add_benelux_folder (#16)

- Add Benelux folder with problems and solutions (d1b90d56d1c9a1a84fd0bdcdf44a81a1968a9682)

Benelux/md/Benelux_en-olympiad_en-bxmo-problems-2010-zz.md

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+## Solutions of Benelux Mathematical Olympiad 2010
+
+Problem 1. A finite set of integers is called bad if its elements add up to 2010. A finite set of integers is a Benelux-set if none of its subsets is bad. Determine the smallest integer $n$ such that the set $\{502,503,504, \ldots, 2009\}$ can be partitioned into $n$ Benelux-sets.
+(A partition of a set $S$ into $n$ subsets is a collection of $n$ pairwise disjoint subsets of $S$, the union of which equals $S$.)
+
+Solution. As $502+1508=2010$, the set $S=\{502,503, \ldots, 2009\}$ is not a Benelux-set, so $n=1$ does not work. We will prove that $n=2$ does work, i.e. that $S$ can be partitioned into 2 Benelux-sets.
+Define the following subsets of $S$ :
+
+$$
+\begin{aligned}
+& A=\{502,503, \ldots, 670\}, \\
+& B=\{671,672, \ldots, 1005\}, \\
+& C=\{1006,1007, \ldots, 1339\}, \\
+& D=\{1340,1341, \ldots, 1508\}, \\
+& E=\{1509,1510, \ldots, 2009\} .
+\end{aligned}
+$$
+
+We will show that $A \cup C \cup E$ and $B \cup D$ are both Benelux-sets.
+Note that there does not exist a bad subset of $S$ of one element, since that element would have to be 2010. Also, there does not exist a bad subset of $S$ of more than three elements, since the sum of four or more elements would be at least $502+503+504+505=2014>2010$. So any possible bad subset of $S$ contains two or three elements.
+Consider a bad subset of two elements $a$ and $b$. As $a, b \geq 502$ and $a+b=2010$, we have $a, b \leq 2010-502=1508$. Furthermore, exactly one of $a$ and $b$ is smaller than 1005 and one is larger than 1005. So one of them, say $a$, is an element of $A \cup B$, and the other is an element of $C \cup D$. Suppose $a \in A$, then $b \geq 2010-670=1340$, so $b \in D$. On the other hand, suppose $a \in B$, then $b \leq 2010-671=1339$, so $b \in C$. Hence $\{a, b\}$ cannot be a subset of $A \cup C \cup E$, nor of $B \cup D$.
+Now consider a bad subset of three elements $a, b$ and $c$. As $a, b, c \geq 502, a+b+c=2010$, and the three elements are pairwise distinct, we have $a, b, c \leq 2010-502-503=1005$. So $a, b, c \in A \cup B$. At least one of the elements, say $a$, is smaller than $\frac{2010}{3}=670$, and at least one of the elements, say $b$, is larger than 670 . So $a \in A$ and $b \in B$. We conclude that $\{a, b, c\}$ cannot be a subset of $A \cup C \cup E$, nor of $B \cup D$.
+This proves that $A \cup C \cup E$ and $B \cup D$ are Benelux-sets, and therefore the smallest $n$ for which $S$ can be partitioned into $n$ Benelux-sets is $n=2$.
+
+Remark. Observe that $A \cup C \cup E_{1}$ and $B \cup D \cup E_{2}$ are also Benelux-sets, where $\left\{E_{1}, E_{2}\right\}$ is any partition of $E$.
+
+Problem 2. Find all polynomials $p(x)$ with real coefficients such that
+
+$$
+p(a+b-2 c)+p(b+c-2 a)+p(c+a-2 b)=3 p(a-b)+3 p(b-c)+3 p(c-a)
+$$
+
+for all $a, b, c \in \mathbb{R}$.
+
+Solution 1. For $a=b=c$, we have $3 p(0)=9 p(0)$, hence $p(0)=0$. Now set $b=c=0$, then we have
+
+$$
+p(a)+p(-2 a)+p(a)=3 p(a)+3 p(-a)
+$$
+
+for all $a \in \mathbb{R}$. So we find a polynomial equation
+
+$$
+p(-2 x)=p(x)+3 p(-x)
+$$
+
+Note that the zero polynomial is a solution to this equation. Now suppose that $p$ is not the zero polynomial, and let $n \geq 0$ be the degree of $p$. Let $a_{n} \neq 0$ be the coefficient of $x^{n}$ in $p(x)$. At the left-hand side of (1), the coefficient of $x^{n}$ is $(-2)^{n} \cdot a_{n}$, while at the right-hand side the coefficient of $x^{n}$ is $a_{n}+3 \cdot(-1)^{n} \cdot a_{n}$. Hence $(-2)^{n}=1+3 \cdot(-1)^{n}$. For $n$ even, we find $2^{n}=4$, so $n=2$, and for $n$ odd, we find $-2^{n}=-2$, so $n=1$. As we already know that $p(0)=0$, we must have $p(x)=a_{2} x^{2}+a_{1} x$, where $a_{1}$ and $a_{2}$ are real numbers (possibly zero).
+The polynomial $p(x)=x$ is a solution to our problem, as
+
+$$
+(a+b-2 c)+(b+c-2 a)+(c+a-2 b)=0=3(a-b)+3(b-c)+3(c-a)
+$$
+
+for all $a, b, c \in \mathbb{R}$. Also, $p(x)=x^{2}$ is a solution, since
+
+$$
+\begin{aligned}
+(a+b-2 c)^{2}+(b+c-2 a)^{2}+(c+a-2 b)^{2} & =6\left(a^{2}+b^{2}+c^{2}\right)-6(a b+b c+c a) \\
+& =3(a-b)^{2}+3(b-c)^{2}+3(c-a)^{2}
+\end{aligned}
+$$
+
+for all $a, b, c \in \mathbb{R}$.
+Now note that if $p(x)$ is a solution to our problem, then so is $\lambda p(x)$ for all $\lambda \in \mathbb{R}$. Also, if $p(x)$ and $q(x)$ are both solutions, then so is $p(x)+q(x)$. We conclude that for all real numbers $a_{2}$ and $a_{1}$ the polynomial $a_{2} x^{2}+a_{1} x$ is a solution. Since we have already shown that there can be no other solutions, these are the only solutions.
+
+Solution 2. For $a=b=c$, we have $3 p(0)=9 p(0)$, hence $p(0)=0$. Now set $b=c=0$, then we have
+
+$$
+p(a)+p(-2 a)+p(a)=3 p(a)+3 p(-a)
+$$
+
+for all $a \in \mathbb{R}$. So we find a polynomial equation
+
+$$
+p(-2 x)=p(x)+3 p(-x)
+$$
+
+Define $q(x)=p(x)+p(-x)$, then we find that
+
+$$
+q(2 x)=p(2 x)+p(-2 x)=(p(-x)+3 p(x))+(p(x)+3 p(-x))=4 q(x)
+$$
+
+Note that the zero polynomial is a solution to this equation. Now suppose that $q$ is not the zero polynomial, and let $m \geq 0$ be the degree of $q$. Let $b_{m} \neq 0$ be the coefficient of $x^{m}$ in $q(x)$. At the left-hand side of (3), the coefficient of $x^{m}$ is $2^{m} \cdot b_{m}$, while at the right-hand side the coefficient of $x^{m}$ is $4 b_{m}$. Hence $m=2$. As $q(x)=p(x)+p(-x)$, the polynomial $q(x)$ does not contain any nonzero terms with odd exponent of $x$. Since also $q(0)=2 p(0)=0$, we conclude that
+
+$$
+q(x)=b_{2} x^{2}
+$$
+
+where $b_{2}$ is a real number (possibly zero).
+From (2) we now deduce that $p(2 x)=p(-x)+3 p(x)=2 p(x)+q(x)$, so
+
+$$
+p(2 x)-2 p(x)=b_{2} x^{2}
+$$
+
+Suppose that that degree $n$ of $p$ is greater than 2 . Let $a_{n} \neq 0$ be the coefficient of $x^{n}$ in $p(x)$. At the left-hand side of (4), the coefficient of $x^{n}$ is $\left(2^{n}-2\right) \cdot a_{n} \neq 0$. But the coefficient of $x^{n}$ at the right-hand side vanishes, yielding a contradiction. So the degree of $p$ is at most 2 . As we already know that $p(0)=0$, we must have $p(x)=a_{2} x^{2}+a_{1} x$, where $a_{1}$ and $a_{2}$ are real numbers (possibly zero).
+We finally check that every polynomial of this form is indeed a solution (see solution 1 ).
+
+Problem 3. On a line $l$ there are three different points $A, B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to $B Q$ intersects $B Q$ in $L$ and $B R$ in $T$. The line through $B$ perpendicular to $A R$ intersects $A R$ in $K$ and $A Q$ in $S$.
+(a) Prove that $P, T, S$ are collinear.
+(b) Prove that $P, K, L$ are collinear.
+
+## Solution 1.
+
+(a) Since $P, R$ and $Q$ are collinear, we have $\triangle P A Q \sim \triangle P B R$, hence
+
+$$
+\frac{|A Q|}{|B R|}=\frac{|A P|}{|B P|}
+$$
+
+Conversely, $P, T$ and $S$ are collinear if it holds that
+
+$$
+\frac{|A S|}{|B T|}=\frac{|A P|}{|B P|}
+$$
+
+So it suffices to prove
+
+$$
+\frac{|B T|}{|B R|}=\frac{|A S|}{|A Q|}
+$$
+
+Since $\angle A B T=90^{\circ}=\angle A L B$ and $\angle T A B=\angle B A L$, we have $\triangle A B T \sim \triangle A L B$. And since $\angle A L B=90^{\circ}=\angle Q A B$ and $\angle L B A=\angle A B Q$, we have $\triangle A L B \sim \triangle Q A B$. Hence $\triangle A B T \sim \triangle Q A B$, so
+
+$$
+\frac{|B T|}{|B A|}=\frac{|A B|}{|A Q|}
+$$
+
+Similarly, we have $\triangle A B R \sim \triangle A K B \sim \triangle S A B$, so
+
+$$
+\frac{|B R|}{|B A|}=\frac{|A B|}{|A S|}
+$$
+
+Combining both results, we get
+
+$$
+\frac{|B T|}{|B R|}=\frac{|B T| /|B A|}{|B R| /|B A|}=\frac{|A B| /|A Q|}{|A B| /|A S|}=\frac{|A S|}{|A Q|}
+$$
+
+which had to be proved.
+(b) Let the line $P K$ intersect $B R$ in $B_{1}$ and $A Q$ in $A_{1}$ and let the line $P L$ intersect $B R$ in $B_{2}$ and $A Q$ in $A_{2}$. Consider the points $A_{1}, A$ and $S$ on the line $A Q$, and the points $B_{1}, B$ and $T$ on the line $B R$. As $A Q \| B R$ and the three lines $A_{1} B_{1}, A B$ and $S T$ are concurrent (in $P$ ), we have
+
+$$
+A_{1} A: A S=B_{1} B: B T
+$$
+
+where all lengths are directed. Similarly, as $A_{1} B_{1}, A R$ and $S B$ are concurrent (in $K$ ), we have
+
+$$
+A_{1} A: A S=B_{1} R: R B
+$$
+
+This gives
+
+$$
+\frac{B B_{1}}{B T}=\frac{R B_{1}}{R B}=\frac{R B+B B_{1}}{R B}=1+\frac{B B_{1}}{R B}=1-\frac{B B_{1}}{B R}
+$$
+
+so
+
+$$
+B B_{1}=\frac{1}{\frac{1}{B T}+\frac{1}{B R}} .
+$$
+
+Similary, using the lines $A_{2} B_{2}, A B$ and $Q R$ (concurrent in $P$ ) and the lines $A_{2} B_{2}$, $A T$ and $Q B$ (concurrent in $L$ ), we find
+
+$$
+B_{2} B: B R=A_{2} A: A Q=B_{2} T: T B
+$$
+
+This gives
+
+$$
+\frac{B B_{2}}{B R}=\frac{T B_{2}}{T B}=\frac{T B+B B_{2}}{T B}=1+\frac{B B_{2}}{T B}=1-\frac{B B_{2}}{B T}
+$$
+
+so
+
+$$
+B B_{2}=\frac{1}{\frac{1}{B R}+\frac{1}{B T}}
+$$
+
+We conclude that $B_{1}=B_{2}$, which implies that $P, K$ and $L$ are collinear.
+
+## Solution 2.
+
+(a) Define $X$ as the intersection of $A T$ and $B S$, and $Y$ as the intersection of $A R$ and $B Q$. To prove that $P, S$ and $T$ are collinear, we will use Menelaos' theorem in $\triangle A B X$, so we have to prove
+
+$$
+\frac{A P}{P B} \frac{B S}{S X} \frac{X T}{T A}=-1
+$$
+
+Note that $B$ is between $P$ and $A, X$ is between $S$ and $B$, and $X$ is between $T$ and $A$, so it suffices to prove that
+
+$$
+\frac{|A P|}{|P B|} \frac{|B S|}{|S X|} \frac{|X T|}{|T A|}=1
+$$
+
+Because $A Q$ and $B R$ are parallel, we have $\triangle A Q P \sim \triangle B R P$, hence
+
+$$
+\frac{|A P|}{|B P|}=\frac{|Q A|}{|R B|}
+$$
+
+Also, since $\angle A S B=\angle K B R$ and $\angle B A S=90^{\circ}=\angle B K R$, we have $\triangle A S B \sim$ $\triangle K B R$, hence
+
+$$
+\frac{|B S|}{|R B|}=\frac{|A S|}{|K B|}, \quad \text { so } \quad|B S|=\frac{|A S|}{|K B|}|R B| \text {. }
+$$
+
+Similarly, we have $\triangle A T B \sim \triangle Q A L$, hence
+
+$$
+\frac{|T A|}{|A Q|}=\frac{|T B|}{|A L|}, \quad \text { so } \quad|T A|=\frac{|T B|}{|A L|}|A Q| \text {. }
+$$
+
+As $\angle A S X=\angle A S B=90^{\circ}-\angle A B S=90^{\circ}-\angle A B K=\angle K A B=\angle Y A B$, and $\angle S A X=90^{\circ}-\angle X A B=90^{\circ}-\angle L A B=\angle A B L=\angle A B Y$, we have $\triangle S X A \sim$ $\triangle A Y B$, hence
+
+$$
+\frac{|S X|}{|A Y|}=\frac{|A S|}{|B A|}, \quad \text { so } \quad|S X|=\frac{|A S|}{|B A|}|A Y| \text {. }
+$$
+
+Similarly, we have $\triangle B X T \sim \triangle A Y B$, hence
+
+$$
+\frac{|X T|}{|Y B|}=\frac{|B T|}{|A B|}, \quad \text { so } \quad|X T|=\frac{|B T|}{|A B|}|Y B| \text {. }
+$$
+
+By combining (5) - (9), we find
+
+$$
+\begin{aligned}
+\frac{|A P|}{|P B|} \frac{|B S|}{|S X|} \frac{|X T|}{|T A|} & =\frac{|Q A|}{|R B|} \cdot \frac{|A S|}{|K B|}|R B| \cdot \frac{|B A|}{|A S||A Y|} \cdot \frac{|B T|}{|A B|}|Y B| \cdot \frac{|A L|}{|T B||A Q|} \\
+& =\frac{|A L|}{|K B|} \frac{|Y B|}{|A Y|} .
+\end{aligned}
+$$
+
+Since $\angle Y L A=90^{\circ}=\angle Y K B$ and $\angle A Y L=\angle B Y K$, we have $\triangle A Y L \sim \triangle B Y K$, hence
+
+$$
+\frac{|A L|}{|B K|}=\frac{|A Y|}{|B Y|}, \quad \text { so } \quad \frac{|A L|}{|B K|} \frac{|B Y|}{|A Y|}=1
+$$
+
+By combining (10) and (11), we find
+
+$$
+\frac{|A P|}{|P B|} \frac{|B S|}{|S X|} \frac{|X T|}{|T A|}=1,
+$$
+
+as we wanted to prove.
+(b) Again, we will use Menelaos' theorem in $\triangle A B X$, so we have to prove
+
+$$
+\frac{A P}{P B} \frac{B K}{K X} \frac{X L}{L A}=-1
+$$
+
+Note that $\frac{A P}{P B}<0$, and $\frac{B K}{K X}<0$ if and only if $\frac{X L}{L A}<0$, so it suffices to prove that
+
+$$
+\frac{|A P|}{|P B|} \frac{|B K|}{|K X|} \frac{|X L|}{|L A|}=1
+$$
+
+As $\angle B X L=\angle A X K$ and $\angle B L X=90^{\circ}=\angle A K X$, we have $\triangle B L X \sim \triangle A K X$, hence
+
+$$
+\frac{|X L|}{|X K|}=\frac{|B L|}{|A K|}
+$$
+
+Since $\angle A L B=90^{\circ}=\angle Q A B$, we have $\triangle A L B \sim \triangle Q A B$, hence
+
+$$
+\frac{|L A|}{|A Q|}=\frac{|L B|}{|A B|}, \quad \text { so } \quad|L A|=\frac{|L B|}{|A B|}|A Q| \text {. }
+$$
+
+Similarly, we have $\triangle A K B \sim \triangle A B R$, hence
+
+$$
+\frac{|B K|}{|R B|}=\frac{|A K|}{|A B|}, \quad \text { so } \quad|B K|=\frac{|A K|}{|A B|}|R B|
+$$
+
+By combining (5) and (12) - (14), we find
+
+$$
+\left.\frac{|A P|}{|P B|}\left|\frac{|B K|}{|K X|} \frac{|X L|}{|L A|}=\frac{|Q A|}{|R B|} \cdot \frac{|B L|}{|A K|} \cdot \frac{|A B|}{|L B||A Q|} \cdot \frac{|A K|}{|A B|}\right| R B \right\rvert\,=1,
+$$
+
+which is what we wanted to prove.
+
+Solution 3. As $\angle A K B=\angle A L B=90^{\circ}$, the points $K$ and $L$ belong to the circle with diameter $A B$. Since $\angle Q A B=\angle A B R=90^{\circ}$, the lines $A Q$ and $B R$ are tangents to this circle.
+Apply Pascal's theorem to the points $A, A, K, L, B$ and $B$, all on the same circle. This yields that the intersection $Q$ of the tangent in $A$ and the line $B L$, the intersection $R$ of the tangent in $B$ and the line $A K$, and the intersection of $K L$ and $A B$ are collinear. So $K L$ passes through the intersection of $A B$ and $Q R$, which is point $P$. Hence $P, K$ and $L$ are collinear. This proves part b.
+Now apply Pascal's theorem to the points $A, A, L, K, B$ and $B$. This yields that the intersection $S$ of the tangent in $A$ and the line $B K$, the intersection $T$ of the tangent in $B$ and the line $A L$, and the intersection $P$ of $K L$ and $A B$ are collinear. This proves part a.
+
+## Solution 4.
+
+(a) W.l.o.g. we may assume that $A=(0,0)$ and $B=(1,0)$ and the line through $P$ is in the upper half plane, so $l$ is the $x$-axis, $a$ is the $y$-axis and $b$ is the line $x=1$. Take $P=(p, 0)(p>1)$ and $Q=(0, q)(q>0)$. Since $P Q$ is given by $\frac{x}{p}+\frac{y}{q}=1$, we find $R=\left(1, \frac{q(p-1)}{p}\right)$.
+Now $A R$ is given by $y=\frac{q(p-1)}{p} x$, hence $B S$, the line perpendicular to $A R$ and passing through $B=(1,0)$, is given by $y=-\frac{p}{q(p-1)}(x-1)$. We find $S=\left(0, \frac{p}{q(p-1)}\right)$.
+Moreover $B Q$ is given by $y=-q(x-1)$, hence $A T$, the line perpendicular to $B Q$ and passing through $A=(0,0)$, is given by $y=\frac{1}{q} x$. We find $T=\left(1, \frac{1}{q}\right)$. Since $\frac{|B T|}{|B P|}=\frac{1 / q}{p-1}=\frac{\frac{p}{q(p-1)}}{p}=\frac{|A S|}{|A P|}$, we conclude that $P, T$ and $S$ are collinear.
+(b) Point $K$ is the intersection of $A R$ and $B S$. Solving for $x$ yields
+
+$$
+\begin{aligned}
+\frac{q(p-1)}{p} x & =-\frac{p}{q(p-1)}(x-1) \\
+\left(\frac{q(p-1)}{p}+\frac{p}{q(p-1)}\right) x & =\frac{p}{q(p-1)} \\
+x & =\frac{\frac{p}{q(p-1)}}{\frac{q(p-1)}{p}+\frac{p}{q(p-1)}}
+\end{aligned}
+$$
+
+so
+
+$$
+K=\left(\frac{\frac{p}{q(p-1)}}{\frac{q(p-1)}{p}+\frac{p}{q(p-1)}}, \frac{1}{\frac{q(p-1)}{p}+\frac{p}{q(p-1)}}\right)
+$$
+
+Point $L$ is the point of intersection of $A T$ and $B Q$. Solving for $x$ yields
+
+$$
+\begin{aligned}
+\frac{1}{q} x & =-q(x-1) \\
+\left(\frac{1}{q}+q\right) x & =q \\
+x & =\frac{q}{\frac{1}{q}+q}
+\end{aligned}
+$$
+
+so
+
+$$
+L=\left(\frac{q}{\frac{1}{q}+q}, \frac{1}{\frac{1}{q}+q}\right)
+$$
+
+Let $K_{0}$ and $L_{0}$ be the projections of $K$ and $L$ on the $x$-axis. We have to show that the following fractions are equal:
+
+$$
+\frac{\left|K_{0} K\right|}{\left|K_{0} P\right|}=\frac{\frac{1}{\frac{q(p-1)}{p}+\frac{p}{q(p-1)}}}{p-\frac{\frac{p}{q-1)}}{\frac{q(p-1)}{p}+\frac{p}{q(p-1)}}} \quad \text { and } \quad \frac{\left|L_{0} L\right|}{\left|L_{0} P\right|}=\frac{\frac{1}{\frac{1}{q}+q}}{p-\frac{q}{\frac{1}{q}+q}}
+$$
+
+Working out cross products twice, this comes down to
+
+$$
+\begin{aligned}
+\frac{1}{\frac{q(p-1)}{p}+\frac{p}{q(p-1)}} \cdot\left(p-\frac{q}{\frac{1}{q}+q}\right) & \stackrel{?}{=} \frac{1}{\frac{1}{q}+q} \cdot\left(p-\frac{\frac{p}{q(p-1)}}{\frac{q(p-1)}{p}+\frac{p}{q(p-1)}}\right) \\
+\left(\frac{1}{q}+q\right) \cdot\left(p-\frac{q}{\frac{1}{q}+q}\right) & \stackrel{?}{=}\left(\frac{q(p-1)}{p}+\frac{p}{q(p-1)}\right) \cdot\left(p-\frac{\frac{p}{q(p-1)}}{\frac{q(p-1)}{p}+\frac{p}{q(p-1)}}\right) \\
+\frac{p}{q}+p q-q & \stackrel{?}{=} q(p-1)+\frac{p^{2}}{q(p-1)}-\frac{p}{q(p-1)} \\
+\frac{p}{q}+p q-q & \stackrel{?}{=} q(p-1)+\frac{p(p-1)}{q(p-1)}
+\end{aligned}
+$$
+
+which is clearly true.
+
+Problem 4. Find all quadruples $(a, b, p, n)$ of positive integers, such that $p$ is a prime and
+
+$$
+a^{3}+b^{3}=p^{n}
+$$
+
+Solution 1. Let $(a, b, p, n)$ be a solution. Note that we can write the given equation as
+
+$$
+(a+b)\left(a^{2}-a b+b^{2}\right)=p^{n}
+$$
+
+As $a$ and $b$ are positive integers, we have $a+b \geq 2$, so $p \mid a+b$. Furthermore, $a^{2}-a b+b^{2}=$ $(a-b)^{2}+a b$, so either $a=b=1$ or $a^{2}-a b+b^{2} \geq 2$. Assume that the latter is the case. Then $p$ is a divisor of both $a+b$ and $a^{2}-a b+b^{2}$, hence also of $(a+b)^{2}-\left(a^{2}-a b+b^{2}\right)=3 a b$. This means that $p$ either is equal to 3 or is a divisor of $a b$. Since $p$ is a divisor of $a+b$, we have $p|a \Leftrightarrow p| b$, hence either $p=3$, or $p \mid a$ and $p \mid b$. If $p \mid a$ and $p \mid b$, then we can write $a=p a^{\prime}, b=p b^{\prime}$ with $a^{\prime}$ and $b^{\prime}$ positive integers, and we have $\left(a^{\prime}\right)^{3}+\left(b^{\prime}\right)^{3}=p^{n-3}$, so $\left(a^{\prime}, b^{\prime}, p, n-3\right)$ then is another solution (note that $\left(a^{\prime}\right)^{3}+\left(b^{\prime}\right)^{3}$ is a positive integer greater than 1 , so $n-3$ is positive).
+Now assume that $\left(a_{0}, b_{0}, p_{0}, n_{0}\right)$ is a solution such that $p \nmid a$. From the reasoning above it follows that either $a_{0}=b_{0}=1$, or $p_{0}=3$. After all, if we do not have $a_{0}=b_{0}=1$ and we have $p_{0} \neq 3$, then $p \mid a$. Also, given an arbitrary solution $(a, b, p, n)$, we can divide everything by $p$ repeatedly until there are no factors $p$ left in $a$.
+Suppose $a_{0}=b_{0}=1$. Then the solution is $(1,1,2,1)$.
+Suppose $p_{0}=3$. Assume that $3^{2} \mid\left(a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}\right)$. As $3^{2} \mid\left(a_{0}+b_{0}\right)^{2}$, we then have $3^{2} \mid\left(a_{0}+b_{0}\right)^{2}-\left(a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}\right)=3 a_{0} b_{0}$, so $3 \mid a_{0} b_{0}$. But $3 \nmid a_{0}$ by assumption, and $3 \mid a_{0}+b_{0}$, so $3 \nmid b_{0}$, which contradicts $3 \mid a_{0} b_{0}$. We conclude that $3^{2} \nmid\left(a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}\right)$. As both $a_{0}+b_{0}$ and $a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}$ must be powers of 3 , we have $a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}=3$. Hence $\left(a_{0}-b_{0}\right)^{2}+a_{0} b_{0}=3$. We must have $\left(a_{0}-b_{0}\right)^{2}=0$ or $\left(a_{0}-b_{0}\right)^{2}=1$. The former does not give a solution; the latter gives $a_{0}=2$ and $b_{0}=1$ or $a_{0}=1$ and $b_{0}=2$.
+So all solutions with $p \nmid a$ are $(1,1,2,1),(2,1,3,2)$ and $(1,2,3,2)$. From the above it follows that all other solutions are of the form $\left(p_{0}^{k} a_{0}, p_{0}^{k} b_{0}, p_{0}, n_{0}+3 k\right)$, where $\left(a_{0}, b_{0}, p_{0}, n_{0}\right)$ is one of these three solutions. Hence we find three families of solutions:
+
+- $\left(2^{k}, 2^{k}, 2,3 k+1\right)$ with $k \in \mathbb{Z}_{\geq 0}$,
+- $\left(2 \cdot 3^{k}, 3^{k}, 3,3 k+2\right)$ with $k \in \mathbb{Z}_{\geq 0}$,
+- $\left(3^{k}, 2 \cdot 3^{k}, 3,3 k+2\right)$ with $k \in \mathbb{Z}_{\geq 0}$.
+
+It is easy to check that all these quadruples are indeed solutions.
+
+Solution 2. Let $(a, b, p, n)$ be a solution. Note that we can write the given equation as
+
+$$
+(a+b)\left(a^{2}-a b+b^{2}\right)=p^{n}
+$$
+
+As $a$ and $b$ are positive integers, we have $a+b \geq 2$ and $a^{2}-a b+b^{2}=(a-b)^{2}+a b \geq 1$. So both factors are positive and therefore must be powers of $p$. Let $k$ be an integer with $1 \leq k \leq n$ such that $a+b=p^{k}$. Then $a^{2}-a b+b^{2}=p^{n-k}$. If we substitute $b=p^{k}-a$, we find
+
+$$
+p^{n-k}=(a+b)^{2}-3 a b=p^{2 k}-3 a\left(p^{k}-a\right) .
+$$
+
+We can rewrite this as:
+
+$$
+3 a^{2}-3 p^{k} a+p^{2 k}-p^{n-k}=0
+$$
+
+from which we see that $a$ is a solution of the following quadratic equation in $x$ :
+
+$$
+3 x^{2}-3 p^{k} x+p^{2 k}-p^{n-k}=0 .
+$$
+
+The discriminant of (15) is
+
+$$
+D=\left(-3 p^{k}\right)^{2}-4 \cdot 3 \cdot\left(p^{2 k}-p^{n-k}\right)=3 \cdot\left(4 p^{n-k}-p^{2 k}\right)=3 p^{n-k} \cdot\left(4-p^{3 k-n}\right)
+$$
+
+As $p^{n-k}=(a+b)^{2}-3 a b<(a+b)^{2}=p^{2 k}$, we have $n-k<2 k$, so $3 k-n>0$. Since $a$ is a solution of (15), the discriminant must be nonnegative. Hence $4-p^{3 k-n} \geq 0$. If $p=2$, this implies $3 k-n=1$ or $3 k-n=2$; if $p=3$, this implies $3 k-n=1$; and if $p>3$, then $p \geq 5$ so $4 \geq p^{3 k-n}$ can never be true.
+Suppose $p=2$ and $3 k-n=1$. Then $D=3 \cdot 2^{2 k-1} \cdot(4-2)=3 \cdot 2^{2 k}$. But this is a not a square, so the solutions of (15) will not be integers, which yields a contradiction.
+Suppose $p=2$ and $3 k-n=2$. Then $D=3 \cdot 2^{2 k-2} \cdot(4-4)=0$, so the only solution of (15) is $x=\frac{3 \cdot 2^{k}}{2 \cdot 3}=2^{k-1}$. Therefore $a=2^{k-1}$ and $b=2^{k}-a=2^{k-1}$, and this gives a solution for all $k \geq 1$, namely ( $\left.2^{k-1}, 2^{k-1}, 2,3 k-2\right)$.
+Suppose $p=3$ and $3 k-n=1$. Then $D=3 \cdot 3^{2 k-1} \cdot(4-3)=3^{2 k}$, so the solutions of (15) are $x=\frac{3^{k+1} \pm 3^{k}}{2 \cdot 3}=\frac{1}{2}\left(3^{k} \pm 3^{k-1}\right)$. Therefore $a=2 \cdot 3^{k-1}$ or $a=3^{k-1}$. For all $k \geq 1$ we find the solutions $\left(2 \cdot 3^{k-1}, 3^{k-1}, 3,3 k-1\right)$ and $\left(3^{k-1}, 2 \cdot 3^{k-1}, 3,3 k-1\right)$.
+We conclude that there are three families of solutions:
+
+- $\left(2^{k-1}, 2^{k-1}, 2,3 k-2\right)$ with $k \in \mathbb{Z}_{\geq 1}$,
+- $\left(2 \cdot 3^{k-1}, 3^{k-1}, 3,3 k-1\right)$ with $k \in \mathbb{Z}_{\geq 1}$,
+- $\left(3^{k-1}, 2 \cdot 3^{k-1}, 3,3 k-1\right)$ with $k \in \mathbb{Z}_{\geq 1}$.
+
+It is easy to check that all these quadruples are indeed solutions.
+

Benelux/md/Benelux_en-olympiad_en-bxmo-problems-2011-zz.md

@@ -0,0 +1,91 @@
+THIRD BENELUX<br>MATHEMATICAL OLYMPIAD<br>Luxembourg, 6-8 May 2011
+
+## PROBLEMS
+
+## AND
+
+## SOLUTIONS
+
+## Problem 1
+
+An ordered pair of integers $(m, n)$ with $1<m<n$ is said to be a Benelux couple if the following two conditions hold: $m$ has the same prime divisors as $n$, and $m+1$ has the same prime divisors as $n+1$.
+(a) Find three Benelux couples $(m, n)$ with $m \leqslant 14$.
+(b) Prove that there exist infinitely many Benelux couples.
+
+## Solution
+
+(a) It is possible to see that $(2,8),(6,48)$ and $(14,224)$ are Benelux couples.
+(b) Let $k \geqslant 2$ be an integer and $m=2^{k}-2$. Define $n=m(m+2)=2^{k}\left(2^{k}-2\right)$. Since $m$ is even, $m$ and $n$ have the same prime factors. Also, $n+1=m(m+2)+1=(m+1)^{2}$, so $m+1$ and $n+1$ have the same prime factors, too. We have thus obtained a Benelux couple $\left(2^{k}-2,2^{k}\left(2^{k}-2\right)\right)$ for each $k \geqslant 2$.
+
+## Problem 2
+
+Let $A B C$ be a triangle with incentre $I$. The angle bisectors $A I, B I$ and $C I$ meet $[B C],[C A]$ and $[A B]$ at $D, E$ and $F$, respectively. The perpendicular bisector of $[A D]$ intersects the lines $B I$ and $C I$ at $M$ and $N$, respectively. Show that $A, I, M$ and $N$ lie on a circle.
+
+## Solution
+
+The quadrilateral $A M D B$ is cyclic. Indeed, $M$ is the intersection of the line $B I$, which bisects the angle $\widehat{A B D}$ in $A B D$ and the perpendicular bisector of $[A D]$. By uniqueness of this intersection point, it follows that $M$ lies on the circumcircle of $A B D$, and thence $A M D B$ is cyclic. Analogously, $A N D C$ is cyclic.
+Moreover, $M N B C$ is cyclic, for $\widehat{N M I}=90^{\circ}-\widehat{E I A}$. Indeed, $A$ and $I$ lie on either side of the midpoint of $[A D]$, for $\widehat{B D A}=\widehat{C A D}+\widehat{B C A}>\widehat{B A D}$. But
+
+$$
+\widehat{E I A}=180^{\circ}-\frac{1}{2} \widehat{B A C}-\widehat{B E A}=180^{\circ}-\frac{1}{2} \widehat{B A C}-\left(\frac{1}{2} \widehat{C B A}+\widehat{B C A}\right)=90^{\circ}-\frac{1}{2} \widehat{B C A}
+$$
+
+It follows that $\widehat{N M I}=\widehat{B C I}$, which implies that $M N B C$ is cyclic, as the points $I$ and $D$ lie on the same side of $M N$.
+
+There are now two ways of completing the proof:
+
+## Solution 1 (using $A M D B$ and $M N B C$ )
+
+Since $A M D B$ is cyclic, $\widehat{M A I}=\widehat{M A D}=\widehat{M B D}$, as, by construction, $B$ and $M$ lie on either side of $A D$. Moreover, $\widehat{M B D}=\widehat{M B C}=\widehat{M N C}$ for $M N B C$ is cyclic. Thus $\widehat{M A I}=\widehat{M N I}$, so $A M I N$ is cyclic, for $M$ and $N$ lie on either side of $A D$.
+
+## Solution 2 (using $A M D B$ and $A N D C$ )
+
+Since $A M D B$ and $A N D C$ are cylic, $\widehat{A M I}+\widehat{A N I}=\widehat{A M B}+\widehat{A N C}=\widehat{A D B}+\widehat{A D C}=180^{\circ}$, because $B$ and $M$, and $C$ and $N$ lie on either side of $A D$. Hence $A M I N$ is cyclic, for $M$ and $N$ lie on either side of $A D$.
+![](https://cdn.mathpix.com/cropped/2024_12_15_2f0d13707b1875872aebg-3.jpg?height=915&width=1014&top_left_y=1530&top_left_x=521)
+
+Remark. It is moreover true that $B M \perp D N$ and $C N \perp D M$. Indeed, symmetry implies that the image $J$ of $I$ under reflection in $M N$ lies on the circumcircle of $D M N$. Moreover, $D I$ is a height of $D M N$, so the fact that $I$ and $J$ are equidistant from the side $[M N]$ implies that $I$ is the orthocentre of $D M N$. This implies the claim.
+
+## Problem 3
+
+If $k$ is an integer, let $\mathrm{c}(k)$ denote the largest cube that is less than or equal to $k$. Find all positive integers $p$ for which the following sequence is bounded:
+
+$$
+a_{0}=p \quad \text { and } \quad a_{n+1}=3 a_{n}-2 c\left(a_{n}\right) \quad \text { for } n \geqslant 0
+$$
+
+## Solution
+
+Since $\mathrm{c}\left(a_{n}\right) \leqslant a_{n}$ for all $n \in \mathbb{N}, a_{n+1} \geqslant a_{n}$ with equality if and only if $\mathrm{c}\left(a_{n}\right)=a_{n}$. Hence the sequence is bounded if and only if it is eventually constant, which is if and only if $a_{n}$ is a perfect cube, for some $n \geqslant 0$. In particular, the sequence is bounded if $p$ is a perfect cube.
+
+We now claim that, if $a_{n}$ is not a cube for some $n$, then neither is $a_{n+1}$. Indeed, if $a_{n}$ is not a cube, $q^{3}<a_{n}<(q+1)^{3}$ for some $q \in \mathbb{N}$, so that $\mathrm{c}\left(a_{n}\right)=q^{3}$. Suppose to the contrary that $a_{n+1}$ is a cube. Then
+
+$$
+a_{n+1}=3 a_{n}-2 \mathrm{c}\left(a_{n}\right)<3(q+1)^{3}-2 q^{3}=q^{3}+9 q^{2}+9 q+3<q^{3}+9 q^{2}+27 q+27=(q+3)^{3}
+$$
+
+Also, since $\mathrm{c}\left(a_{n}\right)<a_{n}, a_{n+1}>a_{n}>q^{3}$, so $q^{3}<a_{n+1}<(q+3)^{3}$. It follows that the only possible values of $a_{n+1}$ are $(q+1)^{3}$ and $(q+2)^{3}$. However, in both of these cases,
+
+$$
+\begin{aligned}
+& 3 a_{n}-2 q^{3}=a_{n+1}=(q+1)^{3} \Longleftrightarrow 3 a_{n}=3\left(q^{3}+q^{2}+q\right)+1 \\
+& 3 a_{n}-2 q^{3}=a_{n+1}=(q+2)^{3} \Longleftrightarrow 3 a_{n}=3\left(q^{3}+2 q^{2}+4 q\right)+8
+\end{aligned}
+$$
+
+a contradiction modulo 3 . This proves that, if $a_{n}$ is not a cube, then neither is $a_{n+1}$. Hence, if $p$ is not a perfect cube, $a_{n}$ is not a cube for any $n \in \mathbb{N}$, and the sequence is not bounded. We conclude that the sequence is bounded if and only if $p$ is a perfect cube.
+
+## Problem 4
+
+Abby and Brian play the following game: They first choose a positive integer $N$. Then they write numbers on a blackboard in turn. Abby starts by writing a 1. Thereafter, when one of them has written the number $n$, the other writes down either $n+1$ or $2 n$, provided that the number is not greater than $N$. The player who writes $N$ on the blackboard wins.
+(a) Determine which player has a winning strategy if $N=2011$.
+(b) Find the number of positive integers $N \leqslant 2011$ for which Brian has a winning strategy.
+
+## Solution
+
+(a) Abby has a winning strategy for odd $N$ : Observe that, whenever any player writes down an odd number, the other player has to write down an even number. By adding 1 to that number, the first player can write down another odd number. Since Abby starts the game by writing down an odd number, she can force Brian to write down even numbers only. Since $N$ is odd, Abby will win the game. In particular, Abby has a winning strategy if $N=2011$.
+(b) - Let $N=4 k$. If any player is forced to write down a number $m \in\{k+1, k+2, \ldots, 2 k\}$, the other player wins the game by writing down $2 m \in\{2 k+2,2 k+4, \ldots, 4 k\}$, for the players will have to write down the remaining numbers one after the other. Since there is an even number of numbers remaining, the latter player wins. This implies that the player who can write down $k$, i.e. has a winning strategy for $N=k$, wins the game for $N=4 k$.
+
+- Similarly, let $N=4 k+2$. If any player is forced to write down a number $m \in\{k+1, k+2, \ldots, 2 k+1\}$, the other player wins the game by writing down $2 m \in\{2 k+2,2 k+4, \ldots, 4 k+2\}$, as in the previous case. Analogously, this implies that the player who has a winning strategy for $N=k$ wins the game for $N=4 k+2$.
+
+Since Abby wins the game for $N=1,3$, while Brian wins the game for $N=2$, Brian wins the game for $N=8,10$, as well, and thus for $N=32,34,40,42$, too. Then Brian wins the game for a further 8 values of $N$ between 128 and 170, and thence for a further 16 values between 512 and 682 , and for no other values with $N \leqslant 2011$. Hence Brian has a winning strategy for precisely 31 values of $N$ with $N \leqslant 2011$.
+

Benelux/md/Benelux_en-olympiad_en-bxmo-problems-2012-zz.md

@@ -0,0 +1,158 @@
+# 4th Benelux Mathematical Olympiad <br> 20-22 April 2012 - Namur, Belgium 
+
+## Solutions
+
+Problem 1. A sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ of natural numbers is defined by the rule
+
+$$
+a_{n+1}=a_{n}+b_{n} \quad(n=1,2, \ldots)
+$$
+
+where $b_{n}$ is the last digit of $a_{n}$. Prove that such a sequence contains infinitely many powers of 2 if and only if $a_{1}$ is not divisible by 5 .
+
+Solution. First we can observe that:
+
+- If $a_{1}$ is divisible by 5 , then $a_{n}=a_{2}=0(\bmod 10) \forall n \geq 2$.
+- If $a_{1}$ is not divisible by 5 , then for $n \geq 2: a_{n}$ is even, the sequence $b_{n}$ is periodic, its period is a cyclic permutation of $(2,4,8,6)$, and $a_{n+4}=a_{n}+20$.
+(a) Let us suppose that $a_{1}$ is divisible by 5 .
+
+Since $2^{k} \neq 0(\bmod 10)$ for any $k \in \mathbb{N}$, the sequence does not contain any power of 2 for $n \geq 2$.
+(b) Let us suppose that $a_{1}$ is not divisible by 5 .
+
+We can remark that the sequence of powers of 2 modulo 20 respects the period $(12,4,8,16)$ starting with $2^{5}=32$. We choose $j$ such that $a_{j}=2(\bmod 10)\left(\right.$ i.e. $\left.b_{j}=2\right)$ and look at the parity of its penultimate digit.
+
+- If $a_{j}=12(\bmod 20)$, then the numbers $a_{j+4 k}, k \in \mathbb{N}$, represent all the numbers congruent to $12(\bmod 20)$ and greater than $a_{j}$, so all powers of 2 congruent to 12 $(\bmod 20)$ and greater than $a_{j}$ appear in the sequence.
+- If $a_{j}=2(\bmod 20)$, then the numbers $a_{j+1+4 k}, k \in \mathbb{N}$, represent all the numbers congruent to $4(\bmod 20)$ and greater than $a_{j+1}$, so all powers of 2 congruent to 4 $(\bmod 20)$ and greater than $a_{j+1}$ appear in the sequence.
+
+Thus, the sequence contains infinitely many powers of 2 .
+
+Alternative 1 for (b). We choose $j$ such that $a_{j}=2(\bmod 10)$ (i.e. $\left.b_{j}=2\right)$.
+
+- If $a_{j}=20 t+12$ for some $t \in \mathbb{N}$, then $a_{j+4 k}=a_{j}+20 k=20(t+k)+12, \forall k \in \mathbb{N}$. We obtain infinitely many powers of 2 by taking $k=\frac{2^{4 s+3}-3}{5}-t$ (with $s \in \mathbb{N}$ large enough to have $k>0)$ since $2^{4 s+3}=3(\bmod 5), \forall s \in \mathbb{N}$.
+- If $a_{j}=20 t+2$ for some $t \in \mathbb{N}$, then $a_{j+1+4 k}=a_{j+1}+20 k=20(t+k)+4, \forall k \in \mathbb{N}$. We obtain infinitely many powers of 2 by taking $k=\frac{2^{4 s}-1}{5}-t$ (with $s \in \mathbb{N}$ large enough to have $k>0)$ since $2^{4 s}=1(\bmod 5), \forall s \in \mathbb{N}$.
+
+Alternative 2 for (b). Choose $j$ such that $a_{j}$ is a multiple of 4 , i.e. $a_{j}=4 q$ (such a $j$ always exists since $a_{n+1}=a_{n}+2$ for infinitely many $n$ ). Then we have $a_{j+4 k}=a_{j}+20 k=4(q+5 k)$. Let us look for $(k, m)$ such that
+
+$$
+a_{j+4 k}=2^{m} \Longleftrightarrow 4(q+5 k)=2^{m} \Longleftrightarrow q+5 k=2^{m-2} \Longleftrightarrow 2^{m-2}=q(\bmod 5)
+$$
+
+Since $q$ could not be a multiple of 5 , we have $q \in\{1,2,3,4\}(\bmod 5)$. Since the sequence $2^{m-2}(\bmod 5)$ is periodic with period $(1,2,4,3)$, we find that $2^{m-2}=q(\bmod 5)$ happens for infinitely many values of $m$. Hence $2^{m-2}=q+5 k$ is solvable for infinitely many pairs $(k, m)$. Noting that $m$ determines $k$ and that $k$ is nonnegative as soon as $m$ is large enough concludes the proof.
+
+Alternative 3 for (b). We shall show that for any $n>1$ there is some $k \geq n$ such that $a_{k}$ is a power of 2. First, we observe that we can always find $m \in\{n, n+1, n+2, n+3\}$ such that $a_{m}$ is divisible by 4 . If $a_{m}$ is not a power of 2 , we write $a_{m}=2^{b} c$ with $b \geq 2$ and $c>1$ odd. Then we have
+
+$$
+a_{m+4 \cdot 2^{b-2}}=a_{m}+20\left(2^{b-2}\right)=2^{b} c+5 \cdot 2^{b}=2^{b+1} \frac{c+5}{2} .
+$$
+
+If $c>5$, we have $\frac{c+5}{2}<c$ and hence the odd factor of $a_{m+4.2^{b-2}}$ is strictly smaller than the odd factor of $a_{m}$. Therefore there is some $m^{\prime}>m$ such that $a_{m^{\prime}}=2^{b^{\prime}} c^{\prime}$ with $c^{\prime}$ odd and $\leq 5$. The case $c^{\prime}=5$ is forbidden. If $c^{\prime}=1$, then $a_{m^{\prime}}$ is a power of 2 . If $c^{\prime}=3$, then $a_{m^{\prime}+4.2^{b^{\prime}-2}}=2^{b^{\prime}+3}$ is a power of 2 .
+
+Problem 2. Find all quadruples $(a, b, c, d)$ of positive real numbers such that $a b c d=1$, $a^{2012}+2012 b=2012 c+d^{2012}$ and $2012 a+b^{2012}=c^{2012}+2012 d$.
+
+Solution. Rewrite the last two equations into
+
+$$
+a^{2012}-d^{2012}=2012(c-b) \text { and } c^{2012}-b^{2012}=2012(a-d)
+$$
+
+and observe that $a=d$ holds if and only if $c=b$ holds. In that case, the last two equations are satisfied, and condition $a b c d=1$ leads to a set of valid quadruples of the form $(a, b, c, d)=\left(t, \frac{1}{t}, \frac{1}{t}, t\right)$ for any $t>0$.
+We show that there are no other solutions. Assume that $a \neq d$ and $c \neq b$. Multiply both sides of (1) to obtain
+
+$$
+\left(a^{2012}-d^{2012}\right)\left(c^{2012}-b^{2012}\right)=2012^{2}(c-b)(a-d)
+$$
+
+and divide the left-hand side by the (nonzero) right-hand side to get
+
+$$
+\frac{a^{2011}+\cdots+a^{2011-i} d^{i}+\cdots+d^{2011}}{2012} \cdot \frac{c^{2011}+\cdots+c^{2011-i} b^{i}+\cdots+b^{2011}}{2012}=1
+$$
+
+Now apply the arithmetic-geometric mean inequality to the first factor
+
+$$
+\frac{a^{2011}+\cdots+a^{2011-i} d^{i}+\cdots+d^{2011}}{2012}>\sqrt[2012]{(a d)^{\frac{2011 \times 2012}{2}}}=(a d)^{\frac{2011}{2}}
+$$
+
+The inequality is strict, since equality holds only if all terms in the mean are equal to each other, which happens only if $a=d$. Similarly, we find
+
+$$
+\frac{c^{2011}+\cdots+c^{2011-i} b^{i}+\cdots b^{2011}}{2012}>\sqrt[2012]{(c b)^{\frac{2011 \times 2012}{2}}}=(c b)^{\frac{2011}{2}} .
+$$
+
+Multiplying both inequalities, we obtain
+
+$$
+(a d)^{\frac{2011}{2}}(c b)^{\frac{2011}{2}}<1
+$$
+
+which is equivalent to $a b c d<1$, a contradiction.
+
+Problem 3. In triangle $A B C$ the midpoint of $B C$ is called $M$. Let $P$ be a variable interior point of the triangle such that $\angle C P M=\angle P A B$. Let $\Gamma$ be the circumcircle of triangle $A B P$. The line $M P$ intersects $\Gamma$ a second time in $Q$. Define $R$ as the reflection of $P$ in the tangent to $\Gamma$ in $B$. Prove that the length $|Q R|$ is independent of the position of $P$ inside the triangle.
+
+Solution. We claim $|Q R|=|B C|$, which will clearly imply that quantity $|Q R|$ is independent from the position of $P$ inside triangle $\triangle A B C$ (and independent from the position of $A$ ).
+This equality will follow from the equality between triangles $\triangle B P C$ and $\triangle R B Q$. This in turn will be shown by means of three equalities (two sides and an angle): $|B P|=|R B|,|P C|=|B Q|$ and $\angle B P C=\angle R B Q$.
+![](https://cdn.mathpix.com/cropped/2024_12_15_ba09b4ac5be87f63f6f9g-4.jpg?height=849&width=1178&top_left_y=995&top_left_x=439)
+(a) $|B P|=|R B|$
+
+Obvious since $R$ is the reflection of $P$ in a line going through $B$.
+(b) $|P C|=|B Q|$
+
+Let $U$ be the fourth vertex of parallelogram $B P C U$. Then $U$ is on line $P Q$ and $\angle B U P=$ $\angle U P C=\alpha$. If $Q$ is on the same $\operatorname{arc} P B$ as $A$, then $\angle B Q P=\alpha$, and $\triangle Q P U$ is isosceles; hence, $|B Q|=|B U|=|P C|$. On the other way, if $Q$ is on the other $\operatorname{arc} P B$, then $\angle B Q P$ and $\alpha$ are supplementary, hence $B Q U=\alpha$, and again $\triangle Q P U$ is isosceles; the same conclusion follows.
+(c) $\angle B P C=\angle R B Q$
+
+Define $T$ to be the midpoint of $P R$. Then line $B T$, tangent to circle $\Gamma$ in $B$, splits $\angle R B Q$ into two parts, $\angle R B T$ and $\angle T B Q$.
+We first show that $\angle R B T=\alpha$. Indeed, by symmetry, $\angle R B T=\angle P B T$ and, since $B T$ is tangent to $\Gamma$, we have that $\angle P B T=\angle P A B$ (because they both intercept the same $\operatorname{arc} \widehat{P B}$ on circle $\Gamma$ ), from which our claim follows.
+
+We then show that $\angle T B Q=\angle B P M$. Indeed, since $\angle T B Q$ and $\angle B P Q$ intercept opposite arcs on circle $\Gamma$, they are supplementary and we have $\angle T B Q=\pi-\angle B P Q=\angle B P M$. We finally conclude that
+
+$$
+\angle R B Q=\angle R B T+\angle T B Q=\alpha+\angle B P M=\angle M P C+\angle B P M=\angle B P C
+$$
+
+We have thus shown $\triangle B P C=\triangle R B Q$, which completes the proof.
+
+Note. Notice that point $A$ does not play any role in the problem except fixing circle $\Gamma$ (and, for that reason, the result is also valid when $P$ is chosen outside of triangle $\triangle A B C$ ).
+
+Alternative 1 for (b). The law of sines in triangle $\triangle B Q M$ gives
+
+$$
+\frac{|B M|}{\sin \angle B Q M}=\frac{|B Q|}{\sin \angle B M Q} .
+$$
+
+Since $Q$ belongs to circle $\Gamma$, we have either $\angle B Q P=\angle B A P=\alpha$, hence $\angle B Q M=\angle M P C$, or these angles are supplementary; in both cases they have equal sines. We also have that $\angle B M Q$ and $\angle C M P$ are supplementary, hence have equal sines. Using these facts along with $|B M|=|M C|$ transforms (2) into
+
+$$
+\frac{|M C|}{\sin \angle M P C}=\frac{|B Q|}{\sin \angle C M P}
+$$
+
+from which the law of sines in triangle $\triangle C P M$ implies that $|B Q|=|P C|$.
+
+Alternative 2 for (b). Let $S$ be the second intersection of line $C P$ with circle $\Gamma$. Then, $\angle B S P=\alpha$, so $B S$ and $M P$ are parallel; since $M$ is the midpoint of segment $B C, P$ is the midpoint of $S C$. If $Q$ is on the same $\operatorname{arc} P B$ as $A$, then the quadrilateral $Q P B S$ is an isosceles trapezoid, and $|Q B|=|S P|=|P C|$. If $Q$ is on the other arc $P B$, then the quadrilateral $P Q B S$ is an isosceles trapezoid, and again $|Q B|=|S P|=|P C|$.
+
+Problem 4. Yesterday, $n \geq 4$ people sat around a round table. Each participant remembers only who his two neighbours were, but not which one sat on his left and which one sat on his right. Today, you would like the same people to sit around the same round table so that each participant has the same two neighbours as yesterday (it is possible that yesterday's lefthand side neighbour is today's right-hand side neighbour). You are allowed to query some of the participants: if anyone is asked, he will answer by pointing at his two neighbours from yesterday.
+(a) Determine the minimal number $f(n)$ of participants you have to query in order to be certain to succeed, if later questions must not depend on the outcome of the previous questions. That is, you have to choose in advance the list of people you are going to query, before effectively asking any question.
+(b) Determine the minimal number $g(n)$ of participants you have to query in order to be certain to succeed, if later questions may depend on the outcome of previous questions. That is, you can wait until you get the first answer to choose whom to ask the second question, and so on.
+
+## Solution.
+
+(a) $f(n)=n-3$.
+
+- Asking $n-4$ questions is not enough since the $n-4$ people queried might be sitting in a consecutive string, in which case the $n-4$ answers allow one to sit $n-2$ people in the same positions as yesterday, but there is still an ambiguity among the two remaining ones.
+- Let us show that $n-3$ questions suffice. Among the 3 people who are not queried, at least 2 must sit next to people who have been queried. If exactly 2 do, then both these people must be neighbours of the third, so that the neighbours of everybody are known and we are done. If all 3 unqueried people sit next to a queried person, then at least one of them has two queried neighbours, and again it follows that the neighbours of everybody are known, so that we are done.
+(b) $g(n)=n-1-\left\lceil\frac{n}{3}\right\rceil\left(=n-1-\left\lfloor\frac{n+2}{3}\right\rfloor=\left\lfloor\frac{2 n}{3}\right\rfloor-1=\left\lceil\frac{2 n-5}{3}\right\rceil\right)$.
+
+Say there is a link between two people if and only if they are neighbours. There are in total $n$ links, which we all need to identify. By asking a person for his neighbours, we can discover at most two new links. More precisely, if at any point we query a participant who has not yet been pointed as a neighbour, we discover exactly two new links (we call this a type-0 query). If we query a participant who has been pointed once as a neighbour, will discover exactly one new link (we call this a type-1 query). Of course, querying a participant who has already been pointed twice provides no information (and we assume in the rest of this solution that it never happens).
+First note that, since $f(4)=1$, we also have $g(4)=1$. We now prove the formula for $g(n)$ for $n \geq 5$.
+
+- Let us show that $n-1-\left\lceil\frac{n}{3}\right\rceil$ questions suffice. Our strategy consists in making sure that the first $\left\lceil\frac{n}{3}\right\rceil$ queries are type- 0 . Let us show that this is always possible. A type-0 query requires a participant that hasn't been queried or pointed before. Since the number of those participants decreases by three at most after each query, we see that it is always possible to perform $\left\lceil\frac{n}{3}\right\rceil$ type-0 queries first. During this phase we discover $2\left\lceil\frac{n}{3}\right\rceil$ links.
+The remaining queries will be either type-0 or type-1, and each of them discovers at least one new link. We perform them until $n-1$ links have been discovered, after which we are done (the last link can be deduced without query). The number of queries in this second phase is therefore at $\operatorname{most}^{1} n-1-2\left\lceil\frac{n}{3}\right\rceil$, and the total is at $\operatorname{most}\left\lceil\frac{n}{3}\right\rceil+\left(n-1-2\left\lceil\frac{n}{3}\right\rceil\right)=n-1-\left\lceil\frac{n}{3}\right\rceil$.
+- We now show that $n-2-\left\lceil\frac{n}{3}\right\rceil=\hat{g}(n)$ questions are not enough.
+(i) Consider the pool of unqueried and unpointed participants ; each type-0 must query this pool. Since, from the point of view of the questioner, all elements of the pool are undistinguishable, we can assume that each type-0 query asks the second leftmost participant in the pool (except if there is only one element left in the pool). One can then check that the pool, which starts as a string of $n$ contiguous participants, will stay contiguous after each type-0 and type1 query. Furthermore, using our assumption, we see that each type-0 query removes three participants from the pool. Therefore there can be at most $\left\lceil\frac{n}{3}\right\rceil$ type-0 queries in the scenarios corresponding to our assumption.
+(ii) Assume there are $k$ type-0 queries. Since there are $\hat{g}(n)$ queries, the number of discovered links is equal to $2 k+(\hat{g}(n)-k)=\hat{g}(n)+k=n-2+k-\left\lceil\frac{n}{3}\right\rceil$. If $k$ is strictly less than $\left\lceil\frac{n}{3}\right\rceil$, we discover strictly less than $n-2$ links, which is clearly insufficient (indeed, there are at least three missing links, and one can check that whatever the configuration of the missing links, there are always several orders compatible with the discovered links).
+(iii) We now analyze the remaining case with $k=\left\lceil\frac{n}{3}\right\rceil$ type-0 queries ${ }^{2}$, in which we discover $n-2$ links. On the one hand, if the missing links are disjoint, there are always two orders compatible with the discovered links (for example when $n=7$ and links are missing between the $(4,5)$ and $(7,1)$ pairs of neighbours, the two orders are $1-2-3-4 \quad 5-6-7$ and $1-2-3-4 \quad 7-6-5)$. On the other hand, a situation where the two missing links would be adjacent would allow the identification of the correct order. However, this never happens in the scenarios corresponding to the assumption we made in (i). Indeed, two adjacent missing links imply that some participant is unqueried and unpointed at the end of the process. Since we perform $k=\left\lceil\frac{n}{3}\right\rceil$ type-0 queries (the maximum), the reasoning from (i) shows that the pool of unqueried and unpointed participants is empty at the end of the process, which contradicts the existence of two adjacent missing links.
+
+[^0]
+[^0]:    ${ }^{1}$ Here we use the assumption $n \geq 5$, since quantity $n-1-2\left\lceil\frac{n}{3}\right\rceil$ is negative when $n=4$.
+    ${ }^{2}$ Note that this cannot happen when $n \in\{4,5,7\}$ since we have $\left\lceil\frac{n}{3}\right\rceil>\hat{g}(n)$ in those cases.
+

Benelux/md/Benelux_en-olympiad_en-bxmo-problems-2013-zz.md

@@ -0,0 +1,186 @@
+# 5th Benelux Mathematical Olympiad 
+
+Dordrecht, 26-28 April 2013
+
+![](https://cdn.mathpix.com/cropped/2024_12_15_38cd9497982ef7e64093g-1.jpg?height=399&width=497&top_left_y=210&top_left_x=1483)
+
+## Solutions
+
+Problem 1. Let $n \geqslant 3$ be an integer. A frog is to jump along the real axis, starting at the point 0 and making $n$ jumps: one of length 1 , one of length $2, \ldots$, one of length $n$. It may perform these $n$ jumps in any order. If at some point the frog is sitting on a number $a \leqslant 0$, its next jump must be to the right (towards the positive numbers). If at some point the frog is sitting on a number $a>0$, its next jump must be to the left (towards the negative numbers). Find the largest positive integer $k$ for which the frog can perform its jumps in such an order that it never lands on any of the numbers $1,2, \ldots, k$.
+
+Solution. We claim that the largest positive integer $k$ with the given property is $\left\lfloor\frac{n-1}{2}\right\rfloor$, where $\lfloor x\rfloor$ is by definition the largest integer not exceeding $x$.
+
+Consider a sequence of $n$ jumps of length $1,2, \ldots n$ such that the frog never lands on any of the numbers $1,2, \ldots, k$, where $k \geqslant 1$. Note that we must have $k<n$ in order for the frog to be able to make its first jump. As the frog jumps to the right only if it is in a number $a \leqslant 0$, and the largest jump has length $n$, it is impossible to reach numbers greater than $n$. On the other hand, suppose the frog is in a number $a>0$, then it must even be in a number $a \geqslant k+1$, since it is not allowed to hit the numbers $1,2, \ldots, k$. So the frog jumps to the left only if it is in a number $a \geqslant k+1$, and therefore it is impossible to reach numbers less than $(k+1)-n=k-n+1$. This means the frog only possibly lands on the numbers $i$ satisfying
+
+$$
+k-n+1 \leqslant i \leqslant 0 \quad \text { or } \quad k+1 \leqslant i \leqslant n \text {. }
+$$
+
+When performing a jump of length $k$, the frog has to remain at either side of the numbers $1,2, \ldots, k$. Indeed, jumping over $1,2, \ldots, k$ requires a jump of at least length $k+1$. In case it starts at a number $a>0$ (in fact $k+1 \leqslant a \leqslant n$ ), it lands in $a-k$ and we must also have $a-k \geqslant k+1$. So $2 k+1 \leqslant a \leqslant n$, therefore $2 k+1 \leqslant n$. In case it starts at a number $a \leqslant 0$ (in fact $k-n+1 \leqslant a \leqslant 0$ ), it lands in $a+k$ and we must also have $a+k \leqslant 0$. Adding $k$ to both sides of $k-n+1 \leqslant a$, we obtain $2 k-n+1 \leqslant a+k \leqslant 0$, so in this case we have $2 k+1 \leqslant n$ as well. We conclude that $k \leqslant \frac{n-1}{2}$. Since $k$ is integer, we even have $k \leqslant\left\lfloor\frac{n-1}{2}\right\rfloor$.
+
+Next we prove that this upperbound is sharp: for $k=\left\lfloor\frac{n-1}{2}\right\rfloor$ the frog really can perform its jumps in such an order that it never lands on any of the numbers $1,2, \ldots, k$.
+
+Suppose $n$ is odd, then $\frac{n-1}{2}$ is an integer and we have $k=\frac{n-1}{2}$, so $n=2 k+1$. We claim that when the frog performs the jumps of length $1, \ldots, 2 k+1$ in the following order, it does never land on $1,2, \ldots, k$ : it starts with a jump of length $k+1$, then it performs two jumps, one of length $k+2$ followed by one of length 1 , next two jumps of length $k+3$ and $2, \ldots$, next two jumps of length $k+(i+1)$ and $i, \ldots$, and finally two jumps of length $k+(k+1)$ and $k$. In this order of the jumps every length between 1 and $n=2 k+1$ does occur: it performs a pair of jumps for $1 \leqslant i \leqslant k$, which are the jumps of length $1,2, \ldots$, $k$ and the jumps of length $k+2, k+3, \ldots, 2 k+1$, and it starts with the jump of length $k+1$.
+
+We now prove the correctness of this jumping scheme. After the first jump the frog lands in $k+1>k$. Now suppose the frog is in 0 or $k+1$ and is about to perform the pair of jumps of length $k+(i+1)$ and $i$. Starting from 0 , it lands in $k+(i+1)>k$, after which it lands in $(k+i+1)-i=k+1>k$. If on the contrary it starts in $k+1$, it lands in $(k+1)-(k+(i+1))=-i<1$, after which it lands in $(-i)+i=0$. We see that, starting in 0 , the frog lands in $k+1$ after the pair of jumps, while starting in $k+1$ the frog lands in 0 , while in both cases the jumps do not touch $1,2, \ldots k$. This proves the correctness of its series of jumps. As the frog (after its first jump) alters between $k+1$ and 0 exactly $k$ times, for odd $k$ it will end up in 0 , while for even $k$ it will end up in $k+1$.
+
+Suppose $n$ is even, then $\frac{n-1}{2}$ is not an integer and we have $k=\frac{n-1}{2}-\frac{1}{2}=\frac{n-2}{2}$, so $n=2 k+2$. Let the frog firstly perform the same series of jumps as in the previous case; they still do not touch $1,2, \ldots, k$. Now let the frog make a final extra jump of length $2 k+2$. It will land in $0+(2 k+2)=2 k+2>k$ if $k$ is odd, or in $(k+1)-(2 k+2)=-k-1<1$ if $k$ is even, and its series of jumps is correct again.
+
+We conclude that the largest positive integer $k$ with the given property is $\left\lfloor\frac{n-1}{2}\right\rfloor$.
+
+Problem 2. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
+
+$$
+f(x+y)+y \leqslant f(f(f(x)))
+$$
+
+holds for all $x, y \in \mathbb{R}$.
+
+Solution. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function satisfying the given inequality (2). Writing $z$ for $x+y$, we find that $f(z)+(z-x) \leqslant f(f(f(x)))$, or equivalently
+
+$$
+f(z)+z \leqslant f(f(f(x)))+x
+$$
+
+for all $x, z \in \mathbb{R}$. Substituting $z=f(f(x))$ yields $f(f(f(x)))+f(f(x)) \leqslant f(f(f(x)))+x$, from which we see that
+
+$$
+f(f(x)) \leqslant x
+$$
+
+for all $x \in \mathbb{R}$. Substituting $f(x)$ for $x$ we get $f(f(f(x))) \leqslant f(x)$, which combined with (3) gives $f(z)+z \leqslant f(f(f(x)))+x \leqslant f(x)+x$. So
+
+$$
+f(z)+z \leqslant f(x)+x
+$$
+
+for all $x, z \in \mathbb{R}$. By symmetry we see that we also have $f(x)+x \leqslant f(z)+z$, from which we conclude that in fact we even have
+
+$$
+f(z)+z=f(x)+x
+$$
+
+for all $x, z \in \mathbb{R}$. So $f(z)+z=f(0)+0$ for all $z \in \mathbb{R}$, and we conclude that $f(z)=c-z$ for some $c \in \mathbb{R}$.
+
+Now we check whether all functions of this form satisfy the given inequality. Let $c \in \mathbb{R}$ be given and consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(z)=c-z$ for all $z \in \mathbb{R}$. Note that $f(f(z))=c-(c-z)=z$ for all $z \in \mathbb{R}$. For the lefthand side of (2) we find
+
+$$
+f(x+y)+y=(c-(x+y))+y=c-x,
+$$
+
+while the righthand side reads
+
+$$
+f(f(f(x)))=f(x)=c-x .
+$$
+
+We see that inequality (2) holds; in fact we even have equality here.
+We conclude that the solutions to (2) are given by the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(z)=c-z$ for all $z \in \mathbb{R}$, where $c$ is an arbitrary real constant.
+
+Problem 3. Let $\triangle A B C$ be a triangle with circumcircle $\Gamma$, and let $I$ be the center of the incircle of $\triangle A B C$. The lines $A I, B I$ and $C I$ intersect $\Gamma$ in $D \neq A, E \neq B$ and $F \neq C$. The tangent lines to $\Gamma$ in $F, D$ and $E$ intersect the lines $A I, B I$ and $C I$ in $R, S$ and $T$, respectively. Prove that
+
+$$
+|A R| \cdot|B S| \cdot|C T|=|I D| \cdot|I E| \cdot|I F|
+$$
+
+Solution. We first prove that $|D B|=|D I|$. (This may also be claimed by referring to the lemma that $D$ is the centre of the circumcircle of $B I C I_{a}$.) By the constant angle theorem and the fact that $A D$ and $B E$ are angle bisectors of triangle $A B C$, we see that
+
+$$
+\angle D B I=\angle D B C+\angle C B I=\angle D A C+\angle C B E=\angle D A B+\angle A B E,
+$$
+
+while
+
+$$
+\angle D I B=180^{\circ}-\angle A I B=\angle I A B+\angle A B I=\angle D A B+\angle A B E .
+$$
+
+So $\triangle B D I$ has equal angles $\angle D B I=\angle D I B$, so $|D B|=|D I|$. This proves our claim. We similarly deduce that $|E C|=|E I|$ and $|F A|=|F I|$.
+
+Rewriting (7) into $\frac{|A R|}{|I F|} \cdot \frac{|B S|}{|I D|} \cdot \frac{|C T|}{|I E|}=1$, we see that it suffices to prove that
+
+$$
+\frac{|A R|}{|A F|} \cdot \frac{|B S|}{|B D|} \cdot \frac{|C T|}{|C E|}=1
+$$
+
+We now prove by angle chasing that $\triangle R F A \sim \triangle A C I$. As $R F$ is tangent to the circumcircle of $\triangle A F C$, we obtain (using also that $C F$ is angle bisector of $\angle A C B$ )
+
+$$
+\angle R F A=\angle F C A=\angle I C A .
+$$
+
+Moreover, from $|F A|=|F I|$ we deduce that $\angle F A I=\angle F I A$, so
+
+$$
+\angle F A R=180^{\circ}-\angle F A I=180^{\circ}-\angle F I A=\angle C I A
+$$
+
+This proves our similarity, which entails that $\frac{|A R|}{|A F|}=\frac{|I A|}{|I C|}$. In the same way we deduce that $\frac{|B S|}{|B D|}=\frac{|I B|}{|I A|}$ and $\frac{|C T|}{|C E|}=\frac{|I C|}{|I B|}$. By these equal ratios we know that
+
+$$
+\frac{|A R|}{|A F|} \cdot \frac{|B S|}{|B D|} \cdot \frac{|C T|}{|C E|}=\frac{|I A|}{|I C|} \cdot \frac{|I B|}{|I A|} \cdot \frac{|I C|}{|I B|}=1
+$$
+
+which proves (8), as required.
+
+## Problem 4.
+
+a) Find all positive integers $g$ with the following property: for each odd prime number $p$ there exists a positive integer $n$ such that $p$ divides the two integers
+
+$$
+g^{n}-n \quad \text { and } \quad g^{n+1}-(n+1)
+$$
+
+b) Find all positive integers $g$ with the following property: for each odd prime number $p$ there exists a positive integer $n$ such that $p$ divides the two integers
+
+$$
+g^{n}-n^{2} \quad \text { and } \quad g^{n+1}-(n+1)^{2}
+$$
+
+## Solution.
+
+a) Let $g$ be a positive integer with the given property. So for each odd prime number $p$ there exists a positive integer $n$ such that $p \mid g^{n}-n$ and $p \mid g^{n+1}-(n+1)$.
+If $g$ has an odd prime factor $p$, then from $p \mid g^{n}-n$ it follows that $p \mid n$, while from $p \mid g^{n+1}-(n+1)$ we deduce that $p \mid n+1$. But $p$ cannot divide both $n$ and $n+1$; contradiction. So $g$ is a power of $2: g=2^{k}$ for some $k \geqslant 0$.
+If $g=2^{0}=1$, then $p \mid 1-n$ and $p \mid 1-(n+1)$, which is again a contradiction.
+Suppose $k \geqslant 2$. Then $g-1$ has an odd prime factor $p$, therefore $g \equiv 1(\bmod p)$ so $0 \equiv g^{n}-n \equiv 1-n(\bmod p)$ and $0 \equiv g^{n+1}-(n+1) \equiv 1-(n+1)(\bmod p)$, which is again a contradiction.
+Now we prove that $g=2^{1}=2$ does satisfy the condition. Let a prime $p>2$ be given. Choose $n=(p-1)^{2}$, then we have $n \equiv(-1)^{2}=1(\bmod p)$. By Fermat's little theorem (using $\operatorname{gcd}(2, p)=1)$ we know that $2^{p-1} \equiv 1(\bmod p)$, so
+
+$$
+2^{n}=2^{(p-1)^{2}}=\left(2^{p-1}\right)^{p-1} \equiv 1 \equiv n \quad(\bmod p)
+$$
+
+Multiplying both sides by 2 , we see that also
+
+$$
+2^{n+1} \equiv 2 n=n+n \equiv n+1 \quad(\bmod p)
+$$
+
+We conclude that only $g=2$ has the given property.
+b) Let $g$ be a positive integer with the given property. So for each odd prime number $p$ there exists a positive integer $n$ such that $p \mid g^{n}-n^{2}$ and $p \mid g^{n+1}-(n+1)^{2}$.
+If $g$ has an odd prime factor $p$, then from $p \mid g^{n}-n^{2}$ it follows that $p \mid n^{2}$, so also $p \mid n$, while from $p \mid g^{n+1}-(n+1)^{2}$ we deduce that $p \mid(n+1)^{2}$, so also $p \mid n+1$.
+
+But $p$ cannot divide both $n$ and $n+1$; contradiction. So $g$ is a power of $2: g=2^{k}$ for some $k \geqslant 0$.
+If $g=2^{0}=1$, then for any odd prime $p$ we have $p \mid 1-n^{2}=(1-n)(1+n)$ and $p \mid 1-(n+1)^{2}=(1-(n+1))(1+(n+1))$. Now take $p=5$. The first statement says that $n \equiv 1$ or $n \equiv-1 \equiv 4(\bmod 5)$, and the second that $n \equiv 0$ or $n \equiv-2 \equiv 3$ $(\bmod 5)$. But this yields a contradiction.
+If $g=2^{1}=2$, then for any odd prime $p$ we have $p \mid 2^{n}-n^{2}$ and $p \mid 2^{n+1}-(n+1)^{2}$. Now take $p=3$. As $3 \nmid 2^{n}$ and $3 \nmid 2^{n+1}$, we know that $3 \nmid n^{2}$ and $3 \nmid(n+1)^{2}$. So these two squares must be 1 modulo 3 (as 2 can never be a square modulo 3 ). Therefore also $2^{n}$ and $2^{n+1}$ must be 1 modulo 3 , which gives $2 \cdot 1 \equiv 2 \cdot 2^{n}=2^{n+1} \equiv 1(\bmod 3)$; contradiction.
+Now suppose $k \geqslant 2$. Then $g-1$ has an odd prime factor $p$, therefore $g \equiv 1(\bmod p)$ so $0 \equiv g^{n}-n^{2} \equiv 1-n^{2}=(1-n)(1+n)(\bmod p)$ and $0 \equiv g^{n+1}-(n+1)^{2} \equiv$ $1-(n+1)^{2}=(1-(n+1))(1+(n+1))(\bmod p)$. Suppose $p \geqslant 5$. The first statement says that $n \equiv 1$ or $n \equiv-1(\bmod p)$, and the second that $n \equiv 0$ or $n \equiv-2(\bmod p)$. But $n$ can only be congruent to at most one of the numbers $-2,-1,0$ and 1 , since $p \geqslant 5$; contradiction. We conclude that $p=3$, so $g-1$ contains only prime factors 3 . Hence $2^{k}-1=3^{\ell}$ for some $\ell>0$. We see that $2^{k}-1 \equiv(-1)^{k}-1(\bmod 3)$, while $3^{\ell} \equiv 0(\bmod 3)$. So $k$ has to be even, say $k=2 m$, and our equation becomes $2^{2 m}-1=3^{\ell}$, or equivalently $\left(2^{m}-1\right)\left(2^{m}+1\right)=3^{\ell}$. Not both factors on the left-hand side can be divisible by 3 , so $2^{m}-1=1$ and $2^{m}+1=3^{\ell}$, so $m=1$. Hence $g=2^{2}=4$.
+Now we show that $g=4$ does have the given property. For this we use that $g=2$ is a solution to part (a): for any odd prime $p$ there exists a positive integer $n$ such that
+
+$$
+n \equiv 2^{n} \quad(\bmod p) \quad \text { and } \quad n+1 \equiv 2^{n+1} \quad(\bmod p) .
+$$
+
+Taking the square of both congruences, we obtain
+
+$$
+n^{2} \equiv\left(2^{n}\right)^{2}=\left(2^{2}\right)^{n}=4^{n} \quad(\bmod p)
+$$
+
+and
+
+$$
+(n+1)^{2} \equiv\left(2^{n+1}\right)^{2}=\left(2^{2}\right)^{n+1}=4^{n+1} \quad(\bmod p)
+$$
+
+as desired.
+We conclude that only $g=4$ has the given property.
+

Benelux/md/Benelux_en-olympiad_en-bxmo-problems-2014-zz.md

@@ -0,0 +1,91 @@
+# 6th Benelux Mathematical Olympiad Solutions 
+
+Brugge, May 2-4 2014
+![](https://cdn.mathpix.com/cropped/2024_12_15_fb9be07ea97b63ba84aeg-1.jpg?height=405&width=497&top_left_y=104&top_left_x=1448)
+
+## Problem 1
+
+Find the smallest possible value of the expression
+
+$$
+\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{b+c+d}{a}\right\rfloor+\left\lfloor\frac{c+d+a}{b}\right\rfloor+\left\lfloor\frac{d+a+b}{c}\right\rfloor,
+$$
+
+in which $a, b, c$ and $d$ vary over the set of positive integers.
+(Here $\lfloor x\rfloor$ denotes the biggest integer which is smaller than or equal to $x$.)
+
+## Solution
+
+The answer is 9 .
+Notice that $\lfloor x\rfloor>x-1$ for all $x \in \mathbb{R}$. Therefore the given expression is strictly greater than
+
+$$
+\frac{a+b+c}{d}+\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+b}{c}-4,
+$$
+
+which can be rewritten as
+
+$$
+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{a}{d}+\frac{d}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{b}{d}+\frac{d}{b}\right)+\left(\frac{c}{d}+\frac{d}{c}\right)-4 .
+$$
+
+Since $t+\frac{1}{t} \geq 2$ for $t>0$, we get that $6 \cdot 2-4=8$ is a strict lower bound for the given expression; since it takes integral values only, we actually get that 9 is a lower bound. It remains to check that 9 can be attained; this happens for $a=b=c=5$ and $d=4$.
+
+## Problem 2
+
+Let $k \geq 1$ be an integer.
+We consider $4 k$ chips, $2 k$ of which are red and $2 k$ of which are blue. A sequence of those $4 k$ chips can be transformed into another sequence by a so-called move, consisting of interchanging a number (possibly one) of consecutive red chips with an equal number of consecutive blue chips. For example, we can move from $r \underline{b b b r} \underline{r} \underline{b} b$ to $r \underline{r} \underline{b} b \underline{b b b} b$ where $r$ denotes a red chip and $b$ denotes a blue chip.
+
+Determine the smallest number $n$ (as a function of $k$ ) such that starting from any initial sequence of the $4 k$ chips, we need at most $n$ moves to reach the state in which the first $2 k$ chips are red.
+
+## Solution
+
+The answer is $n=k$.
+We will first show that $n \geq k$. Let us count the number $C$ of times a red chip is directly to the right of a blue chip. In the final position this number equals 0 . In the position brbrbr $\cdots b r$ this number equals $2 k$. We claim that any move reduces this number by at most 2 . Denote by $R$ the group of red chips and by $B$ the group of blue chips that are interchanged. Any reduction in $C$ must involve a red chip getting rid of its blue left neighbour or a blue chip getting rid of its red right neighbour. This can only happen with the leftmost chip of $R$ (if its left neighbour is blue) and the rightmost chip of $B$ (if its right neighbour is red), but not with the rightmost chip of $R$ (and its right neighbour) or the leftmost chip of $B$ (and its left neighbour). Hence $C$ is reduced by at most 2 in any move. Therefore the number of moves necessary to change $b r b r b r \cdots b r$ into the final position is at least $\frac{2 k}{2}=k$.
+We will now show that $n \leq k$, i.e. that it is always possible to perform at most $k$ moves in order to reach the state in which the first $2 k$ chips are red. Consider the first $2 k$ chips. If at most $k$ of these chips are blue, we can perform one move for each chip, switching it with one of the red chips of the last $2 k$ chips. So then we are done in at most $k$ moves. Now suppose that of the first $2 k$ chips, at least $k+1$ are blue. Then at most $k-1$ chips are red. Hence it is possible to perform at most $k-1$ moves to reach the situation in which the last $2 k$ chips are red. We then perform one final move, switching the first $2 k$ chips and the last $2 k$ chips, ending in the situation in which the first $2 k$ chips are red. Thus, it is always possible to reach the situation in which the first $2 k$ chips are red in at most $k$ steps, hence $n \leq k$.
+We have now shown that $n \geq k$ and $n \leq k$, hence $n=k$ as claimed.
+
+## Problem 3
+
+Find all positive integers $n>1$ with the following property:
+for each two positive divisors $k, \ell<n$ of $n$, at least one of the numbers $2 k-\ell$ and $2 \ell-k$ is a (not necessarily positive) divisor of $n$ as well.
+
+## Solution
+
+If $n$ is prime, then $n$ has the desired property: if $k, \ell<n$ are positive divisors of a prime $n$, we have $k=\ell=1$, in which case $2 k-\ell=1$ is a divisor of $n$ as well.
+
+Assume now that a composite number $n$ has the desired property. Let $p$ be its smallest prime divisor and let $m=n / p$; then $m \geq p \geq 2$. Choosing $(k, \ell)=(1, m)$, we see that at least one of $2 m-1$ and $m-2$ must divide $n$. However, $m<2 m-1<2 m \leq n$; since $m$ and $n$ are the two biggest positive divisors of $n$, we conclude that $2 m-1$ cannot divide $n$. Therefore $n$ must be divisble by $m-2$. It follows that $m-2 \mid m p-p(m-2)=2 p$, hence $m-2 \in\{1,2, p, 2 p\}$.
+We deal with each case separately.
+
+- If $m-2=1$ we have $m=3$. As $p \leq m$, we have $p \in\{2,3\}$, hence $n=6$ or $n=9$.
+- If $m-2=2$ we have $m=4$, hence $n$ is even and $p=2$, so $n=8$.
+- If $m-2=p$, we may assume that $p>2$ (since we already discussed the case $m-2=2$ ). We have $m=p+2$, hence $n=p(p+2)$. Applying the condition in the problem to the pair $(k, \ell)=(1, p)$, we get that $p-2$ or $2 p-1$ divides $n$. If $p-2$ divides $n$, we must have $p=3$ since $p$ is the smallest prime divisor of $n$. If $2 p-1$ divides $n$, we must have $p+2=2 p-1$, since $p+2 \leq 2 p-1<p(p+2)$ and $p+2$ and $p(p+2)$ are the biggest positive divisors of $n$. Therefore $p=3$ in both cases, and we get $n=15$ as a candidate.
+- If finally $m-2=2 p$ we have $m=2 p+2$, so $n$ is even. Hence $p=2$ and $n=12$.
+
+We conclude that if a composite number $n$ satisfies the condition in the problem statement, then $n \in\{6,8,9,12,15\}$. Choosing $(k, \ell)=(1,2)$ shows that $n=8$ does not work; choosing $(k, \ell)=(3,6)$ shows that 12 is not a solution. It is easy to check that $n=6, n=9$ and $n=15$ have the desired property by checking the condition for all possible pairs $(k, \ell)$.
+We conclude that the solutions are given by the prime numbers, $n=6, n=9$ and $n=15$.
+
+## Problem 4
+
+Let $A B C D$ be a square. Consider a variable point $P$ inside the square for which $\angle B A P \geq 60^{\circ}$. Let $Q$ be the intersection of the line $A D$ and the perpendicular to $B P$ in $P$. Let $R$ be the intersection of the line $B Q$ and the perpendicular to $B P$ from $C$.
+(a) Prove that $|B P| \geq|B R|$.
+(b) For which point(s) $P$ does the inequality in (a) become an equality?
+
+## Solution
+
+![](https://cdn.mathpix.com/cropped/2024_12_15_fb9be07ea97b63ba84aeg-4.jpg?height=447&width=927&top_left_y=982&top_left_x=551)
+
+We claim that $\triangle A B P$ and $\triangle R C B$ are similar triangles. Indeed, if we denote the intersection of $B P$ and $C R$ by $S$, then $\angle R C B=\angle S C B=90^{\circ}-\angle S B C=90^{\circ}-\angle P B C=\angle A B P$. Moreover, the right angles in $P$ and $A$ imply that $A$ and $P$ lie on the circle with diameter $[B Q]$, so either $A B P Q$ or $A Q B P$ is a cyclic, convex quadrilateral. In either case, $A$ and $Q$ lie on the same side of $B P$, so $\angle P A B=\angle P Q B$. Since $C R$ and $P Q$ are perpendicular to $B P$, these lines are parallel and hence $\angle P Q B=\angle C R B$. Together with $\angle P A B=\angle P Q B$ and $\angle R C B=\angle A B P$, this implies the claim that $\triangle A B P \sim \triangle R C B$.
+This similarity yields the equality $|A P| /|B R|=|B P| /|B C|$. Since $\angle B A P \geq 60^{\circ}$, we get
+
+$$
+\begin{aligned}
+0 \leq & (|A B|-|A P|)^{2}=|A B|^{2}+|A P|^{2}-2 \cdot|A B| \cdot|A P| \\
+& =|B P|^{2}+2(\cos \angle B A P-1) \cdot|A B| \cdot|A P| \\
+& \leq|B P|^{2}-|A B| \cdot|A P|=|B P|^{2}-|B R| \cdot|B P| .
+\end{aligned}
+$$
+
+This implies that $|B P| \geq|B R|$, as desired.
+In order for equality to occur, one needs equality in each of the inequalities considered above: $|A B|=|A P|$ and $\angle B A P=60^{\circ}$. Hence there is exactly one point $P$ for which we have equality; this is the unique point inside the square such that $\triangle A B P$ is equilateral.
+

Benelux/md/Benelux_en-olympiad_en-bxmo-problems-2015-zz.md

@@ -0,0 +1,158 @@
+## BxMO 2015
+
+Problems with Solutions
+
+## Problem 1.
+
+## Determine the smallest positive integer $q$ with the following property:
+
+for every integer $m$ with $1 \leqslant m \leqslant 1006$, there exists an integer $n$ such that
+
+$$
+\frac{m}{1007} q<n<\frac{m+1}{1008} q .
+$$
+
+Solution 1. For $m=1006$, we have
+
+$$
+q-q / 1007<n<q-q / 1008
+$$
+
+for some integer $n$. If $q \leqslant 1007$, then $q-q / 1007$ and $q-q / 1008$ are both numbers that are at least $q-1$ and smaller than $q$, so there can be no integer $n$ in between. Hence $q>1007$ and $q-q / 1008 \leqslant q-1$, implying that $n<q-1$, and that $q-q / 1007<q-2$. By rearranging terms, we find $q>2014$, and hence $q \geqslant 2015$.
+
+Let us prove that $q=2015$ works. Indeed, $m q / 1007=2 m+m / 1007<2 m+1$ and $(m+1) q / 1008=2(m+1)-(m+1) / 1008>2 m+1$ for $1 \leqslant m \leqslant 1006$. Hence each pair of inequalities can be satisfied by taking $n=2 m+1$. This completes the proof, and shows that the smallest possible of $q$ is indeed 2015.
+
+Solution 2. For $m=1, \ldots, 1006$, there must exist an integer $N$ divisible by $1007 \cdot 1008$ satisfying the double inequality
+
+$$
+1008 m q<N<1007(m+1) q .
+$$
+
+Since $N$ is divisible by 1007 and 1008 , we may write
+
+$$
+N=1008 m q+1008 k=1007(m+1) q-1007 \ell
+$$
+
+where $k, \ell>0$ are integers. Hence
+
+$$
+(1007-m) q=1008 k+1007 \ell \geqslant 2015
+$$
+
+since $k, \ell>0$. Choosing $m=1006$ in this last inequality, it follows that $q \geqslant 2015$. Conversely, for $q=2015=1008+1007,(*)$ can be satisfied by taking $k=\ell=1007-m$. (Indeed, 1007 and 1008 are coprime, and so integer divisible by 1007 and 1008 is also divisible by their product.) Thus $q=2015$ has the desired property, and we are done.
+
+## Problem 2.
+
+Let $A B C$ be an acute triangle with circumcentre $O$. Let $\Gamma_{B}$ be the circle through $A$ and $B$ that is tangent to $A C$, and let $\Gamma_{C}$ be the circle through $A$ and $C$ that is tangent to $A B$. An arbitrary line through $A$ intersects $\Gamma_{B}$ again in $X$ and $\Gamma_{C}$ again in $Y$. Prove that $|O X|=|O Y|$.
+
+The following solutions are valid for the configurations appearing in the diagrams.
+Solution 1. Let $O_{B}$ and $O_{C}$ denote the respective centres of $\Gamma_{B}$ and $\Gamma_{C}$. We shall show that $X O_{B} O$ and $O O_{C} Y$ are congruent. Now $A O_{B} \perp A C$ since $\Gamma_{B}$ is tangent to $A C$ and $O O_{C} \perp A C$ since $O O_{C}$ is the perpendicular bisector of $[A C]$. Hence $A O_{B} \| O O_{C}$, and similarly, $A O_{C} \| O O_{B}$. It follows that $A O_{B} O O_{C}$ is a parallelogram. In particular, $\left|O_{B} X\right|=\left|A O_{B}\right|=\left|O O_{C}\right|$ and $\left|O_{C} Y\right|=\left|O_{C} A\right|=\left|O O_{B}\right|$.
+
+It will therefore suffice to show that the angles $\angle O O_{B} X$ and $\angle Y O_{C} O$ are equal. Noting that $\angle A O_{B} O=\angle A O_{C} O$ since $A O_{B} O O_{C}$ is a parallelogram, this follows by angle chasing:
+
+$$
+\begin{aligned}
+\angle X O_{B} O & =\angle A O_{B} O-\angle A O_{B} X=\angle A O_{C} O-\left(180^{\circ}-2 \angle O_{B} A X\right) \\
+& =\angle A O_{C} O-180^{\circ}+2\left(\angle O_{B} A O_{C}-\angle Y A O_{C}\right) \\
+& =\angle A O_{C} O-180^{\circ}+2\left(180^{\circ}-\angle A O_{C} O\right)-\left(180^{\circ}-\angle A O_{C} Y\right) \\
+& =\angle A O_{C} Y-\angle A O_{C} O=\angle O O_{C} Y .
+\end{aligned}
+$$
+
+![](https://cdn.mathpix.com/cropped/2024_12_15_c84e14b61920ef432fecg-3.jpg?height=966&width=1438&top_left_y=1525&top_left_x=281)
+
+A Variant. As in the main solution, we note that $A O_{B} O O_{C}$ is a parallelogram. Notice that $O_{B}$ and $O_{C}$ lie on the respective perpendicular bisectors of $[A X]$ and $[A Y]$. The following lemma then implies that $O$ lies on the perpendicular bisector of $[X Y]$, completing the proof:
+
+Let $P_{1}, P_{2}, P_{3}$ be three points on a line. Let $O_{1}$ be a point on the perpendicular bisector of $\left[P_{2} P_{3}\right]$, let $O_{2}$ be a point on the perpendicular bisector of $\left[P_{3} P_{1}\right]$ and let $O_{3}$ be the point in the plane such that $O_{1} P_{3} O_{2} O_{3}$ is a parallelogram. Then $O_{3}$ lies on the perpendicular bisector of $\left[P_{1} P_{2}\right]$.
+Proof. One can choose coordinates such that $P_{3}=(0,0)$ and $P_{1}=(2,0)$. Then $P_{2}=(2 a, 0), O_{1}=(a, b)$ and $O_{2}=(1, c)$ for some $a, b, c \in \mathbb{R}$. Hence $O_{3}=(a+1, b+c)$ lies on the perpendicular bisector of $\left[P_{1} P_{2}\right]$.
+
+Solution 2. Let $\alpha=\angle B A C$. Observe that $\angle A X B=180^{\circ}-\alpha=\angle C Y A$ by tangential angles. Let $Z$ be the intersection of $B X$ and $C Y$. Thus $Z X Y$ is isosceles with, in particular, $\angle Z X Y=\angle Z Y X=\alpha$. It will thus suffice to show that $O Z$ bisects $\angle X Z Y$.
+
+Now $\angle B Z C=2 \alpha=\angle B O C$ since $O$ is the circumcentre of $A B C$, and hence $B Z O C$ is cyclic. In particular, since $|O B|=|O C|$, it follows that $\angle O Z Y=\angle O B C=90^{\circ}-\alpha$. But $\angle X Z Y=180^{\circ}-2 \alpha$, and so $\angle X Z Y=2 \angle O Z Y$, which shows that $O Z$ bisects $\angle X Z Y$.
+![](https://cdn.mathpix.com/cropped/2024_12_15_c84e14b61920ef432fecg-4.jpg?height=881&width=901&top_left_y=1408&top_left_x=566)
+
+Solution 3. Consider inversion $\mathscr{I}$ in a circle centred at $A$. Under $\mathscr{I}$,
+
+$$
+B \mapsto B^{\prime}, \quad C \mapsto C^{\prime}, \quad O \mapsto O^{\prime}, \quad X \mapsto X^{\prime}, \quad Y \mapsto Y^{\prime},
+$$
+
+$\Gamma_{B} \mapsto \gamma_{B}$, a line through $B^{\prime}$ parallel to $A C^{\prime}$,
+$\Gamma_{C} \mapsto \gamma_{C}$, a line through $C^{\prime}$ parallel to $A B^{\prime}$,
+Notice that $\mathscr{I}$ sends the circumcircle of $A B C$ to the line $B^{\prime} C^{\prime}$, and hence maps $A O$ to a line perpendicular to $B C$. Further, if $[A D]$ is a diameter of this circumcircle and $D \mapsto D^{\prime}$ under $\mathscr{I}$, then $D^{\prime}$ lies on $B C$. Also, $|A D|=2|A O|$ implies $\left|A O^{\prime}\right|=2\left|A D^{\prime}\right|$, and hence $O^{\prime}$ is the image of $A$ under reflection in $B C$. Finally, $\angle O X A=\angle X^{\prime} O^{\prime} A$ and $\angle O Y A=\angle Y^{\prime} O^{\prime} A$, and hence $|O X|=|O Y|$ if and only if $O^{\prime} A$ bissects $\angle X^{\prime} O^{\prime} Y^{\prime}$ externally. We have thus reduced the problem to the following statement:
+
+In triangle $A B C$, let $O$ be the reflection of $A$ over $B C, \gamma_{B}$ be a line parallel to $A C$ through $B$, and $\gamma_{C}$ be a line parallel to $A B$ through $C$. For an arbitrary line $\ell$ passing through $A$, let $X$ and $Y$ be the intersections of $\ell$ with $\gamma_{B}$ and $\gamma_{C}$, respectively. Prove that $O A$ bisects $\angle X O Y$ externally.
+![](https://cdn.mathpix.com/cropped/2024_12_15_c84e14b61920ef432fecg-5.jpg?height=882&width=643&top_left_y=1196&top_left_x=692)
+
+Let $P=\gamma_{A} \cap \gamma_{B}$. By construction, $P O B C$ is an isosceles trapezoid (and therefore cyclic), $\angle O B X=\angle O B P=\angle O C P=\angle O C Y$. Further, since $X B \| A C$ and $Y C \| A B$, we have $\triangle A B X \sim \triangle Y C A$. Therefore, as $|O B|=|A B|$ and $|A C|=|O C|$ by construction,
+
+$$
+\frac{|O B|}{|B X|}=\frac{|A B|}{|B X|}=\frac{|Y C|}{|C A|}=\frac{|Y C|}{|C O|}
+$$
+
+It follows that $\triangle O B X \sim \triangle Y C O$. Hence
+
+$$
+\frac{|O X|}{|O Y|}=\frac{|X B|}{|O C|}=\frac{|X B|}{|A C|}=\frac{|X A|}{|A Y|}
+$$
+
+and so the result follows from the angle bisector theorem.
+
+## Problem 3.
+
+Does there exist a prime number whose decimal representation is of the form $3811 \cdots 11$ (that is, consisting of the digits 3 and 8 in that order followed by one or more digits 1 )?
+
+Solution. Write
+
+$$
+a(n)=38 \underbrace{11 \cdots 11}_{n \text { digits } 1} .
+$$
+
+There are three cases to consider, depending on the remainder of $n$ upon division by three.
+
+- If $n=3 k+1 \equiv 1(\bmod 3)$, then the sum of the digits of $a(n)$ is equal to $3(k+4)$, i.e. divisible by 3 , and hence so is $a(n)$.
+- If $n=3 k+2 \equiv 2(\bmod 3)$, then note that $a(2)=3811=3700+111$ is divisible by 37 . By induction, as $a(3 k+2)=1000 a(3 k-1)+111$, it follows that $a(3 k+2)$ is divisible by 37 for each $k \geqslant 0$.
+- If $n=3 k \equiv 0(\bmod 3)$, observe that
+
+$$
+9 a(3 k)=342 \underbrace{99 \ldots 99}_{n \text { digits } 9}=\left(7 \cdot 10^{k}\right)^{3}-1
+$$
+
+which is properly divisible by $7 \cdot 10^{k}-1$, a number that is larger than 9 . Hence $a(3 k)$ admits a non-trivial factor and so is not prime.
+
+## Problem 4.
+
+An arithmetic progression is a set of the form $\{a, a+d, \ldots, a+k d\}$, where $a, d, k$ are positive integers and $k \geqslant 2$. Thus an arithmetic progression has at least three elements and the successive elements have difference $d$, called the common difference of the arithmetic progression.
+
+Let $n$ be a positive integer. For each partition of the set $\{1,2, \ldots, 3 n\}$ into arithmetic progressions, we consider the sum $S$ of the respective common differences of these arithmetic progressions. What is the maximal value $S$ that can attain?
+(A partition of a set $A$ is a collection of disjoint subsets of $A$ whose union is $A$.)
+
+Solution. The maximum value is $n^{2}$, which is attained for the partition into $n$ arithmetic progressions $\{1, n+1,2 n+1\}, \ldots,\{n, 2 n, 3 n\}$, each of difference $n$.
+
+Suppose indeed that the set has been partioned into $N$ progressions, of respective lengths $\ell_{i}$, and differences $d_{i}$, for $1 \leqslant i \leqslant N$. Since $\ell_{i} \geqslant 3$,
+
+$$
+2 \sum_{i=1}^{N} d_{i} \leqslant \sum_{i=1}^{N}\left(\ell_{i}-1\right) d_{i}=\sum_{i=1}^{N} a_{i}-\sum_{i=1}^{N} b_{i}
+$$
+
+where $a_{i}$ and $b_{i}$ denote, respectively, the largest and smallest elements of progression $i$. Now
+
+$$
+\begin{aligned}
+& \sum_{i=1}^{N} b_{i} \geqslant 1+2+\cdots+N=N(N+1) / 2 \\
+& \sum_{i=1}^{N} a_{i} \leqslant(3 n-N+1)+\cdots+3 n=N(6 n-N+1) / 2
+\end{aligned}
+$$
+
+and thus
+
+$$
+2 \sum_{i=1}^{N} d_{i} \leqslant N(3 n-N) \leqslant 2 n^{2}
+$$
+
+as $N(3 n-N)$ is an increasing of $N$ on the interval $[0,3 n / 2]$ and since $N \leqslant n$. This completes the proof.
+
+Remark. The maximising partition is in fact unique: from the above, it is immediate that all maximal partitions must have $n_{i}=3$ and hence $n=N$. It follows that the minimal and maximal elements of the arithmetic progressions of a maximal partition are precisely $1,2, \ldots, n$ and $2 n+1,2 n+2, \ldots, 3 n$, respectively. Consider the progression $\{n+1-d, n+1, n+1+d\}$ of difference $d$. Then $n+1-d \geqslant 1$ and $n+1+d \geqslant 2 n+1$,
+which implies $d=n$, and hence $\{1, n+1,2 n+1\}$ is an element of any maximal partition. By induction, it follows similarly that $\{k, n+k, 2 n+k\}$ is an element of any maximal partition for $1 \leqslant k \leqslant n$, since integers allowing for smaller or larger differences have already been used up. This proves the partition $\{1, n+1,2 n+1\}, \ldots,\{n, 2 n, 3 n\}$ is the unique maximising partition, as claimed.
+

Benelux/md/Benelux_en-olympiad_en-bxmo-problems-2016-zz.md

@@ -0,0 +1,86 @@
+# 8th Benelux Mathematical Olympiad 
+
+Soest, 29 April - 1 May 2016
+![](https://cdn.mathpix.com/cropped/2024_12_15_b5441a0bef8d06f43de1g-1.jpg?height=386&width=505&top_left_y=111&top_left_x=1482)
+
+## Solutions
+
+Problem 1. Find the greatest positive integer $N$ with the following property: there exist integers $x_{1}, \ldots, x_{N}$ such that $x_{i}^{2}-x_{i} x_{j}$ is not divisible by 1111 for any $i \neq j$.
+
+Solution. We prove that the greatest $N$ with the required property is $N=1000$. First note that $x_{i}^{2}-x_{i} x_{j}=x_{i}\left(x_{i}-x_{j}\right)$, and that the prime factorisation of 1111 is $11 \cdot 101$.
+
+We first show that we can find 1000 integers $x_{1}, x_{2}, \ldots, x_{1000}$ such that $x_{i}^{2}-x_{i} x_{j}$ is not divisible by 1111 for any $i \neq j$. Consider the set $\{1,2, \ldots, 1110\}$. This set contains 10 integers divisible by 101 , and it contains 100 integers divisible by 11 . None of the integers in the set are divisible by both 11 and 101. If we delete all of these $10+100$ integers from the set, we are left with 1000 integers. Call these $x_{1}, x_{2}, \ldots, x_{1000}$. Now we have $11 \nmid x_{i}$ and $101 \nmid x_{i}$ for all $i$. Suppose there are $i \neq j$ with $1111 \mid x_{i}\left(x_{i}-x_{j}\right)$, then we must have $1111 \mid x_{i}-x_{j}$, which is a contradiction, since $x_{i}, x_{j} \in\{1,2, \ldots, 1110\}$. So this set satisfies the requirement.
+
+We now prove that given 1001 (or more) integers $x_{1}, x_{2}, \ldots, x_{1001}$ there are $i \neq j$ with $1111 \mid x_{i}\left(x_{i}-x_{j}\right)$. Suppose for a contradiction that for all indices $i \neq j$, we have that $x_{i}\left(x_{i}-x_{j}\right)$ is not divisible by 1111, and write $X=\left\{x_{1}, \ldots, x_{1001}\right\}$. We may (after reducing modulo 1111 if necessary) assume that $x_{i} \in\{0,1, \ldots, 1110\}$ for all $i$. Then we know that $x_{i} \neq 0$ for all $i$, and $x_{i} \neq x_{j}$ for all $i \neq j$. Suppose for some $i$ we have $11 \mid x_{i}$. (Since $x_{i} \neq 0$, we know that $101 \nmid x_{i}$.) Then any integer $a \neq x_{i}$ with $a \equiv x_{i} \bmod 101$ cannot be an element of $X$, since $1111 \mid x_{i}\left(x_{i}-a\right)$. In $\{1,2, \ldots, 1110\}$ there are 10 such integers, all of them coprime with $11 \cdot 101$. If there are exactly $k$ different values of $i$ such that $11 \mid x_{i}$, there are $10 k$ different integers from $\{1,2, \ldots, 1110\}$ that cannot be elements of $X$, all of them coprime with $11 \cdot 101$. Similarly, if there are exactly $m$ different values of $i$ such that $101 \mid x_{i}$, then there are 100 m different integers from $\{1,2, \ldots, 1110\}$ that cannot be elements of $X$, all of them coprime with $11 \cdot 101$. (Note that those $10 k$ and 100 m integers can overlap.)
+
+In $\{1,2, \ldots, 1110\}$ there are 100 multiples of 11 , there are 10 multiples of 101 and there is no multiple of $11 \cdot 101$, so there are 1000 integers that are coprime with $11 \cdot 101$. In $X$ we have $1001-k-m$ integers that are coprime with $11 \cdot 101$, so exactly $k+m-1$ of the coprime integers in $\{1,2, \ldots, 1110\}$ are not in $X$. This implies that $10 k \leqslant k+m-1$ and $100 m \leqslant k+m-1$. Adding these two inequalities we find $8 k+98 m \leqslant-2$, a clear contradiction. So $N<1001$.
+
+Problem 2. Let $n$ be a positive integer. Suppose that its positive divisors can be partitioned into pairs (i.e. can be split in groups of two) in such a way that the sum of each pair is a prime number. Prove that these prime numbers are distinct and that none of these are a divisor of $n$.
+
+Solution. Let $d_{1}$ and $d_{2}$ be positive divisors of $n$ that form a pair as given in the problem. If $d_{1}$ and $d_{2}$ have a non-trivial prime divisor $p$ in common, then $p \mid d_{1}+d_{2}$ and $p \leqslant d_{1}<d_{1}+d_{2}$, so $d_{1}+d_{2}$ cannot be prime. Hence $\operatorname{gcd}\left(d_{1}, d_{2}\right)=1$, which implies that $d_{1} d_{2} \mid n$. Suppose the number of positive divisors of $n$ is $2 t$ (it is even since the divisors can be split into pairs). If we now multiply all divisors, then on one hand we have the product of all $d_{1} d_{2}$ where $\left\{d_{1}, d_{2}\right\}$ is a pair, so that product is at most $n^{t}$. On the other hand for every divisor $d$ there is another divisor $\frac{n}{d}$ (since the number of divisors is even, the case $d=\frac{n}{d}$ does not occur), and the product of all those is equal to $n^{t}$. Hence there must be equality in every inequality $d_{1} d_{2} \leqslant n$. So the pairs of divisors given in the problem are all of the form $\left\{d, \frac{n}{d}\right\}$.
+
+Now we prove the two statements in the problem. Suppose that $d, d^{\prime}$ are positive divisors of $n$ such that $d+\frac{n}{d}=d^{\prime}+\frac{n}{d^{\prime}}$. Then $d^{2} d^{\prime}+n d^{\prime}=d\left(d^{\prime}\right)^{2}+n d$, so $d d^{\prime}\left(d-d^{\prime}\right)=n\left(d-d^{\prime}\right)$ and hence $\left(d d^{\prime}-n\right)\left(d-d^{\prime}\right)=0$. Therefore either $d=d^{\prime}$ or $d d^{\prime}=n$, which implies that $\left\{d, \frac{n}{d}\right\}=\left\{d^{\prime}, \frac{n}{d^{\prime}}\right\}$, as required.
+
+Now let $d$ be a positive divisor of $n$. Every prime divisor $p$ of $n$ divides precisely one of $d$ and $\frac{n}{d}$, since $d \frac{n}{d}=n$ and $\operatorname{gcd}\left(d, \frac{n}{d}\right)=1$; so $p \nmid d+\frac{n}{d}$. Therefore $\operatorname{gcd}\left(n, d+\frac{n}{d}\right)=1$. Since $d+\frac{n}{d}>1$ we conclude that $d+\frac{n}{d}$ cannot be a divisor of $n$.
+
+Problem 3. Find all functions $f: \mathbb{R} \rightarrow \mathbb{Z}$ such that
+
+$$
+(f(f(y)-x))^{2}+f(x)^{2}+f(y)^{2}=f(y) \cdot(1+2 f(f(y)))
+$$
+
+for all $x, y \in \mathbb{R}$.
+
+Solution I. Take $x=y=0$ and write $c=f(0)$, then we find $f(c)^{2}+c^{2}+c^{2}=c+2 c f(c)$, so $(f(c)-c)^{2}=c-c^{2}$. The left-hand side is non-negative, so the right-hand side must be non-negative as well, hence $c-c^{2} \geqslant 0$, so $c(1-c) \geqslant 0$. This implies $0 \leqslant c \leqslant 1$, and since $c \in \mathbb{Z}$, we get $c=0$ or $c=1$. In both cases we have $c-c^{2}=0$, so $f(c)-c=0$, hence $f(c)=c$. Now taking $y=0$ we find for all $x \in \mathbb{R}$ that
+
+$$
+(f(c-x))^{2}+f(x)^{2}+c^{2}=c+2 c^{2} .
+$$
+
+If $c=0$, then this equation reduces to $f(-x)^{2}+f(x)^{2}=0$, and since the left-hand side consists of the sum of two squares, both squares must be zero. Therefore $f(x)=0$ for all $x$. This function is indeed a solution of the given functional equation.
+
+Now we consider the other case: $c=1$. Then (1) reduces to $f(1-x)^{2}+f(x)^{2}=2$. The left-hand side consists of two squares of integers, so they must both be 1 . Therefore for all $x$ we have $f(x)=1$ or $f(x)=-1$.
+
+Suppose there is an $a$ with $f(a)=-1$. We take $y=a$ in the functional equation; using that $f(w)^{2}=1$ for any $w$, we find $1+1+1=-1 \cdot(1+2 f(-1))$. Both with $f(-1)=1$ and with $f(-1)=-1$ this gives a contradiction. So there exists no such $a$, and we conclude that $f(x)=1$ for all $x \in \mathbb{R}$. This is also a solution of the given functional equation.
+
+We conclude that there are two solutions: $f(x)=0$ for all $x$, and $f(x)=1$ for all $x$.
+
+Solution II. We proceed as in the first solution up to the point of deriving $f(x)= \pm 1$ for all $x \in \mathbb{R}$. Now we take $x=0$ in the original functional equation, and we find
+
+$$
+(f(f(y)))^{2}+f(0)^{2}+f(y)^{2}=f(y) \cdot(1+2 f(f(y)))
+$$
+
+We can rewrite this as
+
+$$
+(f(y)-f(f(y)))^{2}+f(0)^{2}=f(y)
+$$
+
+The left-hand side is non-negative, so $f(y) \geqslant 0$ for all $y$. If we combine this with $f(x)= \pm 1$ for all $x$, we can conclude that $f(x)=1$ for all $x \in \mathbb{R}$. And this is a solution.
+
+Problem 4. A circle $\omega$ passes through the two vertices $B$ and $C$ of a triangle $A B C$. Furthermore, $\omega$ intersects segment $A C$ in $D \neq C$ and segment $A B$ in $E \neq B$. On the ray from $B$ through $D$ lies a point $K$ such that $|B K|=|A C|$, and on the ray from $C$ through $E$ lies a point $L$ such that $|C L|=|A B|$. Show that the circumcentre $O$ of triangle $A K L$ lies on $\omega$.
+
+Solution I. Let $M$ be the midpoint of the arc $B C$ of $\omega$ that is on the same side of $B C$ as $A$. Then $|B M|=|C M|$. We also have $|B A|=|C L|$ and $\angle A B M=\angle E B M=\angle E C M=$ $\angle L C M$. Hence $\triangle A B M \cong \triangle L C M$. So $|A M|=|L M|$. (In case $M=E$ the triangles $A B M$ and $L C M$ are degenerate, but then the proof still works, since $|A M|=|A B|-|B M|=$ $|L C|-|C M|=|L M|$.) Similarly, we prove that $|A M|=|K M|$, so $M$ is the circumcentre of $\triangle A K L$. This means that $M=O$, and hence we are done as $M$ was defined to be on $\omega$.
+
+Solution II. We consider the configuration where $O$ is in the interior of triangle $A B C$. The proof for other configurations is similar. Furthermore, we exclude the case that $O=D$ or $O=E$; in those cases it is immediate that $O$ is on $\omega$.
+
+We have $\angle A B K=\angle E B D=\angle E C D=\angle L C A$. Together with $|A B|=|C L|$ and $|A C|=$ $|B K|$ this implies $\triangle A B K \cong \triangle L C A$. Hence $|A K|=|A L|, \angle A K B=\angle L A C$ (denote this angle by $\alpha$ ) and $\angle B A K=\angle C L A$ (denote this angle by $\beta$ ). Furthermore, let $\gamma=\angle B E C=$ $\angle B D C$. Then $\angle B A C=180^{\circ}-\angle B D A-\angle A B D=\angle B D C-\angle A B D=\gamma-\angle A B D$. Therefore $\angle K A L=\angle B A K+\angle L A C-\angle B A C=\alpha+\beta-(\gamma-\angle A B D)=\alpha+\beta+\angle A B D-\gamma$. Note that in triangle $A B K$ we have $\alpha+\beta+\angle A B D=180^{\circ}$, hence $\angle K A L=180^{\circ}-\gamma$.
+
+We have $\angle K O L=2\left(180^{\circ}-\angle K A L\right)=2 \gamma$. Since $|A K|=|A L|$, this implies $\angle A O L=\gamma$. As we also have $\angle A E L=\angle B E C=\gamma$, the quadrilateral $O E L A$ is cyclic. Analogously, $O D K A$ is cyclic. Now we have
+
+$$
+\begin{gathered}
+\angle D O E=360^{\circ}-\angle D O A-\angle A O E=180^{\circ}-\angle D O A+180^{\circ}-\angle A O E \\
+=\angle D K A+\angle A L E=\angle B K A+\angle A L C=\alpha+\beta
+\end{gathered}
+$$
+
+On the other hand
+
+$$
+\angle D B E=\angle K B A=180^{\circ}-\angle B A K-\angle A K B=180^{\circ}-\alpha-\beta .
+$$
+
+Hence $\angle D O E+\angle D B E=180^{\circ}$, so $O$ is on the circle containing $D, B$ and $E$, which is $\omega$.
+

Benelux/md/Benelux_en-olympiad_en-bxmo-problems-2018-zz.md

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+![](https://cdn.mathpix.com/cropped/2024_12_15_e80179e8f1d0c8dbe641g-01.jpg?height=335&width=372&top_left_y=341&top_left_x=845)
+
+# 10th Benelux Mathematical Olympiad Luxembourg, 27th-29th April 2018 
+
+## Problems and Solutions
+
+# BxMO 2018: Problems and Solutions 
+
+## Problem 1
+
+(a) Determine the minimal value of
+
+$$
+\left(x+\frac{1}{y}\right)\left(x+\frac{1}{y}-2018\right)+\left(y+\frac{1}{x}\right)\left(y+\frac{1}{x}-2018\right)
+$$
+
+where $x$ and $y$ vary over the positive reals.
+(b) Determine the minimal value of
+
+$$
+\left(x+\frac{1}{y}\right)\left(x+\frac{1}{y}+2018\right)+\left(y+\frac{1}{x}\right)\left(y+\frac{1}{x}+2018\right)
+$$
+
+where $x$ and $y$ vary over the positive reals.
+(Pierre Haas, Luxembourg)
+
+## Solution
+
+Solution 1. By the inequality between arithmetic and quadratic means,
+
+$$
+\left(x+\frac{1}{y}\right)^{2}+\left(y+\frac{1}{x}\right)^{2} \geqslant \frac{1}{2}\left(x+\frac{1}{y}+y+\frac{1}{x}\right)^{2}
+$$
+
+with equality if and only if $x+1 / y=y+1 / x$, which holds if $x=y$. It follows that
+
+$$
+\left(x+\frac{1}{y}\right)\left(x+\frac{1}{y} \pm 2018\right)+\left(y+\frac{1}{x}\right)\left(y+\frac{1}{x} \pm 2018\right) \geqslant \frac{1}{2}\left(x+\frac{1}{y}+y+\frac{1}{x}\right)^{2} \pm 2018\left(x+\frac{1}{y}+y+\frac{1}{x}\right) .
+$$
+
+The parabola $f(X)=\frac{1}{2} X^{2} \pm 2018 X$ attains its minimal value at $X=\mp 2018$, and increases monotonically away from this minimal value. By the inequality between arithmetic and geometric means, $(x+1 / x)+(y+1 / y) \geqslant 4$ with equality iff $x=y=1$. Hence
+(a) $\left(x+\frac{1}{y}\right)\left(x+\frac{1}{y}-2018\right)+\left(y+\frac{1}{x}\right)\left(y+\frac{1}{x}-2018\right) \geqslant f(2018)=-\frac{2018^{2}}{2}$,
+and equality is attained if $x=y=u$, where $u+1 / u=1009$, a quadratic equation with discriminant $1009^{2}-4>0$, and that therefore has two real solutions which are clearly positive.
+(b) $\left(x+\frac{1}{y}\right)\left(x+\frac{1}{y}+2018\right)+\left(y+\frac{1}{x}\right)\left(y+\frac{1}{x}+2018\right) \geqslant f(4)=8080$,
+with equality if and only if $x=y=1$.
+Remark. It is easy to see that equality is attained in the first inequality if and only if $x=y$. Indeed, if $x \leqslant y$, then $1 / y \leqslant 1 / x$, and so $x+1 / y \lessgtr y+1 / x$. Thus $x+1 / y=y+1 / x$ if and only if $x=y$. This is not required for the solution.
+
+Solution 2. By the inequality between arithmetic and geometric means, $x / y+y / x \geqslant 2$, and hence
+
+$$
+\left(x+\frac{1}{y}\right)^{2}+\left(y+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{y^{2}}+y^{2}+\frac{1}{x^{2}}+2\left(\frac{x}{y}+\frac{y}{x}\right) \geqslant x^{2}+\frac{1}{x^{2}}+y^{2}+\frac{1}{y^{2}}+4=\left(x+\frac{1}{x}\right)^{2}+\left(y+\frac{1}{y}\right)^{2}
+$$
+
+with equality if and only if $x=y$. It follows that
+
+$$
+\left(x+\frac{1}{y}\right)\left(x+\frac{1}{y}-K\right)+\left(y+\frac{1}{x}\right)\left(y+\frac{1}{x}-K\right) \geqslant\left[\left(x+\frac{1}{x}\right)^{2}-K\left(x+\frac{1}{x}\right)\right]+\left[\left(y+\frac{1}{y}\right)^{2}-K\left(y+\frac{1}{y}\right)\right] .
+$$
+
+## BxMO 2018: Problems and Solutions
+
+Notice that the parabola $f(X)=X^{2}-K X=(X-K / 2)^{2}-K^{2} / 4$ attains its minimum value at $X=K / 2$, and increases monotonically away from this minimal value. Now $x+1 / x, y+1 / y \geqslant 2$, so it follows that
+
+$$
+\left(x+\frac{1}{y}\right)\left(x+\frac{1}{y}-K\right)+\left(y+\frac{1}{x}\right)\left(y+\frac{1}{x}-K\right) \geqslant\left\{\begin{array}{cl}
+2 f(K / 2)=-K^{2} / 2 & \text { if } K / 2 \geqslant 2 \Longleftrightarrow K \geqslant 4 \\
+2 f(2)=4(2-K) & \text { if } K / 2 \leqslant 2 \Longleftrightarrow K \leqslant 4
+\end{array}\right.
+$$
+
+Equality is attained if (and only if)
+
+$$
+x=y=\frac{K}{4} \pm \sqrt{\frac{K^{2}}{16}-1} \quad(\text { if } K \geqslant 4), \quad x=y=1 \quad(\text { if } K \leqslant 4)
+$$
+
+and so these lower bounds indeed represent the minimal values. Taking (a) $K=2018$ and (b) $K=-2018$ in this result completes the proof.
+
+Solution 3 for part (a). Completing squares,
+
+$$
+\begin{aligned}
+\left(x+\frac{1}{y}\right)\left(x+\frac{1}{y}-2018\right)+\left(y+\frac{1}{x}\right)\left(y+\frac{1}{x}-2018\right) & =\left(x+\frac{1}{y}-1009\right)^{2}+\left(y+\frac{1}{x}-1009\right)^{2}-2 \cdot 1009^{2} \\
+& \geqslant-2 \cdot 1009^{2}
+\end{aligned}
+$$
+
+with equality if $x+1 / y=y+1 / x=1009$, which holds if $x=y=u$, where $u+1 / u=1009$, a quadratic equation with discriminant $1009^{2}-4>0$, and that therefore has two real solutions which are clearly positive.
+
+Remark. As in Solution 1, equality holds if and only if $x=y$, because, if $x \lessgtr y$, then $1 / y \lessgtr 1 / x$, and so $x+1 / y \lessgtr y+1 / x$. Thus $x+1 / y=y+1 / x$ if and only if $x=y$. This is not required for the solution.
+
+Solution 4 for part (b). Using the inequality between arithmetic and geometric means,
+
+$$
+\begin{aligned}
+\left(x+\frac{1}{y}\right)(x+ & \left.\frac{1}{y}+2018\right)+\left(y+\frac{1}{x}\right)\left(y+\frac{1}{x}+2018\right) \\
+& =\left(x+\frac{1}{y}\right)^{2}+\left(y+\frac{1}{x}\right)^{2}+2018\left[\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)\right] \geqslant 4 \frac{x}{y}+4 \frac{y}{x}+2018(2+2)
+\end{aligned}
+$$
+
+But $x / y+y / x \geqslant 2$ by the same inequality, and hence
+
+$$
+\left(x+\frac{1}{y}\right)\left(x+\frac{1}{y}+2018\right)+\left(y+\frac{1}{x}\right)\left(y+\frac{1}{x}+2018\right) \geqslant 8080
+$$
+
+with equality attained if and only if $x=y=1$.
+
+## BxMO 2018: Problems and Solutions
+
+## Problem 2
+
+In the land of Heptanomisma, four different coins and three different banknotes are used, and their denominations are seven different (non-zero) natural numbers. The denomination of the smallest banknote is greater than the sum of the denominations of the four different coins. A tourist has exactly one coin of each denomination and exactly one banknote of each denomination, but he cannot afford the book on numismatics he wishes to buy. However, the mathematically inclined shopkeeper offers to sell the book to the tourist at a price of his choosing, provided that he can pay this price in more than one way.
+(The tourist can pay a price in more than one way if there are two different subsets of his coins and notes, the denominations of which both add up to this price.)
+(a) Prove that the tourist can purchase the book if the denomination of each banknote is smaller than 49.
+(b) Show that the tourist may have to leave the shop empty-handed if the denomination of the largest banknote is 49.
+(Stijn Cambie, Belgium)
+
+## Solution
+
+Let the denominations of the coins and notes be $C_{1}<C_{2}<C_{3}<C_{4}$ and $N_{1}<N_{2}<N_{3}$, respectively. Define $C=C_{1}+C_{2}+C_{3}+C_{4}$ to be the largest amount that can be payed with coins only. The condition of the problem is thus $C<N_{1}$.
+(a) Suppose to the contrary that, whatever the price of the book, the tourist can pay for it in no more than one way.
+
+Solution 1. Consider the hands of one or two notes and any number of (or possibly no) coins. Each of them has one of the $N_{2}+N_{3}$ different values $v$ with $N_{1} \leqslant v<N_{1}+N_{2}+N_{3}$, since $C<N_{1}$. There are $\left(\binom{3}{1}+\binom{3}{2}\right) \cdot 2^{4}=96$ such hands. Hence $95=47+48>N_{2}+N_{3} \geqslant 96$, which is a contradiction.
+
+Solution 2. Consider the hands of exactly one note and any number of (or possibly no) coins, as well as the hand consisting of the two smallest notes only, of value $N_{1}+N_{2}$. Each of these has one of the $N_{3}$ different values $v$ with $N_{1} \leqslant v<N_{3}+N_{1}$, since $C<N_{1}$ and $N_{1}<N_{1}+N_{2}<N_{1}+N_{3}$. Now the number of hands considered above is $3 \cdot 2^{4}+1=49$, so $N_{3} \geqslant 49$, a contradiction.
+Solution 3. Consider the $3 \cdot 2^{4}=48$ hands of exactly two notes and any number of (or possibly no) coins. Each of these has one of the $N_{3}$ different values $v$ with $N_{1}+N_{2} \leqslant v<N_{1}+N_{2}+N_{3}$, since $C<N_{1}$. Hence $48 \leqslant N_{3}<49$, so $N_{3}=48$. Next consider the $3 \cdot 2^{4}=48$ hands of exactly one note and any number of coins. By the above, these cannot have a value greater than or equal to $N_{1}+N_{2}$, since these can be realised using two notes and some coins. Hence they must each have one of the $N_{2}$ different values $v$ with $N_{1} \leqslant v<N_{1}+N_{2}$. This implies $48 \leqslant N_{2}<N_{3}=48$, which is a contradiction.
+(b) Consider the denominations
+
+$$
+C_{1}=3, \quad C_{2}=6, \quad C_{3}=12, \quad C_{4}=24, \quad N_{1}=47, \quad N_{2}=48, \quad N_{3}=49
+$$
+
+These satisfy the conditions of the problem, with $C=45<N_{1}$.
+Solution 1. Any two amounts that can be obtained using some (or possible no) coins are different, and they are multiples of 3 since each of the coin denominations is. Hence any two such amounts differ by at least 3 . Since $N_{3}-N_{1}=2$, it follows that any two hands using one note and some coins each have a different value (and the same is true for hands using more than one note, since at least one note appears at least twice in any two such
+
+## BxMO 2018: Problems and Solutions
+
+hands). Finally, $N_{3}+C<N_{1}+N_{2}$, and hence any hand using one note and some coins has a different value from any hand using two notes and some coins. Hence there is no price that the tourist can pay for in more than one way.
+
+Solution 2. Suppose to the contrary that there are two hands of coins and notes that sum to the same amount. Up to removing coins or notes that appear in both of these hands to obtain two smaller hands summing to the same amount, we may assume that no coin or note appears in both of these hands. Notice that all the denominations, except for $N_{1} \equiv-1(\bmod 3)$ and $N_{3} \equiv 1(\bmod 3)$, are divisible by 3 . Hence, if $N_{1}$ appears in one hand, $N_{3}$ must appear in the same hand, and vice versa. Hence these two hands are two disjoint subsets of $\left\{C_{1}, C_{2}, C_{3}, C_{4}, N_{2}, N_{1}+N_{3}\right\}=\{3,6,12,24,48,96\}=3\{1,2,4,8,16,32\}$ that sum to the same amount, which contradicts the uniqueness of the binary expansion of these two amounts. Hence there is no price that the tourist can pay for in more than one way.
+
+Remark. It is natural to ask whether there are other choices of coins and notes with $N_{3}=49$ in (b) that force the tourist to leave the shop empty-handed. It turns out that the choice in (b) is unique. The argument runs as follows: the inequality $N_{2}+N_{3} \geqslant 96$ implies that $N_{2}=47$ or $N_{2}=48$. Further, strengthening the bound in (a), we must have
+
+$$
+\left(N_{2}+N_{3}+C\right)-\left(N_{1}+N_{2}\right)+1 \geqslant 48 \quad \Longrightarrow \quad C \geqslant 47+N_{1}-N_{3}=N_{1}-2
+$$
+
+But $C<N_{1}$, and so $C=N_{1}-1$ or $C=N_{1}-2$. If $N_{2}=47$, then $C=N_{1}-2$ implies $N_{1}+N_{2}=N_{3}+C$, a contradiction. Thus $C=N_{1}-1$ if $N_{2}=47$, and, similarly, $C=N_{1}-2$ if $N_{2}=48$.
+
+In the first case, $N_{3}+C=48+N_{1}=N_{1}+N_{2}+1$ (*). Hence, by the argument in (a), all the $N_{2}=47$ values $v$ with $N_{1} \leqslant v<N_{1}+N_{2}$ can be represented (using one note and some coins). Now $C_{1} \neq 1$ by (*). Since $N_{1}+1$ can be represented, it follows that $N_{2}=N_{1}+1$ and hence $N_{1}=46$. Then $N_{3}=N_{2}+2=N_{1}+3$ implies that $C_{1}>3$. Considering $50=N_{1}+4$ yields $C_{1}=4$. Now $53=N_{3}+4=N_{2}+6,52=N_{2}+5=N_{1}+6$. Hence $C_{2}=6$ is not possible, so $C_{2}=5$. Then $51=N_{2}+C_{1}=N_{1}+C_{2}$, a contradiction.
+
+In the second case, notice that $N_{2}=48$ implies, by the argument in (a), that all the $N_{2}=48$ values $v$ with $N_{1} \leqslant v<N_{1}+N_{2}$ can be represented. Clearly, $N_{1}+1$ can only be represented if $C_{1}=1$ or $N_{2}=N_{1}+1$. In the former case, $N_{3}=N_{2}+C_{1}$, a contradiction. Hence $N_{1}=N_{2}-1=47$, and so $C=45$. Considering $N_{1}+3, \ldots, N_{1}+12$ then successively yields $C_{1}=3, C_{2}=6, C_{3}=12$. Finally, $C_{4}=C-C_{1}-C_{2}-C_{3}=24$, completing the proof.
+
+One might also ask about other choices of coins and notes, with different values of $N_{3}$, forcing the tourist to leave the shop empty-handed. Numerically, it is easy to tabulate all such choices of coins and notes for small $N_{3}$ :
+
+| $C_{1}$ | $C_{2}$ | $C_{3}$ | $C_{4}$ | $N_{1}$ | $N_{2}$ | $N_{3}$ |
+| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
+| 3 | 6 | 12 | 24 | 47 | 48 | 49 |
+| 1 | 6 | 12 | 24 | 46 | 48 | 50 |
+| 3 | 6 | 12 | 24 | 46 | 48 | 50 |
+| 3 | 6 | 12 | 24 | 48 | 49 | 50 |
+| 3 | 6 | 12 | 25 | 48 | 49 | 50 |
+| 1 | 6 | 12 | 24 | 47 | 49 | 51 |
+| 1 | 6 | 12 | 25 | 47 | 49 | 51 |
+| 3 | 6 | 12 | 24 | 49 | 50 | 51 |
+| 3 | 6 | 12 | 25 | 49 | 50 | 51 |
+| 3 | 6 | 12 | 26 | 49 | 50 | 51 |
+| 3 | 6 | 13 | 25 | 49 | 50 | 51 |
+
+## BxMO 2018: Problems and Solutions
+
+## Problem 3
+
+Let $A B C$ be a triangle with orthocentre $H$, and let $D, E$, and $F$ denote the respective midpoints of line segments $A B, A C$, and $A H$. The reflections of $B$ and $C$ in $F$ are $P$ and $Q$, respectively.
+(a) Show that lines $P E$ and $Q D$ intersect on the circumcircle of triangle $A B C$.
+(b) Prove that lines $P D$ and $Q E$ intersect on line segment $A H$.
+(Merlijn Staps, the Netherlands)
+
+## Solution
+
+(a) Solution 1. Since $F$ is the midpoint of $[A H]$ and $[B P], B H P A$ is a parallelogram. Similarly, $C A Q H$ is a parallelogram, too. Let $O$ denote the circumcentre of triangle $A B C$ and let $R$ be the reflection of $A$ in $O$, so that $R$ lies on the circumcircle of $A B C$. Since $[A R]$ is a diameter of the circumcircle of $A B C, C R \perp A C$. But $B H \perp A C$, and so $C R \| B H$. Similarly, $B R \| C H$, and thus $B R C H$ is a parallelogram. Since $B H P A$ is a parallelogram, $R C P A$ is a parallelogram, too. In particular, the midpoint $E$ of its diagonal $[A C]$ lies on $[P R]$. Similarly, $D$ lies on $[Q R]$, and so $P E$ and $Q D$ meet at $R$.
+![](https://cdn.mathpix.com/cropped/2024_12_15_e80179e8f1d0c8dbe641g-06.jpg?height=932&width=804&top_left_y=1276&top_left_x=626)
+
+Remark. Since $A P$ is parallel to the altitude $B H, A P \perp A C$. Further, $P H \| A B$ since $A P H B$ is a parallelogram. But $C H \perp A B$, so $P H \perp C H$. Hence $A P C H$ is cyclic. Similarly, $A H B Q$ is cyclic, too.
+
+Solution 2. Let $O$ denote the circumcentre of triangle $A B C$. By construction, $D O \perp A B$ and $F E \| C H \perp A B$, and so $D O \| E F$. Similarly, $O E \| D F$, so $D O E F$ is a parallelogram. Since $F$ is the midpoint of $[A H]$ and $[B P], B H P A$ is a parallelogram, and so $A P\|B H\| D F \| O E$ and $|A P|=|B H|=2|D F|=2|O E|$. Extending this argument, it follows that triangles $D O E$ and $Q A P$ and have pairwise parallel sides and the ratio of their sides is $1: 2$. Hence there is a homothety with ratio 2 mapping $D O E$ to $Q A P$. The centre $R$ of this homothety is the reflection of $A$ in $O$, which lies on the circumcircle of $A B C$, and also the intersection of lines $P E, Q D$, and $A O$.
+
+## BxMO 2018: Problems and Solutions
+
+Solution 3. Since $F$ is the midpoint of $[B P]$ and $[C Q], B C P Q$ is a parallelogram. As $D$ and $E$ are the respective midpoints of $[A B]$ and $[A C]$, it follows that $P Q\|B C\| D E$ and $|P Q|=|B C|=2|D E|$. Let $R$ denote the intersection of $P E$ and $Q D$. Then $D$ and $E$ are the midpoints of $[R Q]$ and $[R P]$, respectively. In particular, $B R A Q$ is a parallelogram. But $F$ is the midpoint of $[A H]$ and $[C Q]$, so $A Q C H$ is a parallelogram. Hence $B R C H$ is a parallelogram, too, so $R$ is the reflection of $H$ in the midpoint of $[B C]$, which is well-known to lie on the circumcircle of $A B C$.
+
+Solution 4. Since $F$ is the midpoint of $[B P]$ and $[C Q], B C P Q$ is a parallelogram. Let $R$ be the image of $A$ under the translation that accordingly maps $P, Q$ onto $C, B$, respectively. By construction, $A P C R$ is a parallelogram, and so the midpoint $E$ of $[A C]$ lies on its other diagonal $P R$. Similarly, $D$ lies on $Q R$, and so $R$ is the intersection of $P E$ and $Q D$. Moreover, $F$ is the midpoint of $[A H]$ and $[B P]$, so $A P B H$ is a parallelogram. Hence $C R\|A P\| B H \perp A C$, so $\angle A C R=90^{\circ}$. Similarly, $\angle A B R=90^{\circ}$, so $A B R C$ is cyclic, and thus $R$ lies on the circumcircle of $A B C$, as required.
+
+Solution 5. Let $M$ be the midpoint of $[B C]$. Note that triangle $P H Q$ is the reflection of triangle $B A C$ in $F$. By construction, the sides of triangle $D E M$ are parallel to those of $C B A$, and hence $D E\|B C\| Q P$, and, similarly, $D M \| Q H$ and $E M \| P H$. Hence triangles $Q H P$ and $D M E$ have pairwise parallel sides and the ratio of their sides is $2: 1$. This implies that there is homothety mapping one onto the other. Its centre is the intersection of $Q D$ and $P E$ and is also the reflection of $H$ in $M$, which is well-known to lie on the circumcircle of $A B C$.
+
+Solution 6. Let $G$ be the centroid of $A B C$, and let $\mathscr{H}$ be the well-known homothety with ratio -2 and centre $G$ that maps the nine-point circle of $A B C$ onto its circumcircle. Under $\mathscr{H}, D \mapsto C, E \mapsto B$. Denote by $R$ the image of the Euler point $F$ under $\mathscr{H}$; by construction, $R$ lies on the circumcircle of $A B C$. Further, $\overrightarrow{D F}=\frac{1}{2} \overrightarrow{R C}$ and $\overrightarrow{E F}=\frac{1}{2} \overrightarrow{R B}$. Hence the points of intersection of the pairs of lines $B F, R E$ and $C F, R D$ are $P$ and $Q$, respectively. This completes the proof.
+![](https://cdn.mathpix.com/cropped/2024_12_15_e80179e8f1d0c8dbe641g-07.jpg?height=929&width=775&top_left_y=1743&top_left_x=652)
+
+## BxMO 2018: Problems and Solutions
+
+Solution 7. Let $M, N, K$ denote the respective midpoints of $[B C],[D E],[P Q]$. Thus the intersection $R$ of $P E$ and $Q D$ is the reflection of $K$ in $N$. Under reflection in $F, H \mapsto A$ and $M \mapsto K$. Hence $A K H M$ is a parallelogram, and so $\overrightarrow{M N}=\frac{1}{2} \overrightarrow{M A}=\frac{1}{2} \overrightarrow{H K}$. Hence the intersection of $K N$ and $H M$ is $R$, and $R$ is the reflection of $H$ in $M$, which is well-known to lie on the circumcircle of $A B C$.
+
+Solution 8. Take Cartesian coordinates $B(0,0), C(1,0), A(a, b), H(a, c)$. Then the coordinates of $D, E, F, P, Q$ are successively found to be
+
+$$
+D\left(\frac{a}{2}, \frac{b}{2}\right), \quad E\left(\frac{1+a}{2}, \frac{b}{2}\right), \quad F\left(a, \frac{b+c}{2}\right), \quad P(2 a, b+c), \quad Q(2 a-1, b+c)
+$$
+
+Hence the coordinates of the intersection $R(x, y)$ of $Q D$ and $P E$ satisfy
+
+$$
+\frac{y-\frac{b}{2}}{b+c-\frac{b}{2}}=\frac{x-\frac{a}{2}}{2 a-1-\frac{a}{2}}=\frac{x-\frac{a+1}{2}}{2 a-\frac{a+1}{2}} \quad \Longrightarrow \quad R(1-a,-c) .
+$$
+
+Hence $R$ is the reflection of $H$ in the midpoint $M\left(\frac{1}{2}, 0\right)$ of $[B C]$, and so lies on the circumcircle of $A B C$.
+(b) Solution 1. Lines $A F$ and $Q E$ are medians of triangle $C A Q$, and so, by the properties of the centroid, intersect at a point $S$ of $[A F]$ such that $|A S|=2|S F|$. Similarly, lines $A F$ and $P D$ are medians of triangle $B A P$, and so intersect at the same point $S$. Hence $P D$ and $Q E$ intersect on $[A H]$.
+
+Solution 2. By construction, a homothety with ratio 2 centred at $A$ maps $[D F]$ and $[E F]$ onto $[B H]$ and $[C H]$, respectively, which are mapped in turn onto $[P A]$ and $[Q A]$ under reflection in $F$ by the results of (a). The composition of these maps is a homothety with centre $S$ and ratio -2 that maps $D$ and $E$ onto $P$ and $Q$, respectively. Hence $P D$ and $Q E$ intersect at $S$. Since this homothety leaves $A H$ invariant, $S$ lies on this line. Since the homothety has negative ratio, the centre lies on the line segment $[A F]$, completing the proof.
+
+Remark. This is a projective result: triangles $D F E$ and $P A Q$ are in axial perspective (at $\infty$ ). Hence, by Desargues' theorem, they are in central perspective, and so lines $P D, Q E$, and $A F$ are concurrent.
+
+Solution 3. The intersection $S$ of $P D$ and $Q E$ is the centroid of triangle $P Q R$, where $R$ is the intersection of $P E$ and $Q D$, as in (a). Hence, if $M, N, K$ denote the respective midpoints of $[B C],[D E],[P Q]$, then, by part (a), $\overrightarrow{K S}=2 \overrightarrow{S N}$. Moreover, since $K$ is the reflection of $M$ in $F, d(K, A H)=d(M, A H)=2 d(N, A H)$. It follows that $S$ lies on $A H$; since $K$ and $R$ lie on either side of $A M, S$ lies on [AH].
+
+Solution 4. In Cartesian coordinates and using the results of (a), the coordinates of the intersection $S\left(x^{\prime}, y^{\prime}\right)$ of $Q E$ and $P D$ satisfy
+
+$$
+\frac{y^{\prime}-\frac{b}{2}}{b+c-\frac{b}{2}}=\frac{x^{\prime}-\frac{a}{2}}{2 a-\frac{a}{2}}=\frac{x^{\prime}-\frac{a+1}{2}}{2 a-1-\frac{a+1}{2}}
+$$
+
+Hence $x^{\prime}=a$, so $S$ lies on line $A H$. Further, $y^{\prime}=\frac{1}{3}(2 b+c)$. Without loss of generality, $b>0$. Then $c>0$ by definition, and so $c<y^{\prime}<b$ or $c>y^{\prime}>b$. Hence $S$ lies on line segment $[A H]$.
+
+Remark 1. Solutions 1 and 2 for part (b) have not used the fact that $H$ is the orthocentre of triangle $A B C$, and therefore show that the result of (b) remains true if $H$ is replaced with a general point $X$. The first part of the problem is in some sense independent of $H$, too. The argument of solutions 3,5 , and 7 for part (a) can be extended to yield the following result:
+
+Let $A B C$ be a triangle, and let $X$ be a point of the plane. Let $D, E$, and $Y$ denote the respective midpoints of $[A B],[A C]$, and $[A X]$. The reflections of $B$ and $C$ in $Y$ are $P$ and $Q$, respectively. Lines $P E$ and $Q D$ intersect on the circumcircle $\omega$ of $A B C$ if and only if $X$ lies on the reflection of $\omega$ in $[B C]$.
+
+## BxMO 2018: Problems and Solutions
+
+Remark 2. The intersection points of $P E, Q D$ and $P D, Q E$ are not well-defined if $P, Q, D, E$ lie on a line. In the notation of the analytic solution, this happens when $\frac{b}{2}=b+c$, i.e. $c=-\frac{b}{2}$. Computing the coordinates of $H$ explicitly using $B H \perp A C$ yields $c=a(1-a) / b$, and so $P, Q, D, E$ are aligned if and only if $b^{2}=2 a(a-1)$, which is the equation of a hyperbola passing through $B$ and $C$. (These triangles have an obtuse angle, so this configuration cannot appear for acute-angled triangles.)
+![](https://cdn.mathpix.com/cropped/2024_12_15_e80179e8f1d0c8dbe641g-09.jpg?height=500&width=1106&top_left_y=561&top_left_x=475)
+
+## BxMO 2018: Problems and Solutions
+
+## Problem 4
+
+An integer $n \geqslant 2$ having exactly $s$ positive divisors $1=d_{1}<d_{2}<\cdots<d_{s}=n$ is said to be good if there exists an integer $k$, with $2 \leqslant k \leqslant s$, such that $d_{k}>1+d_{1}+\cdots+d_{k-1}$. An integer $n \geqslant 2$ is said to be bad if it is not good.
+(a) Show that there are infinitely many bad integers.
+(b) Prove that, among any seven consecutive integers all greater than 2 , there are always at least four good integers.
+(c) Show that there are infinitely many sequences of seven consecutive good integers.
+(Gerhard Woeginger, Luxembourg)
+
+## Solution
+
+(a) Solution 1. We note that $n=2^{m}$ has $m+1$ divisors, $d_{k}=2^{k-1}$ for $1 \leqslant k \leqslant m+1$. Thus
+
+$$
+1+d_{1}+\cdots+d_{k-1}=1+\left(2^{k-1}-1\right)=2^{k-1}=d_{k}
+$$
+
+for each $k \geqslant 2$, and hence each power of 2 is a bad integer. This exhibits infinitely many bad integers.
+Remark. It is true more generally that
+If $n=2^{r} m$, where $m$ is a product of (odd) primes each less than $2^{r+1}$, then $n$ is bad.
+This is an immediate corollary of the previous result and the following observation:
+If $n=2^{r} m$ is bad, where $m$ is odd, then so is $p n$ for any odd prime $p<2^{r+1}$.
+Proof. Let $D_{K}>1$ be a divisor of $p n$, so $D_{K}=p$ or $D_{K}=d_{k}$ or $D_{K}=p d_{k}$, where $d_{k}>1$ is a divisor of $n$. In the first case, observe that there exists $t<r+1$ such that $2^{t}<p<2^{t+1}$ by assumption. Then $\left\{1,2, \ldots, 2^{t}\right\} \subseteq\left\{D_{1}, \ldots, D_{K-1}\right\}$, and so
+
+$$
+D_{K}=p<2^{t+1}=1+\left(1+2+\cdots+2^{t}\right) \leqslant 1+D_{1}+\cdots+D_{K-1}
+$$
+
+In the final case, $\left\{1,2, \ldots, 2^{t}, p d_{1}, \ldots, p d_{k-1}\right\} \subseteq\left\{D_{1}, \ldots, D_{K-1}\right\}$, and so
+
+$$
+\begin{aligned}
+D_{K} & =p d_{k} \leqslant p\left(1+d_{1}+\cdots+d_{k-1}\right)=p+p d_{1}+\cdots+p d_{k} \\
+& <2^{t+1}+p d_{1}+\cdots+p d_{k-1}=1+\left(1+\cdots+2^{t}\right)+p d_{1}+\cdots+p d_{k-1}<1+\left(D_{1}+\cdots+D_{K-1}\right)
+\end{aligned}
+$$
+
+In the second case, $\left\{d_{1}, \ldots, d_{k-1}\right\} \subseteq\left\{D_{1}, \ldots, D_{K-1}\right\}$ immediately implies the required inequality, and so $p n$ is indeed bad.
+This result is weak, however: only 57931 (6.99\%) of the 829157 bad numbers not larger than $10^{7}$ are of this form.
+Solution 2. We claim that $n=m$ ! is bad for each integer $m \geqslant 2$. The proof proceeds by induction on $m$, the case $m=2$ being clear. If $D_{K}>1$ is a divisor of $m$ !, then $D_{K}=d_{k}$ or $D_{K}=q$ or $D_{K}=q d_{k}$, where $d_{k}>1$ is a divisor of $(m-1)$ ! and $q>1$ is a divisor of $m$. In the first case, $\left\{d_{1}, \ldots, d_{k-1}\right\} \subseteq\left\{D_{1}, \ldots, D_{K-1}\right\}$, so, invoking the inductive hypothesis, $D_{K}=d_{k} \leqslant 1+\left(d_{1}+\cdots+d_{k-1}\right) \leqslant 1+\left(D_{1}+\cdots+D_{K-1}\right)$. In the second case, $q \leqslant m+1$, so $1,2, \ldots, q-1 \mid m$ ! and $\{1,2, \ldots, q-1\} \subseteq\left\{D_{1}, \ldots, D_{K-1}\right\}$. But $q^{2}-3 q+2 \geqslant 0$ for $q \geqslant 2$, and hence
+
+$$
+D_{K}=q \leqslant \frac{q^{2}-q+2}{2}=1+(1+\cdots+q-1) \leqslant 1+\left(D_{1}+\cdots+D_{K-1}\right)
+$$
+
+In the final case, $\left\{1,2, \ldots, q-1, q d_{1}, \ldots, q d_{k-1}\right\} \subseteq\left\{D_{1}, \ldots, D_{K-1}\right\}$, and so
+
+$$
+D_{K}=q d_{k} \leqslant q\left(1+d_{1}+\cdots+d_{k-1}\right) \leqslant 1+(1+\cdots+q-1)+\left(q d_{1}+\cdots+q d_{k-1}\right)
+$$
+
+and so $D_{K} \leqslant 1+\left(D_{1}+\cdots+D_{K-1}\right)$, completing the inductive step.
+
+## BxMO 2018: Problems and Solutions
+
+(b) If $n$ is odd, then $d_{1}=1, d_{2}>2=1+d_{1}$, and $n$ is good. Among any seven consecutive positive integers, there are either four odd integers and three even ones, or four even ones and three odd ones. In the first case, these four odd integers are good. In the second case, we have to show that one of the consecutive even integers $n=2 m, n+2=2(m+1), n+4=2(m+2), n+6=2(m+3)$ is good. Notice that even integers of the form $n=2(6 \ell \pm 1)$, for $\ell \geqslant 1$, are good, since $d_{2}=2<n$ as $\ell \geqslant 1$, but $d_{3}>4=1+1+2$, since they are divisible by neither 3 nor 4 . But at least one $m, m+1, m+2, m+3$ is congruent to $\pm 1(\bmod 6)($ and larger than 1 by assumption); this completes the proof.
+(c) Solution 1. Let $n=12 q$, an even number. By part (b), $n-3, n-2=2(6 q-1), n-1, n+1, n+2=2(6 q+1), n+3$ are good integers. Take $q>29$ to be prime. Then the divisors of $n$ less than $q$ are precisely the divisors of 12 . Now $q>1+(1+2+3+4+6+12)=29$, and so $n$ is good, too. Since there are infinitely many choices of the prime $q$, there are infinitely many sequences of seven consecutive good integers.
+
+Solution 2. Let $m=2^{3} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 11 \cdot 13$. For any integer $k>0$, the seven consecutive integers $m k+1, m k+2, \ldots, m k+7$ are good. Indeed, the four odd numbers $m k+1, m k+3, m k+5, m k+7$ are good by part (b). Moreover,
+
+| $n$ | $d_{1}$ | $d_{2}$ | $d_{3}$ | $d_{4}$ | $d_{5}$ |  |
+| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
+| $m k+2$ | 1 | 2 | $\geqslant 16$ |  |  | $\Longrightarrow$ |
+| $m k+4$ | 1 | 2 | 4 | $\geqslant 16$ |  | $\Longrightarrow$ |
+| $m k+6$ | 1 | 2 | 3 | 6 | $\geqslant 16$ | $\Longrightarrow$ |
+| $d_{4}>1+d_{1}+d_{2}$, |  |  |  |  |  |  |
+| $m+d_{2}+d_{3}$, |  |  |  |  |  |  |
+| $d_{5}>1+d_{1}+d_{2}+d_{3}+d_{4}$, |  |  |  |  |  |  |
+
+and so $m k+2, m k+4, m k+6$ are good integers, too. Hence there are infinitely many choices of seven consecutive good integers.
+
+Remark. This solution can also be phrased more indirectly in terms of congruence conditions modulo small remainders; the existence of infinitely many appropriate solutions is then guaranteed by the Chinese Remainder Theorem.
+
+Solution 3. Let $m=29$ !. For any integer $k>0$, the seven consecutive integers $m k+1, m k+2, \ldots, m k+7$ are good. Indeed, the divisors of these numbers are either at most 7 or at least 30 . But $30>1+(1+2+\cdots+7)=29$, and so these seven integers are good, giving infinitely many sequences of seven consecutive good integers.
+
+Remark. The ideas underlying these different solutions can be extended to show that there are arbitrarily long runs of consecutive good integers:
+
+For each integer $N$, there exist $N$ consecutive good integers.
+Proof 1 . Let $M=2+(1+2+\cdots+N)$, and let $m=M$ !. Any divisor of any of the $N$ consecutive integers $m+1, m+2, \ldots, m+N$ is either at most $N$ or at least $M$. But $M>1+(1+2+\cdots+N)$, and so these $N$ consecutive integers are good.
+Proof 2. Let $2=p_{1}<p_{2}<\cdots$ denote the prime numbers, and choose positive integers $s$ and $\alpha_{1}, \ldots, \alpha_{s}$ such that $1,2, \ldots, N$ divide $P=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{s}^{\alpha_{s}}$. Let $\sigma(n)$ denote the sum of the divisors of $n$. Choose an integer $t$ such that $p_{t}>\sigma(P)$, let $m=p_{1} p_{2} \cdots p_{t} P$. The $N$ consecutive integers $m+1, m+2, \ldots, m+N$ are all good. Indeed, for $1 \leqslant k \leqslant N$, any divisor of $m+k$ is either at most $k$ or at least $p_{t+1}$. But $p_{t+1}>1+p_{t}>1+\sigma(P)>1+\sigma(k)$, since $k \mid P$. This completes the proof.
+

Benelux/md/Benelux_en-olympiad_en-bxmo-problems-2019-zz.md

@@ -0,0 +1,145 @@
+## 11th Benelux Mathematical Olympiad
+
+Valkenswaard, 26-28 April 2019
+![](https://cdn.mathpix.com/cropped/2024_12_15_be2de868c358eee97d2ag-1.jpg?height=396&width=524&top_left_y=92&top_left_x=1511)
+
+## Solutions
+
+## Problem 1.
+
+a) Let $a, b, c, d$ be real numbers with $0 \leqslant a, b, c, d \leqslant 1$. Prove that
+
+$$
+a b(a-b)+b c(b-c)+c d(c-d)+d a(d-a) \leqslant \frac{8}{27}
+$$
+
+b) Find all quadruples $(a, b, c, d)$ of real numbers with $0 \leqslant a, b, c, d \leqslant 1$ for which equality holds in the above inequality.
+
+Solution. Denote the left-hand side by $S$. We have
+
+$$
+\begin{aligned}
+S & =a b(a-b)+b c(b-c)+c d(c-d)+d a(d-a) \\
+& =a^{2} b-a b^{2}+b^{2} c-b c^{2}+c^{2} d-c d^{2}+d^{2} a-d a^{2} \\
+& =a^{2}(b-d)+b^{2}(c-a)+c^{2}(d-b)+d^{2}(a-c) \\
+& =(b-d)\left(a^{2}-c^{2}\right)+(c-a)\left(b^{2}-d^{2}\right) \\
+& =(b-d)(a-c)(a+c)+(c-a)(b-d)(b+d) \\
+& =(b-d)(a-c)(a+c-b-d) .
+\end{aligned}
+$$
+
+Assume without loss of generality that $a \geqslant c$, then $a-c \geqslant 0$. Now we consider two cases.
+
+- Suppose $b-d \geqslant 0$. Then if $a+c-b-d<0$ we have $S \leqslant 0$, so we're done. If $a+c-b-d \geqslant 0$, we use the AM-GM inequality on $a-c, b-d$ and $a+c-b-d$, which yields
+
+$$
+\sqrt[3]{S}=\sqrt[3]{(b-d)(a-c)(a+c-b-d)} \leqslant \frac{b-d+a-c+a+c-b-d}{3}=\frac{2 a-2 d}{3} \leqslant \frac{2}{3}
+$$
+
+so $S \leqslant \frac{8}{27}$.
+
+- Suppose $b-d<0$. Then if $a+c-b-d \geqslant 0$ we have $S \leqslant 0$, so we're done. If $a+c-b-d<0$, we use the AM-GM inequality on $a-c, d-b$ and $b+d-a-c$, which yields
+
+$$
+\sqrt[3]{S}=\sqrt[3]{(d-b)(a-c)(b+d-a-c)} \leqslant \frac{d-b+a-c+b+d-a-c}{3}=\frac{2 d-2 c}{3} \leqslant \frac{2}{3}
+$$
+
+so $S \leqslant \frac{8}{27}$.
+
+Equality in the first case occurs when $a-c=b-d=a+c-b-d$ and $a=1, d=0$. Then $1-c=b=1+c-b$, which implies $2 c=b=1-c$ and hence $c=\frac{1}{3}$. This results in $(a, b, c, d)=\left(1, \frac{2}{3}, \frac{1}{3}, 0\right)$. Similarly, equality in the second case occurs when $a-c=d-b=b+d-a-c$ and $d=1, c=0$. This yields $(a, b, c, d)=\left(\frac{2}{3}, \frac{1}{3}, 0,1\right)$. The case $a \leqslant c$ gives the other two cyclic variants, which gives all quadruples satisfying the equality.
+
+Problem 2. Pawns and rooks are placed on a $2019 \times 2019$ chessboard, with at most one piece on each of the $2019^{2}$ squares. A rook can see another rook if they are in the same row or column and all squares between them are empty. What is the maximal number $p$ for which $p$ pawns and $p+2019$ rooks can be placed on the chessboard in such a way that no two rooks can see each other?
+
+Solution. Answer: the maximal $p$ equals $1009^{2}$.
+Write $n=2019$ and $k=1009$; then $n=2 k+1$. We first show that we can place $k^{2}$ pawns and $n+k^{2}$ rooks. Each cell of the chess board has coordinates $(x, y)$ with $1 \leqslant x, y \leqslant n$. We colour each cell black or white depending on whether $x+y$ is even or odd.
+
+Let $A$ be cell $(1, k+1), B$ be cell $(k+1,1), C$ be cell $(2 k+1, k+1)$ and $D$ be cell $(k+1,2 k+1)$, and consider the skew square $A B C D$. We place rooks on the cells of this square which have the same colour as $A$, and we place pawns on the other cells of this square. In this way, no rook can see another rook. Now we have placed $p=k^{2}$ pawns and $(k+1)^{2}=k^{2}+(2 k+1)=p+n$ rooks.
+
+Now we show that we can not place more pawns. Observe that in every row the number of rooks exceeds the number of pawns by at most 1 , since there has to be a pawn between every two neighbouring rooks. So the total number of rooks exceeds the number of pawns by at most $n$. On the other hand we are to place $p$ pawns and $p+n$ rooks, so the number of rooks in every row exceeds the number of pawns by exactly 1. This means that the rooks and pawns alternate, with rooks at the two ends. For the columns the same holds.
+
+Consider the $\ell$-th row. Let $a$ be the number of pawns in this row and let $b$ be the number of pawns above the $\ell$-th row. For all these pawns, a rook must be somewhere above it. Counting the rooks directly above these $a+b$ pawns, we conclude that there must be at least $a+b$ rooks in the first $\ell-1$ rows. In every row the number of rooks exceeds the number of pawns by 1 , so in these first $\ell-1$ rows we have at least $a+b-(\ell-1)$ pawns. So $b \geqslant a+b-(\ell-1)$, yielding $a \leqslant \ell-1$. We conclude that the $\ell$-th row contains at most $\ell-1$ pawns. The same holds for the $\ell$-th row counted from below (row $(n+1)-\ell$ ): also in this row, there are at most $\ell-1$ pawns. As $n=2 k+1$, the maximal number $p$ is
+
+$$
+\sum_{\ell=1}^{k}(\ell-1)+\sum_{\ell=1}^{k+1}(\ell-1)=k+2 \cdot \frac{1}{2} k(k-1)=k+k(k-1)=k^{2}
+$$
+
+Problem 3. Two circles $\Gamma_{1}$ and $\Gamma_{2}$ intersect at points $A$ and $Z$ (with $A \neq Z$ ). Let $B$ be the centre of $\Gamma_{1}$ and let $C$ be the centre of $\Gamma_{2}$. The exterior angle bisector of $\angle B A C$ intersects $\Gamma_{1}$ again at $X$ and $\Gamma_{2}$ again at $Y$. Prove that the interior angle bisector of $\angle B Z C$ passes through the circumcentre of $\triangle X Y Z$.
+
+For points $P, Q, R$ that lie on a line $\ell$ in that order, and a point $S$ not on $\ell$, the interior angle bisector of $\angle P Q S$ is the line that divides $\angle P Q S$ into two equal angles, while the exterior angle bisector of $\angle P Q S$ is the line that divides $\angle R Q S$ into two equal angles.
+
+Solution I. We first prove that $\angle A Z X=\angle Y Z A$. Since the triangles $\triangle B A X, \triangle C A Y$ are isosceles, and $X Y$ is the external bisector of $\angle B A C$, we see that
+
+$$
+\angle B X A=\angle X A B=\angle C A Y=\angle A Y C .
+$$
+
+Using these equalities, we find that
+
+$$
+\angle X B A=180^{\circ}-\angle B A X-\angle A X B=180^{\circ}-\angle C Y A-\angle Y A C=\angle A C Y
+$$
+
+Since $B, C$ are the centres of respectively $\Gamma_{1}, \Gamma_{2}$, this implies that
+
+$$
+\angle A Z X=\frac{1}{2} \angle A B X=\frac{1}{2} \angle Y C A=\angle Y Z A .
+$$
+
+Let $O$ be the circumcentre of $\triangle X Y Z$, next we will prove that $\angle B Z O=\angle A Z Y$. Consider the configuration where $\angle Z X A$ is sharp, then $\angle A B Z=2 \angle A X Z=2 \angle Y X Z=\angle Y O Z$. Since $\triangle B Z A$ and $\triangle O Z Y$ are isosceles, this implies $\angle B Z A=\angle O Z Y$. Subtracting $\angle O Z A$ (or adding, depending on the configuration) yields $\angle B Z O=\angle A Z Y$.
+Together with the analogous result $\angle C Z O=\angle A Z X$, we conclude $\angle B Z O=\angle A Z Y=\angle X Z A=$ $\angle O Z C$, so $O$ lies indeed on the internal bisector of $\angle B Z C$.
+
+Solution II. Let $O$ the circumcentre of $\triangle X Y Z$, then we see that $O X=O Z$. Since $B$ is the centre of $\Gamma_{1}$, we also see that $B X=B Z$, so $O B$ is the perpendicular bisector of $X Z$. Therefore $\angle B Z O=\angle B X O$, and analogously we find $\angle C Z O=\angle C Y O$. Note that $B, C$ lie on the same side of $X Y$; we will consider the configuration where $O$ is on the opposite side of $X Y$. Then $\angle B X O=\angle B X A+\angle A X O$. Now the isosceles triangles $\triangle B X A, \triangle C A Y, \triangle O X Y$, and $X Y$ being the external bisector of $\angle B A C$ give
+
+$$
+\begin{aligned}
+& \angle B X A=\angle X A B=\angle C A Y=\angle A Y C, \\
+& \angle A X O=\angle Y X O=\angle O Y X=\angle O Y A,
+\end{aligned}
+$$
+
+so we find that $\angle B Z O=\angle B X O=\angle A Y C+\angle O Y A=\angle O Y C=\angle O Z C$. Hence $O$ lies on the internal bisector of $\angle B Z C$.
+
+Problem 4. An integer $m>1$ is rich if for any positive integer $n$, there exist positive integers $x, y, z$ such that $n=m x^{2}-y^{2}-z^{2}$. An integer $m>1$ is poor if it is not rich.
+a) Find a poor integer.
+b) Find a rich integer.
+a) Solution I. We will show that $m=4$ is poor. If $y$ and $z$ are both even, we have $4 x^{2}-y^{2}-z^{2} \equiv$ $0-0-0=0(\bmod 4)$. If $y$ is even and $z$ is odd or the other way around, then $4 x^{2}-y^{2}-z^{2} \equiv$ $0-0-1 \equiv 3(\bmod 4)$. If $y$ and $z$ are both odd, we have $4 x^{2}-y^{2}-z^{2} \equiv 0-1-1 \equiv 2(\bmod 4)$. Hence it is impossible to write any integer $n \equiv 1(\bmod 4)$ as $n=m x^{2}-y^{2}-z^{2}$. So $m=4$ is poor.
+
+Solution II. We will show that $m=3$ is poor, by proving that it is impossible to write any integer $n \equiv 5(\bmod 8)$ as $n=3 x^{2}-y^{2}-z^{2}$. We consider the equation modulo 8 . If $4 \mid x$, then $n \equiv-y^{2}-z^{2}(\bmod 8)$. So if $n \equiv 5(\bmod 8)$, we need to have $y^{2}+z^{2} \equiv 3(\bmod 8)$. As $y^{2}$ and $z^{2}$ can only be 0,1 or $4 \bmod 8$, this is impossible. If $x \equiv 2(\bmod 4)$, then $3 x^{2} \equiv 4(\bmod 8)$, so for $n \equiv 5(\bmod 8)$ we need to have $y^{2}+z^{2} \equiv 7(\bmod 8)$. Again, this is impossible. Finally, if $x$ is odd, then $3 x^{2} \equiv 3(\bmod 8)$, so for $n \equiv 5(\bmod 8)$ we need to have $y^{2}+z^{2} \equiv 6(\bmod 8)$. This is impossible as well. So $m=3$ is poor.
+b) Solution I. We will show that $m=5$ is rich. For any integer $x \geqslant 2$ we can take $y=2 x-2$ and $z=x+3$, then
+
+$$
+5 x^{2}-y^{2}-z^{2}=5 x^{2}-\left(4 x^{2}-8 x+4\right)-\left(x^{2}+6 x+9\right)=2 x-13
+$$
+
+For $x \geqslant 7$, this gives us all odd positive integers.
+For any integer $x \geqslant 1$ we can take $y=2 x-1$ and $z=x+1$, then
+
+$$
+5 x^{2}-y^{2}-z^{2}=5 x^{2}-\left(4 x^{2}-4 x+1\right)-\left(x^{2}+2 x+1\right)=2 x-2
+$$
+
+For $x \geqslant 2$, this gives us all even positive integers. So $m=5$ is rich.
+Solution II. We will show that $m=2$ is rich. For any integer $x \geqslant 3$ we can take $y=x+1$ and $z=x-2$, then
+
+$$
+2 x^{2}-y^{2}-z^{2}=2 x^{2}-\left(x^{2}+2 x+1\right)-\left(x^{2}-4 x+4\right)=2 x-5 .
+$$
+
+As $x$ can take any positive integer value that is at least 3 , this gives us all odd positive integers.
+For any integer $x \geqslant 5$ we can take $y=x+2$ and $z=x-4$, then
+
+$$
+2 x^{2}-y^{2}-z^{2}=2 x^{2}-\left(x^{2}+4 x+4\right)-\left(x^{2}-8 x+16\right)=4 x-20 .
+$$
+
+For $x \geqslant 6$, this gives us all positive integers divisible by 4 .
+For any integer $x \geqslant 6$ we can take $y=x+3$ and $z=x-5$, then
+
+$$
+2 x^{2}-y^{2}-z^{2}=2 x^{2}-\left(x^{2}+6 x+9\right)-\left(x^{2}-10 x+25\right)=4 x-34
+$$
+
+For $x \geqslant 9$, this gives us all positive integers in the residue class $2(\bmod 4)$. So $m=2$ is rich.
+
+Remark. Considering the equation modulo 8 and 9 , we can show that $m$ is poor when $m \equiv 0,3(\bmod 4)$ or $m \equiv 0(\bmod 3)$. This shows that $6,7,8,9,11,12$ are also poor. Further you can find a construction showing that $m=10$ is rich.
+

Benelux/md/Benelux_en-olympiad_en-bxmo-problems-2020-zz.md

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+# 12th Benelux Mathematical Olympiad <br> Virtual, 2nd-3rd May 2020 
+
+## Problems and Solutions
+
+# BxMO 2020: Problems and Solutions 
+
+## Problem 1
+
+Find all positive integers $d$ with the following property: there exists a polynomial $P$ of degree $d$ with integer coefficients such that $|P(m)|=1$ for at least $d+1$ different integers $m$.
+
+## Solution
+
+Note that $P(x)=c$ for a fixed constant has at most $d$ solutions, since the polynomial $P(x)-c$ of degree $d$ cancels at most $d$ times. This implies that there are integers $m$ satisfying $P(m)=1$, as well as integers $m$ such that $P(m)=-1$.
+Next, we prove the following lemma.
+Lemma. If $a$ and $b$ are integers such that $P(a)=-1$ and $P(b)=1$, then $|b-a| \leq 2$.
+Proof Since $b-a \mid P(b)-P(a)$ by a well-known lemma (corollary of $a-b \mid a^{n}-b^{n}$ for every integer $n \geq 0$ ), the conclusion follows.
+
+Let us first consider the case that $d \geq 4$, and assume that there exists a polynomial P with at least $d+1 \geq 5$ solutions to $|P(m)|=1$. Let $a$ and $b$ be the smallest and largest solution respectively. Since $b-a \geq 4$, we need $P(a)=P(b)$ by the lemma. Without loss of generality (by switching $P$ with $-P$ if necessary) we can assume $P(a)=P(b)=1$. Take a value $m$ such that $P(m)=-1$. Due to the lemma, we need $b-m$ and $m-a$ to be both at most 2 . Since $b-a \geq 4$, there is only one possibility left in which case $b-a=4$ and thus $d=4$. By considering $P(x-m)$, we can assume $P( \pm 2)=P( \pm 1)=1$ and $P(0)=-1$. The unique fourth degree polynomial satisfying these equalities is $P(x)=-0.5\left(x^{2}-4\right)\left(x^{2}-1\right)+1$ which is not a polynomial with integer coefficients.
+
+For $1 \leq d \leq 3$, there exist polynomials satisfying the conditions.
+For $d=1$ we can take $P_{1}(X)=X$ as $P_{1}(-1)=-1$ and $P_{1}(1)=1$.
+For $d=2, P_{2}(X)=2 X(X-2)+1$ satisfies $P_{2}(0)=1, P_{2}(1)=-1$ and $P_{2}(2)=1$.
+For $d=3$, the polynomial ; $P_{3}(X)=(X+1) X(X-2)+1$ satisfies $P_{3}(-1)=1=P_{3}(0)=P_{3}(2)=1$ and $P_{3}(1)=-1$. So $\left|P_{3}(m)\right|=1$ for $m \in\{-1,0,1,2\}$.
+
+Thus the integers with the required property are precisely $d=1,2,3$. This completes the proof.
+
+## BxMO 2020: Problems and Solutions
+
+## Problem 2
+
+Let $N$ be a positive integer. A collection of $4 N^{2}$ unit tiles with two segments drawn on them as shown is assembled into a $2 N \times 2 N$ board. Tiles can be rotated.
+![](https://cdn.mathpix.com/cropped/2024_12_15_aca58b2f26bed6317ca2g-3.jpg?height=141&width=150&top_left_y=563&top_left_x=956)
+
+The segments on the tiles define paths on the board. Determine the least possible number and the largest possible number of such paths.
+
+## Solution
+
+Let $p$ denote the number of paths. Notice that there are two types of paths: (1) those that start and end at a point on the boundary of the board and (2) closed paths in the interior of the board. Let $p_{1}, p_{2}$ denote the respective numbers of paths of either type. There are $8 N$ points on the boundary of the board, and each of these is the starting point or endpoint of exactly one path, so $p_{1}=4 \mathrm{~N}$. Trivially, $p_{2} \geqslant 0$, so $p=p_{1}+p_{2} \geqslant 4 N$.
+
+The paths on the board are made up of $8 N^{2}$ segments in total. There are only 4 possible paths of one segment, in the corners of the board. All other paths on the boundary of the board therefore consist of at least 2 segments. Moreover, all closed paths in the interior of the board consist of at least 4 segments. Hence
+
+$$
+8 N^{2} \geqslant 4 \cdot 1+\left(p_{1}-4\right) \cdot 2+p_{2} \cdot 4 \Longleftrightarrow p_{2} \leqslant N^{2}+(N-1)^{2}, \text { so } p=p_{1}+p_{2} \leqslant N^{2}+(N+1)^{2}
+$$
+
+We have thus shown that $4 N \leqslant p \leqslant N^{2}+(N+1)^{2}$. These minimum and maximum values can indeed be attained, as shown below for $N=3$.
+![](https://cdn.mathpix.com/cropped/2024_12_15_aca58b2f26bed6317ca2g-3.jpg?height=327&width=817&top_left_y=1767&top_left_x=626)
+
+The constructions generalise easily.The constructions clearly attain the required bounds because they satisfy the equalities in our arguments showing the two bounds. Indeed the board on the left has $p_{2}=0$ so give the lower bound. The one on the right has exactly 4 boundary paths with one segment, all other $4 N-4$ boundary paths with 2 segments, and all remaining paths with 4 segments.
+Remark. The same ideas solve the analogous problem for a $(2 N+1) \times(2 N+1)$ assembled from $(2 N+1)^{2}$ such tiles. For this board, $p \geqslant p_{1}=2(2 N+1)$. Next, there are $2(2 N+1)^{2}$ segments, so, again,
+
+$$
+2(2 N+1)^{2} \geqslant 4 \cdot 1+\left(p_{1}-4\right) \cdot 2+p_{2} \cdot 4 \Longleftrightarrow p_{2} \leqslant 2 N^{2}+\frac{1}{2}
+$$
+
+But $p_{2}$ is an integer, so $p_{2} \leqslant 2 N^{2}$, and hence $p \leqslant 2(N+1)^{2}$. We have thus shown that $2(2 N+1) \leqslant p \leqslant 2(N+1)^{2}$. The construction for the lower bound is the same as for the $2 N \times 2 N$ board; the construction for the upper bound, shown for $N=3$, is a small modification of that for the $2 N \times 2 N$ board and again generalises easily.
+![](https://cdn.mathpix.com/cropped/2024_12_15_aca58b2f26bed6317ca2g-3.jpg?height=298&width=301&top_left_y=2441&top_left_x=1597)
+
+## BxMO 2020: Problems and Solutions
+
+## Problem 3
+
+Let $A B C$ be a triangle. The circle $\omega_{A}$ through $A$ is tangent to line $B C$ at $B$. The circle $\omega_{C}$ through $C$ is tangent to line $A B$ at $B$. Let $\omega_{A}$ and $\omega_{C}$ meet again at $D$. Let $M$ be the midpoint of line segment [ $B C]$, and let $E$ be the intersection of lines $M D$ and $A C$. Show that $E$ lies on $\omega_{A}$.
+
+## Solution 1
+
+Let $N$ be the midpoint of $[A B]$. By tangential angles, $\angle C B D=\angle B A D$ and $\angle D B A=\angle D C B$, so triangles $D A B$ and $D B C$ are similar. By definition of the midpoints $M, N$, so are triangles $A N D$ and $B M D$. In particular, $\angle B N D=180^{\circ}-\angle D N A=180^{\circ}-\angle D M B$, so $B N D M$ is cyclic. But $M N \| A C$ by construction, so $\angle D B A=\angle D B N=\angle D M N=\angle E M N=\angle M E A=\angle D E A$, hence $E A D B$ is cyclic, completing the proof.
+![](https://cdn.mathpix.com/cropped/2024_12_15_aca58b2f26bed6317ca2g-4.jpg?height=766&width=1006&top_left_y=1116&top_left_x=542)
+
+## Solution 2
+
+Let $S$ be the point on $\omega_{C}$ such that $B S$ is parallel to $A C$, and let $E^{\prime}$ be the reflection of $S$ in $M$ (such that $B S C E^{\prime}$ is a parallellogram and $E^{\prime}$ lies on $A C$ ). Now we do some (directed) angle chasing:
+
+$$
+\begin{aligned}
+\angle A E^{\prime} B & =\angle C E^{\prime} B\left(\text { since } A, E^{\prime} \text { and } C \text { are collineair }\right) \\
+& =\angle B S C\left(B S C E^{\prime} \text { is a parallellogram }\right) \\
+& =\angle A B C\left(\text { inscribed angles on } \omega_{C}\right) \\
+& =\angle A D B\left(\text { inscribed angles on } \omega_{A}\right) .
+\end{aligned}
+$$
+
+Hence $E^{\prime}$ lies on $\omega_{A}$. Further $\angle A E^{\prime} D=\angle A B D=\angle B S D$, and because $A E^{\prime}$ is parallel to $B S$ we find that $E^{\prime} D$ is parallel to $S D$. This means that $E^{\prime}$ lies on $M D$, so $E^{\prime}=E$ and $E$ lies on $\omega_{A}$.
+
+# BxMO 2020: Problems and Solutions 
+
+## Problem 4
+
+A divisor $d$ of a positive integer $n$ is said to be a close divisor of $n$ if $\sqrt{n}<d<2 \sqrt{n}$. Does there exist a positive integer with exactly 2020 close divisors?
+
+## Solution
+
+Let $m$ be an odd integer with exactly 2020 positive divisors and which is (automatically) not a square. For example, $m=3^{2019}$ suffices, but of course there are many alternatives. Now consider $n=2^{k} m$, for some integer $k$ such that $2^{k}>m$. Any divisor of $n$ is then of the form $2^{\ell} d$ where $d$ is a divisor of $m$. We will now show that for every such divisor $d$, there exists a unique $\ell$ such that $2^{\ell} d$ is a close divisor. Because $\sqrt{n}$ is not an integer, there certainly is a unique integer $a$ such that $\sqrt{n}<2^{a} d<2 \sqrt{n}$. Because $2^{k}>m$, we have $m<\sqrt{n}<2^{k}$. Combining with $1 \leq d \leq m$, we find $1<\frac{\sqrt{n}}{d}<2^{a}<\frac{2 \sqrt{n}}{d}<2 \cdot 2^{k}$ and so we see that $0<a \leqslant k$. So $2^{a} d$ is indeed a close divisor. Because $m$ has exactly 2020 divisors, we find that $n$ has exactly 2020 close divisors.
+

Benelux/md/Benelux_en-olympiad_en-bxmo-problems-2021-zz.md

@@ -0,0 +1,148 @@
+![](https://cdn.mathpix.com/cropped/2024_12_15_7ed0b6fed5333a54e9a5g-1.jpg?height=892&width=1158&top_left_y=339&top_left_x=449)
+
+# 13th Benelux Mathematical Olympiad Virtual, 1st - 2nd May 2021 
+
+## Problems and Solutions
+
+## Problem Selection Committee
+
+Stijn Cambie, Nicolas Radu (Belgium),
+Jeroen Huijben, Ward van der Schoot (the Netherlands),
+Pierre Haas, Pascal Zeihen (Luxembourg).
+
+# BxMO 2021: Problems and Solutions 
+
+## Problem 1
+
+(a) Prove that for all $a, b, c, d \in \mathbb{R}$ with $a+b+c+d=0$,
+
+$$
+\max (a, b)+\max (a, c)+\max (a, d)+\max (b, c)+\max (b, d)+\max (c, d) \geqslant 0
+$$
+
+(b) Find the largest non-negative integer $k$ such that it is possible to replace $k$ of the six maxima in this inequality by minima in such a way that the inequality still holds for all $a, b, c, d \in \mathbb{R}$ with $a+b+c+d=0$.
+
+## Solution 1
+
+The left-hand-side of the inequality is invariant under permutations of $a, b, c, d$. We may therefore suppose that $a \geqslant b \geqslant c \geqslant d$, so that the inequality reduces to
+
+$$
+0 \leqslant 3 a+2 b+c=a+(a+b)+(a+b+c)
+$$
+
+We claim that each of the terms on the right-hand side is non-negative; this will prove the inequality. Indeed, if $a<0$, then $a, b, c, d<0$, and so $a+b+c+d<0$, a contradiction. Also, if $a+b<0$, then, as $b \leqslant a$, $0>b \geqslant c, d$ so $(a+b)+c+d<0$, another contradiction. Finally, if $a+b+c<0$, then, as above, $0>c \geqslant d$, so $(a+b+c)+d<0$, a final contradiction.
+
+Next, we claim that it is impossible to replace $k \geqslant 3$ maxima by minima in the inequality. Indeed, if $k \geqslant 3$, one number, say $d$, appears in two of the terms changed to minima. Take $a=b=c=1, d=-3$, so that $a+b+c+d=0$. Then the sum is at most $4 \cdot 1+2 \cdot(-3)<0$. Hence $k<3$.
+
+Finally, we prove that the required inequality holds if $k=2$ and the terms involving the complementary sets $\{a, b\},\{c, d\}$ are changed to minima. We will assume again that $a \geqslant b \geqslant c \geqslant d$, and prove that the inequality holds for any change of the terms involving permutations of these sets to minima (rather than proving that it holds for all orderings of $a, b, c, d$ for this one change). There are three cases:
+(1) change terms $\{a, b\},\{c, d\}$, so the inequality becomes $2 a+3 b+d \geqslant 0 \Longleftrightarrow a+b+(b-c) \geqslant 0$;
+(2) change terms $\{a, c\},\{b, d\}$, so the inequality becomes $2 a+b+2 c+d \geqslant 0 \Longleftrightarrow a+c \geqslant 0$;
+(3) change terms $\{a, d\},\{b, c\}$, so the inequality becomes $2 a+b+2 c+d \geqslant 0 \Longleftrightarrow a+c \geqslant 0$,
+where we have used $a+b+c+d=0$. In the first case, inequality holds since $a+b \geqslant 0$ as proved earlier and $b \geqslant c$. In the other cases, suppose that $a+c<0$. Then $c<-a$ and hence $d<-a$ as $d \leqslant c$. Hence $c+d<-2 a \leqslant-a-b$ as $a \geqslant b$, so $a+b+c+d<0$, a final contradiction, completing the proof.
+
+## Solution 2
+
+Using the inequality $\max (x, y) \geqslant \frac{1}{2}(x+y)$, we find that
+
+$$
+\max (a, b)+\max (a, c)+\max (a, d)+\max (b, c)+\max (b, d)+\max (c, d) \geqslant \frac{3}{2}(a+b+c+d)=0
+$$
+
+For $k=3$, we take the same counterexample as in Solution 1. Now it remains to prove the inequality where $\max (a, b)$ and $\max (c, d)$ are replaced by $\min (a, b)$ and $\min (c, d)$. We can assume without loss of generality that $\min (a, b)=a$ and $\min (c, d)=c$. Now we find
+
+$$
+\begin{aligned}
+& \min (a, b)+\max (a, c)+\max (a, d)+\max (b, c)+\max (b, d)+\min (c, d) \geqslant \\
+& a+\frac{1}{2}(a+c)+d+b+\frac{1}{2}(b+d)+c=\frac{3}{2}(a+b+c+d)=0
+\end{aligned}
+$$
+
+## BxMO 2021: Problems and Solutions
+
+## Solution 3
+
+We present another proof for the inequality where $\max (a, b)$ and $\max (c, d)$ are replaced by $\min (a, b)$ and $\min (c, d)$. By substituting $\max (x, y)=\frac{1}{2}(x+y+|x-y|)$ and $\min (x, y)=\frac{1}{2}(x+y-|x-y|)$ everywhere and using $a+b+c+d=0$, the inequality may be rewritten as:
+
+$$
+|a-c|+|a-d|+|b-c|+|b-d| \stackrel{?}{\geqslant}|a-b|+|c-d| .
+$$
+
+By the triangle inequality, we have
+
+$$
+|a-c|+|a-d|+|b-c|+|b-d| \geqslant|(a-c)-(b-c)|+|(a-d)-(b-d)|=2|a-b|,
+$$
+
+and similarly, $|a-c|+|a-d|+|b-c|+|b-d| \geqslant 2|c-d|$. Adding these and dividing by 2 yields the desired inequality.
+
+# BxMO 2021: Problems and Solutions 
+
+## Problem 2
+
+Pebbles are placed on a $2021 \times 2021$ board in such a way that each square contains at most one pebble. The pebble set of a square of the board is the collection of all pebbles which are in the same row or column as this square. Determine the least number of pebbles that can be placed on the board in such a way that no two squares have the same pebble set.
+
+## Solution
+
+Let $N \geqslant 1$ be a positive integer. We claim that the least number of pebbles that can be placed on a $(2 N+1) \times(2 N+1)$ chessboard in such a way that no two squares of the board have the same pebble set is $3 N+1$. The problem has $N=1010$, so at least 3031 pebbles are needed.
+
+We begin by placing $(2 N+1)+N=3 N+1$ pebbles on the board as shown. The construction extends to all values of $N$, and one checks that no two squares on the board have the same pebble set.
+![](https://cdn.mathpix.com/cropped/2024_12_15_7ed0b6fed5333a54e9a5g-4.jpg?height=366&width=960&top_left_y=1165&top_left_x=551)
+
+We are left to show that the number of pebbles, $P$, that must be placed on the board is at least $3 N+1$. Suppose that there exists an empty row. Any other row must then contain at least two pebbles. Indeed, if another row is empty, Figure [1], or contains exactly one pebble, Figure [2], then there are two squares (shaded in the figures below) with the same pebble set. So in this case there are at least $P \geqslant 2(2 N+1-1)=4 N \geqslant 3 N+1$ pebbles on the board. The same argument applies, mutatis mutandis, to the columns of the board.
+![](https://cdn.mathpix.com/cropped/2024_12_15_7ed0b6fed5333a54e9a5g-4.jpg?height=292&width=1543&top_left_y=1864&top_left_x=268)
+
+On the other hand, suppose that each each row and each column contains at least one pebble. Let $k$ the number of rows containing precisely one pebble. Counting the number of pebbles per row shows that $P \geqslant k+2(2 N+1-k)$. To count the number of pebbles per column, note that the $k$ pebbles that are the only ones in their row must be in $k$ distinct columns, Figure [3]. Furthermore, at most one of these $k$ columns only has one pebble in it, Figure [4]. So these $k$ columns contain a total of at least $2 k-1$ pebbles, while the remaining columns contain at least one each. Hence $P \geqslant(2 k-1)+(2 N+1-k)$. Adding these two inequalities we get $2 P \geqslant 6 N+2$.
+
+Remark. For a $2 N \times 2 N$ board, at least $3 N$ pebbles are needed. The proof is similar: if there is an empty line, then $P \geqslant 2(2 N-1) \geqslant 3 N$ for $N \geqslant 2$. If there is no empty line, then $2 P \geqslant 6 N-1$. A construction similar to that of the odd boards works. While the construction is unique in the odd case, up to permutation of rows and columns, that is not true for even boards.
+
+## BxMO 2021: Problems and Solutions
+
+## Problem 3
+
+A cyclic quadrilateral $A B X C$ has circumcentre $O$. Let $D$ be a point on line $B X$ such that $|A D|=|B D|$. Let $E$ be a point on line $C X$ such that $|A E|=|C E|$. Prove that the circumcentre of triangle $\triangle D E X$ lies on the perpendicular bisector of $O A$.
+
+## Solution 1
+
+First, note that $\angle A O X=2 \angle A B X=2\left(180^{\circ}-\angle A C X\right)=2 \angle A C E$ as $A B X C$ is cyclic. Secondly, both $\triangle D A B$ and $\triangle E A C$ are isosceles, which implies that $\angle A E X=\angle A E C=180^{\circ}-2 \angle A C E=180^{\circ}-\angle A O X$ and $\angle A D X=$ $\angle A D B=180^{\circ}-2 \angle A B D=180^{\circ}-2 \angle A B X=180^{\circ}-\angle A O X$. From this, see that respectively $A E X O$ and $A D X O$ are cyclic, i.e. $A E D X O$ is cyclic.
+
+Hence, the circumcentre of $\triangle D E X$ is also the circumcentre of $A E D X O$. However, as in a circle any perpendicular bisector of a chord goes through the centre of the circle, we find that the circumcentre of $\triangle D E X$ lies on the perpendicular bisector of $O A$.
+![](https://cdn.mathpix.com/cropped/2024_12_15_7ed0b6fed5333a54e9a5g-5.jpg?height=1086&width=891&top_left_y=1222&top_left_x=611)
+
+## Solution 2
+
+In this solution, we use directed angles $\measuredangle$. We have $\measuredangle A B D=\measuredangle A B X=\measuredangle A C X=\measuredangle A C E$ and since $\triangle A B D$ and $\triangle A C E$ are both isosceles, we see that $\triangle A B D \sim \triangle A C E$ with equal orientation. This means that there exists a spiral symmetry $T$ with centre $A$ such that $T(B)=D$ and $T(C)=E$. Now let $O^{\prime}$ be the centre of $\odot D E X$. Then we find $\measuredangle D O^{\prime} E=2 \measuredangle D X E=2 \measuredangle B X C=\measuredangle B O C$. Moreover, $\triangle D O^{\prime} E$ and $\triangle B O C$ are isosceles, so we have $\triangle D O^{\prime} E \sim \triangle B O C$ with equal orientation. This means that $T$ must send $O$ to $O^{\prime}$, so in particular, $\triangle A O O^{\prime}$ is similar to $\triangle A B D$ and $\triangle A C E$. We conclude that $\left|A O^{\prime}\right|=\left|O O^{\prime}\right|$, from which the result follows.
+
+# BxMO 2021: Problems and Solutions 
+
+## Problem 4
+
+A sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers satisfies $a_{1}>5$ and $a_{n+1}=5+6+\cdots+a_{n}$ for all positive integers $n$. Determine all prime numbers $p$ such that, regardless of the value of $a_{1}$, this sequence must contain a multiple of $p$.
+
+## Solution 1
+
+We claim that the only prime number of which the sequence must contain a multiple is $p=2$. To prove this, we begin by noting that
+
+$$
+a_{n+1}=\frac{a_{n}\left(a_{n}+1\right)}{2}-10=\frac{\left(a_{n}-4\right)\left(a_{n}+5\right)}{2}
+$$
+
+Let $p>2$ be an odd prime, and choose $a_{1} \equiv-4(\bmod p)$, so $2 a_{2} \equiv(-4-4)(-4+5) \equiv-8(\bmod p)$, whence $a_{2} \equiv-4(\bmod p)$, since $p$ is odd. By induction, $a_{n} \equiv-4(\bmod p) \not \equiv 0(\bmod p)$ for all $n$, and so the sequence need not contain a multiple of $p$.
+
+We are left to show that the sequence must contain an even number. Suppose to the contrary that $a_{n}$ is odd for $n=1,2, \ldots$ We observe that
+
+$$
+a_{n+1}-a_{n}=\frac{a_{n}\left(a_{n}+1\right)}{2}-\frac{a_{n-1}\left(a_{n-1}+1\right)}{2}=\frac{a_{n}-a_{n-1}}{2}\left(a_{n}+a_{n-1}+1\right)
+$$
+
+By assumption, $a_{n}+a_{n-1}+1$ is odd for $n=1,2, \ldots$, so this shows that $v_{2}\left(a_{n+1}-a_{n}\right)=v_{2}\left(a_{n}-a_{n-1}\right)-1$, and so there exists $N$ such that $v_{2}\left(a_{N+1}-a_{N}\right)=0$. This is a contradiction, because $a_{n+1}-a_{n}$ is even for $n=1,2, \ldots$ by assumption, and thus completes the proof.
+
+## Solution 2
+
+For odd $p$, proceed as in solution 1 . Now let $p=2$, and suppose that every term of the sequence is odd. We claim that it follows that $a_{n} \equiv 5\left(\bmod 2^{k}\right)$ for every integer $n \geq 1$ and every integer $k \geq 1$. We proceed per induction on $k$. For $k=1$ this simply states that $a_{n}$ is odd for all integers $n \geq 1$, as assumed. Now suppose it is true for $k=r$. Let $k=r+1$. Take any integer $n \geq 1$. Note that, by the induction hypothesis, $a_{n} \equiv 5\left(\bmod 2^{r}\right)$. Therefore there exists an integer $s$ such that $a_{n}=2^{r} s+5$. Now note that
+
+$$
+a_{n+1}=\frac{\left(a_{n}-4\right)\left(a_{n}+5\right)}{2}=\frac{\left(2^{r} s+1\right)\left(2^{r} s+10\right)}{2}=\left(2^{r} s+1\right)\left(2^{r-1} s+5\right) \equiv 2^{r-1} s+5 \quad\left(\bmod 2^{r}\right)
+$$
+
+By the induction hypothesis, $a_{n+1} \equiv 5\left(\bmod 2^{r}\right)$. Therefore $s$ is even, such that $a_{n}$ is of the form $2^{r+1} s+5$ for any integer $n \geq 1$, which concludes the induction. From this property, it follows that $a_{1}-5$ is divisible by $2^{k}$ for every integer $k \geq 1$, which is only possible if $a_{1}=5$. But $a_{1}>5$, so this is a contradiction.
+

Benelux/md/Benelux_en-olympiad_en-bxmo-problems-2022-zz.md

@@ -0,0 +1,176 @@
+![](https://cdn.mathpix.com/cropped/2024_12_15_dfcb8b6cd65e70af5e9eg-1.jpg?height=489&width=635&top_left_y=338&top_left_x=716)
+
+# 14th Benelux Mathematical Olympiad Leuven, 29th April - 1st May 2022 
+
+## Problems and Solutions
+
+## BxMO 2022: Problems and Solutions
+
+## Problem 1
+
+Let $n \geqslant 0$ be an integer, and let $a_{0}, a_{1}, \ldots, a_{n}$ be real numbers. Show that there exists $k \in\{0,1, \ldots, n\}$ such that
+
+$$
+a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n} \leqslant a_{0}+a_{1}+\cdots+a_{k}
+$$
+
+for all real numbers $x \in[0,1]$.
+
+## Solution 1
+
+The case $n=0$ is trivial; for $n>0$, the proof goes by induction on $n$. We need to make one preliminary observation:
+
+Claim. For all reals $a, b, a+b x \leqslant \max \{a, a+b\}$ for all $x \in[0,1]$.
+Proof. If $b \leqslant 0$, then $a+b x \leqslant a$ for all $x \in[0,1]$; otherwise, if $b>0, a+b x \leqslant a+b$ for all $x \in[0,1]$. This proves our claim.
+
+This disposes of the base case $n=1$ of the induction: $a_{0}+a_{1} x \leqslant \max \left\{a_{0}, a_{0}+a_{1}\right\}$ for all $x \in[0,1]$. For $n \geqslant 2$, we note that, for all $x \in[0,1]$,
+
+$$
+\begin{aligned}
+a_{0}+a_{1} x+\cdots+a_{n} x^{n} & =a_{0}+x\left(a_{1}+a_{2} x+\cdots+a_{n} x^{n-1}\right) \\
+& \leqslant a_{0}+x\left(a_{1}+a_{2}+\cdots+a_{k}\right) \leqslant \max \left\{a_{0}, a_{0}+\left(a_{1}+\cdots+a_{k}\right)\right\},
+\end{aligned}
+$$
+
+for some $k \in\{1,2, \ldots, n\}$ by the inductive hypothesis and our earlier claim. This completes the proof by induction.
+
+## Solution 2
+
+Define $s_{i}=a_{0}+a_{1}+\cdots+a_{i}$ for $i \in\{0,1, \ldots, n\}$. Thus $a_{0}=s_{0}$ and $a_{i}=s_{i}-s_{i-1}$ for all $i \in\{1,2, \ldots, n\}$. Hence
+
+$$
+\begin{aligned}
+a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n} & =s_{0}+\left(s_{1}-s_{0}\right) x+\left(s_{2}-s_{1}\right) x^{2}+\ldots+\left(s_{n}-s_{n-1}\right) x^{n} \\
+& =s_{0}(1-x)+s_{1}\left(x-x^{2}\right)+\ldots+s_{n-1}\left(x^{n-1}-x^{n}\right)+s_{n} x^{n}
+\end{aligned}
+$$
+
+Now choose $k \in\{0,1, \ldots, n\}$ such that $s_{k}=\max \left\{s_{0}, s_{1}, \ldots, s_{n}\right\}$. Using the inequality $x^{i-1}-x^{i} \geqslant 0$, valid for all $i \in\{1,2, \ldots, n\}$ and all $x \in[0,1]$, in the right-hand side above, it follows that
+
+$$
+\begin{aligned}
+a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n} \leqslant s_{k}(1- & x)+s_{k}\left(x-x^{2}\right)+\cdots+s_{k}\left(x^{n-1}-x^{n}\right)+s_{k} x^{n} \\
+& =s_{k}\left[(1-x)+\left(x-x^{2}\right)+\cdots+\left(x^{n-1}-x^{n}\right)+x^{n}\right] \\
+& =s_{k}=a_{0}+a_{1}+\cdots+a_{k} .
+\end{aligned}
+$$
+
+This completes the proof.
+
+## BxMO 2022: Problems and Solutions
+
+## Solution 3
+
+The proof proceeds by induction on $n$. The base case $n=0$ is trivial. For $n \geqslant 1$, since $x \in[0,1]$, we have $x^{n} \leqslant x^{n-1}$. Thus, if $a_{n} \geqslant 0$, then $a_{n} x^{n} \leqslant a_{n} x^{n-1}$, while, if $a_{n}<0$, then $a_{n} x^{n}<0$ trivially. This shows that $a_{n} x^{n} \leqslant \max \left\{0, a_{n} x^{n-1}\right\}$, whence
+
+$$
+a_{0}+a_{1} x+\cdots+a_{n-1} x^{n-1}+a_{n} x^{n} \leqslant a_{0}+a_{1} x+\cdots+\max \left\{a_{n-1}, a_{n-1}+a_{n}\right\} x^{n-1}
+$$
+
+By the inductive hypothesis, the polynomial of degree $n-1$ on the right-hand side is bounded above by $a_{0}+\cdots+a_{k}$ for some $k \in\{0,1, \ldots, n-2\}$ or $a_{0}+\cdots+a_{n-2}+\max \left\{a_{n-1}, a_{n-1}+a_{n}\right\}$. But the latter is equal to one of $a_{0}+a_{1}+\cdots+a_{n-1}$ or $a_{0}+a_{1}+\cdots+a_{n}$; both are of the desired form, $a_{0}+a_{1}+\cdots+a_{k}$ for some $k \in\{n-1, n\}$. This completes the proof by induction.
+
+## BxMO 2022: Problems and Solutions
+
+## Problem 2
+
+Let $n$ be a positive integer. There are $n$ ants walking along a line at constant nonzero speeds. Different ants need not walk at the same speed or walk in the same direction. Whenever two or more ants collide, all the ants involved in this collision instantly change directions. (Different ants need not be moving in opposite directions when they collide, since a faster ant may catch up with a slower one that is moving in the same direction.) The ants keep walking indefinitely.
+Assuming that the total number of collisions is finite, determine the largest possible number of collisions in terms of $n$.
+
+## Solution 1
+
+The order of the ants along the line does not change; denote by $v_{1}, v_{2}, \ldots, v_{n}$ the respective speeds of ants $1,2, \ldots, n$ in this order. If $v_{i-1}<v_{i}>v_{i+1}$ for some $i \in\{2, \ldots, n-1\}$, then, at each stage, ant $i$ can catch up with ants $i-1$ or $i+1$ irrespective of the latters' directions of motion, so the number of collisions is infinite. Hence, if the number of collisions is finite, then, up to switching the direction definining the order of the ants, (i) $v_{1} \geqslant \cdots \geqslant v_{n}$ or (ii) $v_{1} \geqslant \cdots \geqslant v_{k-1}>v_{k} \leqslant \cdots \leqslant v_{n}$ for some $k \in\{2, \ldots, n-1\}$. We need the following observation:
+
+Claim. If $v_{1} \geqslant \cdots \geqslant v_{m}$, then ants $m-1$ and $m$ collide at most $m-1$ times.
+Proof. The proof goes by induction on $m$, the case $m=1$ being trivial. Since $v_{m-1} \geqslant v_{m}$, ants $m-1$ and $m$ can only collide if the former is moving towards the latter. Hence, between successive collisions with ant $m$, ant $m-1$ must reverse direction by colliding with ant $m-2$. Since ants $m-1$ and $m-2$ collide at most $(m-1)-1=m-2$ times by the inductive hypothesis, ants $m$ and $m-1$ collide at most $(m-2)+1=m-1$ times.
+
+Hence, in case (i), there are at most $0+1+\cdots+(n-1)=n(n-1) / 2$ collisions. In case (ii), applying the claim to ants $1,2, \ldots, k$ and also to ants $n, n-1, \ldots, k$ by switching their order, the number of collisions is at most $k(k-1) / 2+(n-k+1)(n-k) / 2=n(n-1) / 2-(k-1)(n-k)<n(n-1) / 2$.
+
+Now take a coordinate $x$ along the line, and put ants at $x=1,2, \ldots, n$ with positive initial velocities and speeds $v_{1}=\cdots=v_{n-1}=1, v_{n}=\varepsilon$, for some $\varepsilon$. For $\varepsilon=0$, collisions occur according to the pattern shown below for $n=5$, which clearly extends to all values of $n$ in such a way that ants $m$ and $m+1$ collide exactly $m$ times for $m=1,2, \ldots, n-1$. This yield $1+2+\cdots+(n-1)=n(n-1) / 2$ collisions in total. For all sufficiently small $\varepsilon>0$, the number of collisions remains equal to $n(n-1) / 2$.
+![](https://cdn.mathpix.com/cropped/2024_12_15_dfcb8b6cd65e70af5e9eg-4.jpg?height=292&width=686&top_left_y=2270&top_left_x=679)
+
+This shows that the upper bound obtained above can be attained. If the number of collisions is finite, the largest possible number of collisions is therefore indeed $n(n-1) / 2$.
+
+## BxMO 2022: Problems and Solutions
+
+## Solution 2
+
+We show that there are at most $n(n-1) / 2$ collisions if the number of collisions is finite as in Solution 1.
+To show that the upper bound of $n(n-1) / 2$ collisions can be attained, we construct, inductively, an example of $n$ ants colliding $n(n-1) / 2$ times, the speeds of the ants decrease from left to right, and after all collisions all ants move towards the left, with the possible exception of the rightmost ant. In every case, we will label the ants $1,2, \ldots, n$ from left to right. For $n=1$ this is trivial. For $n \geqslant 2$, we use the construction for $n-1$ ants (now labelled $2,3, \ldots, n$ ). We add ant 1 on the left, moving towards the right, faster than all other ants (so that the speeds of the ants still decrease from left to right), and in such a way that its first collision (with ant 2) happens after all $(n-1)(n-2) / 2$ collisions of the other $n-1$ ants. Now the following events happen (in this order) for $i=1,2, \ldots, n-2$ : ants $i$ and $i+1$ collide, after which ant $i$ moves to the left and ant $i+1$ moves to the right. These collisions do happen because the speeds of the ants decrease from left to right. Then ants $n-1$ and $n$ also collide, resulting in ant $n-1$ moving to the left. This shows that there are (at least) $(n-1)(n-2) / 2+(n-1)=n(n-1) / 2$ collisions. There are in fact no more collisions since the speeds of the ants decrease from left to right; alternatively, this follows from the upper bound proved previously. Since all ants except ant $n$ are moving towards the left after the collisions, this completes the inductive construction.
+
+## BxMO 2022: Problems and Solutions
+
+## Problem 3
+
+Let $A B C$ be a scalene acute triangle. Let $B_{1}$ be the point on ray $\left[A C\right.$ such that $\left|A B_{1}\right|=\left|B B_{1}\right|$. Let $C_{1}$ be the point on ray $\left[A B\right.$ such that $\left|A C_{1}\right|=\left|C C_{1}\right|$. Let $B_{2}$ and $C_{2}$ be the points on line $B C$ such that $\left|A B_{2}\right|=\left|C B_{2}\right|$ and $\left|B C_{2}\right|=\left|A C_{2}\right|$. Prove that $B_{1}, C_{1}, B_{2}, C_{2}$ are concyclic.
+
+## Solution 1
+
+By construction, lines $B_{1} C_{2}$ and $B_{2} C_{1}$ bisect segments $[A B]$ and $[A C]$, respectively, so their intersection $O$ is the circumcentre of $A B C$. Hence $\angle B O C=2 \angle A$ and $\angle C B O=\angle O C B=90^{\circ}-\angle A$. Now, by construction, $\angle O C_{1} B=90^{\circ}-\angle A=\angle O C B$, so $B C_{1} C O$ is cyclic. Similarly, $B C B_{1} O$ is cyclic by construction because $\angle O B_{1} C=180^{\circ}-\angle A B_{1} O=90^{\circ}+\angle A=180^{\circ}-\angle C B O$. In particular, $B C_{1} C B_{1}$ is cyclic, too.
+
+Now $\angle B_{1} C_{1} B_{2}=\angle B_{1} C_{1} B-\angle B_{2} C_{1} B$ and $\angle B_{1} C_{2} B_{2}=\angle B_{1} C B-\angle C B_{1} C_{2}$. But $\angle B_{1} C_{1} B=\angle B_{1} C B$ since $B C_{1} C B_{1}$ is cyclic and $\angle B_{2} C_{1} B=\angle O C_{1} B=90^{\circ}-\angle A=180^{\circ}-\angle O B_{1} C=\angle C B_{1} C_{2}$. Hence $\angle B_{1} C_{1} B_{2}=\angle B_{1} C_{2} B_{2}$, so $B_{1} B_{2} C_{1} C_{2}$ is cyclic, as required.
+![](https://cdn.mathpix.com/cropped/2024_12_15_dfcb8b6cd65e70af5e9eg-6.jpg?height=781&width=1009&top_left_y=1277&top_left_x=521)
+
+## Solution 2
+
+The isosceles triangles $A B_{1} B$ and $A C_{1} C$ have equal base angles $\angle B A B_{1}=\angle C_{1} A C=\angle A$, so are similar. In particular, $|A B| /\left|A B_{1}\right|=|A C| /\left|A C_{1}\right|$. Since $\angle B A C=\angle B_{1} A C_{1}=\angle A$, it follows that triangles $A B C$ and $A B_{1} C_{1}$ are similar, too. In particular, $\angle C B A=\angle A B_{1} C_{1}$.
+
+By construction, lines $B_{1} C_{2}$ and $B_{2} C_{1}$ are the respective perpendicular bisectors of $[A B]$ and $[A C]$, so meet them at their respective midpoints $C^{\prime}$ and $B^{\prime}$. Hence
+
+$$
+\begin{aligned}
+\angle B_{1} C_{2} B_{2} & =\angle C^{\prime} C_{2} B=90^{\circ}-\angle C_{2} B C^{\prime}=90^{\circ}-\angle C B A=90^{\circ}-\angle A B_{1} C_{1}=90^{\circ}-\angle B^{\prime} B_{1} C_{1} \\
+& =\angle B_{1} C_{1} B^{\prime}=\angle B_{1} C_{1} B_{2} .
+\end{aligned}
+$$
+
+Hence $B_{1} C_{2} C_{1} B_{2}$ is cyclic, which completes the proof.
+
+## BxMO 2022: Problems and Solutions
+
+## Solution 3
+
+By construction, lines $B_{1} C_{2}$ and $B_{2} C_{1}$ are the respective perpendicular bisectors of $[A B]$ and $[A C]$, so meet them at their respective midpoints $C^{\prime}$ and $B^{\prime}$. Since $\angle B_{1} C^{\prime} C_{1}=90^{\circ}=\angle C_{1} B^{\prime} B_{1}, B_{1} C_{1} C^{\prime} B^{\prime}$ is cyclic. Together with the fact that $B^{\prime} C^{\prime} \| B C$ by construction, this implies
+
+$$
+\angle B_{1} C_{2} B_{2}=\angle C^{\prime} C_{2} B=\angle C_{2} C^{\prime} B^{\prime}=\angle B_{1} C^{\prime} B^{\prime}=\angle B_{1} C_{1} B^{\prime}=\angle B_{1} C_{1} B_{2},
+$$
+
+whence $B_{1} C_{2} C_{1} B_{2}$ is cyclic. This completes the proof.
+
+# BxMO 2022: Problems and Solutions 
+
+## Problem 4
+
+A subset $A$ of the natural numbers $\mathbb{N}=\{0,1,2, \ldots\}$ is called good if every integer $n>0$ has at most one prime divisor $p$ such that $n-p \in A$.
+(a) Show that the set $S=\{0,1,4,9, \ldots\}$ of perfect squares is good.
+(b) Find an infinite good set disjoint from $S$.
+(Two sets are disjoint if they have no common elements.)
+
+## Solution 1
+
+(a) Suppose to the contrary that $S$ is not good, so there exists $n \in \mathbb{N}$ with two different prime factors $p \neq q$ such that $n-p, n-q$ are perfect squares. In particular $n$ is not prime. Write $n-p=m^{2}$, for some $m \in \mathbb{N}$. As $p \mid n$, it follows that $p \mid m$ and hence $p^{2} \mid m^{2}$ since $p$ is prime. Hence there exists $k \in \mathbb{N}$ such that $n-p=p^{2} k^{2}$. Similarly, there exists $\ell \in \mathbb{N}$ such that $n-q=q^{2} \ell^{2}$. We observe that $k, \ell \neq 0$ since $n$ is not prime.
+Now we have $p-q=(n-q)-(n-p)=(\ell q-k p)(\ell q+k p)$. Since $p-q \neq 0, \ell q-k p \neq 0$, and hence $|p-q|=|k p-\ell q||k p+\ell q| \geqslant|k p+\ell q|=k p+\ell q$. This is a contradiction however, because, since $k, \ell \neq 0$, it is clear that $k p+\ell q \geqslant p+q>|p-q|$. Hence $S$ is good.
+(b) Let $q$ be a prime, and let $Q=\left\{q, q^{3}, q^{5}, \ldots\right\}$ be the (infinite) set of odd powers of $q$, which is disjoint from $S$. We claim that $Q$ is good. Indeed, let $n \in \mathbb{N}$, and let $p \mid n$ be a prime such that $n-p \in Q$, i.e. $n-p=q^{2 k+1}$ for some $k \in \mathbb{N}$. Then $p \mid n-p$, so $p \mid q^{2 k+1}$, and hence $p=q$. Thus $Q$ is good.
+
+## Solution 2
+
+(a) Let $p \mid n$ be a prime such that $n-p=p(n / p-1)=m^{2}$, for some $m \in \mathbb{N}$. Since $p \mid m^{2}$ and $p$ is prime, $p^{2} \mid m^{2}$, and hence $p \mid n / p-1<n / p$, so $p<\sqrt{n}$.
+Now suppose to the contrary that $S$ is not good, so there are primes $p_{1}>p_{2}$ dividing $n$ such that $n-p_{1}<n-p_{2}$ are perfect squares. Then
+
+$$
+n-p_{2} \geqslant\left(\sqrt{n-p_{1}}+1\right)^{2}>n-p_{1}+2 \sqrt{n-p_{1}} \quad \Longrightarrow \quad p_{1}>p_{2}+2 \sqrt{n-p_{1}} \geqslant 2+2 \sqrt{n-p_{1}} .
+$$
+
+The last condition implies that $p_{1}>2 \sqrt{n-1}$. But $p_{1}<\sqrt{n}$ by the first part, so $\sqrt{n}>2 \sqrt{n-1}$, which is a contradiction for $n>1$; the cases $n=0$ and $n=1$ are trivial. Thus $S$ is good.
+(b) We claim that the infinite set $P=\{3,5,7,11, \ldots\}$ of odd primes, which is disjoint from $S$, is good. Indeed, let $n \in \mathbb{N}$ and let $p \mid n$ be a prime such that $n-p=q$, for some odd prime $q$. Then $p \mid n-p$, so $p \mid q$, i.e. $p=q$, and hence $n=2 q$. Since $q$ is the only odd prime divisor of $n=2 q$, $P$ is good.
+The set $P^{\prime}=\{2,3,5,7,11, \ldots\}$ of all primes is also good. The proof is similar: let $n \in \mathbb{N}$ and let $p \mid n$ be a prime such that $n-p=q$, for some prime $q$. Then $p \mid n-p$, so $p \mid q$, i.e. $p=q$, and hence $n=2 q$. If $q=2$, then 2 is the only prime divisor of $n$; if $q \neq 2$, then the only prime divisor of $n$, apart from $q$, is 2 . However, $n-2=2(q-1) \notin P^{\prime}$ since $q-1>1$. Hence $P^{\prime}$ is good.
+
+# BxMO 2022: Problems and Solutions 
+
+## Solution 3
+
+(a) Suppose to the contrary that $S$ is not good, so there exists $n \in \mathbb{N}$ with two different prime factors $p \neq q$ such that $n-p, n-q$ are perfect squares. Write $n-p=m^{2}$, for some $m \in \mathbb{N}$. As $p \mid n$, it follows that $p \mid m$ and hence $p^{2} \mid m^{2}$ since $p$ is prime. Hence there exists $k \in \mathbb{N}$ such that $n-p=p^{2} k^{2}$. Similarly, there exists $\ell \in \mathbb{N}$ such that $n-q=q^{2} \ell^{2}$. By construction, $n$ is not prime, so $n-p, n-q \neq 0$, whence $k, \ell \geqslant 1$.
+Hence $p^{2} k^{2}+p=q^{2} \ell^{2}+q$. Hence $p^{2} k^{2}<p^{2} k^{2}+p=q^{2} \ell^{2}+q<q^{2} \ell^{2}+2 q \ell+1=(q \ell+1)^{2}$. Similarly, $q^{2} \ell^{2}<(p k+1)^{2}$, whence $q \ell-1<p k<q \ell+1$. It follows that $p k=q \ell$, so $p^{2} k^{2}+p=q^{2} \ell^{2}+q$ yields the contradiction $p=q$. Hence $S$ is good.
+(b) Let $A$ be a finite good set such that $0 \notin A$, and let $m=\max A$. Let $a \geqslant 2 m+1$ be an integer. We claim that $A^{\prime}=A \cup\{a\}$ is good. Indeed, suppose to the contrary that there exist $n \in \mathbb{N}$ and primes $p, q \mid n$ with $p \neq q$ such that $n-p, n-q \in A^{\prime}$. If $n<a$, then $n-p, n-q \in A$, which is a contradiction because $A$ is good. Hence $n \geqslant a$. Now $p \mid n-p$, so $n-p \geqslant p$ since $0 \notin A^{\prime}$. Thus $p \leqslant n / 2$ and hence $n-p \geqslant n / 2 \geqslant a / 2>m$. Similarly, $n-q>m$. It follows that $n-p=n-q=a$, which implies the contradiction $p=q$. Hence $A^{\prime}$ is good.
+Now it is clear that any singleton set is good: indeed, if $A=\{a\}$, and $n \in \mathbb{N}$ has prime divisors $p, q$ such that $n-p, n-q \in A$, then $n-p=a=n-q$, so $p=q$. Starting from the singleton $T_{1}=\{2\}$, we use the above construction to obtain, iteratively, good sets $T_{2}, T_{3}, \ldots$ of $2,3, \ldots$ elements. It is clearly possibly to ensure that they are each disjoint from $S$ by not adding a perfect square at any stage. Then $T=T_{1} \cup T_{2} \cup \cdots$ is an infinite good set disjoint from $S$.
+

Benelux/md/Benelux_en-olympiad_en-bxmo-problems-2023-zz.md

@@ -0,0 +1,162 @@
+15th Benelux Mathematical Olympiad Esch-sur-Alzette, 5th - 7th May 2023
+
+## Problems and Solutions
+
+# BxMO 2023: Problems and Solutions 
+
+## Problem 1
+
+Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
+
+$$
+(x-y)(f(x)+f(y)) \leqslant f\left(x^{2}-y^{2}\right) \quad \text { for all } x, y \in \mathbb{R}
+$$
+
+## Solution
+
+Clearly, $f(x)=c x$ is a solution for each $c \in \mathbb{R}$ since $(x-y)(c x+c y)=c\left(x^{2}-y^{2}\right)$. To show that there are no other solutions, we observe that
+(1) $x=y: \quad 0 \leqslant f(0)$;
+$x=1, y=0: \quad f(0)+f(1) \leqslant f(1) \Rightarrow f(0) \leqslant 0$, whence $f(0)=0 ;$
+(2) $y=-x: \quad 2 x(f(x)+f(-x)) \leqslant f(0)=0$;
+$x \rightarrow-x: \quad-2 x(f(-x)+f(x)) \leqslant 0 \Rightarrow 2 x(f(x)+f(-x)) \geqslant 0 ;$
+thus $2 x(f(x)+f(-x))=0$ for all $x$, so $f(-x)=-f(x)$ for all $x \neq 0$, and hence for all $x$, since $f(0)=0$;
+(3) $x \leftrightarrow y: \quad(y-x)(f(y)+f(x)) \leqslant f\left(y^{2}-x^{2}\right)=-f\left(x^{2}-y^{2}\right) \Rightarrow(x-y)(f(x)+f(y)) \geqslant f\left(x^{2}-y^{2}\right)$;
+which is the given inequality with the inequality sign reversed, so $(x-y)(f(x)+f(y))=f\left(x^{2}-y^{2}\right)$ must hold for all $x, y \in \mathbb{R}$;
+(4) $y \leftrightarrow-y: \quad(x-y)(f(x)+f(y))=f\left(x^{2}-y^{2}\right)=f\left(x^{2}-(-y)^{2}\right)=(x+y)(f(x)+f(-y))=(x+y)(f(x)-f(y))$; expanding yields $f(x) y=f(y) x$ for all $x, y \in \mathbb{R}$.
+Taking $y=1$ in the last result, $f(x)=f(1) x$, i.e. $f(x)=c x$, where $c=f(1)$, for all $x \in \mathbb{R}$. Since we have shown above that, conversely, all such functions are solutions, this completes the proof.
+
+Alternative solution. A slight variation of this argument proves that $(x-y)(f(x)+f(y))=f\left(x^{2}-y^{2}\right)$ must hold for all $x, y \in \mathbb{R}$ as above, and then reaches $f(x)=c x$ as follows:
+(4) $y= \pm 1: \quad(x \mp 1)(f(x) \pm f(1))=f\left(x^{2}-1\right)$ using $f(-1)=-f(1)$ from (2);
+hence $(x-1)(f(x)+f(1))=(x+1)(f(x)-f(1)) \Rightarrow f(x)=f(1) x=c x$, where $c=f(1)$, on expanding.
+
+# BxMO 2023: Problems and Solutions 
+
+## Problem 2
+
+Determine all integers $k \geqslant 1$ with the following property: given $k$ different colours, if each integer is coloured in one of these $k$ colours, then there must exist integers $a_{1}<a_{2}<\cdots<a_{2023}$ of the same colour such that the differences $a_{2}-a_{1}, a_{3}-a_{2}, \ldots, a_{2023}-a_{2022}$ are all powers of 2 .
+
+## Solution
+
+We claim that only $k=1$ and $k=2$ satisfy the required property. First, if $k \geqslant 3$, we colour each integer with its residue class modulo 3, so that, whenever two integers have the same colour, their difference is divisible by 3 , so is not a power of 2 . This shows that no $k \geqslant 3$ has the required property.
+
+In the case $k=1$, the sequence defined by $a_{n}=2 n$ for $n=1,2, \ldots, 2023$ clearly has the required property. In the case $k=2$, we call the colours "red" and "blue", and construct, for each $n \geqslant 1$ and by induction, integers $a_{1}<a_{2}<\cdots<a_{n}$ of the same colour such that $a_{m+1}-a_{m}$ is a power of 2 for $m=1,2, \ldots, n-1$. The statement is trivial for $n=1$. For $n>1$, let $a_{1}<a_{2}<\cdots<a_{n}$ be red integers (without loss of generality) having the desired property. Consider the $n+1$ integers $b_{i}=a_{n}+2^{i}$, for $i=1,2, \ldots, n+1$. If one of these, say $b_{j}$, is red, then, as $b_{j}-a_{n}=2^{j}$, the $n+1$ red integers $a_{1}<a_{2}<\cdots<a_{n}<b_{j}$ have the desired property. Otherwise, $b_{1}, b_{2}, \ldots, b_{n+1}$ are all blue, and $b_{i+1}-b_{i}=\left(a_{n}+2^{i+1}\right)-\left(a_{n}+2^{i}\right)=2^{i}$ for $i=1,2, \ldots, n$, so the $n+1$ blue integers $b_{1}<b_{2}<\cdots<b_{n+1}$ have the desired property. This completes the inductive step and hence the proof.
+
+## BxMO 2023: Problems and Solutions
+
+## Problem 3
+
+Let $A B C$ be a triangle with incentre $I$ and circumcircle $\omega$. Let $N$ denote the second point of intersection of line $A I$ and $\omega$. The line through $I$ perpendicular to $A I$ intersects line $B C$, segment $[A B]$, and segment $[A C]$ at the points $D, E$, and $F$, respectively. The circumcircle of triangle $A E F$ meets $\omega$ again at $P$, and lines $P N$ and $B C$ intersect at $Q$. Prove that lines $I Q$ and $D N$ intersect on $\omega$.
+
+## Solution 1
+
+By construction, $A P E F$ and $A P B C$ are cyclic, and so
+
+$$
+\begin{aligned}
+\angle B D E & =\angle C D F=\angle A F D-\angle F C D=\angle A F E-\angle A C B=\left(180^{\circ}-\angle E P A\right)-\left(180^{\circ}-\angle B P A\right) \\
+& =\angle B P A-\angle E P A=\angle B P E .
+\end{aligned}
+$$
+
+Hence $D B E P$ is cyclic, too. It follows that $\angle I D P=\angle E D P=\angle E B P=\angle A B P=\angle A N P=\angle I N P$ since $A P B N$ is cyclic, and so $P D N I$ is also cyclic. In particular, $\angle D P N=\angle D I N=90^{\circ}$. Let $R$ denote the second intersection of $D P$ and $\omega$, so $N Q \perp D R$. Then $\angle N P R=90^{\circ}$, so $R N$ is a diameter of $\omega$. It is well-known that $N$ is the midpoint of the arc $\widehat{B C}$ not containing $A$, whence $R N \perp B C$. Thus $D Q$ and $N Q$ are altitudes of triangle $R D N$, and so $Q$ is its orthocentre. This implies that $R Q \perp D N$, whence, since $R N$ is a diameter of $\omega$, the intersection $X$ of $R Q$ and $D N$ lies on $\omega$.
+![](https://cdn.mathpix.com/cropped/2024_12_15_bda133660d2e7e251cddg-4.jpg?height=1040&width=1294&top_left_y=1450&top_left_x=381)
+
+It is also well-known that $N$ is the centre of the circumcircle $\Omega$ of triangle $B C I$. Since $D I \perp I N$ by construction, $D I$ is tangent to $\Omega$ at $I$. As $D$ lies on the radical axis $B C$ of $\omega$ and $\Omega$, it follows that $|D I|^{2}=|D B||D C|=|D X||D N|$. Hence triangles $D N I$ and $D I X$ are similar; in particular, $\angle D X I=\angle D I N=90^{\circ}$. All of this shows that $R, I, Q, X$ lie on a line perpendicular to $D N$ that intersects $D N$ at $X \in \omega$. This completes the proof.
+
+## BxMO 2023: Problems and Solutions
+
+## Solution 2
+
+Let $K$ be the midpoint of segment [BC]. It is well-known that $N$ is the midpoint of the small arc $\widehat{B C}$ of $\omega$, so $B C \perp K N$. In particular, $\angle D K N=90^{\circ}$. But $\angle D I N=90^{\circ}$ by construction, so $D I K N$ is cyclic, with circumcircle $\Gamma$. Moreover, $\angle P E F=180^{\circ}-\angle P A F=180^{\circ}-\angle P A B=\angle P B C$ and $\angle P F E=\angle P A E=\angle P A B=\angle P C B$ since $A F E P$ and $A C B P$ are cyclic, so triangles $P E F$ and $P B C$ are similar. Now, by construction, $I$ is the midpoint of segment $[E F]$, so, $K$ being the midpoint of $[B C]$, triangles $P I F$ and $P K C$ are similar, too. It follows that $\angle P I D=180^{\circ}-\angle P I F=180^{\circ}-\angle P K C=\angle P K D$, whence $P$ lies on $\Gamma$.
+![](https://cdn.mathpix.com/cropped/2024_12_15_bda133660d2e7e251cddg-5.jpg?height=958&width=1280&top_left_y=800&top_left_x=388)
+
+Let $\Omega$ be the circumcircle of triangle $B C I$. By construction, $Q$ lies on the radical axes $P N$ of $\omega, \Gamma$ and $B C$ of $\omega, \Omega$, so is the radical centre of $\omega, \Gamma, \Omega$. In particular, $I Q$ is the radical axis of $\Gamma, \Omega$, so is perpendicular to the line joining the centres of $\Gamma, \Omega$. Now it is well-known that $N$ is the centre of $\Omega$, and, since $\angle D I N=90^{\circ}$, the centre of $\Gamma$ is the midpoint of segment $[D N]$. This shows that $I Q \perp D N$.
+
+Finally, let $D N$ meet $\omega$ again at $X$. Since $D I \perp I N$ by construction and $N$ is the centre of $\Omega, D I$ is tangent to $\Omega$ at $I$. As $D$ lies on the radical axis $B C$ of $\omega, \Omega$, it follows that $|D I|^{2}=|D B||D C|=|D X||D N|$. Hence triangles $D N I$ and $D I X$ are similar; in particular, $\angle D X I=\angle D I N=90^{\circ}$, i.e. $I X \perp D N$. Since $I Q \perp D N$, it follows that $X$ is the intersection of $I Q$ and $D N$. Since $X$ lies on $\omega$ by construction, this completes the proof.
+
+## Solution 3
+
+Since $A P E F$ and $A P B C$ are cyclic,
+
+$$
+\begin{aligned}
+\angle C P F & =\angle B P A-\angle B P C-\angle F P A=\left(180^{\circ}-\angle B C A\right)-\angle B A C-\angle F E A \\
+& =\left(180^{\circ}-\angle B C A-\angle B A C\right)-\angle B E D=\angle C B A-\angle B E D=\angle C B E-\angle B E D=\angle B D E=\angle C D F,
+\end{aligned}
+$$
+
+so $D P F C$ is cyclic, too. Thence $\angle C P D=\angle C F D=180^{\circ}-\angle I F A=90^{\circ}+\angle I A F=90^{\circ}+\angle C A N=90^{\circ}+\angle C P N$. Hence $\angle D P N=\angle C P D-\angle C P N=90^{\circ}$. Since $\angle D I N=90^{\circ}$ by construction, it follows that DPIN is cyclic, with
+
+## BxMO 2023: Problems and Solutions
+
+circumcircle $\Gamma$. Let $J$ be the second intersection of line $I Q$ and $\Gamma$. Moreover, it is well-known that $N$ is the centre of the circumcircle $\Omega$ of BIC. In particular, $|N I|=|N B|$, and so, since $N I P J$ and $N B P C$ are cyclic,
+
+$$
+\frac{|J Q|}{|J P|}=\frac{|N Q|}{|N I|}=\frac{|N Q|}{|N B|}=\frac{|C Q|}{|C P|}
+$$
+
+Let $S$ now be the point of intersection of $P N$ and $\Omega$ such that $P, N, S$ lie on line $P N$ in this order. By construction, $\angle Q P C=\angle N P C=\angle N A C=\angle B A N=\angle B C N=\angle Q C N$, so triangles $C Q N$ and $P C N$ are similar, whence
+
+$$
+\frac{|C Q|}{|C P|}=\frac{|N C|}{|N P|}=\frac{|N Q|}{|N C|}=\frac{|N C|+|N Q|}{|N C|+|N P|}=\frac{|N S|+|N Q|}{|N S|+|N P|}=\frac{|S Q|}{|S P|}
+$$
+
+Combining (1) and (2) shows that $C, J, S$ lie on a circle of Apollonius, the centre of which lies on the line through $P, Q, N, S$, so, since $|N C|=|N S|$ by construction, is $N$. In other words, $J$ lies on $\Omega$.
+![](https://cdn.mathpix.com/cropped/2024_12_15_bda133660d2e7e251cddg-6.jpg?height=1357&width=1317&top_left_y=938&top_left_x=375)
+
+In particular, $|N I|=|N J|$. Now, by construction, $\angle D I N=\angle D J N=90^{\circ}$, so the right-angled triangles $D I N$ and $D J N$ are congruent, whence $D I N J$ is a kite. In particular, $I J \perp D N$. Since $Q$ lies on $I J$ by definition, this shows that $I Q \perp D N$. We can now conclude as in Solution 2.
+
+## Solution 4
+
+By construction, $P$ is the Miquel point of quadrilateral $B C F E$ (and the resulting complete quadrilateral with points $A$ and $D$ added) because it is the intersection of $\omega$ and the circumcircle of triangle $A E F$. In particular, $D B E P$ is
+
+## BxMO 2023: Problems and Solutions
+
+cyclic. It follows that $\angle I D P=\angle E D P=\angle E B P=\angle A B P=\angle A N P=\angle I N P$ since $A P B N$ is cyclic, and so $P D N I$ is also cyclic.
+![](https://cdn.mathpix.com/cropped/2024_12_15_bda133660d2e7e251cddg-7.jpg?height=963&width=1292&top_left_y=472&top_left_x=382)
+
+Next, let $X$ be the intersection of $D N$ and $\omega$ and let $A N$ meet $B C$ at $Y$. Then $\angle N A C=\angle A / 2=\angle N C B$, so $\angle B Y A=\angle C+\angle N A C=\angle C+\angle N C B=\angle N C A$ and hence
+
+$$
+\begin{aligned}
+\angle D Q P & =\angle N Q Y=\angle Q Y A-\angle Q N Y=\angle B Y A-\angle P N A \\
+& =\angle N C A-\angle P C A=\angle P C N=180^{\circ}-\angle N X P=\angle D X P
+\end{aligned}
+$$
+
+This implies that $D X Q P$ is cyclic. In particular, $Q X \perp D N$. It now suffices to show that $I X \perp D N$, which we do in the same way as in Solution 2.
+
+# BxMO 2023: Problems and Solutions 
+
+## Problem 4
+
+A positive integer $n$ is friendly if every pair of neighbouring digits of $n$, written in base 10, differs by exactly 1. For example, 6787 is friendly, but 211 and 901 are not.
+
+Find all odd natural numbers $m$ for which there exists a friendly integer divisible by $64 m$.
+
+## Solution
+
+Any friendly number divisible by 64 is divisible by 4 , and hence the number formed by its last two digits is a multiple of 4 , so ends in $00,04,08, \ldots$, or 96 . A friendly number divisible by 4 must therefore end in $12,32,56$, or 76 , so cannot be divisible by 5 . In particular, if $5 \mid \mathrm{m}$, then there is no friendly integer divisible by 64 m .
+
+We claim that conversely, if $m$ is odd and $5 \nmid m$, then there exists a friendly integer divisible by $64 m$. First, we notice that $343232=64 \cdot 5363$ is a friendly number divisible by 64 , and hence so is
+
+$$
+N_{k}=343232343232 \cdots 343232=343232 \cdot\left(1+10^{6}+\cdots+10^{6 k}\right) \quad \text { for } k=0,1,2, \ldots
+$$
+
+Now the sequence $N_{0}, N_{1}, N_{2}, \ldots$ eventually repeats modulo $m$, i.e. there exist positive integers $k<\ell$ such that $N_{\ell} \equiv N_{k}(\bmod m)$. Hence $m \mid N_{\ell}-N_{k}=10^{6(k+1)} N_{\ell-k-1}$. Since $m$ is odd and $5 \nmid m,(10, m)=1$, so $m \mid N_{\ell-k-1}$. By construction, $64 \mid N_{\ell-k-1}$. Thus, as $m$ is odd and hence $(64, m)=1$, we conclude that $64 m \mid N_{\ell-k-1}$. This completes the proof.
+
+The solution divides into two parts: (1) showing that, if $5 \mid m$, then there is no friendly integer divisible by $64 m$; (2) showing that, if $5 \nmid m$, then there is a friendly integer divisible by $64 m$.
+
+Alternative solution for part (1). If $5 \mid m$, then $20 \mid 64 m$. The last two digits of a multiple of 20 are $00,20,40,60$, or 80 , so this number is not friendly. Thus, if $m$ is odd and $5 \mid m$, then there is no friendly integer divisible by $64 m$.
+
+Alternative solution for part (2). Notice that $N_{k}=343232 \cdot\left(10^{6(k+1)}-1\right) /\left(10^{6}-1\right)$. Let $M=m\left(10^{6}-1\right)$. Since $5 \nmid m$ and $m$ is odd, $(10, M)=1$, so, taking $k=\varphi(M)-1$, we get $10^{6(k+1)}=10^{6 \varphi(M)} \equiv 1(\bmod M)$ by the Euler-Fermat theorem, i.e. $m \mid\left(10^{6(k+1)}-1\right) /\left(10^{6}-1\right)$, and hence $m \mid N_{k}$.
+
+Alternative constructions of the integers $\boldsymbol{N}_{\boldsymbol{k}}$ for part (2). Direct calculation shows that friendly integers divisible by 64 end in $343232,543232,123456$, or 323456 , so the numbers $N_{k}$ defined in the solution of part (2) above may be replaced by, for instance,
+
+$$
+34543232 \cdot\left(1+10^{8}+\cdots+10^{8 k}\right), 5432123456 \cdot\left(1+10^{10}+\cdots+10^{10 k}\right), 54323456 \cdot\left(1+10^{8}+\cdots+10^{8 k}\right)
+$$
+
+Remark. Interestingly, friendly numbers cannot be divisible by arbitrarily high powers of 2. Direct calculation shows that the 60-digit friendly integer 101232121234323456543434343210121212323434343234565656543232 is divisible by $2^{60}$, but that there is no friendly integer divisible by $2^{61}$. The problem selection committee is not aware of a proof of this fact that eschews direct calculation.
+

Benelux/md/Benelux_en-olympiad_en-bxmo-problems-2024-zz.md

@@ -0,0 +1,291 @@
+# 16th Benelux Mathematical Olympiad Valkenswaard, 26th - 28th April 2024 
+
+## Problems and Solutions
+
+![](https://cdn.mathpix.com/cropped/2024_12_15_e2e131dd99265ed8ccbbg-01.jpg?height=547&width=709&top_left_y=1274&top_left_x=679)
+
+## Problem Selection Committee
+
+Ward van der Schoot (Chair)
+Tijs Buggenhout, Justin Vast (Belgium)
+Kevin van Dijk, Wietze Koops (the Netherlands)
+Bernard Felten, Pierre Haas (Luxembourg)
+
+## BxMO 2024: Problems and Solutions
+
+## Problem 1
+
+(a) Let $a_{0}, a_{1}, \ldots, a_{2024}$ be real numbers such that $\left|a_{i+1}-a_{i}\right| \leqslant 1$ for $i=0,1, \ldots, 2023$.
+
+Find the minimum possible value of
+
+$$
+a_{0} a_{1}+a_{1} a_{2}+\cdots+a_{2023} a_{2024}
+$$
+
+(b) Does there exist a real number $C$ such that
+
+$$
+a_{0} a_{1}-a_{1} a_{2}+a_{2} a_{3}-a_{3} a_{4}+\cdots+a_{2022} a_{2023}-a_{2023} a_{2024} \geqslant C
+$$
+
+for all real numbers $a_{0}, a_{1}, \ldots, a_{2024}$ such that $\left|a_{i+1}-a_{i}\right| \leqslant 1$ for $i=0,1, \ldots, 2023$ ?
+
+## Solution 1
+
+(a) The minimum value is -506 . Note that from $\left|a_{i}-a_{i-1}\right| \leq 1$ it follows that
+
+$$
+a_{i} a_{i-1}=\frac{\left(a_{i}+a_{i-1}\right)^{2}-\left(a_{i}-a_{i-1}\right)^{2}}{4} \geq-\frac{\left(a_{i}-a_{i-1}\right)^{2}}{4} \geq-\frac{1}{4}
+$$
+
+Adding this for $i=1,2, \ldots, 2024$, we obtain that
+
+$$
+a_{0} a_{1}+a_{1} a_{2}+a_{2} a_{3}+\ldots+a_{2023} a_{2024} \geq 2024 \cdot-\frac{1}{4}=-506
+$$
+
+We now show that this value can be attained. Indeed, for the sequence $\left(a_{0}, a_{1}, \ldots, a_{2024}\right)=$ $\left(\frac{1}{2},-\frac{1}{2}, \frac{1}{2},-\frac{1}{2}, \frac{1}{2}, \ldots, \frac{1}{2}\right)$ with alternating $\frac{1}{2}$ 's and $-\frac{1}{2}$ 's, each term $a_{i} a_{i-1}$ is equal to $-\frac{1}{4}$, leading to $a_{0} a_{1}+a_{1} a_{2}+a_{2} a_{3}+\ldots+a_{2023} a_{2024}=2024 \cdot-\frac{1}{4}=-506$.
+(b) No, such a $C$ does not exist. We argue by contradiction. Suppose $C$ has this property, and consider the sequence defined by $a_{0}=C$ and $a_{i}=C-1$ for $i=1,2, \ldots, 2024$ satisfies the condition in the problem. For this sequence, we have $a_{i} a_{i+1}-a_{i+1} a_{i+2}=0$ for $i=2,4, \ldots$, 2022, so the sum
+
+$$
+a_{0} a_{1}-a_{1} a_{2}+a_{2} a_{3}-a_{3} a_{4}+a_{4} a_{5}-a_{5} a_{6}+\ldots+a_{2022} a_{2023}-a_{2023} a_{2024}
+$$
+
+is equal to
+
+$$
+a_{0} a_{1}-a_{1} a_{2}=C(C-1)-(C-1)^{2}=C-1<C,
+$$
+
+contradiction.
+
+## Solution 2
+
+We give an alternative construction for part (b). We choose a real constant $N$, from which we define $a_{i}=N+i$ for each $i=0,1, \ldots, 2024$, which clearly satisfies the requirement $\left|a_{i}-a_{i-1}\right| \leq 1$ for each $i=0,1, \ldots, 1011$. Then, it can be seen that for each $i=0,1, \ldots, 1011$ that
+
+$$
+a_{2 i} a_{2 i+1}-a_{2 i+1} a_{2 i+2}=a_{2 i+1}\left(a_{2 i}-a_{2 i+2}\right)=-2(N+2 i+1) \leqslant-2 N .
+$$
+
+From this, it can be concluded that
+$a_{0} a_{1}-a_{1} a_{2}+a_{2} a_{3}-a_{3} a_{4}+a_{4} a_{5}-a_{5} a_{6}+\ldots+a_{2022} a_{2023}-a_{2023} a_{2024} \leqslant 1012 \cdot-2 N=-2024 N$.
+As $N$ is a constant which can be arbitrarily chosen, there is no constant $C$ which lower bounds the given expression.
+
+## BxMO 2024: Problems and Solutions
+
+## Problem 2
+
+Let $n$ be a positive integer. In a coordinate grid, a path from $(0,0)$ to $(2 n, 2 n)$ consists of $4 n$ consecutive unit steps $(1,0)$ or $(0,1)$. Prove that the number of paths that divide the square with vertices $(0,0)$, $(2 n, 0),(2 n, 2 n),(0,2 n)$ into two regions with even areas is
+
+$$
+\frac{\binom{4 n}{2 n}+\binom{2 n}{n}}{2}
+$$
+
+## Solution 1
+
+Let $X$ denote the set of paths for which $A$ and $B$ have even area and let $Y$ denote the set of paths for which $A$ and $B$ both have odd area. Because $A$ and $B$ together form a square of area $4 n^{2}$, which is even, $|X|+|Y|$ equals the total number of paths from $(0,0)$ to $(2 n, 2 n)$, which is $\binom{4 n}{2 n}$.
+
+Denoting a step to the right by $R$ and a step upwards by $U$, every path from $(0,0)$ to $(2 n, 2 n)$ can be described as a sequence of $4 n$ symbols, $2 n$ of which are $R$ and $2 n$ of which are $U$. We subdivide such a sequence into $2 n$ pairs of consecutive steps that can be $R R, U R, R U$ or $U U$. The number of possible paths for which neither $U R$ nor $R U$ occurs is $\binom{2 n}{n}$, because out of $2 n$ pairs that can be either $R R$ or $U U$ we have to choose $n$ that will be $R R$. These $\binom{2 n}{n}$ all belong to $X$; in fact, for these paths, $A$ and $B$ can be subdivided into $2 \times 2$-square, making their areas divisible by 4 . Now consider the paths that contain at least one $U R$ - or $R U$-pair. If in such a path we replace the first occurrence of a $U R$ or $R U$-pair by a pair of the other type (thus replacing $U R$ by $R U$ or vice versa), the areas of $A$ and $B$ each change by 1 and therefore become even if they were odd and odd if they were even. Because this modification is clearly reversible, we conclude that we can pair up all paths that contain at least one $U R$ - or $R U$-pair into pairs of paths, one of which belongs to $X$ and one of belongs to $Y$. This implies that $|X|-\binom{2 n}{n}=|Y|$. It follows that
+
+$$
+|X|=\frac{|X|+|Y|}{2}+\frac{|X|-|Y|}{2}=\frac{\binom{4 n}{2 n}}{2}+\frac{\binom{2 n}{n}}{2} .
+$$
+
+## Solution 2
+
+Define $Z_{m, n}$ to be the number of routes from $(0,0)$ to $(2 m, 2 n)$ that divide the rectangle with vertices $(0,0),(0,2 n),(2 m, 2 n)$ and $(2 m, 0)$ into two parts of even area. We call such paths good. We claim that
+
+$$
+2 Z_{m, n}=\binom{2 m+2 n}{2 m}+\binom{m+n}{m}
+$$
+
+for all $m, n$, which for $m=n$ establishes the desired result. We prove this formula by induction on $m+n$, noting first that it clearly holds when either $m$ or $n$ is zero, because $Z_{0, n}=Z_{m, 0}=1$ (there is only one path from $(0,0)$ to $(0,2 n)$ or $(2 m, 0)$, which is good). Therefore, suppose that $m, n \geq 1$ and consider a good path from $(0,0)$ to $(2 m, 2 n)$. This path passes through exactly one of $(2 m, 2 n-2)$,
+
+## BxMO 2024: Problems and Solutions
+
+$(2 m-1,2 n-1)$ and $(2 m-2,2 n)$. If it passes through $(2 m, 2 n-2)$ then the subpath from $(0,0)$ to $(2 m, 2 n-2)$ must also be good; moreover, for each good path from $(0,0)$ to $(2 m, 2 n-2)$ there is exactly one corresponding path from $(0,0)$ to $(2 m, 2 n)$, and that path is automatically good because the new area that gets added is even. Thus, there are $Z_{m, n-1}$ good paths from $(0,0)$ to $(2 m, 2 n)$ that pass through $(2 m, 2 n-2)$. Similarly, there are $Z_{m-1, n}$ good paths from $(0,0)$ to $(2 m, 2 n)$ that pass through $(2 m-2,2 n)$. Now notice that any path from $(0,0)$ to $(2 m-1,2 n-1)$ (of which there are $\binom{2 m+2 n-2}{2 m-1}$ ) can be extended in two ways to obtain a path from $(0,0)$ to $(2 m, 2 n)$; because the resulting areas for these paths differ by 1 , exactly one of these paths is good. All in all, we obtain the recursion
+
+$$
+Z_{m, n}=Z_{m, n-1}+Z_{m-1, n}+\binom{2 m+2 n-2}{2 m-1}
+$$
+
+By the inductive hypothesis, we have
+
+$$
+2 Z_{m, n-1}=\binom{2 m+2 n-2}{2 m}+\binom{m+n-1}{m}
+$$
+
+and
+
+$$
+2 Z_{m-1, n}=\binom{2 m+2 n-2}{2 m-2}+\binom{m+n-1}{m-1}
+$$
+
+Therefore, we obtain
+
+$$
+\begin{aligned}
+2 Z_{m, n}= & 2 Z_{m, n-1}+2 Z_{m-1, n}+2\binom{2 m+2 n-2}{2 m-1} \\
+= & \binom{2 m+2 n-2}{2 m}+\binom{m+n-1}{m}+\binom{2 m+2 n-2}{2 m-2}+\binom{m+n-1}{m-1} \\
+& +2\binom{2 m+2 n-2}{2 m-1} .
+\end{aligned}
+$$
+
+To simplify the expression on the right hand side, note that we can reorganize the terms
+
+$$
+\binom{2 m+2 n-2}{2 m}+\binom{2 m+2 n-2}{2 m-2}+2\binom{2 m+2 n-2}{2 m-1}
+$$
+
+as
+
+$$
+\left(\binom{2 m+2 n-2}{2 m}+\binom{2 m+2 n-2}{2 m-1}\right)+\left(\binom{2 m+2 n-2}{2 m-1}+\binom{2 m+2 n-2}{2 m-2}\right)
+$$
+
+which, using the addition rules for binomial coefficients, becomes
+
+$$
+\binom{2 m+2 n-1}{2 m}+\binom{2 m+2 n-1}{2 m-1}=\binom{2 m+2 n}{2 m}
+$$
+
+Similarly, we have
+
+$$
+\binom{m+n-1}{m}+\binom{m+n-1}{m-1}=\binom{m+n}{m}
+$$
+
+Putting everything together, we obtain that
+
+$$
+2 Z_{m, n}=\binom{2 m+2 n}{2 m}+\binom{m+n}{m}
+$$
+
+which completes the induction.
+
+## Remark
+
+The result from Solution 2 can also be proved using a combinatorial argument like in Solution 1.
+
+## BxMO 2024: Problems and Solutions
+
+## Solution 3
+
+We start by proving the following lemma: for a
+
+$$
+(2 m-1,2 n-1)
+$$
+
+-grid, there are equally many paths with a region of even area (called even paths), as there are with odd area (odd paths).
+
+To prove this, we take a path and rotate it around the center of the grid. Then a path spanning a region with area $x$ is mapped on one spanning an area $(2 m-1)(2 n-1)-x$. This gives a bijection between paths creating an even region, and creating an odd region.
+
+Now, for every path from
+to $(2 n, 2 n)$, consider all the coordinates of the grid points it visits in order. There are $\binom{2 n}{n}$ of them which never visit a point with odd coordinates (which we call an odd point). Notice that such paths are all even.
+
+We now construction a bijection between the remaining even paths and the odd paths. For each odd point, there are equally many even as odd paths from $(0,0)$ to that point. Define then a bijection $\phi$ between the sets of odd paths and even paths up to this point for each point. Notice that $\phi$ implicitly depends on the chosen odd point.
+
+Now, for an arbitrary odd path $P$, consider the first odd point it passes through. Map $P$ to another path by changing the path up to this odd point to the $\phi$ of the path up to this point. As $\phi$ maps between even and odd paths, the resulting path is an even path. By the definition of $\phi$, and as each of the remaining paths goes to an odd point, this mapping defines a bijection.
+
+We have hence found a bijection between the odd and even paths in the remaining $\binom{4 n}{2 n}-\binom{2 n}{n}$ paths, which yields the required result like in solution 1.
+
+## BxMO 2024: Problems and Solutions
+
+## Problem 3
+
+Let $A B C$ be a triangle with incentre $I$ and circumcircle $\Omega$ such that $|A C| \neq|B C|$. The internal angle bisector of $\angle C A B$ intersects side [BC] in $D$, and the external angle bisectors of $\angle A B C$ and $\angle B C A$ intersect $\Omega$ again in $E$ and $F$, respectively. Let $G$ be the intersection of lines $A E$ and $F I$ and let $\Gamma$ be the circumcircle of triangle $B D I$. Show that $E$ lies on $\Gamma$ if and only if $G$ lies on $\Gamma$.
+![](https://cdn.mathpix.com/cropped/2024_12_15_e2e131dd99265ed8ccbbg-06.jpg?height=1140&width=1685&top_left_y=612&top_left_x=194)
+
+Solution 1
+
+We first notice the general fact that $E F \perp A I$. This can be proved using the following argument. Denote $S$ for the intersection of $E F$ and $A I$. Then $\angle B I S=(\angle I B A+\angle I A B)=\frac{1}{2}(\angle A B C+\angle B A C)=$ $\frac{1}{2}\left(180^{\circ}-\angle B C A\right)=\angle B C F=\angle B E F=\angle B E S$. Thus $B, I, S, E$ form a cyclic quadrilateral, from which it can be concluded that $E F \perp A I$.
+
+We will now continue to prove the problem by doing both directions separately. Assume first that $E$ lies on $\Gamma$. Then, as angle bisectors are perpendicular to one another, we have that $\angle I D E=\angle I B E=90^{\circ}$. Then, as $E F \perp A I$, it holds that $D$ lies on $E F$. It can then be concluded that $\angle I D F=90^{\circ}=\angle I C F$ (again due to perpendicular bisectors), from which it can be concluded that $I, D, C, F$ form a cyclic quadrilateral. We can now calculate that $\angle G I D=180^{\circ}-\angle F I D=180^{\circ}-\angle F C D=180^{\circ}-\angle F C B=$ $\angle A E F=\angle G E D$, where the second to last step follows from the fact that the arcs $B F$ and $A F$ have the
+
+## BxMO 2024: Problems and Solutions
+
+same length, as $F$ lies on the external bisector of $\angle A C B$. It now follows that $I, D, E, G$ form a cyclic quadrilteral, thus $G$ lies on $\Gamma$.
+
+For the reverse implication, assume that $G$ lies on $\Gamma$. We can then compute that $\angle F A D=\angle F A S=$ $90^{\circ}-\angle A F S=90^{\circ}-\angle A F E=90^{\circ}-\angle A B E=\angle A B I=\angle I B D=\angle I G D=\angle F G D$, from which we can conclude that $A, G, D, F$ form a cyclic quadrilateral. Similarly to $\angle F A D=\angle I B D$, it holds that $\angle G A D=\angle I C D$. From this we can compute that $\angle I F D=\angle G F D=\angle G A D=\angle I C D$, thus $I, D, C, F$ forms a cyclic quadrilateral. Hence, $\angle I D F=180^{\circ}-\angle I C F=90^{\circ}$, so again $D=S$. We then see that $\angle I B E=90^{\circ}=\angle I D E$ from which we can conclude that $I, D, E, B$ form a cyclic quadrilateral. It then follows that $E$ lies on $\Gamma$.
+
+## Solution 2
+
+The external angle bisectors $B E$ and $C F$ meet the internal bisector $I D$ at the $A$-excentre $J$ of triangle $A B C$. If one of BEDI and CFID is cyclic, then, as BECF is cyclic, $|J D||J I|=|J E||J B|=|J C||J F|$ by power of a point, and so the other is cyclic, too. This shows that
+(1) $B E D I$ is cyclic if and only if CFID is cyclic.
+
+Next, $\angle G A D=\angle E A D=\angle B A D-\angle B A E=\angle A / 2-\left(180^{\circ}-\angle A E B-\angle E B A\right)$, with $\angle A E B=\angle A C B=$ $\angle C$ and $\angle E B A=\angle E B I+\angle I B A=90^{\circ}+\angle B / 2$. Hence $\angle G A D=\angle A / 2+\angle B / 2+\angle C-90^{\circ}=\angle C / 2=$ $\angle I C D$. Thus, if one of $C F I D$ and $A G D F$ is cyclic, then $\angle I C D=\angle I F D=\angle G F D=\angle G A D$, and the other is cyclic, too. We have thus established that
+(2) CFID is cyclic if and only if $A G D F$ is cylic.
+
+Finally, let $A^{\prime}$ denote the second intersection of $A I$ with $\Omega$. Then $\angle D A F=\angle A^{\prime} A F=180^{\circ}-\angle F C A^{\prime}$, with $\angle F C A^{\prime}=\angle F C I+\angle I C B+\angle B C A^{\prime}=90^{\circ}+\angle C / 2+\angle B A A^{\prime}=90^{\circ}+\angle C / 2+\angle A / 2$. It follows that $\angle D A F=90^{\circ}-\angle A / 2-\angle C / 2=\angle B / 2=\angle D B I$. Hence, if one of $A G D F$ and $B D I G$ is cyclic, then $\angle D A F=\angle D G F=\angle D G I=\angle D B I$, and so the other is cyclic, too. Hence
+(3) $A G D F$ is cyclic if and only if $B D I G$ is cylic.
+
+These three equivalences prove that $B E D I$ is cyclic if and only if $B D I G$ is cyclic, i.e. $E$ lies on $\omega$ if and only if $G$ does. This completes the proof.
+
+## BxMO 2024: Problems and Solutions
+
+![](https://cdn.mathpix.com/cropped/2024_12_15_e2e131dd99265ed8ccbbg-08.jpg?height=1351&width=1089&top_left_y=221&top_left_x=489)
+
+## Solution 3
+
+This proof only shows $E \in \omega \Longrightarrow G \in \omega$. Note that this argument cannot be used straightforwardly to prove the converse implication.
+
+If $B E D I$ is cyclic, let $A^{\prime}$ be the the second intersection of $A I$ with $\Omega$, so $\angle I A^{\prime} E=\angle A A^{\prime} E=$ $180^{\circ}-\angle E B A$, with $\angle E B A=\angle E B I+\angle I B A=90^{\circ}+\angle B / 2$, so $\angle I A^{\prime} E=90^{\circ}-\angle B / 2$. But $\angle E I A^{\prime}=\angle E B D$ since $E B D I$ is cyclic, with $\angle E B D=\angle E B I-\angle D B I=90^{\circ}-\angle B / 2$. Thus $\angle E A^{\prime} I=\angle E I A^{\prime}$, so $E A^{\prime} I$ is isosceles. Moreover, since $E B D I$ is cyclic and $\angle E B I$ is a right angle, so is $\angle I D E$. It follows that $D$ is the midpoint of $[A I]$. Next, the external bisectors $B E$ and $C F$ and the internal bisector $I D$ meet at the $A$-excentre $J$ of triangle $A B C$. By power of a point, since $B E D I$ and $B E C F$ are cyclic, $|J D||J I|=|J E||J B|=|J C||J F|$, so $C F I D$ is cyclic, too, with $\angle F D I=\angle F C I=90^{\circ}$. We have thus shown that $\angle A D E=\angle F D A=90^{\circ}$, so $D$ is the foot of the altitude from $A$ in triangle $A E F$. Moreover, all of this shows that $I$ is the reflection of $A^{\prime}$, which is the point at which this altitude meets the circumcircle $\Omega$ of $A E F$ again, in the side $[E F]$. Hence $I$ is the orthocentre of triangle $A E F$. By extension, $F I$ is its altitude from $F$, and $G$ is the foot of this altitude, so $\angle E G I=90^{\circ}=\angle I D E=\angle E B I$, and hence $B E D I G$ is cyclic. This shows that if $E$ lies on
+
+## BxMO 2024: Problems and Solutions
+
+$\omega$, then so does $G$.
+
+## Remark
+
+The equivalence $E \in \omega \Longleftrightarrow G \in \omega$ breaks down if triangle $A B C$ is not scalene. Indeed, if $A B C$ is isosceles with $\angle A=\angle B$, then $F=C$ and $G$ is the intersection of $C I$ and $A E$. Angle chasing as above then shows that $\angle G A D=\angle C / 2=\angle G C D$, so $A G D C$ is cyclic. Then $\angle D G I=\angle D G C=\angle D A C=$ $\angle A / 2=\angle B / 2=\angle D B I$, so $B D I G$ is cyclic, i.e. $G \in \omega$. Next, the $A$-excentre $J$ of triangle $A B C$ is the intersection of $B E, I D$, and the tangent to $\Omega$ at $C$, which, being the external bisector of $\angle C$, is perpendicular to $C I$. Hence, if $E \in \omega$, too, i.e. if $B E D I$ were cyclic, then by power of a point, $|J C|^{2}=|J E||J B|=|J D||J I|$, so the circumcircle of $C I D$ would be tangent to $J C$ at $C$, implying $\angle C D I=\angle I C J=90^{\circ}$. This is not however the case, unless $\angle B=\angle C$ and hence $A B C$ is equilateral. This shows that the condition in the problem statement that $A B C$ is scalene cannot be dropped.
+![](https://cdn.mathpix.com/cropped/2024_12_15_e2e131dd99265ed8ccbbg-09.jpg?height=700&width=809&top_left_y=1089&top_left_x=629)
+
+## BxMO 2024: Problems and Solutions
+
+## Problem 4
+
+For each positive integer $n$, let $\operatorname{rad}(n)$ denote the product of the distinct prime factors of $n$. Show that there exist integers $a, b>1$ such that $\operatorname{gcd}(a, b)=1$ and
+
+$$
+\operatorname{rad}(a b(a+b))<\frac{a+b}{2024^{2024}}
+$$
+
+For example, $\operatorname{rad}(20)=\operatorname{rad}\left(2^{2} \cdot 5\right)=2 \cdot 5=10$ and $\operatorname{rad}(18)=\operatorname{rad}\left(2 \cdot 3^{2}\right)=2 \cdot 3=6$.
+
+## Solution 1
+
+We show that the pair $(a, b)$ of the form $a=2^{p(p-1)}, b=3^{p(p-1)}-2^{p(p-1)}$ for sufficiently larger prime number $p$ satisfies the inequality. First, notice that $\operatorname{gcd}(a, b)=\operatorname{gcd}(a, a+b)=1$ indeed. In addition, see that $\operatorname{rad}(a)=2$, and $\operatorname{rad}(a+b)=3$. Because of Euler-Fermat, as $\phi\left(p^{2}\right)=p(p-1)$, it can directly be seen that $p^{2} \mid b$. In this case, $\operatorname{rad}(b) \leqslant \frac{b}{p}$. It then follows that, as rad is multiplicative for coprime numbers, that
+
+$$
+\operatorname{rad}(a b(a+b))=\operatorname{rad}(a) \operatorname{rad}(b) \operatorname{rad}(a+b) \leqslant 2 \cdot 3 \cdot \frac{b}{p} \leqslant \frac{6}{p}(a+b)
+$$
+
+Then, by choosing $p$ such that $\frac{6}{p}<\frac{1}{2024^{2024}}$, we found $a$ and $b$ satifying the inequality.
+
+## Solution 2
+
+We show that the pair $(a, b)$ of the form $a=3^{2^{k}}, b=5^{2^{k}}-3^{2^{k}}$ for sufficiently large $k$ satisifies the inequality. Again, we have that $\operatorname{gcd}(a, b)=\operatorname{gcd}(a, a+b)=1$. Similarly to solution $1, \operatorname{rad}(a(a+b))=$ $\operatorname{rad}(a) \operatorname{rad}(a+b)=3 \cdot 5=15$. Then, we will show that $2^{k+1} \mid b$, from which it would follow that $\operatorname{rad}(b) \leqslant \frac{b}{2^{k}}$. From this, we then see that
+
+$$
+\operatorname{rad}(a b(a+b))=\operatorname{rad}(a(a+b)) \operatorname{rad}(b) \leqslant \frac{15 b}{2^{k}} .
+$$
+
+Like in solution 1, this gives a pair $(a, b)$ satisfying the inequality for sufficiently large $k$.
+There are various ways to show that $2^{k+1} \mid 5^{2^{k}}-3^{2^{k}}$, for example directly by applying the Lifting-The-Exponent Lemma. For a more elementary proof, we can apply induction on $k$. The statement is clearly true for $k=0$, and if the statement holds for $k=n$, then for $k=n+1$ we see that
+
+$$
+5^{2^{k}}-3^{2^{k}}=5^{2^{n+1}}-3^{2^{n+1}}=\left(5^{2^{n}}\right)^{2}-\left(3^{2^{n}}\right)^{2}=\left(5^{2^{n}}-3^{2^{n}}\right)\left(5^{2^{n}}+3^{2^{n}}\right) .
+$$
+
+From the induction hypothesis, the first factor has $n+1$ factors of 2 . As the second factor is a sum of two odd numbers, the second term has at least one factor of 2 . The product thus has at least $n+2=k+1$ factors of 2 , from which the statement follows by induction.
+
+## Solution 3
+
+Choose $a=\left(4^{x}-1\right)^{2}, b=4^{x+1}$ and $a+b=\left(4^{x}+1\right)^{2}$. That is, $a, b$ and $a+b$ are squares, where $b$ only contains the factor 2 . Note that $a$ and $b$ are indeed coprime. Then $\operatorname{rad}(a b c)=2 \operatorname{rad}\left(16^{x}-1\right)$.
+
+Choose then $x=5^{k}$ such that $x>2 \cdot 2024^{2024}$. By Lifting the exponent, we know that $5^{k+1} \mid 16^{5^{k}}-1$. This implies that $2 \operatorname{rad}\left(16^{x}-1\right) \leqslant 2\left(16^{x}-1\right) / 5^{k}<\left(4^{x}+1\right)^{2} / 2024^{2024}$.
+
+This solution also works with Euler-Fermat, by choosing $x=\phi\left(5^{k+1}\right)$ with $5^{k}>2024^{2024}$.
+

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Benelux/segment_script/segment_2024.py

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+import re
+from pprint import pprint
+import os
+import json
+def process_markdown(file_path):
+    with open(file_path, 'r', encoding='utf-8') as file:
+        lines = file.readlines()
+    
+    content_list = []
+    current_group = []
+    skip = False
+
+    # Define the headers to skip
+    skip_headers = [
+        "## Problems and Solutions",
+        "## Problem Selection Committee",
+        "## BxMO 20xx: Problems and Solutions"
+    ]
+
+    # Process each line
+    for line in lines:
+        stripped_line = line.strip()
+
+        # Skip single # headers
+        if stripped_line.startswith('# ') and not stripped_line.startswith('## '):
+            continue
+
+        # Skip the content under ## Problem Selection Committee
+        if (stripped_line == "## Problem Selection Committee") or (stripped_line == "## BxMO 2024: Problems and Solutions"):
+            skip = True
+            continue
+        if skip and stripped_line.startswith('## ') and stripped_line != "## Problem Selection Committee" and stripped_line != "## BxMO 2024: Problems and Solutions":
+            skip = False
+
+        if skip:
+            continue
+
+        # Skip other headers explicitly listed
+        if stripped_line in skip_headers:
+            continue
+
+        # Start a new group when a "## Problem X" pattern is matched
+        if re.match(r"## Problem \d+", stripped_line):
+            if current_group:
+                content_list.append(current_group)
+            current_group = [stripped_line]
+        else:
+            if current_group is not None:
+                current_group.append(stripped_line)
+
+    # Add the last group if it exists
+    if current_group:
+        content_list.append(current_group)
+
+    return content_list
+
+
+def process_group(group):
+    """
+    Process a single group of lines into a dictionary with problem type, label, problem, and solutions.
+    """
+    group_dict = {
+        "problem_type": None,  # Placeholder for problem type if needed later
+        "problem_label": None,
+        "problem": "",
+        "solution": []
+    }
+    
+    problem_pattern = re.compile(r"## Problem (\d+)")
+    solution_pattern = re.compile(r"## Solution (\d+)")
+    
+    current_solution = []
+    in_problem_section = False
+    in_solution_section = False
+    
+    for line in group:
+        stripped_line = line.strip()
+        
+        # Match problem label
+        problem_match = problem_pattern.match(stripped_line)
+        if problem_match:
+            group_dict["problem_label"] = f"Problem {problem_match.group(1)}"
+            in_problem_section = True
+            in_solution_section = False
+            continue
+        
+        # Match solution headers
+        solution_match = solution_pattern.match(stripped_line)
+        if solution_match:
+            if current_solution:
+                group_dict["solution"].append("\n".join(current_solution).strip())
+            current_solution = []
+            in_problem_section = False
+            in_solution_section = True
+            continue
+        
+        # Populate the problem text
+        if in_problem_section:
+            if stripped_line:
+                group_dict["problem"] += stripped_line + " "
+        
+        # Populate the current solution text
+        if in_solution_section:
+            current_solution.append(stripped_line)
+    
+    # Append the last solution if exists
+    if current_solution:
+        group_dict["solution"].append("\n".join(current_solution).strip())
+    
+    # Remove trailing spaces from problem text
+    group_dict["problem"] = group_dict["problem"].strip()
+    
+    return group_dict
+
+
+
+# Example usage
+def process_final(processed_groups):
+    entries = []
+    for p in processed_groups:
+        for s in p['solution']:
+            entry = {}
+            entry['problem'] = p['problem']
+            entry['problem_label'] = p['problem_label']
+            entry['problem_type'] = p['problem_type']
+            entry['solution'] = s
+            entries.append(entry.copy())
+    return entries
+
+def save_to_jsonl(processed_content, original_path):
+    """
+    Save processed content into a JSONL file in a directory two levels up.
+    """
+    # Go up two directories
+    base_dir = os.path.abspath(os.path.join(original_path, "../../"))
+    jsonl_dir = os.path.join(base_dir, "segmented")
+
+    # Create the jsonl directory if it doesn't exist
+    os.makedirs(jsonl_dir, exist_ok=True)
+
+    # Determine the output file name
+    file_name = os.path.basename(original_path).replace('.md', '.jsonl')
+    jsonl_path = os.path.join(jsonl_dir, file_name)
+
+    # Write the processed content to the JSONL file
+    with open(jsonl_path, 'w', encoding='utf-8') as jsonl_file:
+        for group in processed_content:
+            json.dump(group, jsonl_file, ensure_ascii=False)
+            jsonl_file.write('\n')
+
+    print(f"JSONL file saved at: {jsonl_path}")
+    
+markdown_path = r"D:\NeuralTheoremProving\Numina\olympiads-ref\Benelux\md\Benelux_en-olympiad_en-bxmo-problems-2024-zz.md"
+processed_content = process_markdown(markdown_path)
+
+# Convert each group to a dictionary and save to JSONL
+processed_groups = [process_group(group) for group in processed_content]
+output = process_final(processed_groups)
+save_to_jsonl(output, markdown_path)

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl

@@ -0,0 +1,9 @@
+{"problem": ". A finite set of integers is called bad if its elements add up to 2010. A finite set of integers is a Benelux-set if none of its subsets is bad. Determine the smallest integer $n$ such that the set $\\{502,503,504, \\ldots, 2009\\}$ can be partitioned into $n$ Benelux-sets.\n\n(A partition of a set $S$ into $n$ subsets is a collection of $n$ pairwise disjoint subsets of $S$, the union of which equals $S$.)", "solution": "As $502+1508=2010$, the set $S=\\{502,503, \\ldots, 2009\\}$ is not a Benelux-set, so $n=1$ does not work. We will prove that $n=2$ does work, i.e. that $S$ can be partitioned into 2 Benelux-sets.\n\nDefine the following subsets of $S$ :\n\n$$\n\\begin{aligned}\nA & =\\{502,503, \\ldots, 670\\}, \\\\\nB & =\\{671,672, \\ldots, 1005\\}, \\\\\nC & =\\{1006,1007, \\ldots, 1339\\}, \\\\\nD & =\\{1340,1341, \\ldots, 1508\\}, \\\\\nE & =\\{1509,1510, \\ldots, 2009\\} .\n\\end{aligned}\n$$\n\nWe will show that $A \\cup C \\cup E$ and $B \\cup D$ are both Benelux-sets.\n\nNote that there does not exist a bad subset of $S$ of one element, since that element would have to be 2010. Also, there does not exist a bad subset of $S$ of more than three elements, since the sum of four or more elements would be at least $502+503+504+505=2014>2010$. So any possible bad subset of $S$ contains two or three elements.\n\nConsider a bad subset of two elements $a$ and $b$. As $a, b \\geq 502$ and $a+b=2010$, we have $a, b \\leq 2010-502=1508$. Furthermore, exactly one of $a$ and $b$ is smaller than 1005 and one is larger than 1005. So one of them, say $a$, is an element of $A \\cup B$, and the other is an element of $C \\cup D$. Suppose $a \\in A$, then $b \\geq 2010-670=1340$, so $b \\in D$. On the other hand, suppose $a \\in B$, then $b \\leq 2010-671=1339$, so $b \\in C$. Hence $\\{a, b\\}$ cannot be a subset of $A \\cup C \\cup E$, nor of $B \\cup D$.\n\nNow consider a bad subset of three elements $a, b$ and $c$. As $a, b, c \\geq 502, a+b+c=2010$, and the three elements are pairwise distinct, we have $a, b, c \\leq 2010-502-503=1005$. So $a, b, c \\in A \\cup B$. At least one of the elements, say $a$, is smaller than $\\frac{2010}{3}=670$, and at least one of the elements, say $b$, is larger than 670 . So $a \\in A$ and $b \\in B$. We conclude that $\\{a, b, c\\}$ cannot be a subset of $A \\cup C \\cup E$, nor of $B \\cup D$.\n\nThis proves that $A \\cup C \\cup E$ and $B \\cup D$ are Benelux-sets, and therefore the smallest $n$ for which $S$ can be partitioned into $n$ Benelux-sets is $n=2$.\n\nRemark. Observe that $A \\cup C \\cup E_{1}$ and $B \\cup D \\cup E_{2}$ are also Benelux-sets, where $\\left\\{E_{1}, E_{2}\\right\\}$ is any partition of $E$."}
+{"problem": ". Find all polynomials $p(x)$ with real coefficients such that\n\n$$\np(a+b-2 c)+p(b+c-2 a)+p(c+a-2 b)=3 p(a-b)+3 p(b-c)+3 p(c-a)\n$$\n\nfor all $a, b, c \\in \\mathbb{R}$.", "solution": ". For $a=b=c$, we have $3 p(0)=9 p(0)$, hence $p(0)=0$. Now set $b=c=0$, then we have\n\n$$\np(a)+p(-2 a)+p(a)=3 p(a)+3 p(-a)\n$$\n\nfor all $a \\in \\mathbb{R}$. So we find a polynomial equation\n\n$$\np(-2 x)=p(x)+3 p(-x) \\text {. }\n$$\n\nNote that the zero polynomial is a solution to this equation. Now suppose that $p$ is not the zero polynomial, and let $n \\geq 0$ be the degree of $p$. Let $a_{n} \\neq 0$ be the coefficient of $x^{n}$ in $p(x)$. At the left-hand side of (1), the coefficient of $x^{n}$ is $(-2)^{n} \\cdot a_{n}$, while at the right-hand side the coefficient of $x^{n}$ is $a_{n}+3 \\cdot(-1)^{n} \\cdot a_{n}$. Hence $(-2)^{n}=1+3 \\cdot(-1)^{n}$. For $n$ even, we find $2^{n}=4$, so $n=2$, and for $n$ odd, we find $-2^{n}=-2$, so $n=1$. As we already know that $p(0)=0$, we must have $p(x)=a_{2} x^{2}+a_{1} x$, where $a_{1}$ and $a_{2}$ are real numbers (possibly zero).\n\nThe polynomial $p(x)=x$ is a solution to our problem, as\n\n$$\n(a+b-2 c)+(b+c-2 a)+(c+a-2 b)=0=3(a-b)+3(b-c)+3(c-a)\n$$\n\nfor all $a, b, c \\in \\mathbb{R}$. Also, $p(x)=x^{2}$ is a solution, since\n\n$$\n\\begin{aligned}\n(a+b-2 c)^{2}+(b+c-2 a)^{2}+(c+a-2 b)^{2} & =6\\left(a^{2}+b^{2}+c^{2}\\right)-6(a b+b c+c a) \\\\\n& =3(a-b)^{2}+3(b-c)^{2}+3(c-a)^{2}\n\\end{aligned}\n$$\n\nfor all $a, b, c \\in \\mathbb{R}$.\n\nNow note that if $p(x)$ is a solution to our problem, then so is $\\lambda p(x)$ for all $\\lambda \\in \\mathbb{R}$. Also, if $p(x)$ and $q(x)$ are both solutions, then so is $p(x)+q(x)$. We conclude that for all real numbers $a_{2}$ and $a_{1}$ the polynomial $a_{2} x^{2}+a_{1} x$ is a solution. Since we have already shown that there can be no other solutions, these are the only solutions."}
+{"problem": ". Find all polynomials $p(x)$ with real coefficients such that\n\n$$\np(a+b-2 c)+p(b+c-2 a)+p(c+a-2 b)=3 p(a-b)+3 p(b-c)+3 p(c-a)\n$$\n\nfor all $a, b, c \\in \\mathbb{R}$.", "solution": ". For $a=b=c$, we have $3 p(0)=9 p(0)$, hence $p(0)=0$. Now set $b=c=0$, then we have\n\n$$\np(a)+p(-2 a)+p(a)=3 p(a)+3 p(-a)\n$$\n\nfor all $a \\in \\mathbb{R}$. So we find a polynomial equation\n\n$$\np(-2 x)=p(x)+3 p(-x)\n$$\n\nDefine $q(x)=p(x)+p(-x)$, then we find that\n\n$$\nq(2 x)=p(2 x)+p(-2 x)=(p(-x)+3 p(x))+(p(x)+3 p(-x))=4 q(x)\n$$\n\nNote that the zero polynomial is a solution to this equation. Now suppose that $q$ is not the zero polynomial, and let $m \\geq 0$ be the degree of $q$. Let $b_{m} \\neq 0$ be the coefficient of $x^{m}$ in $q(x)$. At the left-hand side of (3), the coefficient of $x^{m}$ is $2^{m} \\cdot b_{m}$, while at the right-hand side the coefficient of $x^{m}$ is $4 b_{m}$. Hence $m=2$. As $q(x)=p(x)+p(-x)$, the polynomial $q(x)$ does not contain any nonzero terms with odd exponent of $x$. Since also $q(0)=2 p(0)=0$, we conclude that\n\n$$\nq(x)=b_{2} x^{2}\n$$\n\nwhere $b_{2}$ is a real number (possibly zero).\n\nFrom (2) we now deduce that $p(2 x)=p(-x)+3 p(x)=2 p(x)+q(x)$, so\n\n$$\np(2 x)-2 p(x)=b_{2} x^{2}\n$$\n\nSuppose that that degree $n$ of $p$ is greater than 2. Let $a_{n} \\neq 0$ be the coefficient of $x^{n}$ in $p(x)$. At the left-hand side of (4), the coefficient of $x^{n}$ is $\\left(2^{n}-2\\right) \\cdot a_{n} \\neq 0$. But the coefficient of $x^{n}$ at the right-hand side vanishes, yielding a contradiction. So the degree of $p$ is at most 2. As we already know that $p(0)=0$, we must have $p(x)=a_{2} x^{2}+a_{1} x$, where $a_{1}$ and $a_{2}$ are real numbers (possibly zero).\n\nWe finally check that every polynomial of this form is indeed a solution (see solution 1)."}
+{"problem": ". On a line $l$ there are three different points $A, B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to $B Q$ intersects $B Q$ in $L$ and $B R$ in $T$. The line through $B$ perpendicular to $A R$ intersects $A R$ in $K$ and $A Q$ in $S$.\n\n(a) Prove that $P, T, S$ are collinear.\n\n(b) Prove that $P, K, L$ are collinear.\n\n#", "solution": ".\n\n(a) Since $P, R$ and $Q$ are collinear, we have $\\triangle P A Q \\sim \\triangle P B R$, hence\n\n$$\n\\frac{|A Q|}{|B R|}=\\frac{|A P|}{|B P|}\n$$\n\nConversely, $P, T$ and $S$ are collinear if it holds that\n\n$$\n\\frac{|A S|}{|B T|}=\\frac{|A P|}{|B P|}\n$$\n\nSo it suffices to prove\n\n$$\n\\frac{|B T|}{|B R|}=\\frac{|A S|}{|A Q|}\n$$\n\nSince $\\angle A B T=90^{\\circ}=\\angle A L B$ and $\\angle T A B=\\angle B A L$, we have $\\triangle A B T \\sim \\triangle A L B$. And since $\\angle A L B=90^{\\circ}=\\angle Q A B$ and $\\angle L B A=\\angle A B Q$, we have $\\triangle A L B \\sim \\triangle Q A B$. Hence $\\triangle A B T \\sim \\triangle Q A B$, so\n\n$$\n\\frac{|B T|}{|B A|}=\\frac{|A B|}{|A Q|}\n$$\n\nSimilarly, we have $\\triangle A B R \\sim \\triangle A K B \\sim \\triangle S A B$, so\n\n$$\n\\frac{|B R|}{|B A|}=\\frac{|A B|}{|A S|}\n$$\n\nCombining both results, we get\n\n$$\n\\frac{|B T|}{|B R|}=\\frac{|B T| /|B A|}{|B R| /|B A|}=\\frac{|A B| /|A Q|}{|A B| /|A S|}=\\frac{|A S|}{|A Q|}\n$$\n\nwhich had to be proved.\n(b) Let the line $P K$ intersect $B R$ in $B_{1}$ and $A Q$ in $A_{1}$ and let the line $P L$ intersect $B R$ in $B_{2}$ and $A Q$ in $A_{2}$. Consider the points $A_{1}, A$ and $S$ on the line $A Q$, and the points $B_{1}, B$ and $T$ on the line $B R$. As $A Q \\| B R$ and the three lines $A_{1} B_{1}, A B$ and $S T$ are concurrent (in $P$ ), we have\n\n$$\nA_{1} A: A S=B_{1} B: B T,\n$$\n\nwhere all lengths are directed. Similarly, as $A_{1} B_{1}, A R$ and $S B$ are concurrent (in $K)$, we have\n\n$$\nA_{1} A: A S=B_{1} R: R B .\n$$\n\nThis gives\n\n$$\n\\frac{B B_{1}}{B T}=\\frac{R B_{1}}{R B}=\\frac{R B+B B_{1}}{R B}=1+\\frac{B B_{1}}{R B}=1-\\frac{B B_{1}}{B R},\n$$\n\nso\n\n$$\nB B_{1}=\\frac{1}{\\frac{1}{B T}+\\frac{1}{B R}} .\n$$\n\nSimilary, using the lines $A_{2} B_{2}, A B$ and $Q R$ (concurrent in $P$ ) and the lines $A_{2} B_{2}$, $A T$ and $Q B$ (concurrent in $L$ ), we find\n\n$$\nB_{2} B: B R=A_{2} A: A Q=B_{2} T: T B .\n$$\n\nThis gives\n\n$$\n\\frac{B B_{2}}{B R}=\\frac{T B_{2}}{T B}=\\frac{T B+B B_{2}}{T B}=1+\\frac{B B_{2}}{T B}=1-\\frac{B B_{2}}{B T},\n$$\n\nSo\n\n$$\nB B_{2}=\\frac{1}{\\frac{1}{B R}+\\frac{1}{B T}} .\n$$\n\nWe conclude that $B_{1}=B_{2}$, which implies that $P, K$ and $L$ are collinear.\n\n#"}
+{"problem": ". On a line $l$ there are three different points $A, B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to $B Q$ intersects $B Q$ in $L$ and $B R$ in $T$. The line through $B$ perpendicular to $A R$ intersects $A R$ in $K$ and $A Q$ in $S$.\n\n(a) Prove that $P, T, S$ are collinear.\n\n(b) Prove that $P, K, L$ are collinear.\n\n#", "solution": ".\n\n(a) Define $X$ as the intersection of $A T$ and $B S$, and $Y$ as the intersection of $A R$ and $B Q$. To prove that $P, S$ and $T$ are collinear, we will use Menelaos' theorem in $\\triangle A B X$, so we have to prove\n\n$$\n\\frac{A P}{P B} \\frac{B S}{S X} \\frac{X T}{T A}=-1\n$$\n\nNote that $B$ is between $P$ and $A, X$ is between $S$ and $B$, and $X$ is between $T$ and $A$, so it suffices to prove that\n\n$$\n\\frac{|A P|}{|P B|} \\frac{|B S|}{|S X|} \\frac{|X T|}{|T A|}=1\n$$\n\nBecause $A Q$ and $B R$ are parallel, we have $\\triangle A Q P \\sim \\triangle B R P$, hence\n\n$$\n\\frac{|A P|}{|B P|}=\\frac{|Q A|}{|R B|}\n$$\n\nAlso, since $\\angle A S B=\\angle K B R$ and $\\angle B A S=90^{\\circ}=\\angle B K R$, we have $\\triangle A S B \\sim$ $\\triangle K B R$, hence\n\n$$\n\\frac{|B S|}{|R B|}=\\frac{|A S|}{|K B|}, \\quad \\text { so } \\quad|B S|=\\frac{|A S|}{|K B|}|R B|\n$$\n\nSimilarly, we have $\\triangle A T B \\sim \\triangle Q A L$, hence\n\n$$\n\\frac{|T A|}{|A Q|}=\\frac{|T B|}{|A L|}, \\quad \\text { so } \\quad|T A|=\\frac{|T B|}{|A L|}|A Q|\n$$\n\nAs $\\angle A S X=\\angle A S B=90^{\\circ}-\\angle A B S=90^{\\circ}-\\angle A B K=\\angle K A B=\\angle Y A B$, and $\\angle S A X=90^{\\circ}-\\angle X A B=90^{\\circ}-\\angle L A B=\\angle A B L=\\angle A B Y$, we have $\\triangle S X A \\sim$ $\\triangle A Y B$, hence\n\n$$\n\\frac{|S X|}{|A Y|}=\\frac{|A S|}{|B A|}, \\quad \\text { so } \\quad|S X|=\\frac{|A S|}{|B A|}|A Y|\n$$\n\nSimilarly, we have $\\triangle B X T \\sim \\triangle A Y B$, hence\n\n$$\n\\frac{|X T|}{|Y B|}=\\frac{|B T|}{|A B|}, \\quad \\text { so } \\quad|X T|=\\frac{|B T|}{|A B|}|Y B|\n$$\n\nBy combining (5) - (9), we find\n\n$$\n\\begin{aligned}\n\\frac{|A P|}{|P B|} \\frac{|B S|}{|S X|} \\frac{|X T|}{|T A|} & =\\frac{|Q A|}{|R B|} \\cdot \\frac{|A S|}{|K B|}|R B| \\cdot \\frac{|B A|}{|A S||A Y|} \\cdot \\frac{|B T|}{|A B|}|Y B| \\cdot \\frac{|A L|}{|T B||A Q|} \\\\\n& =\\frac{|A L|}{|K B|} \\frac{|Y B|}{|A Y|} .\n\\end{aligned}\n$$\n\nSince $\\angle Y L A=90^{\\circ}=\\angle Y K B$ and $\\angle A Y L=\\angle B Y K$, we have $\\triangle A Y L \\sim \\triangle B Y K$, hence\n\n$$\n\\frac{|A L|}{|B K|}=\\frac{|A Y|}{|B Y|}, \\quad \\text { so } \\quad \\frac{|A L|}{|B K|} \\frac{|B Y|}{|A Y|}=1\n$$\n\nBy combining (10) and (11), we find\n\n$$\n\\frac{|A P|}{|P B|} \\frac{|B S|}{|S X|} \\frac{|X T|}{|T A|}=1\n$$\n\nas we wanted to prove.\n\n(b) Again, we will use Menelaos' theorem in $\\triangle A B X$, so we have to prove\n\n$$\n\\frac{A P}{P B} \\frac{B K}{K X} \\frac{X L}{L A}=-1\n$$\n\nNote that $\\frac{A P}{P B}<0$, and $\\frac{B K}{K X}<0$ if and only if $\\frac{X L}{L A}<0$, so it suffices to prove that\n\n$$\n\\frac{|A P|}{|P B|} \\frac{|B K|}{|K X|} \\frac{|X L|}{|L A|}=1\n$$\n\nAs $\\angle B X L=\\angle A X K$ and $\\angle B L X=90^{\\circ}=\\angle A K X$, we have $\\triangle B L X \\sim \\triangle A K X$, hence\n\n$$\n\\frac{|X L|}{|X K|}=\\frac{|B L|}{|A K|}\n$$\n\nSince $\\angle A L B=90^{\\circ}=\\angle Q A B$, we have $\\triangle A L B \\sim \\triangle Q A B$, hence\n\n$$\n\\frac{|L A|}{|A Q|}=\\frac{|L B|}{|A B|}, \\quad \\text { so } \\quad|L A|=\\frac{|L B|}{|A B|}|A Q|\n$$\n\nSimilarly, we have $\\triangle A K B \\sim \\triangle A B R$, hence\n\n$$\n\\frac{|B K|}{|R B|}=\\frac{|A K|}{|A B|}, \\quad \\text { so } \\quad|B K|=\\frac{|A K|}{|A B|}|R B|\n$$\n\nBy combining (5) and (12) - (14), we find\n\n$$\n\\frac{|A P|}{|P B|} \\frac{|B K|}{|K X|} \\frac{|X L|}{|L A|}=\\frac{|Q A|}{|R B|} \\cdot \\frac{|B L|}{|A K|} \\cdot \\frac{|A B|}{|L B||A Q|} \\cdot \\frac{|A K|}{|A B|}|R B|=1\n$$\n\nwhich is what we wanted to prove."}
+{"problem": ". On a line $l$ there are three different points $A, B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to $B Q$ intersects $B Q$ in $L$ and $B R$ in $T$. The line through $B$ perpendicular to $A R$ intersects $A R$ in $K$ and $A Q$ in $S$.\n\n(a) Prove that $P, T, S$ are collinear.\n\n(b) Prove that $P, K, L$ are collinear.\n\n#", "solution": ". As $\\angle A K B=\\angle A L B=90^{\\circ}$, the points $K$ and $L$ belong to the circle with diameter $A B$. Since $\\angle Q A B=\\angle A B R=90^{\\circ}$, the lines $A Q$ and $B R$ are tangents to this circle.\n\nApply Pascal's theorem to the points $A, A, K, L, B$ and $B$, all on the same circle. This yields that the intersection $Q$ of the tangent in $A$ and the line $B L$, the intersection $R$ of the tangent in $B$ and the line $A K$, and the intersection of $K L$ and $A B$ are collinear. So $K L$ passes through the intersection of $A B$ and $Q R$, which is point $P$. Hence $P, K$ and $L$ are collinear. This proves part b.\n\nNow apply Pascal's theorem to the points $A, A, L, K, B$ and $B$. This yields that the intersection $S$ of the tangent in $A$ and the line $B K$, the intersection $T$ of the tangent in $B$ and the line $A L$, and the intersection $P$ of $K L$ and $A B$ are collinear. This proves part a.\n\n#"}
+{"problem": ". On a line $l$ there are three different points $A, B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to $B Q$ intersects $B Q$ in $L$ and $B R$ in $T$. The line through $B$ perpendicular to $A R$ intersects $A R$ in $K$ and $A Q$ in $S$.\n\n(a) Prove that $P, T, S$ are collinear.\n\n(b) Prove that $P, K, L$ are collinear.\n\n#", "solution": ".\n\n(a) W.l.o.g. we may assume that $A=(0,0)$ and $B=(1,0)$ and the line through $P$ is in the upper half plane, so $l$ is the $x$-axis, $a$ is the $y$-axis and $b$ is the line $x=1$. Take $P=(p, 0)(p>1)$ and $Q=(0, q)(q>0)$. Since $P Q$ is given by $\\frac{x}{p}+\\frac{y}{q}=1$, we find $R=\\left(1, \\frac{q(p-1)}{p}\\right)$.\n\nNow $A R$ is given by $y=\\frac{q(p-1)}{p} x$, hence $B S$, the line perpendicular to $A R$ and passing through $B=(1,0)$, is given by $y=-\\frac{p}{q(p-1)}(x-1)$. We find $S=\\left(0, \\frac{p}{q(p-1)}\\right)$.\n\nMoreover $B Q$ is given by $y=-q(x-1)$, hence $A T$, the line perpendicular to $B Q$ and passing through $A=(0,0)$, is given by $y=\\frac{1}{q} x$. We find $T=\\left(1, \\frac{1}{q}\\right)$. Since $\\frac{|B T|}{|B P|}=\\frac{1 / q}{p-1}=\\frac{\\frac{p}{q(p-1)}}{p}=\\frac{|A S|}{|A P|}$, we conclude that $P, T$ and $S$ are collinear.\n\n(b) Point $K$ is the intersection of $A R$ and $B S$. Solving for $x$ yields\n\n$$\n\\begin{aligned}\n\\frac{q(p-1)}{p} x & =-\\frac{p}{q(p-1)}(x-1) \\\\\n\\left(\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}\\right) x & =\\frac{p}{q(p-1)} \\\\\nx & =\\frac{\\frac{p}{q(p-1)}}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}}\n\\end{aligned}\n$$\n\nso\n\n$$\nK=\\left(\\frac{\\frac{p}{q(p-1)}}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}}, \\frac{1}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}}\\right)\n$$\n\nPoint $L$ is the point of intersection of $A T$ and $B Q$. Solving for $x$ yields\n\n$$\n\\begin{aligned}\n\\frac{1}{q} x & =-q(x-1) \\\\\n\\left(\\frac{1}{q}+q\\right) x & =q \\\\\nx & =\\frac{q}{\\frac{1}{q}+q}\n\\end{aligned}\n$$\n\nso\n\n$$\nL=\\left(\\frac{q}{\\frac{1}{q}+q}, \\frac{1}{\\frac{1}{q}+q}\\right)\n$$\n\nLet $K_{0}$ and $L_{0}$ be the projections of $K$ and $L$ on the $x$-axis. We have to show that the following fractions are equal:\n\n$$\n\\frac{\\left|K_{0} K\\right|}{\\left|K_{0} P\\right|}=\\frac{\\frac{1}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}}}{p-\\frac{\\frac{p}{q(p-1)}}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}}} \\quad \\text { and } \\quad \\frac{\\left|L_{0} L\\right|}{\\left|L_{0} P\\right|}=\\frac{\\frac{1}{\\frac{1}{q}+q}}{p-\\frac{q}{\\frac{1}{q}+q}}\n$$\n\nWorking out cross products twice, this comes down to\n\n$$\n\\begin{aligned}\n& \\frac{1}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}} \\cdot\\left(p-\\frac{q}{\\frac{1}{q}+q}\\right) \\stackrel{?}{=} \\frac{1}{\\frac{1}{q}+q} \\cdot\\left(p-\\frac{\\frac{p}{q(p-1)}}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}}\\right) \\\\\n& \\left(\\frac{1}{q}+q\\right) \\cdot\\left(p-\\frac{q}{\\frac{1}{q}+q}\\right) \\stackrel{?}{=}\\left(\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}\\right) \\cdot\\left(p-\\frac{\\frac{p}{q(p-1)}}{\\frac{q(p-1)}{p}+\\frac{p}{q(p-1)}}\\right) \\\\\n& \\frac{p}{q}+p q-q \\stackrel{?}{=} q(p-1)+\\frac{p^{2}}{q(p-1)}-\\frac{p}{q(p-1)} \\\\\n& \\frac{p}{q}+p q-q \\stackrel{?}{=} q(p-1)+\\frac{p(p-1)}{q(p-1)}\n\\end{aligned}\n$$\n\nwhich is clearly true."}
+{"problem": ". Find all quadruples $(a, b, p, n)$ of positive integers, such that $p$ is a prime and\n\n$$\na^{3}+b^{3}=p^{n} \\text {. }\n$$", "solution": ". Let $(a, b, p, n)$ be a solution. Note that we can write the given equation as\n\n$$\n(a+b)\\left(a^{2}-a b+b^{2}\\right)=p^{n} .\n$$\n\nAs $a$ and $b$ are positive integers, we have $a+b \\geq 2$, so $p \\mid a+b$. Furthermore, $a^{2}-a b+b^{2}=$ $(a-b)^{2}+a b$, so either $a=b=1$ or $a^{2}-a b+b^{2} \\geq 2$. Assume that the latter is the case. Then $p$ is a divisor of both $a+b$ and $a^{2}-a b+b^{2}$, hence also of $(a+b)^{2}-\\left(a^{2}-a b+b^{2}\\right)=3 a b$. This means that $p$ either is equal to 3 or is a divisor of $a b$. Since $p$ is a divisor of $a+b$, we have $p|a \\Leftrightarrow p| b$, hence either $p=3$, or $p \\mid a$ and $p \\mid b$. If $p \\mid a$ and $p \\mid b$, then we can write $a=p a^{\\prime}, b=p b^{\\prime}$ with $a^{\\prime}$ and $b^{\\prime}$ positive integers, and we have $\\left(a^{\\prime}\\right)^{3}+\\left(b^{\\prime}\\right)^{3}=p^{n-3}$, so $\\left(a^{\\prime}, b^{\\prime}, p, n-3\\right)$ then is another solution (note that $\\left(a^{\\prime}\\right)^{3}+\\left(b^{\\prime}\\right)^{3}$ is a positive integer greater than 1 , so $n-3$ is positive).\n\nNow assume that $\\left(a_{0}, b_{0}, p_{0}, n_{0}\\right)$ is a solution such that $p \\nmid a$. From the reasoning above it follows that either $a_{0}=b_{0}=1$, or $p_{0}=3$. After all, if we do not have $a_{0}=b_{0}=1$ and we have $p_{0} \\neq 3$, then $p \\mid a$. Also, given an arbitrary solution $(a, b, p, n)$, we can divide everything by $p$ repeatedly until there are no factors $p$ left in $a$.\n\nSuppose $a_{0}=b_{0}=1$. Then the solution is $(1,1,2,1)$.\n\nSuppose $p_{0}=3$. Assume that $3^{2} \\mid\\left(a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}\\right)$. As $3^{2} \\mid\\left(a_{0}+b_{0}\\right)^{2}$, we then have $3^{2} \\mid\\left(a_{0}+b_{0}\\right)^{2}-\\left(a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}\\right)=3 a_{0} b_{0}$, so $3 \\mid a_{0} b_{0}$. But $3 \\nmid a_{0}$ by assumption, and $3 \\mid a_{0}+b_{0}$, so $3 \\nmid b_{0}$, which contradicts $3 \\mid a_{0} b_{0}$. We conclude that $3^{2} \\nmid\\left(a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}\\right)$. As both $a_{0}+b_{0}$ and $a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}$ must be powers of 3 , we have $a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}=3$. Hence $\\left(a_{0}-b_{0}\\right)^{2}+a_{0} b_{0}=3$. We must have $\\left(a_{0}-b_{0}\\right)^{2}=0$ or $\\left(a_{0}-b_{0}\\right)^{2}=1$. The former does not give a solution; the latter gives $a_{0}=2$ and $b_{0}=1$ or $a_{0}=1$ and $b_{0}=2$.\n\nSo all solutions with $p \\nmid a$ are $(1,1,2,1),(2,1,3,2)$ and $(1,2,3,2)$. From the above it follows that all other solutions are of the form $\\left(p_{0}^{k} a_{0}, p_{0}^{k} b_{0}, p_{0}, n_{0}+3 k\\right)$, where $\\left(a_{0}, b_{0}, p_{0}, n_{0}\\right)$ is one of these three solutions. Hence we find three families of solutions:\n\n- $\\left(2^{k}, 2^{k}, 2,3 k+1\\right)$ with $k \\in \\mathbb{Z}_{\\geq 0}$,\n- $\\left(2 \\cdot 3^{k}, 3^{k}, 3,3 k+2\\right)$ with $k \\in \\mathbb{Z}_{\\geq 0}$,\n- $\\left(3^{k}, 2 \\cdot 3^{k}, 3,3 k+2\\right)$ with $k \\in \\mathbb{Z}_{\\geq 0}$.\n\nIt is easy to check that all these quadruples are indeed solutions."}
+{"problem": ". Find all quadruples $(a, b, p, n)$ of positive integers, such that $p$ is a prime and\n\n$$\na^{3}+b^{3}=p^{n} \\text {. }\n$$", "solution": ". Let $(a, b, p, n)$ be a solution. Note that we can write the given equation as\n\n$$\n(a+b)\\left(a^{2}-a b+b^{2}\\right)=p^{n} .\n$$\n\nAs $a$ and $b$ are positive integers, we have $a+b \\geq 2$ and $a^{2}-a b+b^{2}=(a-b)^{2}+a b \\geq 1$. So both factors are positive and therefore must be powers of $p$. Let $k$ be an integer with $1 \\leq k \\leq n$ such that $a+b=p^{k}$. Then $a^{2}-a b+b^{2}=p^{n-k}$. If we substitute $b=p^{k}-a$, we find\n\n$$\np^{n-k}=(a+b)^{2}-3 a b=p^{2 k}-3 a\\left(p^{k}-a\\right) .\n$$\n\nWe can rewrite this as:\n\n$$\n3 a^{2}-3 p^{k} a+p^{2 k}-p^{n-k}=0,\n$$\n\nfrom which we see that $a$ is a solution of the following quadratic equation in $x$ :\n\n$$\n3 x^{2}-3 p^{k} x+p^{2 k}-p^{n-k}=0 .\n$$\n\nThe discriminant of (15) is\n\n$$\nD=\\left(-3 p^{k}\\right)^{2}-4 \\cdot 3 \\cdot\\left(p^{2 k}-p^{n-k}\\right)=3 \\cdot\\left(4 p^{n-k}-p^{2 k}\\right)=3 p^{n-k} \\cdot\\left(4-p^{3 k-n}\\right) .\n$$\n\nAs $p^{n-k}=(a+b)^{2}-3 a b<(a+b)^{2}=p^{2 k}$, we have $n-k<2 k$, so $3 k-n>0$. Since $a$ is a solution of (15), the discriminant must be nonnegative. Hence $4-p^{3 k-n} \\geq 0$. If $p=2$, this implies $3 k-n=1$ or $3 k-n=2$; if $p=3$, this implies $3 k-n=1$; and if $p>3$, then $p \\geq 5$ so $4 \\geq p^{3 k-n}$ can never be true.\n\nSuppose $p=2$ and $3 k-n=1$. Then $D=3 \\cdot 2^{2 k-1} \\cdot(4-2)=3 \\cdot 2^{2 k}$. But this is a not a square, so the solutions of (15) will not be integers, which yields a contradiction.\n\nSuppose $p=2$ and $3 k-n=2$. Then $D=3 \\cdot 2^{2 k-2} \\cdot(4-4)=0$, so the only solution of (15) is $x=\\frac{3 \\cdot 2^{k}}{2 \\cdot 3}=2^{k-1}$. Therefore $a=2^{k-1}$ and $b=2^{k}-a=2^{k-1}$, and this gives a solution for all $k \\geq 1$, namely $\\left(2^{k-1}, 2^{k-1}, 2,3 k-2\\right)$.\n\nSuppose $p=3$ and $3 k-n=1$. Then $D=3 \\cdot 3^{2 k-1} \\cdot(4-3)=3^{2 k}$, so the solutions of (15) are $x=\\frac{3^{k+1} \\pm 3^{k}}{2 \\cdot 3}=\\frac{1}{2}\\left(3^{k} \\pm 3^{k-1}\\right)$. Therefore $a=2 \\cdot 3^{k-1}$ or $a=3^{k-1}$. For all $k \\geq 1$ we find the solutions $\\left(2 \\cdot 3^{k-1}, 3^{k-1}, 3,3 k-1\\right)$ and $\\left(3^{k-1}, 2 \\cdot 3^{k-1}, 3,3 k-1\\right)$.\n\nWe conclude that there are three families of solutions:\n\n- $\\left(2^{k-1}, 2^{k-1}, 2,3 k-2\\right)$ with $k \\in \\mathbb{Z}_{\\geq 1}$,\n- $\\left(2 \\cdot 3^{k-1}, 3^{k-1}, 3,3 k-1\\right)$ with $k \\in \\mathbb{Z}_{\\geq 1}$,\n- $\\left(3^{k-1}, 2 \\cdot 3^{k-1}, 3,3 k-1\\right)$ with $k \\in \\mathbb{Z}_{\\geq 1}$.\n\nIt is easy to check that all these quadruples are indeed solutions."}

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2011-zz.jsonl

@@ -0,0 +1,6 @@
+{"problem": "An ordered pair of integers $(m, n)$ with $1<m<n$ is said to be a Benelux couple if the following two conditions hold : $m$ has the same prime divisors as $n$, and $m+1$ has the same prime divisors as $n+1$.\n\n(a) Find three Benelux couples $(m, n)$ with $m \\leqslant 14$.\n\n(b) Prove that there exist infinitely many Benelux couples.\n\n#", "solution": "(a) It is possible to see that $(2,8),(6,48)$ and $(14,224)$ are Benelux couples.\n\n(b) Let $k \\geqslant 2$ be an integer and $m=2^{k}-2$. Define $n=m(m+2)=2^{k}\\left(2^{k}-2\\right)$. Since $m$ is even, $m$ and $n$ have the same prime factors. Also, $n+1=m(m+2)+1=(m+1)^{2}$, so $m+1$ and $n+1$ have the same prime factors, too. We have thus obtained a Benelux couple $\\left(2^{k}-2,2^{k}\\left(2^{k}-2\\right)\\right.$ ) for each $k \\geqslant 2$.\n\n#"}
+{"problem": "Let $A B C$ be a triangle with incentre $I$. The angle bisectors $A I, B I$ and $C I$ meet $[B C],[C A]$ and $[A B]$ at $D, E$ and $F$, respectively. The perpendicular bisector of $[A D]$ intersects the lines $B I$ and $C I$ at $M$ and $N$, respectively. Show that $A, I, M$ and $N$ lie on a circle.\n\n#", "solution": "The quadrilateral $A M D B$ is cyclic. Indeed, $M$ is the intersection of the line $B I$, which bisects the angle $\\widehat{A B D}$ in $A B D$ and the perpendicular bisector of $[A D]$. By uniqueness of this intersection point, it follows that $M$ lies on the circumcircle of $A B D$, and thence $A M D B$ is cyclic. Analogously, $A N D C$ is cyclic.\n\nMoreover, $M N B C$ is cyclic, for $\\widehat{N M I}=90^{\\circ}-\\widehat{E I A}$. Indeed, $A$ and $I$ lie on either side of the midpoint of $[A D]$, for $\\widehat{B D A}=\\widehat{C A D}+\\widehat{B C A}>\\widehat{B A D}$. But\n\n$$\n\\widehat{E I A}=180^{\\circ}-\\frac{1}{2} \\widehat{B A C}-\\widehat{B E A}=180^{\\circ}-\\frac{1}{2} \\widehat{B A C}-\\left(\\frac{1}{2} \\widehat{C B A}+\\widehat{B C A}\\right)=90^{\\circ}-\\frac{1}{2} \\widehat{B C A}\n$$\n\nIt follows that $\\widehat{N M I}=\\widehat{B C I}$, which implies that $M N B C$ is cyclic, as the points $I$ and $D$ lie on the same side of $M N$.\n\nThere are now two ways of completing the proof:\n\n#"}
+{"problem": "Let $A B C$ be a triangle with incentre $I$. The angle bisectors $A I, B I$ and $C I$ meet $[B C],[C A]$ and $[A B]$ at $D, E$ and $F$, respectively. The perpendicular bisector of $[A D]$ intersects the lines $B I$ and $C I$ at $M$ and $N$, respectively. Show that $A, I, M$ and $N$ lie on a circle.\n\n#", "solution": "(using $A M D B$ and $M N B C$ )\n\nSince $A M D B$ is cyclic, $\\widehat{M A I}=\\widehat{M A D}=\\widehat{M B D}$, as, by construction, $B$ and $M$ lie on either side of $A D$. Moreover, $\\widehat{M B D}=\\widehat{M B C}=\\widehat{M N C}$ for $M N B C$ is cyclic. Thus $\\widehat{M A I}=\\widehat{M N I}$, so $A M I N$ is cyclic, for $M$ and $N$ lie on either side of $A D$.\n\n#"}
+{"problem": "Let $A B C$ be a triangle with incentre $I$. The angle bisectors $A I, B I$ and $C I$ meet $[B C],[C A]$ and $[A B]$ at $D, E$ and $F$, respectively. The perpendicular bisector of $[A D]$ intersects the lines $B I$ and $C I$ at $M$ and $N$, respectively. Show that $A, I, M$ and $N$ lie on a circle.\n\n#", "solution": "(using $A M D B$ and $A N D C$ )\n\nSince $A M D B$ and $A N D C$ are cylic, $\\widehat{A M I}+\\widehat{A N I}=\\widehat{A M B}+\\widehat{A N C}=\\widehat{A D B}+\\widehat{A D C}=180^{\\circ}$, because $B$ and $M$, and $C$ and $N$ lie on either side of $A D$. Hence $A M I N$ is cyclic, for $M$ and $N$ lie on either side of $A D$.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_26f7d25aec405b23419dg-3.jpg?height=919&width=1014&top_left_y=1528&top_left_x=521)\n\nRemark. It is moreover true that $B M \\perp D N$ and $C N \\perp D M$. Indeed, symmetry implies that the image $J$ of $I$ under reflection in $M N$ lies on the circumcircle of $D M N$. Moreover, $D I$ is a height of $D M N$, so the fact that $I$ and $J$ are equidistant from the side $[M N]$ implies that $I$ is the orthocentre of $D M N$. This implies the claim.\n\n#"}
+{"problem": "If $k$ is an integer, let $\\mathrm{c}(k)$ denote the largest cube that is less than or equal to $k$. Find all positive integers $p$ for which the following sequence is bounded:\n\n$$\na_{0}=p \\quad \\text { and } \\quad a_{n+1}=3 a_{n}-2 c\\left(a_{n}\\right) \\quad \\text { for } n \\geqslant 0\n$$\n\n#", "solution": "Since $\\mathrm{c}\\left(a_{n}\\right) \\leqslant a_{n}$ for all $n \\in \\mathbb{N}, a_{n+1} \\geqslant a_{n}$ with equality if and only if $\\mathrm{c}\\left(a_{n}\\right)=a_{n}$. Hence the sequence is bounded if and only if it is eventually constant, which is if and only if $a_{n}$ is a perfect cube, for some $n \\geqslant 0$. In particular, the sequence is bounded if $p$ is a perfect cube.\n\nWe now claim that, if $a_{n}$ is not a cube for some $n$, then neither is $a_{n+1}$. Indeed, if $a_{n}$ is not a cube, $q^{3}<a_{n}<(q+1)^{3}$ for some $q \\in \\mathbb{N}$, so that $\\mathrm{c}\\left(a_{n}\\right)=q^{3}$. Suppose to the contrary that $a_{n+1}$ is a cube. Then\n\n$$\na_{n+1}=3 a_{n}-2 \\mathrm{c}\\left(a_{n}\\right)<3(q+1)^{3}-2 q^{3}=q^{3}+9 q^{2}+9 q+3<q^{3}+9 q^{2}+27 q+27=(q+3)^{3}\n$$\n\nAlso, since $\\mathrm{c}\\left(a_{n}\\right)<a_{n}, a_{n+1}>a_{n}>q^{3}$, so $q^{3}<a_{n+1}<(q+3)^{3}$. It follows that the only possible values of $a_{n+1}$ are $(q+1)^{3}$ and $(q+2)^{3}$. However, in both of these cases,\n\n$$\n\\begin{aligned}\n& 3 a_{n}-2 q^{3}=a_{n+1}=(q+1)^{3} \\Longleftrightarrow 3 a_{n}=3\\left(q^{3}+q^{2}+q\\right)+1 \\\\\n& 3 a_{n}-2 q^{3}=a_{n+1}=(q+2)^{3} \\Longleftrightarrow 3 a_{n}=3\\left(q^{3}+2 q^{2}+4 q\\right)+8\n\\end{aligned}\n$$\n\na contradiction modulo 3. This proves that, if $a_{n}$ is not a cube, then neither is $a_{n+1}$. Hence, if $p$ is not a perfect cube, $a_{n}$ is not a cube for any $n \\in \\mathbb{N}$, and the sequence is not bounded. We conclude that the sequence is bounded if and only if $p$ is a perfect cube.\n\n#"}
+{"problem": "Abby and Brian play the following game: They first choose a positive integer $N$. Then they write numbers on a blackboard in turn. Abby starts by writing a 1. Thereafter, when one of them has written the number $n$, the other writes down either $n+1$ or $2 n$, provided that the number is not greater than $N$. The player who writes $N$ on the blackboard wins.\n\n(a) Determine which player has a winning strategy if $N=2011$.\n\n(b) Find the number of positive integers $N \\leqslant 2011$ for which Brian has a winning strategy.\n\n#", "solution": "(a) Abby has a winning strategy for odd $N$ : Observe that, whenever any player writes down an odd number, the other player has to write down an even number. By adding 1 to that number, the first player can write down another odd number. Since Abby starts the game by writing down an odd number, she can force Brian to write down even numbers only. Since $N$ is odd, Abby will win the game. In particular, Abby has a winning strategy if $N=2011$.\n\n(b) - Let $N=4 k$. If any player is forced to write down a number $m \\in\\{k+1, k+2, \\ldots, 2 k\\}$, the other player wins the game by writing down $2 m \\in\\{2 k+2,2 k+4, \\ldots, 4 k\\}$, for the players will have to write down the remaining numbers one after the other. Since there is an even number of numbers remaining, the latter player wins. This implies that the player who can write down $k$, i.e. has a winning strategy for $N=k$, wins the game for $N=4 k$.\n\n- Similarly, let $N=4 k+2$. If any player is forced to write down a number $m \\in\\{k+1, k+2, \\ldots, 2 k+1\\}$, the other player wins the game by writing down $2 m \\in\\{2 k+2,2 k+4, \\ldots, 4 k+2\\}$, as in the previous case. Analogously, this implies that the player who has a winning strategy for $N=k$ wins the game for $N=4 k+2$.\n\nSince Abby wins the game for $N=1,3$, while Brian wins the game for $N=2$, Brian wins the game for $N=8,10$, as well, and thus for $N=32,34,40,42$, too. Then Brian wins the game for a further 8 values of $N$ between 128 and 170, and thence for a further 16 values between 512 and 682, and for no other values with $N \\leqslant 2011$. Hence Brian has a winning strategy for precisely 31 values of $N$ with $N \\leqslant 2011$."}

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2012-zz.jsonl

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+{"problem": ". A sequence $a_{1}, a_{2}, \\ldots, a_{n}, \\ldots$ of natural numbers is defined by the rule\n\n$$\na_{n+1}=a_{n}+b_{n} \\quad(n=1,2, \\ldots)\n$$\n\nwhere $b_{n}$ is the last digit of $a_{n}$. Prove that such a sequence contains infinitely many powers of 2 if and only if $a_{1}$ is not divisible by 5 .", "solution": "First we can observe that:\n\n- If $a_{1}$ is divisible by 5 , then $a_{n}=a_{2}=0(\\bmod 10) \\forall n \\geq 2$.\n- If $a_{1}$ is not divisible by 5 , then for $n \\geq 2: a_{n}$ is even, the sequence $b_{n}$ is periodic, its period is a cyclic permutation of $(2,4,8,6)$, and $a_{n+4}=a_{n}+20$.\n\n(a) Let us suppose that $a_{1}$ is divisible by 5 .\n\nSince $2^{k} \\neq 0(\\bmod 10)$ for any $k \\in \\mathbb{N}$, the sequence does not contain any power of 2 for $n \\geq 2$.\n\n(b) Let us suppose that $a_{1}$ is not divisible by 5 .\n\nWe can remark that the sequence of powers of 2 modulo 20 respects the period $(12,4,8,16)$ starting with $2^{5}=32$. We choose $j$ such that $a_{j}=2(\\bmod 10)$ (i.e. $\\left.b_{j}=2\\right)$ and look at the parity of its penultimate digit.\n\n- If $a_{j}=12(\\bmod 20)$, then the numbers $a_{j+4 k}, k \\in \\mathbb{N}$, represent all the numbers congruent to $12(\\bmod 20)$ and greater than $a_{j}$, so all powers of 2 congruent to 12 $(\\bmod 20)$ and greater than $a_{j}$ appear in the sequence.\n- If $a_{j}=2(\\bmod 20)$, then the numbers $a_{j+1+4 k}, k \\in \\mathbb{N}$, represent all the numbers congruent to $4(\\bmod 20)$ and greater than $a_{j+1}$, so all powers of 2 congruent to 4 $(\\bmod 20)$ and greater than $a_{j+1}$ appear in the sequence.\n\nThus, the sequence contains infinitely many powers of 2 .\n\nAlternative 1 for $(\\mathbf{b})$. We choose $j$ such that $a_{j}=2(\\bmod 10)\\left(\\right.$ i.e. $\\left.b_{j}=2\\right)$.\n\n- If $a_{j}=20 t+12$ for some $t \\in \\mathbb{N}$, then $a_{j+4 k}=a_{j}+20 k=20(t+k)+12, \\forall k \\in \\mathbb{N}$. We obtain infinitely many powers of 2 by taking $k=\\frac{2^{4 s+3}-3}{5}-t$ (with $s \\in \\mathbb{N}$ large enough to have $k>0)$ since $2^{4 s+3}=3(\\bmod 5), \\forall s \\in \\mathbb{N}$.\n- If $a_{j}=20 t+2$ for some $t \\in \\mathbb{N}$, then $a_{j+1+4 k}=a_{j+1}+20 k=20(t+k)+4, \\forall k \\in \\mathbb{N}$. We obtain infinitely many powers of 2 by taking $k=\\frac{2^{4 s}-1}{5}-t$ (with $s \\in \\mathbb{N}$ large enough to have $k>0)$ since $2^{4 s}=1(\\bmod 5), \\forall s \\in \\mathbb{N}$.\n\nAlternative 2 for (b). Choose $j$ such that $a_{j}$ is a multiple of 4 , i.e. $a_{j}=4 q$ (such a $j$ always exists since $a_{n+1}=a_{n}+2$ for infinitely many $n$ ). Then we have $a_{j+4 k}=a_{j}+20 k=4(q+5 k)$. Let us look for $(k, m)$ such that\n\n$$\na_{j+4 k}=2^{m} \\Longleftrightarrow 4(q+5 k)=2^{m} \\Longleftrightarrow q+5 k=2^{m-2} \\Longleftrightarrow 2^{m-2}=q \\quad(\\bmod 5) .\n$$\n\nSince $q$ could not be a multiple of 5 , we have $q \\in\\{1,2,3,4\\}(\\bmod 5)$. Since the sequence $2^{m-2}(\\bmod 5)$ is periodic with period $(1,2,4,3)$, we find that $2^{m-2}=q(\\bmod 5)$ happens for infinitely many values of $m$. Hence $2^{m-2}=q+5 k$ is solvable for infinitely many pairs $(k, m)$. Noting that $m$ determines $k$ and that $k$ is nonnegative as soon as $m$ is large enough concludes the proof.\n\nAlternative 3 for (b). We shall show that for any $n>1$ there is some $k \\geq n$ such that $a_{k}$ is a power of 2 . First, we observe that we can always find $m \\in\\{n, n+1, n+2, n+3\\}$ such that $a_{m}$ is divisible by 4 . If $a_{m}$ is not a power of 2 , we write $a_{m}=2^{b} c$ with $b \\geq 2$ and $c>1$ odd. Then we have\n\n$$\na_{m+4 \\cdot 2^{b-2}}=a_{m}+20\\left(2^{b-2}\\right)=2^{b} c+5 \\cdot 2^{b}=2^{b+1} \\frac{c+5}{2} .\n$$\n\nIf $c>5$, we have $\\frac{c+5}{2}<c$ and hence the odd factor of $a_{m+4.2^{b-2}}$ is strictly smaller than the odd factor of $a_{m}$. Therefore there is some $m^{\\prime}>m$ such that $a_{m^{\\prime}}=2^{b^{\\prime}} c^{\\prime}$ with $c^{\\prime}$ odd and $\\leq 5$. The case $c^{\\prime}=5$ is forbidden. If $c^{\\prime}=1$, then $a_{m^{\\prime}}$ is a power of 2 . If $c^{\\prime}=3$, then $a_{m^{\\prime}+4.2^{b^{\\prime}-2}}=2^{b^{\\prime}+3}$ is a power of 2 ."}
+{"problem": ". Find all quadruples $(a, b, c, d)$ of positive real numbers such that $a b c d=1$, $a^{2012}+2012 b=2012 c+d^{2012}$ and $2012 a+b^{2012}=c^{2012}+2012 d$.", "solution": "Rewrite the last two equations into\n\n$$\na^{2012}-d^{2012}=2012(c-b) \\text { and } c^{2012}-b^{2012}=2012(a-d)\n$$\n\nand observe that $a=d$ holds if and only if $c=b$ holds. In that case, the last two equations are satisfied, and condition $a b c d=1$ leads to a set of valid quadruples of the form $(a, b, c, d)=\\left(t, \\frac{1}{t}, \\frac{1}{t}, t\\right)$ for any $t>0$.\n\nWe show that there are no other solutions. Assume that $a \\neq d$ and $c \\neq b$. Multiply both sides of (1) to obtain\n\n$$\n\\left(a^{2012}-d^{2012}\\right)\\left(c^{2012}-b^{2012}\\right)=2012^{2}(c-b)(a-d)\n$$\n\nand divide the left-hand side by the (nonzero) right-hand side to get\n\n$$\n\\frac{a^{2011}+\\cdots+a^{2011-i} d^{i}+\\cdots+d^{2011}}{2012} \\cdot \\frac{c^{2011}+\\cdots+c^{2011-i} b^{i}+\\cdots+b^{2011}}{2012}=1\n$$\n\nNow apply the arithmetic-geometric mean inequality to the first factor\n\n$$\n\\frac{a^{2011}+\\cdots+a^{2011-i} d^{i}+\\cdots+d^{2011}}{2012}>\\sqrt[2012]{(a d)^{\\frac{2011 \\times 2012}{2}}}=(a d)^{\\frac{2011}{2}}\n$$\n\nThe inequality is strict, since equality holds only if all terms in the mean are equal to each other, which happens only if $a=d$. Similarly, we find\n\n$$\n\\frac{c^{2011}+\\cdots+c^{2011-i} b^{i}+\\cdots b^{2011}}{2012}>\\sqrt[2012]{(c b)^{\\frac{2011 \\times 2012}{2}}}=(c b)^{\\frac{2011}{2}} .\n$$\n\nMultiplying both inequalities, we obtain\n\n$$\n(a d)^{\\frac{2011}{2}}(c b)^{\\frac{2011}{2}}<1\n$$\n\nwhich is equivalent to $a b c d<1$, a contradiction."}
+{"problem": ". In triangle $A B C$ the midpoint of $B C$ is called $M$. Let $P$ be a variable interior point of the triangle such that $\\angle C P M=\\angle P A B$. Let $\\Gamma$ be the circumcircle of triangle $A B P$. The line $M P$ intersects $\\Gamma$ a second time in $Q$. Define $R$ as the reflection of $P$ in the tangent to $\\Gamma$ in $B$. Prove that the length $|Q R|$ is independent of the position of $P$ inside the triangle.", "solution": "We claim $|Q R|=|B C|$, which will clearly imply that quantity $|Q R|$ is independent from the position of $P$ inside triangle $\\triangle A B C$ (and independent from the position of $A$ ).\n\nThis equality will follow from the equality between triangles $\\triangle B P C$ and $\\triangle R B Q$. This in turn will be shown by means of three equalities (two sides and an angle): $|B P|=|R B|,|P C|=|B Q|$ and $\\angle B P C=\\angle R B Q$.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_bcff5e53df3a142fd72cg-4.jpg?height=848&width=1190&top_left_y=992&top_left_x=433)\n\n(a) $|B P|=|R B|$\n\nObvious since $R$ is the reflection of $P$ in a line going through $B$.\n\n(b) $|P C|=|B Q|$\n\nLet $U$ be the fourth vertex of parallelogram $B P C U$. Then $U$ is on line $P Q$ and $\\angle B U P=$ $\\angle U P C=\\alpha$. If $Q$ is on the same $\\operatorname{arc} P B$ as $A$, then $\\angle B Q P=\\alpha$, and $\\triangle Q P U$ is isosceles; hence, $|B Q|=|B U|=|P C|$. On the other way, if $Q$ is on the other $\\operatorname{arc} P B$, then $\\angle B Q P$ and $\\alpha$ are supplementary, hence $B Q U=\\alpha$, and again $\\triangle Q P U$ is isosceles; the same conclusion follows.\n\n(c) $\\angle B P C=\\angle R B Q$\n\nDefine $T$ to be the midpoint of $P R$. Then line $B T$, tangent to circle $\\Gamma$ in $B$, splits $\\angle R B Q$ into two parts, $\\angle R B T$ and $\\angle T B Q$.\n\nWe first show that $\\angle R B T=\\alpha$. Indeed, by symmetry, $\\angle R B T=\\angle P B T$ and, since $B T$ is tangent to $\\Gamma$, we have that $\\angle P B T=\\angle P A B$ (because they both intercept the same arc $\\widehat{P B}$ on circle $\\Gamma$ ), from which our claim follows.\n\nWe then show that $\\angle T B Q=\\angle B P M$. Indeed, since $\\angle T B Q$ and $\\angle B P Q$ intercept opposite arcs on circle $\\Gamma$, they are supplementary and we have $\\angle T B Q=\\pi-\\angle B P Q=\\angle B P M$. We finally conclude that\n\n$$\n\\angle R B Q=\\angle R B T+\\angle T B Q=\\alpha+\\angle B P M=\\angle M P C+\\angle B P M=\\angle B P C .\n$$\n\nWe have thus shown $\\triangle B P C=\\triangle R B Q$, which completes the proof.\n\nNote. Notice that point $A$ does not play any role in the problem except fixing circle $\\Gamma$ (and, for that reason, the result is also valid when $P$ is chosen outside of triangle $\\triangle A B C$ ).\n\nAlternative 1 for (b). The law of sines in triangle $\\triangle B Q M$ gives\n\n$$\n\\frac{|B M|}{\\sin \\angle B Q M}=\\frac{|B Q|}{\\sin \\angle B M Q}\n$$\n\nSince $Q$ belongs to circle $\\Gamma$, we have either $\\angle B Q P=\\angle B A P=\\alpha$, hence $\\angle B Q M=\\angle M P C$, or these angles are supplementary; in both cases they have equal sines. We also have that $\\angle B M Q$ and $\\angle C M P$ are supplementary, hence have equal sines. Using these facts along with $|B M|=|M C|$ transforms (2) into\n\n$$\n\\frac{|M C|}{\\sin \\angle M P C}=\\frac{|B Q|}{\\sin \\angle C M P}\n$$\n\nfrom which the law of sines in triangle $\\triangle C P M$ implies that $|B Q|=|P C|$.\n\n\n#### Abstract\n\nAlternative 2 for (b). Let $S$ be the second intersection of line $C P$ with circle $\\Gamma$. Then, $\\angle B S P=\\alpha$, so $B S$ and $M P$ are parallel; since $M$ is the midpoint of segment $B C, P$ is the midpoint of $S C$. If $Q$ is on the same $\\operatorname{arc} P B$ as $A$, then the quadrilateral $Q P B S$ is an isosceles trapezoid, and $|Q B|=|S P|=|P C|$. If $Q$ is on the other $\\operatorname{arc} P B$, then the quadrilateral $P Q B S$ is an isosceles trapezoid, and again $|Q B|=|S P|=|P C|$."}
+{"problem": ". Yesterday, $n \\geq 4$ people sat around a round table. Each participant remembers only who his two neighbours were, but not which one sat on his left and which one sat on his right. Today, you would like the same people to sit around the same round table so that each participant has the same two neighbours as yesterday (it is possible that yesterday's lefthand side neighbour is today's right-hand side neighbour). You are allowed to query some of the participants: if anyone is asked, he will answer by pointing at his two neighbours from yesterday.\n\n(a) Determine the minimal number $f(n)$ of participants you have to query in order to be certain to succeed, if later questions must not depend on the outcome of the previous questions. That is, you have to choose in advance the list of people you are going to query, before effectively asking any question.\n\n(b) Determine the minimal number $g(n)$ of participants you have to query in order to be certain to succeed, if later questions may depend on the outcome of previous questions. That is, you can wait until you get the first answer to choose whom to ask the second question, and so on.\n\n#", "solution": "(a) $f(n)=n-3$.\n\n- Asking $n-4$ questions is not enough since the $n-4$ people queried might be sitting in a consecutive string, in which case the $n-4$ answers allow one to sit $n-2$ people in the same positions as yesterday, but there is still an ambiguity among the two remaining ones.\n- Let us show that $n-3$ questions suffice. Among the 3 people who are not queried, at least 2 must sit next to people who have been queried. If exactly 2 do, then both these people must be neighbours of the third, so that the neighbours of everybody are known and we are done. If all 3 unqueried people sit next to a queried person, then at least one of them has two queried neighbours, and again it follows that the neighbours of everybody are known, so that we are done.\n\n(b) $g(n)=n-1-\\left\\lceil\\frac{n}{3}\\right\\rceil\\left(=n-1-\\left\\lfloor\\frac{n+2}{3}\\right\\rfloor=\\left\\lfloor\\frac{2 n}{3}\\right\\rfloor-1=\\left\\lceil\\frac{2 n-5}{3}\\right\\rceil\\right)$.\n\nSay there is a link between two people if and only if they are neighbours. There are in total $n$ links, which we all need to identify. By asking a person for his neighbours, we can discover at most two new links. More precisely, if at any point we query a participant who has not yet been pointed as a neighbour, we discover exactly two new links (we call this a type-0 query). If we query a participant who has been pointed once as a neighbour, will discover exactly one new link (we call this a type-1 query). Of course, querying a participant who has already been pointed twice provides no information (and we assume in the rest of this solution that it never happens).\n\nFirst note that, since $f(4)=1$, we also have $g(4)=1$. We now prove the formula for $g(n)$ for $n \\geq 5$.\n\n- Let us show that $n-1-\\left\\lceil\\frac{n}{3}\\right\\rceil$ questions suffice. Our strategy consists in making sure that the first $\\left\\lceil\\frac{n}{3}\\right\\rceil$ queries are type-0. Let us show that this is always possible. A type-0 query requires a participant that hasn't been queried or pointed before. Since the number of those participants decreases by three at most after each query, we see that it is always possible to perform $\\left\\lceil\\frac{n}{3}\\right\\rceil$ type- 0 queries first. During this phase we discover $2\\left\\lceil\\frac{n}{3}\\right\\rceil$ links.\n\nThe remaining queries will be either type- 0 or type-1, and each of them discovers at least one new link. We perform them until $n-1$ links have been discovered, after which we are done (the last link can be deduced without query). The number of queries in this second phase is therefore at $\\operatorname{most}^{1} n-1-2\\left\\lceil\\frac{n}{3}\\right\\rceil$, and the total is at most $\\left\\lceil\\frac{n}{3}\\right\\rceil+\\left(n-1-2\\left\\lceil\\frac{n}{3}\\right\\rceil\\right)=n-1-\\left\\lceil\\frac{n}{3}\\right\\rceil$.\n\n- We now show that $n-2-\\left\\lceil\\frac{n}{3}\\right\\rceil=\\hat{g}(n)$ questions are not enough.\n\n(i) Consider the pool of unqueried and unpointed participants ; each type-0 must query this pool. Since, from the point of view of the questioner, all elements of the pool are undistinguishable, we can assume that each type-0 query asks the second leftmost participant in the pool (except if there is only one element left in the pool). One can then check that the pool, which starts as a string of $n$ contiguous participants, will stay contiguous after each type- 0 and type1 query. Furthermore, using our assumption, we see that each type-0 query removes three participants from the pool. Therefore there can be at most $\\left\\lceil\\frac{n}{3}\\right\\rceil$ type-0 queries in the scenarios corresponding to our assumption.\n\n(ii) Assume there are $k$ type- 0 queries. Since there are $\\hat{g}(n)$ queries, the number of discovered links is equal to $2 k+(\\hat{g}(n)-k)=\\hat{g}(n)+k=n-2+k-\\left\\lceil\\frac{n}{3}\\right\\rceil$. If $k$ is strictly less than $\\left\\lceil\\frac{n}{3}\\right\\rceil$, we discover strictly less than $n-2$ links, which is clearly insufficient (indeed, there are at least three missing links, and one can check that whatever the configuration of the missing links, there are always several orders compatible with the discovered links).\n\n(iii) We now analyze the remaining case with $k=\\left\\lceil\\frac{n}{3}\\right\\rceil$ type-0 queries ${ }^{2}$, in which we discover $n-2$ links. On the one hand, if the missing links are disjoint, there are always two orders compatible with the discovered links (for example when $n=7$ and links are missing between the $(4,5)$ and $(7,1)$ pairs of neighbours, the two orders are $1-2-3-4 \\quad 5-6-7$ and $1-2-3-4 \\quad 7-6-5)$. On the other hand, a situation where the two missing links would be adjacent would allow the identification of the correct order. However, this never happens in the scenarios corresponding to the assumption we made in (i). Indeed, two adjacent missing links imply that some participant is unqueried and unpointed at the end of the process. Since we perform $k=\\left\\lceil\\frac{n}{3}\\right\\rceil$ type-0 queries (the maximum), the reasoning from (i) shows that the pool of unqueried and unpointed participants is empty at the end of the process, which contradicts the existence of two adjacent missing links.[^0]\n\n\n[^0]:    ${ }^{1}$ Here we use the assumption $n \\geq 5$, since quantity $n-1-2\\left\\lceil\\frac{n}{3}\\right\\rceil$ is negative when $n=4$.\n\n    ${ }^{2}$ Note that this cannot happen when $n \\in\\{4,5,7\\}$ since we have $\\left\\lceil\\frac{n}{3}\\right\\rceil>\\hat{g}(n)$ in those cases."}

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2013-zz.jsonl

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+{"problem": ". Let $n \\geqslant 3$ be an integer. A frog is to jump along the real axis, starting at the point 0 and making $n$ jumps: one of length 1 , one of length $2, \\ldots$, one of length $n$. It may perform these $n$ jumps in any order. If at some point the frog is sitting on a number $a \\leqslant 0$, its next jump must be to the right (towards the positive numbers). If at some point the frog is sitting on a number $a>0$, its next jump must be to the left (towards the negative numbers). Find the largest positive integer $k$ for which the frog can perform its jumps in such an order that it never lands on any of the numbers $1,2, \\ldots, k$.", "solution": "We claim that the largest positive integer $k$ with the given property is $\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor$, where $\\lfloor x\\rfloor$ is by definition the largest integer not exceeding $x$.\n\nConsider a sequence of $n$ jumps of length $1,2, \\ldots n$ such that the frog never lands on any of the numbers $1,2, \\ldots, k$, where $k \\geqslant 1$. Note that we must have $k<n$ in order for the frog to be able to make its first jump. As the frog jumps to the right only if it is in a number $a \\leqslant 0$, and the largest jump has length $n$, it is impossible to reach numbers greater than $n$. On the other hand, suppose the frog is in a number $a>0$, then it must even be in a number $a \\geqslant k+1$, since it is not allowed to hit the numbers $1,2, \\ldots, k$. So the frog jumps to the left only if it is in a number $a \\geqslant k+1$, and therefore it is impossible to reach numbers less than $(k+1)-n=k-n+1$. This means the frog only possibly lands on the numbers $i$ satisfying\n\n$$\nk-n+1 \\leqslant i \\leqslant 0 \\quad \\text { or } \\quad k+1 \\leqslant i \\leqslant n \\text {. }\n$$\n\nWhen performing a jump of length $k$, the frog has to remain at either side of the numbers $1,2, \\ldots, k$. Indeed, jumping over $1,2, \\ldots, k$ requires a jump of at least length $k+1$. In case it starts at a number $a>0$ (in fact $k+1 \\leqslant a \\leqslant n$ ), it lands in $a-k$ and we must also have $a-k \\geqslant k+1$. So $2 k+1 \\leqslant a \\leqslant n$, therefore $2 k+1 \\leqslant n$. In case it starts at a number $a \\leqslant 0$ (in fact $k-n+1 \\leqslant a \\leqslant 0$ ), it lands in $a+k$ and we must also have $a+k \\leqslant 0$. Adding $k$ to both sides of $k-n+1 \\leqslant a$, we obtain $2 k-n+1 \\leqslant a+k \\leqslant 0$, so in this case we have $2 k+1 \\leqslant n$ as well. We conclude that $k \\leqslant \\frac{n-1}{2}$. Since $k$ is integer, we even have $k \\leqslant\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor$.\n\nNext we prove that this upperbound is sharp: for $k=\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor$ the frog really can perform its jumps in such an order that it never lands on any of the numbers $1,2, \\ldots, k$.\n\nSuppose $n$ is odd, then $\\frac{n-1}{2}$ is an integer and we have $k=\\frac{n-1}{2}$, so $n=2 k+1$. We claim that when the frog performs the jumps of length $1, \\ldots, 2 k+1$ in the following order, it does never land on $1,2, \\ldots, k$ : it starts with a jump of length $k+1$, then it performs two jumps, one of length $k+2$ followed by one of length 1 , next two jumps of length $k+3$ and $2, \\ldots$, next two jumps of length $k+(i+1)$ and $i, \\ldots$, and finally two jumps of length $k+(k+1)$ and $k$. In this order of the jumps every length between 1 and $n=2 k+1$ does occur: it performs a pair of jumps for $1 \\leqslant i \\leqslant k$, which are the jumps of length $1,2, \\ldots$, $k$ and the jumps of length $k+2, k+3, \\ldots, 2 k+1$, and it starts with the jump of length $k+1$.\n\nWe now prove the correctness of this jumping scheme. After the first jump the frog lands in $k+1>k$. Now suppose the frog is in 0 or $k+1$ and is about to perform the pair of jumps of length $k+(i+1)$ and $i$. Starting from 0 , it lands in $k+(i+1)>k$, after which it lands in $(k+i+1)-i=k+1>k$. If on the contrary it starts in $k+1$, it lands in $(k+1)-(k+(i+1))=-i<1$, after which it lands in $(-i)+i=0$. We see that, starting in 0 , the frog lands in $k+1$ after the pair of jumps, while starting in $k+1$ the frog lands in 0 , while in both cases the jumps do not touch $1,2, \\ldots k$. This proves the correctness of its series of jumps. As the frog (after its first jump) alters between $k+1$ and 0 exactly $k$ times, for odd $k$ it will end up in 0 , while for even $k$ it will end up in $k+1$.\n\nSuppose $n$ is even, then $\\frac{n-1}{2}$ is not an integer and we have $k=\\frac{n-1}{2}-\\frac{1}{2}=\\frac{n-2}{2}$, so $n=2 k+2$. Let the frog firstly perform the same series of jumps as in the previous case; they still do not touch $1,2, \\ldots, k$. Now let the frog make a final extra jump of length $2 k+2$. It will land in $0+(2 k+2)=2 k+2>k$ if $k$ is odd, or in $(k+1)-(2 k+2)=-k-1<1$ if $k$ is even, and its series of jumps is correct again.\n\nWe conclude that the largest positive integer $k$ with the given property is $\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor$."}
+{"problem": ". Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that\n\n$$\nf(x+y)+y \\leqslant f(f(f(x)))\n$$\n\nholds for all $x, y \\in \\mathbb{R}$.", "solution": "Let $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ be a function satisfying the given inequality (2). Writing $z$ for $x+y$, we find that $f(z)+(z-x) \\leqslant f(f(f(x)))$, or equivalently\n\n$$\nf(z)+z \\leqslant f(f(f(x)))+x\n$$\n\nfor all $x, z \\in \\mathbb{R}$. Substituting $z=f(f(x))$ yields $f(f(f(x)))+f(f(x)) \\leqslant f(f(f(x)))+x$, from which we see that\n\n$$\nf(f(x)) \\leqslant x\n$$\n\nfor all $x \\in \\mathbb{R}$. Substituting $f(x)$ for $x$ we get $f(f(f(x))) \\leqslant f(x)$, which combined with (3) gives $f(z)+z \\leqslant f(f(f(x)))+x \\leqslant f(x)+x$. So\n\n$$\nf(z)+z \\leqslant f(x)+x\n$$\n\nfor all $x, z \\in \\mathbb{R}$. By symmetry we see that we also have $f(x)+x \\leqslant f(z)+z$, from which we conclude that in fact we even have\n\n$$\nf(z)+z=f(x)+x\n$$\n\nfor all $x, z \\in \\mathbb{R}$. So $f(z)+z=f(0)+0$ for all $z \\in \\mathbb{R}$, and we conclude that $f(z)=c-z$ for some $c \\in \\mathbb{R}$.\n\nNow we check whether all functions of this form satisfy the given inequality. Let $c \\in \\mathbb{R}$ be given and consider the function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ defined by $f(z)=c-z$ for all $z \\in \\mathbb{R}$. Note that $f(f(z))=c-(c-z)=z$ for all $z \\in \\mathbb{R}$. For the lefthand side of (2) we find\n\n$$\nf(x+y)+y=(c-(x+y))+y=c-x\n$$\n\nwhile the righthand side reads\n\n$$\nf(f(f(x)))=f(x)=c-x .\n$$\n\nWe see that inequality (2) holds; in fact we even have equality here.\n\nWe conclude that the solutions to (2) are given by the functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ defined by $f(z)=c-z$ for all $z \\in \\mathbb{R}$, where $c$ is an arbitrary real constant."}
+{"problem": ". Let $\\triangle A B C$ be a triangle with circumcircle $\\Gamma$, and let $I$ be the center of the incircle of $\\triangle A B C$. The lines $A I, B I$ and $C I$ intersect $\\Gamma$ in $D \\neq A, E \\neq B$ and $F \\neq C$. The tangent lines to $\\Gamma$ in $F, D$ and $E$ intersect the lines $A I, B I$ and $C I$ in $R, S$ and $T$, respectively. Prove that\n\n$$\n|A R| \\cdot|B S| \\cdot|C T|=|I D| \\cdot|I E| \\cdot|I F| .\n$$", "solution": "We first prove that $|D B|=|D I|$. (This may also be claimed by referring to the lemma that $D$ is the centre of the circumcircle of $B I C I_{a}$.) By the constant angle theorem and the fact that $A D$ and $B E$ are angle bisectors of triangle $A B C$, we see that\n\n$$\n\\angle D B I=\\angle D B C+\\angle C B I=\\angle D A C+\\angle C B E=\\angle D A B+\\angle A B E \\text {, }\n$$\n\nwhile\n\n$$\n\\angle D I B=180^{\\circ}-\\angle A I B=\\angle I A B+\\angle A B I=\\angle D A B+\\angle A B E \\text {. }\n$$\n\nSo $\\triangle B D I$ has equal angles $\\angle D B I=\\angle D I B$, so $|D B|=|D I|$. This proves our claim. We similarly deduce that $|E C|=|E I|$ and $|F A|=|F I|$.\n\nRewriting (7) into $\\frac{|A R|}{|I F|} \\cdot \\frac{|B S|}{|I D|} \\cdot \\frac{|C T|}{|I E|}=1$, we see that it suffices to prove that\n\n$$\n\\frac{|A R|}{|A F|} \\cdot \\frac{|B S|}{|B D|} \\cdot \\frac{|C T|}{|C E|}=1\n$$\n\nWe now prove by angle chasing that $\\triangle R F A \\sim \\triangle A C I$. As $R F$ is tangent to the circumcircle of $\\triangle A F C$, we obtain (using also that $C F$ is angle bisector of $\\angle A C B$ )\n\n$$\n\\angle R F A=\\angle F C A=\\angle I C A .\n$$\n\nMoreover, from $|F A|=|F I|$ we deduce that $\\angle F A I=\\angle F I A$, so\n\n$$\n\\angle F A R=180^{\\circ}-\\angle F A I=180^{\\circ}-\\angle F I A=\\angle C I A .\n$$\n\nThis proves our similarity, which entails that $\\frac{|A R|}{|A F|}=\\frac{|I A|}{|I C|}$. In the same way we deduce that $\\frac{|B S|}{|B D|}=\\frac{|I B|}{|I A|}$ and $\\frac{|C T|}{|C E|}=\\frac{|I C|}{|I B|}$. By these equal ratios we know that\n\n$$\n\\frac{|A R|}{|A F|} \\cdot \\frac{|B S|}{|B D|} \\cdot \\frac{|C T|}{|C E|}=\\frac{|I A|}{|I C|} \\cdot \\frac{|I B|}{|I A|} \\cdot \\frac{|I C|}{|I B|}=1,\n$$\n\nwhich proves (8), as required.\n\n#"}
+{"problem": ".\n\na) Find all positive integers $g$ with the following property: for each odd prime number $p$ there exists a positive integer $n$ such that $p$ divides the two integers\n\n$$\ng^{n}-n \\text { and } g^{n+1}-(n+1) \\text {. }\n$$\n\nb) Find all positive integers $g$ with the following property: for each odd prime number $p$ there exists a positive integer $n$ such that $p$ divides the two integers\n\n$$\ng^{n}-n^{2} \\quad \\text { and } \\quad g^{n+1}-(n+1)^{2} \\text {. }\n$$\n\n#", "solution": "a) Let $g$ be a positive integer with the given property. So for each odd prime number $p$ there exists a positive integer $n$ such that $p \\mid g^{n}-n$ and $p \\mid g^{n+1}-(n+1)$.\n\nIf $g$ has an odd prime factor $p$, then from $p \\mid g^{n}-n$ it follows that $p \\mid n$, while from $p \\mid g^{n+1}-(n+1)$ we deduce that $p \\mid n+1$. But $p$ cannot divide both $n$ and $n+1$; contradiction. So $g$ is a power of $2: g=2^{k}$ for some $k \\geqslant 0$.\n\nIf $g=2^{0}=1$, then $p \\mid 1-n$ and $p \\mid 1-(n+1)$, which is again a contradiction.\n\nSuppose $k \\geqslant 2$. Then $g-1$ has an odd prime factor $p$, therefore $g \\equiv 1(\\bmod p)$ so $0 \\equiv g^{n}-n \\equiv 1-n(\\bmod p)$ and $0 \\equiv g^{n+1}-(n+1) \\equiv 1-(n+1)(\\bmod p)$, which is again a contradiction.\n\nNow we prove that $g=2^{1}=2$ does satisfy the condition. Let a prime $p>2$ be given. Choose $n=(p-1)^{2}$, then we have $n \\equiv(-1)^{2}=1(\\bmod p)$. By Fermat's little theorem $($ using $\\operatorname{gcd}(2, p)=1)$ we know that $2^{p-1} \\equiv 1(\\bmod p)$, so\n\n$$\n2^{n}=2^{(p-1)^{2}}=\\left(2^{p-1}\\right)^{p-1} \\equiv 1 \\equiv n \\quad(\\bmod p)\n$$\n\nMultiplying both sides by 2, we see that also\n\n$$\n2^{n+1} \\equiv 2 n=n+n \\equiv n+1 \\quad(\\bmod p) .\n$$\n\nWe conclude that only $g=2$ has the given property.\n\nb) Let $g$ be a positive integer with the given property. So for each odd prime number $p$ there exists a positive integer $n$ such that $p \\mid g^{n}-n^{2}$ and $p \\mid g^{n+1}-(n+1)^{2}$.\n\nIf $g$ has an odd prime factor $p$, then from $p \\mid g^{n}-n^{2}$ it follows that $p \\mid n^{2}$, so also $p \\mid n$, while from $p \\mid g^{n+1}-(n+1)^{2}$ we deduce that $p \\mid(n+1)^{2}$, so also $p \\mid n+1$.\n\nBut $p$ cannot divide both $n$ and $n+1$; contradiction. So $g$ is a power of $2: g=2^{k}$ for some $k \\geqslant 0$.\n\nIf $g=2^{0}=1$, then for any odd prime $p$ we have $p \\mid 1-n^{2}=(1-n)(1+n)$ and $p \\mid 1-(n+1)^{2}=(1-(n+1))(1+(n+1))$. Now take $p=5$. The first statement says that $n \\equiv 1$ or $n \\equiv-1 \\equiv 4(\\bmod 5)$, and the second that $n \\equiv 0$ or $n \\equiv-2 \\equiv 3$ $(\\bmod 5)$. But this yields a contradiction.\n\nIf $g=2^{1}=2$, then for any odd prime $p$ we have $p \\mid 2^{n}-n^{2}$ and $p \\mid 2^{n+1}-(n+1)^{2}$. Now take $p=3$. As $3 \\nmid 2^{n}$ and $3 \\nmid 2^{n+1}$, we know that $3 \\nmid n^{2}$ and $3 \\nmid(n+1)^{2}$. So these two squares must be 1 modulo 3 (as 2 can never be a square modulo 3 ). Therefore also $2^{n}$ and $2^{n+1}$ must be 1 modulo 3 , which gives $2 \\cdot 1 \\equiv 2 \\cdot 2^{n}=2^{n+1} \\equiv 1(\\bmod 3)$; contradiction.\n\nNow suppose $k \\geqslant 2$. Then $g-1$ has an odd prime factor $p$, therefore $g \\equiv 1(\\bmod p)$ so $0 \\equiv g^{n}-n^{2} \\equiv 1-n^{2}=(1-n)(1+n)(\\bmod p)$ and $0 \\equiv g^{n+1}-(n+1)^{2} \\equiv$ $1-(n+1)^{2}=(1-(n+1))(1+(n+1))(\\bmod p)$. Suppose $p \\geqslant 5$. The first statement says that $n \\equiv 1$ or $n \\equiv-1(\\bmod p)$, and the second that $n \\equiv 0$ or $n \\equiv-2(\\bmod p)$. But $n$ can only be congruent to at most one of the numbers $-2,-1,0$ and 1 , since $p \\geqslant 5$; contradiction. We conclude that $p=3$, so $g-1$ contains only prime factors 3. Hence $2^{k}-1=3^{\\ell}$ for some $\\ell>0$. We see that $2^{k}-1 \\equiv(-1)^{k}-1(\\bmod 3)$, while $3^{\\ell} \\equiv 0(\\bmod 3)$. So $k$ has to be even, say $k=2 m$, and our equation becomes $2^{2 m}-1=3^{\\ell}$, or equivalently $\\left(2^{m}-1\\right)\\left(2^{m}+1\\right)=3^{\\ell}$. Not both factors on the left-hand side can be divisible by 3 , so $2^{m}-1=1$ and $2^{m}+1=3^{\\ell}$, so $m=1$. Hence $g=2^{2}=4$.\n\nNow we show that $g=4$ does have the given property. For this we use that $g=2$ is a solution to part (a): for any odd prime $p$ there exists a positive integer $n$ such that\n\n$$\nn \\equiv 2^{n} \\quad(\\bmod p) \\quad \\text { and } \\quad n+1 \\equiv 2^{n+1} \\quad(\\bmod p)\n$$\n\nTaking the square of both congruences, we obtain\n\n$$\nn^{2} \\equiv\\left(2^{n}\\right)^{2}=\\left(2^{2}\\right)^{n}=4^{n} \\quad(\\bmod p)\n$$\n\nand\n\n$$\n(n+1)^{2} \\equiv\\left(2^{n+1}\\right)^{2}=\\left(2^{2}\\right)^{n+1}=4^{n+1} \\quad(\\bmod p)\n$$\n\nas desired.\n\nWe conclude that only $g=4$ has the given property."}

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2014-zz.jsonl

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+{"problem": "Find the smallest possible value of the expression\n\n$$\n\\left\\lfloor\\frac{a+b+c}{d}\\right\\rfloor+\\left\\lfloor\\frac{b+c+d}{a}\\right\\rfloor+\\left\\lfloor\\frac{c+d+a}{b}\\right\\rfloor+\\left\\lfloor\\frac{d+a+b}{c}\\right\\rfloor,\n$$\n\nin which $a, b, c$ and $d$ vary over the set of positive integers.\n\n(Here $\\lfloor x\\rfloor$ denotes the biggest integer which is smaller than or equal to $x$.)\n\n#", "solution": "The answer is 9 .\n\nNotice that $\\lfloor x\\rfloor>x-1$ for all $x \\in \\mathbb{R}$. Therefore the given expression is strictly greater than\n\n$$\n\\frac{a+b+c}{d}+\\frac{b+c+d}{a}+\\frac{c+d+a}{b}+\\frac{d+a+b}{c}-4,\n$$\n\nwhich can be rewritten as\n\n$$\n\\left(\\frac{a}{b}+\\frac{b}{a}\\right)+\\left(\\frac{a}{c}+\\frac{c}{a}\\right)+\\left(\\frac{a}{d}+\\frac{d}{a}\\right)+\\left(\\frac{b}{c}+\\frac{c}{b}\\right)+\\left(\\frac{b}{d}+\\frac{d}{b}\\right)+\\left(\\frac{c}{d}+\\frac{d}{c}\\right)-4 .\n$$\n\nSince $t+\\frac{1}{t} \\geq 2$ for $t>0$, we get that $6 \\cdot 2-4=8$ is a strict lower bound for the given expression; since it takes integral values only, we actually get that 9 is a lower bound.\n\nIt remains to check that 9 can be attained; this happens for $a=b=c=5$ and $d=4$.\n\n#"}
+{"problem": "Let $k \\geq 1$ be an integer.\n\nWe consider $4 k$ chips, $2 k$ of which are red and $2 k$ of which are blue. A sequence of those $4 k$ chips can be transformed into another sequence by a so-called move, consisting of interchanging a number (possibly one) of consecutive red chips with an equal number of consecutive blue chips. For example, we can move from $r \\underline{b b b r} \\underline{r} b$ to $r \\underline{r r} b r \\underline{b b b}$ where $r$ denotes a red chip and $b$ denotes a blue chip.\n\nDetermine the smallest number $n$ (as a function of $k$ ) such that starting from any initial sequence of the $4 k$ chips, we need at most $n$ moves to reach the state in which the first $2 k$ chips are red.\n\n#", "solution": "The answer is $n=k$.\n\nWe will first show that $n \\geq k$. Let us count the number $C$ of times a red chip is directly to the right of a blue chip. In the final position this number equals 0 . In the position brbrbr $\\cdots b r$ this number equals $2 k$. We claim that any move reduces this number by at most 2 . Denote by $R$ the group of red chips and by $B$ the group of blue chips that are interchanged. Any reduction in $C$ must involve a red chip getting rid of its blue left neighbour or a blue chip getting rid of its red right neighbour. This can only happen with the leftmost chip of $R$ (if its left neighbour is blue) and the rightmost chip of $B$ (if its right neighbour is red), but not with the rightmost chip of $R$ (and its right neighbour) or the leftmost chip of $B$ (and its left neighbour). Hence $C$ is reduced by at most 2 in any move. Therefore the number of moves necessary to change $b r b r b r \\cdots b r$ into the final position is at least $\\frac{2 k}{2}=k$.\n\nWe will now show that $n \\leq k$, i.e. that it is always possible to perform at most $k$ moves in order to reach the state in which the first $2 k$ chips are red. Consider the first $2 k$ chips. If at most $k$ of these chips are blue, we can perform one move for each chip, switching it with one of the red chips of the last $2 k$ chips. So then we are done in at most $k$ moves. Now suppose that of the first $2 k$ chips, at least $k+1$ are blue. Then at most $k-1$ chips are red. Hence it is possible to perform at most $k-1$ moves to reach the situation in which the last $2 k$ chips are red. We then perform one final move, switching the first $2 k$ chips and the last $2 k$ chips, ending in the situation in which the first $2 k$ chips are red. Thus, it is always possible to reach the situation in which the first $2 k$ chips are red in at most $k$ steps, hence $n \\leq k$.\n\nWe have now shown that $n \\geq k$ and $n \\leq k$, hence $n=k$ as claimed.\n\n#"}
+{"problem": "Find all positive integers $n>1$ with the following property:\n\nfor each two positive divisors $k, \\ell<n$ of $n$, at least one of the numbers $2 k-\\ell$ and $2 \\ell-k$ is a (not necessarily positive) divisor of $n$ as well.\n\n#", "solution": "If $n$ is prime, then $n$ has the desired property: if $k, \\ell<n$ are positive divisors of a prime $n$, we have $k=\\ell=1$, in which case $2 k-\\ell=1$ is a divisor of $n$ as well.\n\nAssume now that a composite number $n$ has the desired property. Let $p$ be its smallest prime divisor and let $m=n / p$; then $m \\geq p \\geq 2$. Choosing $(k, \\ell)=(1, m)$, we see that at least one of $2 m-1$ and $m-2$ must divide $n$. However, $m<2 m-1<2 m \\leq n$; since $m$ and $n$ are the two biggest positive divisors of $n$, we conclude that $2 m-1$ cannot divide $n$. Therefore $n$ must be divisble by $m-2$. It follows that $m-2 \\mid m p-p(m-2)=2 p$, hence $m-2 \\in\\{1,2, p, 2 p\\}$. We deal with each case separately.\n\n- If $m-2=1$ we have $m=3$. As $p \\leq m$, we have $p \\in\\{2,3\\}$, hence $n=6$ or $n=9$.\n- If $m-2=2$ we have $m=4$, hence $n$ is even and $p=2$, so $n=8$.\n- If $m-2=p$, we may assume that $p>2$ (since we already discussed the case $m-2=2$ ). We have $m=p+2$, hence $n=p(p+2)$. Applying the condition in the problem to the pair $(k, \\ell)=(1, p)$, we get that $p-2$ or $2 p-1$ divides $n$. If $p-2$ divides $n$, we must have $p=3$ since $p$ is the smallest prime divisor of $n$. If $2 p-1$ divides $n$, we must have $p+2=2 p-1$, since $p+2 \\leq 2 p-1<p(p+2)$ and $p+2$ and $p(p+2)$ are the biggest positive divisors of $n$. Therefore $p=3$ in both cases, and we get $n=15$ as a candidate.\n- If finally $m-2=2 p$ we have $m=2 p+2$, so $n$ is even. Hence $p=2$ and $n=12$.\n\nWe conclude that if a composite number $n$ satisfies the condition in the problem statement, then $n \\in\\{6,8,9,12,15\\}$. Choosing $(k, \\ell)=(1,2)$ shows that $n=8$ does not work; choosing $(k, \\ell)=(3,6)$ shows that 12 is not a solution. It is easy to check that $n=6, n=9$ and $n=15$ have the desired property by checking the condition for all possible pairs $(k, \\ell)$.\n\nWe conclude that the solutions are given by the prime numbers, $n=6, n=9$ and $n=15$.\n\n#"}
+{"problem": "Let $A B C D$ be a square. Consider a variable point $P$ inside the square for which $\\angle B A P \\geq 60^{\\circ}$. Let $Q$ be the intersection of the line $A D$ and the perpendicular to $B P$ in $P$. Let $R$ be the intersection of the line $B Q$ and the perpendicular to $B P$ from $C$.\n\n(a) Prove that $|B P| \\geq|B R|$.\n\n(b) For which point(s) $P$ does the inequality in (a) become an equality?\n\n#", "solution": "![](https://cdn.mathpix.com/cropped/2024_04_17_7136869b0fdb987352a0g-4.jpg?height=446&width=920&top_left_y=980&top_left_x=557)\n\nWe claim that $\\triangle A B P$ and $\\triangle R C B$ are similar triangles. Indeed, if we denote the intersection of $B P$ and $C R$ by $S$, then $\\angle R C B=\\angle S C B=90^{\\circ}-\\angle S B C=90^{\\circ}-\\angle P B C=\\angle A B P$. Moreover, the right angles in $P$ and $A$ imply that $A$ and $P$ lie on the circle with diameter $[B Q]$, so either $A B P Q$ or $A Q B P$ is a cyclic, convex quadrilateral. In either case, $A$ and $Q$ lie on the same side of $B P$, so $\\angle P A B=\\angle P Q B$. Since $C R$ and $P Q$ are perpendicular to $B P$, these lines are parallel and hence $\\angle P Q B=\\angle C R B$. Together with $\\angle P A B=\\angle P Q B$ and $\\angle R C B=\\angle A B P$, this implies the claim that $\\triangle A B P \\sim \\triangle R C B$.\n\nThis similarity yields the equality $|A P| /|B R|=|B P| /|B C|$. Since $\\angle B A P \\geq 60^{\\circ}$, we get\n\n$$\n\\begin{aligned}\n0 \\leq & (|A B|-|A P|)^{2}=|A B|^{2}+|A P|^{2}-2 \\cdot|A B| \\cdot|A P| \\\\\n& =|B P|^{2}+2(\\cos \\angle B A P-1) \\cdot|A B| \\cdot|A P| \\\\\n& \\leq|B P|^{2}-|A B| \\cdot|A P|=|B P|^{2}-|B R| \\cdot|B P| .\n\\end{aligned}\n$$\n\nThis implies that $|B P| \\geq|B R|$, as desired.\n\nIn order for equality to occur, one needs equality in each of the inequalities considered above: $|A B|=|A P|$ and $\\angle B A P=60^{\\circ}$. Hence there is exactly one point $P$ for which we have equality; this is the unique point inside the square such that $\\triangle A B P$ is equilateral."}

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2015-zz.jsonl

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+{"problem": ".\n\n## Determine the smallest positive integer $q$ with the following property:\n\n## for every integer $m$ with $1 \\leqslant m \\leqslant 1006$, there exists an integer $n$ such that\n\n$$\n\\frac{m}{1007} q<n<\\frac{m+1}{1008} q .\n$$", "solution": ". For $m=1006$, we have\n\n$$\nq-q / 1007<n<q-q / 1008\n$$\n\nfor some integer $n$. If $q \\leqslant 1007$, then $q-q / 1007$ and $q-q / 1008$ are both numbers that are at least $q-1$ and smaller than $q$, so there can be no integer $n$ in between. Hence $q>1007$ and $q-q / 1008 \\leqslant q-1$, implying that $n<q-1$, and that $q-q / 1007<q-2$. By rearranging terms, we find $q>2014$, and hence $q \\geqslant 2015$.\n\nLet us prove that $q=2015$ works. Indeed, $m q / 1007=2 m+m / 1007<2 m+1$ and $(m+1) q / 1008=2(m+1)-(m+1) / 1008>2 m+1$ for $1 \\leqslant m \\leqslant 1006$. Hence each pair of inequalities can be satisfied by taking $n=2 m+1$. This completes the proof, and shows that the smallest possible of $q$ is indeed 2015."}
+{"problem": ".\n\n## Determine the smallest positive integer $q$ with the following property:\n\n## for every integer $m$ with $1 \\leqslant m \\leqslant 1006$, there exists an integer $n$ such that\n\n$$\n\\frac{m}{1007} q<n<\\frac{m+1}{1008} q .\n$$", "solution": ". For $m=1, \\ldots, 1006$, there must exist an integer $N$ divisible by $1007 \\cdot 1008$ satisfying the double inequality\n\n$$\n1008 m q<N<1007(m+1) q .\n$$\n\nSince $N$ is divisible by 1007 and 1008, we may write\n\n$$\nN=1008 m q+1008 k=1007(m+1) q-1007 \\ell\n$$\n\nwhere $k, \\ell>0$ are integers. Hence\n\n$$\n(1007-m) q=1008 k+1007 \\ell \\geqslant 2015,\n$$\n\nsince $k, \\ell>0$. Choosing $m=1006$ in this last inequality, it follows that $q \\geqslant 2015$. Conversely, for $q=2015=1008+1007,(*)$ can be satisfied by taking $k=\\ell=1007-m$. (Indeed, 1007 and 1008 are coprime, and so integer divisible by 1007 and 1008 is also divisible by their product.) Thus $q=2015$ has the desired property, and we are done.\n\n#"}
+{"problem": ".\n\nLet $A B C$ be an acute triangle with circumcentre $O$. Let $\\Gamma_{B}$ be the circle through $A$ and $B$ that is tangent to $A C$, and let $\\Gamma_{C}$ be the circle through $A$ and $C$ that is tangent to $A B$. An arbitrary line through $A$ intersects $\\Gamma_{B}$ again in $X$ and $\\Gamma_{C}$ again in $Y$. Prove that $|O X|=|O Y|$.\n\nThe following solutions are valid for the configurations appearing in the diagrams.", "solution": ". Let $O_{B}$ and $O_{C}$ denote the respective centres of $\\Gamma_{B}$ and $\\Gamma_{C}$. We shall show that $X O_{B} O$ and $O O_{C} Y$ are congruent. Now $A O_{B} \\perp A C$ since $\\Gamma_{B}$ is tangent to $A C$ and $O O_{C} \\perp A C$ since $O O_{C}$ is the perpendicular bisector of $[A C]$. Hence $A O_{B} \\| O O_{C}$, and similarly, $A O_{C} \\| O O_{B}$. It follows that $A O_{B} O O_{C}$ is a parallelogram. In particular, $\\left|O_{B} X\\right|=\\left|A O_{B}\\right|=\\left|O O_{C}\\right|$ and $\\left|O_{C} Y\\right|=\\left|O_{C} A\\right|=\\left|O O_{B}\\right|$.\n\nIt will therefore suffice to show that the angles $\\angle O O_{B} X$ and $\\angle Y O_{C} O$ are equal. Noting that $\\angle A O_{B} O=\\angle A O_{C} O$ since $A O_{B} O O_{C}$ is a parallelogram, this follows by angle chasing:\n\n$$\n\\begin{aligned}\n\\angle X O_{B} O & =\\angle A O_{B} O-\\angle A O_{B} X=\\angle A O_{C} O-\\left(180^{\\circ}-2 \\angle O_{B} A X\\right) \\\\\n& =\\angle A O_{C} O-180^{\\circ}+2\\left(\\angle O_{B} A O_{C}-\\angle Y A O_{C}\\right) \\\\\n& =\\angle A O_{C} O-180^{\\circ}+2\\left(180^{\\circ}-\\angle A O_{C} O\\right)-\\left(180^{\\circ}-\\angle A O_{C} Y\\right) \\\\\n& =\\angle A O_{C} Y-\\angle A O_{C} O=\\angle O O_{C} Y\n\\end{aligned}\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_a76495cc78e904fb2ba1g-3.jpg?height=957&width=1426&top_left_y=1533&top_left_x=295)\n\nA Variant. As in the main solution, we note that $A O_{B} O O_{C}$ is a parallelogram. Notice that $O_{B}$ and $O_{C}$ lie on the respective perpendicular bisectors of $[A X]$ and $[A Y]$. The following lemma then implies that $O$ lies on the perpendicular bisector of $[X Y]$, completing the proof:\n\nLet $P_{1}, P_{2}, P_{3}$ be three points on a line. Let $O_{1}$ be a point on the perpendicular bisector of $\\left[P_{2} P_{3}\\right]$, let $O_{2}$ be a point on the perpendicular bisector of $\\left[P_{3} P_{1}\\right]$ and let $O_{3}$ be the point in the plane such that $O_{1} P_{3} O_{2} O_{3}$ is a parallelogram. Then $O_{3}$ lies on the perpendicular bisector of $\\left[P_{1} P_{2}\\right]$.\n\nProof. One can choose coordinates such that $P_{3}=(0,0)$ and $P_{1}=(2,0)$. Then $P_{2}=(2 a, 0), O_{1}=(a, b)$ and $O_{2}=(1, c)$ for some $a, b, c \\in \\mathbb{R}$. Hence $O_{3}=(a+1, b+c)$ lies on the perpendicular bisector of $\\left[P_{1} P_{2}\\right]$."}
+{"problem": ".\n\nLet $A B C$ be an acute triangle with circumcentre $O$. Let $\\Gamma_{B}$ be the circle through $A$ and $B$ that is tangent to $A C$, and let $\\Gamma_{C}$ be the circle through $A$ and $C$ that is tangent to $A B$. An arbitrary line through $A$ intersects $\\Gamma_{B}$ again in $X$ and $\\Gamma_{C}$ again in $Y$. Prove that $|O X|=|O Y|$.\n\nThe following solutions are valid for the configurations appearing in the diagrams.", "solution": ". Let $\\alpha=\\angle B A C$. Observe that $\\angle A X B=180^{\\circ}-\\alpha=\\angle C Y A$ by tangential angles. Let $Z$ be the intersection of $B X$ and $C Y$. Thus $Z X Y$ is isosceles with, in particular, $\\angle Z X Y=\\angle Z Y X=\\alpha$. It will thus suffice to show that $O Z$ bisects $\\angle X Z Y$.\n\nNow $\\angle B Z C=2 \\alpha=\\angle B O C$ since $O$ is the circumcentre of $A B C$, and hence $B Z O C$ is cyclic. In particular, since $|O B|=|O C|$, it follows that $\\angle O Z Y=\\angle O B C=90^{\\circ}-\\alpha$. But $\\angle X Z Y=180^{\\circ}-2 \\alpha$, and so $\\angle X Z Y=2 \\angle O Z Y$, which shows that $O Z$ bisects $\\angle X Z Y . \\square$\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_a76495cc78e904fb2ba1g-4.jpg?height=876&width=897&top_left_y=1405&top_left_x=576)"}
+{"problem": ".\n\nLet $A B C$ be an acute triangle with circumcentre $O$. Let $\\Gamma_{B}$ be the circle through $A$ and $B$ that is tangent to $A C$, and let $\\Gamma_{C}$ be the circle through $A$ and $C$ that is tangent to $A B$. An arbitrary line through $A$ intersects $\\Gamma_{B}$ again in $X$ and $\\Gamma_{C}$ again in $Y$. Prove that $|O X|=|O Y|$.\n\nThe following solutions are valid for the configurations appearing in the diagrams.", "solution": ". Consider inversion $\\mathscr{I}$ in a circle centred at $A$. Under $\\mathscr{I}$,\n\n$$\nB \\mapsto B^{\\prime}, \\quad C \\mapsto C^{\\prime}, \\quad O \\mapsto O^{\\prime}, \\quad X \\mapsto X^{\\prime}, \\quad Y \\mapsto Y^{\\prime},\n$$\n\n$\\Gamma_{B} \\mapsto \\gamma_{B}$, a line through $B^{\\prime}$ parallel to $A C^{\\prime}$,\n\n$\\Gamma_{C} \\mapsto \\gamma_{C}$, a line through $C^{\\prime}$ parallel to $A B^{\\prime}$,\n\nNotice that $\\mathscr{I}$ sends the circumcircle of $A B C$ to the line $B^{\\prime} C^{\\prime}$, and hence maps $A O$ to a line perpendicular to $B C$. Further, if $[A D]$ is a diameter of this circumcircle and $D \\mapsto D^{\\prime}$ under $\\mathscr{I}$, then $D^{\\prime}$ lies on $B C$. Also, $|A D|=2|A O|$ implies $\\left|A O^{\\prime}\\right|=2\\left|A D^{\\prime}\\right|$, and hence $O^{\\prime}$ is the image of $A$ under reflection in $B C$. Finally, $\\angle O X A=\\angle X^{\\prime} O^{\\prime} A$ and $\\angle O Y A=\\angle Y^{\\prime} O^{\\prime} A$, and hence $|O X|=|O Y|$ if and only if $O^{\\prime} A$ bissects $\\angle X^{\\prime} O^{\\prime} Y^{\\prime}$ externally. We have thus reduced the problem to the following statement:\n\nIn triangle $A B C$, let $O$ be the reflection of $A$ over $B C, \\gamma_{B}$ be a line parallel to $A C$ through $B$, and $\\gamma_{C}$ be a line parallel to $A B$ through $C$. For an arbitrary line $\\ell$ passing through $A$, let $X$ and $Y$ be the intersections of $\\ell$ with $\\gamma_{B}$ and $\\gamma_{C}$, respectively. Prove that $O A$ bisects $\\angle X O Y$ externally.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_a76495cc78e904fb2ba1g-5.jpg?height=878&width=634&top_left_y=1201&top_left_x=694)\n\nLet $P=\\gamma_{A} \\cap \\gamma_{B}$. By construction, $P O B C$ is an isosceles trapezoid (and therefore cyclic), $\\angle O B X=\\angle O B P=\\angle O C P=\\angle O C Y$. Further, since $X B \\| A C$ and $Y C \\| A B$, we have $\\triangle A B X \\sim \\triangle Y C A$. Therefore, as $|O B|=|A B|$ and $|A C|=|O C|$ by construction,\n\n$$\n\\frac{|O B|}{|B X|}=\\frac{|A B|}{|B X|}=\\frac{|Y C|}{|C A|}=\\frac{|Y C|}{|C O|}\n$$\n\nIt follows that $\\triangle O B X \\sim \\triangle Y C O$. Hence\n\n$$\n\\frac{|O X|}{|O Y|}=\\frac{|X B|}{|O C|}=\\frac{|X B|}{|A C|}=\\frac{|X A|}{|A Y|}\n$$\n\nand so the result follows from the angle bisector theorem.\n\n#"}
+{"problem": ".\n\nDoes there exist a prime number whose decimal representation is of the form $3811 \\cdots 11$ (that is, consisting of the digits 3 and 8 in that order followed by one or more digits 1 )?", "solution": "Write\n\n$$\na(n)=38 \\underbrace{11 \\cdots 11}_{n \\text { digits } 1}\n$$\n\nThere are three cases to consider, depending on the remainder of $n$ upon division by three.\n\n- If $n=3 k+1 \\equiv 1(\\bmod 3)$, then the sum of the digits of $a(n)$ is equal to $3(k+4)$, i.e. divisible by 3 , and hence so is $a(n)$.\n- If $n=3 k+2 \\equiv 2(\\bmod 3)$, then note that $a(2)=3811=3700+111$ is divisible by 37 . By induction, as $a(3 k+2)=1000 a(3 k-1)+111$, it follows that $a(3 k+2)$ is divisible by 37 for each $k \\geqslant 0$.\n- If $n=3 k \\equiv 0(\\bmod 3)$, observe that\n\n$$\n9 a(3 k)=342 \\underbrace{99 \\cdots 99}_{n \\text { digits } 9}=\\left(7 \\cdot 10^{k}\\right)^{3}-1\n$$\n\nwhich is properly divisible by $7 \\cdot 10^{k}-1$, a number that is larger than 9 . Hence $a(3 k)$ admits a non-trivial factor and so is not prime.\n\n#"}
+{"problem": ".\n\nAn arithmetic progression is a set of the form $\\{a, a+d, \\ldots, a+k d\\}$, where $a, d, k$ are positive integers and $k \\geqslant 2$. Thus an arithmetic progression has at least three elements and the successive elements have difference $d$, called the common difference of the arithmetic progression.\n\nLet $n$ be a positive integer. For each partition of the set $\\{1,2, \\ldots, 3 n\\}$ into arithmetic progressions, we consider the sum $S$ of the respective common differences of these arithmetic progressions. What is the maximal value $S$ that can attain?\n\n(A partition of a set $A$ is a collection of disjoint subsets of $A$ whose union is $A$.)", "solution": "The maximum value is $n^{2}$, which is attained for the partition into $n$ arithmetic progressions $\\{1, n+1,2 n+1\\}, \\ldots,\\{n, 2 n, 3 n\\}$, each of difference $n$.\n\nSuppose indeed that the set has been partioned into $N$ progressions, of respective lengths $\\ell_{i}$, and differences $d_{i}$, for $1 \\leqslant i \\leqslant N$. Since $\\ell_{i} \\geqslant 3$,\n\n$$\n2 \\sum_{i=1}^{N} d_{i} \\leqslant \\sum_{i=1}^{N}\\left(\\ell_{i}-1\\right) d_{i}=\\sum_{i=1}^{N} a_{i}-\\sum_{i=1}^{N} b_{i}\n$$\n\nwhere $a_{i}$ and $b_{i}$ denote, respectively, the largest and smallest elements of progression $i$. Now\n\n$$\n\\begin{aligned}\n& \\sum_{i=1}^{N} b_{i} \\geqslant 1+2+\\cdots+N=N(N+1) / 2 \\\\\n& \\sum_{i=1}^{N} a_{i} \\leqslant(3 n-N+1)+\\cdots+3 n=N(6 n-N+1) / 2\n\\end{aligned}\n$$\n\nand thus\n\n$$\n2 \\sum_{i=1}^{N} d_{i} \\leqslant N(3 n-N) \\leqslant 2 n^{2}\n$$\n\nas $N(3 n-N)$ is an increasing of $N$ on the interval $[0,3 n / 2]$ and since $N \\leqslant n$. This completes the proof.\n\nRemark. The maximising partition is in fact unique: from the above, it is immediate that all maximal partitions must have $n_{i}=3$ and hence $n=N$. It follows that the minimal and maximal elements of the arithmetic progressions of a maximal partition are precisely $1,2, \\ldots, n$ and $2 n+1,2 n+2, \\ldots, 3 n$, respectively. Consider the progression $\\{n+1-d, n+1, n+1+d\\}$ of difference $d$. Then $n+1-d \\geqslant 1$ and $n+1+d \\geqslant 2 n+1$,\nwhich implies $d=n$, and hence $\\{1, n+1,2 n+1\\}$ is an element of any maximal partition. By induction, it follows similarly that $\\{k, n+k, 2 n+k\\}$ is an element of any maximal partition for $1 \\leqslant k \\leqslant n$, since integers allowing for smaller or larger differences have already been used up. This proves the partition $\\{1, n+1,2 n+1\\}, \\ldots,\\{n, 2 n, 3 n\\}$ is the unique maximising partition, as claimed."}

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2016-zz.jsonl

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+{"problem": ". Find the greatest positive integer $N$ with the following property: there exist integers $x_{1}, \\ldots, x_{N}$ such that $x_{i}^{2}-x_{i} x_{j}$ is not divisible by 1111 for any $i \\neq j$.", "solution": "We prove that the greatest $N$ with the required property is $N=1000$. First note that $x_{i}^{2}-x_{i} x_{j}=x_{i}\\left(x_{i}-x_{j}\\right)$, and that the prime factorisation of 1111 is $11 \\cdot 101$.\n\nWe first show that we can find 1000 integers $x_{1}, x_{2}, \\ldots, x_{1000}$ such that $x_{i}^{2}-x_{i} x_{j}$ is not divisible by 1111 for any $i \\neq j$. Consider the set $\\{1,2, \\ldots, 1110\\}$. This set contains 10 integers divisible by 101, and it contains 100 integers divisible by 11 . None of the integers in the set are divisible by both 11 and 101. If we delete all of these $10+100$ integers from the set, we are left with 1000 integers. Call these $x_{1}, x_{2}, \\ldots, x_{1000}$. Now we have $11 \\nmid x_{i}$ and $101 \\nmid x_{i}$ for all $i$. Suppose there are $i \\neq j$ with $1111 \\mid x_{i}\\left(x_{i}-x_{j}\\right)$, then we must have $1111 \\mid x_{i}-x_{j}$, which is a contradiction, since $x_{i}, x_{j} \\in\\{1,2, \\ldots, 1110\\}$. So this set satisfies the requirement.\n\nWe now prove that given 1001 (or more) integers $x_{1}, x_{2}, \\ldots, x_{1001}$ there are $i \\neq j$ with $1111 \\mid x_{i}\\left(x_{i}-x_{j}\\right)$. Suppose for a contradiction that for all indices $i \\neq j$, we have that $x_{i}\\left(x_{i}-x_{j}\\right)$ is not divisible by 1111, and write $X=\\left\\{x_{1}, \\ldots, x_{1001}\\right\\}$. We may (after reducing modulo 1111 if necessary) assume that $x_{i} \\in\\{0,1, \\ldots, 1110\\}$ for all $i$. Then we know that $x_{i} \\neq 0$ for all $i$, and $x_{i} \\neq x_{j}$ for all $i \\neq j$. Suppose for some $i$ we have $11 \\mid x_{i}$. (Since $x_{i} \\neq 0$, we know that $101 \\nmid x_{i}$.) Then any integer $a \\neq x_{i}$ with $a \\equiv x_{i} \\bmod 101$ cannot be an element of $X$, since $1111 \\mid x_{i}\\left(x_{i}-a\\right)$. In $\\{1,2, \\ldots, 1110\\}$ there are 10 such integers, all of them coprime with $11 \\cdot 101$. If there are exactly $k$ different values of $i$ such that $11 \\mid x_{i}$, there are $10 k$ different integers from $\\{1,2, \\ldots, 1110\\}$ that cannot be elements of $X$, all of them coprime with $11 \\cdot 101$. Similarly, if there are exactly $m$ different values of $i$ such that $101 \\mid x_{i}$, then there are $100 \\mathrm{~m}$ different integers from $\\{1,2, \\ldots, 1110\\}$ that cannot be elements of $X$, all of them coprime with $11 \\cdot 101$. (Note that those $10 k$ and $100 \\mathrm{~m}$ integers can overlap.)\n\nIn $\\{1,2, \\ldots, 1110\\}$ there are 100 multiples of 11 , there are 10 multiples of 101 and there is no multiple of $11 \\cdot 101$, so there are 1000 integers that are coprime with $11 \\cdot 101$. In $X$ we have $1001-k-m$ integers that are coprime with $11 \\cdot 101$, so exactly $k+m-1$ of the coprime integers in $\\{1,2, \\ldots, 1110\\}$ are not in $X$. This implies that $10 k \\leqslant k+m-1$ and $100 m \\leqslant k+m-1$. Adding these two inequalities we find $8 k+98 m \\leqslant-2$, a clear contradiction. So $N<1001$."}
+{"problem": ". Let $n$ be a positive integer. Suppose that its positive divisors can be partitioned into pairs (i.e. can be split in groups of two) in such a way that the sum of each pair is a prime number. Prove that these prime numbers are distinct and that none of these are a divisor of $n$.", "solution": "Let $d_{1}$ and $d_{2}$ be positive divisors of $n$ that form a pair as given in the problem. If $d_{1}$ and $d_{2}$ have a non-trivial prime divisor $p$ in common, then $p \\mid d_{1}+d_{2}$ and $p \\leqslant d_{1}<d_{1}+d_{2}$, so $d_{1}+d_{2}$ cannot be prime. Hence $\\operatorname{gcd}\\left(d_{1}, d_{2}\\right)=1$, which implies that $d_{1} d_{2} \\mid n$. Suppose the number of positive divisors of $n$ is $2 t$ (it is even since the divisors can be split into pairs). If we now multiply all divisors, then on one hand we have the product of all $d_{1} d_{2}$ where $\\left\\{d_{1}, d_{2}\\right\\}$ is a pair, so that product is at most $n^{t}$. On the other hand for every divisor $d$ there is another divisor $\\frac{n}{d}$ (since the number of divisors is even, the case $d=\\frac{n}{d}$ does not occur), and the product of all those is equal to $n^{t}$. Hence there must be equality in every inequality $d_{1} d_{2} \\leqslant n$. So the pairs of divisors given in the problem are all of the form $\\left\\{d, \\frac{n}{d}\\right\\}$.\n\nNow we prove the two statements in the problem. Suppose that $d, d^{\\prime}$ are positive divisors of $n$ such that $d+\\frac{n}{d}=d^{\\prime}+\\frac{n}{d^{\\prime}}$. Then $d^{2} d^{\\prime}+n d^{\\prime}=d\\left(d^{\\prime}\\right)^{2}+n d$, so $d d^{\\prime}\\left(d-d^{\\prime}\\right)=n\\left(d-d^{\\prime}\\right)$ and hence $\\left(d d^{\\prime}-n\\right)\\left(d-d^{\\prime}\\right)=0$. Therefore either $d=d^{\\prime}$ or $d d^{\\prime}=n$, which implies that $\\left\\{d, \\frac{n}{d}\\right\\}=\\left\\{d^{\\prime}, \\frac{n}{d^{\\prime}}\\right\\}$, as required.\n\nNow let $d$ be a positive divisor of $n$. Every prime divisor $p$ of $n$ divides precisely one of $d$ and $\\frac{n}{d}$, since $d \\frac{n}{d}=n$ and $\\operatorname{gcd}\\left(d, \\frac{n}{d}\\right)=1$; so $p \\nmid d+\\frac{n}{d}$. Therefore $\\operatorname{gcd}\\left(n, d+\\frac{n}{d}\\right)=1$. Since $d+\\frac{n}{d}>1$ we conclude that $d+\\frac{n}{d}$ cannot be a divisor of $n$."}

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2017-zz.jsonl

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+{"problem": ". Find all functions $f: \\mathbb{Q}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ such that\n\n$$\nf(x y) \\cdot \\operatorname{gcd}\\left(f(x) f(y), f\\left(\\frac{1}{x}\\right) f\\left(\\frac{1}{y}\\right)\\right)=x y f\\left(\\frac{1}{x}\\right) f\\left(\\frac{1}{y}\\right)\n$$\n\nfor all $x, y \\in \\mathbb{Q}_{>0}$, where $\\operatorname{gcd}(a, b)$ denotes the greatest common divisor of $a$ and $b$.", "solution": "Let $f: \\mathbb{Q}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ be a function satisfying\n\n$$\nf(x y) \\cdot \\operatorname{gcd}\\left(f(x) f(y), f\\left(\\frac{1}{x}\\right) f\\left(\\frac{1}{y}\\right)\\right)=x y f\\left(\\frac{1}{x}\\right) f\\left(\\frac{1}{y}\\right)\n$$\n\nfor all $x, y \\in \\mathbb{Q}_{>0}$. Taking $y=\\frac{1}{x}$ in (1), we obtain\n\n$$\nf(1) \\cdot \\operatorname{gcd}\\left(f(x) f\\left(\\frac{1}{x}\\right), f\\left(\\frac{1}{x}\\right) f(x)\\right)=f\\left(\\frac{1}{x}\\right) f(x),\n$$\n\nwhich directly implies that $f(1)=1$. Now if we take $y=1$ in (1), we get\n\n$$\nf(x) \\cdot \\operatorname{gcd}\\left(f(x), f\\left(\\frac{1}{x}\\right)\\right)=x f\\left(\\frac{1}{x}\\right) .\n$$\n\nReplacing $x$ with $\\frac{1}{x}$ in the latter equation, we also get\n\n$$\nf\\left(\\frac{1}{x}\\right) \\cdot \\operatorname{gcd}\\left(f\\left(\\frac{1}{x}\\right), f(x)\\right)=\\frac{1}{x} f(x) .\n$$\n\nComparing (2) and (3), we deduce that\n\n$$\nf(x)^{2}=x^{2} f\\left(\\frac{1}{x}\\right)^{2}\n$$\n\nwhich can be rewritten as\n\n$$\nf(x)=x f\\left(\\frac{1}{x}\\right)\n$$\n\nfor all $x \\in \\mathbb{Q}_{>0}$. We can therefore replace $f(x)$ with $x f\\left(\\frac{1}{x}\\right)$ in (2), and this gives us\n\n$$\n\\operatorname{gcd}\\left(x f\\left(\\frac{1}{x}\\right), f\\left(\\frac{1}{x}\\right)\\right)=1\n$$\n\nWhen $x \\in \\mathbb{Z}_{>0}$, the greatest common divisor of $x f\\left(\\frac{1}{x}\\right)$ and $f\\left(\\frac{1}{x}\\right)$ is $f\\left(\\frac{1}{x}\\right)$, so that we find\n\n$$\nf\\left(\\frac{1}{x}\\right)=1 \\quad \\text { for all } x \\in \\mathbb{Z}_{>0}\n$$\n\nUsing (4), we then also get that\n\n$$\nf(x)=x \\quad \\text { for all } x \\in \\mathbb{Z}_{>0} .\n$$\n\nFinally, given a positive rational number $q=\\frac{m}{n}$ with $m, n \\in \\mathbb{Z}_{>0}$, we can find the value of $f\\left(\\frac{m}{n}\\right)$ by taking $x=m$ and $y=\\frac{1}{n}$ in the initial equation (1):\n\n$$\nf\\left(\\frac{m}{n}\\right)=\\frac{m}{\\operatorname{gcd}(m, n)}\n$$\n\nIf $m$ and $n$ are chosen to be coprime, then we obtain the nice formula\n\n$$\nf\\left(\\frac{m}{n}\\right)=m \\quad \\text { for all } m, n \\in \\mathbb{Z}_{>0} \\text { such that } \\operatorname{gcd}(m, n)=1\n$$\n\nSo the only function $f: \\mathbb{Q}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ that can possibly satisfy the statement is the function defined by (7). We claim that this function indeed satisfies the statement. Indeed, if we write $x=\\frac{a}{b}$ and $y=\\frac{c}{d}$ with $\\operatorname{gcd}(a, b)=1$ and $\\operatorname{gcd}(c, d)=1$ in (1), we obtain the equality\n\n$$\nf\\left(\\frac{a c}{b d}\\right) \\cdot \\operatorname{gcd}(a c, b d)=a c,\n$$\n\nwhich is true since\n\n$$\nf\\left(\\frac{a c}{b d}\\right)=f\\left(\\frac{\\frac{a c}{\\operatorname{gcd}(a c, b d)}}{\\frac{b d}{\\operatorname{gcd}(a c, b d)}}\\right)=\\frac{a c}{\\operatorname{gcd}(a c, b d)} .\n$$"}
+{"problem": ". Let $n \\geqslant 2$ be an integer. Alice and Bob play a game concerning a country made of $n$ islands. Exactly two of those $n$ islands have a factory. Initially there is no bridge in the country. Alice and Bob take turns in the following way. In each turn, the player must build a bridge between two different islands $I_{1}$ and $I_{2}$ such that:\n\n- $I_{1}$ and $I_{2}$ are not already connected by a bridge;\n- at least one of the two islands $I_{1}$ and $I_{2}$ is connected by a series of bridges to an island with a factory (or has a factory itself). (Indeed, access to a factory is needed for the construction.)\n\nAs soon as a player builds a bridge that makes it possible to go from one factory to the other, this player loses the game. (Indeed, it triggers an industrial battle between both factories.) If Alice starts, then determine (for each $n \\geqslant 2$ ) who has a winning strategy.\n\n(Note: It is allowed to construct a bridge passing above another bridge.)", "solution": "The only configurations in which a player can only lose are the ones where there are $k$ islands (including one with a factory) all connected to each other, $n-k$ islands (including the one with the other factory) all connected to each other, and no bridge between these two sets of islands. Indeed, in this situation the player can only connect an island from the first component to an island to the other component and thus loses. Let us call such a situation a ( $k, n-k)$-critical configuration (for $k \\in\\{1,2, \\ldots, n-1\\}$ ). In all other configurations, it is possible to construct a bridge that does not connect the two factories.\n\nLet us count the number of bridges that exist in a $(k, n-k)$-critical configuration. We have $k$ islands connected to each other and $n-k$ islands connected to each other, so the total number of bridges is\n\n$$\nN(k, n-k)=\\left(\\begin{array}{c}\nk \\\\\n2\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k \\\\\n2\n\\end{array}\\right)=\\frac{k(k-1)}{2}+\\frac{(n-k)(n-k-1)}{2} .\n$$\n\nWe are interested in the parity of this number, in order to know which turn it is (Alice's or Bob's turn) when a ( $k, n-k)$-critical configuration appears. We observe easily that the number $\\left(\\begin{array}{c}m \\\\ 2\\end{array}\\right)$ is even when $m \\equiv 0$ or $1(\\bmod 4)$, and odd when $m \\equiv 2$ or $3(\\bmod 4)$. Now let us distinguish some cases:\n\n(a) If $n \\equiv 1(\\bmod 4)$, then for any value of $k \\in\\{1, \\ldots, n-1\\}$ the numbers $k$ and $n-k$ will always be both equal to 0 or 1 modulo 4 , or both equal to 2 or 3 modulo 4 . This means that $\\left(\\begin{array}{l}k \\\\ 2\\end{array}\\right)$ and $\\left(\\begin{array}{c}n-k \\\\ 2\\end{array}\\right)$ always have the same parity in this case, and hence that $N(k, n-k)$ is even for any value of $k$. This implies that a $(k, n-k)$-critical configuration can only appear when it is Alice's turn. So Bob will never face such a situation and he will always be able to choose a bridge that does not connect the factories. Thus Bob has a winning strategy: he just needs to avoid the bridges that make him directly lose. At some point Alice will then be faced with a $(k, n-k)$-critical configuration for some $k$.\n\n(b) If $n \\equiv 3(\\bmod 4)$, then $n-k \\equiv 2$ or $3(\\bmod 4)$ as soon as $k \\equiv 1$ or $0(\\bmod 4)$, and vice versa. So $\\left(\\begin{array}{c}k \\\\ 2\\end{array}\\right)$ and $\\left(\\begin{array}{c}n-k \\\\ 2\\end{array}\\right)$ always have different parities, and $N(k, n-k)$ is odd for any value of $k$. Hence, a $(k, n-k)$-critical configuration can only appear when it is Bob's turn, which means that Alice has the winning strategy.\n(c) If $n$ is even, then a similar reasoning does not work. However, we can see that Bob has an easy winning strategy in this case: the mirror strategy. This strategy can be explained as follows. Before starting the game, let us partition the $n$ islands into $\\frac{n}{2}$ pairs of islands, so that the two islands with a factory belong to the same pair. Given an island $I$, we denote by $m(I)$ the island belonging to the same pair as $I$ (and we call $m(I)$ the mirror of $I$ ). Now the strategy of Bob is simple: if Alice constructs a bridge between islands $I_{1}$ and $I_{2}$, then Bob constructs a bridge between $m\\left(I_{1}\\right)$ and $m\\left(I_{2}\\right)$. In order to show that this is a winning strategy, we just need to observe that the configuration will always be symmetric after Bob's turn:\n\n- If there is a bridge between two islands $X$ and $Y$, then there is a bridge between $m(X)$ and $m(Y)$;\n- If an island $X$ is connected by a series of bridges to an island $F$ with a factory, then $m(X)$ is connected by a series of bridges (the mirror bridges) to $m(F)$, which is the other island with a factory.\n\nHence, when Alice constructs a bridge between $I_{1}$ and $I_{2}$ that does not make her lose (in particular $I_{2} \\neq m\\left(I_{1}\\right)$ ), we conclude that the bridge between $m\\left(I_{1}\\right)$ and $m\\left(I_{2}\\right)$ does not already exist and that it does not make Bob lose as well.\n\nAs a conclusion, Alice has a winning strategy when $n \\equiv 3(\\bmod 4)$ and Bob has a winning strategy when $n \\not \\equiv 3(\\bmod 4)$.[^0]"}
+{"problem": ". In the convex quadrilateral $A B C D$ we have $\\angle B=\\angle C$ and $\\angle D=90^{\\circ}$. Suppose that $|A B|=2|C D|$. Prove that the angle bisector of $\\angle A C B$ is perpendicular to $C D$.", "solution": ". Let $P$ be the reflection of $C$ in $A D$; note that $C, D, P$ are collinear as $\\angle D=90^{\\circ}$. Then $|A B|=2|C D|=|C P|$, and since $\\angle A B C=\\angle B C P$ we have that $A B C P$ is an isosceles trapezoid. Hence $A P \\| B C$. In particular, we have $\\angle P A C=\\angle A C B$ and the bisectors of these two angles are parallel. Since triangle $A C P$ is isosceles with $D$ being the midpoint of segment $C P$, the bisector of $\\angle P A C$ is $A D$. Hence, the bisector of $\\angle A C B$ is parallel to $A D$ and therefore perpendicular to $C D$."}
+{"problem": ". In the convex quadrilateral $A B C D$ we have $\\angle B=\\angle C$ and $\\angle D=90^{\\circ}$. Suppose that $|A B|=2|C D|$. Prove that the angle bisector of $\\angle A C B$ is perpendicular to $C D$.", "solution": ". Let $M$ be the midpoint of segment $A B$ and $T$ be the intersection of lines $A C$ and $M D$. We know that $|M B|=|C D|$ and that $\\angle M B C=\\angle B C D$, so $M B C D$ is an isosceles trapezoid. Hence $M D \\| B C$, so it is a midline of $\\triangle A B C$. It follows that $T$ is the midpoint of segment $A C$. As $\\angle A D C=90^{\\circ}$ this implies that $T$ is the circumcentre of $\\triangle A C D$. Now we have $\\angle A C D=\\angle T C D=\\angle T D C=180^{\\circ}-\\angle B C D$. If we denote by $\\ell$ the perpendicular to $C D$ through $C$, then the latter equality implies that $C B$ is the reflection of $C A$ through $\\ell$. Hence $\\ell$ is the bisector of $\\angle A C B$."}
+{"problem": ". In the convex quadrilateral $A B C D$ we have $\\angle B=\\angle C$ and $\\angle D=90^{\\circ}$. Suppose that $|A B|=2|C D|$. Prove that the angle bisector of $\\angle A C B$ is perpendicular to $C D$.", "solution": ". Let $S$ be the intersection of $B C$ and $A D$ and let $M$ be the midpoint of segment $A B$. (Note that $S$ exists, since if $B C \\| A D$ then $A B C D$ must be a rectangle, contradicting $|A B|=2|C D|$.) We know that $|M B|=|C D|$ and that $\\angle M B C=\\angle B C D$, so $M B C D$ is an isosceles trapezoid. Hence $M D \\| B C$, so it is a midline of $\\triangle A B S$. In particular $D$ is the midpoint of segment $A S$, and triangles $A D C$ and $S D C$ are congruent. Hence $\\angle S C D=\\angle A C D$, so $C D$ is the angle bisector of $\\angle A C S$ and is thus perpendicular to the angle bisector of $\\angle A C B$."}
+{"problem": ". In the convex quadrilateral $A B C D$ we have $\\angle B=\\angle C$ and $\\angle D=90^{\\circ}$. Suppose that $|A B|=2|C D|$. Prove that the angle bisector of $\\angle A C B$ is perpendicular to $C D$.", "solution": ". Let $S$ be the intersection of $B C$ and $A D$ and let $F$ be the intersection of $B C$ with the line through $A$ parallel to $C D$. Then $\\angle A F B=\\angle A F C=180^{\\circ}-\\angle F C D=180^{\\circ}-\\angle A B C=$ $\\angle A B F$, so triangle $A B F$ is isosceles with $|A B|=|A F|$. Hence $|A F|=2|C D|$. Since $A F$ is parallel to $C D$, we find that $C D$ is a midline of $\\triangle A S F$, so $D$ is the midpoint of segment $A S$. Now $C D$ must be the perpendicular bisector of $A S$ and therefore it is also the angle bisector of $\\angle A C S$. This implies that it is perpendicular to the angle bisector of $\\angle A C B$."}
+{"problem": ". A Benelux $n$-square (with $n \\geqslant 2$ ) is an $n \\times n$ grid consisting of $n^{2}$ cells, each of them containing a positive integer, satisfying the following conditions:\n\n- the $n^{2}$ positive integers are pairwise distinct;\n- if for each row and each column we compute the greatest common divisor of the $n$ numbers in that row/column, then we obtain $2 n$ different outcomes.\n\n(a) Prove that, in each Benelux $n$-square (with $n \\geqslant 2$ ), there exists a cell containing a number which is at least $2 n^{2}$.\n\n(b) Call a Benelux $n$-square minimal if all $n^{2}$ numbers in the cells are at most $2 n^{2}$. Determine all $n \\geqslant 2$ for which there exists a minimal Benelux $n$-square.", "solution": "Let us associate to each row and each column of a Benelux $n$-square the greatest common divisor of the $n$ numbers in that row/column, and call it the index of the row/column. By definition of a Benelux $n$-square, we know that these $2 n$ indices are different. In particular, there is a row/column whose index is $k \\geqslant 2 n$. The $n$ numbers in that row/column are distinct multiples of $k$, so the greatest of them is $\\geqslant n k \\geqslant 2 n^{2}$, which proves (a).\n\nLet us now consider a minimal Benelux $n$-square, and first assume that $n \\geqslant 3$. From the previous paragraph it is clear that the $2 n$ indices must exactly be $\\{1,2, \\ldots, 2 n\\}$, otherwise we would get a number $\\geqslant(2 n+1) n$ with the same argument. We can assume without loss of generality that $2 n$ is the index of a column. In order to have all numbers $\\leqslant 2 n^{2}$, the numbers in that column must exactly be the $n$ smallest multiple of $2 n$, i.e. $\\left\\{2 n, 4 n, \\ldots, 2 n^{2}\\right\\}$. We now argue that the index $2 n-1$ must correspond to a column. Indeed, if some row had index $2 n-1$, then the cell at the intersection of this row and the column indexed by $2 n$ would contain a number $x$ which is both a multiple of $2 n$ and $2 n-1$. Since $\\operatorname{gcd}(2 n, 2 n-1)=1$, this would imply $x \\geqslant 2 n(2 n-1)>2 n^{2}$, contradicting the fact that the Benelux $n$-square is minimal. With the same argument, $2 n-2$ cannot be the index of a row because $(2 n-1)(2 n-2)>2 n^{2}$ when $n \\geqslant 3$. We now distinguish the cases $n$ odd and $n$ even.\n\n- Suppose $n$ is odd. Then consider the row/column $A$ whose index is $n$. It must contain multiples of $n$, and we know that all multiples of $2 n$ are in the column indexed by $2 n$. First assume that $A$ is a column. Then we deduce that the numbers in $A$ are exactly $n, 3 n, \\ldots,(2 n-1) n$. But all these numbers are odd, so this implies that the indices of all rows are odd, i.e. they must exactly be $1,3, \\ldots, 2 n-1$. This is a contradiction with the fact that $2 n-1$ is the index of a column. Now assume that $A$ is a row. It has an intersection with the column indexed by $2 n$, but the other $n-1$ numbers in $A$ must belong to $\\{n, 3 n, \\ldots,(2 n-1) n\\}$. This implies that the indices of the corresponding $n-1$ columns are all odd, but this is a contradiction with the fact that $2 n-2$ is the index of a column. Hence, there does not exist a minimal Benelux $n$-square when $n$ is odd.\n- Suppose $n$ is even and assume moreover that $n \\geqslant 8$. With the same argument as earlier, we can deduce that $2 n-3$ and then $2 n-4$ are indices of a column (because\n$(2 n-3)(2 n-4)>2 n^{2}$ when $\\left.n \\geqslant 8\\right)$. Now we do the same reasoning as for $n$ odd but with multiples of $n-1$ instead of $n$. In the column whose index is $2 n-2$, we have $n$ multiples of $2 n-2$. There are exactly $n+1$ multiples of $2 n-2$ which are $\\leqslant 2 n^{2}$, since $(n+1)(2 n-2)=2 n^{2}-2$. Hence, this implies that there is only one multiple of $2 n-2$ which is small enough to appear in the square and which is not already in the column indexed by $2 n-2$. Consider the row/column $A$ whose index is $n-1$. If $A$ is a column, then $A$ contains at most one even number (the remaining multiple of $2 n-2$ ) and this implies that at least $n-1$ indices of rows are odd. This is a contradiction with the fact that $2 n-1$ and $2 n-3$ are indices of a column. Finally, if $A$ is a row, then we get at most one column different from the one indexed by $2 n-2$ which can be indexed by an even number. This is a contradiction with the fact that $2 n$ and $2 n-4$ are indices of columns. So there does not exist a minimal Benelux $n$-square when $n$ is even and $\\geqslant 8$.\n\nThe remaining cases are $n \\in\\{2,4,6\\}$ :\n\n- For $n=6$, we know as above that 12,11 and 10 are indices of columns. But then 9,8 and 7 cannot be indices of rows since $9 \\cdot 11,8 \\cdot 11,7 \\cdot 11$ cannot appear in the square. So they also are indices of columns. We can therefore do the same argumentation as for $n \\geqslant 8$ above to prove that there is no minimal Benelux 6 -square.\n- For $n=4$, we have the following minimal Benelux 4-square (numbers between parentheses are indices):\n\n| 5 | 6 | 7 | 8 | $(1)$ |\n| :---: | :---: | :---: | :---: | :---: |\n| 10 | 30 | 14 | 16 | $(2)$ |\n| 15 | 18 | 21 | 24 | $(3)$ |\n| 20 | 12 | 28 | 32 | $(4)$ |\n| $(5)$ | $(6)$ | $(7)$ | $(8)$ |  |\n\n- For $n=2$, we have the following minimal Benelux 2-square:\n\n| 3 | 4 | $(1)$ |\n| :---: | :---: | :---: |\n| 6 | 8 | $(2)$ |\n| $(3)$ | $(4)$ |  |\n\nSo there exist minimal Benelux $n$-squares only for $n \\in\\{2,4\\}$.\n\n\n[^0]:    Alternative for $n$ even. Another strategy for Bob to win when $n$ is even is to ensure that, before each of Alice's turns, the number of islands connected to the island with the first factory is equal to the number of islands connected to the island with the other factory. Let us say that the country is balanced when this is the case. Remark that the country is balanced at the beginning of the game. Let us show that if the country is balanced before some turn of Alice and if she constructs a bridge without losing, then Bob can always play without losing and such that the country remains balanced. There are two different choices that Alice can make:\n\n    - Alice builds a bridge between an island connected to a factory and an island which is not connected to anything. Then, since $n$ is even and as the country was balanced before Alice's turn, there exists at least one island $I$ which is still not connected to anything. Bob can therefore simply build a bridge between $I$ and an island connected to the other factory. The country will then be balanced again.\n    - Alice builds a bridge between two islands connected to the same factory. We claim that, in this case, is is always possible for Bob to build a bridge between two islands connected to the same factory (so that the country remains balanced). Indeed, the only situation where Bob could not do so is the situation where all $k$ islands connected to the first factory are connected to each other and all $k$ islands connected to the second factory are connected to each other. But there are $2 \\cdot\\left(\\begin{array}{l}k \\\\ 2\\end{array}\\right)$ bridges in such a situation, and so it cannot occur after Alice's turn."}

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2018-zz.jsonl

@@ -0,0 +1,26 @@
+{"problem": "(a) Determine the minimal value of\n\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}-2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}-2018\\right),\n$$\n\nwhere $x$ and $y$ vary over the positive reals.\n\n(b) Determine the minimal value of\n\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}+2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}+2018\\right),\n$$\n\nwhere $x$ and $y$ vary over the positive reals.\n\n(Pierre Haas, Luxembourg)\n\n#", "solution": ""}
+{"problem": "(a) Determine the minimal value of\n\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}-2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}-2018\\right),\n$$\n\nwhere $x$ and $y$ vary over the positive reals.\n\n(b) Determine the minimal value of\n\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}+2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}+2018\\right),\n$$\n\nwhere $x$ and $y$ vary over the positive reals.\n\n(Pierre Haas, Luxembourg)\n\n#", "solution": ". By the inequality between arithmetic and quadratic means,\n\n$$\n\\left(x+\\frac{1}{y}\\right)^{2}+\\left(y+\\frac{1}{x}\\right)^{2} \\geqslant \\frac{1}{2}\\left(x+\\frac{1}{y}+y+\\frac{1}{x}\\right)^{2},\n$$\n\nwith equality if and only if $x+1 / y=y+1 / x$, which holds if $x=y$. It follows that\n\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y} \\pm 2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x} \\pm 2018\\right) \\geqslant \\frac{1}{2}\\left(x+\\frac{1}{y}+y+\\frac{1}{x}\\right)^{2} \\pm 2018\\left(x+\\frac{1}{y}+y+\\frac{1}{x}\\right) .\n$$\n\nThe parabola $f(X)=\\frac{1}{2} X^{2} \\pm 2018 X$ attains its minimal value at $X=\\mp 2018$, and increases monotonically away from this minimal value. By the inequality between arithmetic and geometric means, $(x+1 / x)+(y+1 / y) \\geqslant 4$ with equality iff $x=y=1$. Hence\n\n(a) $\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}-2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}-2018\\right) \\geqslant f(2018)=-\\frac{2018^{2}}{2}$,\n\nand equality is attained if $x=y=u$, where $u+1 / u=1009$, a quadratic equation with discriminant $1009^{2}-4>0$, and that therefore has two real solutions which are clearly positive.\n\n(b) $\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}+2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}+2018\\right) \\geqslant f(4)=8080$,\n\nwith equality if and only if $x=y=1$.\n\nRemark. It is easy to see that equality is attained in the first inequality if and only if $x=y$. Indeed, if $x \\lessgtr y$, then $1 / y \\lessgtr 1 / x$, and so $x+1 / y \\lessgtr y+1 / x$. Thus $x+1 / y=y+1 / x$ if and only if $x=y$. This is not required for the solution."}
+{"problem": "(a) Determine the minimal value of\n\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}-2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}-2018\\right),\n$$\n\nwhere $x$ and $y$ vary over the positive reals.\n\n(b) Determine the minimal value of\n\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}+2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}+2018\\right),\n$$\n\nwhere $x$ and $y$ vary over the positive reals.\n\n(Pierre Haas, Luxembourg)\n\n#", "solution": ". By the inequality between arithmetic and geometric means, $x / y+y / x \\geqslant 2$, and hence\n\n$$\n\\left(x+\\frac{1}{y}\\right)^{2}+\\left(y+\\frac{1}{x}\\right)^{2}=x^{2}+\\frac{1}{y^{2}}+y^{2}+\\frac{1}{x^{2}}+2\\left(\\frac{x}{y}+\\frac{y}{x}\\right) \\geqslant x^{2}+\\frac{1}{x^{2}}+y^{2}+\\frac{1}{y^{2}}+4=\\left(x+\\frac{1}{x}\\right)^{2}+\\left(y+\\frac{1}{y}\\right)^{2}\n$$\n\nwith equality if and only if $x=y$. It follows that\n\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}-K\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}-K\\right) \\geqslant\\left[\\left(x+\\frac{1}{x}\\right)^{2}-K\\left(x+\\frac{1}{x}\\right)\\right]+\\left[\\left(y+\\frac{1}{y}\\right)^{2}-K\\left(y+\\frac{1}{y}\\right)\\right] .\n$$\n\n## BxMO 2018: Problems and Solutions\n\nNotice that the parabola $f(X)=X^{2}-K X=(X-K / 2)^{2}-K^{2} / 4$ attains its minimum value at $X=K / 2$, and increases monotonically away from this minimal value. Now $x+1 / x, y+1 / y \\geqslant 2$, so it follows that\n\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}-K\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}-K\\right) \\geqslant\\left\\{\\begin{array}{cc}\n2 f(K / 2)=-K^{2} / 2 & \\text { if } K / 2 \\geqslant 2 \\Longleftrightarrow K \\geqslant 4 \\\\\n2 f(2)=4(2-K) & \\text { if } K / 2 \\leqslant 2 \\Longleftrightarrow K \\leqslant 4 .\n\\end{array}\\right.\n$$\n\nEquality is attained if (and only if)\n\n$$\nx=y=\\frac{K}{4} \\pm \\sqrt{\\frac{K^{2}}{16}-1} \\quad(\\text { if } K \\geqslant 4), \\quad x=y=1 \\quad(\\text { if } K \\leqslant 4),\n$$\n\nand so these lower bounds indeed represent the minimal values. Taking (a) $K=2018$ and (b) $K=-2018$ in this result completes the proof."}
+{"problem": "(a) Determine the minimal value of\n\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}-2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}-2018\\right),\n$$\n\nwhere $x$ and $y$ vary over the positive reals.\n\n(b) Determine the minimal value of\n\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}+2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}+2018\\right),\n$$\n\nwhere $x$ and $y$ vary over the positive reals.\n\n(Pierre Haas, Luxembourg)\n\n#", "solution": "for part (a). Completing squares,\n\n$$\n\\begin{aligned}\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}-2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}-2018\\right) & =\\left(x+\\frac{1}{y}-1009\\right)^{2}+\\left(y+\\frac{1}{x}-1009\\right)^{2}-2 \\cdot 1009^{2} \\\\\n& \\geqslant-2 \\cdot 1009^{2},\n\\end{aligned}\n$$\n\nwith equality if $x+1 / y=y+1 / x=1009$, which holds if $x=y=u$, where $u+1 / u=1009$, a quadratic equation with discriminant $1009^{2}-4>0$, and that therefore has two real solutions which are clearly positive.\n\nRemark. As in Solution 1, equality holds if and only if $x=y$, because, if $x \\lessgtr y$, then $1 / y \\lessgtr 1 / x$, and so $x+1 / y \\lessgtr y+1 / x$. Thus $x+1 / y=y+1 / x$ if and only if $x=y$. This is not required for the solution."}
+{"problem": "(a) Determine the minimal value of\n\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}-2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}-2018\\right),\n$$\n\nwhere $x$ and $y$ vary over the positive reals.\n\n(b) Determine the minimal value of\n\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}+2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}+2018\\right),\n$$\n\nwhere $x$ and $y$ vary over the positive reals.\n\n(Pierre Haas, Luxembourg)\n\n#", "solution": "for part (b). Using the inequality between arithmetic and geometric means,\n\n$$\n\\begin{aligned}\n& \\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}+2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}+2018\\right) \\\\\n& =\\left(x+\\frac{1}{y}\\right)^{2}+\\left(y+\\frac{1}{x}\\right)^{2}+2018\\left[\\left(x+\\frac{1}{x}\\right)+\\left(y+\\frac{1}{y}\\right)\\right] \\geqslant 4 \\frac{x}{y}+4 \\frac{y}{x}+2018(2+2)\n\\end{aligned}\n$$\n\nBut $x / y+y / x \\geqslant 2$ by the same inequality, and hence\n\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}+2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}+2018\\right) \\geqslant 8080,\n$$\n\nwith equality attained if and only if $x=y=1$.\n\n## BxMO 2018: Problems and Solutions\n\n#"}
+{"problem": "In the land of Heptanomisma, four different coins and three different banknotes are used, and their denominations are seven different (non-zero) natural numbers. The denomination of the smallest banknote is greater than the sum of the denominations of the four different coins. A tourist has exactly one coin of each denomination and exactly one banknote of each denomination, but he cannot afford the book on numismatics he wishes to buy. However, the mathematically inclined shopkeeper offers to sell the book to the tourist at a price of his choosing, provided that he can pay this price in more than one way.\n\n(The tourist can pay a price in more than one way if there are two different subsets of his coins and notes, the denominations of which both add up to this price.)\n\n(a) Prove that the tourist can purchase the book if the denomination of each banknote is smaller than 49.\n\n(b) Show that the tourist may have to leave the shop empty-handed if the denomination of the largest banknote is 49.\n\n(Stijn Cambie, Belgium)\n\n#", "solution": "Let the denominations of the coins and notes be $C_{1}<C_{2}<C_{3}<C_{4}$ and $N_{1}<N_{2}<N_{3}$, respectively. Define $C=C_{1}+C_{2}+C_{3}+C_{4}$ to be the largest amount that can be payed with coins only. The condition of the problem is thus $C<N_{1}$.\n\n(a) Suppose to the contrary that, whatever the price of the book, the tourist can pay for it in no more than one way."}
+{"problem": "In the land of Heptanomisma, four different coins and three different banknotes are used, and their denominations are seven different (non-zero) natural numbers. The denomination of the smallest banknote is greater than the sum of the denominations of the four different coins. A tourist has exactly one coin of each denomination and exactly one banknote of each denomination, but he cannot afford the book on numismatics he wishes to buy. However, the mathematically inclined shopkeeper offers to sell the book to the tourist at a price of his choosing, provided that he can pay this price in more than one way.\n\n(The tourist can pay a price in more than one way if there are two different subsets of his coins and notes, the denominations of which both add up to this price.)\n\n(a) Prove that the tourist can purchase the book if the denomination of each banknote is smaller than 49.\n\n(b) Show that the tourist may have to leave the shop empty-handed if the denomination of the largest banknote is 49.\n\n(Stijn Cambie, Belgium)\n\n#", "solution": ". Consider the hands of one or two notes and any number of (or possibly no) coins. Each of them has one of the $N_{2}+N_{3}$ different values $v$ with $N_{1} \\leqslant v<N_{1}+N_{2}+N_{3}$, since $C<N_{1}$. There are $\\left(\\left(\\begin{array}{l}3 \\\\ 1\\end{array}\\right)+\\left(\\begin{array}{l}3 \\\\ 2\\end{array}\\right)\\right) \\cdot 2^{4}=96$ such hands. Hence $95=47+48>N_{2}+N_{3} \\geqslant 96$, which is a contradiction."}
+{"problem": "In the land of Heptanomisma, four different coins and three different banknotes are used, and their denominations are seven different (non-zero) natural numbers. The denomination of the smallest banknote is greater than the sum of the denominations of the four different coins. A tourist has exactly one coin of each denomination and exactly one banknote of each denomination, but he cannot afford the book on numismatics he wishes to buy. However, the mathematically inclined shopkeeper offers to sell the book to the tourist at a price of his choosing, provided that he can pay this price in more than one way.\n\n(The tourist can pay a price in more than one way if there are two different subsets of his coins and notes, the denominations of which both add up to this price.)\n\n(a) Prove that the tourist can purchase the book if the denomination of each banknote is smaller than 49.\n\n(b) Show that the tourist may have to leave the shop empty-handed if the denomination of the largest banknote is 49.\n\n(Stijn Cambie, Belgium)\n\n#", "solution": ". Consider the hands of exactly one note and any number of (or possibly no) coins, as well as the hand consisting of the two smallest notes only, of value $N_{1}+N_{2}$. Each of these has one of the $N_{3}$ different values $v$ with $N_{1} \\leqslant v<N_{3}+N_{1}$, since $C<N_{1}$ and $N_{1}<N_{1}+N_{2}<N_{1}+N_{3}$. Now the number of hands considered above is $3 \\cdot 2^{4}+1=49$, so $N_{3} \\geqslant 49$, a contradiction."}
+{"problem": "In the land of Heptanomisma, four different coins and three different banknotes are used, and their denominations are seven different (non-zero) natural numbers. The denomination of the smallest banknote is greater than the sum of the denominations of the four different coins. A tourist has exactly one coin of each denomination and exactly one banknote of each denomination, but he cannot afford the book on numismatics he wishes to buy. However, the mathematically inclined shopkeeper offers to sell the book to the tourist at a price of his choosing, provided that he can pay this price in more than one way.\n\n(The tourist can pay a price in more than one way if there are two different subsets of his coins and notes, the denominations of which both add up to this price.)\n\n(a) Prove that the tourist can purchase the book if the denomination of each banknote is smaller than 49.\n\n(b) Show that the tourist may have to leave the shop empty-handed if the denomination of the largest banknote is 49.\n\n(Stijn Cambie, Belgium)\n\n#", "solution": ". Consider the $3 \\cdot 2^{4}=48$ hands of exactly two notes and any number of (or possibly no) coins. Each of these has one of the $N_{3}$ different values $v$ with $N_{1}+N_{2} \\leqslant v<N_{1}+N_{2}+N_{3}$, since $C<N_{1}$. Hence $48 \\leqslant N_{3}<49$, so $N_{3}=48$. Next consider the $3 \\cdot 2^{4}=48$ hands of exactly one note and any number of coins. By the above, these cannot have a value greater than or equal to $N_{1}+N_{2}$, since these can be realised using two notes and some coins. Hence they must each have one of the $N_{2}$ different values $v$ with $N_{1} \\leqslant v<N_{1}+N_{2}$. This implies $48 \\leqslant N_{2}<N_{3}=48$, which is a contradiction.\n\n(b) Consider the denominations\n\n$$\nC_{1}=3, \\quad C_{2}=6, \\quad C_{3}=12, \\quad C_{4}=24, \\quad N_{1}=47, \\quad N_{2}=48, \\quad N_{3}=49 .\n$$\n\nThese satisfy the conditions of the problem, with $C=45<N_{1}$."}
+{"problem": "In the land of Heptanomisma, four different coins and three different banknotes are used, and their denominations are seven different (non-zero) natural numbers. The denomination of the smallest banknote is greater than the sum of the denominations of the four different coins. A tourist has exactly one coin of each denomination and exactly one banknote of each denomination, but he cannot afford the book on numismatics he wishes to buy. However, the mathematically inclined shopkeeper offers to sell the book to the tourist at a price of his choosing, provided that he can pay this price in more than one way.\n\n(The tourist can pay a price in more than one way if there are two different subsets of his coins and notes, the denominations of which both add up to this price.)\n\n(a) Prove that the tourist can purchase the book if the denomination of each banknote is smaller than 49.\n\n(b) Show that the tourist may have to leave the shop empty-handed if the denomination of the largest banknote is 49.\n\n(Stijn Cambie, Belgium)\n\n#", "solution": ". Any two amounts that can be obtained using some (or possible no) coins are different, and they are multiples of 3 since each of the coin denominations is. Hence any two such amounts differ by at least 3 . Since $N_{3}-N_{1}=2$, it follows that any two hands using one note and some coins each have a different value (and the same is true for hands using more than one note, since at least one note appears at least twice in any two such\nhands). Finally, $N_{3}+C<N_{1}+N_{2}$, and hence any hand using one note and some coins has a different value from any hand using two notes and some coins. Hence there is no price that the tourist can pay for in more than one way."}
+{"problem": "In the land of Heptanomisma, four different coins and three different banknotes are used, and their denominations are seven different (non-zero) natural numbers. The denomination of the smallest banknote is greater than the sum of the denominations of the four different coins. A tourist has exactly one coin of each denomination and exactly one banknote of each denomination, but he cannot afford the book on numismatics he wishes to buy. However, the mathematically inclined shopkeeper offers to sell the book to the tourist at a price of his choosing, provided that he can pay this price in more than one way.\n\n(The tourist can pay a price in more than one way if there are two different subsets of his coins and notes, the denominations of which both add up to this price.)\n\n(a) Prove that the tourist can purchase the book if the denomination of each banknote is smaller than 49.\n\n(b) Show that the tourist may have to leave the shop empty-handed if the denomination of the largest banknote is 49.\n\n(Stijn Cambie, Belgium)\n\n#", "solution": ". Suppose to the contrary that there are two hands of coins and notes that sum to the same amount. Up to removing coins or notes that appear in both of these hands to obtain two smaller hands summing to the same amount, we may assume that no coin or note appears in both of these hands. Notice that all the denominations, except for $N_{1} \\equiv-1(\\bmod 3)$ and $N_{3} \\equiv 1(\\bmod 3)$, are divisible by 3 . Hence, if $N_{1}$ appears in one hand, $N_{3}$ must appear in the same hand, and vice versa. Hence these two hands are two disjoint subsets of $\\left\\{C_{1}, C_{2}, C_{3}, C_{4}, N_{2}, N_{1}+N_{3}\\right\\}=\\{3,6,12,24,48,96\\}=3\\{1,2,4,8,16,32\\}$ that sum to the same amount, which contradicts the uniqueness of the binary expansion of these two amounts. Hence there is no price that the tourist can pay for in more than one way.\n\nRemark. It is natural to ask whether there are other choices of coins and notes with $N_{3}=49$ in (b) that force the tourist to leave the shop empty-handed. It turns out that the choice in (b) is unique. The argument runs as follows: the inequality $N_{2}+N_{3} \\geqslant 96$ implies that $N_{2}=47$ or $N_{2}=48$. Further, strengthening the bound in (a), we must have\n\n$$\n\\left(N_{2}+N_{3}+C\\right)-\\left(N_{1}+N_{2}\\right)+1 \\geqslant 48 \\quad \\Longrightarrow \\quad C \\geqslant 47+N_{1}-N_{3}=N_{1}-2\n$$\n\nBut $C<N_{1}$, and so $C=N_{1}-1$ or $C=N_{1}-2$. If $N_{2}=47$, then $C=N_{1}-2$ implies $N_{1}+N_{2}=N_{3}+C$, a contradiction. Thus $C=N_{1}-1$ if $N_{2}=47$, and, similarly, $C=N_{1}-2$ if $N_{2}=48$.\n\nIn the first case, $N_{3}+C=48+N_{1}=N_{1}+N_{2}+1(*)$. Hence, by the argument in (a), all the $N_{2}=47$ values $v$ with $N_{1} \\leqslant v<N_{1}+N_{2}$ can be represented (using one note and some coins). Now $C_{1} \\neq 1$ by (*). Since $N_{1}+1$ can be represented, it follows that $N_{2}=N_{1}+1$ and hence $N_{1}=46$. Then $N_{3}=N_{2}+2=N_{1}+3$ implies that $C_{1}>3$. Considering $50=N_{1}+4$ yields $C_{1}=4$. Now $53=N_{3}+4=N_{2}+6,52=N_{2}+5=N_{1}+6$. Hence $C_{2}=6$ is not possible, so $C_{2}=5$. Then $51=N_{2}+C_{1}=N_{1}+C_{2}$, a contradiction.\n\nIn the second case, notice that $N_{2}=48$ implies, by the argument in (a), that all the $N_{2}=48$ values $v$ with $N_{1} \\leqslant v<N_{1}+N_{2}$ can be represented. Clearly, $N_{1}+1$ can only be represented if $C_{1}=1$ or $N_{2}=N_{1}+1$. In the former case, $N_{3}=N_{2}+C_{1}$, a contradiction. Hence $N_{1}=N_{2}-1=47$, and so $C=45$. Considering $N_{1}+3, \\ldots, N_{1}+12$ then successively yields $C_{1}=3, C_{2}=6, C_{3}=12$. Finally, $C_{4}=C-C_{1}-C_{2}-C_{3}=24$, completing the proof.\n\nOne might also ask about other choices of coins and notes, with different values of $N_{3}$, forcing the tourist to leave the shop empty-handed. Numerically, it is easy to tabulate all such choices of coins and notes for small $N_{3}$ :\n\n| $C_{1}$ | $C_{2}$ | $C_{3}$ | $C_{4}$ | $N_{1}$ | $N_{2}$ | $N_{3}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 3 | 6 | 12 | 24 | 47 | 48 | 49 |\n| 1 | 6 | 12 | 24 | 46 | 48 | 50 |\n| 3 | 6 | 12 | 24 | 46 | 48 | 50 |\n| 3 | 6 | 12 | 24 | 48 | 49 | 50 |\n| 3 | 6 | 12 | 25 | 48 | 49 | 50 |\n| 1 | 6 | 12 | 24 | 47 | 49 | 51 |\n| 1 | 6 | 12 | 25 | 47 | 49 | 51 |\n| 3 | 6 | 12 | 24 | 49 | 50 | 51 |\n| 3 | 6 | 12 | 25 | 49 | 50 | 51 |\n| 3 | 6 | 12 | 26 | 49 | 50 | 51 |\n| 3 | 6 | 13 | 25 | 49 | 50 | 51 |\n\n#"}
+{"problem": "Let $A B C$ be a triangle with orthocentre $H$, and let $D, E$, and $F$ denote the respective midpoints of line segments $A B, A C$, and $A H$. The reflections of $B$ and $C$ in $F$ are $P$ and $Q$, respectively.\n\n(a) Show that lines $P E$ and $Q D$ intersect on the circumcircle of triangle $A B C$.\n\n(b) Prove that lines $P D$ and $Q E$ intersect on line segment $A H$.\n\n(Merlijn Staps, the Netherlands)\n\n#", "solution": "(a) Solution 1. Since $F$ is the midpoint of $[A H]$ and $[B P], B H P A$ is a parallelogram. Similarly, $C A Q H$ is a parallelogram, too. Let $O$ denote the circumcentre of triangle $A B C$ and let $R$ be the reflection of $A$ in $O$, so that $R$ lies on the circumcircle of $A B C$. Since $[A R]$ is a diameter of the circumcircle of $A B C, C R \\perp A C$. But $B H \\perp A C$, and so $C R \\| B H$. Similarly, $B R \\| C H$, and thus $B R C H$ is a parallelogram. Since $B H P A$ is a parallelogram, $R C P A$ is a parallelogram, too. In particular, the midpoint $E$ of its diagonal $[A C]$ lies on $[P R]$. Similarly, $D$ lies on $[Q R]$, and so $P E$ and $Q D$ meet at $R$.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_0352e20d3d113aee8848g-06.jpg?height=928&width=788&top_left_y=1272&top_left_x=631)\n\nRemark. Since $A P$ is parallel to the altitude $B H, A P \\perp A C$. Further, $P H \\| A B$ since $A P H B$ is a parallelogram. But $C H \\perp A B$, so $P H \\perp C H$. Hence $A P C H$ is cyclic. Similarly, $A H B Q$ is cyclic, too."}
+{"problem": "Let $A B C$ be a triangle with orthocentre $H$, and let $D, E$, and $F$ denote the respective midpoints of line segments $A B, A C$, and $A H$. The reflections of $B$ and $C$ in $F$ are $P$ and $Q$, respectively.\n\n(a) Show that lines $P E$ and $Q D$ intersect on the circumcircle of triangle $A B C$.\n\n(b) Prove that lines $P D$ and $Q E$ intersect on line segment $A H$.\n\n(Merlijn Staps, the Netherlands)\n\n#", "solution": ". Let $O$ denote the circumcentre of triangle $A B C$. By construction, $D O \\perp A B$ and $F E \\| C H \\perp A B$, and so $D O \\| E F$. Similarly, $O E \\| D F$, so $D O E F$ is a parallelogram. Since $F$ is the midpoint of $[A H]$ and $[B P], B H P A$ is a parallelogram, and so $A P\\|B H\\| D F \\| O E$ and $|A P|=|B H|=2|D F|=2|O E|$. Extending this argument, it follows that triangles $D O E$ and $Q A P$ and have pairwise parallel sides and the ratio of their sides is $1: 2$. Hence there is a homothety with ratio 2 mapping $D O E$ to $Q A P$. The centre $R$ of this homothety is the reflection of $A$ in $O$, which lies on the circumcircle of $A B C$, and also the intersection of lines $P E, Q D$, and $A O$.\n\n## BxMO 2018: Problems and Solutions"}
+{"problem": "Let $A B C$ be a triangle with orthocentre $H$, and let $D, E$, and $F$ denote the respective midpoints of line segments $A B, A C$, and $A H$. The reflections of $B$ and $C$ in $F$ are $P$ and $Q$, respectively.\n\n(a) Show that lines $P E$ and $Q D$ intersect on the circumcircle of triangle $A B C$.\n\n(b) Prove that lines $P D$ and $Q E$ intersect on line segment $A H$.\n\n(Merlijn Staps, the Netherlands)\n\n#", "solution": ". Since $F$ is the midpoint of $[B P]$ and $[C Q], B C P Q$ is a parallelogram. As $D$ and $E$ are the respective midpoints of $[A B]$ and $[A C]$, it follows that $P Q\\|B C\\| D E$ and $|P Q|=|B C|=2|D E|$. Let $R$ denote the intersection of $P E$ and $Q D$. Then $D$ and $E$ are the midpoints of $[R Q]$ and $[R P]$, respectively. In particular, $B R A Q$ is a parallelogram. But $F$ is the midpoint of $[A H]$ and $[C Q]$, so $A Q C H$ is a parallelogram. Hence $B R C H$ is a parallelogram, too, so $R$ is the reflection of $H$ in the midpoint of $[B C]$, which is well-known to lie on the circumcircle of $A B C$."}
+{"problem": "Let $A B C$ be a triangle with orthocentre $H$, and let $D, E$, and $F$ denote the respective midpoints of line segments $A B, A C$, and $A H$. The reflections of $B$ and $C$ in $F$ are $P$ and $Q$, respectively.\n\n(a) Show that lines $P E$ and $Q D$ intersect on the circumcircle of triangle $A B C$.\n\n(b) Prove that lines $P D$ and $Q E$ intersect on line segment $A H$.\n\n(Merlijn Staps, the Netherlands)\n\n#", "solution": ". Since $F$ is the midpoint of $[B P]$ and $[C Q], B C P Q$ is a parallelogram. Let $R$ be the image of $A$ under the translation that accordingly maps $P, Q$ onto $C, B$, respectively. By construction, $A P C R$ is a parallelogram, and so the midpoint $E$ of $[A C]$ lies on its other diagonal $P R$. Similarly, $D$ lies on $Q R$, and so $R$ is the intersection of $P E$ and $Q D$. Moreover, $F$ is the midpoint of $[A H]$ and $[B P]$, so $A P B H$ is a parallelogram. Hence $C R\\|A P\\| B H \\perp A C$, so $\\angle A C R=90^{\\circ}$. Similarly, $\\angle A B R=90^{\\circ}$, so $A B R C$ is cyclic, and thus $R$ lies on the circumcircle of $A B C$, as required."}
+{"problem": "Let $A B C$ be a triangle with orthocentre $H$, and let $D, E$, and $F$ denote the respective midpoints of line segments $A B, A C$, and $A H$. The reflections of $B$ and $C$ in $F$ are $P$ and $Q$, respectively.\n\n(a) Show that lines $P E$ and $Q D$ intersect on the circumcircle of triangle $A B C$.\n\n(b) Prove that lines $P D$ and $Q E$ intersect on line segment $A H$.\n\n(Merlijn Staps, the Netherlands)\n\n#", "solution": ". Let $M$ be the midpoint of $[B C]$. Note that triangle $P H Q$ is the reflection of triangle $B A C$ in $F$. By construction, the sides of triangle $D E M$ are parallel to those of $C B A$, and hence $D E\\|B C\\| Q P$, and, similarly, $D M \\| Q H$ and $E M \\| P H$. Hence triangles $Q H P$ and $D M E$ have pairwise parallel sides and the ratio of their sides is $2: 1$. This implies that there is homothety mapping one onto the other. Its centre is the intersection of $Q D$ and $P E$ and is also the reflection of $H$ in $M$, which is well-known to lie on the circumcircle of $A B C$."}
+{"problem": "Let $A B C$ be a triangle with orthocentre $H$, and let $D, E$, and $F$ denote the respective midpoints of line segments $A B, A C$, and $A H$. The reflections of $B$ and $C$ in $F$ are $P$ and $Q$, respectively.\n\n(a) Show that lines $P E$ and $Q D$ intersect on the circumcircle of triangle $A B C$.\n\n(b) Prove that lines $P D$ and $Q E$ intersect on line segment $A H$.\n\n(Merlijn Staps, the Netherlands)\n\n#", "solution": ". Let $G$ be the centroid of $A B C$, and let $\\mathscr{H}$ be the well-known homothety with ratio -2 and centre $G$ that maps the nine-point circle of $A B C$ onto its circumcircle. Under $\\mathscr{H}, D \\mapsto C, E \\mapsto B$. Denote by $R$ the image of the Euler point $F$ under $\\mathscr{H}$; by construction, $R$ lies on the circumcircle of $A B C$. Further, $\\overrightarrow{D F}=\\frac{1}{2} \\overrightarrow{R C}$ and $\\overrightarrow{E F}=\\frac{1}{2} \\overrightarrow{R B}$. Hence the points of intersection of the pairs of lines $B F, R E$ and $C F, R D$ are $P$ and $Q$, respectively. This completes the proof.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_0352e20d3d113aee8848g-07.jpg?height=922&width=766&top_left_y=1746&top_left_x=659)\n\n## BxMO 2018: Problems and Solutions"}
+{"problem": "Let $A B C$ be a triangle with orthocentre $H$, and let $D, E$, and $F$ denote the respective midpoints of line segments $A B, A C$, and $A H$. The reflections of $B$ and $C$ in $F$ are $P$ and $Q$, respectively.\n\n(a) Show that lines $P E$ and $Q D$ intersect on the circumcircle of triangle $A B C$.\n\n(b) Prove that lines $P D$ and $Q E$ intersect on line segment $A H$.\n\n(Merlijn Staps, the Netherlands)\n\n#", "solution": ". Let $M, N, K$ denote the respective midpoints of $[B C],[D E],[P Q]$. Thus the intersection $R$ of $P E$ and $Q D$ is the reflection of $K$ in $N$. Under reflection in $F, H \\mapsto A$ and $M \\mapsto K$. Hence $A K H M$ is a parallelogram, and so $\\overrightarrow{M N}=\\frac{1}{2} \\overrightarrow{M A}=\\frac{1}{2} \\overrightarrow{H K}$. Hence the intersection of $K N$ and $H M$ is $R$, and $R$ is the reflection of $H$ in $M$, which is well-known to lie on the circumcircle of $A B C$."}
+{"problem": "Let $A B C$ be a triangle with orthocentre $H$, and let $D, E$, and $F$ denote the respective midpoints of line segments $A B, A C$, and $A H$. The reflections of $B$ and $C$ in $F$ are $P$ and $Q$, respectively.\n\n(a) Show that lines $P E$ and $Q D$ intersect on the circumcircle of triangle $A B C$.\n\n(b) Prove that lines $P D$ and $Q E$ intersect on line segment $A H$.\n\n(Merlijn Staps, the Netherlands)\n\n#", "solution": ". Take Cartesian coordinates $B(0,0), C(1,0), A(a, b), H(a, c)$. Then the coordinates of $D, E, F, P, Q$ are successively found to be\n\n$$\nD\\left(\\frac{a}{2}, \\frac{b}{2}\\right), \\quad E\\left(\\frac{1+a}{2}, \\frac{b}{2}\\right), \\quad F\\left(a, \\frac{b+c}{2}\\right), \\quad P(2 a, b+c), \\quad Q(2 a-1, b+c)\n$$\n\nHence the coordinates of the intersection $R(x, y)$ of $Q D$ and $P E$ satisfy\n\n$$\n\\frac{y-\\frac{b}{2}}{b+c-\\frac{b}{2}}=\\frac{x-\\frac{a}{2}}{2 a-1-\\frac{a}{2}}=\\frac{x-\\frac{a+1}{2}}{2 a-\\frac{a+1}{2}} \\quad \\Longrightarrow \\quad R(1-a,-c)\n$$\n\nHence $R$ is the reflection of $H$ in the midpoint $M\\left(\\frac{1}{2}, 0\\right)$ of $[B C]$, and so lies on the circumcircle of $A B C$.\n\n(b) Solution 1. Lines $A F$ and $Q E$ are medians of triangle $C A Q$, and so, by the properties of the centroid, intersect at a point $S$ of $[A F]$ such that $|A S|=2|S F|$. Similarly, lines $A F$ and $P D$ are medians of triangle $B A P$, and so intersect at the same point $S$. Hence $P D$ and $Q E$ intersect on $[A H]$."}
+{"problem": "Let $A B C$ be a triangle with orthocentre $H$, and let $D, E$, and $F$ denote the respective midpoints of line segments $A B, A C$, and $A H$. The reflections of $B$ and $C$ in $F$ are $P$ and $Q$, respectively.\n\n(a) Show that lines $P E$ and $Q D$ intersect on the circumcircle of triangle $A B C$.\n\n(b) Prove that lines $P D$ and $Q E$ intersect on line segment $A H$.\n\n(Merlijn Staps, the Netherlands)\n\n#", "solution": ". By construction, a homothety with ratio 2 centred at $A$ maps $[D F]$ and $[E F]$ onto $[B H]$ and $[C H]$, respectively, which are mapped in turn onto $[P A]$ and $[Q A]$ under reflection in $F$ by the results of (a). The composition of these maps is a homothety with centre $S$ and ratio -2 that maps $D$ and $E$ onto $P$ and $Q$, respectively. Hence $P D$ and $Q E$ intersect at $S$. Since this homothety leaves $A H$ invariant, $S$ lies on this line. Since the homothety has negative ratio, the centre lies on the line segment $[A F]$, completing the proof.\n\nRemark. This is a projective result: triangles $D F E$ and $P A Q$ are in axial perspective (at $\\infty$ ). Hence, by Desargues' theorem, they are in central perspective, and so lines $P D, Q E$, and $A F$ are concurrent."}
+{"problem": "Let $A B C$ be a triangle with orthocentre $H$, and let $D, E$, and $F$ denote the respective midpoints of line segments $A B, A C$, and $A H$. The reflections of $B$ and $C$ in $F$ are $P$ and $Q$, respectively.\n\n(a) Show that lines $P E$ and $Q D$ intersect on the circumcircle of triangle $A B C$.\n\n(b) Prove that lines $P D$ and $Q E$ intersect on line segment $A H$.\n\n(Merlijn Staps, the Netherlands)\n\n#", "solution": ". The intersection $S$ of $P D$ and $Q E$ is the centroid of triangle $P Q R$, where $R$ is the intersection of $P E$ and $Q D$, as in (a). Hence, if $M, N, K$ denote the respective midpoints of $[B C],[D E],[P Q]$, then, by part (a), $\\overrightarrow{K S}=2 \\overrightarrow{S N}$. Moreover, since $K$ is the reflection of $M$ in $F, d(K, A H)=d(M, A H)=2 d(N, A H)$. It follows that $S$ lies on $A H$; since $K$ and $R$ lie on either side of $A M, S$ lies on $[A H]$."}
+{"problem": "Let $A B C$ be a triangle with orthocentre $H$, and let $D, E$, and $F$ denote the respective midpoints of line segments $A B, A C$, and $A H$. The reflections of $B$ and $C$ in $F$ are $P$ and $Q$, respectively.\n\n(a) Show that lines $P E$ and $Q D$ intersect on the circumcircle of triangle $A B C$.\n\n(b) Prove that lines $P D$ and $Q E$ intersect on line segment $A H$.\n\n(Merlijn Staps, the Netherlands)\n\n#", "solution": ". In Cartesian coordinates and using the results of (a), the coordinates of the intersection $S\\left(x^{\\prime}, y^{\\prime}\\right)$ of $Q E$ and $P D$ satisfy\n\n$$\n\\frac{y^{\\prime}-\\frac{b}{2}}{b+c-\\frac{b}{2}}=\\frac{x^{\\prime}-\\frac{a}{2}}{2 a-\\frac{a}{2}}=\\frac{x^{\\prime}-\\frac{a+1}{2}}{2 a-1-\\frac{a+1}{2}} .\n$$\n\nHence $x^{\\prime}=a$, so $S$ lies on line $A H$. Further, $y^{\\prime}=\\frac{1}{3}(2 b+c)$. Without loss of generality, $b>0$. Then $c>0$ by definition, and so $c<y^{\\prime}<b$ or $c>y^{\\prime}>b$. Hence $S$ lies on line segment $[A H]$.\n\nRemark 1. Solutions 1 and 2 for part (b) have not used the fact that $H$ is the orthocentre of triangle $A B C$, and therefore show that the result of (b) remains true if $H$ is replaced with a general point $X$. The first part of the problem is in some sense independent of $H$, too. The argument of solutions 3, 5, and 7 for part (a) can be extended to yield the following result:\n\nLet $A B C$ be a triangle, and let $X$ be a point of the plane. Let $D, E$, and $Y$ denote the respective midpoints of $[A B],[A C]$, and $[A X]$. The reflections of $B$ and $C$ in $Y$ are $P$ and $Q$, respectively. Lines $P E$ and $Q D$ intersect on the circumcircle $\\omega$ of $A B C$ if and only if $X$ lies on the reflection of $\\omega$ in $[B C]$.\n\n## BxMO 2018: Problems and Solutions\n\nRemark 2. The intersection points of $P E, Q D$ and $P D, Q E$ are not well-defined if $P, Q, D, E$ lie on a line. In the notation of the analytic solution, this happens when $\\frac{b}{2}=b+c$, i.e. $c=-\\frac{b}{2}$. Computing the coordinates of $H$ explicitly using $B H \\perp A C$ yields $c=a(1-a) / b$, and so $P, Q, D, E$ are aligned if and only if $b^{2}=2 a(a-1)$, which is the equation of a hyperbola passing through $B$ and $C$. (These triangles have an obtuse angle, so this configuration cannot appear for acute-angled triangles.)\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_0352e20d3d113aee8848g-09.jpg?height=488&width=1088&top_left_y=567&top_left_x=495)\n\n## BxMO 2018: Problems and Solutions\n\n#"}
+{"problem": "An integer $n \\geqslant 2$ having exactly $s$ positive divisors $1=d_{1}<d_{2}<\\cdots<d_{s}=n$ is said to be good if there exists an integer $k$, with $2 \\leqslant k \\leqslant s$, such that $d_{k}>1+d_{1}+\\cdots+d_{k-1}$. An integer $n \\geqslant 2$ is said to be bad if it is not good.\n\n(a) Show that there are infinitely many bad integers.\n\n(b) Prove that, among any seven consecutive integers all greater than 2, there are always at least four good integers.\n\n(c) Show that there are infinitely many sequences of seven consecutive good integers.\n\n(Gerhard Woeginger, Luxembourg)\n\n#", "solution": "(a) Solution 1. We note that $n=2^{m}$ has $m+1$ divisors, $d_{k}=2^{k-1}$ for $1 \\leqslant k \\leqslant m+1$. Thus\n\n$$\n1+d_{1}+\\cdots+d_{k-1}=1+\\left(2^{k-1}-1\\right)=2^{k-1}=d_{k}\n$$\n\nfor each $k \\geqslant 2$, and hence each power of 2 is a bad integer. This exhibits infinitely many bad integers.\n\nRemark. It is true more generally that\n\nIf $n=2^{r} m$, where $m$ is a product of (odd) primes each less than $2^{r+1}$, then $n$ is bad.\n\nThis is an immediate corollary of the previous result and the following observation:\n\nIf $n=2^{r} m$ is bad, where $m$ is odd, then so is $p n$ for any odd prime $p<2^{r+1}$.\n\nProof. Let $D_{K}>1$ be a divisor of $p n$, so $D_{K}=p$ or $D_{K}=d_{k}$ or $D_{K}=p d_{k}$, where $d_{k}>1$ is a divisor of $n$. In the first case, observe that there exists $t<r+1$ such that $2^{t}<p<2^{t+1}$ by assumption. Then $\\left\\{1,2, \\ldots, 2^{t}\\right\\} \\subseteq\\left\\{D_{1}, \\ldots, D_{K-1}\\right\\}$, and so\n\n$$\nD_{K}=p<2^{t+1}=1+\\left(1+2+\\cdots+2^{t}\\right) \\leqslant 1+D_{1}+\\cdots+D_{K-1} .\n$$\n\nIn the final case, $\\left\\{1,2, \\ldots, 2^{t}, p d_{1}, \\ldots, p d_{k-1}\\right\\} \\subseteq\\left\\{D_{1}, \\ldots, D_{K-1}\\right\\}$, and so\n\n$$\n\\begin{aligned}\nD_{K} & =p d_{k} \\leqslant p\\left(1+d_{1}+\\cdots+d_{k-1}\\right)=p+p d_{1}+\\cdots+p d_{k} \\\\\n& <2^{t+1}+p d_{1}+\\cdots+p d_{k-1}=1+\\left(1+\\cdots+2^{t}\\right)+p d_{1}+\\cdots+p d_{k-1}<1+\\left(D_{1}+\\cdots+D_{K-1}\\right) .\n\\end{aligned}\n$$\n\nIn the second case, $\\left\\{d_{1}, \\ldots, d_{k-1}\\right\\} \\subseteq\\left\\{D_{1}, \\ldots, D_{K-1}\\right\\}$ immediately implies the required inequality, and so $p n$ is indeed bad.\n\nThis result is weak, however: only 57931 (6.99\\%) of the 829157 bad numbers not larger than $10^{7}$ are of this form."}
+{"problem": "An integer $n \\geqslant 2$ having exactly $s$ positive divisors $1=d_{1}<d_{2}<\\cdots<d_{s}=n$ is said to be good if there exists an integer $k$, with $2 \\leqslant k \\leqslant s$, such that $d_{k}>1+d_{1}+\\cdots+d_{k-1}$. An integer $n \\geqslant 2$ is said to be bad if it is not good.\n\n(a) Show that there are infinitely many bad integers.\n\n(b) Prove that, among any seven consecutive integers all greater than 2, there are always at least four good integers.\n\n(c) Show that there are infinitely many sequences of seven consecutive good integers.\n\n(Gerhard Woeginger, Luxembourg)\n\n#", "solution": ". We claim that $n=m$ ! is bad for each integer $m \\geqslant 2$. The proof proceeds by induction on $m$, the case $m=2$ being clear. If $D_{K}>1$ is a divisor of $m$ !, then $D_{K}=d_{k}$ or $D_{K}=q$ or $D_{K}=q d_{k}$, where $d_{k}>1$ is a divisor of $(m-1)$ ! and $q>1$ is a divisor of $m$. In the first case, $\\left\\{d_{1}, \\ldots, d_{k-1}\\right\\} \\subseteq\\left\\{D_{1}, \\ldots, D_{K-1}\\right\\}$, so, invoking the inductive hypothesis, $D_{K}=d_{k} \\leqslant 1+\\left(d_{1}+\\cdots+d_{k-1}\\right) \\leqslant 1+\\left(D_{1}+\\cdots+D_{K-1}\\right)$. In the second case, $q \\leqslant m+1$, so $1,2, \\ldots, q-1 \\mid m$ ! and $\\{1,2, \\ldots, q-1\\} \\subseteq\\left\\{D_{1}, \\ldots, D_{K-1}\\right\\}$. But $q^{2}-3 q+2 \\geqslant 0$ for $q \\geqslant 2$, and hence\n\n$$\nD_{K}=q \\leqslant \\frac{q^{2}-q+2}{2}=1+(1+\\cdots+q-1) \\leqslant 1+\\left(D_{1}+\\cdots+D_{K-1}\\right) .\n$$\n\nIn the final case, $\\left\\{1,2, \\ldots, q-1, q d_{1}, \\ldots, q d_{k-1}\\right\\} \\subseteq\\left\\{D_{1}, \\ldots, D_{K-1}\\right\\}$, and so\n\n$$\nD_{K}=q d_{k} \\leqslant q\\left(1+d_{1}+\\cdots+d_{k-1}\\right) \\leqslant 1+(1+\\cdots+q-1)+\\left(q d_{1}+\\cdots+q d_{k-1}\\right)\n$$\n\nand so $D_{K} \\leqslant 1+\\left(D_{1}+\\cdots+D_{K-1}\\right)$, completing the inductive step.\n(b) If $n$ is odd, then $d_{1}=1, d_{2}>2=1+d_{1}$, and $n$ is good. Among any seven consecutive positive integers, there are either four odd integers and three even ones, or four even ones and three odd ones. In the first case, these four odd integers are good. In the second case, we have to show that one of the consecutive even integers $n=2 m, n+2=2(m+1), n+4=2(m+2), n+6=2(m+3)$ is good. Notice that even integers of the form $n=2(6 \\ell \\pm 1)$, for $\\ell \\geqslant 1$, are good, since $d_{2}=2<n$ as $\\ell \\geqslant 1$, but $d_{3}>4=1+1+2$, since they are divisible by neither 3 nor 4 . But at least one $m, m+1, m+2, m+3$ is congruent to $\\pm 1(\\bmod 6)(\\operatorname{and}$ larger than 1 by assumption); this completes the proof.\n\n(c) Solution 1. Let $n=12 q$, an even number. By part (b), $n-3, n-2=2(6 q-1), n-1, n+1, n+2=2(6 q+1), n+3$ are good integers. Take $q>29$ to be prime. Then the divisors of $n$ less than $q$ are precisely the divisors of 12 . Now $q>1+(1+2+3+4+6+12)=29$, and so $n$ is good, too. Since there are infinitely many choices of the prime $q$, there are infinitely many sequences of seven consecutive good integers."}
+{"problem": "An integer $n \\geqslant 2$ having exactly $s$ positive divisors $1=d_{1}<d_{2}<\\cdots<d_{s}=n$ is said to be good if there exists an integer $k$, with $2 \\leqslant k \\leqslant s$, such that $d_{k}>1+d_{1}+\\cdots+d_{k-1}$. An integer $n \\geqslant 2$ is said to be bad if it is not good.\n\n(a) Show that there are infinitely many bad integers.\n\n(b) Prove that, among any seven consecutive integers all greater than 2, there are always at least four good integers.\n\n(c) Show that there are infinitely many sequences of seven consecutive good integers.\n\n(Gerhard Woeginger, Luxembourg)\n\n#", "solution": ". Let $m=2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13$. For any integer $k>0$, the seven consecutive integers $m k+1, m k+2, \\ldots, m k+7$ are good. Indeed, the four odd numbers $m k+1, m k+3, m k+5, m k+7$ are good by part (b). Moreover,\n\n$$\n\\begin{array}{cccccccc}\nn & d_{1} & d_{2} & d_{3} & d_{4} & d_{5} & & \\\\\nm k+2 & 1 & 2 & \\geqslant 16 & & & \\Longrightarrow & d_{3}>1+d_{1}+d_{2}, \\\\\nm k+4 & 1 & 2 & 4 \\\\\nm k+6 & 1 & 2 & 3 & 6 & & \\Longrightarrow & d_{4}>1+d_{1}+d_{2}+d_{3}, \\\\\nm k & \\geqslant 16 & \\Longrightarrow & d_{5}>1+d_{1}+d_{2}+d_{3}+d_{4},\n\\end{array}\n$$\n\nand so $m k+2, m k+4, m k+6$ are good integers, too. Hence there are infinitely many choices of seven consecutive good integers.\n\nRemark. This solution can also be phrased more indirectly in terms of congruence conditions modulo small remainders; the existence of infinitely many appropriate solutions is then guaranteed by the Chinese Remainder Theorem."}
+{"problem": "An integer $n \\geqslant 2$ having exactly $s$ positive divisors $1=d_{1}<d_{2}<\\cdots<d_{s}=n$ is said to be good if there exists an integer $k$, with $2 \\leqslant k \\leqslant s$, such that $d_{k}>1+d_{1}+\\cdots+d_{k-1}$. An integer $n \\geqslant 2$ is said to be bad if it is not good.\n\n(a) Show that there are infinitely many bad integers.\n\n(b) Prove that, among any seven consecutive integers all greater than 2, there are always at least four good integers.\n\n(c) Show that there are infinitely many sequences of seven consecutive good integers.\n\n(Gerhard Woeginger, Luxembourg)\n\n#", "solution": ". Let $m=29$ !. For any integer $k>0$, the seven consecutive integers $m k+1, m k+2, \\ldots, m k+7$ are good. Indeed, the divisors of these numbers are either at most 7 or at least 30 . But $30>1+(1+2+\\cdots+7)=29$, and so these seven integers are good, giving infinitely many sequences of seven consecutive good integers.\n\nRemark. The ideas underlying these different solutions can be extended to show that there are arbitrarily long runs of consecutive good integers:\n\nFor each integer $N$, there exist $N$ consecutive good integers.\n\nProof 1 . Let $M=2+(1+2+\\cdots+N)$, and let $m=M$ !. Any divisor of any of the $N$ consecutive integers $m+1, m+2, \\ldots, m+N$ is either at most $N$ or at least $M$. But $M>1+(1+2+\\cdots+N)$, and so these $N$ consecutive integers are good.\n\nProof 2. Let $2=p_{1}<p_{2}<\\cdots$ denote the prime numbers, and choose positive integers $s$ and $\\alpha_{1}, \\ldots, \\alpha_{s}$ such that $1,2, \\ldots, N$ divide $P=p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\cdots p_{s}^{\\alpha_{s}}$. Let $\\sigma(n)$ denote the sum of the divisors of $n$. Choose an integer $t$ such that $p_{t}>\\sigma(P)$, let $m=p_{1} p_{2} \\cdots p_{t} P$. The $N$ consecutive integers $m+1, m+2, \\ldots, m+N$ are all good. Indeed, for $1 \\leqslant k \\leqslant N$, any divisor of $m+k$ is either at most $k$ or at least $p_{t+1}$. But $p_{t+1}>1+p_{t}>1+\\sigma(P)>1+\\sigma(k)$, since $k \\mid P$. This completes the proof."}

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2019-zz.jsonl

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+{"problem": ".\n\na) Let $a, b, c, d$ be real numbers with $0 \\leqslant a, b, c, d \\leqslant 1$. Prove that\n\n$$\na b(a-b)+b c(b-c)+c d(c-d)+d a(d-a) \\leqslant \\frac{8}{27} .\n$$\n\nb) Find all quadruples $(a, b, c, d)$ of real numbers with $0 \\leqslant a, b, c, d \\leqslant 1$ for which equality holds in the above inequality.", "solution": "Denote the left-hand side by $S$. We have\n\n$$\n\\begin{aligned}\nS & =a b(a-b)+b c(b-c)+c d(c-d)+d a(d-a) \\\\\n& =a^{2} b-a b^{2}+b^{2} c-b c^{2}+c^{2} d-c d^{2}+d^{2} a-d a^{2} \\\\\n& =a^{2}(b-d)+b^{2}(c-a)+c^{2}(d-b)+d^{2}(a-c) \\\\\n& =(b-d)\\left(a^{2}-c^{2}\\right)+(c-a)\\left(b^{2}-d^{2}\\right) \\\\\n& =(b-d)(a-c)(a+c)+(c-a)(b-d)(b+d) \\\\\n& =(b-d)(a-c)(a+c-b-d) .\n\\end{aligned}\n$$\n\nAssume without loss of generality that $a \\geqslant c$, then $a-c \\geqslant 0$. Now we consider two cases.\n\n- Suppose $b-d \\geqslant 0$. Then if $a+c-b-d<0$ we have $S \\leqslant 0$, so we're done. If $a+c-b-d \\geqslant 0$, we use the AM-GM inequality on $a-c, b-d$ and $a+c-b-d$, which yields\n\n$$\n\\sqrt[3]{S}=\\sqrt[3]{(b-d)(a-c)(a+c-b-d)} \\leqslant \\frac{b-d+a-c+a+c-b-d}{3}=\\frac{2 a-2 d}{3} \\leqslant \\frac{2}{3},\n$$\n\nso $S \\leqslant \\frac{8}{27}$.\n\n- Suppose $b-d<0$. Then if $a+c-b-d \\geqslant 0$ we have $S \\leqslant 0$, so we're done. If $a+c-b-d<0$, we use the AM-GM inequality on $a-c, d-b$ and $b+d-a-c$, which yields\n\n$$\n\\sqrt[3]{S}=\\sqrt[3]{(d-b)(a-c)(b+d-a-c)} \\leqslant \\frac{d-b+a-c+b+d-a-c}{3}=\\frac{2 d-2 c}{3} \\leqslant \\frac{2}{3}\n$$\n\nso $S \\leqslant \\frac{8}{27}$.\n\nEquality in the first case occurs when $a-c=b-d=a+c-b-d$ and $a=1, d=0$. Then $1-c=b=1+c-b$, which implies $2 c=b=1-c$ and hence $c=\\frac{1}{3}$. This results in $(a, b, c, d)=\\left(1, \\frac{2}{3}, \\frac{1}{3}, 0\\right)$. Similarly, equality in the second case occurs when $a-c=d-b=b+d-a-c$ and $d=1, c=0$. This yields $(a, b, c, d)=\\left(\\frac{2}{3}, \\frac{1}{3}, 0,1\\right)$. The case $a \\leqslant c$ gives the other two cyclic variants, which gives all quadruples satisfying the equality."}
+{"problem": ". Pawns and rooks are placed on a $2019 \\times 2019$ chessboard, with at most one piece on each of the $2019^{2}$ squares. A rook can see another rook if they are in the same row or column and all squares between them are empty. What is the maximal number $p$ for which $p$ pawns and $p+2019$ rooks can be placed on the chessboard in such a way that no two rooks can see each other?", "solution": "Answer: the maximal $p$ equals $1009^{2}$.\n\nWrite $n=2019$ and $k=1009$; then $n=2 k+1$. We first show that we can place $k^{2}$ pawns and $n+k^{2}$ rooks. Each cell of the chess board has coordinates $(x, y)$ with $1 \\leqslant x, y \\leqslant n$. We colour each cell black or white depending on whether $x+y$ is even or odd.\n\nLet $A$ be cell $(1, k+1), B$ be cell $(k+1,1), C$ be cell $(2 k+1, k+1)$ and $D$ be cell $(k+1,2 k+1)$, and consider the skew square $A B C D$. We place rooks on the cells of this square which have the same colour as $A$, and we place pawns on the other cells of this square. In this way, no rook can see another rook. Now we have placed $p=k^{2}$ pawns and $(k+1)^{2}=k^{2}+(2 k+1)=p+n$ rooks.\n\nNow we show that we can not place more pawns. Observe that in every row the number of rooks exceeds the number of pawns by at most 1 , since there has to be a pawn between every two neighbouring rooks. So the total number of rooks exceeds the number of pawns by at most $n$. On the other hand we are to place $p$ pawns and $p+n$ rooks, so the number of rooks in every row exceeds the number of pawns by exactly 1 . This means that the rooks and pawns alternate, with rooks at the two ends. For the columns the same holds.\n\nConsider the $\\ell$-th row. Let $a$ be the number of pawns in this row and let $b$ be the number of pawns above the $\\ell$-th row. For all these pawns, a rook must be somewhere above it. Counting the rooks directly above these $a+b$ pawns, we conclude that there must be at least $a+b$ rooks in the first $\\ell-1$ rows. In every row the number of rooks exceeds the number of pawns by 1 , so in these first $\\ell-1$ rows we have at least $a+b-(\\ell-1)$ pawns. So $b \\geqslant a+b-(\\ell-1)$, yielding $a \\leqslant \\ell-1$. We conclude that the $\\ell$-th row contains at most $\\ell-1$ pawns. The same holds for the $\\ell$-th row counted from below (row $(n+1)-\\ell$ ): also in this row, there are at most $\\ell-1$ pawns. As $n=2 k+1$, the maximal number $p$ is\n\n$$\n\\sum_{\\ell=1}^{k}(\\ell-1)+\\sum_{\\ell=1}^{k+1}(\\ell-1)=k+2 \\cdot \\frac{1}{2} k(k-1)=k+k(k-1)=k^{2}\n$$"}

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2020-zz.jsonl

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+{"problem": "Find all positive integers $d$ with the following property: there exists a polynomial $P$ of degree $d$ with integer coefficients such that $|P(m)|=1$ for at least $d+1$ different integers $m$.\n\n#", "solution": "Note that $P(x)=c$ for a fixed constant has at most $d$ solutions, since the polynomial $P(x)-c$ of degree $d$ cancels at most $d$ times. This implies that there are integers $m$ satisfying $P(m)=1$, as well as integers $m$ such that $P(m)=-1$.\n\nNext, we prove the following lemma.\n\nLemma. If $a$ and $b$ are integers such that $P(a)=-1$ and $P(b)=1$, then $|b-a| \\leq 2$.\n\nProof Since $b-a \\mid P(b)-P(a)$ by a well-known lemma (corollary of $a-b \\mid a^{n}-b^{n}$ for every integer $n \\geq 0)$, the conclusion follows.\n\nLet us first consider the case that $d \\geq 4$, and assume that there exists a polynomial $\\mathrm{P}$ with at least $d+1 \\geq 5$ solutions to $|P(m)|=1$. Let $a$ and $b$ be the smallest and largest solution respectively. Since $b-a \\geq 4$, we need $P(a)=P(b)$ by the lemma. Without loss of generality (by switching $P$ with $-P$ if necessary) we can assume $P(a)=P(b)=1$. Take a value $m$ such that $P(m)=-1$. Due to the lemma, we need $b-m$ and $m-a$ to be both at most 2 . Since $b-a \\geq 4$, there is only one possibility left in which case $b-a=4$ and thus $d=4$. By considering $P(x-m)$, we can assume $P( \\pm 2)=P( \\pm 1)=1$ and $P(0)=-1$. The unique fourth degree polynomial satisfying these equalities is $P(x)=-0.5\\left(x^{2}-4\\right)\\left(x^{2}-1\\right)+1$ which is not a polynomial with integer coefficients.\n\nFor $1 \\leq d \\leq 3$, there exist polynomials satisfying the conditions.\n\nFor $d=1$ we can take $P_{1}(X)=X$ as $P_{1}(-1)=-1$ and $P_{1}(1)=1$.\n\nFor $d=2, P_{2}(X)=2 X(X-2)+1$ satisfies $P_{2}(0)=1, P_{2}(1)=-1$ and $P_{2}(2)=1$.\n\nFor $d=3$, the polynomial ; $P_{3}(X)=(X+1) X(X-2)+1$ satisfies $P_{3}(-1)=1=P_{3}(0)=P_{3}(2)=1$ and $P_{3}(1)=-1$. So $\\left|P_{3}(m)\\right|=1$ for $m \\in\\{-1,0,1,2\\}$.\n\nThus the integers with the required property are precisely $d=1,2,3$. This completes the proof.\n\n## BxMO 2020: Problems and Solutions\n\n#"}
+{"problem": "Let $N$ be a positive integer. A collection of $4 N^{2}$ unit tiles with two segments drawn on them as shown is assembled into a $2 N \\times 2 N$ board. Tiles can be rotated.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_74a51d5ecbff9af6115bg-3.jpg?height=140&width=146&top_left_y=564&top_left_x=955)\n\nThe segments on the tiles define paths on the board. Determine the least possible number and the largest possible number of such paths.\n\n#", "solution": "Let $p$ denote the number of paths. Notice that there are two types of paths: (1) those that start and end at a point on the boundary of the board and (2) closed paths in the interior of the board. Let $p_{1}, p_{2}$ denote the respective numbers of paths of either type. There are $8 N$ points on the boundary of the board, and each of these is the starting point or endpoint of exactly one path, so $p_{1}=4 N$. Trivially, $p_{2} \\geqslant 0$, so $p=p_{1}+p_{2} \\geqslant 4 N$.\n\nThe paths on the board are made up of $8 N^{2}$ segments in total. There are only 4 possible paths of one segment, in the corners of the board. All other paths on the boundary of the board therefore consist of at least 2 segments. Moreover, all closed paths in the interior of the board consist of at least 4 segments. Hence\n\n$$\n8 N^{2} \\geqslant 4 \\cdot 1+\\left(p_{1}-4\\right) \\cdot 2+p_{2} \\cdot 4 \\Longleftrightarrow p_{2} \\leqslant N^{2}+(N-1)^{2} \\text {, so } p=p_{1}+p_{2} \\leqslant N^{2}+(N+1)^{2} \\text {. }\n$$\n\nWe have thus shown that $4 N \\leqslant p \\leqslant N^{2}+(N+1)^{2}$. These minimum and maximum values can indeed be attained, as shown below for $N=3$.\n![](https://cdn.mathpix.com/cropped/2024_04_17_74a51d5ecbff9af6115bg-3.jpg?height=310&width=808&top_left_y=1772&top_left_x=630)\n\nThe constructions generalise easily.The constructions clearly attain the required bounds because they satisfy the equalities in our arguments showing the two bounds. Indeed the board on the left has $p_{2}=0$ so give the lower bound. The one on the right has exactly 4 boundary paths with one segment, all other $4 N-4$ boundary paths with 2 segments, and all remaining paths with 4 segments.\n\nRemark. The same ideas solve the analogous problem for a $(2 N+1) \\times(2 N+1)$ assembled from $(2 N+1)^{2}$ such tiles. For this board, $p \\geqslant p_{1}=2(2 N+1)$. Next, there are $2(2 N+1)^{2}$ segments, so, again,\n\n$$\n2(2 N+1)^{2} \\geqslant 4 \\cdot 1+\\left(p_{1}-4\\right) \\cdot 2+p_{2} \\cdot 4 \\Longleftrightarrow p_{2} \\leqslant 2 N^{2}+\\frac{1}{2}\n$$\n\nBut $p_{2}$ is an integer, so $p_{2} \\leqslant 2 N^{2}$, and hence $p \\leqslant 2(N+1)^{2}$. We have thus shown that $2(2 N+1) \\leqslant p \\leqslant 2(N+1)^{2}$. The construction for the lower bound is the same as for the $2 N \\times 2 N$ board; the construction for the upper bound, shown for $N=3$, is a small modification of that for the $2 N \\times 2 N$ board and again generalises easily.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_74a51d5ecbff9af6115bg-3.jpg?height=297&width=299&top_left_y=2441&top_left_x=1598)\n\n## BxMO 2020: Problems and Solutions\n\n#"}
+{"problem": "Let $A B C$ be a triangle. The circle $\\omega_{A}$ through $A$ is tangent to line $B C$ at $B$. The circle $\\omega_{C}$ through $C$ is tangent to line $A B$ at $B$. Let $\\omega_{A}$ and $\\omega_{C}$ meet again at $D$. Let $M$ be the midpoint of line segment $[B C]$, and let $E$ be the intersection of lines $M D$ and $A C$. Show that $E$ lies on $\\omega_{A}$.\n\n#", "solution": "Let $N$ be the midpoint of $[A B]$. By tangential angles, $\\angle C B D=\\angle B A D$ and $\\angle D B A=\\angle D C B$, so triangles $D A B$ and $D B C$ are similar. By definition of the midpoints $M, N$, so are triangles $A N D$ and $B M D$. In particular, $\\angle B N D=180^{\\circ}-\\angle D N A=180^{\\circ}-\\angle D M B$, so $B N D M$ is cyclic. But $M N \\| A C$ by construction, so $\\angle D B A=\\angle D B N=\\angle D M N=\\angle E M N=\\angle M E A=\\angle D E A$, hence $E A D B$ is cyclic, completing the proof.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_74a51d5ecbff9af6115bg-4.jpg?height=760&width=1008&top_left_y=1116&top_left_x=547)\n\n#"}
+{"problem": "Let $A B C$ be a triangle. The circle $\\omega_{A}$ through $A$ is tangent to line $B C$ at $B$. The circle $\\omega_{C}$ through $C$ is tangent to line $A B$ at $B$. Let $\\omega_{A}$ and $\\omega_{C}$ meet again at $D$. Let $M$ be the midpoint of line segment $[B C]$, and let $E$ be the intersection of lines $M D$ and $A C$. Show that $E$ lies on $\\omega_{A}$.\n\n#", "solution": "Let $S$ be the point on $\\omega_{C}$ such that $B S$ is parallel to $A C$, and let $E^{\\prime}$ be the reflection of $S$ in $M$ (such that $B S C E^{\\prime}$ is a parallellogram and $E^{\\prime}$ lies on $A C$ ). Now we do some (directed) angle chasing:\n\n$$\n\\begin{aligned}\n\\angle A E^{\\prime} B & =\\angle C E^{\\prime} B\\left(\\text { since } A, E^{\\prime} \\text { and } C \\text { are collineair }\\right) \\\\\n& =\\angle B S C\\left(B S C E^{\\prime} \\text { is a parallellogram }\\right) \\\\\n& =\\angle A B C\\left(\\text { inscribed angles on } \\omega_{C}\\right) \\\\\n& =\\angle A D B\\left(\\text { inscribed angles on } \\omega_{A}\\right)\n\\end{aligned}\n$$\n\nHence $E^{\\prime}$ lies on $\\omega_{A}$. Further $\\angle A E^{\\prime} D=\\angle A B D=\\angle B S D$, and because $A E^{\\prime}$ is parallel to $B S$ we find that $E^{\\prime} D$ is parallel to $S D$. This means that $E^{\\prime}$ lies on $M D$, so $E^{\\prime}=E$ and $E$ lies on $\\omega_{A}$.\n\n## BxMO 2020: Problems and Solutions\n\n#"}
+{"problem": "A divisor $d$ of a positive integer $n$ is said to be a close divisor of $n$ if $\\sqrt{n}<d<2 \\sqrt{n}$. Does there exist a positive integer with exactly 2020 close divisors?\n\n#", "solution": "Let $m$ be an odd integer with exactly 2020 positive divisors and which is (automatically) not a square. For example, $m=3^{2019}$ suffices, but of course there are many alternatives. Now consider $n=2^{k} m$, for some integer $k$ such that $2^{k}>m$. Any divisor of $n$ is then of the form $2^{\\ell} d$ where $d$ is a divisor of $m$. We will now show that for every such divisor $d$, there exists a unique $\\ell$ such that $2^{\\ell} d$ is a close divisor. Because $\\sqrt{n}$ is not an integer, there certainly is a unique integer $a$ such that $\\sqrt{n}<2^{a} d<2 \\sqrt{n}$. Because $2^{k}>m$, we have $m<\\sqrt{n}<2^{k}$. Combining with $1 \\leq d \\leq m$, we find $1<\\frac{\\sqrt{n}}{d}<2^{a}<\\frac{2 \\sqrt{n}}{d}<2 \\cdot 2^{k}$ and so we see that $0<a \\leqslant k$. So $2^{a} d$ is indeed a close divisor. Because $m$ has exactly 2020 divisors, we find that $n$ has exactly 2020 close divisors."}

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2021-zz.jsonl

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+{"problem": "(a) Prove that for all $a, b, c, d \\in \\mathbb{R}$ with $a+b+c+d=0$,\n\n$$\n\\max (a, b)+\\max (a, c)+\\max (a, d)+\\max (b, c)+\\max (b, d)+\\max (c, d) \\geqslant 0\n$$\n\n(b) Find the largest non-negative integer $k$ such that it is possible to replace $k$ of the six maxima in this inequality by minima in such a way that the inequality still holds for all $a, b, c, d \\in \\mathbb{R}$ with $a+b+c+d=0$.\n\n#", "solution": "The left-hand-side of the inequality is invariant under permutations of $a, b, c, d$. We may therefore suppose that $a \\geqslant b \\geqslant c \\geqslant d$, so that the inequality reduces to\n\n$$\n0 \\leqslant 3 a+2 b+c=a+(a+b)+(a+b+c) .\n$$\n\nWe claim that each of the terms on the right-hand side is non-negative; this will prove the inequality. Indeed, if $a<0$, then $a, b, c, d<0$, and so $a+b+c+d<0$, a contradiction. Also, if $a+b<0$, then, as $b \\leqslant a$, $0>b \\geqslant c, d$ so $(a+b)+c+d<0$, another contradiction. Finally, if $a+b+c<0$, then, as above, $0>c \\geqslant d$, so $(a+b+c)+d<0$, a final contradiction.\n\nNext, we claim that it is impossible to replace $k \\geqslant 3$ maxima by minima in the inequality. Indeed, if $k \\geqslant 3$, one number, say $d$, appears in two of the terms changed to minima. Take $a=b=c=1, d=-3$, so that $a+b+c+d=0$. Then the sum is at most $4 \\cdot 1+2 \\cdot(-3)<0$. Hence $k<3$.\n\nFinally, we prove that the required inequality holds if $k=2$ and the terms involving the complementary sets $\\{a, b\\},\\{c, d\\}$ are changed to minima. We will assume again that $a \\geqslant b \\geqslant c \\geqslant d$, and prove that the inequality holds for any change of the terms involving permutations of these sets to minima (rather than proving that it holds for all orderings of $a, b, c, d$ for this one change). There are three cases:\n\n(1) change terms $\\{a, b\\},\\{c, d\\}$, so the inequality becomes $2 a+3 b+d \\geqslant 0 \\Longleftrightarrow a+b+(b-c) \\geqslant 0$;\n\n(2) change terms $\\{a, c\\},\\{b, d\\}$, so the inequality becomes $2 a+b+2 c+d \\geqslant 0 \\Longleftrightarrow a+c \\geqslant 0$;\n\n(3) change terms $\\{a, d\\},\\{b, c\\}$, so the inequality becomes $2 a+b+2 c+d \\geqslant 0 \\Longleftrightarrow a+c \\geqslant 0$,\n\nwhere we have used $a+b+c+d=0$. In the first case, inequality holds since $a+b \\geqslant 0$ as proved earlier and $b \\geqslant c$. In the other cases, suppose that $a+c<0$. Then $c<-a$ and hence $d<-a$ as $d \\leqslant c$. Hence $c+d<-2 a \\leqslant-a-b$ as $a \\geqslant b$, so $a+b+c+d<0$, a final contradiction, completing the proof.\n\n#"}
+{"problem": "(a) Prove that for all $a, b, c, d \\in \\mathbb{R}$ with $a+b+c+d=0$,\n\n$$\n\\max (a, b)+\\max (a, c)+\\max (a, d)+\\max (b, c)+\\max (b, d)+\\max (c, d) \\geqslant 0\n$$\n\n(b) Find the largest non-negative integer $k$ such that it is possible to replace $k$ of the six maxima in this inequality by minima in such a way that the inequality still holds for all $a, b, c, d \\in \\mathbb{R}$ with $a+b+c+d=0$.\n\n#", "solution": "Using the inequality $\\max (x, y) \\geqslant \\frac{1}{2}(x+y)$, we find that\n\n$$\n\\max (a, b)+\\max (a, c)+\\max (a, d)+\\max (b, c)+\\max (b, d)+\\max (c, d) \\geqslant \\frac{3}{2}(a+b+c+d)=0\n$$\n\nFor $k=3$, we take the same counterexample as in Solution 1. Now it remains to prove the inequality where $\\max (a, b)$ and $\\max (c, d)$ are replaced by $\\min (a, b)$ and $\\min (c, d)$. We can assume without loss of generality that $\\min (a, b)=a$ and $\\min (c, d)=c$. Now we find\n\n$$\n\\begin{aligned}\n& \\min (a, b)+\\max (a, c)+\\max (a, d)+\\max (b, c)+\\max (b, d)+\\min (c, d) \\geqslant \\\\\n& a+\\frac{1}{2}(a+c)+d+b+\\frac{1}{2}(b+d)+c=\\frac{3}{2}(a+b+c+d)=0 .\n\\end{aligned}\n$$\n\n## BxMO 2021: Problems and Solutions\n\n#"}
+{"problem": "(a) Prove that for all $a, b, c, d \\in \\mathbb{R}$ with $a+b+c+d=0$,\n\n$$\n\\max (a, b)+\\max (a, c)+\\max (a, d)+\\max (b, c)+\\max (b, d)+\\max (c, d) \\geqslant 0\n$$\n\n(b) Find the largest non-negative integer $k$ such that it is possible to replace $k$ of the six maxima in this inequality by minima in such a way that the inequality still holds for all $a, b, c, d \\in \\mathbb{R}$ with $a+b+c+d=0$.\n\n#", "solution": "We present another proof for the inequality where $\\max (a, b)$ and $\\max (c, d)$ are replaced by $\\min (a, b)$ and $\\min (c, d)$. By substituting $\\max (x, y)=\\frac{1}{2}(x+y+|x-y|)$ and $\\min (x, y)=\\frac{1}{2}(x+y-|x-y|)$ everywhere and using $a+b+c+d=0$, the inequality may be rewritten as:\n\n$$\n|a-c|+|a-d|+|b-c|+|b-d| \\stackrel{?}{\\geqslant}|a-b|+|c-d| \\text {. }\n$$\n\nBy the triangle inequality, we have\n\n$$\n|a-c|+|a-d|+|b-c|+|b-d| \\geqslant|(a-c)-(b-c)|+|(a-d)-(b-d)|=2|a-b|,\n$$\n\nand similarly, $|a-c|+|a-d|+|b-c|+|b-d| \\geqslant 2|c-d|$. Adding these and dividing by 2 yields the desired inequality.\n\n## BxMO 2021: Problems and Solutions\n\n#"}
+{"problem": "Pebbles are placed on a $2021 \\times 2021$ board in such a way that each square contains at most one pebble. The pebble set of a square of the board is the collection of all pebbles which are in the same row or column as this square. Determine the least number of pebbles that can be placed on the board in such a way that no two squares have the same pebble set.\n\n#", "solution": "Let $N \\geqslant 1$ be a positive integer. We claim that the least number of pebbles that can be placed on a $(2 N+1) \\times(2 N+1)$ chessboard in such a way that no two squares of the board have the same pebble set is $3 N+1$. The problem has $N=1010$, so at least 3031 pebbles are needed.\n\nWe begin by placing $(2 N+1)+N=3 N+1$ pebbles on the board as shown. The construction extends to all values of $N$, and one checks that no two squares on the board have the same pebble set.\n![](https://cdn.mathpix.com/cropped/2024_04_17_e159f221721784e79019g-4.jpg?height=354&width=966&top_left_y=1174&top_left_x=545)\n\nWe are left to show that the number of pebbles, $P$, that must be placed on the board is at least $3 N+1$. Suppose that there exists an empty row. Any other row must then contain at least two pebbles. Indeed, if another row is empty, Figure [1], or contains exactly one pebble, Figure [2], then there are two squares (shaded in the figures below) with the same pebble set. So in this case there are at least $P \\geqslant 2(2 N+1-1)=4 N \\geqslant 3 N+1$ pebbles on the board. The same argument applies, mutatis mutandis, to the columns of the board.\n\n$[1]$\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_e159f221721784e79019g-4.jpg?height=243&width=271&top_left_y=1900&top_left_x=344)\n\n$[2]$\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_e159f221721784e79019g-4.jpg?height=254&width=257&top_left_y=1889&top_left_x=748)\n\n$[3]$\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_e159f221721784e79019g-4.jpg?height=229&width=263&top_left_y=1913&top_left_x=1136)\n\n$[4$\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_e159f221721784e79019g-4.jpg?height=266&width=274&top_left_y=1877&top_left_x=1528)\n\nOn the other hand, suppose that each each row and each column contains at least one pebble. Let $k$ the number of rows containing precisely one pebble. Counting the number of pebbles per row shows that $P \\geqslant k+2(2 N+1-k)$. To count the number of pebbles per column, note that the $k$ pebbles that are the only ones in their row must be in $k$ distinct columns, Figure [3]. Furthermore, at most one of these $k$ columns only has one pebble in it, Figure [4]. So these $k$ columns contain a total of at least $2 k-1$ pebbles, while the remaining columns contain at least one each. Hence $P \\geqslant(2 k-1)+(2 N+1-k)$. Adding these two inequalities we get $2 P \\geqslant 6 N+2$.\n\nRemark. For a $2 N \\times 2 N$ board, at least $3 N$ pebbles are needed. The proof is similar: if there is an empty line, then $P \\geqslant 2(2 N-1) \\geqslant 3 N$ for $N \\geqslant 2$. If there is no empty line, then $2 P \\geqslant 6 N-1$. A construction similar to that of the odd boards works. While the construction is unique in the odd case, up to permutation of rows and columns, that is not true for even boards.\n\n#"}
+{"problem": "A cyclic quadrilateral $A B X C$ has circumcentre $O$. Let $D$ be a point on line $B X$ such that $|A D|=|B D|$. Let $E$ be a point on line $C X$ such that $|A E|=|C E|$. Prove that the circumcentre of triangle $\\triangle D E X$ lies on the perpendicular bisector of $O A$.\n\n#", "solution": "First, note that $\\angle A O X=2 \\angle A B X=2\\left(180^{\\circ}-\\angle A C X\\right)=2 \\angle A C E$ as $A B X C$ is cyclic. Secondly, both $\\triangle D A B$ and $\\triangle E A C$ are isosceles, which implies that $\\angle A E X=\\angle A E C=180^{\\circ}-2 \\angle A C E=180^{\\circ}-\\angle A O X$ and $\\angle A D X=$ $\\angle A D B=180^{\\circ}-2 \\angle A B D=180^{\\circ}-2 \\angle A B X=180^{\\circ}-\\angle A O X$. From this, see that respectively $A E X O$ and $A D X O$ are cyclic, i.e. $A E D X O$ is cyclic.\n\nHence, the circumcentre of $\\triangle D E X$ is also the circumcentre of $A E D X O$. However, as in a circle any perpendicular bisector of a chord goes through the centre of the circle, we find that the circumcentre of $\\triangle D E X$ lies on the perpendicular bisector of $O A$.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_e159f221721784e79019g-5.jpg?height=1088&width=899&top_left_y=1215&top_left_x=607)\n\n#"}
+{"problem": "A cyclic quadrilateral $A B X C$ has circumcentre $O$. Let $D$ be a point on line $B X$ such that $|A D|=|B D|$. Let $E$ be a point on line $C X$ such that $|A E|=|C E|$. Prove that the circumcentre of triangle $\\triangle D E X$ lies on the perpendicular bisector of $O A$.\n\n#", "solution": "In this solution, we use directed angles $\\measuredangle$. We have $\\measuredangle A B D=\\measuredangle A B X=\\measuredangle A C X=\\measuredangle A C E$ and since $\\triangle A B D$ and $\\triangle A C E$ are both isosceles, we see that $\\triangle A B D \\sim \\triangle A C E$ with equal orientation. This means that there exists a spiral symmetry $T$ with centre $A$ such that $T(B)=D$ and $T(C)=E$. Now let $O^{\\prime}$ be the centre of $\\odot D E X$. Then we find $\\measuredangle D O^{\\prime} E=2 \\measuredangle D X E=2 \\measuredangle B X C=\\measuredangle B O C$. Moreover, $\\triangle D O^{\\prime} E$ and $\\triangle B O C$ are isosceles, so we have $\\triangle D O^{\\prime} E \\sim \\triangle B O C$ with equal orientation. This means that $T$ must send $O$ to $O^{\\prime}$, so in particular, $\\triangle A O O^{\\prime}$ is similar to $\\triangle A B D$ and $\\triangle A C E$. We conclude that $\\left|A O^{\\prime}\\right|=\\left|O O^{\\prime}\\right|$, from which the result follows.\n\n#"}
+{"problem": "A sequence $a_{1}, a_{2}, a_{3}, \\ldots$ of positive integers satisfies $a_{1}>5$ and $a_{n+1}=5+6+\\cdots+a_{n}$ for all positive integers $n$. Determine all prime numbers $p$ such that, regardless of the value of $a_{1}$, this sequence must contain a multiple of $p$.\n\n#", "solution": "We claim that the only prime number of which the sequence must contain a multiple is $p=2$. To prove this, we begin by noting that\n\n$$\na_{n+1}=\\frac{a_{n}\\left(a_{n}+1\\right)}{2}-10=\\frac{\\left(a_{n}-4\\right)\\left(a_{n}+5\\right)}{2} .\n$$\n\nLet $p>2$ be an odd prime, and choose $a_{1} \\equiv-4(\\bmod p)$, so $2 a_{2} \\equiv(-4-4)(-4+5) \\equiv-8(\\bmod p)$, whence $a_{2} \\equiv-4(\\bmod p)$, since $p$ is odd. By induction, $a_{n} \\equiv-4(\\bmod p) \\not \\equiv 0(\\bmod p)$ for all $n$, and so the sequence need not contain a multiple of $p$.\n\nWe are left to show that the sequence must contain an even number. Suppose to the contrary that $a_{n}$ is odd for $n=1,2, \\ldots$ We observe that\n\n$$\na_{n+1}-a_{n}=\\frac{a_{n}\\left(a_{n}+1\\right)}{2}-\\frac{a_{n-1}\\left(a_{n-1}+1\\right)}{2}=\\frac{a_{n}-a_{n-1}}{2}\\left(a_{n}+a_{n-1}+1\\right) .\n$$\n\nBy assumption, $a_{n}+a_{n-1}+1$ is odd for $n=1,2, \\ldots$, so this shows that $v_{2}\\left(a_{n+1}-a_{n}\\right)=v_{2}\\left(a_{n}-a_{n-1}\\right)-1$, and so there exists $N$ such that $v_{2}\\left(a_{N+1}-a_{N}\\right)=0$. This is a contradiction, because $a_{n+1}-a_{n}$ is even for $n=1,2, \\ldots$ by assumption, and thus completes the proof.\n\n#"}
+{"problem": "A sequence $a_{1}, a_{2}, a_{3}, \\ldots$ of positive integers satisfies $a_{1}>5$ and $a_{n+1}=5+6+\\cdots+a_{n}$ for all positive integers $n$. Determine all prime numbers $p$ such that, regardless of the value of $a_{1}$, this sequence must contain a multiple of $p$.\n\n#", "solution": "For odd $p$, proceed as in solution 1. Now let $p=2$, and suppose that every term of the sequence is odd. We claim that it follows that $a_{n} \\equiv 5\\left(\\bmod 2^{k}\\right)$ for every integer $n \\geq 1$ and every integer $k \\geq 1$. We proceed per induction on $k$. For $k=1$ this simply states that $a_{n}$ is odd for all integers $n \\geq 1$, as assumed. Now suppose it is true for $k=r$. Let $k=r+1$. Take any integer $n \\geq 1$. Note that, by the induction hypothesis, $a_{n} \\equiv 5\\left(\\bmod 2^{r}\\right)$. Therefore there exists an integer $s$ such that $a_{n}=2^{r} s+5$. Now note that\n\n$$\na_{n+1}=\\frac{\\left(a_{n}-4\\right)\\left(a_{n}+5\\right)}{2}=\\frac{\\left(2^{r} s+1\\right)\\left(2^{r} s+10\\right)}{2}=\\left(2^{r} s+1\\right)\\left(2^{r-1} s+5\\right) \\equiv 2^{r-1} s+5 \\quad\\left(\\bmod 2^{r}\\right)\n$$\n\nBy the induction hypothesis, $a_{n+1} \\equiv 5\\left(\\bmod 2^{r}\\right)$. Therefore $s$ is even, such that $a_{n}$ is of the form $2^{r+1} s+5$ for any integer $n \\geq 1$, which concludes the induction. From this property, it follows that $a_{1}-5$ is divisible by $2^{k}$ for every integer $k \\geq 1$, which is only possible if $a_{1}=5$. But $a_{1}>5$, so this is a contradiction."}

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2022-zz.jsonl

@@ -0,0 +1,11 @@
+{"problem": "Let $n \\geqslant 0$ be an integer, and let $a_{0}, a_{1}, \\ldots, a_{n}$ be real numbers. Show that there exists $k \\in\\{0,1, \\ldots, n\\}$ such that\n\n$$\na_{0}+a_{1} x+a_{2} x^{2}+\\cdots+a_{n} x^{n} \\leqslant a_{0}+a_{1}+\\cdots+a_{k}\n$$\n\nfor all real numbers $x \\in[0,1]$.\n\n#", "solution": "The case $n=0$ is trivial; for $n>0$, the proof goes by induction on $n$. We need to make one preliminary observation:\n\nClaim. For all reals $a, b, a+b x \\leqslant \\max \\{a, a+b\\}$ for all $x \\in[0,1]$.\n\nProof. If $b \\leqslant 0$, then $a+b x \\leqslant a$ for all $x \\in[0,1]$; otherwise, if $b>0, a+b x \\leqslant a+b$ for all $x \\in[0,1]$. This proves our claim.\n\nThis disposes of the base case $n=1$ of the induction: $a_{0}+a_{1} x \\leqslant \\max \\left\\{a_{0}, a_{0}+a_{1}\\right\\}$ for all $x \\in[0,1]$. For $n \\geqslant 2$, we note that, for all $x \\in[0,1]$,\n\n$$\n\\begin{aligned}\na_{0}+a_{1} x+\\cdots+a_{n} x^{n} & =a_{0}+x\\left(a_{1}+a_{2} x+\\cdots+a_{n} x^{n-1}\\right) \\\\\n& \\leqslant a_{0}+x\\left(a_{1}+a_{2}+\\cdots+a_{k}\\right) \\leqslant \\max \\left\\{a_{0}, a_{0}+\\left(a_{1}+\\cdots+a_{k}\\right)\\right\\}\n\\end{aligned}\n$$\n\nfor some $k \\in\\{1,2, \\ldots, n\\}$ by the inductive hypothesis and our earlier claim. This completes the proof by induction.\n\n#"}
+{"problem": "Let $n \\geqslant 0$ be an integer, and let $a_{0}, a_{1}, \\ldots, a_{n}$ be real numbers. Show that there exists $k \\in\\{0,1, \\ldots, n\\}$ such that\n\n$$\na_{0}+a_{1} x+a_{2} x^{2}+\\cdots+a_{n} x^{n} \\leqslant a_{0}+a_{1}+\\cdots+a_{k}\n$$\n\nfor all real numbers $x \\in[0,1]$.\n\n#", "solution": "Define $s_{i}=a_{0}+a_{1}+\\cdots+a_{i}$ for $i \\in\\{0,1, \\ldots, n\\}$. Thus $a_{0}=s_{0}$ and $a_{i}=s_{i}-s_{i-1}$ for all $i \\in\\{1,2, \\ldots, n\\}$. Hence\n\n$$\n\\begin{aligned}\na_{0}+a_{1} x+a_{2} x^{2}+\\cdots+a_{n} x^{n} & =s_{0}+\\left(s_{1}-s_{0}\\right) x+\\left(s_{2}-s_{1}\\right) x^{2}+\\ldots+\\left(s_{n}-s_{n-1}\\right) x^{n} \\\\\n& =s_{0}(1-x)+s_{1}\\left(x-x^{2}\\right)+\\ldots+s_{n-1}\\left(x^{n-1}-x^{n}\\right)+s_{n} x^{n}\n\\end{aligned}\n$$\n\nNow choose $k \\in\\{0,1, \\ldots, n\\}$ such that $s_{k}=\\max \\left\\{s_{0}, s_{1}, \\ldots, s_{n}\\right\\}$. Using the inequality $x^{i-1}-x^{i} \\geqslant 0$, valid for all $i \\in\\{1,2, \\ldots, n\\}$ and all $x \\in[0,1]$, in the right-hand side above, it follows that\n\n$$\n\\begin{aligned}\na_{0}+a_{1} x+a_{2} x^{2}+\\cdots+a_{n} x^{n} \\leqslant s_{k}(1- & x)+s_{k}\\left(x-x^{2}\\right)+\\cdots+s_{k}\\left(x^{n-1}-x^{n}\\right)+s_{k} x^{n} \\\\\n= & s_{k}\\left[(1-x)+\\left(x-x^{2}\\right)+\\cdots+\\left(x^{n-1}-x^{n}\\right)+x^{n}\\right] \\\\\n= & s_{k}=a_{0}+a_{1}+\\cdots+a_{k}\n\\end{aligned}\n$$\n\nThis completes the proof.\n\n## BxMO 2022: Problems and Solutions\n\n#"}
+{"problem": "Let $n \\geqslant 0$ be an integer, and let $a_{0}, a_{1}, \\ldots, a_{n}$ be real numbers. Show that there exists $k \\in\\{0,1, \\ldots, n\\}$ such that\n\n$$\na_{0}+a_{1} x+a_{2} x^{2}+\\cdots+a_{n} x^{n} \\leqslant a_{0}+a_{1}+\\cdots+a_{k}\n$$\n\nfor all real numbers $x \\in[0,1]$.\n\n#", "solution": "The proof proceeds by induction on $n$. The base case $n=0$ is trivial. For $n \\geqslant 1$, since $x \\in[0,1]$, we have $x^{n} \\leqslant x^{n-1}$. Thus, if $a_{n} \\geqslant 0$, then $a_{n} x^{n} \\leqslant a_{n} x^{n-1}$, while, if $a_{n}<0$, then $a_{n} x^{n}<0$ trivially. This shows that $a_{n} x^{n} \\leqslant \\max \\left\\{0, a_{n} x^{n-1}\\right\\}$, whence\n\n$$\na_{0}+a_{1} x+\\cdots+a_{n-1} x^{n-1}+a_{n} x^{n} \\leqslant a_{0}+a_{1} x+\\cdots+\\max \\left\\{a_{n-1}, a_{n-1}+a_{n}\\right\\} x^{n-1} .\n$$\n\nBy the inductive hypothesis, the polynomial of degree $n-1$ on the right-hand side is bounded above by $a_{0}+\\cdots+a_{k}$ for some $k \\in\\{0,1, \\ldots, n-2\\}$ or $a_{0}+\\cdots+a_{n-2}+\\max \\left\\{a_{n-1}, a_{n-1}+a_{n}\\right\\}$. But the latter is equal to one of $a_{0}+a_{1}+\\cdots+a_{n-1}$ or $a_{0}+a_{1}+\\cdots+a_{n}$; both are of the desired form, $a_{0}+a_{1}+\\cdots+a_{k}$ for some $k \\in\\{n-1, n\\}$. This completes the proof by induction.\n\n## BxMO 2022: Problems and Solutions\n\n#"}
+{"problem": "Let $n$ be a positive integer. There are $n$ ants walking along a line at constant nonzero speeds. Different ants need not walk at the same speed or walk in the same direction. Whenever two or more ants collide, all the ants involved in this collision instantly change directions. (Different ants need not be moving in opposite directions when they collide, since a faster ant may catch up with a slower one that is moving in the same direction.) The ants keep walking indefinitely.\n\nAssuming that the total number of collisions is finite, determine the largest possible number of collisions in terms of $n$.\n\n#", "solution": "The order of the ants along the line does not change; denote by $v_{1}, v_{2}, \\ldots, v_{n}$ the respective speeds of ants $1,2, \\ldots, n$ in this order. If $v_{i-1}<v_{i}>v_{i+1}$ for some $i \\in\\{2, \\ldots, n-1\\}$, then, at each stage, ant $i$ can catch up with ants $i-1$ or $i+1$ irrespective of the latters' directions of motion, so the number of collisions is infinite. Hence, if the number of collisions is finite, then, up to switching the direction definining the order of the ants, (i) $v_{1} \\geqslant \\cdots \\geqslant v_{n}$ or (ii) $v_{1} \\geqslant \\cdots \\geqslant v_{k-1}>v_{k} \\leqslant \\cdots \\leqslant v_{n}$ for some $k \\in\\{2, \\ldots, n-1\\}$. We need the following observation:\n\nClaim. If $v_{1} \\geqslant \\cdots \\geqslant v_{m}$, then ants $m-1$ and $m$ collide at most $m-1$ times.\n\nProof. The proof goes by induction on $m$, the case $m=1$ being trivial. Since $v_{m-1} \\geqslant v_{m}$, ants $m-1$ and $m$ can only collide if the former is moving towards the latter. Hence, between successive collisions with ant $m$, ant $m-1$ must reverse direction by colliding with ant $m-2$. Since ants $m-1$ and $m-2$ collide at most $(m-1)-1=m-2$ times by the inductive hypothesis, ants $m$ and $m-1$ collide at most $(m-2)+1=m-1$ times.\n\nHence, in case (i), there are at most $0+1+\\cdots+(n-1)=n(n-1) / 2$ collisions. In case (ii), applying the claim to ants $1,2, \\ldots, k$ and also to ants $n, n-1, \\ldots, k$ by switching their order, the number of collisions is at most $k(k-1) / 2+(n-k+1)(n-k) / 2=n(n-1) / 2-(k-1)(n-k)<n(n-1) / 2$.\n\nNow take a coordinate $x$ along the line, and put ants at $x=1,2, \\ldots, n$ with positive initial velocities and speeds $v_{1}=\\cdots=v_{n-1}=1, v_{n}=\\varepsilon$, for some $\\varepsilon$. For $\\varepsilon=0$, collisions occur according to the pattern shown below for $n=5$, which clearly extends to all values of $n$ in such a way that ants $m$ and $m+1$ collide exactly $m$ times for $m=1,2, \\ldots, n-1$. This yield $1+2+\\cdots+(n-1)=n(n-1) / 2$ collisions in total. For all sufficiently small $\\varepsilon>0$, the number of collisions remains equal to $n(n-1) / 2$.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_a4bb6022e2f5b324b7ecg-4.jpg?height=291&width=686&top_left_y=2270&top_left_x=685)\n\nThis shows that the upper bound obtained above can be attained. If the number of collisions is finite, the largest possible number of collisions is therefore indeed $n(n-1) / 2$.\n\n## BxMO 2022: Problems and Solutions\n\n#"}
+{"problem": "Let $n$ be a positive integer. There are $n$ ants walking along a line at constant nonzero speeds. Different ants need not walk at the same speed or walk in the same direction. Whenever two or more ants collide, all the ants involved in this collision instantly change directions. (Different ants need not be moving in opposite directions when they collide, since a faster ant may catch up with a slower one that is moving in the same direction.) The ants keep walking indefinitely.\n\nAssuming that the total number of collisions is finite, determine the largest possible number of collisions in terms of $n$.\n\n#", "solution": "We show that there are at most $n(n-1) / 2$ collisions if the number of collisions is finite as in Solution 1.\n\nTo show that the upper bound of $n(n-1) / 2$ collisions can be attained, we construct, inductively, an example of $n$ ants colliding $n(n-1) / 2$ times, the speeds of the ants decrease from left to right, and after all collisions all ants move towards the left, with the possible exception of the rightmost ant. In every case, we will label the ants $1,2, \\ldots, n$ from left to right. For $n=1$ this is trivial. For $n \\geqslant 2$, we use the construction for $n-1$ ants (now labelled $2,3, \\ldots, n$ ). We add ant 1 on the left, moving towards the right, faster than all other ants (so that the speeds of the ants still decrease from left to right), and in such a way that its first collision (with ant 2 ) happens after all $(n-1)(n-2) / 2$ collisions of the other $n-1$ ants. Now the following events happen (in this order) for $i=1,2, \\ldots, n-2$ : ants $i$ and $i+1$ collide, after which ant $i$ moves to the left and ant $i+1$ moves to the right. These collisions do happen because the speeds of the ants decrease from left to right. Then ants $n-1$ and $n$ also collide, resulting in ant $n-1$ moving to the left. This shows that there are (at least) $(n-1)(n-2) / 2+(n-1)=n(n-1) / 2$ collisions. There are in fact no more collisions since the speeds of the ants decrease from left to right; alternatively, this follows from the upper bound proved previously. Since all ants except ant $n$ are moving towards the left after the collisions, this completes the inductive construction.\n\n## BxMO 2022: Problems and Solutions\n\n#"}
+{"problem": "Let $A B C$ be a scalene acute triangle. Let $B_{1}$ be the point on ray $\\left[A C\\right.$ such that $\\left|A B_{1}\\right|=\\left|B B_{1}\\right|$. Let $C_{1}$ be the point on ray $\\left[A B\\right.$ such that $\\left|A C_{1}\\right|=\\left|C C_{1}\\right|$. Let $B_{2}$ and $C_{2}$ be the points on line $B C$ such that $\\left|A B_{2}\\right|=\\left|C B_{2}\\right|$ and $\\left|B C_{2}\\right|=\\left|A C_{2}\\right|$. Prove that $B_{1}, C_{1}, B_{2}, C_{2}$ are concyclic.\n\n#", "solution": "By construction, lines $B_{1} C_{2}$ and $B_{2} C_{1}$ bisect segments $[A B]$ and $[A C]$, respectively, so their intersection $O$ is the circumcentre of $A B C$. Hence $\\angle B O C=2 \\angle A$ and $\\angle C B O=\\angle O C B=90^{\\circ}-\\angle A$. Now, by construction, $\\angle O C_{1} B=90^{\\circ}-\\angle A=\\angle O C B$, so $B C_{1} C O$ is cyclic. Similarly, $B C B_{1} O$ is cyclic by construction because $\\angle O B_{1} C=180^{\\circ}-\\angle A B_{1} O=90^{\\circ}+\\angle A=180^{\\circ}-\\angle C B O$. In particular, $B C_{1} C B_{1}$ is cyclic, too.\n\nNow $\\angle B_{1} C_{1} B_{2}=\\angle B_{1} C_{1} B-\\angle B_{2} C_{1} B$ and $\\angle B_{1} C_{2} B_{2}=\\angle B_{1} C B-\\angle C B_{1} C_{2}$. But $\\angle B_{1} C_{1} B=\\angle B_{1} C B$ since $B C_{1} C B_{1}$ is cyclic and $\\angle B_{2} C_{1} B=\\angle O C_{1} B=90^{\\circ}-\\angle A=180^{\\circ}-\\angle O B_{1} C=\\angle C B_{1} C_{2}$. Hence $\\angle B_{1} C_{1} B_{2}=\\angle B_{1} C_{2} B_{2}$, so $B_{1} B_{2} C_{1} C_{2}$ is cyclic, as required.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_a4bb6022e2f5b324b7ecg-6.jpg?height=777&width=1008&top_left_y=1285&top_left_x=524)"}
+{"problem": "Let $A B C$ be a scalene acute triangle. Let $B_{1}$ be the point on ray $\\left[A C\\right.$ such that $\\left|A B_{1}\\right|=\\left|B B_{1}\\right|$. Let $C_{1}$ be the point on ray $\\left[A B\\right.$ such that $\\left|A C_{1}\\right|=\\left|C C_{1}\\right|$. Let $B_{2}$ and $C_{2}$ be the points on line $B C$ such that $\\left|A B_{2}\\right|=\\left|C B_{2}\\right|$ and $\\left|B C_{2}\\right|=\\left|A C_{2}\\right|$. Prove that $B_{1}, C_{1}, B_{2}, C_{2}$ are concyclic.\n\n#", "solution": "The isosceles triangles $A B_{1} B$ and $A C_{1} C$ have equal base angles $\\angle B A B_{1}=\\angle C_{1} A C=\\angle A$, so are similar. In particular, $|A B| /\\left|A B_{1}\\right|=|A C| /\\left|A C_{1}\\right|$. Since $\\angle B A C=\\angle B_{1} A C_{1}=\\angle A$, it follows that triangles $A B C$ and $A B_{1} C_{1}$ are similar, too. In particular, $\\angle C B A=\\angle A B_{1} C_{1}$.\n\nBy construction, lines $B_{1} C_{2}$ and $B_{2} C_{1}$ are the respective perpendicular bisectors of $[A B]$ and $[A C]$, so meet them at their respective midpoints $C^{\\prime}$ and $B^{\\prime}$. Hence\n\n$$\n\\begin{aligned}\n\\angle B_{1} C_{2} B_{2} & =\\angle C^{\\prime} C_{2} B=90^{\\circ}-\\angle C_{2} B C^{\\prime}=90^{\\circ}-\\angle C B A=90^{\\circ}-\\angle A B_{1} C_{1}=90^{\\circ}-\\angle B^{\\prime} B_{1} C_{1} \\\\\n& =\\angle B_{1} C_{1} B^{\\prime}=\\angle B_{1} C_{1} B_{2} .\n\\end{aligned}\n$$\n\nHence $B_{1} C_{2} C_{1} B_{2}$ is cyclic, which completes the proof.\n\n## BxMO 2022: Problems and Solutions\n\n#"}
+{"problem": "Let $A B C$ be a scalene acute triangle. Let $B_{1}$ be the point on ray $\\left[A C\\right.$ such that $\\left|A B_{1}\\right|=\\left|B B_{1}\\right|$. Let $C_{1}$ be the point on ray $\\left[A B\\right.$ such that $\\left|A C_{1}\\right|=\\left|C C_{1}\\right|$. Let $B_{2}$ and $C_{2}$ be the points on line $B C$ such that $\\left|A B_{2}\\right|=\\left|C B_{2}\\right|$ and $\\left|B C_{2}\\right|=\\left|A C_{2}\\right|$. Prove that $B_{1}, C_{1}, B_{2}, C_{2}$ are concyclic.\n\n#", "solution": "By construction, lines $B_{1} C_{2}$ and $B_{2} C_{1}$ are the respective perpendicular bisectors of $[A B]$ and $[A C]$, so meet them at their respective midpoints $C^{\\prime}$ and $B^{\\prime}$. Since $\\angle B_{1} C^{\\prime} C_{1}=90^{\\circ}=\\angle C_{1} B^{\\prime} B_{1}, B_{1} C_{1} C^{\\prime} B^{\\prime}$ is cyclic. Together with the fact that $B^{\\prime} C^{\\prime} \\| B C$ by construction, this implies\n\n$$\n\\angle B_{1} C_{2} B_{2}=\\angle C^{\\prime} C_{2} B=\\angle C_{2} C^{\\prime} B^{\\prime}=\\angle B_{1} C^{\\prime} B^{\\prime}=\\angle B_{1} C_{1} B^{\\prime}=\\angle B_{1} C_{1} B_{2},\n$$\n\nwhence $B_{1} C_{2} C_{1} B_{2}$ is cyclic. This completes the proof.\n\n## BxMO 2022: Problems and Solutions\n\n#"}
+{"problem": "A subset $A$ of the natural numbers $\\mathbb{N}=\\{0,1,2, \\ldots\\}$ is called good if every integer $n>0$ has at most one prime divisor $p$ such that $n-p \\in A$.\n\n(a) Show that the set $S=\\{0,1,4,9, \\ldots\\}$ of perfect squares is good.\n\n(b) Find an infinite good set disjoint from $S$.\n\n(Two sets are disjoint if they have no common elements.)\n\n#", "solution": "(a) Suppose to the contrary that $S$ is not good, so there exists $n \\in \\mathbb{N}$ with two different prime factors $p \\neq q$ such that $n-p, n-q$ are perfect squares. In particular $n$ is not prime. Write $n-p=m^{2}$, for some $m \\in \\mathbb{N}$. As $p \\mid n$, it follows that $p \\mid m$ and hence $p^{2} \\mid m^{2}$ since $p$ is prime. Hence there exists $k \\in \\mathbb{N}$ such that $n-p=p^{2} k^{2}$. Similarly, there exists $\\ell \\in \\mathbb{N}$ such that $n-q=q^{2} \\ell^{2}$. We observe that $k, \\ell \\neq 0$ since $n$ is not prime.\n\nNow we have $p-q=(n-q)-(n-p)=(\\ell q-k p)(\\ell q+k p)$. Since $p-q \\neq 0, \\ell q-k p \\neq 0$, and hence $|p-q|=|k p-\\ell q||k p+\\ell q| \\geqslant|k p+\\ell q|=k p+\\ell q$. This is a contradiction however, because, since $k, \\ell \\neq 0$, it is clear that $k p+\\ell q \\geqslant p+q>|p-q|$. Hence $S$ is good.\n\n(b) Let $q$ be a prime, and let $Q=\\left\\{q, q^{3}, q^{5}, \\ldots\\right\\}$ be the (infinite) set of odd powers of $q$, which is disjoint from $S$. We claim that $Q$ is good. Indeed, let $n \\in \\mathbb{N}$, and let $p \\mid n$ be a prime such that $n-p \\in Q$, i.e. $n-p=q^{2 k+1}$ for some $k \\in \\mathbb{N}$. Then $p \\mid n-p$, so $p \\mid q^{2 k+1}$, and hence $p=q$. Thus $Q$ is good.\n\n#"}
+{"problem": "A subset $A$ of the natural numbers $\\mathbb{N}=\\{0,1,2, \\ldots\\}$ is called good if every integer $n>0$ has at most one prime divisor $p$ such that $n-p \\in A$.\n\n(a) Show that the set $S=\\{0,1,4,9, \\ldots\\}$ of perfect squares is good.\n\n(b) Find an infinite good set disjoint from $S$.\n\n(Two sets are disjoint if they have no common elements.)\n\n#", "solution": "(a) Let $p \\mid n$ be a prime such that $n-p=p(n / p-1)=m^{2}$, for some $m \\in \\mathbb{N}$. Since $p \\mid m^{2}$ and $p$ is prime, $p^{2} \\mid m^{2}$, and hence $p \\mid n / p-1<n / p$, so $p<\\sqrt{n}$.\n\nNow suppose to the contrary that $S$ is not good, so there are primes $p_{1}>p_{2}$ dividing $n$ such that $n-p_{1}<n-p_{2}$ are perfect squares. Then\n\n$$\nn-p_{2} \\geqslant\\left(\\sqrt{n-p_{1}}+1\\right)^{2}>n-p_{1}+2 \\sqrt{n-p_{1}} \\quad \\Longrightarrow \\quad p_{1}>p_{2}+2 \\sqrt{n-p_{1}} \\geqslant 2+2 \\sqrt{n-p_{1}}\n$$\n\nThe last condition implies that $p_{1}>2 \\sqrt{n-1}$. But $p_{1}<\\sqrt{n}$ by the first part, so $\\sqrt{n}>2 \\sqrt{n-1}$, which is a contradiction for $n>1$; the cases $n=0$ and $n=1$ are trivial. Thus $S$ is good.\n\n(b) We claim that the infinite set $P=\\{3,5,7,11, \\ldots\\}$ of odd primes, which is disjoint from $S$, is good. Indeed, let $n \\in \\mathbb{N}$ and let $p \\mid n$ be a prime such that $n-p=q$, for some odd prime $q$. Then $p \\mid n-p$, so $p \\mid q$, i.e. $p=q$, and hence $n=2 q$. Since $q$ is the only odd prime divisor of $n=2 q, P$ is good.\n\nThe set $P^{\\prime}=\\{2,3,5,7,11, \\ldots\\}$ of all primes is also good. The proof is similar: let $n \\in \\mathbb{N}$ and let $p \\mid n$ be a prime such that $n-p=q$, for some prime $q$. Then $p \\mid n-p$, so $p \\mid q$, i.e. $p=q$, and hence $n=2 q$. If $q=2$, then 2 is the only prime divisor of $n$; if $q \\neq 2$, then the only prime divisor of $n$, apart from $q$, is 2 . However, $n-2=2(q-1) \\notin P^{\\prime}$ since $q-1>1$. Hence $P^{\\prime}$ is good.\n\n## BxMO 2022: Problems and Solutions\n\n#"}
+{"problem": "A subset $A$ of the natural numbers $\\mathbb{N}=\\{0,1,2, \\ldots\\}$ is called good if every integer $n>0$ has at most one prime divisor $p$ such that $n-p \\in A$.\n\n(a) Show that the set $S=\\{0,1,4,9, \\ldots\\}$ of perfect squares is good.\n\n(b) Find an infinite good set disjoint from $S$.\n\n(Two sets are disjoint if they have no common elements.)\n\n#", "solution": "(a) Suppose to the contrary that $S$ is not good, so there exists $n \\in \\mathbb{N}$ with two different prime factors $p \\neq q$ such that $n-p, n-q$ are perfect squares. Write $n-p=m^{2}$, for some $m \\in \\mathbb{N}$. As $p \\mid n$, it follows that $p \\mid m$ and hence $p^{2} \\mid m^{2}$ since $p$ is prime. Hence there exists $k \\in \\mathbb{N}$ such that $n-p=p^{2} k^{2}$. Similarly, there exists $\\ell \\in \\mathbb{N}$ such that $n-q=q^{2} \\ell^{2}$. By construction, $n$ is not prime, so $n-p, n-q \\neq 0$, whence $k, \\ell \\geqslant 1$.\n\nHence $p^{2} k^{2}+p=q^{2} \\ell^{2}+q$. Hence $p^{2} k^{2}<p^{2} k^{2}+p=q^{2} \\ell^{2}+q<q^{2} \\ell^{2}+2 q \\ell+1=(q \\ell+1)^{2}$. Similarly, $q^{2} \\ell^{2}<(p k+1)^{2}$, whence $q \\ell-1<p k<q \\ell+1$. It follows that $p k=q \\ell$, so $p^{2} k^{2}+p=q^{2} \\ell^{2}+q$ yields the contradiction $p=q$. Hence $S$ is good.\n\n(b) Let $A$ be a finite good set such that $0 \\notin A$, and let $m=\\max A$. Let $a \\geqslant 2 m+1$ be an integer. We claim that $A^{\\prime}=A \\cup\\{a\\}$ is good. Indeed, suppose to the contrary that there exist $n \\in \\mathbb{N}$ and primes $p, q \\mid n$ with $p \\neq q$ such that $n-p, n-q \\in A^{\\prime}$. If $n<a$, then $n-p, n-q \\in A$, which is a contradiction because $A$ is good. Hence $n \\geqslant a$. Now $p \\mid n-p$, so $n-p \\geqslant p$ since $0 \\notin A^{\\prime}$. Thus $p \\leqslant n / 2$ and hence $n-p \\geqslant n / 2 \\geqslant a / 2>m$. Similarly, $n-q>m$. It follows that $n-p=n-q=a$, which implies the contradiction $p=q$. Hence $A^{\\prime}$ is good.\n\nNow it is clear that any singleton set is good: indeed, if $A=\\{a\\}$, and $n \\in \\mathbb{N}$ has prime divisors $p, q$ such that $n-p, n-q \\in A$, then $n-p=a=n-q$, so $p=q$. Starting from the singleton $T_{1}=\\{2\\}$, we use the above construction to obtain, iteratively, good sets $T_{2}, T_{3}, \\ldots$ of $2,3, \\ldots$ elements. It is clearly possibly to ensure that they are each disjoint from $S$ by not adding a perfect square at any stage. Then $T=T_{1} \\cup T_{2} \\cup \\cdots$ is an infinite good set disjoint from $S$."}

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2023-zz.jsonl

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+{"problem": "Find all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that\n\n$$\n(x-y)(f(x)+f(y)) \\leqslant f\\left(x^{2}-y^{2}\\right) \\quad \\text { for all } x, y \\in \\mathbb{R}\n$$\n\n#", "solution": "Clearly, $f(x)=c x$ is a solution for each $c \\in \\mathbb{R}$ since $(x-y)(c x+c y)=c\\left(x^{2}-y^{2}\\right)$. To show that there are no other solutions, we observe that\n\n(1) $x=$\n\n$$\nx=1, y=0: \\quad f(0)+f(1) \\leqslant f(1) \\Rightarrow f(0) \\leqslant 0 \\text {, whence } f(0)=0 \\text {; }\n$$\n\n(2) $y=-x: \\quad 2 x(f(x)+f(-x)) \\leqslant f(0)=0$;\n\n$$\nx \\rightarrow-x: \\quad-2 x(f(-x)+f(x)) \\leqslant 0 \\Rightarrow 2 x(f(x)+f(-x)) \\geqslant 0 ;\n$$\n\nthus $2 x(f(x)+f(-x))=0$ for all $x$, so $f(-x)=-f(x)$ for all $x \\neq 0$, and hence for all $x$, since $f(0)=0$;\n\n(3) $x \\leftrightarrow y: \\quad(y-x)(f(y)+f(x)) \\leqslant f\\left(y^{2}-x^{2}\\right)=-f\\left(x^{2}-y^{2}\\right) \\Rightarrow(x-y)(f(x)+f(y)) \\geqslant f\\left(x^{2}-y^{2}\\right)$; which is the given inequality with the inequality sign reversed, so $(x-y)(f(x)+f(y))=f\\left(x^{2}-y^{2}\\right)$ must hold for all $x, y \\in \\mathbb{R}$;\n\n(4) $y \\leftrightarrow-y: \\quad(x-y)(f(x)+f(y))=f\\left(x^{2}-y^{2}\\right)=f\\left(x^{2}-(-y)^{2}\\right)=(x+y)(f(x)+f(-y))=(x+y)(f(x)-f(y))$; expanding yields $f(x) y=f(y) x$ for all $x, y \\in \\mathbb{R}$.\n\nTaking $y=1$ in the last result, $f(x)=f(1) x$, i.e. $f(x)=c x$, where $c=f(1)$, for all $x \\in \\mathbb{R}$. Since we have shown above that, conversely, all such functions are solutions, this completes the proof.\n\nAlternative solution. A slight variation of this argument proves that $(x-y)(f(x)+f(y))=f\\left(x^{2}-y^{2}\\right)$ must hold for all $x, y \\in \\mathbb{R}$ as above, and then reaches $f(x)=c x$ as follows:\n\n(4) $y= \\pm 1: \\quad(x \\mp 1)(f(x) \\pm f(1))=f\\left(x^{2}-1\\right)$ using $f(-1)=-f(1)$ from (2);\n\nhence $(x-1)(f(x)+f(1))=(x+1)(f(x)-f(1)) \\Rightarrow f(x)=f(1) x=c x$, where $c=f(1)$, on expanding.\n\n## BxMO 2023: Problems and Solutions\n\n#"}
+{"problem": "Determine all integers $k \\geqslant 1$ with the following property: given $k$ different colours, if each integer is coloured in one of these $k$ colours, then there must exist integers $a_{1}<a_{2}<\\cdots<a_{2023}$ of the same colour such that the differences $a_{2}-a_{1}, a_{3}-a_{2}, \\ldots, a_{2023}-a_{2022}$ are all powers of 2 .\n\n#", "solution": "We claim that only $k=1$ and $k=2$ satisfy the required property. First, if $k \\geqslant 3$, we colour each integer with its residue class modulo 3 , so that, whenever two integers have the same colour, their difference is divisible by 3 , so is not a power of 2 . This shows that no $k \\geqslant 3$ has the required property.\n\nIn the case $k=1$, the sequence defined by $a_{n}=2 n$ for $n=1,2, \\ldots, 2023$ clearly has the required property. In the case $k=2$, we call the colours \"red\" and \"blue\", and construct, for each $n \\geqslant 1$ and by induction, integers $a_{1}<a_{2}<\\cdots<a_{n}$ of the same colour such that $a_{m+1}-a_{m}$ is a power of 2 for $m=1,2, \\ldots, n-1$. The statement is trivial for $n=1$. For $n>1$, let $a_{1}<a_{2}<\\cdots<a_{n}$ be red integers (without loss of generality) having the desired property. Consider the $n+1$ integers $b_{i}=a_{n}+2^{i}$, for $i=1,2, \\ldots, n+1$. If one of these, say $b_{j}$, is red, then, as $b_{j}-a_{n}=2^{j}$, the $n+1$ red integers $a_{1}<a_{2}<\\cdots<a_{n}<b_{j}$ have the desired property. Otherwise, $b_{1}, b_{2}, \\ldots, b_{n+1}$ are all blue, and $b_{i+1}-b_{i}=\\left(a_{n}+2^{i+1}\\right)-\\left(a_{n}+2^{i}\\right)=2^{i}$ for $i=1,2, \\ldots, n$, so the $n+1$ blue integers $b_{1}<b_{2}<\\cdots<b_{n+1}$ have the desired property. This completes the inductive step and hence the proof.\n\n## BxMO 2023: Problems and Solutions\n\n#"}
+{"problem": "Let $A B C$ be a triangle with incentre $I$ and circumcircle $\\omega$. Let $N$ denote the second point of intersection of line $A I$ and $\\omega$. The line through $I$ perpendicular to $A I$ intersects line $B C$, segment $[A B]$, and segment $[A C]$ at the points $D, E$, and $F$, respectively. The circumcircle of triangle $A E F$ meets $\\omega$ again at $P$, and lines $P N$ and $B C$ intersect at $Q$. Prove that lines $I Q$ and $D N$ intersect on $\\omega$.\n\n#", "solution": "By construction, $A P E F$ and $A P B C$ are cyclic, and so\n\n$$\n\\begin{aligned}\n\\angle B D E & =\\angle C D F=\\angle A F D-\\angle F C D=\\angle A F E-\\angle A C B=\\left(180^{\\circ}-\\angle E P A\\right)-\\left(180^{\\circ}-\\angle B P A\\right) \\\\\n& =\\angle B P A-\\angle E P A=\\angle B P E .\n\\end{aligned}\n$$\n\nHence $D B E P$ is cyclic, too. It follows that $\\angle I D P=\\angle E D P=\\angle E B P=\\angle A B P=\\angle A N P=\\angle I N P$ since $A P B N$ is cyclic, and so $P D N I$ is also cyclic. In particular, $\\angle D P N=\\angle D I N=90^{\\circ}$. Let $R$ denote the second intersection of $D P$ and $\\omega$, so $N Q \\perp D R$. Then $\\angle N P R=90^{\\circ}$, so $R N$ is a diameter of $\\omega$. It is well-known that $N$ is the midpoint of the arc $\\overparen{B C}$ not containing $A$, whence $R N \\perp B C$. Thus $D Q$ and $N Q$ are altitudes of triangle $R D N$, and so $Q$ is its orthocentre. This implies that $R Q \\perp D N$, whence, since $R N$ is a diameter of $\\omega$, the intersection $X$ of $R Q$ and $D N$ lies on $\\omega$.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_a8b8dd435251ab822ad2g-4.jpg?height=1028&width=1288&top_left_y=1456&top_left_x=384)\n\nIt is also well-known that $N$ is the centre of the circumcircle $\\Omega$ of triangle $B C I$. Since $D I \\perp I N$ by construction, $D I$ is tangent to $\\Omega$ at $I$. As $D$ lies on the radical axis $B C$ of $\\omega$ and $\\Omega$, it follows that $|D I|^{2}=|D B||D C|=|D X||D N|$. Hence triangles $D N I$ and $D I X$ are similar; in particular, $\\angle D X I=\\angle D I N=90^{\\circ}$. All of this shows that $R, I, Q, X$ lie on a line perpendicular to $D N$ that intersects $D N$ at $X \\in \\omega$. This completes the proof.\n\n## BxMO 2023: Problems and Solutions\n\n#"}
+{"problem": "Let $A B C$ be a triangle with incentre $I$ and circumcircle $\\omega$. Let $N$ denote the second point of intersection of line $A I$ and $\\omega$. The line through $I$ perpendicular to $A I$ intersects line $B C$, segment $[A B]$, and segment $[A C]$ at the points $D, E$, and $F$, respectively. The circumcircle of triangle $A E F$ meets $\\omega$ again at $P$, and lines $P N$ and $B C$ intersect at $Q$. Prove that lines $I Q$ and $D N$ intersect on $\\omega$.\n\n#", "solution": "Let $K$ be the midpoint of segment $[B C]$. It is well-known that $N$ is the midpoint of the small $\\operatorname{arc} \\widehat{B C}$ of $\\omega$, so $B C \\perp K N$. In particular, $\\angle D K N=90^{\\circ}$. But $\\angle D I N=90^{\\circ}$ by construction, so $D I K N$ is cyclic, with circumcircle $\\Gamma$.\n\nMoreover, $\\angle P E F=180^{\\circ}-\\angle P A F=180^{\\circ}-\\angle P A B=\\angle P B C$ and $\\angle P F E=\\angle P A E=\\angle P A B=\\angle P C B$ since $A F E P$ and $A C B P$ are cyclic, so triangles $P E F$ and $P B C$ are similar. Now, by construction, $I$ is the midpoint of segment $[E F]$, so, $K$ being the midpoint of $[B C]$, triangles $P I F$ and $P K C$ are similar, too. It follows that $\\angle P I D=180^{\\circ}-\\angle P I F=180^{\\circ}-\\angle P K C=\\angle P K D$, whence $P$ lies on $\\Gamma$.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_a8b8dd435251ab822ad2g-5.jpg?height=957&width=1288&top_left_y=801&top_left_x=384)\n\nLet $\\Omega$ be the circumcircle of triangle $B C I$. By construction, $Q$ lies on the radical axes $P N$ of $\\omega, \\Gamma$ and $B C$ of $\\omega, \\Omega$, so is the radical centre of $\\omega, \\Gamma, \\Omega$. In particular, $I Q$ is the radical axis of $\\Gamma, \\Omega$, so is perpendicular to the line joining the centres of $\\Gamma, \\Omega$. Now it is well-known that $N$ is the centre of $\\Omega$, and, since $\\angle D I N=90^{\\circ}$, the centre of $\\Gamma$ is the midpoint of segment $[D N]$. This shows that $I Q \\perp D N$.\n\nFinally, let $D N$ meet $\\omega$ again at $X$. Since $D I \\perp I N$ by construction and $N$ is the centre of $\\Omega, D I$ is tangent to $\\Omega$ at $I$. As $D$ lies on the radical axis $B C$ of $\\omega, \\Omega$, it follows that $|D I|^{2}=|D B||D C|=|D X||D N|$. Hence triangles $D N I$ and $D I X$ are similar; in particular, $\\angle D X I=\\angle D I N=90^{\\circ}$, i.e. $I X \\perp D N$. Since $I Q \\perp D N$, it follows that $X$ is the intersection of $I Q$ and $D N$. Since $X$ lies on $\\omega$ by construction, this completes the proof.\n\n#"}
+{"problem": "Let $A B C$ be a triangle with incentre $I$ and circumcircle $\\omega$. Let $N$ denote the second point of intersection of line $A I$ and $\\omega$. The line through $I$ perpendicular to $A I$ intersects line $B C$, segment $[A B]$, and segment $[A C]$ at the points $D, E$, and $F$, respectively. The circumcircle of triangle $A E F$ meets $\\omega$ again at $P$, and lines $P N$ and $B C$ intersect at $Q$. Prove that lines $I Q$ and $D N$ intersect on $\\omega$.\n\n#", "solution": "Since $A P E F$ and $A P B C$ are cyclic,\n\n$$\n\\begin{aligned}\n\\angle C P F & =\\angle B P A-\\angle B P C-\\angle F P A=\\left(180^{\\circ}-\\angle B C A\\right)-\\angle B A C-\\angle F E A \\\\\n& =\\left(180^{\\circ}-\\angle B C A-\\angle B A C\\right)-\\angle B E D=\\angle C B A-\\angle B E D=\\angle C B E-\\angle B E D=\\angle B D E=\\angle C D F,\n\\end{aligned}\n$$\n\nso $D P F C$ is cyclic, too. Thence $\\angle C P D=\\angle C F D=180^{\\circ}-\\angle I F A=90^{\\circ}+\\angle I A F=90^{\\circ}+\\angle C A N=90^{\\circ}+\\angle C P N$. Hence $\\angle D P N=\\angle C P D-\\angle C P N=90^{\\circ}$. Since $\\angle D I N=90^{\\circ}$ by construction, it follows that $D P I N$ is cyclic, with\ncircumcircle $\\Gamma$. Let $J$ be the second intersection of line $I Q$ and $\\Gamma$. Moreover, it is well-known that $N$ is the centre of the circumcircle $\\Omega$ of $B I C$. In particular, $|N I|=|N B|$, and so, since $N I P J$ and $N B P C$ are cyclic,\n\n$$\n\\frac{|J Q|}{|J P|}=\\frac{|N Q|}{|N I|}=\\frac{|N Q|}{|N B|}=\\frac{|C Q|}{|C P|}\n$$\n\nLet $S$ now be the point of intersection of $P N$ and $\\Omega$ such that $P, N, S$ lie on line $P N$ in this order. By construction, $\\angle Q P C=\\angle N P C=\\angle N A C=\\angle B A N=\\angle B C N=\\angle Q C N$, so triangles $C Q N$ and $P C N$ are similar, whence\n\n$$\n\\frac{|C Q|}{|C P|}=\\frac{|N C|}{|N P|}=\\frac{|N Q|}{|N C|}=\\frac{|N C|+|N Q|}{|N C|+|N P|}=\\frac{|N S|+|N Q|}{|N S|+|N P|}=\\frac{|S Q|}{|S P|}\n$$\n\nCombining (1) and (2) shows that $C, J, S$ lie on a circle of Apollonius, the centre of which lies on the line through $P, Q, N, S$, so, since $|N C|=|N S|$ by construction, is $N$. In other words, $J$ lies on $\\Omega$.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_a8b8dd435251ab822ad2g-6.jpg?height=1357&width=1319&top_left_y=932&top_left_x=377)\n\nIn particular, $|N I|=|N J|$. Now, by construction, $\\angle D I N=\\angle D J N=90^{\\circ}$, so the right-angled triangles $D I N$ and $D J N$ are congruent, whence $D I N J$ is a kite. In particular, $I J \\perp D N$. Since $Q$ lies on $I J$ by definition, this shows that $I Q \\perp D N$. We can now conclude as in Solution 2.\n\n#"}
+{"problem": "Let $A B C$ be a triangle with incentre $I$ and circumcircle $\\omega$. Let $N$ denote the second point of intersection of line $A I$ and $\\omega$. The line through $I$ perpendicular to $A I$ intersects line $B C$, segment $[A B]$, and segment $[A C]$ at the points $D, E$, and $F$, respectively. The circumcircle of triangle $A E F$ meets $\\omega$ again at $P$, and lines $P N$ and $B C$ intersect at $Q$. Prove that lines $I Q$ and $D N$ intersect on $\\omega$.\n\n#", "solution": "By construction, $P$ is the Miquel point of quadrilateral $B C F E$ (and the resulting complete quadrilateral with points $A$ and $D$ added) because it is the intersection of $\\omega$ and the circumcircle of triangle $A E F$. In particular, $D B E P$ is\n\n## BxMO 2023: Problems and Solutions\n\ncyclic. It follows that $\\angle I D P=\\angle E D P=\\angle E B P=\\angle A B P=\\angle A N P=\\angle I N P$ since $A P B N$ is cyclic, and so $P D N I$ is also cyclic.\n\n![](https://cdn.mathpix.com/cropped/2024_04_17_a8b8dd435251ab822ad2g-7.jpg?height=959&width=1307&top_left_y=474&top_left_x=383)\n\nNext, let $X$ be the intersection of $D N$ and $\\omega$ and let $A N$ meet $B C$ at $Y$. Then $\\angle N A C=\\angle A / 2=\\angle N C B$, so $\\angle B Y A=\\angle C+\\angle N A C=\\angle C+\\angle N C B=\\angle N C A$ and hence\n\n$$\n\\begin{aligned}\n\\angle D Q P & =\\angle N Q Y=\\angle Q Y A-\\angle Q N Y=\\angle B Y A-\\angle P N A \\\\\n& =\\angle N C A-\\angle P C A=\\angle P C N=180^{\\circ}-\\angle N X P=\\angle D X P\n\\end{aligned}\n$$\n\nThis implies that $D X Q P$ is cyclic. In particular, $Q X \\perp D N$. It now suffices to show that $I X \\perp D N$, which we do in the same way as in Solution 2.\n\n## BxMO 2023: Problems and Solutions\n\n#"}
+{"problem": "A positive integer $n$ is friendly if every pair of neighbouring digits of $n$, written in base 10, differs by exactly 1 . For example, 6787 is friendly, but 211 and 901 are not.\n\nFind all odd natural numbers $m$ for which there exists a friendly integer divisible by $64 m$.\n\n#", "solution": "Any friendly number divisible by 64 is divisible by 4 , and hence the number formed by its last two digits is a multiple of 4 , so ends in $00,04,08, \\ldots$, or 96 . A friendly number divisible by 4 must therefore end in $12,32,56$, or 76 , so cannot be divisible by 5 . In particular, if $5 \\mid m$, then there is no friendly integer divisible by $64 m$.\n\nWe claim that conversely, if $m$ is odd and $5 \\nmid m$, then there exists a friendly integer divisible by $64 m$. First, we notice that $343232=64 \\cdot 5363$ is a friendly number divisible by 64 , and hence so is\n\n$$\nN_{k}=343232343232 \\cdots 343232=343232 \\cdot\\left(1+10^{6}+\\cdots+10^{6 k}\\right) \\quad \\text { for } k=0,1,2, \\ldots\n$$\n\nNow the sequence $N_{0}, N_{1}, N_{2}, \\ldots$ eventually repeats modulo $m$, i.e. there exist positive integers $k<\\ell$ such that $N_{\\ell} \\equiv N_{k}(\\bmod m)$. Hence $m \\mid N_{\\ell}-N_{k}=10^{6(k+1)} N_{\\ell-k-1}$. Since $m$ is odd and $5 \\nmid m,(10, m)=1$, so $m \\mid N_{\\ell-k-1}$. By construction, $64 \\mid N_{\\ell-k-1}$. Thus, as $m$ is odd and hence $(64, m)=1$, we conclude that $64 m \\mid N_{\\ell-k-1}$. This completes the proof.\n\nThe solution divides into two parts: (1) showing that, if $5 \\mid m$, then there is no friendly integer divisible by $64 m$; (2) showing that, if $5 \\nmid m$, then there is a friendly integer divisible by $64 m$.\n\nAlternative solution for part (1). If $5 \\mid m$, then $20 \\mid 64 m$. The last two digits of a multiple of 20 are $00,20,40,60$, or 80 , so this number is not friendly. Thus, if $m$ is odd and $5 \\mid m$, then there is no friendly integer divisible by $64 m$.\n\nAlternative solution for part (2). Notice that $N_{k}=343232 \\cdot\\left(10^{6(k+1)}-1\\right) /\\left(10^{6}-1\\right)$. Let $M=m\\left(10^{6}-1\\right)$. Since $5 \\nmid m$ and $m$ is odd, $(10, M)=1$, so, taking $k=\\varphi(M)-1$, we get $10^{6(k+1)}=10^{6 \\varphi(M)} \\equiv 1(\\bmod M)$ by the Euler-Fermat theorem, i.e. $m \\mid\\left(10^{6(k+1)}-1\\right) /\\left(10^{6}-1\\right)$, and hence $m \\mid N_{k}$.\n\nAlternative constructions of the integers $N_{\\boldsymbol{k}}$ for part (2). Direct calculation shows that friendly integers divisible by 64 end in $343232,543232,123456$, or 323456 , so the numbers $N_{k}$ defined in the solution of part (2) above may be replaced by, for instance,\n\n$$\n34543232 \\cdot\\left(1+10^{8}+\\cdots+10^{8 k}\\right), 5432123456 \\cdot\\left(1+10^{10}+\\cdots+10^{10 k}\\right), 54323456 \\cdot\\left(1+10^{8}+\\cdots+10^{8 k}\\right)\n$$\n\nRemark. Interestingly, friendly numbers cannot be divisible by arbitrarily high powers of 2 . Direct calculation shows that the 60-digit friendly integer 101232121234323456543434343210121212323434343234565656543232 is divisible by $2^{60}$, but that there is no friendly integer divisible by $2^{61}$. The problem selection committee is not aware of a proof of this fact that eschews direct calculation."}

Benelux/segmented/Benelux_en-olympiad_en-bxmo-problems-2024-zz.jsonl

@@ -0,0 +1,10 @@
+{"problem": "(a) Let $a_{0}, a_{1}, \\ldots, a_{2024}$ be real numbers such that $\\left|a_{i+1}-a_{i}\\right| \\leqslant 1$ for $i=0,1, \\ldots, 2023$. Find the minimum possible value of $$ a_{0} a_{1}+a_{1} a_{2}+\\cdots+a_{2023} a_{2024} $$ (b) Does there exist a real number $C$ such that $$ a_{0} a_{1}-a_{1} a_{2}+a_{2} a_{3}-a_{3} a_{4}+\\cdots+a_{2022} a_{2023}-a_{2023} a_{2024} \\geqslant C $$ for all real numbers $a_{0}, a_{1}, \\ldots, a_{2024}$ such that $\\left|a_{i+1}-a_{i}\\right| \\leqslant 1$ for $i=0,1, \\ldots, 2023$ ?", "problem_label": "Problem 1", "problem_type": null, "solution": "(a) The minimum value is -506 . Note that from $\\left|a_{i}-a_{i-1}\\right| \\leq 1$ it follows that\n\n$$\na_{i} a_{i-1}=\\frac{\\left(a_{i}+a_{i-1}\\right)^{2}-\\left(a_{i}-a_{i-1}\\right)^{2}}{4} \\geq-\\frac{\\left(a_{i}-a_{i-1}\\right)^{2}}{4} \\geq-\\frac{1}{4}\n$$\n\nAdding this for $i=1,2, \\ldots, 2024$, we obtain that\n\n$$\na_{0} a_{1}+a_{1} a_{2}+a_{2} a_{3}+\\ldots+a_{2023} a_{2024} \\geq 2024 \\cdot-\\frac{1}{4}=-506\n$$\n\nWe now show that this value can be attained. Indeed, for the sequence $\\left(a_{0}, a_{1}, \\ldots, a_{2024}\\right)=$ $\\left(\\frac{1}{2},-\\frac{1}{2}, \\frac{1}{2},-\\frac{1}{2}, \\frac{1}{2}, \\ldots, \\frac{1}{2}\\right)$ with alternating $\\frac{1}{2}$ 's and $-\\frac{1}{2}$ 's, each term $a_{i} a_{i-1}$ is equal to $-\\frac{1}{4}$, leading to $a_{0} a_{1}+a_{1} a_{2}+a_{2} a_{3}+\\ldots+a_{2023} a_{2024}=2024 \\cdot-\\frac{1}{4}=-506$.\n(b) No, such a $C$ does not exist. We argue by contradiction. Suppose $C$ has this property, and consider the sequence defined by $a_{0}=C$ and $a_{i}=C-1$ for $i=1,2, \\ldots, 2024$ satisfies the condition in the problem. For this sequence, we have $a_{i} a_{i+1}-a_{i+1} a_{i+2}=0$ for $i=2,4, \\ldots$, 2022, so the sum\n\n$$\na_{0} a_{1}-a_{1} a_{2}+a_{2} a_{3}-a_{3} a_{4}+a_{4} a_{5}-a_{5} a_{6}+\\ldots+a_{2022} a_{2023}-a_{2023} a_{2024}\n$$\n\nis equal to\n\n$$\na_{0} a_{1}-a_{1} a_{2}=C(C-1)-(C-1)^{2}=C-1<C,\n$$\n\ncontradiction."}
+{"problem": "(a) Let $a_{0}, a_{1}, \\ldots, a_{2024}$ be real numbers such that $\\left|a_{i+1}-a_{i}\\right| \\leqslant 1$ for $i=0,1, \\ldots, 2023$. Find the minimum possible value of $$ a_{0} a_{1}+a_{1} a_{2}+\\cdots+a_{2023} a_{2024} $$ (b) Does there exist a real number $C$ such that $$ a_{0} a_{1}-a_{1} a_{2}+a_{2} a_{3}-a_{3} a_{4}+\\cdots+a_{2022} a_{2023}-a_{2023} a_{2024} \\geqslant C $$ for all real numbers $a_{0}, a_{1}, \\ldots, a_{2024}$ such that $\\left|a_{i+1}-a_{i}\\right| \\leqslant 1$ for $i=0,1, \\ldots, 2023$ ?", "problem_label": "Problem 1", "problem_type": null, "solution": "(a) The minimum value is -506 . Note that from $\\left|a_{i}-a_{i-1}\\right| \\leq 1$ it follows that\n\n$$\na_{i} a_{i-1}=\\frac{\\left(a_{i}+a_{i-1}\\right)^{2}-\\left(a_{i}-a_{i-1}\\right)^{2}}{4} \\geq-\\frac{\\left(a_{i}-a_{i-1}\\right)^{2}}{4} \\geq-\\frac{1}{4}\n$$\n\nAdding this for $i=1,2, \\ldots, 2024$, we obtain that\n\n$$\na_{0} a_{1}+a_{1} a_{2}+a_{2} a_{3}+\\ldots+a_{2023} a_{2024} \\geq 2024 \\cdot-\\frac{1}{4}=-506\n$$\n\nWe now show that this value can be attained. Indeed, for the sequence $\\left(a_{0}, a_{1}, \\ldots, a_{2024}\\right)=$ $\\left(\\frac{1}{2},-\\frac{1}{2}, \\frac{1}{2},-\\frac{1}{2}, \\frac{1}{2}, \\ldots, \\frac{1}{2}\\right)$ with alternating $\\frac{1}{2}$ 's and $-\\frac{1}{2}$ 's, each term $a_{i} a_{i-1}$ is equal to $-\\frac{1}{4}$, leading to $a_{0} a_{1}+a_{1} a_{2}+a_{2} a_{3}+\\ldots+a_{2023} a_{2024}=2024 \\cdot-\\frac{1}{4}=-506$.\n(b) We give an alternative construction for part (b). We choose a real constant $N$, from which we define $a_{i}=N+i$ for each $i=0,1, \\ldots, 2024$, which clearly satisfies the requirement $\\left|a_{i}-a_{i-1}\\right| \\leq 1$ for each $i=0,1, \\ldots, 1011$. Then, it can be seen that for each $i=0,1, \\ldots, 1011$ that\n\n$$\na_{2 i} a_{2 i+1}-a_{2 i+1} a_{2 i+2}=a_{2 i+1}\\left(a_{2 i}-a_{2 i+2}\\right)=-2(N+2 i+1) \\leqslant-2 N .\n$$\n\nFrom this, it can be concluded that\n$a_{0} a_{1}-a_{1} a_{2}+a_{2} a_{3}-a_{3} a_{4}+a_{4} a_{5}-a_{5} a_{6}+\\ldots+a_{2022} a_{2023}-a_{2023} a_{2024} \\leqslant 1012 \\cdot-2 N=-2024 N$.\nAs $N$ is a constant which can be arbitrarily chosen, there is no constant $C$ which lower bounds the given expression."}
+{"problem": "Let $n$ be a positive integer. In a coordinate grid, a path from $(0,0)$ to $(2 n, 2 n)$ consists of $4 n$ consecutive unit steps $(1,0)$ or $(0,1)$. Prove that the number of paths that divide the square with vertices $(0,0)$, $(2 n, 0),(2 n, 2 n),(0,2 n)$ into two regions with even areas is $$ \\frac{\\binom{4 n}{2 n}+\\binom{2 n}{n}}{2} $$", "problem_label": "Problem 2", "problem_type": null, "solution": "Let $X$ denote the set of paths for which $A$ and $B$ have even area and let $Y$ denote the set of paths for which $A$ and $B$ both have odd area. Because $A$ and $B$ together form a square of area $4 n^{2}$, which is even, $|X|+|Y|$ equals the total number of paths from $(0,0)$ to $(2 n, 2 n)$, which is $\\binom{4 n}{2 n}$.\n\nDenoting a step to the right by $R$ and a step upwards by $U$, every path from $(0,0)$ to $(2 n, 2 n)$ can be described as a sequence of $4 n$ symbols, $2 n$ of which are $R$ and $2 n$ of which are $U$. We subdivide such a sequence into $2 n$ pairs of consecutive steps that can be $R R, U R, R U$ or $U U$. The number of possible paths for which neither $U R$ nor $R U$ occurs is $\\binom{2 n}{n}$, because out of $2 n$ pairs that can be either $R R$ or $U U$ we have to choose $n$ that will be $R R$. These $\\binom{2 n}{n}$ all belong to $X$; in fact, for these paths, $A$ and $B$ can be subdivided into $2 \\times 2$-square, making their areas divisible by 4 . Now consider the paths that contain at least one $U R$ - or $R U$-pair. If in such a path we replace the first occurrence of a $U R$ or $R U$-pair by a pair of the other type (thus replacing $U R$ by $R U$ or vice versa), the areas of $A$ and $B$ each change by 1 and therefore become even if they were odd and odd if they were even. Because this modification is clearly reversible, we conclude that we can pair up all paths that contain at least one $U R$ - or $R U$-pair into pairs of paths, one of which belongs to $X$ and one of belongs to $Y$. This implies that $|X|-\\binom{2 n}{n}=|Y|$. It follows that\n\n$$\n|X|=\\frac{|X|+|Y|}{2}+\\frac{|X|-|Y|}{2}=\\frac{\\binom{4 n}{2 n}}{2}+\\frac{\\binom{2 n}{n}}{2} .\n$$"}
+{"problem": "Let $n$ be a positive integer. In a coordinate grid, a path from $(0,0)$ to $(2 n, 2 n)$ consists of $4 n$ consecutive unit steps $(1,0)$ or $(0,1)$. Prove that the number of paths that divide the square with vertices $(0,0)$, $(2 n, 0),(2 n, 2 n),(0,2 n)$ into two regions with even areas is $$ \\frac{\\binom{4 n}{2 n}+\\binom{2 n}{n}}{2} $$", "problem_label": "Problem 2", "problem_type": null, "solution": "Define $Z_{m, n}$ to be the number of routes from $(0,0)$ to $(2 m, 2 n)$ that divide the rectangle with vertices $(0,0),(0,2 n),(2 m, 2 n)$ and $(2 m, 0)$ into two parts of even area. We call such paths good. We claim that\n\n$$\n2 Z_{m, n}=\\binom{2 m+2 n}{2 m}+\\binom{m+n}{m}\n$$\n\nfor all $m, n$, which for $m=n$ establishes the desired result. We prove this formula by induction on $m+n$, noting first that it clearly holds when either $m$ or $n$ is zero, because $Z_{0, n}=Z_{m, 0}=1$ (there is only one path from $(0,0)$ to $(0,2 n)$ or $(2 m, 0)$, which is good). Therefore, suppose that $m, n \\geq 1$ and consider a good path from $(0,0)$ to $(2 m, 2 n)$. This path passes through exactly one of $(2 m, 2 n-2)$,\n\n## Remark\n\nThe result from Solution 2 can also be proved using a combinatorial argument like in Solution 1."}
+{"problem": "Let $n$ be a positive integer. In a coordinate grid, a path from $(0,0)$ to $(2 n, 2 n)$ consists of $4 n$ consecutive unit steps $(1,0)$ or $(0,1)$. Prove that the number of paths that divide the square with vertices $(0,0)$, $(2 n, 0),(2 n, 2 n),(0,2 n)$ into two regions with even areas is $$ \\frac{\\binom{4 n}{2 n}+\\binom{2 n}{n}}{2} $$", "problem_label": "Problem 2", "problem_type": null, "solution": "We start by proving the following lemma: for a\n\n$$\n(2 m-1,2 n-1)\n$$\n\n-grid, there are equally many paths with a region of even area (called even paths), as there are with odd area (odd paths).\n\nTo prove this, we take a path and rotate it around the center of the grid. Then a path spanning a region with area $x$ is mapped on one spanning an area $(2 m-1)(2 n-1)-x$. This gives a bijection between paths creating an even region, and creating an odd region.\n\nNow, for every path from\nto $(2 n, 2 n)$, consider all the coordinates of the grid points it visits in order. There are $\\binom{2 n}{n}$ of them which never visit a point with odd coordinates (which we call an odd point). Notice that such paths are all even.\n\nWe now construction a bijection between the remaining even paths and the odd paths. For each odd point, there are equally many even as odd paths from $(0,0)$ to that point. Define then a bijection $\\phi$ between the sets of odd paths and even paths up to this point for each point. Notice that $\\phi$ implicitly depends on the chosen odd point.\n\nNow, for an arbitrary odd path $P$, consider the first odd point it passes through. Map $P$ to another path by changing the path up to this odd point to the $\\phi$ of the path up to this point. As $\\phi$ maps between even and odd paths, the resulting path is an even path. By the definition of $\\phi$, and as each of the remaining paths goes to an odd point, this mapping defines a bijection.\n\nWe have hence found a bijection between the odd and even paths in the remaining $\\binom{4 n}{2 n}-\\binom{2 n}{n}$ paths, which yields the required result like in solution 1."}
+{"problem": "Let $A B C$ be a triangle with incentre $I$ and circumcircle $\\Omega$ such that $|A C| \\neq|B C|$. The internal angle bisector of $\\angle C A B$ intersects side [BC] in $D$, and the external angle bisectors of $\\angle A B C$ and $\\angle B C A$ intersect $\\Omega$ again in $E$ and $F$, respectively. Let $G$ be the intersection of lines $A E$ and $F I$ and let $\\Gamma$ be the circumcircle of triangle $B D I$. Show that $E$ lies on $\\Gamma$ if and only if $G$ lies on $\\Gamma$. ![](https://cdn.mathpix.com/cropped/2024_12_15_e2e131dd99265ed8ccbbg-06.jpg?height=1140&width=1685&top_left_y=612&top_left_x=194) Solution 1 We first notice the general fact that $E F \\perp A I$. This can be proved using the following argument. Denote $S$ for the intersection of $E F$ and $A I$. Then $\\angle B I S=(\\angle I B A+\\angle I A B)=\\frac{1}{2}(\\angle A B C+\\angle B A C)=$ $\\frac{1}{2}\\left(180^{\\circ}-\\angle B C A\\right)=\\angle B C F=\\angle B E F=\\angle B E S$. Thus $B, I, S, E$ form a cyclic quadrilateral, from which it can be concluded that $E F \\perp A I$. We will now continue to prove the problem by doing both directions separately. Assume first that $E$ lies on $\\Gamma$. Then, as angle bisectors are perpendicular to one another, we have that $\\angle I D E=\\angle I B E=90^{\\circ}$. Then, as $E F \\perp A I$, it holds that $D$ lies on $E F$. It can then be concluded that $\\angle I D F=90^{\\circ}=\\angle I C F$ (again due to perpendicular bisectors), from which it can be concluded that $I, D, C, F$ form a cyclic quadrilateral. We can now calculate that $\\angle G I D=180^{\\circ}-\\angle F I D=180^{\\circ}-\\angle F C D=180^{\\circ}-\\angle F C B=$ $\\angle A E F=\\angle G E D$, where the second to last step follows from the fact that the arcs $B F$ and $A F$ have the", "problem_label": "Problem 3", "problem_type": null, "solution": "The external angle bisectors $B E$ and $C F$ meet the internal bisector $I D$ at the $A$-excentre $J$ of triangle $A B C$. If one of BEDI and CFID is cyclic, then, as BECF is cyclic, $|J D||J I|=|J E||J B|=|J C||J F|$ by power of a point, and so the other is cyclic, too. This shows that\n(1) $B E D I$ is cyclic if and only if CFID is cyclic.\n\nNext, $\\angle G A D=\\angle E A D=\\angle B A D-\\angle B A E=\\angle A / 2-\\left(180^{\\circ}-\\angle A E B-\\angle E B A\\right)$, with $\\angle A E B=\\angle A C B=$ $\\angle C$ and $\\angle E B A=\\angle E B I+\\angle I B A=90^{\\circ}+\\angle B / 2$. Hence $\\angle G A D=\\angle A / 2+\\angle B / 2+\\angle C-90^{\\circ}=\\angle C / 2=$ $\\angle I C D$. Thus, if one of $C F I D$ and $A G D F$ is cyclic, then $\\angle I C D=\\angle I F D=\\angle G F D=\\angle G A D$, and the other is cyclic, too. We have thus established that\n(2) CFID is cyclic if and only if $A G D F$ is cylic.\n\nFinally, let $A^{\\prime}$ denote the second intersection of $A I$ with $\\Omega$. Then $\\angle D A F=\\angle A^{\\prime} A F=180^{\\circ}-\\angle F C A^{\\prime}$, with $\\angle F C A^{\\prime}=\\angle F C I+\\angle I C B+\\angle B C A^{\\prime}=90^{\\circ}+\\angle C / 2+\\angle B A A^{\\prime}=90^{\\circ}+\\angle C / 2+\\angle A / 2$. It follows that $\\angle D A F=90^{\\circ}-\\angle A / 2-\\angle C / 2=\\angle B / 2=\\angle D B I$. Hence, if one of $A G D F$ and $B D I G$ is cyclic, then $\\angle D A F=\\angle D G F=\\angle D G I=\\angle D B I$, and so the other is cyclic, too. Hence\n(3) $A G D F$ is cyclic if and only if $B D I G$ is cylic.\n\nThese three equivalences prove that $B E D I$ is cyclic if and only if $B D I G$ is cyclic, i.e. $E$ lies on $\\omega$ if and only if $G$ does. This completes the proof."}
+{"problem": "Let $A B C$ be a triangle with incentre $I$ and circumcircle $\\Omega$ such that $|A C| \\neq|B C|$. The internal angle bisector of $\\angle C A B$ intersects side [BC] in $D$, and the external angle bisectors of $\\angle A B C$ and $\\angle B C A$ intersect $\\Omega$ again in $E$ and $F$, respectively. Let $G$ be the intersection of lines $A E$ and $F I$ and let $\\Gamma$ be the circumcircle of triangle $B D I$. Show that $E$ lies on $\\Gamma$ if and only if $G$ lies on $\\Gamma$. ![](https://cdn.mathpix.com/cropped/2024_12_15_e2e131dd99265ed8ccbbg-06.jpg?height=1140&width=1685&top_left_y=612&top_left_x=194) Solution 1 We first notice the general fact that $E F \\perp A I$. This can be proved using the following argument. Denote $S$ for the intersection of $E F$ and $A I$. Then $\\angle B I S=(\\angle I B A+\\angle I A B)=\\frac{1}{2}(\\angle A B C+\\angle B A C)=$ $\\frac{1}{2}\\left(180^{\\circ}-\\angle B C A\\right)=\\angle B C F=\\angle B E F=\\angle B E S$. Thus $B, I, S, E$ form a cyclic quadrilateral, from which it can be concluded that $E F \\perp A I$. We will now continue to prove the problem by doing both directions separately. Assume first that $E$ lies on $\\Gamma$. Then, as angle bisectors are perpendicular to one another, we have that $\\angle I D E=\\angle I B E=90^{\\circ}$. Then, as $E F \\perp A I$, it holds that $D$ lies on $E F$. It can then be concluded that $\\angle I D F=90^{\\circ}=\\angle I C F$ (again due to perpendicular bisectors), from which it can be concluded that $I, D, C, F$ form a cyclic quadrilateral. We can now calculate that $\\angle G I D=180^{\\circ}-\\angle F I D=180^{\\circ}-\\angle F C D=180^{\\circ}-\\angle F C B=$ $\\angle A E F=\\angle G E D$, where the second to last step follows from the fact that the arcs $B F$ and $A F$ have the", "problem_label": "Problem 3", "problem_type": null, "solution": "This proof only shows $E \\in \\omega \\Longrightarrow G \\in \\omega$. Note that this argument cannot be used straightforwardly to prove the converse implication.\n\nIf $B E D I$ is cyclic, let $A^{\\prime}$ be the the second intersection of $A I$ with $\\Omega$, so $\\angle I A^{\\prime} E=\\angle A A^{\\prime} E=$ $180^{\\circ}-\\angle E B A$, with $\\angle E B A=\\angle E B I+\\angle I B A=90^{\\circ}+\\angle B / 2$, so $\\angle I A^{\\prime} E=90^{\\circ}-\\angle B / 2$. But $\\angle E I A^{\\prime}=\\angle E B D$ since $E B D I$ is cyclic, with $\\angle E B D=\\angle E B I-\\angle D B I=90^{\\circ}-\\angle B / 2$. Thus $\\angle E A^{\\prime} I=\\angle E I A^{\\prime}$, so $E A^{\\prime} I$ is isosceles. Moreover, since $E B D I$ is cyclic and $\\angle E B I$ is a right angle, so is $\\angle I D E$. It follows that $D$ is the midpoint of $[A I]$. Next, the external bisectors $B E$ and $C F$ and the internal bisector $I D$ meet at the $A$-excentre $J$ of triangle $A B C$. By power of a point, since $B E D I$ and $B E C F$ are cyclic, $|J D||J I|=|J E||J B|=|J C||J F|$, so $C F I D$ is cyclic, too, with $\\angle F D I=\\angle F C I=90^{\\circ}$. We have thus shown that $\\angle A D E=\\angle F D A=90^{\\circ}$, so $D$ is the foot of the altitude from $A$ in triangle $A E F$. Moreover, all of this shows that $I$ is the reflection of $A^{\\prime}$, which is the point at which this altitude meets the circumcircle $\\Omega$ of $A E F$ again, in the side $[E F]$. Hence $I$ is the orthocentre of triangle $A E F$. By extension, $F I$ is its altitude from $F$, and $G$ is the foot of this altitude, so $\\angle E G I=90^{\\circ}=\\angle I D E=\\angle E B I$, and hence $B E D I G$ is cyclic. This shows that if $E$ lies on\n\n## Remark\n\nThe equivalence $E \\in \\omega \\Longleftrightarrow G \\in \\omega$ breaks down if triangle $A B C$ is not scalene. Indeed, if $A B C$ is isosceles with $\\angle A=\\angle B$, then $F=C$ and $G$ is the intersection of $C I$ and $A E$. Angle chasing as above then shows that $\\angle G A D=\\angle C / 2=\\angle G C D$, so $A G D C$ is cyclic. Then $\\angle D G I=\\angle D G C=\\angle D A C=$ $\\angle A / 2=\\angle B / 2=\\angle D B I$, so $B D I G$ is cyclic, i.e. $G \\in \\omega$. Next, the $A$-excentre $J$ of triangle $A B C$ is the intersection of $B E, I D$, and the tangent to $\\Omega$ at $C$, which, being the external bisector of $\\angle C$, is perpendicular to $C I$. Hence, if $E \\in \\omega$, too, i.e. if $B E D I$ were cyclic, then by power of a point, $|J C|^{2}=|J E||J B|=|J D||J I|$, so the circumcircle of $C I D$ would be tangent to $J C$ at $C$, implying $\\angle C D I=\\angle I C J=90^{\\circ}$. This is not however the case, unless $\\angle B=\\angle C$ and hence $A B C$ is equilateral. This shows that the condition in the problem statement that $A B C$ is scalene cannot be dropped.\n![](https://cdn.mathpix.com/cropped/2024_12_15_e2e131dd99265ed8ccbbg-09.jpg?height=700&width=809&top_left_y=1089&top_left_x=629)"}
+{"problem": "For each positive integer $n$, let $\\operatorname{rad}(n)$ denote the product of the distinct prime factors of $n$. Show that there exist integers $a, b>1$ such that $\\operatorname{gcd}(a, b)=1$ and $$ \\operatorname{rad}(a b(a+b))<\\frac{a+b}{2024^{2024}} $$ For example, $\\operatorname{rad}(20)=\\operatorname{rad}\\left(2^{2} \\cdot 5\\right)=2 \\cdot 5=10$ and $\\operatorname{rad}(18)=\\operatorname{rad}\\left(2 \\cdot 3^{2}\\right)=2 \\cdot 3=6$.", "problem_label": "Problem 4", "problem_type": null, "solution": "We show that the pair $(a, b)$ of the form $a=2^{p(p-1)}, b=3^{p(p-1)}-2^{p(p-1)}$ for sufficiently larger prime number $p$ satisfies the inequality. First, notice that $\\operatorname{gcd}(a, b)=\\operatorname{gcd}(a, a+b)=1$ indeed. In addition, see that $\\operatorname{rad}(a)=2$, and $\\operatorname{rad}(a+b)=3$. Because of Euler-Fermat, as $\\phi\\left(p^{2}\\right)=p(p-1)$, it can directly be seen that $p^{2} \\mid b$. In this case, $\\operatorname{rad}(b) \\leqslant \\frac{b}{p}$. It then follows that, as rad is multiplicative for coprime numbers, that\n\n$$\n\\operatorname{rad}(a b(a+b))=\\operatorname{rad}(a) \\operatorname{rad}(b) \\operatorname{rad}(a+b) \\leqslant 2 \\cdot 3 \\cdot \\frac{b}{p} \\leqslant \\frac{6}{p}(a+b)\n$$\n\nThen, by choosing $p$ such that $\\frac{6}{p}<\\frac{1}{2024^{2024}}$, we found $a$ and $b$ satifying the inequality."}
+{"problem": "For each positive integer $n$, let $\\operatorname{rad}(n)$ denote the product of the distinct prime factors of $n$. Show that there exist integers $a, b>1$ such that $\\operatorname{gcd}(a, b)=1$ and $$ \\operatorname{rad}(a b(a+b))<\\frac{a+b}{2024^{2024}} $$ For example, $\\operatorname{rad}(20)=\\operatorname{rad}\\left(2^{2} \\cdot 5\\right)=2 \\cdot 5=10$ and $\\operatorname{rad}(18)=\\operatorname{rad}\\left(2 \\cdot 3^{2}\\right)=2 \\cdot 3=6$.", "problem_label": "Problem 4", "problem_type": null, "solution": "We show that the pair $(a, b)$ of the form $a=3^{2^{k}}, b=5^{2^{k}}-3^{2^{k}}$ for sufficiently large $k$ satisifies the inequality. Again, we have that $\\operatorname{gcd}(a, b)=\\operatorname{gcd}(a, a+b)=1$. Similarly to solution $1, \\operatorname{rad}(a(a+b))=$ $\\operatorname{rad}(a) \\operatorname{rad}(a+b)=3 \\cdot 5=15$. Then, we will show that $2^{k+1} \\mid b$, from which it would follow that $\\operatorname{rad}(b) \\leqslant \\frac{b}{2^{k}}$. From this, we then see that\n\n$$\n\\operatorname{rad}(a b(a+b))=\\operatorname{rad}(a(a+b)) \\operatorname{rad}(b) \\leqslant \\frac{15 b}{2^{k}} .\n$$\n\nLike in solution 1, this gives a pair $(a, b)$ satisfying the inequality for sufficiently large $k$.\nThere are various ways to show that $2^{k+1} \\mid 5^{2^{k}}-3^{2^{k}}$, for example directly by applying the Lifting-The-Exponent Lemma. For a more elementary proof, we can apply induction on $k$. The statement is clearly true for $k=0$, and if the statement holds for $k=n$, then for $k=n+1$ we see that\n\n$$\n5^{2^{k}}-3^{2^{k}}=5^{2^{n+1}}-3^{2^{n+1}}=\\left(5^{2^{n}}\\right)^{2}-\\left(3^{2^{n}}\\right)^{2}=\\left(5^{2^{n}}-3^{2^{n}}\\right)\\left(5^{2^{n}}+3^{2^{n}}\\right) .\n$$\n\nFrom the induction hypothesis, the first factor has $n+1$ factors of 2 . As the second factor is a sum of two odd numbers, the second term has at least one factor of 2 . The product thus has at least $n+2=k+1$ factors of 2 , from which the statement follows by induction."}
+{"problem": "For each positive integer $n$, let $\\operatorname{rad}(n)$ denote the product of the distinct prime factors of $n$. Show that there exist integers $a, b>1$ such that $\\operatorname{gcd}(a, b)=1$ and $$ \\operatorname{rad}(a b(a+b))<\\frac{a+b}{2024^{2024}} $$ For example, $\\operatorname{rad}(20)=\\operatorname{rad}\\left(2^{2} \\cdot 5\\right)=2 \\cdot 5=10$ and $\\operatorname{rad}(18)=\\operatorname{rad}\\left(2 \\cdot 3^{2}\\right)=2 \\cdot 3=6$.", "problem_label": "Problem 4", "problem_type": null, "solution": "Choose $a=\\left(4^{x}-1\\right)^{2}, b=4^{x+1}$ and $a+b=\\left(4^{x}+1\\right)^{2}$. That is, $a, b$ and $a+b$ are squares, where $b$ only contains the factor 2 . Note that $a$ and $b$ are indeed coprime. Then $\\operatorname{rad}(a b c)=2 \\operatorname{rad}\\left(16^{x}-1\\right)$.\n\nChoose then $x=5^{k}$ such that $x>2 \\cdot 2024^{2024}$. By Lifting the exponent, we know that $5^{k+1} \\mid 16^{5^{k}}-1$. This implies that $2 \\operatorname{rad}\\left(16^{x}-1\\right) \\leqslant 2\\left(16^{x}-1\\right) / 5^{k}<\\left(4^{x}+1\\right)^{2} / 2024^{2024}$.\n\nThis solution also works with Euler-Fermat, by choosing $x=\\phi\\left(5^{k+1}\\right)$ with $5^{k}>2024^{2024}$."}