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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 |
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