sqrt{c a}+\sqrt{a b} . \quad[1 \text { mark. }]
$$
This is equivalent to
$$
\begin{aligned}
& \sqrt{a(a+b+c)+b c}+\sqrt{b(a+b+c)+c a}+\sqrt{c(a+b+c)+a b} \\
& \geq a+b+c+\sqrt{b c}+\sqrt{c a}+\sqrt{a b}, \quad[1 \text { mark. }]
\end{aligned}
$$
which in turn is equivalent to
$$
\sqrt{(a+b)(a+c)}+\sqrt{(b+c)(b+a)}+\sqrt{(c+a)(c+b)} \geq a+b+c+\sqrt{b c}+\sqrt{c a}+\sqrt{a b}
$$
[1 mark.] (This is a homogeneous version of the original inequality.) By the Cauchy-Schwarz inequality (or since $b+c \geq 2 \sqrt{b c}$ ), we have
$$
\left[(\sqrt{a})^{2}+(\sqrt{b})^{2}\right]\left[(\sqrt{a})^{2}+(\sqrt{c})^{2}\right] \geq(\sqrt{a} \sqrt{a}+\sqrt{b} \sqrt{c})^{2}
$$
or
$$
\sqrt{(a+b)(a+c)} \geq a+\sqrt{b c} . \quad[2 \text { marks. }]
$$
Taking the cyclic sum of this inequality over $a, b, c$, we get the desired inequality. [2 marks.]
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $x, y, z$ be positive numbers such that
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1
$$
Show that
$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z}
$$
|
This is also another way of presenting the idea in the first solution.
We make the substitution $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$. Then it is enough to show that
$$
\sqrt{\frac{1}{a}+\frac{1}{b c}}+\sqrt{\frac{1}{b}+\frac{1}{c a}}+\sqrt{\frac{1}{c}+\frac{1}{a b}} \geq \sqrt{\frac{1}{a b c}}+\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}},
$$
where $a+b+c=1$. Multiplying this inequality by $\sqrt{a b c}$, we find that it can be written
$$
\sqrt{a+b c}+\sqrt{b+c a}+\sqrt{c+a b} \geq 1+\sqrt{b c}+\sqrt{c a}+\sqrt{a b} . \quad[1 \text { mark. }]
$$
This is equivalent to
$$
\begin{aligned}
& \sqrt{a(a+b+c)+b c}+\sqrt{b(a+b+c)+c a}+\sqrt{c(a+b+c)+a b} \\
& \geq a+b+c+\sqrt{b c}+\sqrt{c a}+\sqrt{a b}, \quad[1 \text { mark. }]
\end{aligned}
$$
which in turn is equivalent to
$$
\sqrt{(a+b)(a+c)}+\sqrt{(b+c)(b+a)}+\sqrt{(c+a)(c+b)} \geq a+b+c+\sqrt{b c}+\sqrt{c a}+\sqrt{a b}
$$
[1 mark.] (This is a homogeneous version of the original inequality.) By the Cauchy-Schwarz inequality (or since $b+c \geq 2 \sqrt{b c}$ ), we have
$$
\left[(\sqrt{a})^{2}+(\sqrt{b})^{2}\right]\left[(\sqrt{a})^{2}+(\sqrt{c})^{2}\right] \geq(\sqrt{a} \sqrt{a}+\sqrt{b} \sqrt{c})^{2}
$$
or
$$
\sqrt{(a+b)(a+c)} \geq a+\sqrt{b c} . \quad[2 \text { marks. }]
$$
Taking the cyclic sum of this inequality over $a, b, c$, we get the desired inequality. [2 marks.]
|
{
"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl",
"problem_match": "\n4. ",
"solution_match": "# Solution 4."
}
|
2ce20b58-70b3-5744-a593-c218e443a610
| 607,117
|
Let $k \geq 14$ be an integer, and let $p_{k}$ be the largest prime number which is strictly less than $k$. You may assume that $p_{k} \geq 3 k / 4$. Let $n$ be a composite integer. Prove:
(a) if $n=2 p_{k}$, then $n$ does not divide $(n-k)$ !;
(b) if $n>2 p_{k}$, then $n$ divides $(n-k)$ !.
|
(a) Note that $n-k=2 p_{k}-k<2 p_{k}-p_{k}=p_{k}$, so $p_{k} \nmid(n-k)$ !, so $2 p_{k} \nless(n-k)$ !. [1 mark]
(b) Note that $n>2 p_{k} \geq 3 k / 2$ implies $k<2 n / 3$, so $n-k>n / 3$. So if we can find integers $a, b \geq 3$ such that $n=a b$ and $a \neq b$, then both $a$ and $b$ will appear separately in the product $(n-k)!=1 \times 2 \times \cdots \times(n-k)$, which means $n \mid(n-k)!$. Observe that $k \geq 14$ implies $p_{k} \geq 13$, so that $n>2 p_{k} \geq 26$.
If $n=2^{\alpha}$ for some integer $\alpha \geq 5$, then take $a=2^{2}, b=2^{\alpha-2}$. [ 1 mark] Otherwise, since $n \geq 26>16$, we can take $a$ to be an odd prime factor of $n$ and $b=n / a$ [1 mark], unless $b<3$ or $b=a$.
Case (i): $b<3$. Since $n$ is composite, this means $b=2$, so that $2 a=n>2 p_{k}$. As $a$ is a prime number and $p_{k}$ is the largest prime number which is strictly less than $k$, it follows that $a \geq k$. From $n-k=2 a-k \geq$ $2 a-a=a>2$ we see that $n=2 a$ divides into $(n-k)$. [ 2 marks]
Case (ii): $b=a$. Then $n=a^{2}$ and $a>6$ since $n \geq 26$. Thus $n-k>n / 3=a^{2} / 3>2 a$, so that both $a$ and $2 a$ appear among $\{1,2, \ldots, n-k\}$. Hence $n=a^{2}$ divides into $(n-k)!$. [2 marks]
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $k \geq 14$ be an integer, and let $p_{k}$ be the largest prime number which is strictly less than $k$. You may assume that $p_{k} \geq 3 k / 4$. Let $n$ be a composite integer. Prove:
(a) if $n=2 p_{k}$, then $n$ does not divide $(n-k)$ !;
(b) if $n>2 p_{k}$, then $n$ divides $(n-k)$ !.
|
(a) Note that $n-k=2 p_{k}-k<2 p_{k}-p_{k}=p_{k}$, so $p_{k} \nmid(n-k)$ !, so $2 p_{k} \nless(n-k)$ !. [1 mark]
(b) Note that $n>2 p_{k} \geq 3 k / 2$ implies $k<2 n / 3$, so $n-k>n / 3$. So if we can find integers $a, b \geq 3$ such that $n=a b$ and $a \neq b$, then both $a$ and $b$ will appear separately in the product $(n-k)!=1 \times 2 \times \cdots \times(n-k)$, which means $n \mid(n-k)!$. Observe that $k \geq 14$ implies $p_{k} \geq 13$, so that $n>2 p_{k} \geq 26$.
If $n=2^{\alpha}$ for some integer $\alpha \geq 5$, then take $a=2^{2}, b=2^{\alpha-2}$. [ 1 mark] Otherwise, since $n \geq 26>16$, we can take $a$ to be an odd prime factor of $n$ and $b=n / a$ [1 mark], unless $b<3$ or $b=a$.
Case (i): $b<3$. Since $n$ is composite, this means $b=2$, so that $2 a=n>2 p_{k}$. As $a$ is a prime number and $p_{k}$ is the largest prime number which is strictly less than $k$, it follows that $a \geq k$. From $n-k=2 a-k \geq$ $2 a-a=a>2$ we see that $n=2 a$ divides into $(n-k)$. [ 2 marks]
Case (ii): $b=a$. Then $n=a^{2}$ and $a>6$ since $n \geq 26$. Thus $n-k>n / 3=a^{2} / 3>2 a$, so that both $a$ and $2 a$ appear among $\{1,2, \ldots, n-k\}$. Hence $n=a^{2}$ divides into $(n-k)!$. [2 marks]
|
{
"resource_path": "APMO/segmented/en-apmo2003_sol.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution."
}
|
9a6b26c4-8d6d-578f-afec-47d1fe749184
| 607,121
|
Let $a, b, c$ be the sides of a triangle, with $a+b+c=1$, and let $n \geq 2$ be an integer. Show that
$$
\sqrt[n]{a^{n}+b^{n}}+\sqrt[n]{b^{n}+c^{n}}+\sqrt[n]{c^{n}+a^{n}}<1+\frac{\sqrt[n]{2}}{2}
$$
|
Without loss of generality, assume $a \leq b \leq c$. As $a+b>c$, we have
$$
\frac{\sqrt[n]{2}}{2}=\frac{\sqrt[n]{2}}{2}(a+b+c)>\frac{\sqrt[n]{2}}{2}(c+c)=\sqrt[n]{2 c^{n}} \geq \sqrt[n]{b^{n}+c^{n}} \quad \quad[2 \text { marks }]
$$
As $a \leq c$ and $n \geq 2$, we have
$$
\begin{aligned}
\left(c^{n}+a^{n}\right)-\left(c+\frac{a}{2}\right)^{n} & =a^{n}-\sum_{k=1}^{n}\binom{n}{k} c^{n-k}\left(\frac{a}{2}\right)^{k} \\
& \leq\left[1-\sum_{k=1}^{n}\binom{n}{k}\left(\frac{1}{2}\right)^{k}\right] a^{n} \quad\left(\text { since } c^{n-k} \geq a^{n-k}\right) \\
& =\left[\left(1-\frac{n}{2}\right)-\sum_{k=2}^{n}\binom{n}{k}\left(\frac{1}{2}\right)^{k}\right] a^{n}<0
\end{aligned}
$$
Thus
$$
\sqrt[n]{c^{n}+a^{n}}<c+\frac{a}{2} . \quad[3 \text { marks }]
$$
Likewise
$$
\sqrt[n]{b^{n}+a^{n}}<b+\frac{a}{2} \quad \quad[1 \text { mark }]
$$
Adding (1), (2) and (3), we get
$$
\sqrt[n]{a^{n}+b^{n}}+\sqrt[n]{b^{n}+c^{n}}+\sqrt[n]{c^{n}+a^{n}}<\frac{\sqrt[n]{2}}{2}+c+\frac{a}{2}+b+\frac{a}{2}=1+\frac{\sqrt[n]{2}}{2} . \quad[1 \text { mark }]
$$
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a, b, c$ be the sides of a triangle, with $a+b+c=1$, and let $n \geq 2$ be an integer. Show that
$$
\sqrt[n]{a^{n}+b^{n}}+\sqrt[n]{b^{n}+c^{n}}+\sqrt[n]{c^{n}+a^{n}}<1+\frac{\sqrt[n]{2}}{2}
$$
|
Without loss of generality, assume $a \leq b \leq c$. As $a+b>c$, we have
$$
\frac{\sqrt[n]{2}}{2}=\frac{\sqrt[n]{2}}{2}(a+b+c)>\frac{\sqrt[n]{2}}{2}(c+c)=\sqrt[n]{2 c^{n}} \geq \sqrt[n]{b^{n}+c^{n}} \quad \quad[2 \text { marks }]
$$
As $a \leq c$ and $n \geq 2$, we have
$$
\begin{aligned}
\left(c^{n}+a^{n}\right)-\left(c+\frac{a}{2}\right)^{n} & =a^{n}-\sum_{k=1}^{n}\binom{n}{k} c^{n-k}\left(\frac{a}{2}\right)^{k} \\
& \leq\left[1-\sum_{k=1}^{n}\binom{n}{k}\left(\frac{1}{2}\right)^{k}\right] a^{n} \quad\left(\text { since } c^{n-k} \geq a^{n-k}\right) \\
& =\left[\left(1-\frac{n}{2}\right)-\sum_{k=2}^{n}\binom{n}{k}\left(\frac{1}{2}\right)^{k}\right] a^{n}<0
\end{aligned}
$$
Thus
$$
\sqrt[n]{c^{n}+a^{n}}<c+\frac{a}{2} . \quad[3 \text { marks }]
$$
Likewise
$$
\sqrt[n]{b^{n}+a^{n}}<b+\frac{a}{2} \quad \quad[1 \text { mark }]
$$
Adding (1), (2) and (3), we get
$$
\sqrt[n]{a^{n}+b^{n}}+\sqrt[n]{b^{n}+c^{n}}+\sqrt[n]{c^{n}+a^{n}}<\frac{\sqrt[n]{2}}{2}+c+\frac{a}{2}+b+\frac{a}{2}=1+\frac{\sqrt[n]{2}}{2} . \quad[1 \text { mark }]
$$
|
{
"resource_path": "APMO/segmented/en-apmo2003_sol.jsonl",
"problem_match": "\n4. ",
"solution_match": "# Solution."
}
|
b79cc3ed-c8e3-5cf0-ac83-ddeed2f10e82
| 607,122
|
Let $O$ be the circumcentre and $H$ the orthocentre of an acute triangle $A B C$. Prove that the area of one of the triangles $\mathrm{AOH}, \mathrm{BOH}$ and COH is equal to the sum of the areas of the other two.
|
Suppose, without loss of generality, that $B$ and $C$ lies in the same side of line $O H$. Such line is the Euler line of $A B C$, so the centroid $G$ lies in this line.
Let $M$ be the midpoint of $B C$. Then the distance between $M$ and the line $O H$ is the average of the distances from $B$ and C to OH , and the sum of the areas of triangles BOH and COH is
$$
[B O H]+[C O H]=\frac{O H \cdot d(B, O H)}{2}+\frac{O H \cdot d(C, O H)}{2}=\frac{O H \cdot 2 d(M, O H)}{2} .
$$
Since $A G=2 G M, d(A, O H)=2 d(M, O H)$. Hence
$$
[B O H]+[C O H]=\frac{O H \cdot d(A, O H)}{2}=[A O H]
$$
and the result follows.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $O$ be the circumcentre and $H$ the orthocentre of an acute triangle $A B C$. Prove that the area of one of the triangles $\mathrm{AOH}, \mathrm{BOH}$ and COH is equal to the sum of the areas of the other two.
|
Suppose, without loss of generality, that $B$ and $C$ lies in the same side of line $O H$. Such line is the Euler line of $A B C$, so the centroid $G$ lies in this line.
Let $M$ be the midpoint of $B C$. Then the distance between $M$ and the line $O H$ is the average of the distances from $B$ and C to OH , and the sum of the areas of triangles BOH and COH is
$$
[B O H]+[C O H]=\frac{O H \cdot d(B, O H)}{2}+\frac{O H \cdot d(C, O H)}{2}=\frac{O H \cdot 2 d(M, O H)}{2} .
$$
Since $A G=2 G M, d(A, O H)=2 d(M, O H)$. Hence
$$
[B O H]+[C O H]=\frac{O H \cdot d(A, O H)}{2}=[A O H]
$$
and the result follows.
|
{
"resource_path": "APMO/segmented/en-apmo2004_sol.jsonl",
"problem_match": "# Problem 2",
"solution_match": "# Solution 1"
}
|
3d8f831e-00b1-5b0f-b0b0-17aa6bd4254d
| 604,841
|
Let $O$ be the circumcentre and $H$ the orthocentre of an acute triangle $A B C$. Prove that the area of one of the triangles $\mathrm{AOH}, \mathrm{BOH}$ and COH is equal to the sum of the areas of the other two.
|
One can use barycentric coordinates: it is well known that
$$
\begin{gathered}
A=(1: 0: 0), \quad B=(0: 1: 0), \quad C=(0: 0: 1), \\
O=(\sin 2 A: \sin 2 B: \sin 2 C) \quad \text { and } \quad H=(\tan A: \tan B: \tan C) .
\end{gathered}
$$
Then the (signed) area of $A O H$ is proportional to
$$
\left|\begin{array}{ccc}
1 & 0 & 0 \\
\sin 2 A & \sin 2 B & \sin 2 C \\
\tan A & \tan B & \tan C
\end{array}\right|
$$
Adding all three expressions we find that the sum of the signed sums of the areas is a constant times
$$
\left|\begin{array}{ccc}
1 & 0 & 0 \\
\sin 2 A & \sin 2 B & \sin 2 C \\
\tan A & \tan B & \tan C
\end{array}\right|+\left|\begin{array}{ccc}
0 & 1 & 0 \\
\sin 2 A & \sin 2 B & \sin 2 C \\
\tan A & \tan B & \tan C
\end{array}\right|+\left|\begin{array}{ccc}
0 & 0 & 1 \\
\sin 2 A & \sin 2 B & \sin 2 C \\
\tan A & \tan B & \tan C
\end{array}\right|
$$
By multilinearity of the determinant, this sum equals
$$
\left|\begin{array}{ccc}
1 & 1 & 1 \\
\sin 2 A & \sin 2 B & \sin 2 C \\
\tan A & \tan B & \tan C
\end{array}\right|
$$
which contains, in its rows, the coordinates of the centroid, the circumcenter, and the orthocenter. Since these three points lie in the Euler line of $A B C$, the signed sum of the areas is 0 , which means that one of the areas of $A O H, B O H, C O H$ is the sum of the other two areas.
Comment: Both solutions can be adapted to prove a stronger result: if the centroid $G$ of triangle $A B C$ belongs to line $X Y$ then one of the areas of triangles $A X Y, B X Y$, and $C X Y$ is equal to the sum of the other two.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $O$ be the circumcentre and $H$ the orthocentre of an acute triangle $A B C$. Prove that the area of one of the triangles $\mathrm{AOH}, \mathrm{BOH}$ and COH is equal to the sum of the areas of the other two.
|
One can use barycentric coordinates: it is well known that
$$
\begin{gathered}
A=(1: 0: 0), \quad B=(0: 1: 0), \quad C=(0: 0: 1), \\
O=(\sin 2 A: \sin 2 B: \sin 2 C) \quad \text { and } \quad H=(\tan A: \tan B: \tan C) .
\end{gathered}
$$
Then the (signed) area of $A O H$ is proportional to
$$
\left|\begin{array}{ccc}
1 & 0 & 0 \\
\sin 2 A & \sin 2 B & \sin 2 C \\
\tan A & \tan B & \tan C
\end{array}\right|
$$
Adding all three expressions we find that the sum of the signed sums of the areas is a constant times
$$
\left|\begin{array}{ccc}
1 & 0 & 0 \\
\sin 2 A & \sin 2 B & \sin 2 C \\
\tan A & \tan B & \tan C
\end{array}\right|+\left|\begin{array}{ccc}
0 & 1 & 0 \\
\sin 2 A & \sin 2 B & \sin 2 C \\
\tan A & \tan B & \tan C
\end{array}\right|+\left|\begin{array}{ccc}
0 & 0 & 1 \\
\sin 2 A & \sin 2 B & \sin 2 C \\
\tan A & \tan B & \tan C
\end{array}\right|
$$
By multilinearity of the determinant, this sum equals
$$
\left|\begin{array}{ccc}
1 & 1 & 1 \\
\sin 2 A & \sin 2 B & \sin 2 C \\
\tan A & \tan B & \tan C
\end{array}\right|
$$
which contains, in its rows, the coordinates of the centroid, the circumcenter, and the orthocenter. Since these three points lie in the Euler line of $A B C$, the signed sum of the areas is 0 , which means that one of the areas of $A O H, B O H, C O H$ is the sum of the other two areas.
Comment: Both solutions can be adapted to prove a stronger result: if the centroid $G$ of triangle $A B C$ belongs to line $X Y$ then one of the areas of triangles $A X Y, B X Y$, and $C X Y$ is equal to the sum of the other two.
|
{
"resource_path": "APMO/segmented/en-apmo2004_sol.jsonl",
"problem_match": "# Problem 2",
"solution_match": "# Solution 2"
}
|
3d8f831e-00b1-5b0f-b0b0-17aa6bd4254d
| 604,841
|
Let a set $S$ of 2004 points in the plane be given, no three of which are collinear. Let $\mathcal{L}$ denote the set of all lines (extended indefinitely in both directions) determined by pairs of points from the set. Show that it is possible to colour the points of $S$ with at most two colours, such that for any points $p, q$ of $S$, the number of lines in $\mathcal{L}$ which separate $p$ from $q$ is odd if and only if $p$ and $q$ have the same colour.
Note: A line $\ell$ separates two points $p$ and $q$ if $p$ and $q$ lie on opposite sides of $\ell$ with neither point on $\ell$.
|
Choose any point $p$ from $S$ and color it, say, blue. Let $n(q, r)$ be the number of lines from $\mathcal{L}$ that separates $q$ and $r$. Then color any other point $q$ blue if $n(p, q)$ is odd and red if $n(p, q)$ is even.
Now it remains to show that $q$ and $r$ have the same color if and only if $n(q, r)$ is odd for all $q \neq p$ and $r \neq p$, which is equivalent to proving that $n(p, q)+n(p, r)+n(q, r)$ is always odd. For this purpose, consider the seven numbered regions defined by lines $p q$, $p r$, and $q r$ :
Any line that do not pass through any of points $p, q, r$ meets the sides $p q, q r, p r$ of triangle $p q r$ in an even number of points (two sides or no sides), so these lines do not affect the parity of $n(p, q)+n(p, r)+n(q, r)$. Hence the only lines that need to be considered are the ones that pass through one of vertices $p, q, r$ and cuts the opposite side in the triangle $p q r$.
Let $n_{i}$ be the number of points in region $i, p, q$, and $r$ excluded, as depicted in the diagram. Then the lines through $p$ that separate $q$ and $r$ are the lines passing through $p$ and points from regions 1,4 , and 7 . The same applies for $p, q$ and regions 2,5 , and 7 ; and $p, r$ and regions 3,6 , and 7. Therefore
$$
\begin{aligned}
n(p, q)+n(q, r)+n(p, r) & \equiv\left(n_{2}+n_{5}+n_{7}\right)+\left(n_{1}+n_{4}+n_{7}\right)+\left(n_{3}+n_{6}+n_{7}\right) \\
& \equiv n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}=2004-3 \equiv 1 \quad(\bmod 2)
\end{aligned}
$$
and the result follows.
Comment: The problem statement is also true if 2004 is replaced by any even number and is not true if 2004 is replaced by any odd number greater than 1.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let a set $S$ of 2004 points in the plane be given, no three of which are collinear. Let $\mathcal{L}$ denote the set of all lines (extended indefinitely in both directions) determined by pairs of points from the set. Show that it is possible to colour the points of $S$ with at most two colours, such that for any points $p, q$ of $S$, the number of lines in $\mathcal{L}$ which separate $p$ from $q$ is odd if and only if $p$ and $q$ have the same colour.
Note: A line $\ell$ separates two points $p$ and $q$ if $p$ and $q$ lie on opposite sides of $\ell$ with neither point on $\ell$.
|
Choose any point $p$ from $S$ and color it, say, blue. Let $n(q, r)$ be the number of lines from $\mathcal{L}$ that separates $q$ and $r$. Then color any other point $q$ blue if $n(p, q)$ is odd and red if $n(p, q)$ is even.
Now it remains to show that $q$ and $r$ have the same color if and only if $n(q, r)$ is odd for all $q \neq p$ and $r \neq p$, which is equivalent to proving that $n(p, q)+n(p, r)+n(q, r)$ is always odd. For this purpose, consider the seven numbered regions defined by lines $p q$, $p r$, and $q r$ :
Any line that do not pass through any of points $p, q, r$ meets the sides $p q, q r, p r$ of triangle $p q r$ in an even number of points (two sides or no sides), so these lines do not affect the parity of $n(p, q)+n(p, r)+n(q, r)$. Hence the only lines that need to be considered are the ones that pass through one of vertices $p, q, r$ and cuts the opposite side in the triangle $p q r$.
Let $n_{i}$ be the number of points in region $i, p, q$, and $r$ excluded, as depicted in the diagram. Then the lines through $p$ that separate $q$ and $r$ are the lines passing through $p$ and points from regions 1,4 , and 7 . The same applies for $p, q$ and regions 2,5 , and 7 ; and $p, r$ and regions 3,6 , and 7. Therefore
$$
\begin{aligned}
n(p, q)+n(q, r)+n(p, r) & \equiv\left(n_{2}+n_{5}+n_{7}\right)+\left(n_{1}+n_{4}+n_{7}\right)+\left(n_{3}+n_{6}+n_{7}\right) \\
& \equiv n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}=2004-3 \equiv 1 \quad(\bmod 2)
\end{aligned}
$$
and the result follows.
Comment: The problem statement is also true if 2004 is replaced by any even number and is not true if 2004 is replaced by any odd number greater than 1.
|
{
"resource_path": "APMO/segmented/en-apmo2004_sol.jsonl",
"problem_match": "# Problem 3",
"solution_match": "# Solution\n\n"
}
|
0a9dc559-c5b2-5327-9624-e2ad03392ef8
| 604,863
|
For a real number $x$, let $\lfloor x\rfloor$ stand for the largest integer that is less than or equal to $x$. Prove that
$$
\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor
$$
is even for every positive integer $n$.
|
Consider four cases:
- $n \leq 5$. Then $\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor=0$ is an even number.
- $n$ and $n+1$ are both composite (in particular, $n \geq 8$ ). Then $n=a b$ and $n+1=c d$ for $a, b, c, d \geq 2$. Moreover, since $n$ and $n+1$ are coprime, $a, b, c, d$ are all distinct and smaller than $n$, and one can choose $a, b, c, d$ such that exactly one of these four numbers is even. Hence $\frac{(n-1)!}{n(n+1)}$ is an integer. As $n \geq 8>6,(n-1)!$ has at least three even factors, so $\frac{(n-1)!}{n(n+1)}$ is an even integer.
- $n \geq 7$ is an odd prime. By Wilson's theorem, $(n-1)!\equiv-1(\bmod n)$, that is, $\frac{(n-1)!+1}{n}$ is an integer, as $\frac{(n-1)!+n+1}{n}=\frac{(n-1)!+1}{n}+1$ is. As before, $\frac{(n-1)!}{n+1}$ is an even integer; therefore $\frac{(n-1)!+n+1}{n+1}=\frac{(n-1)!}{n+1}+1$ is an odd integer.
Also, $n$ and $n+1$ are coprime and $n$ divides the odd integer $\frac{(n-1)!+n+1}{n+1}$, so $\frac{(n-1)!+n+1}{n(n+1)}$ is also an odd integer. Then
$$
\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor=\frac{(n-1)!+n+1}{n(n+1)}-1
$$
is even.
- $n+1 \geq 7$ is an odd prime. Again, since $n$ is composite, $\frac{(n-1)!}{n}$ is an even integer, and $\frac{(n-1)!+n}{n}$ is an odd integer. By Wilson's theorem, $n!\equiv-1(\bmod n+1) \Longleftrightarrow(n-1)!\equiv 1$ $(\bmod n+1)$. This means that $n+1$ divides $(n-1)!+n$, and since $n$ and $n+1$ are coprime, $n+1$ also divides $\frac{(n-1)!+n}{n}$. Then $\frac{(n-1)!+n}{n(n+1)}$ is also an odd integer and
$$
\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor=\frac{(n-1)!+n}{n(n+1)}-1
$$
is even.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
For a real number $x$, let $\lfloor x\rfloor$ stand for the largest integer that is less than or equal to $x$. Prove that
$$
\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor
$$
is even for every positive integer $n$.
|
Consider four cases:
- $n \leq 5$. Then $\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor=0$ is an even number.
- $n$ and $n+1$ are both composite (in particular, $n \geq 8$ ). Then $n=a b$ and $n+1=c d$ for $a, b, c, d \geq 2$. Moreover, since $n$ and $n+1$ are coprime, $a, b, c, d$ are all distinct and smaller than $n$, and one can choose $a, b, c, d$ such that exactly one of these four numbers is even. Hence $\frac{(n-1)!}{n(n+1)}$ is an integer. As $n \geq 8>6,(n-1)!$ has at least three even factors, so $\frac{(n-1)!}{n(n+1)}$ is an even integer.
- $n \geq 7$ is an odd prime. By Wilson's theorem, $(n-1)!\equiv-1(\bmod n)$, that is, $\frac{(n-1)!+1}{n}$ is an integer, as $\frac{(n-1)!+n+1}{n}=\frac{(n-1)!+1}{n}+1$ is. As before, $\frac{(n-1)!}{n+1}$ is an even integer; therefore $\frac{(n-1)!+n+1}{n+1}=\frac{(n-1)!}{n+1}+1$ is an odd integer.
Also, $n$ and $n+1$ are coprime and $n$ divides the odd integer $\frac{(n-1)!+n+1}{n+1}$, so $\frac{(n-1)!+n+1}{n(n+1)}$ is also an odd integer. Then
$$
\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor=\frac{(n-1)!+n+1}{n(n+1)}-1
$$
is even.
- $n+1 \geq 7$ is an odd prime. Again, since $n$ is composite, $\frac{(n-1)!}{n}$ is an even integer, and $\frac{(n-1)!+n}{n}$ is an odd integer. By Wilson's theorem, $n!\equiv-1(\bmod n+1) \Longleftrightarrow(n-1)!\equiv 1$ $(\bmod n+1)$. This means that $n+1$ divides $(n-1)!+n$, and since $n$ and $n+1$ are coprime, $n+1$ also divides $\frac{(n-1)!+n}{n}$. Then $\frac{(n-1)!+n}{n(n+1)}$ is also an odd integer and
$$
\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor=\frac{(n-1)!+n}{n(n+1)}-1
$$
is even.
|
{
"resource_path": "APMO/segmented/en-apmo2004_sol.jsonl",
"problem_match": "# Problem 4",
"solution_match": "# Solution\n\n"
}
|
110f7ed3-2dd7-5e0e-a16f-cc0238f9e851
| 604,876
|
Prove that
$$
\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9(a b+b c+c a)
$$
for all real numbers $a, b, c>0$.
|
Let $p=a+b+c, q=a b+b c+c a$, and $r=a b c$. The inequality simplifies to
$$
a^{2} b^{2} c^{2}+2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+4\left(a^{2}+b^{2}+c^{2}\right)+8-9(a b+b c+c a) \geq 0
$$
Since $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=q^{2}-2 p r$ and $a^{2}+b^{2}+c^{2}=p^{2}-2 q$,
$$
r^{2}+2 q^{2}-4 p r+4 p^{2}-8 q+8-9 q \geq 0
$$
which simplifies to
$$
r^{2}+2 q^{2}+4 p^{2}-17 q-4 p r+8 \geq 0
$$
Bearing in mind that equality occurs for $a=b=c=1$, which means that, for instance, $p=3 r$, one can rewrite $(I)$ as
$$
\left(r-\frac{p}{3}\right)^{2}-\frac{10}{3} p r+\frac{35}{9} p^{2}+2 q^{2}-17 q+8 \geq 0
$$
Since $(a b-b c)^{2}+(b c-c a)^{2}+(c a-a b)^{2} \geq 0$ is equivalent to $q^{2} \geq 3 p r$, rewrite $(I I)$ as
$$
\left(r-\frac{p}{3}\right)^{2}+\frac{10}{9}\left(q^{2}-3 p r\right)+\frac{35}{9} p^{2}+\frac{8}{9} q^{2}-17 q+8 \geq 0
$$
Finally, $a=b=c=1$ implies $q=3$; then rewrite (III) as
$$
\left(r-\frac{p}{3}\right)^{2}+\frac{10}{9}\left(q^{2}-3 p r\right)+\frac{35}{9}\left(p^{2}-3 q\right)+\frac{8}{9}(q-3)^{2} \geq 0
$$
This final inequality is true because $q^{2} \geq 3 p r$ and $p^{2}-3 q=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \geq 0$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Prove that
$$
\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9(a b+b c+c a)
$$
for all real numbers $a, b, c>0$.
|
Let $p=a+b+c, q=a b+b c+c a$, and $r=a b c$. The inequality simplifies to
$$
a^{2} b^{2} c^{2}+2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+4\left(a^{2}+b^{2}+c^{2}\right)+8-9(a b+b c+c a) \geq 0
$$
Since $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=q^{2}-2 p r$ and $a^{2}+b^{2}+c^{2}=p^{2}-2 q$,
$$
r^{2}+2 q^{2}-4 p r+4 p^{2}-8 q+8-9 q \geq 0
$$
which simplifies to
$$
r^{2}+2 q^{2}+4 p^{2}-17 q-4 p r+8 \geq 0
$$
Bearing in mind that equality occurs for $a=b=c=1$, which means that, for instance, $p=3 r$, one can rewrite $(I)$ as
$$
\left(r-\frac{p}{3}\right)^{2}-\frac{10}{3} p r+\frac{35}{9} p^{2}+2 q^{2}-17 q+8 \geq 0
$$
Since $(a b-b c)^{2}+(b c-c a)^{2}+(c a-a b)^{2} \geq 0$ is equivalent to $q^{2} \geq 3 p r$, rewrite $(I I)$ as
$$
\left(r-\frac{p}{3}\right)^{2}+\frac{10}{9}\left(q^{2}-3 p r\right)+\frac{35}{9} p^{2}+\frac{8}{9} q^{2}-17 q+8 \geq 0
$$
Finally, $a=b=c=1$ implies $q=3$; then rewrite (III) as
$$
\left(r-\frac{p}{3}\right)^{2}+\frac{10}{9}\left(q^{2}-3 p r\right)+\frac{35}{9}\left(p^{2}-3 q\right)+\frac{8}{9}(q-3)^{2} \geq 0
$$
This final inequality is true because $q^{2} \geq 3 p r$ and $p^{2}-3 q=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \geq 0$.
|
{
"resource_path": "APMO/segmented/en-apmo2004_sol.jsonl",
"problem_match": "# Problem 5",
"solution_match": "# Solution 1"
}
|
85c2e20c-c2c5-5955-a346-fe489616b284
| 604,886
|
Prove that
$$
\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9(a b+b c+c a)
$$
for all real numbers $a, b, c>0$.
|
We prove the stronger inequality
$$
\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 3(a+b+c)^{2}
$$
which implies the proposed inequality because $(a+b+c)^{2} \geq 3(a b+b c+c a)$ is equivalent to $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0$, which is immediate.
The inequality $(*)$ is equivalent to
$$
\left(\left(b^{2}+2\right)\left(c^{2}+2\right)-3\right) a^{2}-6(b+c) a+2\left(b^{2}+2\right)\left(c^{2}+2\right)-3(b+c)^{2} \geq 0
$$
Seeing this inequality as a quadratic inequality in $a$ with positive leading coefficient $\left(b^{2}+2\right)\left(c^{2}+\right.$ 2) $-3=b^{2} c^{2}+2 b^{2}+2 c^{2}+1$, it suffices to prove that its discriminant is non-positive, which is equivalent to
$$
(3(b+c))^{2}-\left(\left(b^{2}+2\right)\left(c^{2}+2\right)-3\right)\left(2\left(b^{2}+2\right)\left(c^{2}+2\right)-3(b+c)^{2}\right) \leq 0
$$
This simplifies to
$$
-2\left(b^{2}+2\right)\left(c^{2}+2\right)+3(b+c)^{2}+6 \leq 0
$$
Now we look $(* *)$ as a quadratic inequality in $b$ with negative leading coefficient $-2 c^{2}-1$ :
$$
\left(-2 c^{2}-1\right) b^{2}+6 c b-c^{2}-2 \leq 0
$$
If suffices to show that the discriminant of $(* *)$ is non-positive, which is equivalent to
$$
9 c^{2}-\left(2 c^{2}+1\right)\left(c^{2}+2\right) \leq 0
$$
It simplifies to $-2\left(c^{2}-1\right)^{2} \leq 0$, which is true. The equality occurs for $c^{2}=1$, that is, $c=1$, for which $b=\frac{6 c}{2\left(2 c^{2}+1\right)}=1$, and $a=\frac{6(b+c)}{2\left(\left(b^{2}+2\right)\left(c^{2}+2\right)-3\right)}=1$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Prove that
$$
\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9(a b+b c+c a)
$$
for all real numbers $a, b, c>0$.
|
We prove the stronger inequality
$$
\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 3(a+b+c)^{2}
$$
which implies the proposed inequality because $(a+b+c)^{2} \geq 3(a b+b c+c a)$ is equivalent to $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0$, which is immediate.
The inequality $(*)$ is equivalent to
$$
\left(\left(b^{2}+2\right)\left(c^{2}+2\right)-3\right) a^{2}-6(b+c) a+2\left(b^{2}+2\right)\left(c^{2}+2\right)-3(b+c)^{2} \geq 0
$$
Seeing this inequality as a quadratic inequality in $a$ with positive leading coefficient $\left(b^{2}+2\right)\left(c^{2}+\right.$ 2) $-3=b^{2} c^{2}+2 b^{2}+2 c^{2}+1$, it suffices to prove that its discriminant is non-positive, which is equivalent to
$$
(3(b+c))^{2}-\left(\left(b^{2}+2\right)\left(c^{2}+2\right)-3\right)\left(2\left(b^{2}+2\right)\left(c^{2}+2\right)-3(b+c)^{2}\right) \leq 0
$$
This simplifies to
$$
-2\left(b^{2}+2\right)\left(c^{2}+2\right)+3(b+c)^{2}+6 \leq 0
$$
Now we look $(* *)$ as a quadratic inequality in $b$ with negative leading coefficient $-2 c^{2}-1$ :
$$
\left(-2 c^{2}-1\right) b^{2}+6 c b-c^{2}-2 \leq 0
$$
If suffices to show that the discriminant of $(* *)$ is non-positive, which is equivalent to
$$
9 c^{2}-\left(2 c^{2}+1\right)\left(c^{2}+2\right) \leq 0
$$
It simplifies to $-2\left(c^{2}-1\right)^{2} \leq 0$, which is true. The equality occurs for $c^{2}=1$, that is, $c=1$, for which $b=\frac{6 c}{2\left(2 c^{2}+1\right)}=1$, and $a=\frac{6(b+c)}{2\left(\left(b^{2}+2\right)\left(c^{2}+2\right)-3\right)}=1$.
|
{
"resource_path": "APMO/segmented/en-apmo2004_sol.jsonl",
"problem_match": "# Problem 5",
"solution_match": "# Solution 2"
}
|
85c2e20c-c2c5-5955-a346-fe489616b284
| 604,886
|
Prove that
$$
\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9(a b+b c+c a)
$$
for all real numbers $a, b, c>0$.
|
Let $A, B, C$ angles in $(0, \pi / 2)$ such that $a=\sqrt{2} \tan A, b=\sqrt{2} \tan B$, and $c=\sqrt{2} \tan C$. Then the inequality is equivalent to
$$
4 \sec ^{2} A \sec ^{2} B \sec ^{2} C \geq 9(\tan A \tan B+\tan B \tan C+\tan C \tan A)
$$
Substituting $\sec x=\frac{1}{\cos x}$ for $x \in\{A, B, C\}$ and clearing denominators, the inequality is equivalent to
$$
\cos A \cos B \cos C(\sin A \sin B \cos C+\cos A \sin B \sin C+\sin A \cos B \sin C) \leq \frac{4}{9}
$$
Since
$$
\begin{aligned}
& \cos (A+B+C)=\cos A \cos (B+C)-\sin A \sin (B+C) \\
= & \cos A \cos B \cos C-\cos A \sin B \sin C-\sin A \cos B \sin C-\sin A \sin B \cos C,
\end{aligned}
$$
we rewrite our inequality as
$$
\cos A \cos B \cos C(\cos A \cos B \cos C-\cos (A+B+C)) \leq \frac{4}{9}
$$
The cosine function is concave down on $(0, \pi / 2)$. Therefore, if $\theta=\frac{A+B+C}{3}$, by the AM-GM inequality and Jensen's inequality,
$$
\cos A \cos B \cos C \leq\left(\frac{\cos A+\cos B+\cos C}{3}\right)^{3} \leq \cos ^{3} \frac{A+B+C}{3}=\cos ^{3} \theta
$$
Therefore, since $\cos A \cos B \cos C-\cos (A+B+C)=\sin A \sin B \cos C+\cos A \sin B \sin C+$ $\sin A \cos B \sin C>0$, and recalling that $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$,
$\cos A \cos B \cos C(\cos A \cos B \cos C-\cos (A+B+C)) \leq \cos ^{3} \theta\left(\cos ^{3} \theta-\cos 3 \theta\right)=3 \cos ^{4} \theta\left(1-\cos ^{2} \theta\right)$. Finally, by AM-GM (notice that $1-\cos ^{2} \theta=\sin ^{2} \theta>0$ ),
$3 \cos ^{4} \theta\left(1-\cos ^{2} \theta\right)=\frac{3}{2} \cos ^{2} \theta \cdot \cos ^{2} \theta\left(2-2 \cos ^{2} \theta\right) \leq \frac{3}{2}\left(\frac{\cos ^{2} \theta+\cos ^{2} \theta+\left(2-2 \cos ^{2} \theta\right)}{3}\right)^{3}=\frac{4}{9}$,
and the result follows.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Prove that
$$
\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9(a b+b c+c a)
$$
for all real numbers $a, b, c>0$.
|
Let $A, B, C$ angles in $(0, \pi / 2)$ such that $a=\sqrt{2} \tan A, b=\sqrt{2} \tan B$, and $c=\sqrt{2} \tan C$. Then the inequality is equivalent to
$$
4 \sec ^{2} A \sec ^{2} B \sec ^{2} C \geq 9(\tan A \tan B+\tan B \tan C+\tan C \tan A)
$$
Substituting $\sec x=\frac{1}{\cos x}$ for $x \in\{A, B, C\}$ and clearing denominators, the inequality is equivalent to
$$
\cos A \cos B \cos C(\sin A \sin B \cos C+\cos A \sin B \sin C+\sin A \cos B \sin C) \leq \frac{4}{9}
$$
Since
$$
\begin{aligned}
& \cos (A+B+C)=\cos A \cos (B+C)-\sin A \sin (B+C) \\
= & \cos A \cos B \cos C-\cos A \sin B \sin C-\sin A \cos B \sin C-\sin A \sin B \cos C,
\end{aligned}
$$
we rewrite our inequality as
$$
\cos A \cos B \cos C(\cos A \cos B \cos C-\cos (A+B+C)) \leq \frac{4}{9}
$$
The cosine function is concave down on $(0, \pi / 2)$. Therefore, if $\theta=\frac{A+B+C}{3}$, by the AM-GM inequality and Jensen's inequality,
$$
\cos A \cos B \cos C \leq\left(\frac{\cos A+\cos B+\cos C}{3}\right)^{3} \leq \cos ^{3} \frac{A+B+C}{3}=\cos ^{3} \theta
$$
Therefore, since $\cos A \cos B \cos C-\cos (A+B+C)=\sin A \sin B \cos C+\cos A \sin B \sin C+$ $\sin A \cos B \sin C>0$, and recalling that $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$,
$\cos A \cos B \cos C(\cos A \cos B \cos C-\cos (A+B+C)) \leq \cos ^{3} \theta\left(\cos ^{3} \theta-\cos 3 \theta\right)=3 \cos ^{4} \theta\left(1-\cos ^{2} \theta\right)$. Finally, by AM-GM (notice that $1-\cos ^{2} \theta=\sin ^{2} \theta>0$ ),
$3 \cos ^{4} \theta\left(1-\cos ^{2} \theta\right)=\frac{3}{2} \cos ^{2} \theta \cdot \cos ^{2} \theta\left(2-2 \cos ^{2} \theta\right) \leq \frac{3}{2}\left(\frac{\cos ^{2} \theta+\cos ^{2} \theta+\left(2-2 \cos ^{2} \theta\right)}{3}\right)^{3}=\frac{4}{9}$,
and the result follows.
|
{
"resource_path": "APMO/segmented/en-apmo2004_sol.jsonl",
"problem_match": "# Problem 5",
"solution_match": "# Solution 3"
}
|
85c2e20c-c2c5-5955-a346-fe489616b284
| 604,886
|
Prove that for every irrational real number $a$, there are irrational real numbers $b$ and $b^{\prime}$ so that $a+b$ and $a b^{\prime}$ are both rational while $a b$ and $a+b^{\prime}$ are both irrational.
|
Let $a$ be an irrational number. If $a^{2}$ is irrational, we let $b=-a$. Then, $a+b=0$ is rational and $a b=-a^{2}$ is irrational. If $a^{2}$ is rational, we let $b=a^{2}-a$. Then, $a+b=a^{2}$ is rational and $a b=a^{2}(a-1)$. Since
$$
a=\frac{a b}{a^{2}}+1
$$
is irrational, so is $a b$.
Now, we let $b^{\prime}=\frac{1}{a}$ or $b^{\prime}=\frac{2}{a}$. Then $a b^{\prime}=1$ or 2 , which is rational. Note that
$$
a+b^{\prime}=\frac{a^{2}+1}{a} \quad \text { or } \quad a+b^{\prime}=\frac{a^{2}+2}{a} .
$$
Since,
$$
\frac{a^{2}+2}{a}-\frac{a^{2}+1}{a}=\frac{1}{a}
$$
at least one of them is irrational.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Prove that for every irrational real number $a$, there are irrational real numbers $b$ and $b^{\prime}$ so that $a+b$ and $a b^{\prime}$ are both rational while $a b$ and $a+b^{\prime}$ are both irrational.
|
Let $a$ be an irrational number. If $a^{2}$ is irrational, we let $b=-a$. Then, $a+b=0$ is rational and $a b=-a^{2}$ is irrational. If $a^{2}$ is rational, we let $b=a^{2}-a$. Then, $a+b=a^{2}$ is rational and $a b=a^{2}(a-1)$. Since
$$
a=\frac{a b}{a^{2}}+1
$$
is irrational, so is $a b$.
Now, we let $b^{\prime}=\frac{1}{a}$ or $b^{\prime}=\frac{2}{a}$. Then $a b^{\prime}=1$ or 2 , which is rational. Note that
$$
a+b^{\prime}=\frac{a^{2}+1}{a} \quad \text { or } \quad a+b^{\prime}=\frac{a^{2}+2}{a} .
$$
Since,
$$
\frac{a^{2}+2}{a}-\frac{a^{2}+1}{a}=\frac{1}{a}
$$
at least one of them is irrational.
|
{
"resource_path": "APMO/segmented/en-apmo2005_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "(Solution)"
}
|
197b3c87-50d9-56a2-8740-d7d83df5ffba
| 260,545
|
Prove that there exists a triangle which can be cut into 2005 congruent triangles.
|
Suppose that one side of a triangle has length $n$. Then it can be cut into $n^{2}$ congruent triangles which are similar to the original one and whose corresponding sides to the side of length $n$ have lengths 1 .
Since $2005=5 \times 401$ where 5 and 401 are primes and both primes are of the type $4 k+1$, it is representable as a sum of two integer squares. Indeed, it is easy to see that
$$
\begin{aligned}
2005 & =5 \times 401=\left(2^{2}+1\right)\left(20^{2}+1\right) \\
& =40^{2}+20^{2}+2^{2}+1 \\
& =(40-1)^{2}+2 \times 40+20^{2}+2^{2} \\
& =39^{2}+22^{2}
\end{aligned}
$$
Let $A B C$ be a right-angled triangle with the legs $A B$ and $B C$ having lengths 39 and 22 , respectively. We draw the altitude $B K$, which divides $A B C$ into two similar triangles. Now we divide $A B K$ into $39^{2}$ congruent triangles as described above and $B C K$ into $22^{2}$ congruent triangles. Since $A B K$ is similar to $B K C$, all 2005 triangles will be congruent.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Prove that there exists a triangle which can be cut into 2005 congruent triangles.
|
Suppose that one side of a triangle has length $n$. Then it can be cut into $n^{2}$ congruent triangles which are similar to the original one and whose corresponding sides to the side of length $n$ have lengths 1 .
Since $2005=5 \times 401$ where 5 and 401 are primes and both primes are of the type $4 k+1$, it is representable as a sum of two integer squares. Indeed, it is easy to see that
$$
\begin{aligned}
2005 & =5 \times 401=\left(2^{2}+1\right)\left(20^{2}+1\right) \\
& =40^{2}+20^{2}+2^{2}+1 \\
& =(40-1)^{2}+2 \times 40+20^{2}+2^{2} \\
& =39^{2}+22^{2}
\end{aligned}
$$
Let $A B C$ be a right-angled triangle with the legs $A B$ and $B C$ having lengths 39 and 22 , respectively. We draw the altitude $B K$, which divides $A B C$ into two similar triangles. Now we divide $A B K$ into $39^{2}$ congruent triangles as described above and $B C K$ into $22^{2}$ congruent triangles. Since $A B K$ is similar to $B K C$, all 2005 triangles will be congruent.
|
{
"resource_path": "APMO/segmented/en-apmo2005_sol.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "(Solution)"
}
|
b44bcc5e-7b01-51d5-b263-7753936a317f
| 55,289
|
Prove that every positive integer can be written as a finite sum of distinct integral powers of the golden mean $\tau=\frac{1+\sqrt{5}}{2}$. Here, an integral power of $\tau$ is of the form $\tau^{i}$, where $i$ is an integer (not necessarily positive).
|
We will prove this statement by induction using the equality
$$
\tau^{2}=\tau+1
$$
If $n=1$, then $1=\tau^{0}$. Suppose that $n-1$ can be written as a finite sum of integral powers of $\tau$, say
$$
n-1=\sum_{i=-k}^{k} a_{i} \tau^{i}
$$
where $a_{i} \in\{0,1\}$ and $n \geq 2$. We will write (1) as
$$
n-1=a_{k} \cdots a_{1} a_{0} \cdot a_{-1} a_{-2} \cdots a_{-k}
$$
For example,
$$
1=1.0=0.11=0.1011=0.101011
$$
Firstly, we will prove that we may assume that in (2) we have $a_{i} a_{i+1}=0$ for all $i$ with $-k \leq i \leq k-1$. Indeed, if we have several occurrences of 11 , then we take the leftmost such occurrence. Since we may assume that it is preceded by a 0 , we can replace 011 with 100 using the identity $\tau^{i+1}+\tau^{i}=\tau^{i+2}$. By doing so repeatedly, if necessary, we will eliminate all occurrences of two 1's standing together. Now we have the representation
$$
n-1=\sum_{i=-K}^{K} b_{i} \tau^{i}
$$
where $b_{i} \in\{0,1\}$ and $b_{i} b_{i+1}=0$.
If $b_{0}=0$ in (3), then we just add $1=\tau^{0}$ to both sides of (3) and we are done.
Suppose now that there is 1 in the unit position of (3), that is $b_{0}=1$. If there are two 0 's to the right of it, i.e.
$$
n-1=\cdots 1.00 \cdots
$$
then we can replace 1.00 with 0.11 because $1=\tau^{-1}+\tau^{-2}$, and we are done because we obtain 0 in the unit position. Thus we may assume that
$$
n-1=\cdots 1.010 \cdots
$$
Again, if we have $n-1=\cdots 1.0100 \cdots$, we may rewrite it as
$$
n-1=\cdots 1.0100 \cdots=\cdots 1.0011 \cdots=\cdots 0.1111 \cdots
$$
and obtain 0 in the unit position. Therefore, we may assume that
$$
n-1=\cdots 1.01010 \cdots
$$
Since the number of 1's is finite, eventually we will obtain an occurrence of 100 at the end, i.e.
$$
n-1=\cdots 1.01010 \cdots 100
$$
Then we can shift all 1's to the right to obtain 0 in the unit position, i.e.
$$
n-1=\cdots 0.11 \cdots 11
$$
and we are done.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Prove that every positive integer can be written as a finite sum of distinct integral powers of the golden mean $\tau=\frac{1+\sqrt{5}}{2}$. Here, an integral power of $\tau$ is of the form $\tau^{i}$, where $i$ is an integer (not necessarily positive).
|
We will prove this statement by induction using the equality
$$
\tau^{2}=\tau+1
$$
If $n=1$, then $1=\tau^{0}$. Suppose that $n-1$ can be written as a finite sum of integral powers of $\tau$, say
$$
n-1=\sum_{i=-k}^{k} a_{i} \tau^{i}
$$
where $a_{i} \in\{0,1\}$ and $n \geq 2$. We will write (1) as
$$
n-1=a_{k} \cdots a_{1} a_{0} \cdot a_{-1} a_{-2} \cdots a_{-k}
$$
For example,
$$
1=1.0=0.11=0.1011=0.101011
$$
Firstly, we will prove that we may assume that in (2) we have $a_{i} a_{i+1}=0$ for all $i$ with $-k \leq i \leq k-1$. Indeed, if we have several occurrences of 11 , then we take the leftmost such occurrence. Since we may assume that it is preceded by a 0 , we can replace 011 with 100 using the identity $\tau^{i+1}+\tau^{i}=\tau^{i+2}$. By doing so repeatedly, if necessary, we will eliminate all occurrences of two 1's standing together. Now we have the representation
$$
n-1=\sum_{i=-K}^{K} b_{i} \tau^{i}
$$
where $b_{i} \in\{0,1\}$ and $b_{i} b_{i+1}=0$.
If $b_{0}=0$ in (3), then we just add $1=\tau^{0}$ to both sides of (3) and we are done.
Suppose now that there is 1 in the unit position of (3), that is $b_{0}=1$. If there are two 0 's to the right of it, i.e.
$$
n-1=\cdots 1.00 \cdots
$$
then we can replace 1.00 with 0.11 because $1=\tau^{-1}+\tau^{-2}$, and we are done because we obtain 0 in the unit position. Thus we may assume that
$$
n-1=\cdots 1.010 \cdots
$$
Again, if we have $n-1=\cdots 1.0100 \cdots$, we may rewrite it as
$$
n-1=\cdots 1.0100 \cdots=\cdots 1.0011 \cdots=\cdots 0.1111 \cdots
$$
and obtain 0 in the unit position. Therefore, we may assume that
$$
n-1=\cdots 1.01010 \cdots
$$
Since the number of 1's is finite, eventually we will obtain an occurrence of 100 at the end, i.e.
$$
n-1=\cdots 1.01010 \cdots 100
$$
Then we can shift all 1's to the right to obtain 0 in the unit position, i.e.
$$
n-1=\cdots 0.11 \cdots 11
$$
and we are done.
|
{
"resource_path": "APMO/segmented/en-apmo2006_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "(Solution)"
}
|
2c596044-e4f8-5e79-a5e6-6136742a352d
| 261,486
|
Let $p \geq 5$ be a prime and let $r$ be the number of ways of placing $p$ checkers on a $p \times p$ checkerboard so that not all checkers are in the same row (but they may all be in the same column). Show that $r$ is divisible by $p^{5}$. Here, we assume that all the checkers are identical.
|
Note that $r=\binom{p^{2}}{p}-p$. Hence, it suffices to show that
$$
\left(p^{2}-1\right)\left(p^{2}-2\right) \cdots\left(p^{2}-(p-1)\right)-(p-1)!\equiv 0 \quad\left(\bmod p^{4}\right)
$$
Now, let
$$
f(x):=(x-1)(x-2) \cdots(x-(p-1))=x^{p-1}+s_{p-2} x^{p-2}+\cdots+s_{1} x+s_{0} .
$$
Then the congruence equation (1) is same as $f\left(p^{2}\right)-s_{0} \equiv 0\left(\bmod p^{4}\right)$. Therefore, it suffices to show that $s_{1} p^{2} \equiv 0\left(\bmod p^{4}\right)$ or $s_{1} \equiv 0\left(\bmod p^{2}\right)$.
Since $a^{p-1} \equiv 1(\bmod p)$ for all $1 \leq a \leq p-1$, we can factor
$$
x^{p-1}-1 \equiv(x-1)(x-2) \cdots(x-(p-1)) \quad(\bmod p)
$$
Comparing the coefficients of the left hand side of (3) with those of the right hand side of (2), we obtain $p \mid s_{i}$ for all $1 \leq i \leq p-2$ and $s_{0} \equiv-1(\bmod p)$. On the other hand, plugging $p$ for $x$ in (2), we get
$$
f(p)=(p-1)!=s_{0}=p^{p-1}+s_{p-2} p^{p-2}+\cdots+s_{1} p+s_{0}
$$
which implies
$$
p^{p-1}+s_{p-2} p^{p-2}+\cdots+s_{2} p^{2}=-s_{1} p
$$
Since $p \geq 5, p \mid s_{2}$ and hence $s_{1} \equiv 0\left(\bmod p^{2}\right)$ as desired.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $p \geq 5$ be a prime and let $r$ be the number of ways of placing $p$ checkers on a $p \times p$ checkerboard so that not all checkers are in the same row (but they may all be in the same column). Show that $r$ is divisible by $p^{5}$. Here, we assume that all the checkers are identical.
|
Note that $r=\binom{p^{2}}{p}-p$. Hence, it suffices to show that
$$
\left(p^{2}-1\right)\left(p^{2}-2\right) \cdots\left(p^{2}-(p-1)\right)-(p-1)!\equiv 0 \quad\left(\bmod p^{4}\right)
$$
Now, let
$$
f(x):=(x-1)(x-2) \cdots(x-(p-1))=x^{p-1}+s_{p-2} x^{p-2}+\cdots+s_{1} x+s_{0} .
$$
Then the congruence equation (1) is same as $f\left(p^{2}\right)-s_{0} \equiv 0\left(\bmod p^{4}\right)$. Therefore, it suffices to show that $s_{1} p^{2} \equiv 0\left(\bmod p^{4}\right)$ or $s_{1} \equiv 0\left(\bmod p^{2}\right)$.
Since $a^{p-1} \equiv 1(\bmod p)$ for all $1 \leq a \leq p-1$, we can factor
$$
x^{p-1}-1 \equiv(x-1)(x-2) \cdots(x-(p-1)) \quad(\bmod p)
$$
Comparing the coefficients of the left hand side of (3) with those of the right hand side of (2), we obtain $p \mid s_{i}$ for all $1 \leq i \leq p-2$ and $s_{0} \equiv-1(\bmod p)$. On the other hand, plugging $p$ for $x$ in (2), we get
$$
f(p)=(p-1)!=s_{0}=p^{p-1}+s_{p-2} p^{p-2}+\cdots+s_{1} p+s_{0}
$$
which implies
$$
p^{p-1}+s_{p-2} p^{p-2}+\cdots+s_{2} p^{2}=-s_{1} p
$$
Since $p \geq 5, p \mid s_{2}$ and hence $s_{1} \equiv 0\left(\bmod p^{2}\right)$ as desired.
|
{
"resource_path": "APMO/segmented/en-apmo2006_sol.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "(Solution)"
}
|
440e1b89-c0aa-59a4-b99a-ecabe3f3d54a
| 261,496
|
Let $A, B$ be two distinct points on a given circle $O$ and let $P$ be the midpoint of the line segment $A B$. Let $O_{1}$ be the circle tangent to the line $A B$ at $P$ and tangent to the circle $O$. Let $\ell$ be the tangent line, different from the line $A B$, to $O_{1}$ passing through $A$. Let $C$ be the intersection point, different from $A$, of $\ell$ and $O$. Let $Q$ be the midpoint of the line segment $B C$ and $O_{2}$ be the circle tangent to the line $B C$ at $Q$ and tangent to the line segment $A C$. Prove that the circle $O_{2}$ is tangent to the circle $O$.
|
Let $S$ be the tangent point of the circles $O$ and $O_{1}$ and let $T$ be the intersection point, different from $S$, of the circle $O$ and the line $S P$. Let $X$ be the tangent point of $\ell$ to $O_{1}$ and let $M$ be the midpoint of the line segment $X P$. Since $\angle T B P=\angle A S P$, the triangle $T B P$ is similar to the triangle $A S P$. Therefore,
$$
\frac{P T}{P B}=\frac{P A}{P S}
$$
Since the line $\ell$ is tangent to the circle $O_{1}$ at $X$, we have
$$
\angle S P X=90^{\circ}-\angle X S P=90^{\circ}-\angle A P M=\angle P A M
$$
which implies that the triangle $P A M$ is similar to the triangle $S P X$. Consequently,
$$
\frac{X S}{X P}=\frac{M P}{M A}=\frac{X P}{2 M A} \quad \text { and } \quad \frac{X P}{P S}=\frac{M A}{A P}
$$
From this and the above observation follows
$$
\frac{X S}{X P} \cdot \frac{P T}{P B}=\frac{X P}{2 M A} \cdot \frac{P A}{P S}=\frac{X P}{2 M A} \cdot \frac{M A}{X P}=\frac{1}{2} .
$$
Let $A^{\prime}$ be the intersection point of the circle $O$ and the perpendicular bisector of the chord $B C$ such that $A, A^{\prime}$ are on the same side of the line $B C$, and $N$ be the intersection point of the lines $A^{\prime} Q$ and $C T$. Since
$$
\angle N C Q=\angle T C B=\angle T C A=\angle T B A=\angle T B P
$$
and
$$
\angle C A^{\prime} Q=\frac{\angle C A B}{2}=\frac{\angle X A P}{2}=\angle P A M=\angle S P X,
$$
the triangle $N C Q$ is similar to the triangle $T B P$ and the triangle $C A^{\prime} Q$ is similar to the triangle $S P X$. Therefore
$$
\frac{Q N}{Q C}=\frac{P T}{P B} \quad \text { and } \quad \frac{Q C}{Q A^{\prime}}=\frac{X S}{X P} .
$$
and hence $Q A^{\prime}=2 Q N$ by (1). This implies that $N$ is the midpoint of the line segment $Q A^{\prime}$. Let the circle $O_{2}$ touch the line segment $A C$ at $Y$. Since
$$
\angle A C N=\angle A C T=\angle B C T=\angle Q C N
$$
and $|C Y|=|C Q|$, the triangles $Y C N$ and $Q C N$ are congruent and hence $N Y \perp A C$ and $N Y=N Q=N A^{\prime}$. Therefore, $N$ is the center of the circle $O_{2}$, which completes the proof.
Remark: Analytic solutions are possible: For example, one can prove for a triangle $A B C$ inscribed in a circle $O$ that $A B=k(2+2 t), A C=k(1+2 t), B C=k(1+4 t)$ for some positive numbers $k, t$ if and only if there exists a circle $O_{1}$ such that $O_{1}$ is tangent to the side $A B$ at its midpoint, the side $A C$ and the circle $O$.
One obtains $A B=k^{\prime}\left(1+4 t^{\prime}\right), A C=k^{\prime}\left(1+2 t^{\prime}\right), B C=k^{\prime}\left(2+2 t^{\prime}\right)$ by substituting $t=1 / 4 t^{\prime}$ and $k=2 k^{\prime} t^{\prime}$. So, there exists a circle $O_{2}$ such that $O_{2}$ is tangent to the side $B C$ at its midpoint, the side $A C$ and the circle $O$.
In the above, $t=\tan ^{2} \alpha$ and $k=\frac{4 R \tan \alpha}{\left(1+\tan ^{2} \alpha\right)\left(1+4 \tan ^{2} \alpha\right)}$, where $R$ is the radius of $O$ and $\angle A=2 \alpha$. Furthermore, $t^{\prime}=\tan ^{2} \gamma$ and $k^{\prime}=\frac{4 R \tan \gamma}{\left(1+\tan ^{2} \gamma\right)\left(1+4 \tan ^{2} \gamma\right)}$, where $\angle C=2 \gamma$. Observe that $\sqrt{t t^{\prime}}=\tan \alpha \cdot \tan \gamma=\frac{X S}{X P} \cdot \frac{P T}{P B}=\frac{1}{2}$, which implies $t t^{\prime}=\frac{1}{4}$. It is now routine easy to check that $k=2 k^{\prime} t^{\prime}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A, B$ be two distinct points on a given circle $O$ and let $P$ be the midpoint of the line segment $A B$. Let $O_{1}$ be the circle tangent to the line $A B$ at $P$ and tangent to the circle $O$. Let $\ell$ be the tangent line, different from the line $A B$, to $O_{1}$ passing through $A$. Let $C$ be the intersection point, different from $A$, of $\ell$ and $O$. Let $Q$ be the midpoint of the line segment $B C$ and $O_{2}$ be the circle tangent to the line $B C$ at $Q$ and tangent to the line segment $A C$. Prove that the circle $O_{2}$ is tangent to the circle $O$.
|
Let $S$ be the tangent point of the circles $O$ and $O_{1}$ and let $T$ be the intersection point, different from $S$, of the circle $O$ and the line $S P$. Let $X$ be the tangent point of $\ell$ to $O_{1}$ and let $M$ be the midpoint of the line segment $X P$. Since $\angle T B P=\angle A S P$, the triangle $T B P$ is similar to the triangle $A S P$. Therefore,
$$
\frac{P T}{P B}=\frac{P A}{P S}
$$
Since the line $\ell$ is tangent to the circle $O_{1}$ at $X$, we have
$$
\angle S P X=90^{\circ}-\angle X S P=90^{\circ}-\angle A P M=\angle P A M
$$
which implies that the triangle $P A M$ is similar to the triangle $S P X$. Consequently,
$$
\frac{X S}{X P}=\frac{M P}{M A}=\frac{X P}{2 M A} \quad \text { and } \quad \frac{X P}{P S}=\frac{M A}{A P}
$$
From this and the above observation follows
$$
\frac{X S}{X P} \cdot \frac{P T}{P B}=\frac{X P}{2 M A} \cdot \frac{P A}{P S}=\frac{X P}{2 M A} \cdot \frac{M A}{X P}=\frac{1}{2} .
$$
Let $A^{\prime}$ be the intersection point of the circle $O$ and the perpendicular bisector of the chord $B C$ such that $A, A^{\prime}$ are on the same side of the line $B C$, and $N$ be the intersection point of the lines $A^{\prime} Q$ and $C T$. Since
$$
\angle N C Q=\angle T C B=\angle T C A=\angle T B A=\angle T B P
$$
and
$$
\angle C A^{\prime} Q=\frac{\angle C A B}{2}=\frac{\angle X A P}{2}=\angle P A M=\angle S P X,
$$
the triangle $N C Q$ is similar to the triangle $T B P$ and the triangle $C A^{\prime} Q$ is similar to the triangle $S P X$. Therefore
$$
\frac{Q N}{Q C}=\frac{P T}{P B} \quad \text { and } \quad \frac{Q C}{Q A^{\prime}}=\frac{X S}{X P} .
$$
and hence $Q A^{\prime}=2 Q N$ by (1). This implies that $N$ is the midpoint of the line segment $Q A^{\prime}$. Let the circle $O_{2}$ touch the line segment $A C$ at $Y$. Since
$$
\angle A C N=\angle A C T=\angle B C T=\angle Q C N
$$
and $|C Y|=|C Q|$, the triangles $Y C N$ and $Q C N$ are congruent and hence $N Y \perp A C$ and $N Y=N Q=N A^{\prime}$. Therefore, $N$ is the center of the circle $O_{2}$, which completes the proof.
Remark: Analytic solutions are possible: For example, one can prove for a triangle $A B C$ inscribed in a circle $O$ that $A B=k(2+2 t), A C=k(1+2 t), B C=k(1+4 t)$ for some positive numbers $k, t$ if and only if there exists a circle $O_{1}$ such that $O_{1}$ is tangent to the side $A B$ at its midpoint, the side $A C$ and the circle $O$.
One obtains $A B=k^{\prime}\left(1+4 t^{\prime}\right), A C=k^{\prime}\left(1+2 t^{\prime}\right), B C=k^{\prime}\left(2+2 t^{\prime}\right)$ by substituting $t=1 / 4 t^{\prime}$ and $k=2 k^{\prime} t^{\prime}$. So, there exists a circle $O_{2}$ such that $O_{2}$ is tangent to the side $B C$ at its midpoint, the side $A C$ and the circle $O$.
In the above, $t=\tan ^{2} \alpha$ and $k=\frac{4 R \tan \alpha}{\left(1+\tan ^{2} \alpha\right)\left(1+4 \tan ^{2} \alpha\right)}$, where $R$ is the radius of $O$ and $\angle A=2 \alpha$. Furthermore, $t^{\prime}=\tan ^{2} \gamma$ and $k^{\prime}=\frac{4 R \tan \gamma}{\left(1+\tan ^{2} \gamma\right)\left(1+4 \tan ^{2} \gamma\right)}$, where $\angle C=2 \gamma$. Observe that $\sqrt{t t^{\prime}}=\tan \alpha \cdot \tan \gamma=\frac{X S}{X P} \cdot \frac{P T}{P B}=\frac{1}{2}$, which implies $t t^{\prime}=\frac{1}{4}$. It is now routine easy to check that $k=2 k^{\prime} t^{\prime}$.
|
{
"resource_path": "APMO/segmented/en-apmo2006_sol.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "(Solution)"
}
|
311403b9-0e44-541c-b940-2241ae909e73
| 261,504
|
Let $S$ be a set of 9 distinct integers all of whose prime factors are at most 3. Prove that $S$ contains 3 distinct integers such that their product is a perfect cube.
|
Without loss of generality, we may assume that $S$ contains only positive integers. Let
$$
S=\left\{2^{a_{i}} 3^{b_{i}} \mid a_{i}, b_{i} \in \mathbb{Z}, a_{i}, b_{i} \geq 0,1 \leq i \leq 9\right\}
$$
It suffices to show that there are $1 \leq i_{1}, i_{2}, i_{3} \leq 9$ such that
$$
a_{i_{1}}+a_{i_{2}}+a_{i_{3}} \equiv b_{i_{1}}+b_{i_{2}}+b_{i_{3}} \equiv 0 \quad(\bmod 3) .
$$
For $n=2^{a} 3^{b} \in S$, let's call $(a(\bmod 3), b(\bmod 3))$ the type of $n$. Then there are 9 possible types:
$$
(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)
$$
Let $N(i, j)$ be the number of integers in $S$ of type $(i, j)$. We obtain 3 distinct integers whose product is a perfect cube when
(1) $N(i, j) \geq 3$ for some $i, j$, or
(2) $N(i, 0) N(i, 1) N(i, 2) \neq 0$ for some $i=0,1,2$, or
(3) $N(0, j) N(1, j) N(2, j) \neq 0$ for some $j=0,1,2$, or
(4) $N\left(i_{1}, j_{1}\right) N\left(i_{2}, j_{2}\right) N\left(i_{3}, j_{3}\right) \neq 0$, where $\left\{i_{1}, i_{2}, i_{3}\right\}=\left\{j_{1}, j_{2}, j_{3}\right\}=\{0,1,2\}$.
Assume that none of the conditions (1) (3) holds. Since $N(i, j) \leq 2$ for all $(i, j)$, there are at least five $N(i, j)$ 's that are nonzero. Furthermore, among those nonzero $N(i, j)$ 's, no three have the same $i$ nor the same $j$. Using these facts, one may easily conclude that the condition (4) should hold. (For example, if one places each nonzero $N(i, j)$ in the $(i, j)$-th box of a regular $3 \times 3$ array of boxes whose rows and columns are indexed by 0,1 and 2 , then one can always find three boxes, occupied by at least one nonzero $N(i, j)$, whose rows and columns are all distinct. This implies (4).)
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $S$ be a set of 9 distinct integers all of whose prime factors are at most 3. Prove that $S$ contains 3 distinct integers such that their product is a perfect cube.
|
Without loss of generality, we may assume that $S$ contains only positive integers. Let
$$
S=\left\{2^{a_{i}} 3^{b_{i}} \mid a_{i}, b_{i} \in \mathbb{Z}, a_{i}, b_{i} \geq 0,1 \leq i \leq 9\right\}
$$
It suffices to show that there are $1 \leq i_{1}, i_{2}, i_{3} \leq 9$ such that
$$
a_{i_{1}}+a_{i_{2}}+a_{i_{3}} \equiv b_{i_{1}}+b_{i_{2}}+b_{i_{3}} \equiv 0 \quad(\bmod 3) .
$$
For $n=2^{a} 3^{b} \in S$, let's call $(a(\bmod 3), b(\bmod 3))$ the type of $n$. Then there are 9 possible types:
$$
(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)
$$
Let $N(i, j)$ be the number of integers in $S$ of type $(i, j)$. We obtain 3 distinct integers whose product is a perfect cube when
(1) $N(i, j) \geq 3$ for some $i, j$, or
(2) $N(i, 0) N(i, 1) N(i, 2) \neq 0$ for some $i=0,1,2$, or
(3) $N(0, j) N(1, j) N(2, j) \neq 0$ for some $j=0,1,2$, or
(4) $N\left(i_{1}, j_{1}\right) N\left(i_{2}, j_{2}\right) N\left(i_{3}, j_{3}\right) \neq 0$, where $\left\{i_{1}, i_{2}, i_{3}\right\}=\left\{j_{1}, j_{2}, j_{3}\right\}=\{0,1,2\}$.
Assume that none of the conditions (1) (3) holds. Since $N(i, j) \leq 2$ for all $(i, j)$, there are at least five $N(i, j)$ 's that are nonzero. Furthermore, among those nonzero $N(i, j)$ 's, no three have the same $i$ nor the same $j$. Using these facts, one may easily conclude that the condition (4) should hold. (For example, if one places each nonzero $N(i, j)$ in the $(i, j)$-th box of a regular $3 \times 3$ array of boxes whose rows and columns are indexed by 0,1 and 2 , then one can always find three boxes, occupied by at least one nonzero $N(i, j)$, whose rows and columns are all distinct. This implies (4).)
|
{
"resource_path": "APMO/segmented/en-apmo2007_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution."
}
|
cb1a75c0-029c-534e-9491-d72a61bf40f2
| 605,094
|
Let $S$ be a set of 9 distinct integers all of whose prime factors are at most 3. Prove that $S$ contains 3 distinct integers such that their product is a perfect cube.
|
Up to $(\dagger)$, we do the same as above and get 9 possible types:
$$
(a(\bmod 3), b(\bmod 3))=(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)
$$
for $n=2^{a} 3^{b} \in S$.
Note that (i) among any 5 integers, there exist 3 whose sum is $0(\bmod 3)$, and that (ii) if $i, j, k \in\{0,1,2\}$, then $i+j+k \equiv 0(\bmod 3)$ if and only if $i=j=k$ or $\{i, j, k\}=\{0,1,2\}$.
Let's define
$T$ : the set of types of the integers in $S$;
$N(i)$ : the number of integers in $S$ of the type $(i, \cdot)$;
$M(i)$ : the number of integers $j \in\{0,1,2\}$ such that $(i, j) \in T$.
If $N(i) \geq 5$ for some $i$, the result follows from (i). Otherwise, for some permutation $(i, j, k)$ of $(0,1,2)$,
$$
N(i) \geq 3, \quad N(j) \geq 3, \quad N(k) \geq 1
$$
If $M(i)$ or $M(j)$ is 1 or 3 , the result follows from (ii). Otherwise $M(i)=M(j)=2$. Then either
$$
(i, x),(i, y),(j, x),(j, y) \in T \quad \text { or } \quad(i, x),(i, y),(j, x),(j, z) \in T
$$
for some permutation $(x, y, z)$ of $(0,1,2)$. Since $N(k) \geq 1$, at least one of $(k, x),(k, y)$ and $(k, z)$ contained in $T$. Therefore, in any case, the result follows from (ii). (For example, if $(k, y) \in T$, then take $(i, y),(j, y),(k, y)$ or $(i, x),(j, z),(k, y)$ from T.)
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $S$ be a set of 9 distinct integers all of whose prime factors are at most 3. Prove that $S$ contains 3 distinct integers such that their product is a perfect cube.
|
Up to $(\dagger)$, we do the same as above and get 9 possible types:
$$
(a(\bmod 3), b(\bmod 3))=(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)
$$
for $n=2^{a} 3^{b} \in S$.
Note that (i) among any 5 integers, there exist 3 whose sum is $0(\bmod 3)$, and that (ii) if $i, j, k \in\{0,1,2\}$, then $i+j+k \equiv 0(\bmod 3)$ if and only if $i=j=k$ or $\{i, j, k\}=\{0,1,2\}$.
Let's define
$T$ : the set of types of the integers in $S$;
$N(i)$ : the number of integers in $S$ of the type $(i, \cdot)$;
$M(i)$ : the number of integers $j \in\{0,1,2\}$ such that $(i, j) \in T$.
If $N(i) \geq 5$ for some $i$, the result follows from (i). Otherwise, for some permutation $(i, j, k)$ of $(0,1,2)$,
$$
N(i) \geq 3, \quad N(j) \geq 3, \quad N(k) \geq 1
$$
If $M(i)$ or $M(j)$ is 1 or 3 , the result follows from (ii). Otherwise $M(i)=M(j)=2$. Then either
$$
(i, x),(i, y),(j, x),(j, y) \in T \quad \text { or } \quad(i, x),(i, y),(j, x),(j, z) \in T
$$
for some permutation $(x, y, z)$ of $(0,1,2)$. Since $N(k) \geq 1$, at least one of $(k, x),(k, y)$ and $(k, z)$ contained in $T$. Therefore, in any case, the result follows from (ii). (For example, if $(k, y) \in T$, then take $(i, y),(j, y),(k, y)$ or $(i, x),(j, z),(k, y)$ from T.)
|
{
"resource_path": "APMO/segmented/en-apmo2007_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSecond solution."
}
|
cb1a75c0-029c-534e-9491-d72a61bf40f2
| 605,094
|
Let $A B C$ be an acute angled triangle with $\angle B A C=60^{\circ}$ and $A B>A C$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $A B C$. Prove that
$$
2 \angle A H I=3 \angle A B C .
$$
|
Let $D$ be the intersection point of the lines $A H$ and $B C$. Let $K$ be the intersection point of the circumcircle $O$ of the triangle $A B C$ and the line $A H$. Let the line through $I$ perpendicular to $B C$ meet $B C$ and the minor arc $B C$ of the circumcircle $O$ at $E$ and $N$, respectively. We have
$\angle B I C=180^{\circ}-(\angle I B C+\angle I C B)=180^{\circ}-\frac{1}{2}(\angle A B C+\angle A C B)=90^{\circ}+\frac{1}{2} \angle B A C=120^{\circ}$
and also $\angle B N C=180^{\circ}-\angle B A C=120^{\circ}=\angle B I C$. Since $I N \perp B C$, the quadrilateral $B I C N$ is a kite and thus $I E=E N$.
Now, since $H$ is the orthocenter of the triangle $A B C, H D=D K$. Also because $E D \perp I N$ and $E D \perp H K$, we conclude that $I H K N$ is an isosceles trapezoid with $I H=N K$.
Hence
$$
\angle A H I=180^{\circ}-\angle I H K=180^{\circ}-\angle A K N=\angle A B N .
$$
Since $I E=E N$ and $B E \perp I N$, the triangles $I B E$ and $N B E$ are congruent. Therefore
$$
\angle N B E=\angle I B E=\angle I B C=\angle I B A=\frac{1}{2} \angle A B C
$$
and thus
$$
\angle A H I=\angle A B N=\frac{3}{2} \angle A B C .
$$
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute angled triangle with $\angle B A C=60^{\circ}$ and $A B>A C$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $A B C$. Prove that
$$
2 \angle A H I=3 \angle A B C .
$$
|
Let $D$ be the intersection point of the lines $A H$ and $B C$. Let $K$ be the intersection point of the circumcircle $O$ of the triangle $A B C$ and the line $A H$. Let the line through $I$ perpendicular to $B C$ meet $B C$ and the minor arc $B C$ of the circumcircle $O$ at $E$ and $N$, respectively. We have
$\angle B I C=180^{\circ}-(\angle I B C+\angle I C B)=180^{\circ}-\frac{1}{2}(\angle A B C+\angle A C B)=90^{\circ}+\frac{1}{2} \angle B A C=120^{\circ}$
and also $\angle B N C=180^{\circ}-\angle B A C=120^{\circ}=\angle B I C$. Since $I N \perp B C$, the quadrilateral $B I C N$ is a kite and thus $I E=E N$.
Now, since $H$ is the orthocenter of the triangle $A B C, H D=D K$. Also because $E D \perp I N$ and $E D \perp H K$, we conclude that $I H K N$ is an isosceles trapezoid with $I H=N K$.
Hence
$$
\angle A H I=180^{\circ}-\angle I H K=180^{\circ}-\angle A K N=\angle A B N .
$$
Since $I E=E N$ and $B E \perp I N$, the triangles $I B E$ and $N B E$ are congruent. Therefore
$$
\angle N B E=\angle I B E=\angle I B C=\angle I B A=\frac{1}{2} \angle A B C
$$
and thus
$$
\angle A H I=\angle A B N=\frac{3}{2} \angle A B C .
$$
|
{
"resource_path": "APMO/segmented/en-apmo2007_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "\nSolution."
}
|
bf88f2c2-e0fd-5cc3-9f01-ea337ecb05fb
| 605,119
|
Let $x, y$ and $z$ be positive real numbers such that $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$. Prove that
$$
\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \geq 1
$$
|
We first note that
$$
\begin{aligned}
\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}} & =\frac{x^{2}-x(y+z)+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{x(y+z)}{\sqrt{2 x^{2}(y+z)}} \\
& =\frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\sqrt{\frac{y+z}{2}} \\
& \geq \frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\frac{\sqrt{y}+\sqrt{z}}{2} .
\end{aligned}
$$
Similarly, we have
$$
\begin{aligned}
& \frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}} \geq \frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}+\frac{\sqrt{z}+\sqrt{x}}{2}, \\
& \frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \geq \frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}}+\frac{\sqrt{x}+\sqrt{y}}{2} .
\end{aligned}
$$
We now add (1) (3) to get
$$
\begin{aligned}
& \frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \\
& \geq \frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}+\frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}}+\sqrt{x}+\sqrt{y}+\sqrt{z} \\
& =\frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}+\frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}}+1 .
\end{aligned}
$$
Thus, it suffices to show that
$$
\frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}+\frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}} \geq 0 .
$$
Now, assume without loss of generality, that $x \geq y \geq z$. Then we have
$$
\frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}} \geq 0
$$
and
$$
\begin{aligned}
& \frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}}+\frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}=\frac{(y-z)(x-z)}{\sqrt{2 z^{2}(x+y)}}-\frac{(y-z)(x-y)}{\sqrt{2 y^{2}(z+x)}} \\
& \geq \frac{(y-z)(x-y)}{\sqrt{2 z^{2}(x+y)}}-\frac{(y-z)(x-y)}{\sqrt{2 y^{2}(z+x)}}=(y-z)(x-y)\left(\frac{1}{\sqrt{2 z^{2}(x+y)}}-\frac{1}{\sqrt{2 y^{2}(z+x)}}\right)
\end{aligned}
$$
The last quantity is non-negative due to the fact that
$$
y^{2}(z+x)=y^{2} z+y^{2} x \geq y z^{2}+z^{2} x=z^{2}(x+y)
$$
This completes the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $x, y$ and $z$ be positive real numbers such that $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$. Prove that
$$
\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \geq 1
$$
|
We first note that
$$
\begin{aligned}
\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}} & =\frac{x^{2}-x(y+z)+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{x(y+z)}{\sqrt{2 x^{2}(y+z)}} \\
& =\frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\sqrt{\frac{y+z}{2}} \\
& \geq \frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\frac{\sqrt{y}+\sqrt{z}}{2} .
\end{aligned}
$$
Similarly, we have
$$
\begin{aligned}
& \frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}} \geq \frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}+\frac{\sqrt{z}+\sqrt{x}}{2}, \\
& \frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \geq \frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}}+\frac{\sqrt{x}+\sqrt{y}}{2} .
\end{aligned}
$$
We now add (1) (3) to get
$$
\begin{aligned}
& \frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \\
& \geq \frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}+\frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}}+\sqrt{x}+\sqrt{y}+\sqrt{z} \\
& =\frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}+\frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}}+1 .
\end{aligned}
$$
Thus, it suffices to show that
$$
\frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}+\frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}} \geq 0 .
$$
Now, assume without loss of generality, that $x \geq y \geq z$. Then we have
$$
\frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}} \geq 0
$$
and
$$
\begin{aligned}
& \frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}}+\frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}=\frac{(y-z)(x-z)}{\sqrt{2 z^{2}(x+y)}}-\frac{(y-z)(x-y)}{\sqrt{2 y^{2}(z+x)}} \\
& \geq \frac{(y-z)(x-y)}{\sqrt{2 z^{2}(x+y)}}-\frac{(y-z)(x-y)}{\sqrt{2 y^{2}(z+x)}}=(y-z)(x-y)\left(\frac{1}{\sqrt{2 z^{2}(x+y)}}-\frac{1}{\sqrt{2 y^{2}(z+x)}}\right)
\end{aligned}
$$
The last quantity is non-negative due to the fact that
$$
y^{2}(z+x)=y^{2} z+y^{2} x \geq y z^{2}+z^{2} x=z^{2}(x+y)
$$
This completes the proof.
|
{
"resource_path": "APMO/segmented/en-apmo2007_sol.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "\nSolution."
}
|
2976be20-c465-5a41-b660-ed93dce21add
| 605,165
|
Let $x, y$ and $z$ be positive real numbers such that $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$. Prove that
$$
\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \geq 1
$$
|
By Cauchy-Schwarz inequality,
$$
\begin{aligned}
& \left(\frac{x^{2}}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}}{\sqrt{2 z^{2}(x+y)}}\right) \\
& \quad \times(\sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}) \geq(\sqrt{x}+\sqrt{y}+\sqrt{z})^{2}=1
\end{aligned}
$$
and
$$
\begin{aligned}
& \left(\frac{y z}{\sqrt{2 x^{2}(y+z)}}+\frac{z x}{\sqrt{2 y^{2}(z+x)}}+\frac{x y}{\sqrt{2 z^{2}(x+y)}}\right) \\
& \quad \times(\sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}) \geq\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right)^{2}
\end{aligned}
$$
We now combine (5) and (6) to find
$$
\begin{aligned}
& \left(\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}}\right) \\
& \quad \times(\sqrt{2(x+y)}+\sqrt{2(y+z)}+\sqrt{2(z+x)}) \\
& \geq 1+\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right)^{2} \geq 2\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right) .
\end{aligned}
$$
Thus, it suffices to show that
$$
2\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right) \geq \sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}
$$
Consider the following inequality using AM-GM inequality
$$
\left[\sqrt{\frac{y z}{x}}+\left(\frac{1}{2} \sqrt{\frac{z x}{y}}+\frac{1}{2} \sqrt{\frac{x y}{z}}\right)\right]^{2} \geq 4 \sqrt{\frac{y z}{x}}\left(\frac{1}{2} \sqrt{\frac{z x}{y}}+\frac{1}{2} \sqrt{\frac{x y}{z}}\right)=2(y+z)
$$
or equivalently
$$
\sqrt{\frac{y z}{x}}+\left(\frac{1}{2} \sqrt{\frac{z x}{y}}+\frac{1}{2} \sqrt{\frac{x y}{z}}\right) \geq \sqrt{2(y+z)} .
$$
Similarly, we have
$$
\begin{aligned}
& \sqrt{\frac{z x}{y}}+\left(\frac{1}{2} \sqrt{\frac{x y}{z}}+\frac{1}{2} \sqrt{\frac{y z}{x}}\right) \geq \sqrt{2(z+x)} \\
& \sqrt{\frac{x y}{z}}+\left(\frac{1}{2} \sqrt{\frac{y z}{x}}+\frac{1}{2} \sqrt{\frac{z x}{y}}\right) \geq \sqrt{2(x+y)}
\end{aligned}
$$
Adding the last three inequalities, we get
$$
2\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right) \geq \sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)} .
$$
This completes the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $x, y$ and $z$ be positive real numbers such that $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$. Prove that
$$
\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \geq 1
$$
|
By Cauchy-Schwarz inequality,
$$
\begin{aligned}
& \left(\frac{x^{2}}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}}{\sqrt{2 z^{2}(x+y)}}\right) \\
& \quad \times(\sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}) \geq(\sqrt{x}+\sqrt{y}+\sqrt{z})^{2}=1
\end{aligned}
$$
and
$$
\begin{aligned}
& \left(\frac{y z}{\sqrt{2 x^{2}(y+z)}}+\frac{z x}{\sqrt{2 y^{2}(z+x)}}+\frac{x y}{\sqrt{2 z^{2}(x+y)}}\right) \\
& \quad \times(\sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}) \geq\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right)^{2}
\end{aligned}
$$
We now combine (5) and (6) to find
$$
\begin{aligned}
& \left(\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}}\right) \\
& \quad \times(\sqrt{2(x+y)}+\sqrt{2(y+z)}+\sqrt{2(z+x)}) \\
& \geq 1+\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right)^{2} \geq 2\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right) .
\end{aligned}
$$
Thus, it suffices to show that
$$
2\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right) \geq \sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}
$$
Consider the following inequality using AM-GM inequality
$$
\left[\sqrt{\frac{y z}{x}}+\left(\frac{1}{2} \sqrt{\frac{z x}{y}}+\frac{1}{2} \sqrt{\frac{x y}{z}}\right)\right]^{2} \geq 4 \sqrt{\frac{y z}{x}}\left(\frac{1}{2} \sqrt{\frac{z x}{y}}+\frac{1}{2} \sqrt{\frac{x y}{z}}\right)=2(y+z)
$$
or equivalently
$$
\sqrt{\frac{y z}{x}}+\left(\frac{1}{2} \sqrt{\frac{z x}{y}}+\frac{1}{2} \sqrt{\frac{x y}{z}}\right) \geq \sqrt{2(y+z)} .
$$
Similarly, we have
$$
\begin{aligned}
& \sqrt{\frac{z x}{y}}+\left(\frac{1}{2} \sqrt{\frac{x y}{z}}+\frac{1}{2} \sqrt{\frac{y z}{x}}\right) \geq \sqrt{2(z+x)} \\
& \sqrt{\frac{x y}{z}}+\left(\frac{1}{2} \sqrt{\frac{y z}{x}}+\frac{1}{2} \sqrt{\frac{z x}{y}}\right) \geq \sqrt{2(x+y)}
\end{aligned}
$$
Adding the last three inequalities, we get
$$
2\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right) \geq \sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)} .
$$
This completes the proof.
|
{
"resource_path": "APMO/segmented/en-apmo2007_sol.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "\nSecond solution."
}
|
2976be20-c465-5a41-b660-ed93dce21add
| 605,165
|
Let $A B C$ be a triangle with $\angle A<60^{\circ}$. Let $X$ and $Y$ be the points on the sides $A B$ and $A C$, respectively, such that $C A+A X=C B+B X$ and $B A+A Y=B C+C Y$. Let $P$ be the point in the plane such that the lines $P X$ and $P Y$ are perpendicular to $A B$ and $A C$, respectively. Prove that $\angle B P C<120^{\circ}$.
|
Let $I$ be the incenter of $\triangle A B C$, and let the feet of the perpendiculars from $I$ to $A B$ and to $A C$ be $D$ and $E$, respectively. (Without loss of generality, we may assume that $A C$ is the longest side. Then $X$ lies on the line segment $A D$. Although $P$ may or may not lie inside $\triangle A B C$, the proof below works for both cases. Note that $P$ is on the line perpendicular to $A B$ passing through $X$.) Let $O$ be the midpoint of $I P$, and let the feet of the perpendiculars from $O$ to $A B$ and to $A C$ be $M$ and $N$, respectively. Then $M$ and $N$ are the midpoints of $D X$ and $E Y$, respectively.
The conditions on the points $X$ and $Y$ yield the equations
$$
A X=\frac{A B+B C-C A}{2} \quad \text { and } \quad A Y=\frac{B C+C A-A B}{2} .
$$
From $A D=A E=\frac{C A+A B-B C}{2}$, we obtain
$$
B D=A B-A D=A B-\frac{C A+A B-B C}{2}=\frac{A B+B C-C A}{2}=A X .
$$
Since $M$ is the midpoint of $D X$, it follows that $M$ is the midpoint of $A B$. Similarly, $N$ is the midpoint of $A C$. Therefore, the perpendicular bisectors of $A B$ and $A C$ meet at $O$, that is, $O$ is the circumcenter of $\triangle A B C$. Since $\angle B A C<60^{\circ}, O$ lies on the same side of $B C$ as the point $A$ and
$$
\angle B O C=2 \angle B A C
$$
We can compute $\angle B I C$ as follows:
$$
\begin{aligned}
\angle B I C & =180^{\circ}-\angle I B C-\angle I C B=180^{\circ}-\frac{1}{2} \angle A B C-\frac{1}{2} \angle A C B \\
& =180^{\circ}-\frac{1}{2}(\angle A B C+\angle A C B)=180^{\circ}-\frac{1}{2}\left(180^{\circ}-\angle B A C\right)=90^{\circ}+\frac{1}{2} \angle B A C
\end{aligned}
$$
It follows from $\angle B A C<60^{\circ}$ that
$$
2 \angle B A C<90^{\circ}+\frac{1}{2} \angle B A C, \quad \text { i.e., } \quad \angle B O C<\angle B I C \text {. }
$$
From this it follows that $I$ lies inside the circumcircle of the isosceles triangle $B O C$ because $O$ and $I$ lie on the same side of $B C$. However, as $O$ is the midpoint of $I P, P$ must lie outside the circumcircle of triangle $B O C$ and on the same side of $B C$ as $O$. Therefore
$$
\angle B P C<\angle B O C=2 \angle B A C<120^{\circ} .
$$
Remark. If one assumes that $\angle A$ is smaller than the other two, then it is clear that the line $P X$ (or the line perpendicular to $A B$ at $X$ if $P=X$ ) runs through the excenter $I_{C}$ of the excircle tangent to the side $A B$. Since $2 \angle A C I_{C}=\angle A C B$ and $B C<A C$, we have $2 \angle P C B>\angle C$. Similarly, $2 \angle P B C>\angle B$. Therefore,
$$
\angle B P C=180^{\circ}-(\angle P B C+\angle P C B)<180^{\circ}-\left(\frac{\angle B+\angle C}{2}\right)=90+\frac{\angle A}{2}<120^{\circ}
$$
In this way, a special case of the problem can be easily proved.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $\angle A<60^{\circ}$. Let $X$ and $Y$ be the points on the sides $A B$ and $A C$, respectively, such that $C A+A X=C B+B X$ and $B A+A Y=B C+C Y$. Let $P$ be the point in the plane such that the lines $P X$ and $P Y$ are perpendicular to $A B$ and $A C$, respectively. Prove that $\angle B P C<120^{\circ}$.
|
Let $I$ be the incenter of $\triangle A B C$, and let the feet of the perpendiculars from $I$ to $A B$ and to $A C$ be $D$ and $E$, respectively. (Without loss of generality, we may assume that $A C$ is the longest side. Then $X$ lies on the line segment $A D$. Although $P$ may or may not lie inside $\triangle A B C$, the proof below works for both cases. Note that $P$ is on the line perpendicular to $A B$ passing through $X$.) Let $O$ be the midpoint of $I P$, and let the feet of the perpendiculars from $O$ to $A B$ and to $A C$ be $M$ and $N$, respectively. Then $M$ and $N$ are the midpoints of $D X$ and $E Y$, respectively.
The conditions on the points $X$ and $Y$ yield the equations
$$
A X=\frac{A B+B C-C A}{2} \quad \text { and } \quad A Y=\frac{B C+C A-A B}{2} .
$$
From $A D=A E=\frac{C A+A B-B C}{2}$, we obtain
$$
B D=A B-A D=A B-\frac{C A+A B-B C}{2}=\frac{A B+B C-C A}{2}=A X .
$$
Since $M$ is the midpoint of $D X$, it follows that $M$ is the midpoint of $A B$. Similarly, $N$ is the midpoint of $A C$. Therefore, the perpendicular bisectors of $A B$ and $A C$ meet at $O$, that is, $O$ is the circumcenter of $\triangle A B C$. Since $\angle B A C<60^{\circ}, O$ lies on the same side of $B C$ as the point $A$ and
$$
\angle B O C=2 \angle B A C
$$
We can compute $\angle B I C$ as follows:
$$
\begin{aligned}
\angle B I C & =180^{\circ}-\angle I B C-\angle I C B=180^{\circ}-\frac{1}{2} \angle A B C-\frac{1}{2} \angle A C B \\
& =180^{\circ}-\frac{1}{2}(\angle A B C+\angle A C B)=180^{\circ}-\frac{1}{2}\left(180^{\circ}-\angle B A C\right)=90^{\circ}+\frac{1}{2} \angle B A C
\end{aligned}
$$
It follows from $\angle B A C<60^{\circ}$ that
$$
2 \angle B A C<90^{\circ}+\frac{1}{2} \angle B A C, \quad \text { i.e., } \quad \angle B O C<\angle B I C \text {. }
$$
From this it follows that $I$ lies inside the circumcircle of the isosceles triangle $B O C$ because $O$ and $I$ lie on the same side of $B C$. However, as $O$ is the midpoint of $I P, P$ must lie outside the circumcircle of triangle $B O C$ and on the same side of $B C$ as $O$. Therefore
$$
\angle B P C<\angle B O C=2 \angle B A C<120^{\circ} .
$$
Remark. If one assumes that $\angle A$ is smaller than the other two, then it is clear that the line $P X$ (or the line perpendicular to $A B$ at $X$ if $P=X$ ) runs through the excenter $I_{C}$ of the excircle tangent to the side $A B$. Since $2 \angle A C I_{C}=\angle A C B$ and $B C<A C$, we have $2 \angle P C B>\angle C$. Similarly, $2 \angle P B C>\angle B$. Therefore,
$$
\angle B P C=180^{\circ}-(\angle P B C+\angle P C B)<180^{\circ}-\left(\frac{\angle B+\angle C}{2}\right)=90+\frac{\angle A}{2}<120^{\circ}
$$
In this way, a special case of the problem can be easily proved.
|
{
"resource_path": "APMO/segmented/en-apmo2008_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "(Solution)"
}
|
416c31d1-d8c4-560f-9c8e-280509bf2b72
| 605,211
|
Students in a class form groups each of which contains exactly three members such that any two distinct groups have at most one member in common. Prove that, when the class size is 46 , there is a set of 10 students in which no group is properly contained.
|
We let $C$ be the set of all 46 students in the class and let
$$
s:=\max \{|S|: S \subseteq C \text { such that } S \text { contains no group properly }\}
$$
Then it suffices to prove that $s \geq 10$. (If $|S|=s>10$, we may choose a subset of $S$ consisting of 10 students.)
Suppose that $s \leq 9$ and let $S$ be a set of size $s$ in which no group is properly contained. Take any student, say $v$, from outside $S$. Because of the maximality of $s$, there should be a group containing the student $v$ and two other students in $S$. The number of ways to choose two students from $S$ is
$$
\binom{s}{2} \leq\binom{ 9}{2}=36
$$
On the other hand, there are at least $37=46-9$ students outside of $S$. Thus, among those 37 students outside, there is at least one student, say $u$, who does not belong to any group containing two students in $S$ and one outside. This is because no two distinct groups have two members in common. But then, $S$ can be enlarged by including $u$, which is a contradiction.
Remark. One may choose a subset $S$ of $C$ that contains no group properly. Then, assuming $|S|<10$, prove that there is a student outside $S$, say $u$, who does not belong to any group containing two students in $S$. After enlarging $S$ by including $u$, prove that the enlarged $S$ still contains no group properly.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Students in a class form groups each of which contains exactly three members such that any two distinct groups have at most one member in common. Prove that, when the class size is 46 , there is a set of 10 students in which no group is properly contained.
|
We let $C$ be the set of all 46 students in the class and let
$$
s:=\max \{|S|: S \subseteq C \text { such that } S \text { contains no group properly }\}
$$
Then it suffices to prove that $s \geq 10$. (If $|S|=s>10$, we may choose a subset of $S$ consisting of 10 students.)
Suppose that $s \leq 9$ and let $S$ be a set of size $s$ in which no group is properly contained. Take any student, say $v$, from outside $S$. Because of the maximality of $s$, there should be a group containing the student $v$ and two other students in $S$. The number of ways to choose two students from $S$ is
$$
\binom{s}{2} \leq\binom{ 9}{2}=36
$$
On the other hand, there are at least $37=46-9$ students outside of $S$. Thus, among those 37 students outside, there is at least one student, say $u$, who does not belong to any group containing two students in $S$ and one outside. This is because no two distinct groups have two members in common. But then, $S$ can be enlarged by including $u$, which is a contradiction.
Remark. One may choose a subset $S$ of $C$ that contains no group properly. Then, assuming $|S|<10$, prove that there is a student outside $S$, say $u$, who does not belong to any group containing two students in $S$. After enlarging $S$ by including $u$, prove that the enlarged $S$ still contains no group properly.
|
{
"resource_path": "APMO/segmented/en-apmo2008_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "(Solution)"
}
|
fc1b8cc1-65cf-5aa3-a777-4dd71b58a19f
| 605,225
|
Let $\Gamma$ be the circumcircle of a triangle $A B C$. A circle passing through points $A$ and $C$ meets the sides $B C$ and $B A$ at $D$ and $E$, respectively. The lines $A D$ and $C E$ meet $\Gamma$ again at $G$ and $H$, respectively. The tangent lines of $\Gamma$ at $A$ and $C$ meet the line $D E$ at $L$ and $M$, respectively. Prove that the lines $L H$ and $M G$ meet at $\Gamma$.
|
Let $M G$ meet $\Gamma$ at $P$. Since $\angle M C D=\angle C A E$ and $\angle M D C=\angle C A E$, we have $M C=M D$. Thus
$$
M D^{2}=M C^{2}=M G \cdot M P
$$
and hence $M D$ is tangent to the circumcircle of $\triangle D G P$. Therefore $\angle D G P=\angle E D P$.
Let $\Gamma^{\prime}$ be the circumcircle of $\triangle B D E$. If $B=P$, then, since $\angle B G D=\angle B D E$, the tangent lines of $\Gamma^{\prime}$ and $\Gamma$ at $B$ should coincide, that is $\Gamma^{\prime}$ is tangent to $\Gamma$ from inside. Let $B \neq P$. If $P$ lies in the same side of the line $B C$ as $G$, then we have
$$
\angle E D P+\angle A B P=180^{\circ}
$$
because $\angle D G P+\angle A B P=180^{\circ}$. That is, the quadrilateral $B P D E$ is cyclic, and hence $P$ is on the intersection of $\Gamma^{\prime}$ with $\Gamma$.
Otherwise,
$$
\angle E D P=\angle D G P=\angle A G P=\angle A B P=\angle E B P .
$$
Therefore the quadrilateral $P B D E$ is cyclic, and hence $P$ again is on the intersection of $\Gamma^{\prime}$ with $\Gamma$.
Similarly, if $L H$ meets $\Gamma$ at $Q$, we either have $Q=B$, in which case $\Gamma^{\prime}$ is tangent to $\Gamma$ from inside, or $Q \neq B$. In the latter case, $Q$ is on the intersection of $\Gamma^{\prime}$ with $\Gamma$. In either case, we have $P=Q$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $\Gamma$ be the circumcircle of a triangle $A B C$. A circle passing through points $A$ and $C$ meets the sides $B C$ and $B A$ at $D$ and $E$, respectively. The lines $A D$ and $C E$ meet $\Gamma$ again at $G$ and $H$, respectively. The tangent lines of $\Gamma$ at $A$ and $C$ meet the line $D E$ at $L$ and $M$, respectively. Prove that the lines $L H$ and $M G$ meet at $\Gamma$.
|
Let $M G$ meet $\Gamma$ at $P$. Since $\angle M C D=\angle C A E$ and $\angle M D C=\angle C A E$, we have $M C=M D$. Thus
$$
M D^{2}=M C^{2}=M G \cdot M P
$$
and hence $M D$ is tangent to the circumcircle of $\triangle D G P$. Therefore $\angle D G P=\angle E D P$.
Let $\Gamma^{\prime}$ be the circumcircle of $\triangle B D E$. If $B=P$, then, since $\angle B G D=\angle B D E$, the tangent lines of $\Gamma^{\prime}$ and $\Gamma$ at $B$ should coincide, that is $\Gamma^{\prime}$ is tangent to $\Gamma$ from inside. Let $B \neq P$. If $P$ lies in the same side of the line $B C$ as $G$, then we have
$$
\angle E D P+\angle A B P=180^{\circ}
$$
because $\angle D G P+\angle A B P=180^{\circ}$. That is, the quadrilateral $B P D E$ is cyclic, and hence $P$ is on the intersection of $\Gamma^{\prime}$ with $\Gamma$.
Otherwise,
$$
\angle E D P=\angle D G P=\angle A G P=\angle A B P=\angle E B P .
$$
Therefore the quadrilateral $P B D E$ is cyclic, and hence $P$ again is on the intersection of $\Gamma^{\prime}$ with $\Gamma$.
Similarly, if $L H$ meets $\Gamma$ at $Q$, we either have $Q=B$, in which case $\Gamma^{\prime}$ is tangent to $\Gamma$ from inside, or $Q \neq B$. In the latter case, $Q$ is on the intersection of $\Gamma^{\prime}$ with $\Gamma$. In either case, we have $P=Q$.
|
{
"resource_path": "APMO/segmented/en-apmo2008_sol.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "(Solution)"
}
|
c0f9b626-0f74-5cf8-8e79-7f33add8bbf1
| 260,671
|
Let $a, b, c$ be integers satisfying $0<a<c-1$ and $1<b<c$. For each $k$, $0 \leq k \leq a$, let $r_{k}, 0 \leq r_{k}<c$, be the remainder of $k b$ when divided by $c$. Prove that the two sets $\left\{r_{0}, r_{1}, r_{2}, \ldots, r_{a}\right\}$ and $\{0,1,2, \ldots, a\}$ are different.
|
Suppose that two sets are equal. Then $\operatorname{gcd}(b, c)=1$ and the polynomial
$$
f(x):=\left(1+x^{b}+x^{2 b}+\cdots+x^{a b}\right)-\left(1+x+x^{2}+\cdots+x^{a-1}+x^{a}\right)
$$
is divisible by $x^{c}-1$. (This is because: $m=n+c q \Longrightarrow x^{m}-x^{n}=x^{n+c q}-x^{n}=x^{n}\left(x^{c q}-1\right)$ and $\left(x^{c q}-1\right)=\left(x^{c}-1\right)\left(\left(x^{c}\right)^{q-1}+\left(x^{c}\right)^{q-2}+\cdots+1\right)$.) From
$$
f(x)=\frac{x^{(a+1) b}-1}{x^{b}-1}-\frac{x^{a+1}-1}{x-1}=\frac{F(x)}{(x-1)\left(x^{b}-1\right)}
$$
where $F(x)=x^{a b+b+1}+x^{b}+x^{a+1}-x^{a b+b}-x^{a+b+1}-x$, we have
$$
F(x) \equiv 0 \quad\left(\bmod x^{c}-1\right)
$$
Since $x^{c} \equiv 1\left(\bmod x^{c}-1\right)$, we may conclude that
$$
\{a b+b+1, b, a+1\} \equiv\{a b+b, a+b+1,1\} \quad(\bmod c)
$$
Thus,
$$
b \equiv a b+b, a+b+1 \text { or } 1(\bmod c)
$$
But neither $b \equiv 1(\bmod c)$ nor $b \equiv a+b+1(\bmod c)$ are possible by the given conditions. Therefore, $b \equiv a b+b(\bmod c)$. But this is also impossible because $\operatorname{gcd}(b, c)=1$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a, b, c$ be integers satisfying $0<a<c-1$ and $1<b<c$. For each $k$, $0 \leq k \leq a$, let $r_{k}, 0 \leq r_{k}<c$, be the remainder of $k b$ when divided by $c$. Prove that the two sets $\left\{r_{0}, r_{1}, r_{2}, \ldots, r_{a}\right\}$ and $\{0,1,2, \ldots, a\}$ are different.
|
Suppose that two sets are equal. Then $\operatorname{gcd}(b, c)=1$ and the polynomial
$$
f(x):=\left(1+x^{b}+x^{2 b}+\cdots+x^{a b}\right)-\left(1+x+x^{2}+\cdots+x^{a-1}+x^{a}\right)
$$
is divisible by $x^{c}-1$. (This is because: $m=n+c q \Longrightarrow x^{m}-x^{n}=x^{n+c q}-x^{n}=x^{n}\left(x^{c q}-1\right)$ and $\left(x^{c q}-1\right)=\left(x^{c}-1\right)\left(\left(x^{c}\right)^{q-1}+\left(x^{c}\right)^{q-2}+\cdots+1\right)$.) From
$$
f(x)=\frac{x^{(a+1) b}-1}{x^{b}-1}-\frac{x^{a+1}-1}{x-1}=\frac{F(x)}{(x-1)\left(x^{b}-1\right)}
$$
where $F(x)=x^{a b+b+1}+x^{b}+x^{a+1}-x^{a b+b}-x^{a+b+1}-x$, we have
$$
F(x) \equiv 0 \quad\left(\bmod x^{c}-1\right)
$$
Since $x^{c} \equiv 1\left(\bmod x^{c}-1\right)$, we may conclude that
$$
\{a b+b+1, b, a+1\} \equiv\{a b+b, a+b+1,1\} \quad(\bmod c)
$$
Thus,
$$
b \equiv a b+b, a+b+1 \text { or } 1(\bmod c)
$$
But neither $b \equiv 1(\bmod c)$ nor $b \equiv a+b+1(\bmod c)$ are possible by the given conditions. Therefore, $b \equiv a b+b(\bmod c)$. But this is also impossible because $\operatorname{gcd}(b, c)=1$.
|
{
"resource_path": "APMO/segmented/en-apmo2008_sol.jsonl",
"problem_match": "\nProblem 5.",
"solution_match": "(Solution)"
}
|
3830ebdd-4bb9-511f-b0cd-2a146406404f
| 260,687
|
Consider the following operation on positive real numbers written on a blackboard:
Choose a number $r$ written on the blackboard, erase that number, and then write a pair of positive real numbers $a$ and $b$ satisfying the condition $2 r^{2}=a b$ on the board.
Assume that you start out with just one positive real number $r$ on the blackboard, and apply this operation $k^{2}-1$ times to end up with $k^{2}$ positive real numbers, not necessarily distinct. Show that there exists a number on the board which does not exceed $k r$.
|
Using AM-GM inequality, we obtain
$$
\frac{1}{r^{2}}=\frac{2}{a b}=\frac{2 a b}{a^{2} b^{2}} \leq \frac{a^{2}+b^{2}}{a^{2} b^{2}} \leq \frac{1}{a^{2}}+\frac{1}{b^{2}}
$$
Consequently, if we let $S_{\ell}$ be the sum of the squares of the reciprocals of the numbers written on the board after $\ell$ operations, then $S_{\ell}$ increases as $\ell$ increases, that is,
$$
S_{0} \leq S_{1} \leq \cdots \leq S_{k^{2}-1}
$$
Therefore if we let $s$ be the smallest real number written on the board after $k^{2}-1$ operations, then $\frac{1}{s^{2}} \geq \frac{1}{t^{2}}$ for any number $t$ among $k^{2}$ numbers on the board and hence
$$
k^{2} \times \frac{1}{s^{2}} \geq S_{k^{2}-1} \geq S_{0}=\frac{1}{r^{2}}
$$
which implies that $s \leq k r$ as desired.
Remark. The nature of the problem does not change at all if the numbers on the board are restricted to be positive integers. But that may mislead some contestants to think the problem is a number theoretic problem rather than a combinatorial problem.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Consider the following operation on positive real numbers written on a blackboard:
Choose a number $r$ written on the blackboard, erase that number, and then write a pair of positive real numbers $a$ and $b$ satisfying the condition $2 r^{2}=a b$ on the board.
Assume that you start out with just one positive real number $r$ on the blackboard, and apply this operation $k^{2}-1$ times to end up with $k^{2}$ positive real numbers, not necessarily distinct. Show that there exists a number on the board which does not exceed $k r$.
|
Using AM-GM inequality, we obtain
$$
\frac{1}{r^{2}}=\frac{2}{a b}=\frac{2 a b}{a^{2} b^{2}} \leq \frac{a^{2}+b^{2}}{a^{2} b^{2}} \leq \frac{1}{a^{2}}+\frac{1}{b^{2}}
$$
Consequently, if we let $S_{\ell}$ be the sum of the squares of the reciprocals of the numbers written on the board after $\ell$ operations, then $S_{\ell}$ increases as $\ell$ increases, that is,
$$
S_{0} \leq S_{1} \leq \cdots \leq S_{k^{2}-1}
$$
Therefore if we let $s$ be the smallest real number written on the board after $k^{2}-1$ operations, then $\frac{1}{s^{2}} \geq \frac{1}{t^{2}}$ for any number $t$ among $k^{2}$ numbers on the board and hence
$$
k^{2} \times \frac{1}{s^{2}} \geq S_{k^{2}-1} \geq S_{0}=\frac{1}{r^{2}}
$$
which implies that $s \leq k r$ as desired.
Remark. The nature of the problem does not change at all if the numbers on the board are restricted to be positive integers. But that may mislead some contestants to think the problem is a number theoretic problem rather than a combinatorial problem.
|
{
"resource_path": "APMO/segmented/en-apmo2009_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "(Solution)"
}
|
c4b367ea-94e5-56c6-9195-665d13c451fa
| 260,905
|
Let three circles $\Gamma_{1}, \Gamma_{2}, \Gamma_{3}$, which are non-overlapping and mutually external, be given in the plane. For each point $P$ in the plane, outside the three circles, construct six points $A_{1}, B_{1}, A_{2}, B_{2}, A_{3}, B_{3}$ as follows: For each $i=1,2,3, A_{i}, B_{i}$ are distinct points on the circle $\Gamma_{i}$ such that the lines $P A_{i}$ and $P B_{i}$ are both tangents to $\Gamma_{i}$. Call the point $P$ exceptional if, from the construction, three lines $A_{1} B_{1}, A_{2} B_{2}, A_{3} B_{3}$ are concurrent. Show that every exceptional point of the plane, if exists, lies on the same circle.
|
Let $O_{i}$ be the center and $r_{i}$ the radius of circle $\Gamma_{i}$ for each $i=1,2,3$. Let $P$ be an exceptional point, and let the three corresponding lines $A_{1} B_{1}, A_{2} B_{2}, A_{3} B_{3}$ concur at $Q$. Construct the circle with diameter $P Q$. Call the circle $\Gamma$, its center $O$ and its radius $r$. We now claim that all exceptional points lie on $\Gamma$.
Let $P O_{1}$ intersect $A_{1} B_{1}$ in $X_{1}$. As $P O_{1} \perp A_{1} B_{1}$, we see that $X_{1}$ lies on $\Gamma$. As $P A_{1}$ is a tangent to $\Gamma_{1}$, triangle $P A_{1} O_{1}$ is right-angled and similar to triangle $A_{1} X_{1} O_{1}$. It follows that
$$
\frac{O_{1} X_{1}}{O_{1} A_{1}}=\frac{O_{1} A_{1}}{O_{1} P}, \quad \text { i.e., } \quad O_{1} X_{1} \cdot O_{1} P=O_{1} A_{1}^{2}=r_{1}^{2}
$$
On the other hand, $O_{1} X_{1} \cdot O_{1} P$ is also the power of $O_{1}$ with respect to $\Gamma$, so that
$$
r_{1}^{2}=O_{1} X_{1} \cdot O_{1} P=\left(O_{1} O-r\right)\left(O_{1} O+r\right)=O_{1} O^{2}-r^{2}
$$
and hence
$$
r^{2}=O O_{1}^{2}-r_{1}^{2}=\left(O O_{1}-r_{1}\right)\left(O O_{1}+r_{1}\right)
$$
Thus, $r^{2}$ is the power of $O$ with respect to $\Gamma_{1}$. By the same token, $r^{2}$ is also the power of $O$ with respect to $\Gamma_{2}$ and $\Gamma_{3}$. Hence $O$ must be the radical center of the three given circles. Since $r$, as the square root of the power of $O$ with respect to the three given circles, does not depend on $P$, it follows that all exceptional points lie on $\Gamma$.
Remark. In the event of the radical point being at infinity (and hence the three radical axes being parallel), there are no exceptional points in the plane, which is consistent with the statement of the problem.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let three circles $\Gamma_{1}, \Gamma_{2}, \Gamma_{3}$, which are non-overlapping and mutually external, be given in the plane. For each point $P$ in the plane, outside the three circles, construct six points $A_{1}, B_{1}, A_{2}, B_{2}, A_{3}, B_{3}$ as follows: For each $i=1,2,3, A_{i}, B_{i}$ are distinct points on the circle $\Gamma_{i}$ such that the lines $P A_{i}$ and $P B_{i}$ are both tangents to $\Gamma_{i}$. Call the point $P$ exceptional if, from the construction, three lines $A_{1} B_{1}, A_{2} B_{2}, A_{3} B_{3}$ are concurrent. Show that every exceptional point of the plane, if exists, lies on the same circle.
|
Let $O_{i}$ be the center and $r_{i}$ the radius of circle $\Gamma_{i}$ for each $i=1,2,3$. Let $P$ be an exceptional point, and let the three corresponding lines $A_{1} B_{1}, A_{2} B_{2}, A_{3} B_{3}$ concur at $Q$. Construct the circle with diameter $P Q$. Call the circle $\Gamma$, its center $O$ and its radius $r$. We now claim that all exceptional points lie on $\Gamma$.
Let $P O_{1}$ intersect $A_{1} B_{1}$ in $X_{1}$. As $P O_{1} \perp A_{1} B_{1}$, we see that $X_{1}$ lies on $\Gamma$. As $P A_{1}$ is a tangent to $\Gamma_{1}$, triangle $P A_{1} O_{1}$ is right-angled and similar to triangle $A_{1} X_{1} O_{1}$. It follows that
$$
\frac{O_{1} X_{1}}{O_{1} A_{1}}=\frac{O_{1} A_{1}}{O_{1} P}, \quad \text { i.e., } \quad O_{1} X_{1} \cdot O_{1} P=O_{1} A_{1}^{2}=r_{1}^{2}
$$
On the other hand, $O_{1} X_{1} \cdot O_{1} P$ is also the power of $O_{1}$ with respect to $\Gamma$, so that
$$
r_{1}^{2}=O_{1} X_{1} \cdot O_{1} P=\left(O_{1} O-r\right)\left(O_{1} O+r\right)=O_{1} O^{2}-r^{2}
$$
and hence
$$
r^{2}=O O_{1}^{2}-r_{1}^{2}=\left(O O_{1}-r_{1}\right)\left(O O_{1}+r_{1}\right)
$$
Thus, $r^{2}$ is the power of $O$ with respect to $\Gamma_{1}$. By the same token, $r^{2}$ is also the power of $O$ with respect to $\Gamma_{2}$ and $\Gamma_{3}$. Hence $O$ must be the radical center of the three given circles. Since $r$, as the square root of the power of $O$ with respect to the three given circles, does not depend on $P$, it follows that all exceptional points lie on $\Gamma$.
Remark. In the event of the radical point being at infinity (and hence the three radical axes being parallel), there are no exceptional points in the plane, which is consistent with the statement of the problem.
|
{
"resource_path": "APMO/segmented/en-apmo2009_sol.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "(Solution)"
}
|
fcc2aa8b-17ae-5861-a9ef-182d467cfc53
| 260,914
|
Prove that for any positive integer $k$, there exists an arithmetic sequence
$$
\frac{a_{1}}{b_{1}}, \quad \frac{a_{2}}{b_{2}}, \ldots, \quad \frac{a_{k}}{b_{k}}
$$
of rational numbers, where $a_{i}, b_{i}$ are relatively prime positive integers for each $i=1,2, \ldots, k$, such that the positive integers $a_{1}, b_{1}, a_{2}, b_{2}, \ldots, a_{k}, b_{k}$ are all distinct.
|
For $k=1$, there is nothing to prove. Henceforth assume $k \geq 2$.
Let $p_{1}, p_{2}, \ldots, p_{k}$ be $k$ distinct primes such that
$$
k<p_{k}<\cdots<p_{2}<p_{1}
$$
and let $N=p_{1} p_{2} \cdots p_{k}$. By Chinese Remainder Theorem, there exists a positive integer $x$ satisfying
$$
x \equiv-i \quad\left(\bmod p_{i}\right)
$$
for all $i=1,2, \ldots, k$ and $x>N^{2}$. Consider the following sequence:
$$
\frac{x+1}{N}, \frac{x+2}{N}, \quad, \ldots, \frac{x+k}{N}
$$
This sequence is obviously an arithmetic sequence of positive rational numbers of length $k$. For each $i=1,2, \ldots, k$, the numerator $x+i$ is divisible by $p_{i}$ but not by $p_{j}$ for $j \neq i$, for otherwise $p_{j}$ divides $|i-j|$, which is not possible because $p_{j}>k>|i-j|$. Let
$$
a_{i}:=\frac{x+i}{p_{i}}, \quad b_{i}:=\frac{N}{p_{i}} \quad \text { for all } i=1,2, \ldots, k
$$
Then
$$
\frac{x+i}{N}=\frac{a_{i}}{b_{i}}, \quad \operatorname{gcd}\left(a_{i}, b_{i}\right)=1 \quad \text { for all } i=1,2, \ldots, k
$$
and all $b_{i}$ 's are distinct from each other. Moreover, $x>N^{2}$ implies
$$
a_{i}=\frac{x+i}{p_{i}}>\frac{N^{2}}{p_{i}}>N>\frac{N}{p_{j}}=b_{j} \quad \text { for all } i, j=1,2, \ldots, k
$$
and hence all $a_{i}$ 's are distinct from $b_{i}$ 's. It only remains to show that all $a_{i}$ 's are distinct from each other. This follows from
$$
a_{j}=\frac{x+j}{p_{j}}>\frac{x+i}{p_{j}}>\frac{x+i}{p_{i}}=a_{i} \quad \text { for all } i<j
$$
by our choice of $p_{1}, p_{2}, \ldots, p_{k}$. Thus, the arithmetic sequence
$$
\frac{a_{1}}{b_{1}}, \quad \frac{a_{2}}{b_{2}}, \ldots, \quad \frac{a_{k}}{b_{k}}
$$
of positive rational numbers satisfies the conditions of the problem.
Remark. Here is a much easier solution :
For any positive integer $k \geq 2$, consider the sequence
$$
\frac{(k!)^{2}+1}{k!}, \frac{(k!)^{2}+2}{k!}, \ldots, \frac{(k!)^{2}+k}{k!}
$$
Note that $\operatorname{gcd}\left(k!,(k!)^{2}+i\right)=i$ for all $i=1,2, \ldots, k$. So, taking
$$
a_{i}:=\frac{(k!)^{2}+i}{i}, \quad b_{i}:=\frac{k!}{i} \quad \text { for all } i=1,2, \ldots, k
$$
we have $\operatorname{gcd}\left(a_{i}, b_{i}\right)=1$ and
$$
a_{i}=\frac{(k!)^{2}+i}{i}>a_{j}=\frac{(k!)^{2}+j}{j}>b_{i}=\frac{k!}{i}>b_{j}=\frac{k!}{j}
$$
for any $1 \leq i<j \leq k$. Therefore this sequence satisfies every condition given in the problem.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Prove that for any positive integer $k$, there exists an arithmetic sequence
$$
\frac{a_{1}}{b_{1}}, \quad \frac{a_{2}}{b_{2}}, \ldots, \quad \frac{a_{k}}{b_{k}}
$$
of rational numbers, where $a_{i}, b_{i}$ are relatively prime positive integers for each $i=1,2, \ldots, k$, such that the positive integers $a_{1}, b_{1}, a_{2}, b_{2}, \ldots, a_{k}, b_{k}$ are all distinct.
|
For $k=1$, there is nothing to prove. Henceforth assume $k \geq 2$.
Let $p_{1}, p_{2}, \ldots, p_{k}$ be $k$ distinct primes such that
$$
k<p_{k}<\cdots<p_{2}<p_{1}
$$
and let $N=p_{1} p_{2} \cdots p_{k}$. By Chinese Remainder Theorem, there exists a positive integer $x$ satisfying
$$
x \equiv-i \quad\left(\bmod p_{i}\right)
$$
for all $i=1,2, \ldots, k$ and $x>N^{2}$. Consider the following sequence:
$$
\frac{x+1}{N}, \frac{x+2}{N}, \quad, \ldots, \frac{x+k}{N}
$$
This sequence is obviously an arithmetic sequence of positive rational numbers of length $k$. For each $i=1,2, \ldots, k$, the numerator $x+i$ is divisible by $p_{i}$ but not by $p_{j}$ for $j \neq i$, for otherwise $p_{j}$ divides $|i-j|$, which is not possible because $p_{j}>k>|i-j|$. Let
$$
a_{i}:=\frac{x+i}{p_{i}}, \quad b_{i}:=\frac{N}{p_{i}} \quad \text { for all } i=1,2, \ldots, k
$$
Then
$$
\frac{x+i}{N}=\frac{a_{i}}{b_{i}}, \quad \operatorname{gcd}\left(a_{i}, b_{i}\right)=1 \quad \text { for all } i=1,2, \ldots, k
$$
and all $b_{i}$ 's are distinct from each other. Moreover, $x>N^{2}$ implies
$$
a_{i}=\frac{x+i}{p_{i}}>\frac{N^{2}}{p_{i}}>N>\frac{N}{p_{j}}=b_{j} \quad \text { for all } i, j=1,2, \ldots, k
$$
and hence all $a_{i}$ 's are distinct from $b_{i}$ 's. It only remains to show that all $a_{i}$ 's are distinct from each other. This follows from
$$
a_{j}=\frac{x+j}{p_{j}}>\frac{x+i}{p_{j}}>\frac{x+i}{p_{i}}=a_{i} \quad \text { for all } i<j
$$
by our choice of $p_{1}, p_{2}, \ldots, p_{k}$. Thus, the arithmetic sequence
$$
\frac{a_{1}}{b_{1}}, \quad \frac{a_{2}}{b_{2}}, \ldots, \quad \frac{a_{k}}{b_{k}}
$$
of positive rational numbers satisfies the conditions of the problem.
Remark. Here is a much easier solution :
For any positive integer $k \geq 2$, consider the sequence
$$
\frac{(k!)^{2}+1}{k!}, \frac{(k!)^{2}+2}{k!}, \ldots, \frac{(k!)^{2}+k}{k!}
$$
Note that $\operatorname{gcd}\left(k!,(k!)^{2}+i\right)=i$ for all $i=1,2, \ldots, k$. So, taking
$$
a_{i}:=\frac{(k!)^{2}+i}{i}, \quad b_{i}:=\frac{k!}{i} \quad \text { for all } i=1,2, \ldots, k
$$
we have $\operatorname{gcd}\left(a_{i}, b_{i}\right)=1$ and
$$
a_{i}=\frac{(k!)^{2}+i}{i}>a_{j}=\frac{(k!)^{2}+j}{j}>b_{i}=\frac{k!}{i}>b_{j}=\frac{k!}{j}
$$
for any $1 \leq i<j \leq k$. Therefore this sequence satisfies every condition given in the problem.
|
{
"resource_path": "APMO/segmented/en-apmo2009_sol.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "(Solution)"
}
|
65d2cd32-d74b-5aa1-9ed5-cdb332a0343a
| 605,340
|
Let $A B C$ be a triangle with $\angle B A C \neq 90^{\circ}$. Let $O$ be the circumcenter of the triangle $A B C$ and let $\Gamma$ be the circumcircle of the triangle $B O C$. Suppose that $\Gamma$ intersects the line segment $A B$ at $P$ different from $B$, and the line segment $A C$ at $Q$ different from $C$. Let $O N$ be a diameter of the circle $\Gamma$. Prove that the quadrilateral $A P N Q$ is a parallelogram.
|
From the assumption that the circle $\Gamma$ intersects both of the line segments $A B$ and $A C$, it follows that the 4 points $N, C, Q, O$ are located on $\Gamma$ in the order of $N, C, Q, O$ or in the order of $N, C, O, Q$. The following argument for the proof of the assertion of the problem is valid in either case. Since $\angle N Q C$ and $\angle N O C$ are subtended by the same arc $\widehat{N C}$ of $\Gamma$ at the points $Q$ and $O$, respectively, on $\Gamma$, we have $\angle N Q C=\angle N O C$. We also have $\angle B O C=2 \angle B A C$, since $\angle B O C$ and $\angle B A C$ are subtended by the same arc $\widehat{B C}$ of the circum-circle of the triangle $A B C$ at the center $O$ of the circle and at the point $A$ on the circle, respectively. From $O B=O C$ and the fact that $O N$ is a diameter of $\Gamma$, it follows that the triangles $O B N$ and $O C N$ are congruent, and therefore we obtain $2 \angle N O C=\angle B O C$. Consequently, we have $\angle N Q C=\frac{1}{2} \angle B O C=\angle B A C$, which shows that the 2 lines $A P, Q N$ are parallel.
In the same manner, we can show that the 2 lines $A Q, P N$ are also parallel. Thus, the quadrilateral $A P N Q$ is a parallelogram.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $\angle B A C \neq 90^{\circ}$. Let $O$ be the circumcenter of the triangle $A B C$ and let $\Gamma$ be the circumcircle of the triangle $B O C$. Suppose that $\Gamma$ intersects the line segment $A B$ at $P$ different from $B$, and the line segment $A C$ at $Q$ different from $C$. Let $O N$ be a diameter of the circle $\Gamma$. Prove that the quadrilateral $A P N Q$ is a parallelogram.
|
From the assumption that the circle $\Gamma$ intersects both of the line segments $A B$ and $A C$, it follows that the 4 points $N, C, Q, O$ are located on $\Gamma$ in the order of $N, C, Q, O$ or in the order of $N, C, O, Q$. The following argument for the proof of the assertion of the problem is valid in either case. Since $\angle N Q C$ and $\angle N O C$ are subtended by the same arc $\widehat{N C}$ of $\Gamma$ at the points $Q$ and $O$, respectively, on $\Gamma$, we have $\angle N Q C=\angle N O C$. We also have $\angle B O C=2 \angle B A C$, since $\angle B O C$ and $\angle B A C$ are subtended by the same arc $\widehat{B C}$ of the circum-circle of the triangle $A B C$ at the center $O$ of the circle and at the point $A$ on the circle, respectively. From $O B=O C$ and the fact that $O N$ is a diameter of $\Gamma$, it follows that the triangles $O B N$ and $O C N$ are congruent, and therefore we obtain $2 \angle N O C=\angle B O C$. Consequently, we have $\angle N Q C=\frac{1}{2} \angle B O C=\angle B A C$, which shows that the 2 lines $A P, Q N$ are parallel.
In the same manner, we can show that the 2 lines $A Q, P N$ are also parallel. Thus, the quadrilateral $A P N Q$ is a parallelogram.
|
{
"resource_path": "APMO/segmented/en-apmo2010_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution:"
}
|
28b834bc-001b-53aa-b86b-5010753075aa
| 260,815
|
For a positive integer $k$, call an integer a pure $k$-th power if it can be represented as $m^{k}$ for some integer $m$. Show that for every positive integer $n$ there exist $n$ distinct positive integers such that their sum is a pure 2009-th power, and their product is a pure 2010-th power.
|
For the sake of simplicity, let us set $k=2009$.
First of all, choose $n$ distinct positive integers $b_{1}, \cdots, b_{n}$ suitably so that their product is a pure $k+1$-th power (for example, let $b_{i}=i^{k+1}$ for $i=1, \cdots, n$ ). Then we have $b_{1} \cdots b_{n}=t^{k+1}$ for some positive integer $t$. Set $b_{1}+\cdots+b_{n}=s$.
Now we set $a_{i}=b_{i} s^{k^{2}-1}$ for $i=1, \cdots, n$, and show that $a_{1}, \cdots, a_{n}$ satisfy the required conditions. Since $b_{1}, \cdots, b_{n}$ are distinct positive integers, it is clear that so are $a_{1}, \cdots, a_{n}$. From
$$
\begin{aligned}
a_{1}+\cdots+a_{n} & =s^{k^{2}-1}\left(b_{1}+\cdots+b_{n}\right)=s^{k^{2}}=\left(s^{k}\right)^{2009} \\
a_{1} \cdots a_{n} & =\left(s^{k^{2}-1}\right)^{n} b_{1} \cdots b_{n}=\left(s^{k^{2}-1}\right)^{n} t^{k+1}=\left(s^{(k-1) n} t\right)^{2010}
\end{aligned}
$$
we can see that $a_{1}, \cdots, a_{n}$ satisfy the conditions on the sum and the product as well. This ends the proof of the assertion.
Remark: We can find the appropriate exponent $k^{2}-1$ needed for the construction of the $a_{i}$ 's by solving the simultaneous congruence relations: $x \equiv 0(\bmod k+1), x \equiv-1(\bmod k)$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
For a positive integer $k$, call an integer a pure $k$-th power if it can be represented as $m^{k}$ for some integer $m$. Show that for every positive integer $n$ there exist $n$ distinct positive integers such that their sum is a pure 2009-th power, and their product is a pure 2010-th power.
|
For the sake of simplicity, let us set $k=2009$.
First of all, choose $n$ distinct positive integers $b_{1}, \cdots, b_{n}$ suitably so that their product is a pure $k+1$-th power (for example, let $b_{i}=i^{k+1}$ for $i=1, \cdots, n$ ). Then we have $b_{1} \cdots b_{n}=t^{k+1}$ for some positive integer $t$. Set $b_{1}+\cdots+b_{n}=s$.
Now we set $a_{i}=b_{i} s^{k^{2}-1}$ for $i=1, \cdots, n$, and show that $a_{1}, \cdots, a_{n}$ satisfy the required conditions. Since $b_{1}, \cdots, b_{n}$ are distinct positive integers, it is clear that so are $a_{1}, \cdots, a_{n}$. From
$$
\begin{aligned}
a_{1}+\cdots+a_{n} & =s^{k^{2}-1}\left(b_{1}+\cdots+b_{n}\right)=s^{k^{2}}=\left(s^{k}\right)^{2009} \\
a_{1} \cdots a_{n} & =\left(s^{k^{2}-1}\right)^{n} b_{1} \cdots b_{n}=\left(s^{k^{2}-1}\right)^{n} t^{k+1}=\left(s^{(k-1) n} t\right)^{2010}
\end{aligned}
$$
we can see that $a_{1}, \cdots, a_{n}$ satisfy the conditions on the sum and the product as well. This ends the proof of the assertion.
Remark: We can find the appropriate exponent $k^{2}-1$ needed for the construction of the $a_{i}$ 's by solving the simultaneous congruence relations: $x \equiv 0(\bmod k+1), x \equiv-1(\bmod k)$.
|
{
"resource_path": "APMO/segmented/en-apmo2010_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "\nSolution:"
}
|
3720b38c-4edb-5765-88ad-6277b0bfb140
| 605,389
|
Let $A B C$ be an acute triangle satisfying the condition $A B>B C$ and $A C>B C$. Denote by $O$ and $H$ the circumcenter and the orthocenter, respectively, of the triangle $A B C$. Suppose that the circumcircle of the triangle $A H C$ intersects the line $A B$ at $M$ different from $A$, and that the circumcircle of the triangle $A H B$ intersects the line $A C$ at $N$ different from $A$. Prove that the circumcenter of the triangle $M N H$ lies on the line $O H$.
|
In the sequel, we denote $\angle B A C=\alpha, \angle C B A=\beta, \angle A C B=\gamma$. Let $O^{\prime}$ be the circumcenter of the triangle $M N H$. The lengths of line segments starting from the point $H$ will be treated as signed quantities.
Let us denote by $M^{\prime}, N^{\prime}$ the point of intersection of $C H, B H$, respectively, with the circumcircle of the triangle $A B C$ (distinct from $C, B$, respectively.) From the fact that 4 points $A, M, H, C$ lie on the same circle, we see that $\angle M H M^{\prime}=\alpha$ holds. Furthermore, $\angle B M^{\prime} C, \angle B N^{\prime} C$ and $\alpha$ are all subtended by the same arc $\widehat{B C}$ of the circumcircle of the triangle $A B C$ at points on the circle, and therefore, we have $\angle B M^{\prime} C=\alpha$, and $\angle B N^{\prime} C=\alpha$ as well. We also have $\angle A B H=\angle A C N^{\prime}$ as they are subtended by the same $\operatorname{arc} A N^{\prime}$ of the circumcircle of the triangle $A B C$ at points on the circle. Since $H M^{\prime} \perp B M, H N^{\prime} \perp A C$, we conclude that
$$
\angle M^{\prime} H B=90^{\circ}-\angle A B H=90^{\circ}-\angle A C N^{\prime}=\alpha
$$
is valid as well. Putting these facts together, we obtain the fact that the quadrilateral $H B M^{\prime} M$ is a rhombus. In a similar manner, we can conclude that the quadrilateral $H C N^{\prime} N$ is also a rhombus. Since both of these rhombuses are made up of 4 right triangles with an angle of magnitude $\alpha$, we also see that these rhombuses are similar.
Let us denote by $P, Q$ the feet of the perpendicular lines on $H M$ and $H N$, respectively, drawn from the point $O^{\prime}$. Since $O^{\prime}$ is the circumcenter of the triangle $M N H, P, Q$ are respectively, the midpoints of the line segments $H M, H N$. Furthermore, if we denote by $R, S$ the feet of the perpendicular lines on $H M$ and $H N$, respectively, drawn from the point $O$, then since $O$ is the circumcenter of both the triangle $M^{\prime} B C$ and the triangle $N^{\prime} B C$, we see that $R$ is the intersection point of $H M$ and the perpendicular bisector of $B M^{\prime}$, and $S$ is the intersection point of $H N$ and the perpendicular bisector of $C N^{\prime}$.
We note that the similarity map $\phi$ between the rhombuses $H B M^{\prime} M$ and $H C N^{\prime} N$ carries the perpendicular bisector of $B M^{\prime}$ onto the perpendicular bisector of $C N^{\prime}$, and straight line $H M$ onto the straight line $H N$, and hence $\phi$ maps $R$ onto $S$, and $P$ onto $Q$. Therefore, we get $H P: H R=H Q: H S$. If we now denote by $X, Y$ the intersection points of the line $H O^{\prime}$ with the line through $R$ and perpendicular to $H P$, and with the line through $S$ and perpendicular to $H Q$, respectively, then we get
$$
H O^{\prime}: H X=H P: H R=H Q: H S=H O^{\prime}: H Y
$$
so that we must have $H X=H Y$, and therefore, $X=Y$. But it is obvious that the point of intersection of the line through $R$ and perpendicular to $H P$ with the line through $S$ and perpendicular to $H Q$ must be $O$, and therefore, we conclude that $X=Y=O$ and that the points $H, O^{\prime}, O$ are collinear.
Alternate Solution: Deduction of the fact that both of the quadrilaterals $H B M^{\prime} M$ and $H C N^{\prime} N$ are rhombuses is carried out in the same way as in the preceding proof.
We then see that the point $M$ is located in a symmetric position with the point $B$ with respect to the line $C H$, we conclude that we have $\angle C M B=\beta$. Similarly, we have $\angle C N B=\gamma$. If we now put $x=\angle A H O^{\prime}$, then we get
$$
\angle O^{\prime}=\beta-\alpha-x, \angle M N H=90^{\circ}-\beta-\alpha+x
$$
from which it follows that
$$
\angle A N M=180^{\circ}-\angle M N H-\left(90^{\circ}-\alpha\right)=\beta-x
$$
Similarly, we get
$$
\angle N M A=\gamma+x
$$
Using the laws of sines, we then get
$$
\begin{aligned}
\frac{\sin (\gamma+x)}{\sin (\beta-x)} & =\frac{A N}{A M}=\frac{A C}{A M} \cdot \frac{A B}{A C} \cdot \frac{A N}{A B} \\
& =\frac{\sin \beta}{\sin (\beta-\alpha)} \cdot \frac{\sin \gamma}{\sin \beta} \cdot \frac{\sin (\gamma-\alpha)}{\sin \gamma}=\frac{\sin (\gamma-\alpha)}{\sin (\beta-\alpha)}
\end{aligned}
$$
On the other hand, if we let $y=\angle A H O$, we then get
$$
\angle O H B=180^{\circ}-\gamma-y, \angle C H O=180^{\circ}-\beta+y,
$$
and since
$$
\angle H B O=\gamma-\alpha, \angle O C H=\beta-\alpha,
$$
using the laws of sines and observing that $O B=O C$, we get
$$
\begin{aligned}
\frac{\sin (\gamma-\alpha)}{\sin (\beta-\alpha)}=\frac{\sin \angle H B O}{\sin \angle O C H} & =\frac{\sin \left(180^{\circ}-\gamma-y\right) \cdot \frac{O H}{O B}}{\sin \left(180^{\circ}-\beta+y\right) \cdot \frac{O H}{O C}} \\
& =\frac{\sin \left(180^{\circ}-\gamma-y\right)}{\sin \left(180^{\circ}-\beta+y\right)}=\frac{\sin (\gamma+y)}{\sin (\beta-y)}
\end{aligned}
$$
We then get $\sin (\gamma+x) \sin (\beta-y)=\sin (\beta-x) \sin (\gamma+y)$. Expanding both sides of the last identity by using the addition formula for the sine function and after factoring and using again the addition formula we obtain that $\sin (x-y) \sin (\beta+\gamma)=0$. This implies that $x-y$ must be an integral multiple of $180^{\circ}$, and hence we conclude that $H, O, O^{\prime}$ are collinear.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle satisfying the condition $A B>B C$ and $A C>B C$. Denote by $O$ and $H$ the circumcenter and the orthocenter, respectively, of the triangle $A B C$. Suppose that the circumcircle of the triangle $A H C$ intersects the line $A B$ at $M$ different from $A$, and that the circumcircle of the triangle $A H B$ intersects the line $A C$ at $N$ different from $A$. Prove that the circumcenter of the triangle $M N H$ lies on the line $O H$.
|
In the sequel, we denote $\angle B A C=\alpha, \angle C B A=\beta, \angle A C B=\gamma$. Let $O^{\prime}$ be the circumcenter of the triangle $M N H$. The lengths of line segments starting from the point $H$ will be treated as signed quantities.
Let us denote by $M^{\prime}, N^{\prime}$ the point of intersection of $C H, B H$, respectively, with the circumcircle of the triangle $A B C$ (distinct from $C, B$, respectively.) From the fact that 4 points $A, M, H, C$ lie on the same circle, we see that $\angle M H M^{\prime}=\alpha$ holds. Furthermore, $\angle B M^{\prime} C, \angle B N^{\prime} C$ and $\alpha$ are all subtended by the same arc $\widehat{B C}$ of the circumcircle of the triangle $A B C$ at points on the circle, and therefore, we have $\angle B M^{\prime} C=\alpha$, and $\angle B N^{\prime} C=\alpha$ as well. We also have $\angle A B H=\angle A C N^{\prime}$ as they are subtended by the same $\operatorname{arc} A N^{\prime}$ of the circumcircle of the triangle $A B C$ at points on the circle. Since $H M^{\prime} \perp B M, H N^{\prime} \perp A C$, we conclude that
$$
\angle M^{\prime} H B=90^{\circ}-\angle A B H=90^{\circ}-\angle A C N^{\prime}=\alpha
$$
is valid as well. Putting these facts together, we obtain the fact that the quadrilateral $H B M^{\prime} M$ is a rhombus. In a similar manner, we can conclude that the quadrilateral $H C N^{\prime} N$ is also a rhombus. Since both of these rhombuses are made up of 4 right triangles with an angle of magnitude $\alpha$, we also see that these rhombuses are similar.
Let us denote by $P, Q$ the feet of the perpendicular lines on $H M$ and $H N$, respectively, drawn from the point $O^{\prime}$. Since $O^{\prime}$ is the circumcenter of the triangle $M N H, P, Q$ are respectively, the midpoints of the line segments $H M, H N$. Furthermore, if we denote by $R, S$ the feet of the perpendicular lines on $H M$ and $H N$, respectively, drawn from the point $O$, then since $O$ is the circumcenter of both the triangle $M^{\prime} B C$ and the triangle $N^{\prime} B C$, we see that $R$ is the intersection point of $H M$ and the perpendicular bisector of $B M^{\prime}$, and $S$ is the intersection point of $H N$ and the perpendicular bisector of $C N^{\prime}$.
We note that the similarity map $\phi$ between the rhombuses $H B M^{\prime} M$ and $H C N^{\prime} N$ carries the perpendicular bisector of $B M^{\prime}$ onto the perpendicular bisector of $C N^{\prime}$, and straight line $H M$ onto the straight line $H N$, and hence $\phi$ maps $R$ onto $S$, and $P$ onto $Q$. Therefore, we get $H P: H R=H Q: H S$. If we now denote by $X, Y$ the intersection points of the line $H O^{\prime}$ with the line through $R$ and perpendicular to $H P$, and with the line through $S$ and perpendicular to $H Q$, respectively, then we get
$$
H O^{\prime}: H X=H P: H R=H Q: H S=H O^{\prime}: H Y
$$
so that we must have $H X=H Y$, and therefore, $X=Y$. But it is obvious that the point of intersection of the line through $R$ and perpendicular to $H P$ with the line through $S$ and perpendicular to $H Q$ must be $O$, and therefore, we conclude that $X=Y=O$ and that the points $H, O^{\prime}, O$ are collinear.
Alternate Solution: Deduction of the fact that both of the quadrilaterals $H B M^{\prime} M$ and $H C N^{\prime} N$ are rhombuses is carried out in the same way as in the preceding proof.
We then see that the point $M$ is located in a symmetric position with the point $B$ with respect to the line $C H$, we conclude that we have $\angle C M B=\beta$. Similarly, we have $\angle C N B=\gamma$. If we now put $x=\angle A H O^{\prime}$, then we get
$$
\angle O^{\prime}=\beta-\alpha-x, \angle M N H=90^{\circ}-\beta-\alpha+x
$$
from which it follows that
$$
\angle A N M=180^{\circ}-\angle M N H-\left(90^{\circ}-\alpha\right)=\beta-x
$$
Similarly, we get
$$
\angle N M A=\gamma+x
$$
Using the laws of sines, we then get
$$
\begin{aligned}
\frac{\sin (\gamma+x)}{\sin (\beta-x)} & =\frac{A N}{A M}=\frac{A C}{A M} \cdot \frac{A B}{A C} \cdot \frac{A N}{A B} \\
& =\frac{\sin \beta}{\sin (\beta-\alpha)} \cdot \frac{\sin \gamma}{\sin \beta} \cdot \frac{\sin (\gamma-\alpha)}{\sin \gamma}=\frac{\sin (\gamma-\alpha)}{\sin (\beta-\alpha)}
\end{aligned}
$$
On the other hand, if we let $y=\angle A H O$, we then get
$$
\angle O H B=180^{\circ}-\gamma-y, \angle C H O=180^{\circ}-\beta+y,
$$
and since
$$
\angle H B O=\gamma-\alpha, \angle O C H=\beta-\alpha,
$$
using the laws of sines and observing that $O B=O C$, we get
$$
\begin{aligned}
\frac{\sin (\gamma-\alpha)}{\sin (\beta-\alpha)}=\frac{\sin \angle H B O}{\sin \angle O C H} & =\frac{\sin \left(180^{\circ}-\gamma-y\right) \cdot \frac{O H}{O B}}{\sin \left(180^{\circ}-\beta+y\right) \cdot \frac{O H}{O C}} \\
& =\frac{\sin \left(180^{\circ}-\gamma-y\right)}{\sin \left(180^{\circ}-\beta+y\right)}=\frac{\sin (\gamma+y)}{\sin (\beta-y)}
\end{aligned}
$$
We then get $\sin (\gamma+x) \sin (\beta-y)=\sin (\beta-x) \sin (\gamma+y)$. Expanding both sides of the last identity by using the addition formula for the sine function and after factoring and using again the addition formula we obtain that $\sin (x-y) \sin (\beta+\gamma)=0$. This implies that $x-y$ must be an integral multiple of $180^{\circ}$, and hence we conclude that $H, O, O^{\prime}$ are collinear.
|
{
"resource_path": "APMO/segmented/en-apmo2010_sol.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "\nSolution:"
}
|
0a291d96-de84-5662-9c32-37cc21e9831c
| 605,417
|
Let $A B C$ be an acute triangle with altitudes $A D, B E$ and $C F$, and let $O$ be the center of its circumcircle. Show that the segments $O A, O F, O B, O D, O C, O E$ dissect the triangle $A B C$ into three pairs of triangles that have equal areas.
|
Let $M$ and $N$ be midpoints of sides $B C$ and $A C$, respectively. Notice that $\angle M O C=\frac{1}{2} \angle B O C=\angle E A B, \angle O M C=90^{\circ}=\angle A E B$, so triangles $O M C$ and $A E B$ are similar and we get $\frac{O M}{A E}=\frac{O C}{A B}$. For triangles $O N A$ and $B D A$ we also have $\frac{O N}{B D}=\frac{O A}{B A}$. Then $\frac{O M}{A E}=\frac{O N}{B D}$ or $B D \cdot O M=A E \cdot O N$.
Denote by $S(\Phi)$ the area of the figure $\Phi$. So, we see that $S(O B D)=\frac{1}{2} B D \cdot O M=$ $\frac{1}{2} A E \cdot O N=S(O A E)$. Analogously, $S(O C D)=S(O A F)$ and $S(O C E)=S(O B F)$.
Alternative solution. Let $R$ be the circumradius of triangle $A B C$, and as usual write $A, B, C$ for angles $\angle C A B, \angle A B C, \angle B C A$ respectively, and $a, b, c$ for sides $B C, C A, A B$ respectively. Then the area of triangle $O C D$ is
$$
S(O C D)=\frac{1}{2} \cdot O C \cdot C D \cdot \sin (\angle O C D)=\frac{1}{2} R \cdot C D \cdot \sin (\angle O C D)
$$
Now $C D=b \cos C$, and
$$
\angle O C D=\frac{180^{\circ}-2 A}{2}=90^{\circ}-A
$$
(since triangle $O B C$ is isosceles, and $\angle B O C=2 A$ ). So
$$
S(O C D)=\frac{1}{2} R b \cos C \sin \left(90^{\circ}-A\right)=\frac{1}{2} R b \cos C \cos A
$$
A similar calculation gives
$$
\begin{aligned}
S(O A F) & =\frac{1}{2} O A \cdot A F \cdot \sin (\angle O A F) \\
& =\frac{1}{2} R \cdot(b \cos A) \sin \left(90^{\circ}-C\right) \\
& =\frac{1}{2} R b \cos A \cos C
\end{aligned}
$$
so $O C D$ and $O A F$ have the same area. In the same way we find that $O B D$ and $O A E$ have the same area, as do $O C E$ and $O B F$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle with altitudes $A D, B E$ and $C F$, and let $O$ be the center of its circumcircle. Show that the segments $O A, O F, O B, O D, O C, O E$ dissect the triangle $A B C$ into three pairs of triangles that have equal areas.
|
Let $M$ and $N$ be midpoints of sides $B C$ and $A C$, respectively. Notice that $\angle M O C=\frac{1}{2} \angle B O C=\angle E A B, \angle O M C=90^{\circ}=\angle A E B$, so triangles $O M C$ and $A E B$ are similar and we get $\frac{O M}{A E}=\frac{O C}{A B}$. For triangles $O N A$ and $B D A$ we also have $\frac{O N}{B D}=\frac{O A}{B A}$. Then $\frac{O M}{A E}=\frac{O N}{B D}$ or $B D \cdot O M=A E \cdot O N$.
Denote by $S(\Phi)$ the area of the figure $\Phi$. So, we see that $S(O B D)=\frac{1}{2} B D \cdot O M=$ $\frac{1}{2} A E \cdot O N=S(O A E)$. Analogously, $S(O C D)=S(O A F)$ and $S(O C E)=S(O B F)$.
Alternative solution. Let $R$ be the circumradius of triangle $A B C$, and as usual write $A, B, C$ for angles $\angle C A B, \angle A B C, \angle B C A$ respectively, and $a, b, c$ for sides $B C, C A, A B$ respectively. Then the area of triangle $O C D$ is
$$
S(O C D)=\frac{1}{2} \cdot O C \cdot C D \cdot \sin (\angle O C D)=\frac{1}{2} R \cdot C D \cdot \sin (\angle O C D)
$$
Now $C D=b \cos C$, and
$$
\angle O C D=\frac{180^{\circ}-2 A}{2}=90^{\circ}-A
$$
(since triangle $O B C$ is isosceles, and $\angle B O C=2 A$ ). So
$$
S(O C D)=\frac{1}{2} R b \cos C \sin \left(90^{\circ}-A\right)=\frac{1}{2} R b \cos C \cos A
$$
A similar calculation gives
$$
\begin{aligned}
S(O A F) & =\frac{1}{2} O A \cdot A F \cdot \sin (\angle O A F) \\
& =\frac{1}{2} R \cdot(b \cos A) \sin \left(90^{\circ}-C\right) \\
& =\frac{1}{2} R b \cos A \cos C
\end{aligned}
$$
so $O C D$ and $O A F$ have the same area. In the same way we find that $O B D$ and $O A E$ have the same area, as do $O C E$ and $O B F$.
|
{
"resource_path": "APMO/segmented/en-apmo2013_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution."
}
|
bde58b6c-5e7d-500f-a39d-4e26409f0c00
| 261,160
|
For $2 k$ real numbers $a_{1}, a_{2}, \ldots, a_{k}, b_{1}, b_{2}, \ldots, b_{k}$ define the sequence of numbers $X_{n}$ by
$$
X_{n}=\sum_{i=1}^{k}\left[a_{i} n+b_{i}\right] \quad(n=1,2, \ldots)
$$
If the sequence $X_{n}$ forms an arithmetic progression, show that $\sum_{i=1}^{k} a_{i}$ must be an integer. Here $[r]$ denotes the greatest integer less than or equal to $r$.
|
Let us write $A=\sum_{i=1}^{k} a_{i}$ and $B=\sum_{i=1}^{k} b_{i}$. Summing the corresponding terms of the following inequalities over $i$,
$$
a_{i} n+b_{i}-1<\left[a_{i} n+b_{i}\right] \leq a_{i} n+b_{i}
$$
we obtain $A n+B-k<X_{n}<A n+B$. Now suppose that $\left\{X_{n}\right\}$ is an arithmetic progression with the common difference $d$, then we have $n d=X_{n+1}-X_{1}$ and $A+B-k<X_{1} \leq A+B$ Combining with the inequalities obtained above, we get
$$
A(n+1)+B-k<n d+X_{1}<A(n+1)+B,
$$
or
$$
A n-k \leq A n+\left(A+B-X_{1}\right)-k<n d<A n+\left(A+B-X_{1}\right)<A n+k,
$$
from which we conclude that $|A-d|<\frac{k}{n}$ must hold. Since this inequality holds for any positive integer $n$, we must have $A=d$. Since $\left\{X_{n}\right\}$ is a sequence of integers, $d$ must be an integer also, and thus we conclude that $A$ is also an integer.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
For $2 k$ real numbers $a_{1}, a_{2}, \ldots, a_{k}, b_{1}, b_{2}, \ldots, b_{k}$ define the sequence of numbers $X_{n}$ by
$$
X_{n}=\sum_{i=1}^{k}\left[a_{i} n+b_{i}\right] \quad(n=1,2, \ldots)
$$
If the sequence $X_{n}$ forms an arithmetic progression, show that $\sum_{i=1}^{k} a_{i}$ must be an integer. Here $[r]$ denotes the greatest integer less than or equal to $r$.
|
Let us write $A=\sum_{i=1}^{k} a_{i}$ and $B=\sum_{i=1}^{k} b_{i}$. Summing the corresponding terms of the following inequalities over $i$,
$$
a_{i} n+b_{i}-1<\left[a_{i} n+b_{i}\right] \leq a_{i} n+b_{i}
$$
we obtain $A n+B-k<X_{n}<A n+B$. Now suppose that $\left\{X_{n}\right\}$ is an arithmetic progression with the common difference $d$, then we have $n d=X_{n+1}-X_{1}$ and $A+B-k<X_{1} \leq A+B$ Combining with the inequalities obtained above, we get
$$
A(n+1)+B-k<n d+X_{1}<A(n+1)+B,
$$
or
$$
A n-k \leq A n+\left(A+B-X_{1}\right)-k<n d<A n+\left(A+B-X_{1}\right)<A n+k,
$$
from which we conclude that $|A-d|<\frac{k}{n}$ must hold. Since this inequality holds for any positive integer $n$, we must have $A=d$. Since $\left\{X_{n}\right\}$ is a sequence of integers, $d$ must be an integer also, and thus we conclude that $A$ is also an integer.
|
{
"resource_path": "APMO/segmented/en-apmo2013_sol.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "\nSolution."
}
|
47f9799b-15a3-5677-a00a-5c4e96c7ffcd
| 605,479
|
Let $A B C D$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $A C$ such that $P B$ and $P D$ are tangent to $\omega$. The tangent at $C$ intersects $P D$ at $Q$ and the line $A D$ at $R$. Let $E$ be the second point of intersection between $A Q$ and $\omega$. Prove that $B, E, R$ are collinear.
|
To show $B, E, R$ are collinear, it is equivalent to show the lines $A D, B E, C Q$ are concurrent. Let $C Q$ intersect $A D$ at $R$ and $B E$ intersect $A D$ at $R^{\prime}$. We shall show $R D / R A=R^{\prime} D / R^{\prime} A$ so that $R=R^{\prime}$.
Since $\triangle P A D$ is similar to $\triangle P D C$ and $\triangle P A B$ is similar to $\triangle P B C$, we have $A D / D C=$ $P A / P D=P A / P B=A B / B C$. Hence, $A B \cdot D C=B C \cdot A D$. By Ptolemy's theorem, $A B \cdot D C=B C \cdot A D=\frac{1}{2} C A \cdot D B$. Similarly $C A \cdot E D=C E \cdot A D=\frac{1}{2} A E \cdot D C$.
Thus
$$
\frac{D B}{A B}=\frac{2 D C}{C A}
$$
and
$$
\frac{D C}{C A}=\frac{2 E D}{A E}
$$
Since the triangles $R D C$ and $R C A$ are similar, we have $\frac{R D}{R C}=\frac{D C}{C A}=\frac{R C}{R A}$. Thus using (4)
$$
\frac{R D}{R A}=\frac{R D \cdot R A}{R A^{2}}=\left(\frac{R C}{R A}\right)^{2}=\left(\frac{D C}{C A}\right)^{2}=\left(\frac{2 E D}{A E}\right)^{2}
$$
Using the similar triangles $A B R^{\prime}$ and $E D R^{\prime}$, we have $R^{\prime} D / R^{\prime} B=E D / A B$. Using the similar triangles $D B R^{\prime}$ and $E A R^{\prime}$ we have $R^{\prime} A / R^{\prime} B=E A / D B$. Thus using (3) and (4),
$$
\frac{R^{\prime} D}{R^{\prime} A}=\frac{E D \cdot D B}{E A \cdot A B}=\left(\frac{2 E D}{A E}\right)^{2}
$$
It follows from (5) and (6) that $R=R^{\prime}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $A C$ such that $P B$ and $P D$ are tangent to $\omega$. The tangent at $C$ intersects $P D$ at $Q$ and the line $A D$ at $R$. Let $E$ be the second point of intersection between $A Q$ and $\omega$. Prove that $B, E, R$ are collinear.
|
To show $B, E, R$ are collinear, it is equivalent to show the lines $A D, B E, C Q$ are concurrent. Let $C Q$ intersect $A D$ at $R$ and $B E$ intersect $A D$ at $R^{\prime}$. We shall show $R D / R A=R^{\prime} D / R^{\prime} A$ so that $R=R^{\prime}$.
Since $\triangle P A D$ is similar to $\triangle P D C$ and $\triangle P A B$ is similar to $\triangle P B C$, we have $A D / D C=$ $P A / P D=P A / P B=A B / B C$. Hence, $A B \cdot D C=B C \cdot A D$. By Ptolemy's theorem, $A B \cdot D C=B C \cdot A D=\frac{1}{2} C A \cdot D B$. Similarly $C A \cdot E D=C E \cdot A D=\frac{1}{2} A E \cdot D C$.
Thus
$$
\frac{D B}{A B}=\frac{2 D C}{C A}
$$
and
$$
\frac{D C}{C A}=\frac{2 E D}{A E}
$$
Since the triangles $R D C$ and $R C A$ are similar, we have $\frac{R D}{R C}=\frac{D C}{C A}=\frac{R C}{R A}$. Thus using (4)
$$
\frac{R D}{R A}=\frac{R D \cdot R A}{R A^{2}}=\left(\frac{R C}{R A}\right)^{2}=\left(\frac{D C}{C A}\right)^{2}=\left(\frac{2 E D}{A E}\right)^{2}
$$
Using the similar triangles $A B R^{\prime}$ and $E D R^{\prime}$, we have $R^{\prime} D / R^{\prime} B=E D / A B$. Using the similar triangles $D B R^{\prime}$ and $E A R^{\prime}$ we have $R^{\prime} A / R^{\prime} B=E A / D B$. Thus using (3) and (4),
$$
\frac{R^{\prime} D}{R^{\prime} A}=\frac{E D \cdot D B}{E A \cdot A B}=\left(\frac{2 E D}{A E}\right)^{2}
$$
It follows from (5) and (6) that $R=R^{\prime}$.
|
{
"resource_path": "APMO/segmented/en-apmo2013_sol.jsonl",
"problem_match": "\nProblem 5.",
"solution_match": "\nSolution."
}
|
fbec6dda-212e-5638-8659-c762b5b6e97f
| 261,186
|
For a positive integer $m$ denote by $S(m)$ and $P(m)$ the sum and product, respectively, of the digits of $m$. Show that for each positive integer $n$, there exist positive integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying the following conditions:
$$
S\left(a_{1}\right)<S\left(a_{2}\right)<\cdots<S\left(a_{n}\right) \text { and } S\left(a_{i}\right)=P\left(a_{i+1}\right) \quad(i=1,2, \ldots, n)
$$
(We let $\left.a_{n+1}=a_{1}.\right)$ (Problem Committee of the Japan Mathematical Olympiad Foundation)
|
Let $k$ be a sufficiently large positive integer. Choose for each $i=2,3, \ldots, n$, $a_{i}$ to be a positive integer among whose digits the number 2 appears exactly $k+i-2$ times and the number 1 appears exactly $2^{k+i-1}-2(k+i-2)$ times, and nothing else. Then, we have $S\left(a_{i}\right)=2^{k+i-1}$ and $P\left(a_{i}\right)=2^{k+i-2}$ for each $i, 2 \leq i \leq n$. Then, we let $a_{1}$ be a positive integer among whose digits the number 2 appears exactly $k+n-1$ times and the number 1 appears exactly $2^{k}-2(k+n-1)$ times, and nothing else. Then, we see that $a_{1}$ satisfies $S\left(a_{1}\right)=2^{k}$ and $P\left(a_{1}\right)=2^{k+n-1}$. Such a choice of $a_{1}$ is possible if we take $k$ to be large enough to satisfy $2^{k}>2(k+n-1)$ and we see that the numbers $a_{1}, \ldots, a_{n}$ chosen this way satisfy the given requirements.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
For a positive integer $m$ denote by $S(m)$ and $P(m)$ the sum and product, respectively, of the digits of $m$. Show that for each positive integer $n$, there exist positive integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying the following conditions:
$$
S\left(a_{1}\right)<S\left(a_{2}\right)<\cdots<S\left(a_{n}\right) \text { and } S\left(a_{i}\right)=P\left(a_{i+1}\right) \quad(i=1,2, \ldots, n)
$$
(We let $\left.a_{n+1}=a_{1}.\right)$ (Problem Committee of the Japan Mathematical Olympiad Foundation)
|
Let $k$ be a sufficiently large positive integer. Choose for each $i=2,3, \ldots, n$, $a_{i}$ to be a positive integer among whose digits the number 2 appears exactly $k+i-2$ times and the number 1 appears exactly $2^{k+i-1}-2(k+i-2)$ times, and nothing else. Then, we have $S\left(a_{i}\right)=2^{k+i-1}$ and $P\left(a_{i}\right)=2^{k+i-2}$ for each $i, 2 \leq i \leq n$. Then, we let $a_{1}$ be a positive integer among whose digits the number 2 appears exactly $k+n-1$ times and the number 1 appears exactly $2^{k}-2(k+n-1)$ times, and nothing else. Then, we see that $a_{1}$ satisfies $S\left(a_{1}\right)=2^{k}$ and $P\left(a_{1}\right)=2^{k+n-1}$. Such a choice of $a_{1}$ is possible if we take $k$ to be large enough to satisfy $2^{k}>2(k+n-1)$ and we see that the numbers $a_{1}, \ldots, a_{n}$ chosen this way satisfy the given requirements.
|
{
"resource_path": "APMO/segmented/en-apmo2014_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution."
}
|
926b6029-7a9d-56f1-9d99-850851b2b42f
| 605,529
|
Let $n$ and $b$ be positive integers. We say $n$ is $b$-discerning if there exists a set consisting of $n$ different positive integers less than $b$ that has no two different subsets $U$ and $V$ such that the sum of all elements in $U$ equals the sum of all elements in $V$.
(a) Prove that 8 is a 100 -discerning.
(b) Prove that 9 is not 100-discerning.
(Senior Problems Committee of the Australian Mathematical Olympiad Committee)
|
(a) Take $S=\{3,6,12,24,48,95,96,97\}$, i.e.
$$
S=\left\{3 \cdot 2^{k}: 0 \leq k \leq 5\right\} \cup\left\{3 \cdot 2^{5}-1,3 \cdot 2^{5}+1\right\}
$$
As $k$ ranges between 0 to 5 , the sums obtained from the numbers $3 \cdot 2^{k}$ are $3 t$, where $1 \leq t \leq 63$. These are 63 numbers that are divisible by 3 and are at most $3 \cdot 63=189$.
Sums of elements of $S$ are also the numbers $95+97=192$ and all the numbers that are sums of 192 and sums obtained from the numbers $3 \cdot 2^{k}$ with $0 \leq k \leq 5$. These are 64 numbers that are all divisible by 3 and at least equal to 192. In addition, sums of elements of $S$ are the numbers 95 and all the numbers that are sums of 95 and sums obtained from the numbers $3 \cdot 2^{k}$ with $0 \leq k \leq 5$. These are 64 numbers that are all congruent to $-1 \bmod$ 3.
Finally, sums of elements of $S$ are the numbers 97 and all the numbers that are sums of 97 and sums obtained from the numbers $3 \cdot 2^{k}$ with $0 \leq k \leq 5$. These are 64 numbers that are all congruent to $1 \bmod 3$.
Hence there are at least $63+64+64+64=255$ different sums from elements of $S$. On the other hand, $S$ has $2^{8}-1=255$ non-empty subsets. Therefore $S$ has no two different subsets with equal sums of elements. Therefore, 8 is 100 -discerning.
(b) Suppose that 9 is 100 -discerning. Then there is a set $S=\left\{s_{1}, \ldots, s_{9}\right\}, s_{i}<100$ that has no two different subsets with equal sums of elements. Assume that $0<s_{1}<\cdots<s_{9}<$ 100.
Let $X$ be the set of all subsets of $S$ having at least 3 and at most 6 elements and let $Y$ be the set of all subsets of $S$ having exactly 2 or 3 or 4 elements greater than $s_{3}$.
The set $X$ consists of
$$
\binom{9}{3}+\binom{9}{4}+\binom{9}{5}+\binom{9}{6}=84+126+126+84=420
$$
subsets of $S$. The set in $X$ with the largest sums of elements is $\left\{s_{4}, \ldots, s_{9}\right\}$ and the smallest sums is in $\left\{s_{1}, s_{2}, s_{3}\right\}$. Thus the sum of the elements of each of the 420 sets in $X$ is at least $s_{1}+s_{2}+s_{3}$ and at most $s_{4}+\cdots+s_{9}$, which is one of $\left(s_{4}+\cdots+s_{9}\right)-\left(s_{1}+s_{2}+s_{3}\right)+1$ integers. From the pigeonhole principle it follows that $\left(s_{4}+\cdots+s_{9}\right)-\left(s_{1}+s_{2}+s_{3}\right)+1 \geq 420$, i.e.,
$$
\left(s_{4}+\cdots+s_{9}\right)-\left(s_{1}+s_{2}+s_{3}\right) \geq 419
$$
Now let us calculate the number of subsets in $Y$. Observe that $\left\{s_{4}, \ldots, s_{9}\right\}$ has $\binom{6}{2}$ 2-element subsets, $\binom{6}{3}$ 3-element subsets and $\binom{6}{4}$ 4-element subsets, while $\left\{s_{1}, s_{2}, s_{3}\right\}$ has exactly 8 subsets. Hence the number of subsets of $S$ in $Y$ equals
$$
8\left(\binom{6}{2}+\binom{6}{3}+\binom{6}{4}\right)=8(15+20+15)=400
$$
The set in $Y$ with the largest sum of elements is $\left\{s_{1}, s_{2}, s_{3}, s_{6}, s_{7}, s_{8}, s_{9}\right\}$ and the smallest sum is in $\left\{s_{4}, s_{5}\right\}$. Again, by the pigeonhole principle it follows that $\left(s_{1}+s_{2}+s_{3}+s_{6}+s_{7}+\right.$ $\left.s_{8}+s_{9}\right)-\left(s_{4}+s_{5}\right)+1 \geq 400$, i.e.,
$$
\left(s_{1}+s_{2}+s_{3}+s_{6}+s_{7}+s_{8}+s_{9}\right)-\left(s_{4}+s_{5}\right) \geq 399
$$
Adding (1) and (2) yields $2\left(s_{6}+s_{7}+s_{8}+s_{9}\right) \geq 818$, so that $s_{9}+98+97+96 \geq$ $s_{9}+s_{8}+s_{7}+s_{6} \geq 409$, i.e. $s_{9} \geq 118$, a contradiction with $s_{9}<100$. Therefore, 9 is not 100-discerning.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n$ and $b$ be positive integers. We say $n$ is $b$-discerning if there exists a set consisting of $n$ different positive integers less than $b$ that has no two different subsets $U$ and $V$ such that the sum of all elements in $U$ equals the sum of all elements in $V$.
(a) Prove that 8 is a 100 -discerning.
(b) Prove that 9 is not 100-discerning.
(Senior Problems Committee of the Australian Mathematical Olympiad Committee)
|
(a) Take $S=\{3,6,12,24,48,95,96,97\}$, i.e.
$$
S=\left\{3 \cdot 2^{k}: 0 \leq k \leq 5\right\} \cup\left\{3 \cdot 2^{5}-1,3 \cdot 2^{5}+1\right\}
$$
As $k$ ranges between 0 to 5 , the sums obtained from the numbers $3 \cdot 2^{k}$ are $3 t$, where $1 \leq t \leq 63$. These are 63 numbers that are divisible by 3 and are at most $3 \cdot 63=189$.
Sums of elements of $S$ are also the numbers $95+97=192$ and all the numbers that are sums of 192 and sums obtained from the numbers $3 \cdot 2^{k}$ with $0 \leq k \leq 5$. These are 64 numbers that are all divisible by 3 and at least equal to 192. In addition, sums of elements of $S$ are the numbers 95 and all the numbers that are sums of 95 and sums obtained from the numbers $3 \cdot 2^{k}$ with $0 \leq k \leq 5$. These are 64 numbers that are all congruent to $-1 \bmod$ 3.
Finally, sums of elements of $S$ are the numbers 97 and all the numbers that are sums of 97 and sums obtained from the numbers $3 \cdot 2^{k}$ with $0 \leq k \leq 5$. These are 64 numbers that are all congruent to $1 \bmod 3$.
Hence there are at least $63+64+64+64=255$ different sums from elements of $S$. On the other hand, $S$ has $2^{8}-1=255$ non-empty subsets. Therefore $S$ has no two different subsets with equal sums of elements. Therefore, 8 is 100 -discerning.
(b) Suppose that 9 is 100 -discerning. Then there is a set $S=\left\{s_{1}, \ldots, s_{9}\right\}, s_{i}<100$ that has no two different subsets with equal sums of elements. Assume that $0<s_{1}<\cdots<s_{9}<$ 100.
Let $X$ be the set of all subsets of $S$ having at least 3 and at most 6 elements and let $Y$ be the set of all subsets of $S$ having exactly 2 or 3 or 4 elements greater than $s_{3}$.
The set $X$ consists of
$$
\binom{9}{3}+\binom{9}{4}+\binom{9}{5}+\binom{9}{6}=84+126+126+84=420
$$
subsets of $S$. The set in $X$ with the largest sums of elements is $\left\{s_{4}, \ldots, s_{9}\right\}$ and the smallest sums is in $\left\{s_{1}, s_{2}, s_{3}\right\}$. Thus the sum of the elements of each of the 420 sets in $X$ is at least $s_{1}+s_{2}+s_{3}$ and at most $s_{4}+\cdots+s_{9}$, which is one of $\left(s_{4}+\cdots+s_{9}\right)-\left(s_{1}+s_{2}+s_{3}\right)+1$ integers. From the pigeonhole principle it follows that $\left(s_{4}+\cdots+s_{9}\right)-\left(s_{1}+s_{2}+s_{3}\right)+1 \geq 420$, i.e.,
$$
\left(s_{4}+\cdots+s_{9}\right)-\left(s_{1}+s_{2}+s_{3}\right) \geq 419
$$
Now let us calculate the number of subsets in $Y$. Observe that $\left\{s_{4}, \ldots, s_{9}\right\}$ has $\binom{6}{2}$ 2-element subsets, $\binom{6}{3}$ 3-element subsets and $\binom{6}{4}$ 4-element subsets, while $\left\{s_{1}, s_{2}, s_{3}\right\}$ has exactly 8 subsets. Hence the number of subsets of $S$ in $Y$ equals
$$
8\left(\binom{6}{2}+\binom{6}{3}+\binom{6}{4}\right)=8(15+20+15)=400
$$
The set in $Y$ with the largest sum of elements is $\left\{s_{1}, s_{2}, s_{3}, s_{6}, s_{7}, s_{8}, s_{9}\right\}$ and the smallest sum is in $\left\{s_{4}, s_{5}\right\}$. Again, by the pigeonhole principle it follows that $\left(s_{1}+s_{2}+s_{3}+s_{6}+s_{7}+\right.$ $\left.s_{8}+s_{9}\right)-\left(s_{4}+s_{5}\right)+1 \geq 400$, i.e.,
$$
\left(s_{1}+s_{2}+s_{3}+s_{6}+s_{7}+s_{8}+s_{9}\right)-\left(s_{4}+s_{5}\right) \geq 399
$$
Adding (1) and (2) yields $2\left(s_{6}+s_{7}+s_{8}+s_{9}\right) \geq 818$, so that $s_{9}+98+97+96 \geq$ $s_{9}+s_{8}+s_{7}+s_{6} \geq 409$, i.e. $s_{9} \geq 118$, a contradiction with $s_{9}<100$. Therefore, 9 is not 100-discerning.
|
{
"resource_path": "APMO/segmented/en-apmo2014_sol.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "# Solution."
}
|
d80210e6-a097-5993-aa27-92adb8bca0bf
| 605,580
|
Circles $\omega$ and $\Omega$ meet at points $A$ and $B$. Let $M$ be the midpoint of the $\operatorname{arc} A B$ of circle $\omega$ ( $M$ lies inside $\Omega$ ). A chord $M P$ of circle $\omega$ intersects $\Omega$ at $Q(Q$ lies inside $\omega)$. Let $\ell_{P}$ be the tangent line to $\omega$ at $P$, and let $\ell_{Q}$ be the tangent line to $\Omega$ at $Q$. Prove that the circumcircle of the triangle formed by the lines $\ell_{P}, \ell_{Q}$, and $A B$ is tangent to $\Omega$. (Ilya Bogdanov, Russia and Medeubek Kungozhin, Kazakhstan)
|
Denote $X=A B \cap \ell_{P}, Y=A B \cap \ell_{Q}$, and $Z=\ell_{P} \cap \ell_{Q}$. Without loss of generality we have $A X<B X$. Let $F=M P \cap A B$.
Denote by $R$ the second point of intersection of $P Q$ and $\Omega$; by $S$ the point of $\Omega$ such that $S R \| A B$; and by $T$ the point of $\Omega$ such that $R T \| \ell_{P}$. Since $M$ is the midpoint of arc $A B$, the tangent $\ell_{M}$ at $M$ to $\omega$ is parallel to $A B$, so $\angle(A B, P M)=\angle\left(P M, \ell_{P}\right)$. Therefore we have $\angle P R T=\angle M P X=\angle P F X=\angle P R S$. Thus the point $Q$ is the midpoint of the $\operatorname{arc} T Q S$ of $\Omega$, hence $S T \| \ell_{Q}$. So the corresponding sides of the triangles $R S T$ and $X Y Z$ are parallel, and there exist a homothety $h$ mapping $R S T$ to $X Y Z$.
Let $D$ be the second point of intersection of $X R$ and $\Omega$. We claim that $D$ is the center of the homothety $h$; since $D \in \Omega$, this implies that the circumcircles of triangles $R S T$ and $X Y Z$ are tangent, as required. So, it remains to prove this claim. In order to do this, it suffices to show that $D \in S Y$.
By $\angle P F X=\angle X P F$ we have $X F^{2}=X P^{2}=X A \cdot X B=X D \cdot X R$. Therefore, $\frac{X F}{X D}=\frac{X R}{X F}$, so the triangles $X D F$ and $X F R$ are similar, hence $\angle D F X=\angle X R F=\angle D R Q=$ $\angle D Q Y$; thus the points $D, Y, Q$, and $F$ are concyclic. It follows that $\angle Y D Q=\angle Y F Q=$ $\angle S R Q=180^{\circ}-\angle S D Q$ which means exactly that the points $Y, D$, and $S$ are collinear, with $D$ between $S$ and $Y$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Circles $\omega$ and $\Omega$ meet at points $A$ and $B$. Let $M$ be the midpoint of the $\operatorname{arc} A B$ of circle $\omega$ ( $M$ lies inside $\Omega$ ). A chord $M P$ of circle $\omega$ intersects $\Omega$ at $Q(Q$ lies inside $\omega)$. Let $\ell_{P}$ be the tangent line to $\omega$ at $P$, and let $\ell_{Q}$ be the tangent line to $\Omega$ at $Q$. Prove that the circumcircle of the triangle formed by the lines $\ell_{P}, \ell_{Q}$, and $A B$ is tangent to $\Omega$. (Ilya Bogdanov, Russia and Medeubek Kungozhin, Kazakhstan)
|
Denote $X=A B \cap \ell_{P}, Y=A B \cap \ell_{Q}$, and $Z=\ell_{P} \cap \ell_{Q}$. Without loss of generality we have $A X<B X$. Let $F=M P \cap A B$.
Denote by $R$ the second point of intersection of $P Q$ and $\Omega$; by $S$ the point of $\Omega$ such that $S R \| A B$; and by $T$ the point of $\Omega$ such that $R T \| \ell_{P}$. Since $M$ is the midpoint of arc $A B$, the tangent $\ell_{M}$ at $M$ to $\omega$ is parallel to $A B$, so $\angle(A B, P M)=\angle\left(P M, \ell_{P}\right)$. Therefore we have $\angle P R T=\angle M P X=\angle P F X=\angle P R S$. Thus the point $Q$ is the midpoint of the $\operatorname{arc} T Q S$ of $\Omega$, hence $S T \| \ell_{Q}$. So the corresponding sides of the triangles $R S T$ and $X Y Z$ are parallel, and there exist a homothety $h$ mapping $R S T$ to $X Y Z$.
Let $D$ be the second point of intersection of $X R$ and $\Omega$. We claim that $D$ is the center of the homothety $h$; since $D \in \Omega$, this implies that the circumcircles of triangles $R S T$ and $X Y Z$ are tangent, as required. So, it remains to prove this claim. In order to do this, it suffices to show that $D \in S Y$.
By $\angle P F X=\angle X P F$ we have $X F^{2}=X P^{2}=X A \cdot X B=X D \cdot X R$. Therefore, $\frac{X F}{X D}=\frac{X R}{X F}$, so the triangles $X D F$ and $X F R$ are similar, hence $\angle D F X=\angle X R F=\angle D R Q=$ $\angle D Q Y$; thus the points $D, Y, Q$, and $F$ are concyclic. It follows that $\angle Y D Q=\angle Y F Q=$ $\angle S R Q=180^{\circ}-\angle S D Q$ which means exactly that the points $Y, D$, and $S$ are collinear, with $D$ between $S$ and $Y$.
|
{
"resource_path": "APMO/segmented/en-apmo2014_sol.jsonl",
"problem_match": "\nProblem 5.",
"solution_match": "\nSolution."
}
|
4cd3c0ca-6b01-5351-8e48-0a03b74760c3
| 605,596
|
Let $A B C$ be a triangle, and let $D$ be a point on side $B C$. A line through $D$ intersects side $A B$ at $X$ and ray $A C$ at $Y$. The circumcircle of triangle $B X D$ intersects the circumcircle $\omega$ of triangle $A B C$ again at point $Z \neq B$. The lines $Z D$ and $Z Y$ intersect $\omega$ again at $V$ and $W$, respectively. Prove that $A B=V W$.
|
Suppose $X Y$ intersects $\omega$ at points $P$ and $Q$, where $Q$ lies between $X$ and $Y$. We will show that $V$ and $W$ are the reflections of $A$ and $B$ with respect to the perpendicular bisector of $P Q$. From this, it follows that $A V W B$ is an isosceles trapezoid and hence $A B=V W$.
First, note that
$$
\angle B Z D=\angle A X Y=\angle A P Q+\angle B A P=\angle A P Q+\angle B Z P,
$$
so $\angle A P Q=\angle P Z V=\angle P Q V$, and hence $V$ is the reflection of $A$ with respect to the perpendicular bisector of $P Q$.
Now, suppose $W^{\prime}$ is the reflection of $B$ with respect to the perpendicular bisector of $P Q$, and let $Z^{\prime}$ be the intersection of $Y W^{\prime}$ and $\omega$. It suffices to show that $B, X, D, Z^{\prime}$ are concyclic. Note that
$$
\angle Y D C=\angle P D B=\angle P C B+\angle Q P C=\angle W^{\prime} P Q+\angle Q P C=\angle W^{\prime} P C=\angle Y Z^{\prime} C .
$$
So $D, C, Y, Z^{\prime}$ are concyclic. Next, $\angle B Z^{\prime} D=\angle C Z^{\prime} B-\angle C Z^{\prime} D=180^{\circ}-\angle B X D$ and due to the previous concyclicity we are done.
Alternative solution 1. Using cyclic quadrilaterals $B X D Z$ and $A B Z V$ in turn, we have $\angle Z D Y=\angle Z B A=\angle Z C Y$. So $Z D C Y$ is cyclic.
Using cyclic quadrilaterals $A B Z C$ and $Z D C Y$ in turn, we have $\angle A Z B=\angle A C B=\angle W Z V$ (or $180^{\circ}-\angle W Z V$ if $Z$ lies between $W$ and $C$ ).
So $A B=V W$ because they subtend equal (or supplementary) angles in $\omega$.
Alternative solution 2. Using cyclic quadrilaterals $B X D Z$ and $A B Z V$ in turn, we have $\angle Z D Y=\angle Z B A=\angle Z C Y$. So $Z D C Y$ is cyclic.
Using cyclic quadrilaterals $B X D Z$ and $A B Z V$ in turn, we have $\angle D X A=\angle V Z B=$ $180^{\circ}-B A V$. So $X D \| A V$.
Using cyclic quadrilaterals $Z D C Y$ and $B C W Z$ in turn, we have $\angle Y D C=\angle Y Z C=$ $\angle W B C$. So $X D \| B W$.
Hence $B W \| A V$ which implies that $A V W B$ is an isosceles trapezium with $A B=V W$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle, and let $D$ be a point on side $B C$. A line through $D$ intersects side $A B$ at $X$ and ray $A C$ at $Y$. The circumcircle of triangle $B X D$ intersects the circumcircle $\omega$ of triangle $A B C$ again at point $Z \neq B$. The lines $Z D$ and $Z Y$ intersect $\omega$ again at $V$ and $W$, respectively. Prove that $A B=V W$.
|
Suppose $X Y$ intersects $\omega$ at points $P$ and $Q$, where $Q$ lies between $X$ and $Y$. We will show that $V$ and $W$ are the reflections of $A$ and $B$ with respect to the perpendicular bisector of $P Q$. From this, it follows that $A V W B$ is an isosceles trapezoid and hence $A B=V W$.
First, note that
$$
\angle B Z D=\angle A X Y=\angle A P Q+\angle B A P=\angle A P Q+\angle B Z P,
$$
so $\angle A P Q=\angle P Z V=\angle P Q V$, and hence $V$ is the reflection of $A$ with respect to the perpendicular bisector of $P Q$.
Now, suppose $W^{\prime}$ is the reflection of $B$ with respect to the perpendicular bisector of $P Q$, and let $Z^{\prime}$ be the intersection of $Y W^{\prime}$ and $\omega$. It suffices to show that $B, X, D, Z^{\prime}$ are concyclic. Note that
$$
\angle Y D C=\angle P D B=\angle P C B+\angle Q P C=\angle W^{\prime} P Q+\angle Q P C=\angle W^{\prime} P C=\angle Y Z^{\prime} C .
$$
So $D, C, Y, Z^{\prime}$ are concyclic. Next, $\angle B Z^{\prime} D=\angle C Z^{\prime} B-\angle C Z^{\prime} D=180^{\circ}-\angle B X D$ and due to the previous concyclicity we are done.
Alternative solution 1. Using cyclic quadrilaterals $B X D Z$ and $A B Z V$ in turn, we have $\angle Z D Y=\angle Z B A=\angle Z C Y$. So $Z D C Y$ is cyclic.
Using cyclic quadrilaterals $A B Z C$ and $Z D C Y$ in turn, we have $\angle A Z B=\angle A C B=\angle W Z V$ (or $180^{\circ}-\angle W Z V$ if $Z$ lies between $W$ and $C$ ).
So $A B=V W$ because they subtend equal (or supplementary) angles in $\omega$.
Alternative solution 2. Using cyclic quadrilaterals $B X D Z$ and $A B Z V$ in turn, we have $\angle Z D Y=\angle Z B A=\angle Z C Y$. So $Z D C Y$ is cyclic.
Using cyclic quadrilaterals $B X D Z$ and $A B Z V$ in turn, we have $\angle D X A=\angle V Z B=$ $180^{\circ}-B A V$. So $X D \| A V$.
Using cyclic quadrilaterals $Z D C Y$ and $B C W Z$ in turn, we have $\angle Y D C=\angle Y Z C=$ $\angle W B C$. So $X D \| B W$.
Hence $B W \| A V$ which implies that $A V W B$ is an isosceles trapezium with $A B=V W$.
|
{
"resource_path": "APMO/segmented/en-apmo2015_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution."
}
|
372115e8-55ec-5e65-a7c6-4e677231b56e
| 261,024
|
Let $n$ be a positive integer. Consider $2 n$ distinct lines on the plane, no two of which are parallel. Of the $2 n$ lines, $n$ are colored blue, the other $n$ are colored red. Let $\mathcal{B}$ be the set of all points on the plane that lie on at least one blue line, and $\mathcal{R}$ the set of all points on the plane that lie on at least one red line. Prove that there exists a circle that intersects $\mathcal{B}$ in exactly $2 n-1$ points, and also intersects $\mathcal{R}$ in exactly $2 n-1$ points.
|
Consider a line $\ell$ on the plane and a point $P$ on it such that $\ell$ is not parallel to any of the $2 n$ lines. Rotate $\ell$ about $P$ counterclockwise until it is parallel to one of the $2 n$ lines. Take note of that line and keep rotating until all the $2 n$ lines are met. The $2 n$ lines are now ordered according to which line is met before or after. Say the lines are in order $\ell_{1}, \ldots, \ell_{2 n}$. Clearly there must be $k \in\{1, \ldots, 2 n-1\}$ such that $\ell_{k}$ and $\ell_{k+1}$ are of different colors.
Now we set up a system of $X-$ and $Y$ - axes on the plane. Consider the two angular bisectors of $\ell_{k}$ and $\ell_{k+1}$. If we rotate $\ell_{k+1}$ counterclockwise, the line will be parallel to one of the bisectors before the other. Let the bisector that is parallel to the rotation of $\ell_{k+1}$ first be the $X$-axis, and the other the $Y$-axis. From now on, we will be using the directed angle notation: for lines $s$ and $s^{\prime}$, we define $\angle\left(s, s^{\prime}\right)$ to be a real number in $[0, \pi)$ denoting the angle in radians such that when $s$ is rotated counterclockwise by $\angle\left(s, s^{\prime}\right)$ radian, it becomes parallel to $s^{\prime}$. Using this
notation, we notice that there is no $i=1, \ldots, 2 n$ such that $\angle\left(X, l_{i}\right)$ is between $\angle\left(X, \ell_{k}\right)$ and $\angle\left(X, \ell_{k+1}\right)$.
Because the $2 n$ lines are distinct, the set $S$ of all the intersections between $\ell_{i}$ and $\ell_{j}(i \neq j)$ is a finite set of points. Consider a rectangle with two opposite vertices lying on $\ell_{k}$ and the other two lying on $\ell_{k+1}$. With respect to the origin (the intersection of $\ell_{k}$ and $\ell_{k+1}$ ), we can enlarge the rectangle as much as we want, while all the vertices remain on the lines. Thus, there is one of these rectangles $R$ which contains all the points in $S$ in its interior. Since each side of $R$ is parallel to either $X$ - or $Y$ - axis, $R$ is a part of the four lines $x= \pm a, y= \pm b$. where $a, b>0$.
Consider the circle $\mathcal{C}$ tangent to the right of the $x=a$ side of the rectangle, and to both $\ell_{k}$ and $\ell_{k+1}$. We claim that this circle intersects $\mathcal{B}$ in exactly $2 n-1$ points, and also intersects $\mathcal{R}$ in exactly $2 n-1$ points. Since $\mathcal{C}$ is tangent to both $\ell_{k}$ and $\ell_{k+1}$ and the two lines have different colors, it is enough to show that $\mathcal{C}$ intersects with each of the other $2 n-2$ lines in exactly 2 points. Note that no two lines intersect on the circle because all the intersections between lines are in $S$ which is in the interior of $R$.
Consider any line $L$ among these $2 n-2$ lines. Let $L$ intersect with $\ell_{k}$ and $\ell_{k+1}$ at the points $M$ and $N$, respectively ( $M$ and $N$ are not necessarily distinct). Notice that both $M$ and $N$ must be inside $R$. There are two cases:
(i) $L$ intersects $R$ on the $x=-a$ side once and another time on $x=a$ side;
(ii) $L$ intersects $y=-b$ and $y=b$ sides.
However, if (ii) happens, $\angle\left(\ell_{k}, L\right)$ and $\angle\left(L, \ell_{k+1}\right)$ would be both positive, and then $\angle(X, L)$ would be between $\angle\left(X, \ell_{k}\right)$ and $\angle\left(X, \ell_{k+1}\right)$, a contradiction. Thus, only (i) can happen. Then $L$ intersects $\mathcal{C}$ in exactly two points, and we are done.
Alternative solution. By rotating the diagram we can ensure that no line is vertical. Let $\ell_{1}, \ell_{2}, \ldots, \ell_{2 n}$ be the lines listed in order of increasing gradient. Then there is a $k$ such that lines $\ell_{k}$ and $\ell_{k+1}$ are oppositely coloured. By rotating our coordinate system and cyclicly relabelling our lines we can ensure that $\ell_{1}, \ell_{2}, \ldots, \ell_{2 n}$ are listed in order of increasing gradient, $\ell_{1}$ and $\ell_{2 n}$ are oppositely coloured, and no line is vertical.
Let $\mathcal{D}$ be a circle centred at the origin and of sufficiently large radius so that
- All intersection points of all pairs of lines lie strictly inside $\mathcal{D}$; and
- Each line $\ell_{i}$ intersects $\mathcal{D}$ in two points $A_{i}$ and $B_{i}$ say, such that $A_{i}$ is on the right semicircle (the part of the circle in the positive $x$ half plane) and $B_{i}$ is on the left semicircle.
Note that the anticlockwise order of the points $A_{i}, B_{i}$ around $\mathcal{D}$ is $A_{1}, A_{2}, \ldots, A_{n}, B_{1}, B_{2}, \ldots, B_{n}$.
(If $A_{i+1}$ occurred before $A_{i}$ then rays $r_{i}$ and $r_{i+1}$ (as defined below) would intersect outside $\mathcal{D}$.)
For each $i$, let $r_{i}$ be the ray that is the part of the line $\ell_{i}$ starting from point $A_{i}$ and that extends to the right. Let $\mathcal{C}$ be any circle tangent to $r_{1}$ and $r_{2 n}$, that lies entirely to the right of $\mathcal{D}$. Then $\mathcal{C}$ intersects each of $r_{2}, r_{3}, \ldots, r_{2 n-1}$ twice and is tangent to $r_{1}$ and $r_{2 n}$. Thus $\mathcal{C}$ has the required properties.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $n$ be a positive integer. Consider $2 n$ distinct lines on the plane, no two of which are parallel. Of the $2 n$ lines, $n$ are colored blue, the other $n$ are colored red. Let $\mathcal{B}$ be the set of all points on the plane that lie on at least one blue line, and $\mathcal{R}$ the set of all points on the plane that lie on at least one red line. Prove that there exists a circle that intersects $\mathcal{B}$ in exactly $2 n-1$ points, and also intersects $\mathcal{R}$ in exactly $2 n-1$ points.
|
Consider a line $\ell$ on the plane and a point $P$ on it such that $\ell$ is not parallel to any of the $2 n$ lines. Rotate $\ell$ about $P$ counterclockwise until it is parallel to one of the $2 n$ lines. Take note of that line and keep rotating until all the $2 n$ lines are met. The $2 n$ lines are now ordered according to which line is met before or after. Say the lines are in order $\ell_{1}, \ldots, \ell_{2 n}$. Clearly there must be $k \in\{1, \ldots, 2 n-1\}$ such that $\ell_{k}$ and $\ell_{k+1}$ are of different colors.
Now we set up a system of $X-$ and $Y$ - axes on the plane. Consider the two angular bisectors of $\ell_{k}$ and $\ell_{k+1}$. If we rotate $\ell_{k+1}$ counterclockwise, the line will be parallel to one of the bisectors before the other. Let the bisector that is parallel to the rotation of $\ell_{k+1}$ first be the $X$-axis, and the other the $Y$-axis. From now on, we will be using the directed angle notation: for lines $s$ and $s^{\prime}$, we define $\angle\left(s, s^{\prime}\right)$ to be a real number in $[0, \pi)$ denoting the angle in radians such that when $s$ is rotated counterclockwise by $\angle\left(s, s^{\prime}\right)$ radian, it becomes parallel to $s^{\prime}$. Using this
notation, we notice that there is no $i=1, \ldots, 2 n$ such that $\angle\left(X, l_{i}\right)$ is between $\angle\left(X, \ell_{k}\right)$ and $\angle\left(X, \ell_{k+1}\right)$.
Because the $2 n$ lines are distinct, the set $S$ of all the intersections between $\ell_{i}$ and $\ell_{j}(i \neq j)$ is a finite set of points. Consider a rectangle with two opposite vertices lying on $\ell_{k}$ and the other two lying on $\ell_{k+1}$. With respect to the origin (the intersection of $\ell_{k}$ and $\ell_{k+1}$ ), we can enlarge the rectangle as much as we want, while all the vertices remain on the lines. Thus, there is one of these rectangles $R$ which contains all the points in $S$ in its interior. Since each side of $R$ is parallel to either $X$ - or $Y$ - axis, $R$ is a part of the four lines $x= \pm a, y= \pm b$. where $a, b>0$.
Consider the circle $\mathcal{C}$ tangent to the right of the $x=a$ side of the rectangle, and to both $\ell_{k}$ and $\ell_{k+1}$. We claim that this circle intersects $\mathcal{B}$ in exactly $2 n-1$ points, and also intersects $\mathcal{R}$ in exactly $2 n-1$ points. Since $\mathcal{C}$ is tangent to both $\ell_{k}$ and $\ell_{k+1}$ and the two lines have different colors, it is enough to show that $\mathcal{C}$ intersects with each of the other $2 n-2$ lines in exactly 2 points. Note that no two lines intersect on the circle because all the intersections between lines are in $S$ which is in the interior of $R$.
Consider any line $L$ among these $2 n-2$ lines. Let $L$ intersect with $\ell_{k}$ and $\ell_{k+1}$ at the points $M$ and $N$, respectively ( $M$ and $N$ are not necessarily distinct). Notice that both $M$ and $N$ must be inside $R$. There are two cases:
(i) $L$ intersects $R$ on the $x=-a$ side once and another time on $x=a$ side;
(ii) $L$ intersects $y=-b$ and $y=b$ sides.
However, if (ii) happens, $\angle\left(\ell_{k}, L\right)$ and $\angle\left(L, \ell_{k+1}\right)$ would be both positive, and then $\angle(X, L)$ would be between $\angle\left(X, \ell_{k}\right)$ and $\angle\left(X, \ell_{k+1}\right)$, a contradiction. Thus, only (i) can happen. Then $L$ intersects $\mathcal{C}$ in exactly two points, and we are done.
Alternative solution. By rotating the diagram we can ensure that no line is vertical. Let $\ell_{1}, \ell_{2}, \ldots, \ell_{2 n}$ be the lines listed in order of increasing gradient. Then there is a $k$ such that lines $\ell_{k}$ and $\ell_{k+1}$ are oppositely coloured. By rotating our coordinate system and cyclicly relabelling our lines we can ensure that $\ell_{1}, \ell_{2}, \ldots, \ell_{2 n}$ are listed in order of increasing gradient, $\ell_{1}$ and $\ell_{2 n}$ are oppositely coloured, and no line is vertical.
Let $\mathcal{D}$ be a circle centred at the origin and of sufficiently large radius so that
- All intersection points of all pairs of lines lie strictly inside $\mathcal{D}$; and
- Each line $\ell_{i}$ intersects $\mathcal{D}$ in two points $A_{i}$ and $B_{i}$ say, such that $A_{i}$ is on the right semicircle (the part of the circle in the positive $x$ half plane) and $B_{i}$ is on the left semicircle.
Note that the anticlockwise order of the points $A_{i}, B_{i}$ around $\mathcal{D}$ is $A_{1}, A_{2}, \ldots, A_{n}, B_{1}, B_{2}, \ldots, B_{n}$.
(If $A_{i+1}$ occurred before $A_{i}$ then rays $r_{i}$ and $r_{i+1}$ (as defined below) would intersect outside $\mathcal{D}$.)
For each $i$, let $r_{i}$ be the ray that is the part of the line $\ell_{i}$ starting from point $A_{i}$ and that extends to the right. Let $\mathcal{C}$ be any circle tangent to $r_{1}$ and $r_{2 n}$, that lies entirely to the right of $\mathcal{D}$. Then $\mathcal{C}$ intersects each of $r_{2}, r_{3}, \ldots, r_{2 n-1}$ twice and is tangent to $r_{1}$ and $r_{2 n}$. Thus $\mathcal{C}$ has the required properties.
|
{
"resource_path": "APMO/segmented/en-apmo2015_sol.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "\nSolution."
}
|
73d35a90-ea20-5e0d-a185-baa0a5765d98
| 261,046
|
We say that a triangle $A B C$ is great if the following holds: for any point $D$ on the side $B C$, if $P$ and $Q$ are the feet of the perpendiculars from $D$ to the lines $A B$ and $A C$, respectively, then the reflection of $D$ in the line $P Q$ lies on the circumcircle of the triangle $A B C$.
Prove that triangle $A B C$ is great if and only if $\angle A=90^{\circ}$ and $A B=A C$.
|
For every point $D$ on the side $B C$, let $D^{\prime}$ be the reflection of $D$ in the line $P Q$. We will first prove that if the triangle satisfies the condition then it is isosceles and right-angled at $A$.
Choose $D$ to be the point where the angle bisector from $A$ meets $B C$. Note that $P$ and $Q$ lie on the rays $A B$ and $A C$ respectively. Furthermore, $P$ and $Q$ are reflections of each other in the line $A D$, from which it follows that $P Q \perp A D$. Therefore, $D^{\prime}$ lies on the line $A D$ and we may deduce that either $D^{\prime}=A$ or $D^{\prime}$ is the second point of the angle bisector at $A$ and the circumcircle of $A B C$. However, since $A P D Q$ is a cyclic quadrilateral, the segment $P Q$ intersects the segment $A D$. Therefore, $D^{\prime}$ lies on the ray $D A$ and therefore $D^{\prime}=A$. By angle chasing we obtain
$$
\angle P D^{\prime} Q=\angle P D Q=180^{\circ}-\angle B A C
$$
and since $D^{\prime}=A$ we also know $\angle P D^{\prime} Q=\angle B A C$. This implies that $\angle B A C=90^{\circ}$.
Now we choose $D$ to be the midpoint of $B C$. Since $\angle B A C=90^{\circ}$, we can deduce that $D Q P$ is the medial triangle of triangle $A B C$. Therefore, $P Q \| B C$ from which it follows that $D D^{\prime} \perp B C$. But the distance from $D^{\prime}$ to $B C$ is equal to both the circumradius of triangle $A B C$ and to the distance from $A$ to $B C$. This can only happen if $A=D^{\prime}$. This implies that $A B C$ is isosceles and right-angled at $A$.
We will now prove that if $A B C$ is isosceles and right-angled at $A$ then the required property in the problem holds. Let $D$ be any point on side $B C$. Then $D^{\prime} P=D P$ and we also have $D P=B P$. Hence, $D^{\prime} P=B P$ and similarly $D^{\prime} Q=C Q$. Note that $A P D Q D^{\prime}$ is cyclic with diameter $P Q$. Therefore, $\angle A P D^{\prime}=\angle A Q D^{\prime}$, from which we obtain $\angle B P D^{\prime}=\angle C Q D^{\prime}$. So triangles $D^{\prime} P B$ and $D^{\prime} Q C$ are similar. It follows that $\angle P D^{\prime} Q=\angle P D^{\prime} C+\angle C D^{\prime} Q=$ $\angle P D^{\prime} C+\angle B D^{\prime} P=\angle B D^{\prime} C$ and $\frac{D^{\prime} P}{D^{\prime} Q}=\frac{D^{\prime} B}{D^{\prime} C}$. So we also obtain that triangles $D^{\prime} P Q$ and $D^{\prime} B C$ are similar. But since $D P Q$ and $D^{\prime} P Q$ are congruent, we may deduce that $\angle B D^{\prime} C=$ $\angle P D^{\prime} Q=\angle P D Q=90^{\circ}$. Therefore, $D^{\prime}$ lies on the circle with diameter $B C$, which is the circumcircle of triangle $A B C$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
We say that a triangle $A B C$ is great if the following holds: for any point $D$ on the side $B C$, if $P$ and $Q$ are the feet of the perpendiculars from $D$ to the lines $A B$ and $A C$, respectively, then the reflection of $D$ in the line $P Q$ lies on the circumcircle of the triangle $A B C$.
Prove that triangle $A B C$ is great if and only if $\angle A=90^{\circ}$ and $A B=A C$.
|
For every point $D$ on the side $B C$, let $D^{\prime}$ be the reflection of $D$ in the line $P Q$. We will first prove that if the triangle satisfies the condition then it is isosceles and right-angled at $A$.
Choose $D$ to be the point where the angle bisector from $A$ meets $B C$. Note that $P$ and $Q$ lie on the rays $A B$ and $A C$ respectively. Furthermore, $P$ and $Q$ are reflections of each other in the line $A D$, from which it follows that $P Q \perp A D$. Therefore, $D^{\prime}$ lies on the line $A D$ and we may deduce that either $D^{\prime}=A$ or $D^{\prime}$ is the second point of the angle bisector at $A$ and the circumcircle of $A B C$. However, since $A P D Q$ is a cyclic quadrilateral, the segment $P Q$ intersects the segment $A D$. Therefore, $D^{\prime}$ lies on the ray $D A$ and therefore $D^{\prime}=A$. By angle chasing we obtain
$$
\angle P D^{\prime} Q=\angle P D Q=180^{\circ}-\angle B A C
$$
and since $D^{\prime}=A$ we also know $\angle P D^{\prime} Q=\angle B A C$. This implies that $\angle B A C=90^{\circ}$.
Now we choose $D$ to be the midpoint of $B C$. Since $\angle B A C=90^{\circ}$, we can deduce that $D Q P$ is the medial triangle of triangle $A B C$. Therefore, $P Q \| B C$ from which it follows that $D D^{\prime} \perp B C$. But the distance from $D^{\prime}$ to $B C$ is equal to both the circumradius of triangle $A B C$ and to the distance from $A$ to $B C$. This can only happen if $A=D^{\prime}$. This implies that $A B C$ is isosceles and right-angled at $A$.
We will now prove that if $A B C$ is isosceles and right-angled at $A$ then the required property in the problem holds. Let $D$ be any point on side $B C$. Then $D^{\prime} P=D P$ and we also have $D P=B P$. Hence, $D^{\prime} P=B P$ and similarly $D^{\prime} Q=C Q$. Note that $A P D Q D^{\prime}$ is cyclic with diameter $P Q$. Therefore, $\angle A P D^{\prime}=\angle A Q D^{\prime}$, from which we obtain $\angle B P D^{\prime}=\angle C Q D^{\prime}$. So triangles $D^{\prime} P B$ and $D^{\prime} Q C$ are similar. It follows that $\angle P D^{\prime} Q=\angle P D^{\prime} C+\angle C D^{\prime} Q=$ $\angle P D^{\prime} C+\angle B D^{\prime} P=\angle B D^{\prime} C$ and $\frac{D^{\prime} P}{D^{\prime} Q}=\frac{D^{\prime} B}{D^{\prime} C}$. So we also obtain that triangles $D^{\prime} P Q$ and $D^{\prime} B C$ are similar. But since $D P Q$ and $D^{\prime} P Q$ are congruent, we may deduce that $\angle B D^{\prime} C=$ $\angle P D^{\prime} Q=\angle P D Q=90^{\circ}$. Therefore, $D^{\prime}$ lies on the circle with diameter $B C$, which is the circumcircle of triangle $A B C$.
|
{
"resource_path": "APMO/segmented/en-apmo2016_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution."
}
|
81c2b023-cfa5-5fa3-b87d-f6b983a0117d
| 260,693
|
Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$ and $A C$, and let $M$ be the intersection of line $A B$ and the line through $R$ parallel to $A C$. Prove that line $M N$ is tangent to $\omega$.
|
We present two approaches. The first one introduces an auxiliary point and studies similarities in the figure. The second one reduces the problem to computations involving a particular exradius of a triangle. The second approach has two variants.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$ and $A C$, and let $M$ be the intersection of line $A B$ and the line through $R$ parallel to $A C$. Prove that line $M N$ is tangent to $\omega$.
|
We present two approaches. The first one introduces an auxiliary point and studies similarities in the figure. The second one reduces the problem to computations involving a particular exradius of a triangle. The second approach has two variants.
|
{
"resource_path": "APMO/segmented/en-apmo2016_sol.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "\nSolution."
}
|
8817de03-1ca7-53cd-bcd6-18d6c28edfed
| 260,709
|
Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$ and $A C$, and let $M$ be the intersection of line $A B$ and the line through $R$ parallel to $A C$. Prove that line $M N$ is tangent to $\omega$.
|
.
Let the line through $N$ tangent to $\omega$ at point $X \neq E$ intersect $A B$ at point $M^{\prime}$. It suffices to show that $M^{\prime} R \| A C$, since this would yield $M^{\prime}=M$.
Suppose that the line $P O$ intersects $A C$ at $Q$ and the circumcircle of $A M^{\prime} O$ at $Y$, respectively. Then
$$
\angle A Y M^{\prime}=\angle A O M^{\prime}=90^{\circ}-\angle M^{\prime} O P
$$
By angle chasing we have $\angle E O Q=\angle F O P=90^{\circ}-\angle A O F=\angle M^{\prime} A O=\angle M^{\prime} Y P$ and by symmetry $\angle E Q O=\angle M^{\prime} P Y$. Therefore $\triangle M^{\prime} Y P \sim \triangle E O Q$.
On the other hand, we have
$$
\begin{aligned}
\angle M^{\prime} O P & =\angle M^{\prime} O F+\angle F O P=\frac{1}{2}(\angle F O X+\angle F O P+\angle E O Q)= \\
& =\frac{1}{2}\left(\frac{180^{\circ}-\angle X O E}{2}\right)=90^{\circ}-\frac{\angle X O E}{2} .
\end{aligned}
$$
Since we know that $\angle A Y M^{\prime}$ and $\angle M^{\prime} O P$ are complementary this implies
$$
\angle A Y M^{\prime}=\frac{\angle X O E}{2}=\angle N O E
$$
Therefore, $\angle A Y M^{\prime}$ and $\angle N O E$ are congruent angles, and this means that $A$ and $N$ are corresponding points in the similarity of triangles $\triangle M^{\prime} Y P$ and $\triangle E O Q$. It follows that
$$
\frac{A M^{\prime}}{M^{\prime} P}=\frac{N E}{E Q}=\frac{N R}{R P}
$$
We conclude that $M^{\prime} R \| A C$, as desired.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$ and $A C$, and let $M$ be the intersection of line $A B$ and the line through $R$ parallel to $A C$. Prove that line $M N$ is tangent to $\omega$.
|
.
Let the line through $N$ tangent to $\omega$ at point $X \neq E$ intersect $A B$ at point $M^{\prime}$. It suffices to show that $M^{\prime} R \| A C$, since this would yield $M^{\prime}=M$.
Suppose that the line $P O$ intersects $A C$ at $Q$ and the circumcircle of $A M^{\prime} O$ at $Y$, respectively. Then
$$
\angle A Y M^{\prime}=\angle A O M^{\prime}=90^{\circ}-\angle M^{\prime} O P
$$
By angle chasing we have $\angle E O Q=\angle F O P=90^{\circ}-\angle A O F=\angle M^{\prime} A O=\angle M^{\prime} Y P$ and by symmetry $\angle E Q O=\angle M^{\prime} P Y$. Therefore $\triangle M^{\prime} Y P \sim \triangle E O Q$.
On the other hand, we have
$$
\begin{aligned}
\angle M^{\prime} O P & =\angle M^{\prime} O F+\angle F O P=\frac{1}{2}(\angle F O X+\angle F O P+\angle E O Q)= \\
& =\frac{1}{2}\left(\frac{180^{\circ}-\angle X O E}{2}\right)=90^{\circ}-\frac{\angle X O E}{2} .
\end{aligned}
$$
Since we know that $\angle A Y M^{\prime}$ and $\angle M^{\prime} O P$ are complementary this implies
$$
\angle A Y M^{\prime}=\frac{\angle X O E}{2}=\angle N O E
$$
Therefore, $\angle A Y M^{\prime}$ and $\angle N O E$ are congruent angles, and this means that $A$ and $N$ are corresponding points in the similarity of triangles $\triangle M^{\prime} Y P$ and $\triangle E O Q$. It follows that
$$
\frac{A M^{\prime}}{M^{\prime} P}=\frac{N E}{E Q}=\frac{N R}{R P}
$$
We conclude that $M^{\prime} R \| A C$, as desired.
|
{
"resource_path": "APMO/segmented/en-apmo2016_sol.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "# Solution 1"
}
|
8817de03-1ca7-53cd-bcd6-18d6c28edfed
| 260,709
|
Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$ and $A C$, and let $M$ be the intersection of line $A B$ and the line through $R$ parallel to $A C$. Prove that line $M N$ is tangent to $\omega$.
|
a.
As in Solution 1, we introduce point $M^{\prime}$ and reduce the problem to proving $\frac{P R}{R N}=\frac{P M^{\prime}}{M^{\prime} A}$. Menelaus theorem in triangle $A N P$ with transversal line $F R E$ yields
$$
\frac{P R}{R N} \cdot \frac{N E}{E A} \cdot \frac{A F}{F P}=1
$$
Since $A F=E A$, we have $\frac{F P}{N E}=\frac{P R}{R N}$, so that it suffices to prove
$$
\frac{F P}{N E}=\frac{P M^{\prime}}{M^{\prime} A}
$$
This is a computation regarding the triangle $A M^{\prime} N$ and its excircle opposite $A$. Indeed, setting $a=M^{\prime} N, b=N A, c=M^{\prime} A, s=\frac{a+b+c}{2}, x=s-a, y=s-b$ and $z=s-c$, then $A E=A F=s, M^{\prime} F=z$ and $N E=y$. From $\triangle O F P \sim \triangle A F O$ we have $F P=\frac{r_{a}^{2}}{s}$, where $r_{a}=O F$ is the exradius opposite $A$. Combining the following two standard formulas for the area of a triangle
$$
\left|A M^{\prime} N\right|^{2}=x y z s \quad \text { (Heron's formula) and } \quad\left|A M^{\prime} N\right|=r_{a}(s-a),
$$
we have $r_{a}^{2}=\frac{y z s}{x}$. Therefore, $F P=\frac{y z}{x}$. We can now write everything in (1) in terms of $x, y, z$. We conclude that we have to verify
$$
\frac{\frac{y z}{x}}{y}=\frac{z+\frac{y z}{x}}{x+y}
$$
which is easily seen to be true.
Note: Antoher approach using Menalaus theorem is to construct the tangent from $M$ to create a point $N^{\prime}$ in $A C$ and then prove, using the theorem, that $P, R$ and $N^{\prime}$ are collinear. This also reduces to an algebraic identity.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$ and $A C$, and let $M$ be the intersection of line $A B$ and the line through $R$ parallel to $A C$. Prove that line $M N$ is tangent to $\omega$.
|
a.
As in Solution 1, we introduce point $M^{\prime}$ and reduce the problem to proving $\frac{P R}{R N}=\frac{P M^{\prime}}{M^{\prime} A}$. Menelaus theorem in triangle $A N P$ with transversal line $F R E$ yields
$$
\frac{P R}{R N} \cdot \frac{N E}{E A} \cdot \frac{A F}{F P}=1
$$
Since $A F=E A$, we have $\frac{F P}{N E}=\frac{P R}{R N}$, so that it suffices to prove
$$
\frac{F P}{N E}=\frac{P M^{\prime}}{M^{\prime} A}
$$
This is a computation regarding the triangle $A M^{\prime} N$ and its excircle opposite $A$. Indeed, setting $a=M^{\prime} N, b=N A, c=M^{\prime} A, s=\frac{a+b+c}{2}, x=s-a, y=s-b$ and $z=s-c$, then $A E=A F=s, M^{\prime} F=z$ and $N E=y$. From $\triangle O F P \sim \triangle A F O$ we have $F P=\frac{r_{a}^{2}}{s}$, where $r_{a}=O F$ is the exradius opposite $A$. Combining the following two standard formulas for the area of a triangle
$$
\left|A M^{\prime} N\right|^{2}=x y z s \quad \text { (Heron's formula) and } \quad\left|A M^{\prime} N\right|=r_{a}(s-a),
$$
we have $r_{a}^{2}=\frac{y z s}{x}$. Therefore, $F P=\frac{y z}{x}$. We can now write everything in (1) in terms of $x, y, z$. We conclude that we have to verify
$$
\frac{\frac{y z}{x}}{y}=\frac{z+\frac{y z}{x}}{x+y}
$$
which is easily seen to be true.
Note: Antoher approach using Menalaus theorem is to construct the tangent from $M$ to create a point $N^{\prime}$ in $A C$ and then prove, using the theorem, that $P, R$ and $N^{\prime}$ are collinear. This also reduces to an algebraic identity.
|
{
"resource_path": "APMO/segmented/en-apmo2016_sol.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "# Solution 2"
}
|
8817de03-1ca7-53cd-bcd6-18d6c28edfed
| 260,709
|
Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$ and $A C$, and let $M$ be the intersection of line $A B$ and the line through $R$ parallel to $A C$. Prove that line $M N$ is tangent to $\omega$.
|
b.
As in Solution 1, we introduce point $M^{\prime}$. Let the line through $M^{\prime}$ and parallel to $A N$ intersect $E F$ at $R^{\prime}$. Let $P^{\prime}$ be the intersection of lines $N R^{\prime}$ and $A M$. It suffices to show that $P^{\prime} O \| F E$, since this would yield $P=P^{\prime}$, and then $R=R^{\prime}$ and $M=M^{\prime}$. Hence it is enough to prove that
$$
\frac{A F}{F P^{\prime}}=\frac{A D}{D O}
$$
where $D$ is the intersection of $A O$ and $E F$. Once again, this reduces to a computation regarding the triangle $A M^{\prime} N$ and its excircle opposite $A$.
Let $u=P^{\prime} F$ and $x, y, z, s$ as in Solution 2a. Note that since $A E=A F$ and $M^{\prime} R^{\prime} \| A E$, we have $M^{\prime} R^{\prime}=M^{\prime} F=z$. Since $M^{\prime} R^{\prime} \| A N$, we have $\frac{P^{\prime} M^{\prime}}{P^{\prime} A}=\frac{M^{\prime} R^{\prime}}{N A}$, that is,
$$
\frac{u+z}{u+x+y+z}=\frac{z}{x+z}
$$
From this last equation we obtain $u=\frac{y z}{x}$. Hence $\frac{A F}{F P^{\prime}}=\frac{x s}{y z}$. Also, as in Solution 2a, we have $r_{a}^{2}=\frac{y z s}{x}$.
Finally, using similar triangles $O D F, F D A$ and $O F A$, and the above equalities, we have
$$
\frac{A D}{D O}=\frac{A D}{D F} \cdot \frac{D F}{D O}=\frac{A F}{O F} \cdot \frac{A F}{O F}=\frac{s^{2}}{r_{a}^{2}}=\frac{s^{2}}{\frac{y z s}{x}}=\frac{x s}{y z}=\frac{A F}{F P^{\prime}}
$$
as required.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$ and $A C$, and let $M$ be the intersection of line $A B$ and the line through $R$ parallel to $A C$. Prove that line $M N$ is tangent to $\omega$.
|
b.
As in Solution 1, we introduce point $M^{\prime}$. Let the line through $M^{\prime}$ and parallel to $A N$ intersect $E F$ at $R^{\prime}$. Let $P^{\prime}$ be the intersection of lines $N R^{\prime}$ and $A M$. It suffices to show that $P^{\prime} O \| F E$, since this would yield $P=P^{\prime}$, and then $R=R^{\prime}$ and $M=M^{\prime}$. Hence it is enough to prove that
$$
\frac{A F}{F P^{\prime}}=\frac{A D}{D O}
$$
where $D$ is the intersection of $A O$ and $E F$. Once again, this reduces to a computation regarding the triangle $A M^{\prime} N$ and its excircle opposite $A$.
Let $u=P^{\prime} F$ and $x, y, z, s$ as in Solution 2a. Note that since $A E=A F$ and $M^{\prime} R^{\prime} \| A E$, we have $M^{\prime} R^{\prime}=M^{\prime} F=z$. Since $M^{\prime} R^{\prime} \| A N$, we have $\frac{P^{\prime} M^{\prime}}{P^{\prime} A}=\frac{M^{\prime} R^{\prime}}{N A}$, that is,
$$
\frac{u+z}{u+x+y+z}=\frac{z}{x+z}
$$
From this last equation we obtain $u=\frac{y z}{x}$. Hence $\frac{A F}{F P^{\prime}}=\frac{x s}{y z}$. Also, as in Solution 2a, we have $r_{a}^{2}=\frac{y z s}{x}$.
Finally, using similar triangles $O D F, F D A$ and $O F A$, and the above equalities, we have
$$
\frac{A D}{D O}=\frac{A D}{D F} \cdot \frac{D F}{D O}=\frac{A F}{O F} \cdot \frac{A F}{O F}=\frac{s^{2}}{r_{a}^{2}}=\frac{s^{2}}{\frac{y z s}{x}}=\frac{x s}{y z}=\frac{A F}{F P^{\prime}}
$$
as required.
|
{
"resource_path": "APMO/segmented/en-apmo2016_sol.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "# Solution 2"
}
|
8817de03-1ca7-53cd-bcd6-18d6c28edfed
| 260,709
|
Let $A B C$ be a triangle with $A B<A C$. Let $D$ be the intersection point of the internal bisector of angle $B A C$ and the circumcircle of $A B C$. Let $Z$ be the intersection point of the perpendicular bisector of $A C$ with the external bisector of angle $\angle B A C$. Prove that the midpoint of the segment $A B$ lies on the circumcircle of triangle $A D Z$.
|
. Let $N$ be the midpoint of $A C$. Let $M$ be the intersection point of the circumcircle of triangle $A D Z$ and the segment $A B$. We will show that $M$ is the midpoint of $A B$. To do this, let $D^{\prime}$ the reflection of $D$ with respect to $M$. It suffices to show that $A D B D^{\prime}$ is a parallelogram.
The internal and external bisectors of an angle in a triangle are perpendicular. This implies that $Z D$ is a diameter of the circumcircle of $A Z D$ and thus $\angle Z M D=90^{\circ}$. This means that $Z M$ is the perpendicular bisector of $D^{\prime} D$ and thus $Z D^{\prime}=Z D$. By construction, $Z$ is in the perpendicular bisector of $A C$ and thus $Z A=Z C$.
Now, let $\alpha$ be the angle $\angle B A D=\angle D A C$. In the cyclic quadrilateral $A Z D M$ we get $\angle M Z D=\angle M A D=\alpha$, and thus $\angle D^{\prime} Z D=2 \alpha$. By angle chasing we get
$$
\angle A Z N=90^{\circ}-\angle Z A N=\angle D A C=\alpha,
$$
which implies that $\angle A Z C=2 \alpha$. Therefore,
$$
\angle D^{\prime} Z A=\angle D^{\prime} Z D-\angle A Z D=2 \alpha-\angle A Z D=\angle A Z C-\angle A Z D=\angle D Z C .
$$
Combining $\angle D^{\prime} Z A=\angle D Z C, Z D^{\prime}=Z D$ and $Z A=Z C$, we obtain by the $S A S$ criterion that the triangles $D^{\prime} Z A$ and $D Z C$ are congruent. In particular, $D^{\prime} A=D C$ and $\angle D^{\prime} A Z=$ $\angle D C Z$. From here $D B=D C=D^{\prime} A$.
Finally, let $\beta=\angle A B C=\angle A D C$. We get the first of the following equalities by the sum of angles around point $A$ and the second one by the sum of internal angles of quadrilateral $A Z C D$
$$
\begin{aligned}
& 360^{\circ}=\angle D^{\prime} A Z+\angle Z A D+\angle D A B+\angle B A D^{\prime}=\angle D^{\prime} A Z+90^{\circ}+\alpha+\angle B A D^{\prime} \\
& 360^{\circ}=\angle D C Z+\angle Z A D+\angle C Z A+\angle A D C=\angle D C Z+90^{\circ}+2 \alpha+\beta
\end{aligned}
$$
By canceling equal terms we conclude that $\angle B A D^{\prime}=\alpha+\beta$. Also, $\angle A B D=\alpha+\beta$. Therefore, the segments $D^{\prime} A$ and $D B$ are parallel and have the same length. We conclude that $A D B D^{\prime}$ is a parallelogram. As the diagonals of a parallelogram intersect at their midpoints, we obtain that $M$ is the midpoint of $A B$ as desired.
Variant of solution. The solution above is indirect in the sense that it assumes that $M$ is in the circumcircle of $A Z D$ and then shows that $M$ is the midpoint of $A B$. We point out that the same ideas in the solution can be used to give a direct solution. Here we present a sketch on how to proceed in this manner.
Now we know that $M$ is the midpoint of the side $A B$. We construct the point $D^{\prime}$ in the same way. Now we have directly that $A D B D^{\prime}$ is a parallelogram and thus $D^{\prime} A=D B=D C$. By construction $Z A=Z C$. Also, the two sums of angles equal to $360^{\circ}$ in the previous solution let us conclude that $\angle D^{\prime} A Z=\angle D C Z$. Once again, we use (differently) the $S A S$ criterion and obtain that the triangles $D^{\prime} A Z$ and $D C Z$ are congruent. Thus, $D^{\prime} Z=D Z$.
We finish the problem by noting that $Z M$ is a median of the isosceles triangle $D^{\prime} Z D$, so it is also a perpendicular bisector. This shows that $\angle D M Z=90^{\circ}=\angle D A Z$, and therefore $M$ lies in the circumcircle of $D A Z$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $A B<A C$. Let $D$ be the intersection point of the internal bisector of angle $B A C$ and the circumcircle of $A B C$. Let $Z$ be the intersection point of the perpendicular bisector of $A C$ with the external bisector of angle $\angle B A C$. Prove that the midpoint of the segment $A B$ lies on the circumcircle of triangle $A D Z$.
|
. Let $N$ be the midpoint of $A C$. Let $M$ be the intersection point of the circumcircle of triangle $A D Z$ and the segment $A B$. We will show that $M$ is the midpoint of $A B$. To do this, let $D^{\prime}$ the reflection of $D$ with respect to $M$. It suffices to show that $A D B D^{\prime}$ is a parallelogram.
The internal and external bisectors of an angle in a triangle are perpendicular. This implies that $Z D$ is a diameter of the circumcircle of $A Z D$ and thus $\angle Z M D=90^{\circ}$. This means that $Z M$ is the perpendicular bisector of $D^{\prime} D$ and thus $Z D^{\prime}=Z D$. By construction, $Z$ is in the perpendicular bisector of $A C$ and thus $Z A=Z C$.
Now, let $\alpha$ be the angle $\angle B A D=\angle D A C$. In the cyclic quadrilateral $A Z D M$ we get $\angle M Z D=\angle M A D=\alpha$, and thus $\angle D^{\prime} Z D=2 \alpha$. By angle chasing we get
$$
\angle A Z N=90^{\circ}-\angle Z A N=\angle D A C=\alpha,
$$
which implies that $\angle A Z C=2 \alpha$. Therefore,
$$
\angle D^{\prime} Z A=\angle D^{\prime} Z D-\angle A Z D=2 \alpha-\angle A Z D=\angle A Z C-\angle A Z D=\angle D Z C .
$$
Combining $\angle D^{\prime} Z A=\angle D Z C, Z D^{\prime}=Z D$ and $Z A=Z C$, we obtain by the $S A S$ criterion that the triangles $D^{\prime} Z A$ and $D Z C$ are congruent. In particular, $D^{\prime} A=D C$ and $\angle D^{\prime} A Z=$ $\angle D C Z$. From here $D B=D C=D^{\prime} A$.
Finally, let $\beta=\angle A B C=\angle A D C$. We get the first of the following equalities by the sum of angles around point $A$ and the second one by the sum of internal angles of quadrilateral $A Z C D$
$$
\begin{aligned}
& 360^{\circ}=\angle D^{\prime} A Z+\angle Z A D+\angle D A B+\angle B A D^{\prime}=\angle D^{\prime} A Z+90^{\circ}+\alpha+\angle B A D^{\prime} \\
& 360^{\circ}=\angle D C Z+\angle Z A D+\angle C Z A+\angle A D C=\angle D C Z+90^{\circ}+2 \alpha+\beta
\end{aligned}
$$
By canceling equal terms we conclude that $\angle B A D^{\prime}=\alpha+\beta$. Also, $\angle A B D=\alpha+\beta$. Therefore, the segments $D^{\prime} A$ and $D B$ are parallel and have the same length. We conclude that $A D B D^{\prime}$ is a parallelogram. As the diagonals of a parallelogram intersect at their midpoints, we obtain that $M$ is the midpoint of $A B$ as desired.
Variant of solution. The solution above is indirect in the sense that it assumes that $M$ is in the circumcircle of $A Z D$ and then shows that $M$ is the midpoint of $A B$. We point out that the same ideas in the solution can be used to give a direct solution. Here we present a sketch on how to proceed in this manner.
Now we know that $M$ is the midpoint of the side $A B$. We construct the point $D^{\prime}$ in the same way. Now we have directly that $A D B D^{\prime}$ is a parallelogram and thus $D^{\prime} A=D B=D C$. By construction $Z A=Z C$. Also, the two sums of angles equal to $360^{\circ}$ in the previous solution let us conclude that $\angle D^{\prime} A Z=\angle D C Z$. Once again, we use (differently) the $S A S$ criterion and obtain that the triangles $D^{\prime} A Z$ and $D C Z$ are congruent. Thus, $D^{\prime} Z=D Z$.
We finish the problem by noting that $Z M$ is a median of the isosceles triangle $D^{\prime} Z D$, so it is also a perpendicular bisector. This shows that $\angle D M Z=90^{\circ}=\angle D A Z$, and therefore $M$ lies in the circumcircle of $D A Z$.
|
{
"resource_path": "APMO/segmented/en-apmo2017_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "\nSolution 1"
}
|
62bc85d1-87c1-5509-a2b1-505f944e2474
| 605,781
|
Let $A B C$ be a triangle with $A B<A C$. Let $D$ be the intersection point of the internal bisector of angle $B A C$ and the circumcircle of $A B C$. Let $Z$ be the intersection point of the perpendicular bisector of $A C$ with the external bisector of angle $\angle B A C$. Prove that the midpoint of the segment $A B$ lies on the circumcircle of triangle $A D Z$.
|
. We proceed directly. As above, we name
$$
\alpha=\angle D A C=\angle A Z N=\angle C Z N=\angle D C B .
$$
Let $L$ be the midpoint of the segment $B C$. Since $M$ and $N$ are midpoints of $A B$ and $A C$, we have that $M N=C L$ and that the segment $M N$ is parallel to $B C$. Thus, $\angle A N M=\angle A C B$.
Therefore, $\angle Z N M=\angle A C B+90^{\circ}$. Also, $\angle D C Z=\angle D C B+\angle A C B+\angle Z C A=\angle A C B+90^{\circ}$. We conclude that $\angle Z N M=\angle D C Z$.
Now, the triangles $Z N C$ and $C L D$ are similar since $\angle D C L=90^{\circ}=\angle C N Z$ and $\angle D C L=$ $\alpha=\angle C Z N$. We use this fact to obtain the following chain of equalities
$$
\frac{C D}{M N}=\frac{C D}{C L}=\frac{C Z}{Z N}
$$
We combine the equality above with $\angle Z N M=\angle D C Z$ to conclude that the triangles $Z N M$ and $Z C D$ are similar. In particular, $\angle M Z N=\angle D Z C$ and $\frac{Z M}{Z N}=\frac{Z D}{Z C}$. Since we also have $\angle M Z D=\angle M Z N+\angle N Z D=\angle D Z C+\angle N Z D=\angle N Z C$, we conclude that the triangles $M Z D$ and $N Z C$ are similar. This yields that $\angle Z M D=90^{\circ}$ and therefore $M$ is in the circumcircle of triangle $D A Z$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $A B<A C$. Let $D$ be the intersection point of the internal bisector of angle $B A C$ and the circumcircle of $A B C$. Let $Z$ be the intersection point of the perpendicular bisector of $A C$ with the external bisector of angle $\angle B A C$. Prove that the midpoint of the segment $A B$ lies on the circumcircle of triangle $A D Z$.
|
. We proceed directly. As above, we name
$$
\alpha=\angle D A C=\angle A Z N=\angle C Z N=\angle D C B .
$$
Let $L$ be the midpoint of the segment $B C$. Since $M$ and $N$ are midpoints of $A B$ and $A C$, we have that $M N=C L$ and that the segment $M N$ is parallel to $B C$. Thus, $\angle A N M=\angle A C B$.
Therefore, $\angle Z N M=\angle A C B+90^{\circ}$. Also, $\angle D C Z=\angle D C B+\angle A C B+\angle Z C A=\angle A C B+90^{\circ}$. We conclude that $\angle Z N M=\angle D C Z$.
Now, the triangles $Z N C$ and $C L D$ are similar since $\angle D C L=90^{\circ}=\angle C N Z$ and $\angle D C L=$ $\alpha=\angle C Z N$. We use this fact to obtain the following chain of equalities
$$
\frac{C D}{M N}=\frac{C D}{C L}=\frac{C Z}{Z N}
$$
We combine the equality above with $\angle Z N M=\angle D C Z$ to conclude that the triangles $Z N M$ and $Z C D$ are similar. In particular, $\angle M Z N=\angle D Z C$ and $\frac{Z M}{Z N}=\frac{Z D}{Z C}$. Since we also have $\angle M Z D=\angle M Z N+\angle N Z D=\angle D Z C+\angle N Z D=\angle N Z C$, we conclude that the triangles $M Z D$ and $N Z C$ are similar. This yields that $\angle Z M D=90^{\circ}$ and therefore $M$ is in the circumcircle of triangle $D A Z$.
|
{
"resource_path": "APMO/segmented/en-apmo2017_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "\nSolution 2"
}
|
62bc85d1-87c1-5509-a2b1-505f944e2474
| 605,781
|
Let $A B C$ be a triangle with $A B<A C$. Let $D$ be the intersection point of the internal bisector of angle $B A C$ and the circumcircle of $A B C$. Let $Z$ be the intersection point of the perpendicular bisector of $A C$ with the external bisector of angle $\angle B A C$. Prove that the midpoint of the segment $A B$ lies on the circumcircle of triangle $A D Z$.
|
. Let $m$ be the perpendicular bisector of $A D$; thus $m$ passes through the center $O$ of the circumcircle of triangle $A B C$. Since $A D$ is the internal angle bisector of $A$ and $O M$ and $O N$ are perpendicular to $A B$ and $A C$ respectively, we obtain that $O M$ and $O N$ form equal angles with $A D$. This implies that they are symmetric with respect to $M$.
Therefore, the line $M O$ passes through the point $Z^{\prime}$ symmetrical to $Z$ with respect to $m$. Since $\angle D A Z=90^{\circ}$, then also $\angle A D Z^{\prime}=90^{\circ}$. Moreover, since $A Z=D Z^{\prime}$, we have that $\angle A Z Z^{\prime}=\angle D Z^{\prime} Z=90^{\circ}$, so $A Z Z^{\prime} D$ is a rectangle. Since $\angle A M Z^{\prime}=\angle A M O=90^{\circ}$, we conclude that $M$ is in the circle with diameter $A Z^{\prime}$, which is the circumcircle of the rectangle. Thus $M$ lies on the circumcircle of the triangle $A D Z$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $A B<A C$. Let $D$ be the intersection point of the internal bisector of angle $B A C$ and the circumcircle of $A B C$. Let $Z$ be the intersection point of the perpendicular bisector of $A C$ with the external bisector of angle $\angle B A C$. Prove that the midpoint of the segment $A B$ lies on the circumcircle of triangle $A D Z$.
|
. Let $m$ be the perpendicular bisector of $A D$; thus $m$ passes through the center $O$ of the circumcircle of triangle $A B C$. Since $A D$ is the internal angle bisector of $A$ and $O M$ and $O N$ are perpendicular to $A B$ and $A C$ respectively, we obtain that $O M$ and $O N$ form equal angles with $A D$. This implies that they are symmetric with respect to $M$.
Therefore, the line $M O$ passes through the point $Z^{\prime}$ symmetrical to $Z$ with respect to $m$. Since $\angle D A Z=90^{\circ}$, then also $\angle A D Z^{\prime}=90^{\circ}$. Moreover, since $A Z=D Z^{\prime}$, we have that $\angle A Z Z^{\prime}=\angle D Z^{\prime} Z=90^{\circ}$, so $A Z Z^{\prime} D$ is a rectangle. Since $\angle A M Z^{\prime}=\angle A M O=90^{\circ}$, we conclude that $M$ is in the circle with diameter $A Z^{\prime}$, which is the circumcircle of the rectangle. Thus $M$ lies on the circumcircle of the triangle $A D Z$.
|
{
"resource_path": "APMO/segmented/en-apmo2017_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "\nSolution 3"
}
|
62bc85d1-87c1-5509-a2b1-505f944e2474
| 605,781
|
Let $A(n)$ denote the number of sequences $a_{1} \geq a_{2} \geq \ldots \geq a_{k}$ of positive integers for which $a_{1}+\cdots+a_{k}=n$ and each $a_{i}+1$ is a power of two $(i=1,2, \ldots, k)$. Let $B(n)$ denote the number of sequences $b_{1} \geq b_{2} \geq \ldots \geq b_{m}$ of positive integers for which $b_{1}+\cdots+b_{m}=n$ and each inequality $b_{j} \geq 2 b_{j+1}$ holds $(j=1,2, \ldots, m-1)$.
Prove that $A(n)=B(n)$ for every positive integer $n$.
|
We say that a sequence $a_{1} \geq a_{2} \geq \ldots \geq a_{k}$ of positive integers has type $A$ if $a_{i}+1$ is a power of two for $i=1,2, \ldots, k$. We say that a sequence $b_{1} \geq b_{2} \geq \ldots \geq b_{m}$ of positive integer has type $B$ if $b_{j} \geq 2 b_{j+1}$ for $j=1,2, \ldots, m-1$.
Recall that the binary representation of a positive integer expresses it as a sum of distinct powers of two in a unique way. Furthermore, we have the following formula for every positive integer $N$
$$
2^{N}-1=2^{N-1}+2^{N-2}+\cdots+2^{1}+2^{0}
$$
Given a sequence $a_{1} \geq a_{2} \geq \ldots \geq a_{k}$ of type $A$, use the preceding formula to express each term as a sum of powers of two. Write these powers of two in left-aligned rows, in decreasing order of size. By construction, $a_{i}$ is the sum of the numbers in the $i$ th row. For example, we obtain the following array when we start with the type A sequence $15,15,7,3,3,3,1$.
| 8 | 4 | 2 | 1 |
| :--- | :--- | :--- | :--- |
| 8 | 4 | 2 | 1 |
| 4 | 2 | 1 | |
| 2 | 1 | | |
| 2 | 1 | | |
| 2 | 1 | | |
| 1 | | | |
| 27 | 13 | 5 | 2 |
Define the sequence $b_{1}, b_{2}, \ldots, b_{n}$ by setting $b_{j}$ to be the sum of the numbers in the $j$ th column of the array. For example, we obtain the sequence $27,13,5,2$ from the array above. We now show that this new sequence has type B. This is clear from the fact that each column in the array contains at least as many entries as the column to the right of it and that each number larger than 1 in the array is twice the number to the right of it. Furthermore, it is clear that $a_{1}+a_{2}+\cdots+a_{k}=b_{1}+b_{2}+\cdots+b_{m}$, since both are equal to the sum of all the entries in the array.
We now show that we can do this operation backwards. Suppose that we are given a type B sequence $b_{1} \geq b_{2} \geq \ldots \geq b_{m}$. We construct an array inductively as follows:
- We fill $b_{m}$ left-aligned rows with the numbers $2^{m-1}, 2^{m-2}, \ldots, 2^{1}, 2^{0}$.
- Then we fill $b_{m-1}-2 b_{m}$ left aligned rows with the numbers $2^{m-2}, 2^{m-3}, \ldots, 2^{1}, 2^{0}$.
- Then we fill $b_{m-2}-2 b_{m-1}$ left aligned rows with the numbers $2^{m-3}, 2^{m-4}, \ldots, 2^{1}, 2^{0}$, and so on.
- In the last step we fill $b_{1}-2 b_{2}$ left-aligned rows with the number 1 .
For example, if we start with the type B sequence $27,13,5,2$, we obtain once again the array above. We define the sequence $a_{1}, a_{2}, \ldots, a_{k}$ by setting $a_{i}$ to be the sum of the numbers in the $i$ th row of the array. By construction, this sequence has type A. Furthermore, it is clear that $a_{1}+\cdots+a_{k}=b_{1}+\cdots+b_{m}$, since once again both sums are equal to the sum of all the entries in the array.
We have defined an operation that starts with a sequence of type A , produces an array whose row sums are given by the sequence, and outputs a sequence of type B corresponding to the column sums. We have also defined an operation that starts with a sequence of type B, produces an array whose column sums are given by the sequence, and outputs a sequence of type A corresponding to the row sums. The arrays produced in both cases comprise leftaligned rows of the form $2^{N-1}, 2^{N-2}, \ldots, 2^{1}, 2^{0}$, with non-increasing lengths. Let us refer to arrays obeying these properties as marvelous.
To show that these two operations are inverses of each other, it then suffices to prove that marvelous arrays are uniquely defined by either their row sums or their column sums. The former is obviously true and the latter arises from the observation that each step in the above inductive algorithm was forced in order to create a marvelous array with the prescribed column sums.
Thus, we have produced a bijection between the sequences of type A with sum $n$ and the sequences of type B with sum $n$. So we can conclude that $A(n)=B(n)$ for every positive integer $n$.
Remark The solution above provides a bijection between type A and type B sequences via an algorithm. There are alternative ways to provide such a bijection. For example, given the numbers $a_{1} \geq \ldots \geq a_{k}$ we may define the $b_{i}$ 's as
$$
b_{j}=\sum_{i}\left\lfloor\frac{a_{i}+1}{2^{j}}\right\rfloor .
$$
Conversely, given the numbers $b_{1} \geq \ldots \geq b_{m}$, one may define the $a_{i}$ 's by taking, as in the solution, $b_{m}$ numbers equal to $2^{m}-1, b_{m-1}-2 b_{m}$ numbers equal to $2^{m-1}-1, \ldots$, and $b_{1}-2 b_{2}$ numbers equal to $2^{1}-1$. One now needs to verify that these maps are mutually inverse.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $A(n)$ denote the number of sequences $a_{1} \geq a_{2} \geq \ldots \geq a_{k}$ of positive integers for which $a_{1}+\cdots+a_{k}=n$ and each $a_{i}+1$ is a power of two $(i=1,2, \ldots, k)$. Let $B(n)$ denote the number of sequences $b_{1} \geq b_{2} \geq \ldots \geq b_{m}$ of positive integers for which $b_{1}+\cdots+b_{m}=n$ and each inequality $b_{j} \geq 2 b_{j+1}$ holds $(j=1,2, \ldots, m-1)$.
Prove that $A(n)=B(n)$ for every positive integer $n$.
|
We say that a sequence $a_{1} \geq a_{2} \geq \ldots \geq a_{k}$ of positive integers has type $A$ if $a_{i}+1$ is a power of two for $i=1,2, \ldots, k$. We say that a sequence $b_{1} \geq b_{2} \geq \ldots \geq b_{m}$ of positive integer has type $B$ if $b_{j} \geq 2 b_{j+1}$ for $j=1,2, \ldots, m-1$.
Recall that the binary representation of a positive integer expresses it as a sum of distinct powers of two in a unique way. Furthermore, we have the following formula for every positive integer $N$
$$
2^{N}-1=2^{N-1}+2^{N-2}+\cdots+2^{1}+2^{0}
$$
Given a sequence $a_{1} \geq a_{2} \geq \ldots \geq a_{k}$ of type $A$, use the preceding formula to express each term as a sum of powers of two. Write these powers of two in left-aligned rows, in decreasing order of size. By construction, $a_{i}$ is the sum of the numbers in the $i$ th row. For example, we obtain the following array when we start with the type A sequence $15,15,7,3,3,3,1$.
| 8 | 4 | 2 | 1 |
| :--- | :--- | :--- | :--- |
| 8 | 4 | 2 | 1 |
| 4 | 2 | 1 | |
| 2 | 1 | | |
| 2 | 1 | | |
| 2 | 1 | | |
| 1 | | | |
| 27 | 13 | 5 | 2 |
Define the sequence $b_{1}, b_{2}, \ldots, b_{n}$ by setting $b_{j}$ to be the sum of the numbers in the $j$ th column of the array. For example, we obtain the sequence $27,13,5,2$ from the array above. We now show that this new sequence has type B. This is clear from the fact that each column in the array contains at least as many entries as the column to the right of it and that each number larger than 1 in the array is twice the number to the right of it. Furthermore, it is clear that $a_{1}+a_{2}+\cdots+a_{k}=b_{1}+b_{2}+\cdots+b_{m}$, since both are equal to the sum of all the entries in the array.
We now show that we can do this operation backwards. Suppose that we are given a type B sequence $b_{1} \geq b_{2} \geq \ldots \geq b_{m}$. We construct an array inductively as follows:
- We fill $b_{m}$ left-aligned rows with the numbers $2^{m-1}, 2^{m-2}, \ldots, 2^{1}, 2^{0}$.
- Then we fill $b_{m-1}-2 b_{m}$ left aligned rows with the numbers $2^{m-2}, 2^{m-3}, \ldots, 2^{1}, 2^{0}$.
- Then we fill $b_{m-2}-2 b_{m-1}$ left aligned rows with the numbers $2^{m-3}, 2^{m-4}, \ldots, 2^{1}, 2^{0}$, and so on.
- In the last step we fill $b_{1}-2 b_{2}$ left-aligned rows with the number 1 .
For example, if we start with the type B sequence $27,13,5,2$, we obtain once again the array above. We define the sequence $a_{1}, a_{2}, \ldots, a_{k}$ by setting $a_{i}$ to be the sum of the numbers in the $i$ th row of the array. By construction, this sequence has type A. Furthermore, it is clear that $a_{1}+\cdots+a_{k}=b_{1}+\cdots+b_{m}$, since once again both sums are equal to the sum of all the entries in the array.
We have defined an operation that starts with a sequence of type A , produces an array whose row sums are given by the sequence, and outputs a sequence of type B corresponding to the column sums. We have also defined an operation that starts with a sequence of type B, produces an array whose column sums are given by the sequence, and outputs a sequence of type A corresponding to the row sums. The arrays produced in both cases comprise leftaligned rows of the form $2^{N-1}, 2^{N-2}, \ldots, 2^{1}, 2^{0}$, with non-increasing lengths. Let us refer to arrays obeying these properties as marvelous.
To show that these two operations are inverses of each other, it then suffices to prove that marvelous arrays are uniquely defined by either their row sums or their column sums. The former is obviously true and the latter arises from the observation that each step in the above inductive algorithm was forced in order to create a marvelous array with the prescribed column sums.
Thus, we have produced a bijection between the sequences of type A with sum $n$ and the sequences of type B with sum $n$. So we can conclude that $A(n)=B(n)$ for every positive integer $n$.
Remark The solution above provides a bijection between type A and type B sequences via an algorithm. There are alternative ways to provide such a bijection. For example, given the numbers $a_{1} \geq \ldots \geq a_{k}$ we may define the $b_{i}$ 's as
$$
b_{j}=\sum_{i}\left\lfloor\frac{a_{i}+1}{2^{j}}\right\rfloor .
$$
Conversely, given the numbers $b_{1} \geq \ldots \geq b_{m}$, one may define the $a_{i}$ 's by taking, as in the solution, $b_{m}$ numbers equal to $2^{m}-1, b_{m-1}-2 b_{m}$ numbers equal to $2^{m-1}-1, \ldots$, and $b_{1}-2 b_{2}$ numbers equal to $2^{1}-1$. One now needs to verify that these maps are mutually inverse.
|
{
"resource_path": "APMO/segmented/en-apmo2017_sol.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "\nSolution."
}
|
af37d72c-51b6-59d4-84ab-161c2b46f22e
| 260,410
|
Call a rational number $r$ powerful if $r$ can be expressed in the form $\frac{p^{k}}{q}$ for some relatively prime positive integers $p, q$ and some integer $k>1$. Let $a, b, c$ be positive
rational numbers such that $a b c=1$. Suppose there exist positive integers $x, y, z$ such that $a^{x}+b^{y}+c^{z}$ is an integer. Prove that $a, b, c$ are all powerful.
|
Let $a=\frac{a_{1}}{b_{1}}, b=\frac{a_{2}}{b_{2}}$, where $\operatorname{gcd}\left(a_{1}, b_{1}\right)=\operatorname{gcd}\left(a_{2}, b_{2}\right)=1$. Then $c=\frac{b_{1} b_{2}}{a_{1} a_{2}}$. The condition that $a^{x}+b^{y}+c^{z}$ is an integer becomes
$$
\frac{a_{1}^{x+z} a_{2}^{z} b_{2}^{y}+a_{1}^{z} a_{2}^{y+z} b_{1}^{x}+b_{1}^{x+z} b_{2}^{y+z}}{a_{1}^{z} a_{2}^{z} b_{1}^{x} b_{2}^{y}} \in \mathbb{Z}
$$
which can be restated as
$$
a_{1}^{z} a_{2}^{z} b_{1}^{x} b_{2}^{y} \mid a_{1}^{x+z} a_{2}^{z} b_{2}^{y}+a_{1}^{z} a_{2}^{y+z} b_{1}^{x}+b_{1}^{x+z} b_{2}^{y+z} .
$$
In particular, $a_{1}^{z}$ divides the right-hand side. Since it divides the first and second terms in the sum, we conclude that $a_{1}^{z} \mid b_{1}^{x+z} b_{2}^{y+z}$. Since $\operatorname{gcd}\left(a_{1}, b_{1}\right)=1$, we have $a_{1}^{z} \mid b_{2}^{y+z}$.
Let $p$ be a prime that divides $a_{1}$. Let $m, n \geq 1$ be integers such that $p^{n} \| a_{1}$ (i.e. $p^{n} \mid a_{1}$ but $\left.p^{n+1} \nmid a_{1}\right)$ and $p^{m} \| b_{2}$. The fact that $a_{1}^{z} \mid b_{2}^{y+z}$ implies $n z \leq m(y+z)$. Since $\operatorname{gcd}\left(a_{1}, b_{1}\right)=$ $\operatorname{gcd}\left(a_{2}, b_{2}\right)=1$, we have $p$ does not divide $b_{1}$ and does not divide $a_{2}$. Thus
$$
p^{n z} \| a_{1}^{z} a_{2}^{y+z} b_{1}^{x} \text { and } p^{m(y+z)} \| b_{1}^{x+z} b_{2}^{y+z}
$$
On the other hand, (1) implies that
$$
p^{n z+m y} \mid a_{1}^{z} a_{2}^{y+z} b_{1}^{x}+b_{1}^{x+z} b_{2}^{y+z} .
$$
If $n z<m(y+z)$, then (2) gives $p^{n z} \| a_{1}^{z} a_{2}^{y+z} b_{1}^{x}+b_{1}^{x+z} b_{2}^{y+z}$, which contradicts (3). Thus $n z=m(y+z)$ so $n$ is divisible by $k:=\frac{y+z}{\operatorname{gcd}(z, y+z)}>1$. Thus each exponent in the prime decomposition of $a_{1}$ must be divisible by $k$. Hence $a_{1}$ is a perfect $k$-power which means $a$ is powerful. Similarly, $b$ and $c$ are also powerful.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Call a rational number $r$ powerful if $r$ can be expressed in the form $\frac{p^{k}}{q}$ for some relatively prime positive integers $p, q$ and some integer $k>1$. Let $a, b, c$ be positive
rational numbers such that $a b c=1$. Suppose there exist positive integers $x, y, z$ such that $a^{x}+b^{y}+c^{z}$ is an integer. Prove that $a, b, c$ are all powerful.
|
Let $a=\frac{a_{1}}{b_{1}}, b=\frac{a_{2}}{b_{2}}$, where $\operatorname{gcd}\left(a_{1}, b_{1}\right)=\operatorname{gcd}\left(a_{2}, b_{2}\right)=1$. Then $c=\frac{b_{1} b_{2}}{a_{1} a_{2}}$. The condition that $a^{x}+b^{y}+c^{z}$ is an integer becomes
$$
\frac{a_{1}^{x+z} a_{2}^{z} b_{2}^{y}+a_{1}^{z} a_{2}^{y+z} b_{1}^{x}+b_{1}^{x+z} b_{2}^{y+z}}{a_{1}^{z} a_{2}^{z} b_{1}^{x} b_{2}^{y}} \in \mathbb{Z}
$$
which can be restated as
$$
a_{1}^{z} a_{2}^{z} b_{1}^{x} b_{2}^{y} \mid a_{1}^{x+z} a_{2}^{z} b_{2}^{y}+a_{1}^{z} a_{2}^{y+z} b_{1}^{x}+b_{1}^{x+z} b_{2}^{y+z} .
$$
In particular, $a_{1}^{z}$ divides the right-hand side. Since it divides the first and second terms in the sum, we conclude that $a_{1}^{z} \mid b_{1}^{x+z} b_{2}^{y+z}$. Since $\operatorname{gcd}\left(a_{1}, b_{1}\right)=1$, we have $a_{1}^{z} \mid b_{2}^{y+z}$.
Let $p$ be a prime that divides $a_{1}$. Let $m, n \geq 1$ be integers such that $p^{n} \| a_{1}$ (i.e. $p^{n} \mid a_{1}$ but $\left.p^{n+1} \nmid a_{1}\right)$ and $p^{m} \| b_{2}$. The fact that $a_{1}^{z} \mid b_{2}^{y+z}$ implies $n z \leq m(y+z)$. Since $\operatorname{gcd}\left(a_{1}, b_{1}\right)=$ $\operatorname{gcd}\left(a_{2}, b_{2}\right)=1$, we have $p$ does not divide $b_{1}$ and does not divide $a_{2}$. Thus
$$
p^{n z} \| a_{1}^{z} a_{2}^{y+z} b_{1}^{x} \text { and } p^{m(y+z)} \| b_{1}^{x+z} b_{2}^{y+z}
$$
On the other hand, (1) implies that
$$
p^{n z+m y} \mid a_{1}^{z} a_{2}^{y+z} b_{1}^{x}+b_{1}^{x+z} b_{2}^{y+z} .
$$
If $n z<m(y+z)$, then (2) gives $p^{n z} \| a_{1}^{z} a_{2}^{y+z} b_{1}^{x}+b_{1}^{x+z} b_{2}^{y+z}$, which contradicts (3). Thus $n z=m(y+z)$ so $n$ is divisible by $k:=\frac{y+z}{\operatorname{gcd}(z, y+z)}>1$. Thus each exponent in the prime decomposition of $a_{1}$ must be divisible by $k$. Hence $a_{1}$ is a perfect $k$-power which means $a$ is powerful. Similarly, $b$ and $c$ are also powerful.
|
{
"resource_path": "APMO/segmented/en-apmo2017_sol.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "\nSolution."
}
|
38e76064-44d1-57d3-9c7e-f0a4c4f83e4c
| 260,417
|
Let $H$ be the orthocenter of the triangle $A B C$. Let $M$ and $N$ be the midpoints of the sides $A B$ and $A C$, respectively. Assume that $H$ lies inside the quadrilateral $B M N C$ and that the circumcircles of triangles $B M H$ and $C N H$ are tangent to each other. The line through $H$ parallel to $B C$ intersects the circumcircles of the triangles $B M H$ and $C N H$ in the points $K$ and $L$, respectively. Let $F$ be the intersection point of $M K$ and $N L$ and let $J$ be the incenter of triangle $M H N$. Prove that $F J=F A$.
|
Lemma 1. In a triangle $A B C$, let $D$ be the intersection of the interior angle bisector at $A$ with the circumcircle of $A B C$, and let $I$ be the incenter of $\triangle A B C$. Then
$$
D I=D B=D C
$$
Proof.
$$
\angle D B I=\frac{\angle B A C}{2}+\frac{\widehat{B}}{2}=\angle D I B \quad \Rightarrow \quad D I=D B
$$
Analogously $D I=D C$.
We start solving the problem. First we state some position considerations. Since there is an arc of the circumcircle of $B H M$ outside the triangle $A B C$, it must happen that $K$ and $N$ lie on opposite sides of $A M$. Similarly, $L$ and $M$ lie on opposite sides of $A N$. Also, $K$ and $L$ lie on the same side of $M N$, and opposite to $A$. Therefore, $F$ lies inside the triangle $A M N$.
Now, since $H$ is the orthocenter of $\triangle A B C$ and the circumcircles of $B M H$ and $C N H$ are tangent we have
$$
\angle A B H=90^{\circ}-\angle B A C=\angle A C H \quad \Rightarrow \quad \angle M H N=\angle M B H+\angle N C H=180^{\circ}-2 \angle B A C .
$$
So $\angle M B H=\angle M K H=\angle N C H=\angle N L H=90^{\circ}-\angle B A C$ and, since $M N \| K L$, we have
$$
\angle F M N=\angle F N M=90^{\circ}-\angle B A C \Rightarrow \angle M F N=2 \angle B A C .
$$
The relations (1) and (2) yield that the quadrilateral $M F N H$ is cyclic, with the vertices in this order around the circumference. Since $F M=F N, \angle M F N=2 \angle B A C$ and $F$ is the correct side of $M N$ we have that the point $F$ is the circumcenter of triangle $A M N$, and thus $F A=F M=F N$.
Since the quadrilateral $M F N H$ is cyclic, $F M=F N$ and $H$ lies on the correct side of $M N$, we have that $H, J$ and $F$ are collinear. According to Lemma $1, F J=F M=F N$. So $F J=F A$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $H$ be the orthocenter of the triangle $A B C$. Let $M$ and $N$ be the midpoints of the sides $A B$ and $A C$, respectively. Assume that $H$ lies inside the quadrilateral $B M N C$ and that the circumcircles of triangles $B M H$ and $C N H$ are tangent to each other. The line through $H$ parallel to $B C$ intersects the circumcircles of the triangles $B M H$ and $C N H$ in the points $K$ and $L$, respectively. Let $F$ be the intersection point of $M K$ and $N L$ and let $J$ be the incenter of triangle $M H N$. Prove that $F J=F A$.
|
Lemma 1. In a triangle $A B C$, let $D$ be the intersection of the interior angle bisector at $A$ with the circumcircle of $A B C$, and let $I$ be the incenter of $\triangle A B C$. Then
$$
D I=D B=D C
$$
Proof.
$$
\angle D B I=\frac{\angle B A C}{2}+\frac{\widehat{B}}{2}=\angle D I B \quad \Rightarrow \quad D I=D B
$$
Analogously $D I=D C$.
We start solving the problem. First we state some position considerations. Since there is an arc of the circumcircle of $B H M$ outside the triangle $A B C$, it must happen that $K$ and $N$ lie on opposite sides of $A M$. Similarly, $L$ and $M$ lie on opposite sides of $A N$. Also, $K$ and $L$ lie on the same side of $M N$, and opposite to $A$. Therefore, $F$ lies inside the triangle $A M N$.
Now, since $H$ is the orthocenter of $\triangle A B C$ and the circumcircles of $B M H$ and $C N H$ are tangent we have
$$
\angle A B H=90^{\circ}-\angle B A C=\angle A C H \quad \Rightarrow \quad \angle M H N=\angle M B H+\angle N C H=180^{\circ}-2 \angle B A C .
$$
So $\angle M B H=\angle M K H=\angle N C H=\angle N L H=90^{\circ}-\angle B A C$ and, since $M N \| K L$, we have
$$
\angle F M N=\angle F N M=90^{\circ}-\angle B A C \Rightarrow \angle M F N=2 \angle B A C .
$$
The relations (1) and (2) yield that the quadrilateral $M F N H$ is cyclic, with the vertices in this order around the circumference. Since $F M=F N, \angle M F N=2 \angle B A C$ and $F$ is the correct side of $M N$ we have that the point $F$ is the circumcenter of triangle $A M N$, and thus $F A=F M=F N$.
Since the quadrilateral $M F N H$ is cyclic, $F M=F N$ and $H$ lies on the correct side of $M N$, we have that $H, J$ and $F$ are collinear. According to Lemma $1, F J=F M=F N$. So $F J=F A$.
|
{
"resource_path": "APMO/segmented/en-apmo2018_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "# Solution."
}
|
85a4a617-2773-5189-93bd-57b19ded3478
| 605,867
|
Let $H$ be the orthocenter of the triangle $A B C$. Let $M$ and $N$ be the midpoints of the sides $A B$ and $A C$, respectively. Assume that $H$ lies inside the quadrilateral $B M N C$ and that the circumcircles of triangles $B M H$ and $C N H$ are tangent to each other. The line through $H$ parallel to $B C$ intersects the circumcircles of the triangles $B M H$ and $C N H$ in the points $K$ and $L$, respectively. Let $F$ be the intersection point of $M K$ and $N L$ and let $J$ be the incenter of triangle $M H N$. Prove that $F J=F A$.
|
: According to Solution 1, we have $\angle M H N=180^{\circ}-2 \angle B A C$ and since the point $J$ is the incenter of $\triangle M H N$, we have $\angle M J N=90^{\circ}+\frac{1}{2} \angle M H N=180^{\circ}-\angle B A C$. So the quadrilateral $A M J N$ is cyclic.
According to Solution 1, the point $F$ is the circumcenter of $\triangle A M N$. So $F J=F A$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $H$ be the orthocenter of the triangle $A B C$. Let $M$ and $N$ be the midpoints of the sides $A B$ and $A C$, respectively. Assume that $H$ lies inside the quadrilateral $B M N C$ and that the circumcircles of triangles $B M H$ and $C N H$ are tangent to each other. The line through $H$ parallel to $B C$ intersects the circumcircles of the triangles $B M H$ and $C N H$ in the points $K$ and $L$, respectively. Let $F$ be the intersection point of $M K$ and $N L$ and let $J$ be the incenter of triangle $M H N$. Prove that $F J=F A$.
|
: According to Solution 1, we have $\angle M H N=180^{\circ}-2 \angle B A C$ and since the point $J$ is the incenter of $\triangle M H N$, we have $\angle M J N=90^{\circ}+\frac{1}{2} \angle M H N=180^{\circ}-\angle B A C$. So the quadrilateral $A M J N$ is cyclic.
According to Solution 1, the point $F$ is the circumcenter of $\triangle A M N$. So $F J=F A$.
|
{
"resource_path": "APMO/segmented/en-apmo2018_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution 2"
}
|
85a4a617-2773-5189-93bd-57b19ded3478
| 605,867
|
Let $f(x)$ and $g(x)$ be given by
$$
f(x)=\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x-4}+\cdots+\frac{1}{x-2018}
$$
and
$$
g(x)=\frac{1}{x-1}+\frac{1}{x-3}+\frac{1}{x-5}+\cdots+\frac{1}{x-2017} .
$$
Prove that
$$
|f(x)-g(x)|>2
$$
for any non-integer real number $x$ satisfying $0<x<2018$.
|
There are two cases: $2 n-1<x<2 n$ and $2 n<x<2 n+1$. Note that $f(2018-x)=-f(x)$ and $g(2018-x)=-g(x)$, that is, a half turn about the point $(1009,0)$ preserves the graphs of $f$ and $g$. So it suffices to consider only the case $2 n-1<x<2 n$.
Let $d(x)=g(x)-f(x)$. We will show that $d(x)>2$ whenever $2 n-1<x<2 n$ and $n \in\{1,2, \ldots, 1009\}$.
For any non-integer $x$ with $0<x<2018$, we have
$$
d(x+2)-d(x)=\left(\frac{1}{x+1}-\frac{1}{x+2}\right)+\left(\frac{1}{x-2018}-\frac{1}{x-2017}\right)>0+0=0
$$
Hence it suffices to prove $d(x)>2$ for $1<x<2$. Since $x<2$, it follows that $\frac{1}{x-2 i-1}>$ $\frac{1}{x-2 i}$ for $i=2,3, \ldots, 1008$. We also have $\frac{1}{x-2018}<0$. Hence it suffices to prove the following
for $1<x<2$.
$$
\begin{aligned}
& \frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x}-\frac{1}{x-2}>2 \\
\Leftrightarrow & \left(\frac{1}{x-1}+\frac{1}{2-x}\right)+\left(\frac{1}{x-3}-\frac{1}{x}\right)>2 \\
\Leftrightarrow & \frac{1}{(x-1)(2-x)}+\frac{3}{x(x-3)}>2 .
\end{aligned}
$$
By the $G M-H M$ inequality (alternatively, by considering the maximum of the quadratic $(x-1)(2-x))$ we have
$$
\frac{1}{x-1} \cdot \frac{1}{2-x}>\left(\frac{2}{(x-1)+(2-x)}\right)^{2}=4
$$
To find a lower bound for $\frac{3}{x(x-3)}$, note that $x(x-3)<0$ for $1<x<2$. So we seek an upper bound for $x(x-3)$. From the shape of the quadratic, this occurs at $x=1$ or $x=2$, both of which yield $\frac{3}{x(x-3)}>-\frac{3}{2}$.
It follows that $d(x)>4-\frac{3}{2}>2$, as desired.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let $f(x)$ and $g(x)$ be given by
$$
f(x)=\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x-4}+\cdots+\frac{1}{x-2018}
$$
and
$$
g(x)=\frac{1}{x-1}+\frac{1}{x-3}+\frac{1}{x-5}+\cdots+\frac{1}{x-2017} .
$$
Prove that
$$
|f(x)-g(x)|>2
$$
for any non-integer real number $x$ satisfying $0<x<2018$.
|
There are two cases: $2 n-1<x<2 n$ and $2 n<x<2 n+1$. Note that $f(2018-x)=-f(x)$ and $g(2018-x)=-g(x)$, that is, a half turn about the point $(1009,0)$ preserves the graphs of $f$ and $g$. So it suffices to consider only the case $2 n-1<x<2 n$.
Let $d(x)=g(x)-f(x)$. We will show that $d(x)>2$ whenever $2 n-1<x<2 n$ and $n \in\{1,2, \ldots, 1009\}$.
For any non-integer $x$ with $0<x<2018$, we have
$$
d(x+2)-d(x)=\left(\frac{1}{x+1}-\frac{1}{x+2}\right)+\left(\frac{1}{x-2018}-\frac{1}{x-2017}\right)>0+0=0
$$
Hence it suffices to prove $d(x)>2$ for $1<x<2$. Since $x<2$, it follows that $\frac{1}{x-2 i-1}>$ $\frac{1}{x-2 i}$ for $i=2,3, \ldots, 1008$. We also have $\frac{1}{x-2018}<0$. Hence it suffices to prove the following
for $1<x<2$.
$$
\begin{aligned}
& \frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x}-\frac{1}{x-2}>2 \\
\Leftrightarrow & \left(\frac{1}{x-1}+\frac{1}{2-x}\right)+\left(\frac{1}{x-3}-\frac{1}{x}\right)>2 \\
\Leftrightarrow & \frac{1}{(x-1)(2-x)}+\frac{3}{x(x-3)}>2 .
\end{aligned}
$$
By the $G M-H M$ inequality (alternatively, by considering the maximum of the quadratic $(x-1)(2-x))$ we have
$$
\frac{1}{x-1} \cdot \frac{1}{2-x}>\left(\frac{2}{(x-1)+(2-x)}\right)^{2}=4
$$
To find a lower bound for $\frac{3}{x(x-3)}$, note that $x(x-3)<0$ for $1<x<2$. So we seek an upper bound for $x(x-3)$. From the shape of the quadratic, this occurs at $x=1$ or $x=2$, both of which yield $\frac{3}{x(x-3)}>-\frac{3}{2}$.
It follows that $d(x)>4-\frac{3}{2}>2$, as desired.
|
{
"resource_path": "APMO/segmented/en-apmo2018_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "\nSolution 1"
}
|
e9416f41-7b24-51c7-8d9d-4e9f3fd5dc23
| 605,885
|
Let $f(x)$ and $g(x)$ be given by
$$
f(x)=\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x-4}+\cdots+\frac{1}{x-2018}
$$
and
$$
g(x)=\frac{1}{x-1}+\frac{1}{x-3}+\frac{1}{x-5}+\cdots+\frac{1}{x-2017} .
$$
Prove that
$$
|f(x)-g(x)|>2
$$
for any non-integer real number $x$ satisfying $0<x<2018$.
|
As in Solution 1, we may assume $2 n-1<x<2 n$ for some $1 \leq n \leq 1009$. Let $d(x)=$ $f(x)-g(x)$, and note that
$$
d(x)=\frac{1}{x}+\sum_{m=1}^{1009} \frac{1}{(x-2 m)(x-2 m+1)}
$$
We split the sum into three parts: the terms before $m=n$, after $m=n$, and the term $m=n$. The first two are
$$
\begin{aligned}
0 & \leq \sum_{m=1}^{n-1} \frac{1}{(x-2 m)(x-2 m+1)} \\
& \leq \sum_{m=1}^{n-1} \frac{1}{(2 n-1-2 m)(2 n-2 m)}=\sum_{i=1}^{n-1} \frac{1}{(2 i)(2 i-1)} \leq \sum_{i=1}^{1008} \frac{1}{2 i-1}-\frac{1}{2 i} \\
0 & \leq \sum_{m=n+1}^{1009} \frac{1}{(2 m-x)(2 m-1-x)} \\
& \leq \sum_{m=n+1}^{1009} \frac{1}{(2 m-2 n+1)(2 m-2 n)}=\sum_{i=1}^{1009-n} \frac{1}{(2 i+1)(2 i)} \leq \sum_{i=1}^{1008} \frac{1}{2 i}-\frac{1}{2 i+1} .
\end{aligned}
$$
When we add the two sums the terms telescope and we are left with
$$
0 \leq \sum_{1 \leq m \leq 1009, m \neq n} \frac{1}{(x-2 m)(x-2 m+1)} \leq 1-\frac{1}{2017}<1
$$
For the term $m=n$, we write
$$
0<-(x-2 n)(x-2 n+1)=0.25-(x-2 n+0.5)^{2} \leq 0.25
$$
whence
$$
-4 \geq \frac{1}{(x-2 n)(x-2 n+1)}
$$
Finally, $\frac{1}{x}<1$ since $x>2 n-1 \geq 1$. Combining these we get
$$
d(x)=\frac{1}{x}+\sum_{m=1}^{1009} \frac{1}{(x-2 m)(x-2 m+1)}<1+1-4<-2 .
$$
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let $f(x)$ and $g(x)$ be given by
$$
f(x)=\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x-4}+\cdots+\frac{1}{x-2018}
$$
and
$$
g(x)=\frac{1}{x-1}+\frac{1}{x-3}+\frac{1}{x-5}+\cdots+\frac{1}{x-2017} .
$$
Prove that
$$
|f(x)-g(x)|>2
$$
for any non-integer real number $x$ satisfying $0<x<2018$.
|
As in Solution 1, we may assume $2 n-1<x<2 n$ for some $1 \leq n \leq 1009$. Let $d(x)=$ $f(x)-g(x)$, and note that
$$
d(x)=\frac{1}{x}+\sum_{m=1}^{1009} \frac{1}{(x-2 m)(x-2 m+1)}
$$
We split the sum into three parts: the terms before $m=n$, after $m=n$, and the term $m=n$. The first two are
$$
\begin{aligned}
0 & \leq \sum_{m=1}^{n-1} \frac{1}{(x-2 m)(x-2 m+1)} \\
& \leq \sum_{m=1}^{n-1} \frac{1}{(2 n-1-2 m)(2 n-2 m)}=\sum_{i=1}^{n-1} \frac{1}{(2 i)(2 i-1)} \leq \sum_{i=1}^{1008} \frac{1}{2 i-1}-\frac{1}{2 i} \\
0 & \leq \sum_{m=n+1}^{1009} \frac{1}{(2 m-x)(2 m-1-x)} \\
& \leq \sum_{m=n+1}^{1009} \frac{1}{(2 m-2 n+1)(2 m-2 n)}=\sum_{i=1}^{1009-n} \frac{1}{(2 i+1)(2 i)} \leq \sum_{i=1}^{1008} \frac{1}{2 i}-\frac{1}{2 i+1} .
\end{aligned}
$$
When we add the two sums the terms telescope and we are left with
$$
0 \leq \sum_{1 \leq m \leq 1009, m \neq n} \frac{1}{(x-2 m)(x-2 m+1)} \leq 1-\frac{1}{2017}<1
$$
For the term $m=n$, we write
$$
0<-(x-2 n)(x-2 n+1)=0.25-(x-2 n+0.5)^{2} \leq 0.25
$$
whence
$$
-4 \geq \frac{1}{(x-2 n)(x-2 n+1)}
$$
Finally, $\frac{1}{x}<1$ since $x>2 n-1 \geq 1$. Combining these we get
$$
d(x)=\frac{1}{x}+\sum_{m=1}^{1009} \frac{1}{(x-2 m)(x-2 m+1)}<1+1-4<-2 .
$$
|
{
"resource_path": "APMO/segmented/en-apmo2018_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "# Solution 2"
}
|
e9416f41-7b24-51c7-8d9d-4e9f3fd5dc23
| 605,885
|
Let $f(x)$ and $g(x)$ be given by
$$
f(x)=\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x-4}+\cdots+\frac{1}{x-2018}
$$
and
$$
g(x)=\frac{1}{x-1}+\frac{1}{x-3}+\frac{1}{x-5}+\cdots+\frac{1}{x-2017} .
$$
Prove that
$$
|f(x)-g(x)|>2
$$
for any non-integer real number $x$ satisfying $0<x<2018$.
|
First notice that
$$
f(x)-g(x)=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-2}-\cdots-\frac{1}{x-2017}+\frac{1}{x-2018} .
$$
As in Solution 1, we may deal only with the case $2 n<x<2 n+1$. Then $x-2 k+1$ and $x-2 k$ never differ in sign for any integer $k$. Then
$$
\begin{aligned}
-\frac{1}{x-2 k+1}+\frac{1}{x-2 k} & =\frac{1}{(x-2 k+1)(x-2 k)}>0 \quad \text { for } k=1,2, \ldots, n-1, n+2, \ldots, 1009 \\
\frac{1}{x-2 n}-\frac{1}{x-2 n-1} & =\frac{1}{(x-2 n)(2 n+1-x)} \geq\left(\frac{2}{x-2 n+2 n+1-x}\right)^{2}=4
\end{aligned}
$$
Therefore, summing all inequalities and collecting the remaining terms we find $f(x)-g(x)>$ $4+\frac{1}{x-2}>4-1=3$ for $0<x<1$ and, for $n>0$,
$$
\begin{aligned}
f(x)-g(x) & >\frac{1}{x}-\frac{1}{x-2 n+1}+4+\frac{1}{x-2 n-2} \\
& =\frac{1}{x}-\frac{1}{x-2 n+1}+4-\frac{1}{2 n+2-x} \\
& >\frac{1}{x}-\frac{1}{2 n-2 n+1}+4-\frac{1}{2 n+2-2 n-1} \\
& =2+\frac{1}{x}>2 .
\end{aligned}
$$
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let $f(x)$ and $g(x)$ be given by
$$
f(x)=\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x-4}+\cdots+\frac{1}{x-2018}
$$
and
$$
g(x)=\frac{1}{x-1}+\frac{1}{x-3}+\frac{1}{x-5}+\cdots+\frac{1}{x-2017} .
$$
Prove that
$$
|f(x)-g(x)|>2
$$
for any non-integer real number $x$ satisfying $0<x<2018$.
|
First notice that
$$
f(x)-g(x)=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-2}-\cdots-\frac{1}{x-2017}+\frac{1}{x-2018} .
$$
As in Solution 1, we may deal only with the case $2 n<x<2 n+1$. Then $x-2 k+1$ and $x-2 k$ never differ in sign for any integer $k$. Then
$$
\begin{aligned}
-\frac{1}{x-2 k+1}+\frac{1}{x-2 k} & =\frac{1}{(x-2 k+1)(x-2 k)}>0 \quad \text { for } k=1,2, \ldots, n-1, n+2, \ldots, 1009 \\
\frac{1}{x-2 n}-\frac{1}{x-2 n-1} & =\frac{1}{(x-2 n)(2 n+1-x)} \geq\left(\frac{2}{x-2 n+2 n+1-x}\right)^{2}=4
\end{aligned}
$$
Therefore, summing all inequalities and collecting the remaining terms we find $f(x)-g(x)>$ $4+\frac{1}{x-2}>4-1=3$ for $0<x<1$ and, for $n>0$,
$$
\begin{aligned}
f(x)-g(x) & >\frac{1}{x}-\frac{1}{x-2 n+1}+4+\frac{1}{x-2 n-2} \\
& =\frac{1}{x}-\frac{1}{x-2 n+1}+4-\frac{1}{2 n+2-x} \\
& >\frac{1}{x}-\frac{1}{2 n-2 n+1}+4-\frac{1}{2 n+2-2 n-1} \\
& =2+\frac{1}{x}>2 .
\end{aligned}
$$
|
{
"resource_path": "APMO/segmented/en-apmo2018_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "# Solution 3"
}
|
e9416f41-7b24-51c7-8d9d-4e9f3fd5dc23
| 605,885
|
Let $A B C$
be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $B C$. A variable point $P$ is selected in the line segment $A M$. The circumcircles of triangles $B P M$ and $C P M$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $D P$ and $E P$ intersect (a second time) the circumcircles to triangles $C P M$ and $B P M$ at $X$ and $Y$, respectively. Prove that as $P$ varies, the circumcircle of $\triangle A X Y$ passes through a fixed point $T$ distinct from $A$.
|
Let $N$ be the radical center of the circumcircles of triangles $A B C, B M P$ and $C M P$. The pairwise radical axes of these circles are $B D, C E$ and $P M$, and hence they concur at $N$. Now, note that in directed angles:
$$
\angle M C E=\angle M P E=\angle M P Y=\angle M B Y .
$$
It follows that $B Y$ is parallel to $C E$, and analogously that $C X$ is parallel to $B D$. Then, if $L$ is the intersection of $B Y$ and $C X$, it follows that $B N C L$ is a parallelogram. Since $B M=M C$ we deduce that $L$ is the reflection of $N$ with respect to $M$, and therefore $L \in A M$. Using power of a point from $L$ to the circumcircles of triangles $B P M$ and $C P M$, we have
$$
L Y \cdot L B=L P \cdot L M=L X \cdot L C
$$
Hence, $B Y X C$ is cyclic. Using the cyclic quadrilateral we find in directed angles:
$$
\angle L X Y=\angle L B C=\angle B C N=\angle N D E .
$$
Since $C X \| B N$, it follows that $X Y \| D E$.
Let $Q$ and $R$ be two points in $\Gamma$ such that $C Q, B R$, and $A M$ are all parallel. Then in directed angles:
$$
\angle Q D B=\angle Q C B=\angle A M B=\angle P M B=\angle P D B .
$$
Then $D, P, Q$ are collinear. Analogously $E, P, R$ are collinear. From here we get $\angle P R Q=$ $\angle P D E=\angle P X Y$, since $X Y$ and $D E$ are parallel. Therefore $Q R Y X$ is cyclic. Let $S$ be the radical center of the circumcircle of triangle $A B C$ and the circles $B C Y X$ and $Q R Y X$. This point lies in the lines $B C, Q R$ and $X Y$ because these are the radical axes of the circles. Let $T$ be the second intersection of $A S$ with $\Gamma$. By power of a point from $S$ to the circumcircle of $A B C$ and the circle $B C X Y$ we have
$$
S X \cdot S Y=S B \cdot S C=S T \cdot S A
$$
Therefore $T$ is in the circumcircle of triangle $A X Y$. Since $Q$ and $R$ are fixed regardless of the choice of $P$, then $S$ is also fixed, since it is the intersection of $Q R$ and $B C$. This implies $T$ is also fixed, and therefore, the circumcircle of triangle $A X Y$ goes through $T \neq A$ for any choice of $P$.
Now we show an alternative way to prove that $B C X Y$ and $Q R X T$ are cyclic.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$
be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $B C$. A variable point $P$ is selected in the line segment $A M$. The circumcircles of triangles $B P M$ and $C P M$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $D P$ and $E P$ intersect (a second time) the circumcircles to triangles $C P M$ and $B P M$ at $X$ and $Y$, respectively. Prove that as $P$ varies, the circumcircle of $\triangle A X Y$ passes through a fixed point $T$ distinct from $A$.
|
Let $N$ be the radical center of the circumcircles of triangles $A B C, B M P$ and $C M P$. The pairwise radical axes of these circles are $B D, C E$ and $P M$, and hence they concur at $N$. Now, note that in directed angles:
$$
\angle M C E=\angle M P E=\angle M P Y=\angle M B Y .
$$
It follows that $B Y$ is parallel to $C E$, and analogously that $C X$ is parallel to $B D$. Then, if $L$ is the intersection of $B Y$ and $C X$, it follows that $B N C L$ is a parallelogram. Since $B M=M C$ we deduce that $L$ is the reflection of $N$ with respect to $M$, and therefore $L \in A M$. Using power of a point from $L$ to the circumcircles of triangles $B P M$ and $C P M$, we have
$$
L Y \cdot L B=L P \cdot L M=L X \cdot L C
$$
Hence, $B Y X C$ is cyclic. Using the cyclic quadrilateral we find in directed angles:
$$
\angle L X Y=\angle L B C=\angle B C N=\angle N D E .
$$
Since $C X \| B N$, it follows that $X Y \| D E$.
Let $Q$ and $R$ be two points in $\Gamma$ such that $C Q, B R$, and $A M$ are all parallel. Then in directed angles:
$$
\angle Q D B=\angle Q C B=\angle A M B=\angle P M B=\angle P D B .
$$
Then $D, P, Q$ are collinear. Analogously $E, P, R$ are collinear. From here we get $\angle P R Q=$ $\angle P D E=\angle P X Y$, since $X Y$ and $D E$ are parallel. Therefore $Q R Y X$ is cyclic. Let $S$ be the radical center of the circumcircle of triangle $A B C$ and the circles $B C Y X$ and $Q R Y X$. This point lies in the lines $B C, Q R$ and $X Y$ because these are the radical axes of the circles. Let $T$ be the second intersection of $A S$ with $\Gamma$. By power of a point from $S$ to the circumcircle of $A B C$ and the circle $B C X Y$ we have
$$
S X \cdot S Y=S B \cdot S C=S T \cdot S A
$$
Therefore $T$ is in the circumcircle of triangle $A X Y$. Since $Q$ and $R$ are fixed regardless of the choice of $P$, then $S$ is also fixed, since it is the intersection of $Q R$ and $B C$. This implies $T$ is also fixed, and therefore, the circumcircle of triangle $A X Y$ goes through $T \neq A$ for any choice of $P$.
Now we show an alternative way to prove that $B C X Y$ and $Q R X T$ are cyclic.
|
{
"resource_path": "APMO/segmented/en-apmo2019_sol.jsonl",
"problem_match": "# Problem 3.",
"solution_match": "\nSolution."
}
|
936de2fe-2844-5df4-aa45-a257faed6359
| 606,018
|
Let $A B C$
be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $B C$. A variable point $P$ is selected in the line segment $A M$. The circumcircles of triangles $B P M$ and $C P M$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $D P$ and $E P$ intersect (a second time) the circumcircles to triangles $C P M$ and $B P M$ at $X$ and $Y$, respectively. Prove that as $P$ varies, the circumcircle of $\triangle A X Y$ passes through a fixed point $T$ distinct from $A$.
|
. Let the lines $D P$ and $E P$ meet the circumcircle of $A B C$ again at $Q$ and $R$, respectively. Then $\angle D Q C \angle D B C=\angle D P M$, so $Q C \| P M$. Similarly, $R B \| P M$.
Now, $\angle Q C B=\angle P M B=\angle P X C=\angle(Q X, C X)$, which is half of the arc $Q C$ in the circumcircle $\omega_{C}$ of $Q X C$. So $\omega_{C}$ is tangent to $B S$; analogously, $\omega_{B}$, the circumcicle of $R Y B$, is also tangent to $B C$. Since $B R \| C Q$, the inscribed trapezoid $B R Q C$ is isosceles, and by symmetry $Q R$ is also tangent to both circles, and the common perpendicular bisector of $B R$ and $C Q$ passes through the centers of $\omega_{B}$ and $\omega_{C}$. Since $M B=M C$ and $P M\|B R\| C Q$, the line $P M$ is the radical axis of $\omega_{B}$ and $\omega_{C}$.
However, $P M$ is also the radical axis of the circumcircles $\gamma_{B}$ of $P M B$ and $\gamma_{C}$ of $P M C$. Let $C X$ and $P M$ meet at $Z$. Let $p(K, \omega)$ denote the power of a point $K$ with respect to a circumference $\omega$. We have
$$
p\left(Z, \gamma_{B}\right)=p\left(Z, \gamma_{C}\right)=Z X \cdot Z C=p\left(Z, \omega_{B}\right)=p\left(Z, \omega_{C}\right)
$$
Point $Z$ is thus the radical center of $\gamma_{B}, \gamma_{C}, \omega_{B}, \omega_{C}$. Thus, the radical axes $B Y, C X, P M$ meet at $Z$. From here,
$$
\begin{aligned}
& Z Y \cdot Z B=Z C \cdot Z X \Rightarrow B C X Y \text { cyclic } \\
& P Y \cdot P R=P X \cdot P Q \Rightarrow Q R X T \text { cyclic. }
\end{aligned}
$$
We may now finish as in Solution 1.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$
be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $B C$. A variable point $P$ is selected in the line segment $A M$. The circumcircles of triangles $B P M$ and $C P M$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $D P$ and $E P$ intersect (a second time) the circumcircles to triangles $C P M$ and $B P M$ at $X$ and $Y$, respectively. Prove that as $P$ varies, the circumcircle of $\triangle A X Y$ passes through a fixed point $T$ distinct from $A$.
|
. Let the lines $D P$ and $E P$ meet the circumcircle of $A B C$ again at $Q$ and $R$, respectively. Then $\angle D Q C \angle D B C=\angle D P M$, so $Q C \| P M$. Similarly, $R B \| P M$.
Now, $\angle Q C B=\angle P M B=\angle P X C=\angle(Q X, C X)$, which is half of the arc $Q C$ in the circumcircle $\omega_{C}$ of $Q X C$. So $\omega_{C}$ is tangent to $B S$; analogously, $\omega_{B}$, the circumcicle of $R Y B$, is also tangent to $B C$. Since $B R \| C Q$, the inscribed trapezoid $B R Q C$ is isosceles, and by symmetry $Q R$ is also tangent to both circles, and the common perpendicular bisector of $B R$ and $C Q$ passes through the centers of $\omega_{B}$ and $\omega_{C}$. Since $M B=M C$ and $P M\|B R\| C Q$, the line $P M$ is the radical axis of $\omega_{B}$ and $\omega_{C}$.
However, $P M$ is also the radical axis of the circumcircles $\gamma_{B}$ of $P M B$ and $\gamma_{C}$ of $P M C$. Let $C X$ and $P M$ meet at $Z$. Let $p(K, \omega)$ denote the power of a point $K$ with respect to a circumference $\omega$. We have
$$
p\left(Z, \gamma_{B}\right)=p\left(Z, \gamma_{C}\right)=Z X \cdot Z C=p\left(Z, \omega_{B}\right)=p\left(Z, \omega_{C}\right)
$$
Point $Z$ is thus the radical center of $\gamma_{B}, \gamma_{C}, \omega_{B}, \omega_{C}$. Thus, the radical axes $B Y, C X, P M$ meet at $Z$. From here,
$$
\begin{aligned}
& Z Y \cdot Z B=Z C \cdot Z X \Rightarrow B C X Y \text { cyclic } \\
& P Y \cdot P R=P X \cdot P Q \Rightarrow Q R X T \text { cyclic. }
\end{aligned}
$$
We may now finish as in Solution 1.
|
{
"resource_path": "APMO/segmented/en-apmo2019_sol.jsonl",
"problem_match": "# Problem 3.",
"solution_match": "\nSolution 2"
}
|
936de2fe-2844-5df4-aa45-a257faed6359
| 606,018
|
Consider a $2018 \times 2019$ board with integers in each unit square. Two unit squares are said to be neighbours if they share a common edge. In each turn, you choose some unit squares. Then for each chosen unit square the average of all its neighbours is calculated. Finally, after these calculations are done, the number in each chosen unit square is replaced by the corresponding average. Is it always possible to make the numbers in all squares become the same after finitely many turns?
Answer: No
|
Let $n$ be a positive integer relatively prime to 2 and 3 . We may study the whole process modulo $n$ by replacing divisions by $2,3,4$ with multiplications by the corresponding inverses modulo $n$. If at some point the original process makes all the numbers equal, then the process modulo $n$ will also have all the numbers equal. Our aim is to choose $n$ and an initial configuration modulo $n$ for which no process modulo $n$ reaches a board with all numbers equal modulo $n$. We split this goal into two lemmas.
Lemma 1. There is a $2 \times 3$ board that stays constant modulo 5 and whose entries are not all equal.
Proof. Here is one such a board:
The fact that the board remains constant regardless of the choice of squares can be checked square by square.
Lemma 2. If there is an $r \times s$ board with $r \geq 2, s \geq 2$, that stays constant modulo 5 , then there is also a $k r \times l s$ board with the same property.
Proof. We prove by a case by case analysis that repeateadly reflecting the $r \times s$ with respect to an edge preserves the property:
- If a cell had 4 neighbors, after reflections it still has the same neighbors.
- If a cell with $a$ had 3 neighbors $b, c, d$, we have by hypothesis that $a \equiv 3^{-1}(b+c+d) \equiv$ $2(b+c+d)(\bmod 5)$. A reflection may add $a$ as a neighbor of the cell and now
$$
4^{-1}(a+b+c+d) \equiv 4(a+b+c+d) \equiv 4 a+2 a \equiv a \quad(\bmod 5)
$$
- If a cell with $a$ had 2 neighbors $b, c$, we have by hypothesis that $a \equiv 2^{-1}(b+c) \equiv 3(b+c)$ $(\bmod 5)$. If the reflections add one $a$ as neighbor, now
$$
3^{-1}(a+b+c) \equiv 2(3(b+c)+b+c) \equiv 8(b+c) \equiv 3(b+c) \equiv a \quad(\bmod 5)
$$
- If a cell with $a$ had 2 neighbors $b, c$, we have by hypothesis that $a \equiv 2^{-1}(b+c)(\bmod 5)$. If the reflections add two $a$ 's as neighbors, now
$$
4^{-1}(2 a+b+c) \equiv\left(2^{-1} a+2^{-1} a\right) \equiv a \quad(\bmod 5)
$$
In the three cases, any cell is still preserved modulo 5 after an operation. Hence we can fill in the $k r \times l s$ board by $k \times l$ copies by reflection.
Since 2|2018 and 3|2019, we can get through reflections the following board:
By the lemmas above, the board is invariant modulo 5, so the answer is no.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Consider a $2018 \times 2019$ board with integers in each unit square. Two unit squares are said to be neighbours if they share a common edge. In each turn, you choose some unit squares. Then for each chosen unit square the average of all its neighbours is calculated. Finally, after these calculations are done, the number in each chosen unit square is replaced by the corresponding average. Is it always possible to make the numbers in all squares become the same after finitely many turns?
Answer: No
|
Let $n$ be a positive integer relatively prime to 2 and 3 . We may study the whole process modulo $n$ by replacing divisions by $2,3,4$ with multiplications by the corresponding inverses modulo $n$. If at some point the original process makes all the numbers equal, then the process modulo $n$ will also have all the numbers equal. Our aim is to choose $n$ and an initial configuration modulo $n$ for which no process modulo $n$ reaches a board with all numbers equal modulo $n$. We split this goal into two lemmas.
Lemma 1. There is a $2 \times 3$ board that stays constant modulo 5 and whose entries are not all equal.
Proof. Here is one such a board:
The fact that the board remains constant regardless of the choice of squares can be checked square by square.
Lemma 2. If there is an $r \times s$ board with $r \geq 2, s \geq 2$, that stays constant modulo 5 , then there is also a $k r \times l s$ board with the same property.
Proof. We prove by a case by case analysis that repeateadly reflecting the $r \times s$ with respect to an edge preserves the property:
- If a cell had 4 neighbors, after reflections it still has the same neighbors.
- If a cell with $a$ had 3 neighbors $b, c, d$, we have by hypothesis that $a \equiv 3^{-1}(b+c+d) \equiv$ $2(b+c+d)(\bmod 5)$. A reflection may add $a$ as a neighbor of the cell and now
$$
4^{-1}(a+b+c+d) \equiv 4(a+b+c+d) \equiv 4 a+2 a \equiv a \quad(\bmod 5)
$$
- If a cell with $a$ had 2 neighbors $b, c$, we have by hypothesis that $a \equiv 2^{-1}(b+c) \equiv 3(b+c)$ $(\bmod 5)$. If the reflections add one $a$ as neighbor, now
$$
3^{-1}(a+b+c) \equiv 2(3(b+c)+b+c) \equiv 8(b+c) \equiv 3(b+c) \equiv a \quad(\bmod 5)
$$
- If a cell with $a$ had 2 neighbors $b, c$, we have by hypothesis that $a \equiv 2^{-1}(b+c)(\bmod 5)$. If the reflections add two $a$ 's as neighbors, now
$$
4^{-1}(2 a+b+c) \equiv\left(2^{-1} a+2^{-1} a\right) \equiv a \quad(\bmod 5)
$$
In the three cases, any cell is still preserved modulo 5 after an operation. Hence we can fill in the $k r \times l s$ board by $k \times l$ copies by reflection.
Since 2|2018 and 3|2019, we can get through reflections the following board:
By the lemmas above, the board is invariant modulo 5, so the answer is no.
|
{
"resource_path": "APMO/segmented/en-apmo2019_sol.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "\nSolution."
}
|
5cc1ae6c-1b53-5f93-935e-a9744bfa473a
| 261,423
|
Let $\Gamma$ be the circumcircle of $\triangle A B C$. Let $D$ be a point on the side $B C$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $B A$ through $D$ at point $E$. The segment $C E$ intersects $\Gamma$ again at $F$. Suppose $B, D, F, E$ are concyclic. Prove that $A C, B F, D E$ are concurrent.
|
From the conditions, we have
Let $P$ be the intersection of $A C$ and $B F$. Then we have
$$
\angle P A E=\angle C B A=\angle B A C=\angle B F C .
$$
This implies $A, P, F, E$ are concyclic. It follows that
$$
\angle F P E=\angle F A E=\angle F B A,
$$
and hence $A B$ and $E P$ are parallel. So $E, P, D$ are collinear, and the result follows.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $\Gamma$ be the circumcircle of $\triangle A B C$. Let $D$ be a point on the side $B C$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $B A$ through $D$ at point $E$. The segment $C E$ intersects $\Gamma$ again at $F$. Suppose $B, D, F, E$ are concyclic. Prove that $A C, B F, D E$ are concurrent.
|
From the conditions, we have
Let $P$ be the intersection of $A C$ and $B F$. Then we have
$$
\angle P A E=\angle C B A=\angle B A C=\angle B F C .
$$
This implies $A, P, F, E$ are concyclic. It follows that
$$
\angle F P E=\angle F A E=\angle F B A,
$$
and hence $A B$ and $E P$ are parallel. So $E, P, D$ are collinear, and the result follows.
|
{
"resource_path": "APMO/segmented/en-apmo2020_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution 1"
}
|
bfe99559-b2a2-5e18-958e-8e267a77beb3
| 260,974
|
Let $\Gamma$ be the circumcircle of $\triangle A B C$. Let $D$ be a point on the side $B C$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $B A$ through $D$ at point $E$. The segment $C E$ intersects $\Gamma$ again at $F$. Suppose $B, D, F, E$ are concyclic. Prove that $A C, B F, D E$ are concurrent.
|
Let $E^{\prime}$ be any point on the extension of $E A$. From $\angle A E D=\angle E^{\prime} A B=\angle A C D$, points $A, D, C, E$ are concyclic.
Let $P$ be the intersection of $B F$ and $D E$. From $\angle A F P=\angle A C B=\angle A E P$, the points $A, P, F, E$ are concyclic. In addition, from $\angle E P A=\angle E F A=\angle D B A$, points $A, B, D, P$ are concyclic.
By considering the radical centre of $(B D F E),(A P F E)$ and $(B D P A)$, we find that the lines $B D, A P, E F$ are concurrent at $C$. The result follows.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $\Gamma$ be the circumcircle of $\triangle A B C$. Let $D$ be a point on the side $B C$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $B A$ through $D$ at point $E$. The segment $C E$ intersects $\Gamma$ again at $F$. Suppose $B, D, F, E$ are concyclic. Prove that $A C, B F, D E$ are concurrent.
|
Let $E^{\prime}$ be any point on the extension of $E A$. From $\angle A E D=\angle E^{\prime} A B=\angle A C D$, points $A, D, C, E$ are concyclic.
Let $P$ be the intersection of $B F$ and $D E$. From $\angle A F P=\angle A C B=\angle A E P$, the points $A, P, F, E$ are concyclic. In addition, from $\angle E P A=\angle E F A=\angle D B A$, points $A, B, D, P$ are concyclic.
By considering the radical centre of $(B D F E),(A P F E)$ and $(B D P A)$, we find that the lines $B D, A P, E F$ are concurrent at $C$. The result follows.
|
{
"resource_path": "APMO/segmented/en-apmo2020_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution 2"
}
|
bfe99559-b2a2-5e18-958e-8e267a77beb3
| 260,974
|
Show that $r=2$ is the largest real number $r$ which satisfies the following condition:
If a sequence $a_{1}, a_{2}, \ldots$ of positive integers fulfills the inequalities
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+r a_{n+1}}
$$
for every positive integer $n$, then there exists a positive integer $M$ such that $a_{n+2}=a_{n}$ for every $n \geq M$.
|
. First, let us assume that $r>2$, and take a positive integer $a \geq 1 /(r-2)$.
Then, if we let $a_{n}=a+\lfloor n / 2\rfloor$ for $n=1,2, \ldots$, the sequence $a_{n}$ satisfies the inequalities
$$
\sqrt{a_{n}^{2}+r a_{n+1}} \geq \sqrt{a_{n}^{2}+r a_{n}} \geq \sqrt{a_{n}^{2}+\left(2+\frac{1}{a}\right) a_{n}} \geq a_{n}+1=a_{n+2}
$$
but since $a_{n+2}>a_{n}$ for any $n$, we see that $r$ does not satisfy the condition given in the problem.
Now we show that $r=2$ does satisfy the condition of the problem. Suppose $a_{1}, a_{2}, \ldots$ is a sequence of positive integers satisfying the inequalities given in the problem, and there exists a positive integer $m$ for which $a_{m+2}>a_{m}$ is satisfied.
By induction we prove the following assertion:
$$
a_{m+2 k} \leq a_{m+2 k-1}=a_{m+1} \text { holds for every positive integer } k
$$
The truth of $(\dagger)$ for $k=1$ follows from the inequalities below
$$
2 a_{m+2}-1=a_{m+2}^{2}-\left(a_{m+2}-1\right)^{2} \leq a_{m}^{2}+2 a_{m+1}-\left(a_{m+2}-1\right)^{2} \leq 2 a_{m+1}
$$
Let us assume that $(\dagger)$ holds for some positive integer $k$. From
$$
a_{m+1}^{2} \leq a_{m+2 k+1}^{2} \leq a_{m+2 k-1}^{2}+2 a_{m+2 k} \leq a_{m+1}^{2}+2 a_{m+1}<\left(a_{m+1}+1\right)^{2}
$$
it follows that $a_{m+2 k+1}=a_{m+1}$ must hold. Furthermore, since $a_{m+2 k} \leq a_{m+1}$, we have
$$
a_{m+2 k+2}^{2} \leq a_{m+2 k}^{2}+2 a_{m+2 k+1} \leq a_{m+1}^{2}+2 a_{m+1}<\left(a_{m+1}+1\right)^{2}
$$
from which it follows that $a_{m+2 k+2} \leq a_{m+1}$, which proves the assertion $(\dagger)$.
We can conclude that for the value of $m$ with which we started our argument above, $a_{m+2 k+1}=a_{m+1}$ holds for every positive integer $k$. Therefore, in order to finish the proof, it is enough to show that $a_{m+2 k}$ becomes constant after some value of $k$. Since every $a_{m+2 k}$ is a positive integer less than or equal to $a_{m+1}$, there exists $k=K$ for which $a_{m+2 K}$ takes the maximum value. By the monotonicity of $a_{m+2 k}$, it then follows that $a_{m+2 k}=a_{m+2 K}$ for all $k \geq K$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Show that $r=2$ is the largest real number $r$ which satisfies the following condition:
If a sequence $a_{1}, a_{2}, \ldots$ of positive integers fulfills the inequalities
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+r a_{n+1}}
$$
for every positive integer $n$, then there exists a positive integer $M$ such that $a_{n+2}=a_{n}$ for every $n \geq M$.
|
. First, let us assume that $r>2$, and take a positive integer $a \geq 1 /(r-2)$.
Then, if we let $a_{n}=a+\lfloor n / 2\rfloor$ for $n=1,2, \ldots$, the sequence $a_{n}$ satisfies the inequalities
$$
\sqrt{a_{n}^{2}+r a_{n+1}} \geq \sqrt{a_{n}^{2}+r a_{n}} \geq \sqrt{a_{n}^{2}+\left(2+\frac{1}{a}\right) a_{n}} \geq a_{n}+1=a_{n+2}
$$
but since $a_{n+2}>a_{n}$ for any $n$, we see that $r$ does not satisfy the condition given in the problem.
Now we show that $r=2$ does satisfy the condition of the problem. Suppose $a_{1}, a_{2}, \ldots$ is a sequence of positive integers satisfying the inequalities given in the problem, and there exists a positive integer $m$ for which $a_{m+2}>a_{m}$ is satisfied.
By induction we prove the following assertion:
$$
a_{m+2 k} \leq a_{m+2 k-1}=a_{m+1} \text { holds for every positive integer } k
$$
The truth of $(\dagger)$ for $k=1$ follows from the inequalities below
$$
2 a_{m+2}-1=a_{m+2}^{2}-\left(a_{m+2}-1\right)^{2} \leq a_{m}^{2}+2 a_{m+1}-\left(a_{m+2}-1\right)^{2} \leq 2 a_{m+1}
$$
Let us assume that $(\dagger)$ holds for some positive integer $k$. From
$$
a_{m+1}^{2} \leq a_{m+2 k+1}^{2} \leq a_{m+2 k-1}^{2}+2 a_{m+2 k} \leq a_{m+1}^{2}+2 a_{m+1}<\left(a_{m+1}+1\right)^{2}
$$
it follows that $a_{m+2 k+1}=a_{m+1}$ must hold. Furthermore, since $a_{m+2 k} \leq a_{m+1}$, we have
$$
a_{m+2 k+2}^{2} \leq a_{m+2 k}^{2}+2 a_{m+2 k+1} \leq a_{m+1}^{2}+2 a_{m+1}<\left(a_{m+1}+1\right)^{2}
$$
from which it follows that $a_{m+2 k+2} \leq a_{m+1}$, which proves the assertion $(\dagger)$.
We can conclude that for the value of $m$ with which we started our argument above, $a_{m+2 k+1}=a_{m+1}$ holds for every positive integer $k$. Therefore, in order to finish the proof, it is enough to show that $a_{m+2 k}$ becomes constant after some value of $k$. Since every $a_{m+2 k}$ is a positive integer less than or equal to $a_{m+1}$, there exists $k=K$ for which $a_{m+2 K}$ takes the maximum value. By the monotonicity of $a_{m+2 k}$, it then follows that $a_{m+2 k}=a_{m+2 K}$ for all $k \geq K$.
|
{
"resource_path": "APMO/segmented/en-apmo2020_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "\nSolution 1"
}
|
6153a381-e07e-53b5-80de-558f60948be5
| 606,096
|
Show that $r=2$ is the largest real number $r$ which satisfies the following condition:
If a sequence $a_{1}, a_{2}, \ldots$ of positive integers fulfills the inequalities
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+r a_{n+1}}
$$
for every positive integer $n$, then there exists a positive integer $M$ such that $a_{n+2}=a_{n}$ for every $n \geq M$.
|
We only give an alternative proof of the assertion $(\dagger)$ in solution 1 . Let $\left\{a_{n}\right\}$ be a sequence satisfying the inequalities given in the problem. We will use the following key observations:
(a) If $a_{n+1} \leq a_{n}$ for some $n \geq 1$, then
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+2 a_{n+1}}<\sqrt{a_{n}^{2}+2 a_{n}+1}=a_{n}+1,
$$
hence $a_{n}=a_{n+2}$.
(b) If $a_{n} \leq a_{n+1}$ for some $n \geq 1$, then
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+2 a_{n+1}}<\sqrt{a_{n+1}^{2}+2 a_{n+1}+1}=a_{n+1}+1
$$
hence $a_{n} \leq a_{n+2} \leq a_{n+1}$.
Now let $m$ be a positive integer such that $a_{m+2}>a_{m}$. By the observations above, we must have $a_{m}<a_{m+2} \leq a_{m+1}$. Thus the assertion ( $\dagger$ ) is true for $k=1$. Assume that the assertion holds for some positive integer $k$. Using observation (a), we get $a_{m+2 k+1}=a_{m+2 k-1}=a_{m+1}$. Thus $a_{m+2 k} \leq a_{m+2 k+1}$, and then using observation (b), we get $a_{m+2 k+2} \leq a_{m+2 k+1}=a_{m+1}$, which proves the assertion ( $\dagger$ ).
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Show that $r=2$ is the largest real number $r$ which satisfies the following condition:
If a sequence $a_{1}, a_{2}, \ldots$ of positive integers fulfills the inequalities
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+r a_{n+1}}
$$
for every positive integer $n$, then there exists a positive integer $M$ such that $a_{n+2}=a_{n}$ for every $n \geq M$.
|
We only give an alternative proof of the assertion $(\dagger)$ in solution 1 . Let $\left\{a_{n}\right\}$ be a sequence satisfying the inequalities given in the problem. We will use the following key observations:
(a) If $a_{n+1} \leq a_{n}$ for some $n \geq 1$, then
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+2 a_{n+1}}<\sqrt{a_{n}^{2}+2 a_{n}+1}=a_{n}+1,
$$
hence $a_{n}=a_{n+2}$.
(b) If $a_{n} \leq a_{n+1}$ for some $n \geq 1$, then
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+2 a_{n+1}}<\sqrt{a_{n+1}^{2}+2 a_{n+1}+1}=a_{n+1}+1
$$
hence $a_{n} \leq a_{n+2} \leq a_{n+1}$.
Now let $m$ be a positive integer such that $a_{m+2}>a_{m}$. By the observations above, we must have $a_{m}<a_{m+2} \leq a_{m+1}$. Thus the assertion ( $\dagger$ ) is true for $k=1$. Assume that the assertion holds for some positive integer $k$. Using observation (a), we get $a_{m+2 k+1}=a_{m+2 k-1}=a_{m+1}$. Thus $a_{m+2 k} \leq a_{m+2 k+1}$, and then using observation (b), we get $a_{m+2 k+2} \leq a_{m+2 k+1}=a_{m+1}$, which proves the assertion ( $\dagger$ ).
|
{
"resource_path": "APMO/segmented/en-apmo2020_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "# Solution 2"
}
|
6153a381-e07e-53b5-80de-558f60948be5
| 606,096
|
Let $\mathbb{Z}$ denote the set of all integers. Find all polynomials $P(x)$ with integer coefficients that satisfy the following property:
For any infinite sequence $a_{1}, a_{2}, \ldots$ of integers in which each integer in $\mathbb{Z}$ appears exactly once, there exist indices $i<j$ and an integer $k$ such that $a_{i}+a_{i+1}+\cdots+a_{j}=P(k)$.
|
Part 1: All polynomials with $\operatorname{deg} P=1$ satisfy the given property.
Suppose $P(x)=c x+d$, and assume without loss of generality that $c>d \geq 0$. Denote $s_{i}=a_{1}+a_{2}+$ $\cdots+a_{i}(\bmod c)$. It suffices to show that there exist indices $i$ and $j$ such that $j-i \geq 2$ and $s_{j}-s_{i} \equiv d$ $(\bmod c)$.
Consider $c+1$ indices $e_{1}, e_{2}, \ldots, e_{c+1}>1$ such that $a_{e_{l}} \equiv d(\bmod c)$. By the pigeonhole principle, among the $n+1$ pairs $\left(s_{e_{1}-1}, s_{e_{1}}\right),\left(s_{e_{2}-1}, s_{e_{2}}\right), \ldots,\left(s_{e_{n+1}-1}, s_{e_{n+1}}\right)$, some two are equal, say $\left(s_{m-1}, s_{m}\right)$ and $\left(s_{n-1}, s_{n}\right)$. We can then take $i=m-1$ and $j=n$.
Part 2: All polynomials with $\operatorname{deg} P \neq 1$ do not satisfy the given property.
Lemma: If $\operatorname{deg} P \neq 1$, then for any positive integers $A, B$, and $C$, there exists an integer $y$ with $|y|>C$ such that no value in the range of $P$ falls within the interval $[y-A, y+B]$.
Proof of Lemma: The claim is immediate when $P$ is constant or when $\operatorname{deg} P$ is even since $P$ is bounded from below. Let $P(x)=a_{n} x^{n}+\cdots+a_{1} x+a_{0}$ be of odd degree greater than 1 , and assume without
loss of generality that $a_{n}>0$. Since $P(x+1)-P(x)=a_{n} n x^{n-1}+\ldots$, and $n-1>0$, the gap between $P(x)$ and $P(x+1)$ grows arbitrarily for large $x$. The claim follows.
Suppose $\operatorname{deg} P \neq 1$. We will inductively construct a sequence $\left\{a_{i}\right\}$ such that for any indices $i<j$ and any integer $k$ it holds that $a_{i}+a_{i+1}+\cdots+a_{j} \neq P(k)$. Suppose that we have constructed the sequence up to $a_{i}$, and $m$ is an integer with smallest magnitude yet to appear in the sequence. We will add two more terms to the sequence. Take $a_{i+2}=m$. Consider all the new sums of at least two consecutive terms; each of them contains $a_{i+1}$. Hence all such sums are in the interval $\left[a_{i+1}-A, a_{i+1}+B\right]$ for fixed constants $A, B$. The lemma allows us to choose $a_{i+1}$ so that all such sums avoid the range of $P$.
Alternate Solution for Part 1: Again, suppose $P(x)=c x+d$, and assume without loss of generality that $c>d \geq 0$. Let $S_{i}=\left\{a_{j}+a_{j+1}+\cdots+a_{i}(\bmod c) \mid j=1,2, \ldots, i\right\}$. Then $S_{i+1}=\left\{s_{i}+a_{i+1}\right.$ $\left.(\bmod c) \mid s_{i} \in S_{i}\right\} \cup\left\{a_{i+1}(\bmod c)\right\}$. Hence $\left|S_{i+1}\right|=\left|S_{i}\right|$ or $S_{i+1}=\left|S_{i}\right|+1$, with the former occuring exactly when $0 \in S_{i}$. Since $\left|S_{i}\right| \leq c$, the latter can only occur finitely many times, so there exists $I$ such that $0 \in S_{i}$ for all $i \geq I$. Let $t>I$ be an index with $a_{t} \equiv d(\bmod c)$. Then we can find a sum of at least two consecutive terms ending at $a_{t}$ and congruent to $d(\bmod c)$.
Alternate Construction when $P(x)$ is constant or of even degree
If $P(x)$ is of even degree, then $P$ is bounded from below or from above. In case of $P$ is constant or bounded from above, then there exists a positive integer $c$ such that $P(x)<c$. Let $\left\{a_{i}\right\}$ be the sequence
$$
0,1,-1,2,3,-2,4,5,-3, \cdots
$$
which is given by $a_{3 n+1}=2 n, a_{3 n+2}=2 n+1, a_{3 n+3}=-(n+1)$ for all $n \geq 0$. Notice that for any $i<j$ we have $a_{i}+\cdots+a_{j} \geq 0$. Then for the sequence $\left\{b_{n}\right\}$ defined by $b_{n}=a_{n}+c$, clearly $b_{i}+\cdots+b_{j} \geq\left(a_{i}+\cdots+a_{j}\right)+2 c>c$ which is out side the range of $P(x)$.
Now if $P$ is bounded from below, there exist a positive integer $c$ such that $P(x)>-c$. In this case, take $b_{n}$ to be $b_{n}=-a_{n}-c$. Then for all $i<j$ we have $b_{i}+\cdots+b_{j} \leq-\left(a_{1}+\cdots a_{n}\right)-2 c<-c$ which is again out side the range of $P(x)$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $\mathbb{Z}$ denote the set of all integers. Find all polynomials $P(x)$ with integer coefficients that satisfy the following property:
For any infinite sequence $a_{1}, a_{2}, \ldots$ of integers in which each integer in $\mathbb{Z}$ appears exactly once, there exist indices $i<j$ and an integer $k$ such that $a_{i}+a_{i+1}+\cdots+a_{j}=P(k)$.
|
Part 1: All polynomials with $\operatorname{deg} P=1$ satisfy the given property.
Suppose $P(x)=c x+d$, and assume without loss of generality that $c>d \geq 0$. Denote $s_{i}=a_{1}+a_{2}+$ $\cdots+a_{i}(\bmod c)$. It suffices to show that there exist indices $i$ and $j$ such that $j-i \geq 2$ and $s_{j}-s_{i} \equiv d$ $(\bmod c)$.
Consider $c+1$ indices $e_{1}, e_{2}, \ldots, e_{c+1}>1$ such that $a_{e_{l}} \equiv d(\bmod c)$. By the pigeonhole principle, among the $n+1$ pairs $\left(s_{e_{1}-1}, s_{e_{1}}\right),\left(s_{e_{2}-1}, s_{e_{2}}\right), \ldots,\left(s_{e_{n+1}-1}, s_{e_{n+1}}\right)$, some two are equal, say $\left(s_{m-1}, s_{m}\right)$ and $\left(s_{n-1}, s_{n}\right)$. We can then take $i=m-1$ and $j=n$.
Part 2: All polynomials with $\operatorname{deg} P \neq 1$ do not satisfy the given property.
Lemma: If $\operatorname{deg} P \neq 1$, then for any positive integers $A, B$, and $C$, there exists an integer $y$ with $|y|>C$ such that no value in the range of $P$ falls within the interval $[y-A, y+B]$.
Proof of Lemma: The claim is immediate when $P$ is constant or when $\operatorname{deg} P$ is even since $P$ is bounded from below. Let $P(x)=a_{n} x^{n}+\cdots+a_{1} x+a_{0}$ be of odd degree greater than 1 , and assume without
loss of generality that $a_{n}>0$. Since $P(x+1)-P(x)=a_{n} n x^{n-1}+\ldots$, and $n-1>0$, the gap between $P(x)$ and $P(x+1)$ grows arbitrarily for large $x$. The claim follows.
Suppose $\operatorname{deg} P \neq 1$. We will inductively construct a sequence $\left\{a_{i}\right\}$ such that for any indices $i<j$ and any integer $k$ it holds that $a_{i}+a_{i+1}+\cdots+a_{j} \neq P(k)$. Suppose that we have constructed the sequence up to $a_{i}$, and $m$ is an integer with smallest magnitude yet to appear in the sequence. We will add two more terms to the sequence. Take $a_{i+2}=m$. Consider all the new sums of at least two consecutive terms; each of them contains $a_{i+1}$. Hence all such sums are in the interval $\left[a_{i+1}-A, a_{i+1}+B\right]$ for fixed constants $A, B$. The lemma allows us to choose $a_{i+1}$ so that all such sums avoid the range of $P$.
Alternate Solution for Part 1: Again, suppose $P(x)=c x+d$, and assume without loss of generality that $c>d \geq 0$. Let $S_{i}=\left\{a_{j}+a_{j+1}+\cdots+a_{i}(\bmod c) \mid j=1,2, \ldots, i\right\}$. Then $S_{i+1}=\left\{s_{i}+a_{i+1}\right.$ $\left.(\bmod c) \mid s_{i} \in S_{i}\right\} \cup\left\{a_{i+1}(\bmod c)\right\}$. Hence $\left|S_{i+1}\right|=\left|S_{i}\right|$ or $S_{i+1}=\left|S_{i}\right|+1$, with the former occuring exactly when $0 \in S_{i}$. Since $\left|S_{i}\right| \leq c$, the latter can only occur finitely many times, so there exists $I$ such that $0 \in S_{i}$ for all $i \geq I$. Let $t>I$ be an index with $a_{t} \equiv d(\bmod c)$. Then we can find a sum of at least two consecutive terms ending at $a_{t}$ and congruent to $d(\bmod c)$.
Alternate Construction when $P(x)$ is constant or of even degree
If $P(x)$ is of even degree, then $P$ is bounded from below or from above. In case of $P$ is constant or bounded from above, then there exists a positive integer $c$ such that $P(x)<c$. Let $\left\{a_{i}\right\}$ be the sequence
$$
0,1,-1,2,3,-2,4,5,-3, \cdots
$$
which is given by $a_{3 n+1}=2 n, a_{3 n+2}=2 n+1, a_{3 n+3}=-(n+1)$ for all $n \geq 0$. Notice that for any $i<j$ we have $a_{i}+\cdots+a_{j} \geq 0$. Then for the sequence $\left\{b_{n}\right\}$ defined by $b_{n}=a_{n}+c$, clearly $b_{i}+\cdots+b_{j} \geq\left(a_{i}+\cdots+a_{j}\right)+2 c>c$ which is out side the range of $P(x)$.
Now if $P$ is bounded from below, there exist a positive integer $c$ such that $P(x)>-c$. In this case, take $b_{n}$ to be $b_{n}=-a_{n}-c$. Then for all $i<j$ we have $b_{i}+\cdots+b_{j} \leq-\left(a_{1}+\cdots a_{n}\right)-2 c<-c$ which is again out side the range of $P(x)$.
|
{
"resource_path": "APMO/segmented/en-apmo2020_sol.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "# Solution:"
}
|
9da51615-88b2-5097-a788-7bb706eea4af
| 606,135
|
Prove that for each real number $r>2$, there are exactly two or three positive real numbers $x$ satisfying the equation $x^{2}=r\lfloor x\rfloor$.
Note: $\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$
|
Let $r>2$ be a real number. Let $x$ be a positive real number such that $x^{2}=r\lfloor x\rfloor$ with $\lfloor x\rfloor=k$. Since $x>0$ and $x^{2}=r k$, we also have $k>0$. From $k \leq x<k+1$, we get $k^{2} \leq x^{2}=r k<$ $k^{2}+2 k+1 \leq k^{2}+3 k$, hence $k \leq r<k+3$, or $r-3<k \leq r$. There are at most three positive integers in the interval $(r-3, r]$. Thus there are at most three possible values for $k$. Consequently, there are at most three positive solutions to the given equation.
Now suppose that $k$ is a positive integer in the interval $[r-2, r]$. There are at least two such positive integer. Observe that $k \leq \sqrt{r k} \leq \sqrt{(k+2) k}<k+1$ and so $r k=r\lfloor\sqrt{r k}\rfloor$. We conclude that the equation $x^{2}=r\lfloor x\rfloor$ has at least two positive solutions, namely $x=\sqrt{r k}$ with $k \in[r-2, r]$.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Prove that for each real number $r>2$, there are exactly two or three positive real numbers $x$ satisfying the equation $x^{2}=r\lfloor x\rfloor$.
Note: $\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$
|
Let $r>2$ be a real number. Let $x$ be a positive real number such that $x^{2}=r\lfloor x\rfloor$ with $\lfloor x\rfloor=k$. Since $x>0$ and $x^{2}=r k$, we also have $k>0$. From $k \leq x<k+1$, we get $k^{2} \leq x^{2}=r k<$ $k^{2}+2 k+1 \leq k^{2}+3 k$, hence $k \leq r<k+3$, or $r-3<k \leq r$. There are at most three positive integers in the interval $(r-3, r]$. Thus there are at most three possible values for $k$. Consequently, there are at most three positive solutions to the given equation.
Now suppose that $k$ is a positive integer in the interval $[r-2, r]$. There are at least two such positive integer. Observe that $k \leq \sqrt{r k} \leq \sqrt{(k+2) k}<k+1$ and so $r k=r\lfloor\sqrt{r k}\rfloor$. We conclude that the equation $x^{2}=r\lfloor x\rfloor$ has at least two positive solutions, namely $x=\sqrt{r k}$ with $k \in[r-2, r]$.
|
{
"resource_path": "APMO/segmented/en-apmo2021_sol.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution "
}
|
6c726acd-7c5f-56e3-8c41-96274f3501a1
| 606,157
|
Let $A B C D$ be a cyclic convex quadrilateral and $\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals $A C$ and $B D$, let $L$ be the center of the circle tangent to sides $A B, B C$, and $C D$, and let $M$ be the midpoint of the arc $B C$ of $\Gamma$ not containing $A$ and $D$. Prove that the excenter of triangle $B C E$ opposite $E$ lies on the line $L M$.
|
Let $L$ be the intersection of the bisectors of $\angle A B C$ and $\angle B C D$. Let $N$ be the $E$-excenter of $\triangle B C E$. Let $\angle B A C=\angle B D C=\alpha, \angle D B C=\beta$ and $\angle A C B=\gamma$.
We have the following:
$$
\begin{array}{r}
\angle C B L=\frac{1}{2} \angle A B C=90^{\circ}-\frac{1}{2} \alpha-\frac{1}{2} \gamma \text { and } \angle B C L=90^{\circ}-\frac{1}{2} \alpha-\frac{1}{2} \beta, \\
\angle C B N=90^{\circ}-\frac{1}{2} \beta \text { and } \angle B C N=90^{\circ}-\frac{1}{2} \gamma, \\
\angle M B L=\angle M B C+\angle C B L=90^{\circ}-\frac{1}{2} \gamma \text { and } \angle M C L=90^{\circ}-\frac{1}{2} \beta, \\
\angle L C N=\angle L B N=180^{\circ}-\frac{1}{2}(\alpha+\beta+\gamma) .
\end{array}
$$
Applying the sine rule to $\triangle M B L$ and $\triangle M C L$ we obtain
$$
\frac{M B}{M L}=\frac{M C}{M L}=\frac{\sin \angle B L M}{\sin \angle M B L}=\frac{\sin \angle C L M}{\sin \angle M C L}
$$
It follows that
$$
\frac{\sin \angle B L M}{\sin \angle C L M}=\frac{\sin \angle M B L}{\sin \angle M C L}=\frac{\cos (\gamma / 2)}{\cos (\beta / 2)}
$$
Now
$$
\frac{\sin \angle B L M}{\sin \angle M L C} \cdot \frac{\sin \angle L C N}{\sin \angle N C B} \cdot \frac{\sin \angle N B C}{\sin \angle N B L}=\frac{\cos (\gamma / 2)}{\cos (\beta / 2)} \cdot \frac{\sin \left(90^{\circ}-\frac{1}{2} \beta\right)}{\sin \left(90^{\circ}-\frac{1}{2} \gamma\right)}=1
$$
Hence $L M, B N, C N$ are concurrent and therefore $L, M, N$ are collinear.
## Alternative proof
We proceed similarly as above until the equation (1).
We use the following lemma.
Lemma: If $\pi>\alpha, \beta, \gamma, \delta>0, \alpha+\beta=\gamma+\delta<\pi$, and $\frac{\sin \alpha}{\sin \beta}=\frac{\sin \gamma}{\sin \delta}$, then $\alpha=\gamma$ and $\beta=\delta$.
Proof of Lemma: Let $\theta=\alpha+\beta=\gamma+\delta$. Then $\frac{\sin (\theta-\beta)}{\sin \beta}=\frac{\sin (\theta-\delta)}{\sin \delta}$.
$$
\begin{gathered}
\Longleftrightarrow \sin (\theta-\beta) \sin \delta=\sin (\theta-\delta) \sin \beta \\
\Longleftrightarrow(\sin \theta \cos \beta-\sin \beta \cos \theta) \sin \delta=(\sin \theta \cos \delta-\sin \delta \cos \theta) \sin \beta \\
\Longleftrightarrow \sin \theta \cos \beta \sin \delta=\sin \theta \cos \delta \sin \beta \\
\Longleftrightarrow \sin \theta \sin (\beta-\delta)=0
\end{gathered}
$$
Since $0<\theta<\pi$, then $\sin \theta \neq 0$. Therefore, $\sin (\beta-\delta)=0$, and we must have $\beta=\delta$.
Applying the sine rule to $\triangle N B L$ and $\triangle N C L$ we obtain
$$
\begin{aligned}
& \frac{N B}{N L}=\frac{\sin \angle B L N}{\sin \angle L B N} \\
& \frac{N C}{N L}=\frac{\sin \angle C L N}{\sin \angle L C N}
\end{aligned}
$$
Since $\angle L B N=\angle L C N$, it follows that
$$
\frac{\sin \angle B L N}{\sin \angle C L N}=\frac{N B}{N C}=\frac{\sin \angle B C N}{\sin \angle C B N}=\frac{\cos (\gamma / 2)}{\cos (\beta / 2)}=\frac{\sin \angle B L M}{\sin \angle C L M}
$$
By the lemma, it is concluded that $\angle B L M=\angle B L N$ and $\angle C L M=\angle C L N$. Therefore, $L, M, N$ are collinear.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a cyclic convex quadrilateral and $\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals $A C$ and $B D$, let $L$ be the center of the circle tangent to sides $A B, B C$, and $C D$, and let $M$ be the midpoint of the arc $B C$ of $\Gamma$ not containing $A$ and $D$. Prove that the excenter of triangle $B C E$ opposite $E$ lies on the line $L M$.
|
Let $L$ be the intersection of the bisectors of $\angle A B C$ and $\angle B C D$. Let $N$ be the $E$-excenter of $\triangle B C E$. Let $\angle B A C=\angle B D C=\alpha, \angle D B C=\beta$ and $\angle A C B=\gamma$.
We have the following:
$$
\begin{array}{r}
\angle C B L=\frac{1}{2} \angle A B C=90^{\circ}-\frac{1}{2} \alpha-\frac{1}{2} \gamma \text { and } \angle B C L=90^{\circ}-\frac{1}{2} \alpha-\frac{1}{2} \beta, \\
\angle C B N=90^{\circ}-\frac{1}{2} \beta \text { and } \angle B C N=90^{\circ}-\frac{1}{2} \gamma, \\
\angle M B L=\angle M B C+\angle C B L=90^{\circ}-\frac{1}{2} \gamma \text { and } \angle M C L=90^{\circ}-\frac{1}{2} \beta, \\
\angle L C N=\angle L B N=180^{\circ}-\frac{1}{2}(\alpha+\beta+\gamma) .
\end{array}
$$
Applying the sine rule to $\triangle M B L$ and $\triangle M C L$ we obtain
$$
\frac{M B}{M L}=\frac{M C}{M L}=\frac{\sin \angle B L M}{\sin \angle M B L}=\frac{\sin \angle C L M}{\sin \angle M C L}
$$
It follows that
$$
\frac{\sin \angle B L M}{\sin \angle C L M}=\frac{\sin \angle M B L}{\sin \angle M C L}=\frac{\cos (\gamma / 2)}{\cos (\beta / 2)}
$$
Now
$$
\frac{\sin \angle B L M}{\sin \angle M L C} \cdot \frac{\sin \angle L C N}{\sin \angle N C B} \cdot \frac{\sin \angle N B C}{\sin \angle N B L}=\frac{\cos (\gamma / 2)}{\cos (\beta / 2)} \cdot \frac{\sin \left(90^{\circ}-\frac{1}{2} \beta\right)}{\sin \left(90^{\circ}-\frac{1}{2} \gamma\right)}=1
$$
Hence $L M, B N, C N$ are concurrent and therefore $L, M, N$ are collinear.
## Alternative proof
We proceed similarly as above until the equation (1).
We use the following lemma.
Lemma: If $\pi>\alpha, \beta, \gamma, \delta>0, \alpha+\beta=\gamma+\delta<\pi$, and $\frac{\sin \alpha}{\sin \beta}=\frac{\sin \gamma}{\sin \delta}$, then $\alpha=\gamma$ and $\beta=\delta$.
Proof of Lemma: Let $\theta=\alpha+\beta=\gamma+\delta$. Then $\frac{\sin (\theta-\beta)}{\sin \beta}=\frac{\sin (\theta-\delta)}{\sin \delta}$.
$$
\begin{gathered}
\Longleftrightarrow \sin (\theta-\beta) \sin \delta=\sin (\theta-\delta) \sin \beta \\
\Longleftrightarrow(\sin \theta \cos \beta-\sin \beta \cos \theta) \sin \delta=(\sin \theta \cos \delta-\sin \delta \cos \theta) \sin \beta \\
\Longleftrightarrow \sin \theta \cos \beta \sin \delta=\sin \theta \cos \delta \sin \beta \\
\Longleftrightarrow \sin \theta \sin (\beta-\delta)=0
\end{gathered}
$$
Since $0<\theta<\pi$, then $\sin \theta \neq 0$. Therefore, $\sin (\beta-\delta)=0$, and we must have $\beta=\delta$.
Applying the sine rule to $\triangle N B L$ and $\triangle N C L$ we obtain
$$
\begin{aligned}
& \frac{N B}{N L}=\frac{\sin \angle B L N}{\sin \angle L B N} \\
& \frac{N C}{N L}=\frac{\sin \angle C L N}{\sin \angle L C N}
\end{aligned}
$$
Since $\angle L B N=\angle L C N$, it follows that
$$
\frac{\sin \angle B L N}{\sin \angle C L N}=\frac{N B}{N C}=\frac{\sin \angle B C N}{\sin \angle C B N}=\frac{\cos (\gamma / 2)}{\cos (\beta / 2)}=\frac{\sin \angle B L M}{\sin \angle C L M}
$$
By the lemma, it is concluded that $\angle B L M=\angle B L N$ and $\angle C L M=\angle C L N$. Therefore, $L, M, N$ are collinear.
|
{
"resource_path": "APMO/segmented/en-apmo2021_sol.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "# Solution 1"
}
|
1df0f04f-7786-5944-8c9d-b08d52c547dd
| 606,205
|
Let $A B C D$ be a cyclic convex quadrilateral and $\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals $A C$ and $B D$, let $L$ be the center of the circle tangent to sides $A B, B C$, and $C D$, and let $M$ be the midpoint of the arc $B C$ of $\Gamma$ not containing $A$ and $D$. Prove that the excenter of triangle $B C E$ opposite $E$ lies on the line $L M$.
|
Denote by $N$ the excenter of triangle $B C E$ opposite $E$. Since $B L$ bisects $\angle A B C$, we have $\angle C B L=$ $\frac{\angle A B C}{2}$. Since $M$ is the midpoint of arc $B C$, we have $\angle M B C=\frac{1}{2}(\angle M B C+\angle M C B)$ It follows by angle chasing that
$$
\begin{aligned}
\angle M B L & =\angle M B C+\angle C B L=\frac{1}{2}(\angle M B C+\angle M C B+\angle A B C) \\
& =\frac{1}{2}(\angle M B A+\angle M C B)=90^{\circ}-\frac{\angle B C E}{2}=\angle B C N .
\end{aligned}
$$
Denote by $X$ and $Y$ the second intersections of lines $B M$ and $C M$ with the circumcircle of $B C L$, respectively. Since $\angle M B C=\angle M C B$, we have $B C \| X Y$. It suffices to show that $B N \| X L$ and $C N \| Y L$. Indeed, from this it follows that $\triangle B C N \sim \triangle X Y L$, and therefore a homothety with center $M$ that maps $B$ to $X$ and $C$ to $Y$ also maps $N$ to $L$, implying that $N$ lies on the line $L M$.
By symmetry, it suffices to show that $C N \| Y L$, which is equivalent to showing that $\angle B C N=\angle X Y L$. But we have $\angle B C N=\angle M B L=\angle X B L=\angle X Y L$, completing the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a cyclic convex quadrilateral and $\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals $A C$ and $B D$, let $L$ be the center of the circle tangent to sides $A B, B C$, and $C D$, and let $M$ be the midpoint of the arc $B C$ of $\Gamma$ not containing $A$ and $D$. Prove that the excenter of triangle $B C E$ opposite $E$ lies on the line $L M$.
|
Denote by $N$ the excenter of triangle $B C E$ opposite $E$. Since $B L$ bisects $\angle A B C$, we have $\angle C B L=$ $\frac{\angle A B C}{2}$. Since $M$ is the midpoint of arc $B C$, we have $\angle M B C=\frac{1}{2}(\angle M B C+\angle M C B)$ It follows by angle chasing that
$$
\begin{aligned}
\angle M B L & =\angle M B C+\angle C B L=\frac{1}{2}(\angle M B C+\angle M C B+\angle A B C) \\
& =\frac{1}{2}(\angle M B A+\angle M C B)=90^{\circ}-\frac{\angle B C E}{2}=\angle B C N .
\end{aligned}
$$
Denote by $X$ and $Y$ the second intersections of lines $B M$ and $C M$ with the circumcircle of $B C L$, respectively. Since $\angle M B C=\angle M C B$, we have $B C \| X Y$. It suffices to show that $B N \| X L$ and $C N \| Y L$. Indeed, from this it follows that $\triangle B C N \sim \triangle X Y L$, and therefore a homothety with center $M$ that maps $B$ to $X$ and $C$ to $Y$ also maps $N$ to $L$, implying that $N$ lies on the line $L M$.
By symmetry, it suffices to show that $C N \| Y L$, which is equivalent to showing that $\angle B C N=\angle X Y L$. But we have $\angle B C N=\angle M B L=\angle X B L=\angle X Y L$, completing the proof.
|
{
"resource_path": "APMO/segmented/en-apmo2021_sol.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "# Solution 2"
}
|
1df0f04f-7786-5944-8c9d-b08d52c547dd
| 606,205
|
Given a $32 \times 32$ table, we put a mouse (facing up) at the bottom left cell and a piece of cheese at several other cells. The mouse then starts moving. It moves forward except that when it reaches a piece of cheese, it eats a part of it, turns right, and continues moving forward. We say that a subset of cells containing cheese is good if, during this process, the mouse tastes each piece of cheese exactly once and then falls off the table. Show that:
(a) No good subset consists of 888 cells.
(b) There exists a good subset consisting of at least 666 cells.
|
(a) For the sake of contradiction, assume a good subset consisting of 888 cells exists. We call those cheese-cells and the other ones gap-cells. Observe that since each cheese-cell is visited once, each gap-cell is visited at most twice (once vertically and once horizontally). Define a finite sequence $s$ whose $i$-th element is $C$ if the $i$-th step of the mouse was onto a cheese-cell, and $G$ if it was onto a gap-cell. By assumption, $s$ contains $888 C$ 's. Note that $s$ does not contain a contiguous block of 4 (or more) $C$ 's. Hence $s$ contains at least $888 / 3=296$ such $C$-blocks and thus at least $295 G^{\prime}$ 's. But since each gap-cell is traversed at most twice, this implies there are at least $\lceil 295 / 2\rceil=148$ gap-cells, for a total of $888+148=1036>32^{2}$ cells, a contradiction.
(b) Let $L_{i}, X_{i}$ be two $2^{i} \times 2^{i}$ tiles that allow the mouse to "turn left" and "cross", respectively. In detail, the "turn left" tiles allow the mouse to enter at its bottom left cell facing up and to leave at its bottom left cell facing left. The "cross" tiles allow the mouse to enter at its top right facing down and leave at its bottom left facing left, while also to enter at its bottom left facing up and leave at its top right facing right.
(a) Basic tiles
(b) Inductive construction
(c) $16 \times 16$
Note that given two $2^{i} \times 2^{i}$ tiles $L_{i}, X_{i}$ we can construct larger $2^{i+1} \times 2^{i+1}$ tiles $L_{i+1}, X_{i+1}$ inductively as shown on in (b). The construction works because the path intersects itself (or the other path) only inside the smaller $X$-tiles where it works by induction.
For a tile $T$, let $|T|$ be the number of pieces of cheese in it. By straightforward induction, $\left|L_{i}\right|=\left|X_{i}\right|+1$ and $\left|L_{i+1}\right|=4 \cdot\left|L_{i}\right|-1$. From the initial condition $\left|L_{1}\right|=3$. We now easily compute $\left|L_{2}\right|=11,\left|L_{3}\right|=43,\left|L_{4}\right|=171$, and $\left|L_{5}\right|=683$. Hence we get the desired subset.
## Another proof of (a).
Let $X_{N}$ be the largest possible density of cheese-cells in a good subset on an $N \times N$ table. We will show that $X_{N} \leq 4 / 5+o(1)$. Specifically, this gives $X_{32} \leq 817 / 1024$. We look at the (discrete analogue) of the winding number of the trajectory of the mouse. Since the mouse enters and leaves the table, for every 4 right turns in its trajectory there has to be a self-crossing. But each self-crossing requires a different empty square, hence $X_{N} \leq 4 / 5$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Given a $32 \times 32$ table, we put a mouse (facing up) at the bottom left cell and a piece of cheese at several other cells. The mouse then starts moving. It moves forward except that when it reaches a piece of cheese, it eats a part of it, turns right, and continues moving forward. We say that a subset of cells containing cheese is good if, during this process, the mouse tastes each piece of cheese exactly once and then falls off the table. Show that:
(a) No good subset consists of 888 cells.
(b) There exists a good subset consisting of at least 666 cells.
|
(a) For the sake of contradiction, assume a good subset consisting of 888 cells exists. We call those cheese-cells and the other ones gap-cells. Observe that since each cheese-cell is visited once, each gap-cell is visited at most twice (once vertically and once horizontally). Define a finite sequence $s$ whose $i$-th element is $C$ if the $i$-th step of the mouse was onto a cheese-cell, and $G$ if it was onto a gap-cell. By assumption, $s$ contains $888 C$ 's. Note that $s$ does not contain a contiguous block of 4 (or more) $C$ 's. Hence $s$ contains at least $888 / 3=296$ such $C$-blocks and thus at least $295 G^{\prime}$ 's. But since each gap-cell is traversed at most twice, this implies there are at least $\lceil 295 / 2\rceil=148$ gap-cells, for a total of $888+148=1036>32^{2}$ cells, a contradiction.
(b) Let $L_{i}, X_{i}$ be two $2^{i} \times 2^{i}$ tiles that allow the mouse to "turn left" and "cross", respectively. In detail, the "turn left" tiles allow the mouse to enter at its bottom left cell facing up and to leave at its bottom left cell facing left. The "cross" tiles allow the mouse to enter at its top right facing down and leave at its bottom left facing left, while also to enter at its bottom left facing up and leave at its top right facing right.
(a) Basic tiles
(b) Inductive construction
(c) $16 \times 16$
Note that given two $2^{i} \times 2^{i}$ tiles $L_{i}, X_{i}$ we can construct larger $2^{i+1} \times 2^{i+1}$ tiles $L_{i+1}, X_{i+1}$ inductively as shown on in (b). The construction works because the path intersects itself (or the other path) only inside the smaller $X$-tiles where it works by induction.
For a tile $T$, let $|T|$ be the number of pieces of cheese in it. By straightforward induction, $\left|L_{i}\right|=\left|X_{i}\right|+1$ and $\left|L_{i+1}\right|=4 \cdot\left|L_{i}\right|-1$. From the initial condition $\left|L_{1}\right|=3$. We now easily compute $\left|L_{2}\right|=11,\left|L_{3}\right|=43,\left|L_{4}\right|=171$, and $\left|L_{5}\right|=683$. Hence we get the desired subset.
## Another proof of (a).
Let $X_{N}$ be the largest possible density of cheese-cells in a good subset on an $N \times N$ table. We will show that $X_{N} \leq 4 / 5+o(1)$. Specifically, this gives $X_{32} \leq 817 / 1024$. We look at the (discrete analogue) of the winding number of the trajectory of the mouse. Since the mouse enters and leaves the table, for every 4 right turns in its trajectory there has to be a self-crossing. But each self-crossing requires a different empty square, hence $X_{N} \leq 4 / 5$.
|
{
"resource_path": "APMO/segmented/en-apmo2021_sol.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "# Solution."
}
|
57846a27-b487-59b0-aea9-f0c09db3dc6f
| 606,226
|
Let $A B C$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $C B$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $A D$ and let $F$ be the second intersection point of the circumcircle of $\triangle A C D$ and the circumcircle of $\triangle B D E$. Prove that as $D$ varies, the line $E F$ passes through a fixed point.
|
.1
Let the line $E F$ intersect the line $B C$ at $P$ and the circumcircle of $\triangle A C D$ at $G$ distinct from $F$. We will prove that $P$ is the fixed point.
First, notice that $\triangle B E D$ is isosceles with $E B=E D$. This implies $\angle E B C=\angle E D P$.
Then, $\angle D A G=\angle D F G=\angle E B C=\angle E D P$ which implies $A G \| D C$. Hence, $A G C D$ is an isosceles trapezoid.
Also, $A G \| D C$ and $A E=E D$. This implies $\triangle A E G \cong \triangle D E P$ and $A G=D P$.
Since $B$ is the foot of the perpendicular from $A$ onto the side $C D$ of the isosceles trapezoid $A G C D$, we have $P B=P D+D B=A G+D B=B C$, which does not depend on the choice of $D$. Hence, the initial statement is proven.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $C B$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $A D$ and let $F$ be the second intersection point of the circumcircle of $\triangle A C D$ and the circumcircle of $\triangle B D E$. Prove that as $D$ varies, the line $E F$ passes through a fixed point.
|
.1
Let the line $E F$ intersect the line $B C$ at $P$ and the circumcircle of $\triangle A C D$ at $G$ distinct from $F$. We will prove that $P$ is the fixed point.
First, notice that $\triangle B E D$ is isosceles with $E B=E D$. This implies $\angle E B C=\angle E D P$.
Then, $\angle D A G=\angle D F G=\angle E B C=\angle E D P$ which implies $A G \| D C$. Hence, $A G C D$ is an isosceles trapezoid.
Also, $A G \| D C$ and $A E=E D$. This implies $\triangle A E G \cong \triangle D E P$ and $A G=D P$.
Since $B$ is the foot of the perpendicular from $A$ onto the side $C D$ of the isosceles trapezoid $A G C D$, we have $P B=P D+D B=A G+D B=B C$, which does not depend on the choice of $D$. Hence, the initial statement is proven.
|
{
"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "# Solution 2"
}
|
f0000b7c-a3e2-50a5-a918-f70d20c0d7f2
| 606,272
|
Let $A B C$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $C B$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $A D$ and let $F$ be the second intersection point of the circumcircle of $\triangle A C D$ and the circumcircle of $\triangle B D E$. Prove that as $D$ varies, the line $E F$ passes through a fixed point.
|
.2
Set up a coordinate system where $B C$ is along the positive $x$-axis, $B A$ is along the positive $y$-axis, and $B$ is the origin. Take $A=(0, a), B=(0,0), C=(c, 0), D=(-d, 0)$ where $a, b, c, d>0$. Then $E=\left(-\frac{d}{2}, \frac{a}{2}\right)$. The general equation of a circle is
$$
x^{2}+y^{2}+2 f x+2 g y+h=0
$$
Substituting the coordinates of $A, D, C$ into (1) and solving for $f, g, h$, we find that the equation of the circumcircle of $\triangle A D C$ is
$$
x^{2}+y^{2}+(d-c) x+\left(\frac{c d}{a}-a\right) y-c d=0
$$
Similarly, the equation of the circumcircle of $\triangle B D E$ is
$$
x^{2}+y^{2}+d x+\left(\frac{d^{2}}{2 a}-\frac{a}{2}\right) y=0
$$
Then (3)-(2) gives the equation of the line $D F$ which is
$$
c x+\frac{a^{2}+d^{2}-2 c d}{2 a} y+c d=0
$$
Solving (3) and (4) simultaneously, we get
$$
F=\left(\frac{c\left(d^{2}-a^{2}-2 c d\right)}{a^{2}+(d-2 c)^{2}}, \frac{2 a c(c-d)}{a^{2}+(d-2 c)^{2}}\right)
$$
and the other solution $D=(-d, 0)$.
From this we obtain the equation of the line $E F$ which is $a x+(d-2 c) y+a c=0$. It passes through $P(-c, 0)$ which is independent of $d$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $C B$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $A D$ and let $F$ be the second intersection point of the circumcircle of $\triangle A C D$ and the circumcircle of $\triangle B D E$. Prove that as $D$ varies, the line $E F$ passes through a fixed point.
|
.2
Set up a coordinate system where $B C$ is along the positive $x$-axis, $B A$ is along the positive $y$-axis, and $B$ is the origin. Take $A=(0, a), B=(0,0), C=(c, 0), D=(-d, 0)$ where $a, b, c, d>0$. Then $E=\left(-\frac{d}{2}, \frac{a}{2}\right)$. The general equation of a circle is
$$
x^{2}+y^{2}+2 f x+2 g y+h=0
$$
Substituting the coordinates of $A, D, C$ into (1) and solving for $f, g, h$, we find that the equation of the circumcircle of $\triangle A D C$ is
$$
x^{2}+y^{2}+(d-c) x+\left(\frac{c d}{a}-a\right) y-c d=0
$$
Similarly, the equation of the circumcircle of $\triangle B D E$ is
$$
x^{2}+y^{2}+d x+\left(\frac{d^{2}}{2 a}-\frac{a}{2}\right) y=0
$$
Then (3)-(2) gives the equation of the line $D F$ which is
$$
c x+\frac{a^{2}+d^{2}-2 c d}{2 a} y+c d=0
$$
Solving (3) and (4) simultaneously, we get
$$
F=\left(\frac{c\left(d^{2}-a^{2}-2 c d\right)}{a^{2}+(d-2 c)^{2}}, \frac{2 a c(c-d)}{a^{2}+(d-2 c)^{2}}\right)
$$
and the other solution $D=(-d, 0)$.
From this we obtain the equation of the line $E F$ which is $a x+(d-2 c) y+a c=0$. It passes through $P(-c, 0)$ which is independent of $d$.
|
{
"resource_path": "APMO/segmented/en-apmo2022_sol.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "# Solution 2"
}
|
f0000b7c-a3e2-50a5-a918-f70d20c0d7f2
| 606,272
|
Let $n \geq 5$ be an integer. Consider $n$ squares with side lengths $1,2, \ldots, n$, respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices.
Show that it is possible to arrange these squares in a way such that every square touches exactly two other squares.
|
Set aside the squares with sidelengths $n-3, n-2, n-1$, and $n$ and suppose we can split the remaining squares into two sets $A$ and $B$ such that the sum of the sidelengths of the squares in $A$ is 1 or 2 units larger than the sum of the sidelengths of the squares in $B$.
String the squares of each set $A, B$ along two parallel diagonals, one for each diagonal. Now use the four largest squares along two perpendicular diagonals to finish the construction: one will have sidelengths $n$ and $n-3$, and the other, sidelengths $n-1$ and $n-2$. If the sum of the sidelengths of the squares in $A$ is 1 unit larger than the sum of the sidelengths of the squares in $B$, attach the squares with sidelengths $n-3$ and $n-1$ to the $A$-diagonal, and the other two squares to the $B$-diagonal. The resulting configuration, in which the $A$ and $B$-diagonals are represented by unit squares, and the sidelengths $a_{i}$ of squares from $A$ and $b_{j}$ of squares from $B$ are indicated within each square, follows:
Since $\left(a_{1}+a_{2}+\cdots+a_{k}\right) \sqrt{2}+\frac{((n-3)+(n-2)) \sqrt{2}}{2}=\left(b_{1}+b_{2}+\cdots+b_{\ell}+2\right) \sqrt{2}+\frac{(n+(n-1)) \sqrt{2}}{2}$, this case is done.
If the sum of the sidelengths of the squares in $A$ is 1 unit larger than the sum of the sidelengths of the squares in $B$, attach the squares with sidelengths $n-3$ and $n-2$ to the $A$-diagonal, and the other two squares to the $B$-diagonal. The resulting configuration follows:
Since $\left(a_{1}+a_{2}+\cdots+a_{k}\right) \sqrt{2}+\frac{((n-3)+(n-1)) \sqrt{2}}{2}=\left(b_{1}+b_{2}+\cdots+b_{\ell}+1\right) \sqrt{2}+\frac{(n+(n-2)) \sqrt{2}}{2}$, this case is also done.
In both cases, the distance between the $A$-diagonal and the $B$-diagonal is $\frac{((n-3)+n) \sqrt{2}}{2}=\frac{(2 n-3) \sqrt{2}}{2}$. Since $a_{i}, b_{j} \leq n-4, \frac{\left(a_{i}+b_{j}\right) \sqrt{2}}{2}<\frac{(2 n-4) \sqrt{2}}{2}<\frac{(2 n-3) \sqrt{2}}{2}$, and therefore the $A$ - and $B$-diagonals do not overlap.
Finally, we prove that it is possible to split the squares of sidelengths 1 to $n-4$ into two sets $A$ and $B$ such that the sum of the sidelengths of the squares in $A$ is 1 or 2 units larger than the sum of the sidelengths of the squares in $B$. One can do that in several ways; we present two possibilities:
- Direct construction: Split the numbers from 1 to $n-4$ into several sets of four consecutive numbers $\{t, t+1, t+2, t+3\}$, beginning with the largest numbers; put squares of sidelengths $t$ and $t+3$ in $A$ and squares of sidelengths $t+1$ and $t+2$ in $B$. Notice that $t+(t+3)=$ $(t+1)+(t+2)$. In the end, at most four numbers remain.
- If only 1 remains, put the corresponding square in $A$, so the sum of the sidelengths of the squares in $A$ is one unit larger that those in $B$;
- If 1 and 2 remains, put the square of sidelength 2 in $A$ and the square of sidelength 1 in $B$ (the difference is 1 );
- If 1,2 , and 3 remains, put the squares of sidelengths 1 and 3 in $A$, and the square of sidelength 2 in $B$ (the difference is 2 );
- If $1,2,3$, and 4 remains, put the squares of sidelengths 2 and 4 in $A$, and the squares of sidelengths 1 and 3 in $B$ (the difference is 2 ).
- Indirect construction: Starting with $A$ and $B$ as empty sets, add the squares of sidelengths $n-4, n-3, \ldots, 2$ to either $A$ or $B$ in that order such that at each stage the difference between the sum of the sidelengths in $A$ and the sum of the sidelengths of B is minimized. By induction it is clear that after adding an integer $j$ to one of the sets, this difference is at most $j$. In particular, the difference is 0,1 or 2 at the end. Finally adding the final 1 to one of the sets can ensure that the final difference is 1 or 2 . If necessary, flip $A$ and $B$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n \geq 5$ be an integer. Consider $n$ squares with side lengths $1,2, \ldots, n$, respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices.
Show that it is possible to arrange these squares in a way such that every square touches exactly two other squares.
|
Set aside the squares with sidelengths $n-3, n-2, n-1$, and $n$ and suppose we can split the remaining squares into two sets $A$ and $B$ such that the sum of the sidelengths of the squares in $A$ is 1 or 2 units larger than the sum of the sidelengths of the squares in $B$.
String the squares of each set $A, B$ along two parallel diagonals, one for each diagonal. Now use the four largest squares along two perpendicular diagonals to finish the construction: one will have sidelengths $n$ and $n-3$, and the other, sidelengths $n-1$ and $n-2$. If the sum of the sidelengths of the squares in $A$ is 1 unit larger than the sum of the sidelengths of the squares in $B$, attach the squares with sidelengths $n-3$ and $n-1$ to the $A$-diagonal, and the other two squares to the $B$-diagonal. The resulting configuration, in which the $A$ and $B$-diagonals are represented by unit squares, and the sidelengths $a_{i}$ of squares from $A$ and $b_{j}$ of squares from $B$ are indicated within each square, follows:
Since $\left(a_{1}+a_{2}+\cdots+a_{k}\right) \sqrt{2}+\frac{((n-3)+(n-2)) \sqrt{2}}{2}=\left(b_{1}+b_{2}+\cdots+b_{\ell}+2\right) \sqrt{2}+\frac{(n+(n-1)) \sqrt{2}}{2}$, this case is done.
If the sum of the sidelengths of the squares in $A$ is 1 unit larger than the sum of the sidelengths of the squares in $B$, attach the squares with sidelengths $n-3$ and $n-2$ to the $A$-diagonal, and the other two squares to the $B$-diagonal. The resulting configuration follows:
Since $\left(a_{1}+a_{2}+\cdots+a_{k}\right) \sqrt{2}+\frac{((n-3)+(n-1)) \sqrt{2}}{2}=\left(b_{1}+b_{2}+\cdots+b_{\ell}+1\right) \sqrt{2}+\frac{(n+(n-2)) \sqrt{2}}{2}$, this case is also done.
In both cases, the distance between the $A$-diagonal and the $B$-diagonal is $\frac{((n-3)+n) \sqrt{2}}{2}=\frac{(2 n-3) \sqrt{2}}{2}$. Since $a_{i}, b_{j} \leq n-4, \frac{\left(a_{i}+b_{j}\right) \sqrt{2}}{2}<\frac{(2 n-4) \sqrt{2}}{2}<\frac{(2 n-3) \sqrt{2}}{2}$, and therefore the $A$ - and $B$-diagonals do not overlap.
Finally, we prove that it is possible to split the squares of sidelengths 1 to $n-4$ into two sets $A$ and $B$ such that the sum of the sidelengths of the squares in $A$ is 1 or 2 units larger than the sum of the sidelengths of the squares in $B$. One can do that in several ways; we present two possibilities:
- Direct construction: Split the numbers from 1 to $n-4$ into several sets of four consecutive numbers $\{t, t+1, t+2, t+3\}$, beginning with the largest numbers; put squares of sidelengths $t$ and $t+3$ in $A$ and squares of sidelengths $t+1$ and $t+2$ in $B$. Notice that $t+(t+3)=$ $(t+1)+(t+2)$. In the end, at most four numbers remain.
- If only 1 remains, put the corresponding square in $A$, so the sum of the sidelengths of the squares in $A$ is one unit larger that those in $B$;
- If 1 and 2 remains, put the square of sidelength 2 in $A$ and the square of sidelength 1 in $B$ (the difference is 1 );
- If 1,2 , and 3 remains, put the squares of sidelengths 1 and 3 in $A$, and the square of sidelength 2 in $B$ (the difference is 2 );
- If $1,2,3$, and 4 remains, put the squares of sidelengths 2 and 4 in $A$, and the squares of sidelengths 1 and 3 in $B$ (the difference is 2 ).
- Indirect construction: Starting with $A$ and $B$ as empty sets, add the squares of sidelengths $n-4, n-3, \ldots, 2$ to either $A$ or $B$ in that order such that at each stage the difference between the sum of the sidelengths in $A$ and the sum of the sidelengths of B is minimized. By induction it is clear that after adding an integer $j$ to one of the sets, this difference is at most $j$. In particular, the difference is 0,1 or 2 at the end. Finally adding the final 1 to one of the sets can ensure that the final difference is 1 or 2 . If necessary, flip $A$ and $B$.
|
{
"resource_path": "APMO/segmented/en-apmo2023_sol.jsonl",
"problem_match": "# Problem 1",
"solution_match": "# Solution 1"
}
|
771f559a-5e7e-5f34-b318-cd602a03c47f
| 606,314
|
Let $n \geq 5$ be an integer. Consider $n$ squares with side lengths $1,2, \ldots, n$, respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices.
Show that it is possible to arrange these squares in a way such that every square touches exactly two other squares.
|
Solve the problem by induction in $n$. Construct examples for $n=5,6,7,8,9,10$ (one can use the constructions from the previous solution, for instance). For $n>10$, set aside the six larger squares and arrange them in the following fashion:
By the induction hypothesis, one can arrange the remaining $n-6$ squares away from the six larger squares, so we are done.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n \geq 5$ be an integer. Consider $n$ squares with side lengths $1,2, \ldots, n$, respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices.
Show that it is possible to arrange these squares in a way such that every square touches exactly two other squares.
|
Solve the problem by induction in $n$. Construct examples for $n=5,6,7,8,9,10$ (one can use the constructions from the previous solution, for instance). For $n>10$, set aside the six larger squares and arrange them in the following fashion:
By the induction hypothesis, one can arrange the remaining $n-6$ squares away from the six larger squares, so we are done.
|
{
"resource_path": "APMO/segmented/en-apmo2023_sol.jsonl",
"problem_match": "# Problem 1",
"solution_match": "# Solution 2"
}
|
771f559a-5e7e-5f34-b318-cd602a03c47f
| 606,314
|
Let $A B C D$ be a parallelogram. Let $W, X, Y$, and $Z$ be points on sides $A B, B C, C D$, and $D A$, respectively, such that the incenters of triangles $A W Z, B X W, C Y X$ and $D Z Y$ form a parallelogram. Prove that $W X Y Z$ is a parallelogram.
|
Let the four incenters be $I_{1}, I_{2}, I_{3}$, and $I_{4}$ with inradii $r_{1}, r_{2}, r_{3}$, and $r_{4}$ respectively (in the order given in the question). Without loss of generality, let $I_{1}$ be closer to $A B$ than $I_{2}$. Let the acute angle between $I_{1} I_{2}$ and $A B$ (and hence also the angle between $I_{3} I_{4}$ and $C D$ ) be $\theta$. Then
$$
r_{2}-r_{1}=I_{1} I_{2} \sin \theta=I_{3} I_{4} \sin \theta=r_{4}-r_{3}
$$
which implies $r_{1}+r_{4}=r_{2}+r_{3}$. Similar arguments show that $r_{1}+r_{2}=r_{3}+r_{4}$. Thus we obtain $r_{1}=r_{3}$ and $r_{2}=r_{4}$.
Now let's consider the possible positions of $W, X, Y, Z$. Suppose $A Z \neq C X$. Without loss of generality assume $A Z>C X$. Since the incircles of $A W Z$ and $C Y X$ are symmetric about the centre of the parallelogram $A B C D$, this implies $C Y>A W$. Using similar arguments, we have
$$
C Y>A W \Longrightarrow B W>D Y \Longrightarrow D Z>B X \Longrightarrow C X>A Z
$$
which is a contradiction. Therefore $A Z=C X \Longrightarrow A W=C Y$ and $W X Y Z$ is a parallelogram.
Comment: There are several ways to prove that $r_{1}=r_{3}$ and $r_{2}=r_{4}$. The proposer shows the following three alternative approaches:
Using parallel lines: Let $O$ be the centre of parallelogram $A B C D$ and $P$ be the centre of parallelogram $I_{1} I_{2} I_{3} I_{4}$. Since $A I_{1}$ and $C I_{3}$ are angle bisectors, we must have $A I_{1} \| C I_{3}$. Let $\ell_{1}$ be the line through $O$ parallel to $A I_{1}$. Since $A O=O C, \ell_{1}$ is halfway between $A I_{1}$ and $C I_{3}$. Hence $P$ must lie on $\ell_{1}$.
Similarly, $P$ must also lie on $\ell_{2}$, the line through $O$ parallel to $B I_{2}$. Thus $P$ is the intersection of $\ell_{1}$ and $\ell_{2}$, which must be $O$. So the four incentres and hence the four incircles must be symmetric about $O$, which implies $r_{1}=r_{3}$ and $r_{2}=r_{4}$.
Using a rotation: Let the bisectors of $\angle D A B$ and $\angle A B C$ meet at $X$ and the bisectors of $\angle B C D$ and $\angle C D A$ meet at $Y$. Then $I_{1}$ is on $A X, I_{2}$ is on $B X, I_{3}$ is on $C Y$, and $I_{4}$ is on $D Y$. Let $O$ be the centre of $A B C D$. Then a 180 degree rotation about $O$ takes $\triangle A X B$ to $\triangle C Y D$. Under the same transformation $I_{1} I_{2}$ is mapped to a parallel segment $I_{1}^{\prime} I_{2}^{\prime}$ with $I_{1}^{\prime}$ on $C Y$ and $I_{2}^{\prime}$ on $D Y$. Since $I_{1} I_{2} I_{3} I_{4}$ is a parallelogram, $I_{3} I_{4}=I_{1} I_{2}$ and $I_{3} I_{4} \| I_{1} I_{2}$. Hence $I_{1}^{\prime} I_{2}^{\prime}$ and $I_{3} I_{4}$ are parallel, equal length segments on sides $C Y, D Y$ and we conclude that $I_{1}^{\prime}=I_{3}, I_{2}^{\prime}=I_{4}$. Hence the centre of $I_{1} I_{2} I_{3} I_{4}$ is also $O$ and we establish that by rotational symmetry that $r_{1}=r_{3}$ and $r_{2}=r_{4}$.
Using congruent triangles: Let $A I_{1}$ and $B I_{2}$ intersect at $E$ and let $C I_{3}$ and $D I_{4}$ intersect at $F$. Note that $\triangle A B E$ and $\triangle C D F$ are congruent, since $A B=C D$ and corresponding pairs of angles are equal (equal opposite angles parallelogram $A B C D$ are each bisected).
Since $A I_{1} \| C I_{3}$ and $I_{1} I_{2} \| I_{4} I_{3}, \angle I_{2} I_{1} E=\angle I_{4} I_{3} F$. Similarly $\angle I_{1} I_{2} E=\angle I_{3} I_{4} F$. Furthermore $I_{1} I_{2}=I_{3} I_{4}$. Hence triangles $I_{2} I_{1} E$ and $I 4 I_{3} F$ are also congruent.
Hence $A B E I_{1} I_{2}$ and $D C F I_{3} I_{4}$ are congruent. Therefore, the perpendicular distance from $I_{1}$ to $A B$ equals the perpendicular distance from $I_{3}$ to $C D$, that is, $r_{1}=r_{3}$. Similarly $r_{2}=r_{4}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a parallelogram. Let $W, X, Y$, and $Z$ be points on sides $A B, B C, C D$, and $D A$, respectively, such that the incenters of triangles $A W Z, B X W, C Y X$ and $D Z Y$ form a parallelogram. Prove that $W X Y Z$ is a parallelogram.
|
Let the four incenters be $I_{1}, I_{2}, I_{3}$, and $I_{4}$ with inradii $r_{1}, r_{2}, r_{3}$, and $r_{4}$ respectively (in the order given in the question). Without loss of generality, let $I_{1}$ be closer to $A B$ than $I_{2}$. Let the acute angle between $I_{1} I_{2}$ and $A B$ (and hence also the angle between $I_{3} I_{4}$ and $C D$ ) be $\theta$. Then
$$
r_{2}-r_{1}=I_{1} I_{2} \sin \theta=I_{3} I_{4} \sin \theta=r_{4}-r_{3}
$$
which implies $r_{1}+r_{4}=r_{2}+r_{3}$. Similar arguments show that $r_{1}+r_{2}=r_{3}+r_{4}$. Thus we obtain $r_{1}=r_{3}$ and $r_{2}=r_{4}$.
Now let's consider the possible positions of $W, X, Y, Z$. Suppose $A Z \neq C X$. Without loss of generality assume $A Z>C X$. Since the incircles of $A W Z$ and $C Y X$ are symmetric about the centre of the parallelogram $A B C D$, this implies $C Y>A W$. Using similar arguments, we have
$$
C Y>A W \Longrightarrow B W>D Y \Longrightarrow D Z>B X \Longrightarrow C X>A Z
$$
which is a contradiction. Therefore $A Z=C X \Longrightarrow A W=C Y$ and $W X Y Z$ is a parallelogram.
Comment: There are several ways to prove that $r_{1}=r_{3}$ and $r_{2}=r_{4}$. The proposer shows the following three alternative approaches:
Using parallel lines: Let $O$ be the centre of parallelogram $A B C D$ and $P$ be the centre of parallelogram $I_{1} I_{2} I_{3} I_{4}$. Since $A I_{1}$ and $C I_{3}$ are angle bisectors, we must have $A I_{1} \| C I_{3}$. Let $\ell_{1}$ be the line through $O$ parallel to $A I_{1}$. Since $A O=O C, \ell_{1}$ is halfway between $A I_{1}$ and $C I_{3}$. Hence $P$ must lie on $\ell_{1}$.
Similarly, $P$ must also lie on $\ell_{2}$, the line through $O$ parallel to $B I_{2}$. Thus $P$ is the intersection of $\ell_{1}$ and $\ell_{2}$, which must be $O$. So the four incentres and hence the four incircles must be symmetric about $O$, which implies $r_{1}=r_{3}$ and $r_{2}=r_{4}$.
Using a rotation: Let the bisectors of $\angle D A B$ and $\angle A B C$ meet at $X$ and the bisectors of $\angle B C D$ and $\angle C D A$ meet at $Y$. Then $I_{1}$ is on $A X, I_{2}$ is on $B X, I_{3}$ is on $C Y$, and $I_{4}$ is on $D Y$. Let $O$ be the centre of $A B C D$. Then a 180 degree rotation about $O$ takes $\triangle A X B$ to $\triangle C Y D$. Under the same transformation $I_{1} I_{2}$ is mapped to a parallel segment $I_{1}^{\prime} I_{2}^{\prime}$ with $I_{1}^{\prime}$ on $C Y$ and $I_{2}^{\prime}$ on $D Y$. Since $I_{1} I_{2} I_{3} I_{4}$ is a parallelogram, $I_{3} I_{4}=I_{1} I_{2}$ and $I_{3} I_{4} \| I_{1} I_{2}$. Hence $I_{1}^{\prime} I_{2}^{\prime}$ and $I_{3} I_{4}$ are parallel, equal length segments on sides $C Y, D Y$ and we conclude that $I_{1}^{\prime}=I_{3}, I_{2}^{\prime}=I_{4}$. Hence the centre of $I_{1} I_{2} I_{3} I_{4}$ is also $O$ and we establish that by rotational symmetry that $r_{1}=r_{3}$ and $r_{2}=r_{4}$.
Using congruent triangles: Let $A I_{1}$ and $B I_{2}$ intersect at $E$ and let $C I_{3}$ and $D I_{4}$ intersect at $F$. Note that $\triangle A B E$ and $\triangle C D F$ are congruent, since $A B=C D$ and corresponding pairs of angles are equal (equal opposite angles parallelogram $A B C D$ are each bisected).
Since $A I_{1} \| C I_{3}$ and $I_{1} I_{2} \| I_{4} I_{3}, \angle I_{2} I_{1} E=\angle I_{4} I_{3} F$. Similarly $\angle I_{1} I_{2} E=\angle I_{3} I_{4} F$. Furthermore $I_{1} I_{2}=I_{3} I_{4}$. Hence triangles $I_{2} I_{1} E$ and $I 4 I_{3} F$ are also congruent.
Hence $A B E I_{1} I_{2}$ and $D C F I_{3} I_{4}$ are congruent. Therefore, the perpendicular distance from $I_{1}$ to $A B$ equals the perpendicular distance from $I_{3}$ to $C D$, that is, $r_{1}=r_{3}$. Similarly $r_{2}=r_{4}$.
|
{
"resource_path": "APMO/segmented/en-apmo2023_sol.jsonl",
"problem_match": "# Problem 3",
"solution_match": "# Solution\n\n"
}
|
5f56a53f-2cd7-5611-aa33-44b1c7c296f2
| 606,319
|
There are $n$ line segments on the plane, no three intersecting at a point, and each pair intersecting once in their respective interiors. Tony and his $2 n-1$ friends each stand at a distinct endpoint of a line segment. Tony wishes to send Christmas presents to each of his friends as follows:
First, he chooses an endpoint of each segment as a "sink". Then he places the present at the endpoint of the segment he is at. The present moves as follows:
- If it is on a line segment, it moves towards the sink.
- When it reaches an intersection of two segments, it changes the line segment it travels on and starts moving towards the new sink.
If the present reaches an endpoint, the friend on that endpoint can receive their present. Prove Tony can send presents to exactly $n$ of his $2 n-1$ friends.
|
First part: at most n friends can receive a present.
Similarly to the first solution, consider a circle that encompasses all line segments, extend the lines, and use the endpoints of the chords instead of the line segments, and prove that each chord connects vertices $k$ and $k+n$. We also consider, even in the first part, $n$ presents leaving from $n$ outcoming vertices.
First we prove that a present always goes to a sink. If it does not, then it loops; let it first enter the loop at point $P$ after turning from chord $a$ to chord $b$. Therefore after it loops once,
[^0]it must turn to chord $b$ at $P$. But $P$ is the intersection of $a$ and $b$, so the present should turn from chord $a$ to chord $b$, which can only be done in one way - the same way it came in first. This means that some part of chord $a$ before the present enters the loop at $P$ is part of the loop, which contradicts the fact that $P$ is the first point in the loop. So no present enters a loop, and every present goes to a sink.
There are no loops
No two paths cross
The present paths also do not cross: in fact, every time two paths share a point $P$, intersection of chords $a$ and $b$, one path comes from $a$ to $b$ and the other path comes from $b$ to $a$, and they touch at $P$. This implies the following sequence of facts:
- Every path divides the circle into two regions with paths connecting vertices within each region.
- All $n$ presents will be delivered to $n$ different persons; that is, all sinks receive a present. This implies that every vertex is an endpoint of a path.
- The number of chord endpoints inside each region is even, because they are connected within their own region.
Now consider the path starting at vertex 1 , with Tony. It divides the circle into two regions with an even number of vertices in their interior. Then there is an even number of vertices between Tony and the recipient of his present, that is, their vertex is an even numbered one. Second part: all even-numbered vertices can receive a present.
The construction is the same as the in the previous solution: direct each chord from $i$ to $i+n$ if $1 \leq i \leq k$ and from $i+n$ to $i$ otherwise; in other words, the sinks are $k+1, k+2, \ldots, k+n$. Then, since the paths do not cross, $k$ will send a present to $k+1, k-1$ will send a present to $k+2$, and so on, until 1 sends a present to $(k+1)+(k-1)=2 k$.
[^0]: ${ }^{1}$ The paths do not depend uniquely on $k$ and $n$; different chord configurations and vertex labelings may change the paths.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
There are $n$ line segments on the plane, no three intersecting at a point, and each pair intersecting once in their respective interiors. Tony and his $2 n-1$ friends each stand at a distinct endpoint of a line segment. Tony wishes to send Christmas presents to each of his friends as follows:
First, he chooses an endpoint of each segment as a "sink". Then he places the present at the endpoint of the segment he is at. The present moves as follows:
- If it is on a line segment, it moves towards the sink.
- When it reaches an intersection of two segments, it changes the line segment it travels on and starts moving towards the new sink.
If the present reaches an endpoint, the friend on that endpoint can receive their present. Prove Tony can send presents to exactly $n$ of his $2 n-1$ friends.
|
First part: at most n friends can receive a present.
Similarly to the first solution, consider a circle that encompasses all line segments, extend the lines, and use the endpoints of the chords instead of the line segments, and prove that each chord connects vertices $k$ and $k+n$. We also consider, even in the first part, $n$ presents leaving from $n$ outcoming vertices.
First we prove that a present always goes to a sink. If it does not, then it loops; let it first enter the loop at point $P$ after turning from chord $a$ to chord $b$. Therefore after it loops once,
[^0]it must turn to chord $b$ at $P$. But $P$ is the intersection of $a$ and $b$, so the present should turn from chord $a$ to chord $b$, which can only be done in one way - the same way it came in first. This means that some part of chord $a$ before the present enters the loop at $P$ is part of the loop, which contradicts the fact that $P$ is the first point in the loop. So no present enters a loop, and every present goes to a sink.
There are no loops
No two paths cross
The present paths also do not cross: in fact, every time two paths share a point $P$, intersection of chords $a$ and $b$, one path comes from $a$ to $b$ and the other path comes from $b$ to $a$, and they touch at $P$. This implies the following sequence of facts:
- Every path divides the circle into two regions with paths connecting vertices within each region.
- All $n$ presents will be delivered to $n$ different persons; that is, all sinks receive a present. This implies that every vertex is an endpoint of a path.
- The number of chord endpoints inside each region is even, because they are connected within their own region.
Now consider the path starting at vertex 1 , with Tony. It divides the circle into two regions with an even number of vertices in their interior. Then there is an even number of vertices between Tony and the recipient of his present, that is, their vertex is an even numbered one. Second part: all even-numbered vertices can receive a present.
The construction is the same as the in the previous solution: direct each chord from $i$ to $i+n$ if $1 \leq i \leq k$ and from $i+n$ to $i$ otherwise; in other words, the sinks are $k+1, k+2, \ldots, k+n$. Then, since the paths do not cross, $k$ will send a present to $k+1, k-1$ will send a present to $k+2$, and so on, until 1 sends a present to $(k+1)+(k-1)=2 k$.
[^0]: ${ }^{1}$ The paths do not depend uniquely on $k$ and $n$; different chord configurations and vertex labelings may change the paths.
|
{
"resource_path": "APMO/segmented/en-apmo2023_sol.jsonl",
"problem_match": "# Problem 5",
"solution_match": "# Solution 2"
}
|
86f2f1e7-71d9-54b7-9207-e735b0c224a2
| 606,325
|
Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $B Q X$ and $C P X$ intersect at a point $Y \neq X$. Prove that points $A, X$, and $Y$ are collinear.
|
Let $\ell$ be the radical axis of circles $B Q X$ and $C P X$. Since $X$ and $Y$ are on $\ell$, it is sufficient to show that $A$ is on $\ell$. Let line $A X$ intersect segments $B C$ and $D E$ at $Z$ and $Z^{\prime}$, respectively. Then it is sufficient to show that $Z$ is on $\ell$. By $B C \| D E$, we obtain
$$
\frac{B Z}{Z C}=\frac{D Z^{\prime}}{Z^{\prime} E}=\frac{P Z}{Z Q},
$$
thus $B Z \cdot Q Z=C Z \cdot P Z$, which implies that $Z$ is on $\ell$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $B Q X$ and $C P X$ intersect at a point $Y \neq X$. Prove that points $A, X$, and $Y$ are collinear.
|
Let $\ell$ be the radical axis of circles $B Q X$ and $C P X$. Since $X$ and $Y$ are on $\ell$, it is sufficient to show that $A$ is on $\ell$. Let line $A X$ intersect segments $B C$ and $D E$ at $Z$ and $Z^{\prime}$, respectively. Then it is sufficient to show that $Z$ is on $\ell$. By $B C \| D E$, we obtain
$$
\frac{B Z}{Z C}=\frac{D Z^{\prime}}{Z^{\prime} E}=\frac{P Z}{Z Q},
$$
thus $B Z \cdot Q Z=C Z \cdot P Z$, which implies that $Z$ is on $\ell$.
|
{
"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl",
"problem_match": "# Problem 1",
"solution_match": "# Solution 1"
}
|
c2662896-d3ab-5e68-bb60-20bb487f6879
| 606,331
|
Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $B Q X$ and $C P X$ intersect at a point $Y \neq X$. Prove that points $A, X$, and $Y$ are collinear.
|
Let circle $B Q X$ intersect line $A B$ at a point $S$ which is different from $B$. Then $\angle D E X=$ $\angle X Q C=\angle B S X$, thus $S$ is on circle $D E X$. Similarly, let circle $C P X$ intersect line $A C$ at a point $T$ which is different from $C$. Then $T$ is on circle $D E X$. The power of $A$ with respect to the circle $D E X$ is $A S \cdot A D=A T \cdot A E$. Since $\frac{A D}{A B}=\frac{A E}{A C}, A S \cdot A B=A T \cdot A C$. Then $A$ is in the radical axis of circles $B Q X$ and $C P X$, which implies that three points $A, X$ and $Y$ are collinear.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $B Q X$ and $C P X$ intersect at a point $Y \neq X$. Prove that points $A, X$, and $Y$ are collinear.
|
Let circle $B Q X$ intersect line $A B$ at a point $S$ which is different from $B$. Then $\angle D E X=$ $\angle X Q C=\angle B S X$, thus $S$ is on circle $D E X$. Similarly, let circle $C P X$ intersect line $A C$ at a point $T$ which is different from $C$. Then $T$ is on circle $D E X$. The power of $A$ with respect to the circle $D E X$ is $A S \cdot A D=A T \cdot A E$. Since $\frac{A D}{A B}=\frac{A E}{A C}, A S \cdot A B=A T \cdot A C$. Then $A$ is in the radical axis of circles $B Q X$ and $C P X$, which implies that three points $A, X$ and $Y$ are collinear.
|
{
"resource_path": "APMO/segmented/en-apmo2024_sol.jsonl",
"problem_match": "# Problem 1",
"solution_match": "# Solution 2"
}
|
c2662896-d3ab-5e68-bb60-20bb487f6879
| 606,331
|
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