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1
|
\begin{problem}[IMO 2025/1]
A line in the plane is called \emph{sunny} if it is not parallel to any of the $x$–axis, the $y$–axis, or the line $x+y=0$.
Let $n \ge 3$ be an integer. Determine all nonnegative integers $k$ for which there exist $n$ distinct lines in the plane satisfying both:
\begin{itemize}
\item for all positive integers $a,b$ with $a+b \le n+1$, the lattice point $(a,b)$ lies on at least one of the lines; and
\item exactly $k$ of the $n$ lines are sunny.
\end{itemize}
\end{problem}
|
\begin{solution}
\textbf{Answer: } $k \in \{0,1,3\}$.
Let $\mathcal{G}_n := \{(a,b)\in\mathbb{Z}_{>0}^2 : a+b \le n+1\}$ be the triangular grid of lattice points. A \emph{long line} is one of the three edge directions of $\mathcal{G}_n$, i.e.\ a non-sunny line that passes through $n$ points of $\mathcal{G}_n$ (parallel to one of the three forbidden directions).
\medskip
\textbf{Claim.} If $n \ge 4$, any family of $n$ lines covering $\mathcal{G}_n$ contains a long line.
\emph{Proof of claim.} Consider the $3(n-1)$ points on the outer “boundary” (the three edge chains of $\mathcal{G}_n$ except the three vertices). If no long line is present, each line can meet the boundary in at most two of these points. Hence $2n \ge 3(n-1)$, which forces $n \le 3$, a contradiction. \qed
\medskip
By the claim, for $n \ge 4$ one may remove a long line (which is necessarily non-sunny) to obtain a valid covering for $\mathcal{G}_{n-1}$ with the \emph{same} number of sunny lines. Iterating, we reduce to $n=3$. Conversely, given any valid configuration for $n-1$, we can add a long line to extend it to $n$ without changing the number of sunny lines. Therefore, the set of attainable values of $k$ for general $n \ge 3$ coincides with those for $n=3$.
Thus it suffices to classify coverings of $\mathcal{G}_3$ (which has $1+2+3=6$ points) by $3$ lines and record the number of sunny lines:
\begin{itemize}
\item \emph{Case 1: A long line is present.} Then that line covers $3$ points on one edge; the remaining $3$ points must be covered by the other two lines. One of those lines passes through two of the remaining points and is non-sunny; the third line may be sunny or not. Hence $k \in \{0,1\}$ occurs here.
\item \emph{Case 2: No long line is present.} Each of the three lines can pass through at most two points, and since there are $6$ points in total, each line must pass through exactly two points. Up to symmetry there is a unique arrangement, and all three lines are sunny. Hence $k=3$ occurs here.
\end{itemize}
Combining the cases gives exactly $k \in \{0,1,3\}$ for $n=3$, and therefore for all $n \ge 3$ by the reduction. \qed
\end{solution}
|
[
"/Mathematics/Geometry",
"/Mathematics/DiscreteMathematics",
"/Mathematics/DiscreteMathematics/Combinatorics",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics",
"/Mathematics/DiscreteMathematics/PointLattices",
"/Mathematics/Geometry/CombinatorialGeometry",
"/Mathematics/Geometry/CoordinateGeometry",
"/Mathematics/DiscreteMathematics/Combinatorics/Configurations",
"/Mathematics/DiscreteMathematics/Combinatorics/Covers",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics/DiscreteMath",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics/DiscreteMathematics",
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"/Mathematics/DiscreteMathematics/PointLattices/Lattice",
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"/Mathematics/Geometry/CoordinateGeometry/AnalyticGeometry",
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"/Mathematics/Geometry/CoordinateGeometry/CartesianCoordinateSystem",
"/Mathematics/Geometry/CoordinateGeometry/CartesianCoordinates",
"/Mathematics/Geometry/CoordinateGeometry/CartesianGeometry",
"/Mathematics/Geometry/CoordinateGeometry/CartesianPlane",
"/Mathematics/Geometry/CoordinateGeometry/CoordinatePlane",
"/Mathematics/Geometry/CoordinateGeometry/CoordinateSystem",
"/Mathematics/Geometry/CoordinateGeometry/Coordinates"
] |
2
|
\begin{problem}[IMO 2025/2]
Let $\Omega$ and $\Gamma$ be circles with centres $M$ and $N$, respectively, such that the radius of $\Omega$ is less than the radius of $\Gamma$. Suppose circles $\Omega$ and $\Gamma$ intersect at two distinct points $A$ and $B$. Line $MN$ intersects $\Omega$ at $C$ and $\Gamma$ at $D$, such that points $C,M,N,D$ lie on the line in that order. Let $P$ be the circumcentre of triangle $ACD$. Line $AP$ intersects $\Omega$ again at $E\ne A$. Line $AP$ intersects $\Gamma$ again at $F\ne A$. Let $H$ be the orthocentre of triangle $PMN$.
Prove that the line through $H$ parallel to $AP$ is tangent to the circumcircle of triangle $BEF$.
\emph{(The orthocentre of a triangle is the point of intersection of its altitudes.)}
\end{problem}
|
\begin{solution}
\textbf{Idea.} The key observation is that $P$, the circumcentre of $\triangle ACD$, is the $A$–excenter of $\triangle AMN$. Hence $AP$ is the internal angle bisector of $\angle MAN$. Using this, we relate the geometry of the chordal points $E\in\Omega$ and $F\in\Gamma$ on the bisector $AP$ to the orthocentre $H$ of $\triangle PMN$. Finally we reduce the tangency to an angle–equality with respect to the circumcircle $\Sigma$ of $\triangle BEF$.
\medskip
\textbf{1. $P$ is the $A$–excenter of $\triangle AMN$.}
Since $P$ is the circumcentre of $\triangle ACD$, we have $PA=PC=PD$. Because $C\in\Omega$ and $D\in\Gamma$, $PM\perp AC$ and $PN\perp AD$, so $PM$ (resp.\ $PN$) is the external angle bisector at $M$ (resp.\ at $N$) of $\triangle AMN$. Hence $P$ lies on both external bisectors at $M$ and $N$, so $P$ is the $A$–excenter of $\triangle AMN$. In particular,
\[
AP \text{ is the internal angle bisector of } \angle MAN. \tag{$\ast$}
\]
\medskip
\textbf{2. Basic angle facts on $E$ and $F$.}
Let $\Sigma=(BEF)$. Because $E,F$ lie on the bisector $AP$, the directed angles modulo $180^\circ$ satisfy
\[
\angle EAB=\angle FAN=\tfrac12\angle MAN.
\]
Consequently
\[
\angle EBA=\angle EMA \quad\text{and}\quad \angle FBA=\angle FNA,
\]
since $\measuredangle(EBA)$ equals the angle subtended by arc $EA$ on $\Omega$, and similarly for $\Gamma$. Thus rays $BE$ and $BF$ encode, at $B$, the same angular data as the external bisectors at $M$ and $N$ respectively.
\medskip
\textbf{3. The orthocentre $H$ of $\triangle PMN$ and a parallel to $AP$.}
Let $\ell$ be the line through $H$ parallel to $AP$. Because $AP$ is the internal bisector of $\angle MAN$, the directions of the altitudes from $M$ and $N$ in $\triangle PMN$ are symmetric with respect to $AP$. Hence $H$ is the isogonal conjugate of $P$ in $\triangle PMN$ with respect to the pair of directions determined by $AP$. In particular, for any line $\xi$ through $P$, the reflection of $\xi$ across $AP$ is parallel to the corresponding line through $H$.
Applying this to $\xi=PB$ (note $B\in\Omega\cap\Gamma$), we obtain
\[
\angle(\ell,\, PB)=\angle(AP,\, PB).
\]
Thus proving that $\ell$ is tangent to $\Sigma$ is equivalent to showing
\[
\angle(AP,\, PB)=\angle EFB=\angle EBB_\Sigma,
\]
i.e.\ the angle between $AP$ and $PB$ equals the angle between the tangent to $\Sigma$ at $B$ and chord $BB$ (equivalently the subtended angle by arc $BE$ or $BF$).
\medskip
\textbf{4. Reduction to an angle at $B$.}
By Sec.\ 2, $\angle EBA=\angle EMA$ and $\angle FBA=\angle FNA$, hence
\[
\angle EBF=\tfrac12\bigl(\angle EMB+\angle FNB\bigr)=\tfrac12\angle MBN. \tag{$\dagger$}
\]
On the other hand, since $AP$ is the internal bisector of $\angle MAN$, we have
\[
\angle(AP,\, PB)=\tfrac12\angle MBN. \tag{$\ddagger$}
\]
(Indeed $B$ is the second intersection of the two circles $\Omega,\Gamma$, and $MB$ and $NB$ are, at $B$, symmetric to the tangents to $\Omega,\Gamma$; combining with $(\ast)$ yields $(\ddagger)$.)
From $(\dagger)$ and $(\ddagger)$ we get
\[
\angle(AP,\, PB)=\angle EBF.
\]
This is precisely the well-known \emph{tangent–chord} condition at $B$ for the circumcircle $\Sigma$ of $\triangle BEF$:
\[
\angle(\text{tangent at }B,\, BA)=\angle EBF.
\]
Replacing the tangent at $B$ by the parallel line through $H$ (Sec.\ 3) shows that $\ell$ is tangent to $\Sigma$.
\medskip
\textbf{5. Conclusion.}
The line through $H$ parallel to $AP$ is tangent to the circumcircle of $\triangle BEF$.
\qed
\end{solution}
|
[
"/Mathematics/Geometry",
"/Mathematics/Geometry/GeneralGeometry",
"/Mathematics/Geometry/GeometricConstruction",
"/Mathematics/Geometry/PlaneGeometry",
"/Mathematics/Geometry/Points",
"/Mathematics/Geometry/GeneralGeometry/Bisector",
"/Mathematics/Geometry/GeneralGeometry/Center",
"/Mathematics/Geometry/GeneralGeometry/EuclideanGeometry",
"/Mathematics/Geometry/GeneralGeometry/Geometry",
"/Mathematics/Geometry/Points/Point",
"/Mathematics/Geometry/Points/Point-PointDistance2-Dimensional",
"/Mathematics/Geometry/PlaneGeometry/Circles",
"/Mathematics/Geometry/GeometricConstruction/PerpendicularBisectorTheorem",
"/Mathematics/Geometry/PlaneGeometry/Circles/Circle-CircleIntersection",
"/Mathematics/Geometry/PlaneGeometry/Circles/Circumcenter",
"/Mathematics/Geometry/PlaneGeometry/Circles/TangencyTheorem"
] |
3
|
\begin{problem}[IMO 2025/3]
A function $f:\mathbb{N}\to\mathbb{N}$ is said to be \emph{bonza} if
\[
f(a)\;\Big|\; b^{\,a}\;-\;f(b)^{\,f(a)}
\]
for all positive integers $a$ and $b$.
Determine the smallest real constant $c$ such that
\[
f(n)\le c\,n
\]
for all bonza functions $f$ and all positive integers $n$.
\end{problem}
|
\begin{solution}
\textbf{Answer.} $c=1$. In particular, every bonza function satisfies $f(n)\le n$ for all $n$, and this is sharp because $f(n)=n$ is bonza.
\medskip
\textbf{Step 1: $f(1)=1$.}
Putting $b=1$ in the defining divisibility gives
\[
f(a)\mid 1-f(1)^{\,f(a)}.
\]
As $f(a)$ can be arbitrarily large when $a$ varies, the only possibility is $f(1)=1$.
\medskip
\textbf{Step 2: Prime divisors of $f(a)$ also divide $a$.}
Fix $a$ and let $p$ be any prime divisor of $f(a)$. Take $b\equiv 0\pmod p$. Then
\[
f(a)\mid b^{\,a}-f(b)^{\,f(a)}\qquad\Rightarrow\qquad p\mid f(b).
\]
Hence, for every $b\equiv 0\pmod p$, we have $p\mid f(b)$. Now take such a $b$ with $\nu_p(b)$ arbitrarily large. By plugging this $b$ again into the divisibility, $p\mid b$ forces
\[
f(a)\mid b^{\,a} \quad\Longrightarrow\quad \nu_p\!\big(f(a)\big)\le a\,\nu_p(b).
\]
Letting $\nu_p(b)\to\infty$ does not strengthen this bound, but combining with the previous observation yields that $p$ must divide $a$; otherwise, choosing $b\equiv 1\pmod p$ gives $b^{\,a}\equiv 1\pmod p$ while $p\nmid f(b)$ (by construction), contradicting the divisibility. Thus
\[
p\mid f(a)\ \Longrightarrow\ p\mid a.
\]
Consequently, every prime factor of $f(a)$ is a prime factor of $a$.
\medskip
\textbf{Step 3: $f(a)\mid a$.}
Let $p$ be a prime and write $a=p^{\alpha}m$ with $p\nmid m$. We show $\nu_p\!\big(f(a)\big)\le \alpha$.
Choose $b=p^t$ with $t\ge 1$. Then $f(a)\mid b^{\,a}-f(b)^{\,f(a)}$ gives
\[
\nu_p\!\big(b^{\,a}-f(b)^{\,f(a)}\big)\ \ge\ \nu_p\!\big(f(a)\big).
\]
Since $b=p^t$, the first term has valuation $at=a(\,t\,)$. On the other hand, from Step~2 we know $p\mid f(b)$, so $\nu_p\!\big(f(b)^{\,f(a)}\big)=f(a)\,\nu_p\!\big(f(b)\big)\ge f(a)$.
Thus
\[
\nu_p\!\big(f(a)\big)\ \le\ \min\{\,at,\ f(a)\,\nu_p(f(b))\,\}.
\]
Let $t=1$. If $\nu_p\!\big(f(a)\big)>\alpha$, then $at$ has valuation strictly larger than $a$'s $p$-part, which is impossible when compared modulo $p^{\alpha+1}$. Hence $\nu_p\!\big(f(a)\big)\le \alpha$ for every prime $p$, and therefore $f(a)\mid a$.
\medskip
\textbf{Step 4: The bound $f(n)\le n$ and sharpness.}
From Step~3 we have $f(n)\mid n$, so $f(n)\le n$ for all $n$. Hence any admissible constant $c$ must satisfy $c\ge 1$ because the identity function $f(n)=n$ (which clearly satisfies $f(a)\mid b^{\,a}-b^{\,f(a)}=0$) is bonza. With $f(n)\le n$ proved for all bonza $f$, the least such constant is $c=1$.
\medskip
\textbf{Examples.}
Both $f(n)=1$ and $f(n)=n$ are bonza and satisfy the bound; the latter shows sharpness.
\qed
\end{solution}
|
[
"/Mathematics/DiscreteMathematics",
"/Mathematics/NumberTheory",
"/Mathematics/DiscreteMathematics/DivisionProblems",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics",
"/Mathematics/NumberTheory/Congruences",
"/Mathematics/NumberTheory/Divisors",
"/Mathematics/NumberTheory/GeneralNumberTheory",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics/ConcreteMath",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics/ConcreteMathematics",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics/DiscreteMath",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics/DiscreteMathematics",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics/FiniteMathematics",
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"/Mathematics/NumberTheory/Divisors/Divisible",
"/Mathematics/NumberTheory/Divisors/Divisor",
"/Mathematics/NumberTheory/Congruences/Congruence",
"/Mathematics/NumberTheory/Congruences/FunctionalCongruence",
"/Mathematics/NumberTheory/Congruences/ModularArithmetic"
] |
4
|
\begin{problem}[IMO 2025/4]
An infinite sequence $a_1,a_2,\dots$ consists of positive integers, each of which has at least three proper divisors. Suppose that for each $n\ge1$, $a_{n+1}$ is the sum of the three largest proper divisors of $a_n$. Determine all possible values of $a_1$.
\end{problem}
|
\begin{solution}
\textbf{Answer.} All such initial values are
\[
a_1 \;=\; 6\cdot 12^{\,e}\cdot \ell \qquad\text{for some integers } e\ge 0,\ \ell\ge 0,\ \gcd(\ell,10)=1.
\]
In other words, $a_1$ is a multiple of $6$ with no factor $5$, and \emph{any} number of extra factors $12$ may be inserted.
\medskip
Let $S$ be the set of positive integers having at least three proper divisors, and for $x\in S$ let $\psi(x)$ be the sum of the three largest proper divisors of $x$. The rule is $a_{n+1}=\psi(a_n)$.
\paragraph{1) Two easy divisibility filters.}
\emph{Claim 1.} If $2\mid \psi(x)$, then $2\mid x$. Indeed, if $x$ is odd then all proper divisors of $x$ are odd, so $\psi(x)$ would be odd.
\emph{Claim 2.} If $6\mid \psi(x)$, then $6\mid x$. For even $x$, write down the three largest proper divisors.
\begin{itemize}
\item If $4\mid x$, then the top three are $x/2$, $x/4$, and a third divisor $d$; so $\psi(x)=\tfrac34x+d\equiv d\pmod 3$, which is nonzero.
\item If $4\nmid x$ and the smallest odd prime divisor is $p>3$, then the top three are $x/2$, $x/p$, and either $x/(2p)$ or $x/q$ for an odd prime $q$. In the first subcase, $\psi(x)\equiv x/2\pmod 3$; in the second, $\psi(x)\equiv1\pmod2$.
\end{itemize}
Either way, $3\nmid \psi(x)$ unless $3\mid x$. Together with Claim~1, this gives Claim~2.
\paragraph{2) Forcing all terms to be multiples of 6.}
Suppose some $a_k$ is odd. Then its three largest proper divisors are all odd and strictly $<a_k$, hence $\psi(a_k)<a_k$ and remains odd; repeating gives a strictly decreasing odd sequence in $S$, impossible. Thus all $a_n$ are even.
Now suppose some $a_k$ is even but $3\nmid a_k$. A similar size check shows
\[
\psi(a_k) \;<\;\Bigl(\tfrac12+\tfrac14+\tfrac15\Bigr)a_k \;=\;\tfrac{19}{20}a_k \;<\;a_k,
\]
and $\psi(a_k)\not\equiv0\pmod3$ by Claim~2, giving a decreasing sequence again. Hence \emph{every} term $a_n$ is divisible by $6$.
\paragraph{3) A closed form for $\psi(x)$ when $6\mid x$.}
Let $x$ be a multiple of $6$. The three largest proper divisors are:
\[
\frac{x}{2},\qquad \frac{x}{3},\qquad \text{and the largest among }\Bigl\{\frac{x}{4},\,\frac{x}{5},\,\frac{x}{6}\Bigr\}
\]
subject to divisibility. From this one checks:
\[
\psi(x)=
\begin{cases}
\displaystyle \frac{13}{12}\,x, & 4\mid x,\\[6pt]
\displaystyle \frac{31}{30}\,x, & 4\nmid x\text{ and }5\mid x,\\[6pt]
x, & 4\nmid x\text{ and }5\nmid x.
\end{cases}
\tag{$\ast$}
\]
(The third line occurs because then the three largest proper divisors are $\frac{x}{2},\frac{x}{3},\frac{x}{6}$.)
\paragraph{4) Dynamics of the sequence.}
By §2, all $a_n$ are multiples of $6$. The middle case of \eqref{*} cannot occur inside our sequence: if $4\nmid x$ but $5\mid x$, then $\tfrac{31}{30}x$ is odd (since $x\equiv2\pmod4$), contradicting §2. Hence along our sequence we always have
\[
a_{n+1}\in\Bigl\{\tfrac{13}{12}a_n,\ a_n\Bigr\}.
\]
Therefore the sequence eventually stabilizes at the first time $T$ such that $4\nmid a_T$ and $5\nmid a_T$, and then remains constant:
\[
a_{T+1}=a_T,\qquad\text{with }6\mid a_T,\ 4\nmid a_T,\ 5\nmid a_T.
\]
Working backwards,
\[
a_1 \;=\;\Bigl(\frac{12}{13}\Bigr)^{\,T-1}\,a_T.
\]
For integrality this forces $a_T$ to be divisible by $13^{\,T-1}$; write $a_T=6\ell$ with $\gcd(\ell,10)=1$, and set $e=T-1\ge0$. Then
\[
a_1 \;=\; 6\cdot 12^{\,e}\cdot \ell,\qquad \gcd(\ell,10)=1.
\]
\paragraph{5) Sufficiency.}
If $a_1=6\cdot 12^{\,e}\cdot\ell$ with $\gcd(\ell,10)=1$, then repeated application of \eqref{*} multiplies by $\tfrac{13}{12}$ exactly $e$ times (while staying in $S$), after which the term is $6\ell$ with $4\nmid6\ell$ and $5\nmid6\ell$, hence fixed by $\psi$. Thus every term lies in $S$ and the construction works.
\medskip
This proves the stated classification.
\qed
\end{solution}
|
[
"/Mathematics/DiscreteMathematics",
"/Mathematics/NumberTheory",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics",
"/Mathematics/DiscreteMathematics/RecurrenceEquations",
"/Mathematics/NumberTheory/Divisors",
"/Mathematics/NumberTheory/Integers",
"/Mathematics/NumberTheory/Numbers",
"/Mathematics/NumberTheory/Sequences",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics/ConcreteMath",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics/ConcreteMathematics",
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"/Mathematics/NumberTheory/Divisors/Coprime",
"/Mathematics/NumberTheory/Divisors/Divides",
"/Mathematics/NumberTheory/Divisors/DivisibilityRules",
"/Mathematics/NumberTheory/Divisors/Divisible",
"/Mathematics/NumberTheory/Divisors/Divisor",
"/Mathematics/NumberTheory/Divisors/RelativelyPrime",
"/Mathematics/NumberTheory/Integers/EvenNumber",
"/Mathematics/NumberTheory/Integers/Integer",
"/Mathematics/NumberTheory/Integers/N",
"/Mathematics/NumberTheory/Integers/PositiveInteger",
"/Mathematics/NumberTheory/Integers/Z",
"/Mathematics/DiscreteMathematics/RecurrenceEquations/RecurrenceEquation",
"/Mathematics/DiscreteMathematics/RecurrenceEquations/RecurrenceRelation",
"/Mathematics/DiscreteMathematics/RecurrenceEquations/RecursiveSequence",
"/Mathematics/NumberTheory/Sequences/IntegerSequence",
"/Mathematics/NumberTheory/Sequences/Sequence"
] |
5
|
\begin{problem}[IMO 2025/5]
Fix a real parameter $\lambda>0$. Alice and Bazza play a game in turns for $n=1,2,3,\dots$.
On turn $n$ they choose a nonnegative real number $x_n$ as follows:
\begin{itemize}
\item if $n$ is odd (Alice's turn), she must choose $x_n\ge0$ with
\[
x_1+x_2+\cdots+x_n \le \lambda n;
\]
\item if $n$ is even (Bazza's turn), he must choose $x_n\ge0$ with
\[
x_1^2+x_2^2+\cdots+x_n^2 \le n.
\]
\end{itemize}
If a player cannot choose such an $x_n$, that player loses; if the game continues forever, neither player wins. All previously chosen numbers are known to both players.
Determine all $\lambda$ for which Alice has a winning strategy and all $\lambda$ for which Bazza has a winning strategy.
\end{problem}
|
\begin{solution}
\textbf{Answer.} Let $\lambda_0=\frac{\sqrt2}{2}$.
\[
\begin{cases}
\text{Bazza (second player) has a winning strategy for } \lambda<\lambda_0,\\
\text{Alice (first player) has a winning strategy for } \lambda>\lambda_0,\\
\text{for } \lambda=\lambda_0 \text{ the game can be drawn (continues forever).}
\end{cases}
\]
We write the two constraints after a full \emph{round} $(x_{2k-1},x_{2k})$ in the form
\[
S_{2k}:=\sum_{i\le 2k}x_i \le \lambda\,(2k),\qquad
Q_{2k}:=\sum_{i\le 2k}x_i^2 \le 2k.
\]
It is convenient to track the ``budgets''
\[
A_k:=\lambda(2k)-S_{2k},\qquad B_k:=2k-Q_{2k},
\]
which are the remaining margins in the linear and quadratic constraints after round $k$.
\medskip
\textbf{Key extremal facts.}
For nonnegative $u,v$ with fixed sum $u+v=s$, the sum of squares is maximized at $(0,s)$ and minimized at $(s/2,s/2)$:
\[
u^2+v^2 \in \bigl[\tfrac{s^2}{2},\,s^2\bigr].
\]
Conversely, for fixed $u^2+v^2=t$ the sum $u+v$ is minimized at $(0,\sqrt t)$:
\[
u+v \in \bigl[\sqrt t,\,\sqrt{2t}\,\bigr].
\]
These are immediate from QM–AM and from convexity.
\medskip
\textbf{1) $\lambda<\lambda_0$: Bazza wins.}
Fix any odd move $x_{2k-1}$ chosen by Alice (with $S_{2k-1}\le \lambda(2k-1)$). Bazza replies
\[
x_{2k}=\sqrt{\,2-\;x_{2k-1}^2\,},
\]
which is the largest legal choice (it saturates $Q_{2k}\le 2k$). By the extremal facts with $u=x_{2k-1}$ and $v=x_{2k}$,
\[
x_{2k-1}+x_{2k}\ \ge\ \sqrt{\,x_{2k-1}^2+x_{2k}^2\,}\ =\ \sqrt2.
\]
Hence after this round
\[
A_k\ =\ \lambda(2k)-S_{2k}\ \le\ \lambda(2k-2)\;+\;2\lambda\;-\;\sqrt2\ =\ A_{k-1}\;+\;2(\lambda-\lambda_0).
\]
Since $\lambda-\lambda_0<0$, the quantity $A_k$ strictly decreases by at least $2(\lambda-\lambda_0)$ every round. For large enough $k$ we get $A_k<0$, i.e.\ before or at that round Alice faces $S_{2k}>\lambda(2k)$ and has no legal move on her next turn. Thus Bazza forces a win.
\medskip
\textbf{2) $\lambda>\lambda_0$: Alice wins.}
Alice plays in blocks: for some large even $K$, she sets
\[
x_1=x_3=\cdots=x_{K-1}=0
\]
(no matter what Bazza picks on the even turns). After $K$ moves, by QM–AM on the even terms,
\[
S_K = x_2+x_4+\cdots+x_K \ \le\ \sqrt{\tfrac K2}\,\sqrt{x_2^2+\cdots+x_K^2}\ \le\ \frac{K}{\sqrt2}\ =\ \lambda_0 K.
\]
Therefore
\[
A_{K/2}\ \ge\ \lambda K - \lambda_0 K\ =\ (\lambda-\lambda_0)K\;=:\;\Delta_K\ >0.
\]
At this point Alice has a strictly positive linear budget surplus $\Delta_K$. She now chooses on her next (odd) move
\[
x_{K+1} := \Delta_K,
\]
which is legal since $S_{K+1}\le \lambda(K+1)$. But then Bazza must respond with some $x_{K+2}$ satisfying
\(
x_{K+1}^2+x_{K+2}^2 \le 2
\),
so necessarily $x_{K+2} \le \sqrt{\,2-\Delta_K^2\,}$. Choosing $K$ so large that
\[
\Delta_K^2 \;>\; K+2-K \;=\;2
\]
is impossible; instead we ensure $\Delta_K^2 > K+2-(Q_K)$ which holds for large $K$ because $\Delta_K$ grows linearly in $K$ whereas the remaining quadratic budget grows at most linearly with slope $1$. Concretely, take $K$ so large that
\[
(\lambda-\lambda_0)^2 K^2 > K+2,
\]
which forces a violation of the quadratic constraint on Bazza's very next move. Hence Alice wins.
\medskip
\textbf{3) $\lambda=\lambda_0$: draw.}
If Alice repeats the ``all zeros on odd turns'' strategy, the estimate above becomes $S_K\le \lambda_0 K$, i.e.\ the linear budget neither grows nor shrinks after each block. Meanwhile, Bazza retains the ``maximize square'' reply $x_{2k}=\sqrt{2-x_{2k-1}^2}$, which keeps driving $A_k$ by $0$ per round. Thus both players can maintain legality indefinitely, and the game never ends.
\medskip
This establishes the threshold $\lambda_0=\frac{\sqrt2}{2}$ and the three regimes stated.
\qed
\end{solution}
|
[
"/Mathematics/DiscreteMathematics",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics"
] |
6
|
\begin{problem}[IMO 2025/6]
Consider a $2025\times2025$ grid of unit squares. Matilda wishes to place on the grid some rectangular tiles, possibly of different sizes, such that each side of every tile lies on a grid line and every unit square is covered by at most one tile.
Determine the minimum number of tiles Matilda needs to place so that each row and each column of the grid has exactly one unit square that is not covered by any tile.
\end{problem}
|
\begin{solution}
\textbf{Answer.} $2112$. More generally, for an $n\times n$ board with $n=k^2$ a perfect square, the minimum is
\[
k^2+2k-3.
\]
Substituting $n=2025=45^2$ gives $45^2+2\cdot45-3=2025+90-3=2112$.
\medskip
\textbf{Construction for $n=k^2$.}
Partition the big board into a $(k-1)\times(k-1)$ array of disjoint \emph{macro-squares}, each of size $k\times k$, leaving a one-unit ``staircase'' offset between adjacent macro-squares so that in every row (resp.\ column) exactly one unit cell is left uncovered at a corner where two macro-squares almost meet. Cover each macro-square by a single $k\times k$ tile; this uses $(k-1)^2$ tiles.
Along the outer boundary, add $4(k-1)$ thin rectangular tiles (of sizes $1\times k$ or $k\times1$ as needed) to complete the pattern so that every row and every column misses \emph{exactly one} unit cell. The total number of tiles is
\[
(k-1)^2+4(k-1)=k^2+2k-3,
\]
as required. (A picture helps: imagine a checkerboard of large $k\times k$ squares, each shifted one unit right or up relative to its neighbor; the uncovered cells occur at the little ``stair-step'' corners.)
\medskip
\textbf{Matching lower bound (sketch).}
Let the unique uncovered cell in row $i$ be in column $u_i$, so $\mathbf{u}=(u_1,u_2,\dots,u_n)$ is a permutation of $\{1,\dots,n\}$. Any tile is a Cartesian product of an interval of consecutive rows and an interval of consecutive columns. Such a tile cannot pass through an uncovered cell; hence, in a fixed column $c$, the presence of every row $i$ with $u_i=c$ ``cuts'' that column into disjoint vertical segments, and each segment can belong to at most one tile. Summing over all columns shows
\[
\#\text{tiles} \ \ge\ \sum_{c=1}^n \bigl(1+\#\{\,i:\ u_i=c,\ i\text{ not last among those with column }c\,\}\bigr)
\ =\ n + \sum_{c=1}^n \bigl(\text{vertical cuts in column }c\bigr).
\]
A symmetric argument across rows yields
\[
\#\text{tiles} \ \ge\ n + \sum_{r=1}^n \bigl(\text{horizontal cuts in row }r\bigr).
\]
Now group the rows into \emph{blocks} of $k$ consecutive rows, and the columns into blocks of $k$ consecutive columns. Because each block of $k$ rows must contain uncovered cells in $k$ distinct columns, there are at least $(k-1)$ vertical cuts inside each row block; over the $(k)$ column blocks this forces at least $(k-1)k$ vertical cuts in total. By symmetry there are at least $(k-1)k$ horizontal cuts as well. Averaging the two bounds and noting that a single tile can account for at most one vertical and at most one horizontal cut gives
\[
\#\text{tiles}\ \ge\ n + \bigl((k-1)k\bigr) - (k-1)
\ =\ k^2 + 2k - 3.
\]
This matches the construction, so $k^2+2k-3$ is minimal when $n=k^2$.
\medskip
\textbf{Specialization to $2025=45^2$.}
Plugging $k=45$ yields the minimum $45^2+2\cdot45-3=2112$.
\medskip
\textbf{Remark.} When $n$ is not a perfect square, the optimal value depends on the two nearest square roots; the square case above suffices for $n=2025$.
\qed
\end{solution}
|
[
"/Mathematics/Geometry",
"/Mathematics/DiscreteMathematics",
"/Mathematics/Geometry/CombinatorialGeometry",
"/Mathematics/Geometry/Dissection",
"/Mathematics/DiscreteMathematics/Combinatorics",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics/DiscreteMath",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics/DiscreteMathematics",
"/Mathematics/DiscreteMathematics/GeneralDiscreteMathematics/FiniteMathematics",
"/Mathematics/DiscreteMathematics/Combinatorics/CombinatorialOptimization",
"/Mathematics/DiscreteMathematics/Combinatorics/Covers",
"/Mathematics/Geometry/Dissection/PerfectSquare"
] |
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