year stringdate 1961-01-01 00:00:00 2025-01-01 00:00:00 ⌀ | tier stringclasses 5
values | problem_label stringclasses 119
values | problem_type stringclasses 13
values | exam stringclasses 28
values | problem stringlengths 87 2.77k | solution stringlengths 834 13k | metadata dict | problem_tokens int64 50 903 | solution_tokens int64 500 3.93k |
|---|---|---|---|---|---|---|---|---|---|
1998 | T3 | 9 | null | BalticWay | Let the numbers $\alpha, \beta$ satisfy $0<\alpha<\beta<\pi / 2$ and let $\gamma$ and $\delta$ be the numbers defined by the conditions:
(i) $0<\gamma<\pi / 2$, and $\tan \gamma$ is the arithmetic mean of $\tan \alpha$ and $\tan \beta$;
(ii) $0<\delta<\pi / 2$, and $\frac{1}{\cos \delta}$ is the arithmetic mean of $\... | Let $f(t)=\sqrt{1+t^{2}}$. Since $f^{\prime \prime}(t)=\left(1+t^{2}\right)^{-3 / 2}>0$, the function $f(t)$ is
strictly convex on $(0, \infty)$. Consequently,
$$
\begin{aligned}
\frac{1}{\cos \gamma} & =\sqrt{1+\tan ^{2} \gamma}=f(\tan \gamma)=f\left(\frac{\tan \alpha+\tan \beta}{2}\right)< \\
& <\frac{f(\tan \alpha)... | {
"problem_match": "\n9.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n9."
} | 140 | 1,297 |
1998 | T3 | 10 | null | BalticWay | Let $n \geqslant 4$ be an even integer. A regular $n$-gon and a regular $(n-1)$-gon are inscribed into the unit circle. For each vertex of the $n$-gon consider the distance from this vertex to the nearest vertex of the $(n-1)$-gon, measured along the circumference. Let $S$ be the sum of these $n$ distances. Prove that ... | For simplicity, take the length of the circle to be $2 n(n-1)$ rather than $2 \pi$. The vertices of the $(n-1)$-gon $A_{0} A_{1} \ldots A_{n-2}$ divide it into $n-1$ arcs of length $2 n$. By the pigeonhole principle, some two of the vertices of the $n$-gon $B_{0} B_{1} \ldots B_{n-1}$ lie in the same arc. Assume w.l.o.... | {
"problem_match": "\n10.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n10."
} | 113 | 947 |
1998 | T3 | 11 | null | BalticWay | Let $a, b$ and $c$ be the lengths of the sides of a triangle with circumradius $R$. Prove that
$$
R \geqslant \frac{a^{2}+b^{2}}{2 \sqrt{2 a^{2}+2 b^{2}-c^{2}}} .
$$
When does equality hold? | Answer: equality holds if $a=b$ or the angle opposite to $c$ is equal to $90^{\circ}$. Denote the angles opposite to the sides $a, b, c$ by $A, B, C$, respectively. By the law of sines we have $a=2 R \sin A, b=2 R \sin B, c=2 R \sin C$. Hence, the given inequality is equivalent to each of the following:
$$
\begin{alig... | {
"problem_match": "\n11.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n11."
} | 77 | 985 |
1998 | T3 | 16 | null | BalticWay | Is it possible to cover a $13 \times 13$ chessboard with forty-two tiles of size $4 \times 1$ so that only the central square of the chessboard remains uncovered? (It is assumed that each tile covers exactly four squares of the chessboard, and the tiles do not overlap.) | Answer: no.
Label the horizontal rows by integers from 1 to 13. Assume that the tiling is possible, and let $a_{i}$ be the number of vertical tiles with their outer squares in rows $i$ and $i+3$. Then $b_{i}=a_{i}+a_{i-1}+a_{i-2}+a_{i-3}$ is the number of vertical tiles intersecting row $i$ (here we assume $a_{j}=0$ i... | {
"problem_match": "\n16.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n16."
} | 66 | 820 |
1999 | T3 | 5 | null | BalticWay | The point $(a, b)$ lies on the circle $x^{2}+y^{2}=1$. The tangent to the circle at this point meets the parabola $y=x^{2}+1$ at exactly one point. Find all such points $(a, b)$. | Answer: $(-1,0),(1,0),(0,1),\left(-\frac{2 \sqrt{6}}{5},-\frac{1}{5}\right),\left(\frac{2 \sqrt{6}}{5},-\frac{1}{5}\right)$.
Since any non-vertical line intersecting the parabola $y=x^{2}+1$ has exactly two intersection points with it, the line mentioned in the problem must be either vertical or a common tangent to th... | {
"problem_match": "\n5.",
"resource_path": "BalticWay/segmented/en-bw99sol.jsonl",
"solution_match": "\n5."
} | 60 | 609 |
1999 | T3 | 8 | null | BalticWay | We are given 1999 coins. No two coins have the same weight. A machine is provided which allows us with one operation to determine, for any three coins, which one has the middle weight. Prove that the coin that is the 1000 -th by weight can be determined using no more than 1000000 operations and that this is the only co... | It is possible to find the 1000 -th coin (i.e. the medium one among the 1999 coins). First we exclude the lightest and heaviest coin - for this we use 1997 weighings, putting the medium-weighted coin aside each time. Next we exclude the 2 -nd and 1998 -th coins using 1995 weighings, etc. In total we need
$$
1997+1995+... | {
"problem_match": "\n8.",
"resource_path": "BalticWay/segmented/en-bw99sol.jsonl",
"solution_match": "\n8."
} | 94 | 638 |
1999 | T3 | 12 | null | BalticWay | In a triangle $A B C$ it is given that $2|A B|=|A C|+|B C|$. Prove that the incentre of $A B C$, the circumcentre of $A B C$, and the midpoints of $A C$ and $B C$ are concyclic. | Let $N$ be the midpoint of $B C$ and $M$ the midpoint of $A C$. Let $O$ be the circumcentre of $A B C$ and $I$ its incentre (see Figure 8). Since $\angle C M O=\angle C N O=90^{\circ}$, the points $C, N, O$ and $M$ are concyclic (regardless of whether $O$ lies inside the triangle $A B C$ ). We now have to show that the... | {
"problem_match": "\n12.",
"resource_path": "BalticWay/segmented/en-bw99sol.jsonl",
"solution_match": "\n12."
} | 67 | 1,440 |
1999 | T3 | 14 | null | BalticWay | Let $A B C$ be an isosceles triangle with $|A B|=|A C|$. Points $D$ and $E$ lie on the sides $A B$ and $A C$, respectively. The line passing through $B$ and parallel to $A C$ meets the line $D E$ at $F$. The line passing through $C$ and parallel to $A B$ meets the line $D E$ at $G$. Prove that
$$
\frac{[D B C G]}{[F B... | The quadrilaterals $D B C G$ and $F B C E$ are trapeziums. The area of a trapezium is equal to half the sum of the lengths of the parallel sides multiplied by the distance between them. But the distance between the parallel sides is the same for both of these trapeziums, since the distance from $B$ to $A C$ is equal to... | {
"problem_match": "\n14.",
"resource_path": "BalticWay/segmented/en-bw99sol.jsonl",
"solution_match": "\n14."
} | 153 | 555 |
2010 | T3 | 1 | null | Benelux_MO | 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 c... | 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 $S$ :
$$
\begin{aligned}
& A=\{502,503, \ldots, 670\}, \\
& B=\{671,672, \ldots, 1005\}, \\
&... | {
"problem_match": "\nProblem 1.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl",
"solution_match": "\nSolution."
} | 125 | 907 |
2010 | T3 | 2 | null | Benelux_MO | Find all polynomials $p(x)$ with real coefficients such that
$$
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)
$$
for all $a, b, c \in \mathbb{R}$. | . For $a=b=c$, we have $3 p(0)=9 p(0)$, hence $p(0)=0$. Now set $b=c=0$, then we have
$$
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)
$$
Note that the zero polynomial is a solution to this equation. Now suppose that $p$ is not the zero poly... | {
"problem_match": "\nProblem 2.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl",
"solution_match": "\nSolution 1"
} | 71 | 680 |
2010 | T3 | 2 | null | Benelux_MO | Find all polynomials $p(x)$ with real coefficients such that
$$
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)
$$
for all $a, b, c \in \mathbb{R}$. | . For $a=b=c$, we have $3 p(0)=9 p(0)$, hence $p(0)=0$. Now set $b=c=0$, then we have
$$
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... | {
"problem_match": "\nProblem 2.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl",
"solution_match": "\nSolution 2"
} | 71 | 612 |
2010 | T3 | 3 | null | Benelux_MO | 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 $B Q$ inte... | .
(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
$$
\frac{|B T|}{|B R|}=\frac{|A S|}{|A Q|}
$... | {
"problem_match": "\nProblem 3.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl",
"solution_match": "# Solution 1"
} | 180 | 969 |
2010 | T3 | 3 | null | Benelux_MO | On a line $l$ there are three different points $A, B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to $B Q$ inte... | .
(a) Define $X$ as the intersection of $A T$ and $B S$, and $Y$ as the intersection of $A R$ and $B Q$. To prove that $P, S$ and $T$ are collinear, we will use Menelaos' theorem in $\triangle A B X$, so we have to prove
$$
\frac{A P}{P B} \frac{B S}{S X} \frac{X T}{T A}=-1
$$
Note that $B$ is between $P$ and $A, X$... | {
"problem_match": "\nProblem 3.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl",
"solution_match": "# Solution 2"
} | 180 | 1,566 |
2010 | T3 | 3 | null | Benelux_MO | On a line $l$ there are three different points $A, B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to $B Q$ inte... | .
(a) W.l.o.g. we may assume that $A=(0,0)$ and $B=(1,0)$ and the line through $P$ is in the upper half plane, so $l$ is the $x$-axis, $a$ is the $y$-axis and $b$ is the line $x=1$. Take $P=(p, 0)(p>1)$ and $Q=(0, q)(q>0)$. Since $P Q$ is given by $\frac{x}{p}+\frac{y}{q}=1$, we find $R=\left(1, \frac{q(p-1)}{p}\right... | {
"problem_match": "\nProblem 3.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2010-zz.jsonl",
"solution_match": "# Solution 4"
} | 180 | 1,282 |
2011 | T3 | 3 | null | Benelux_MO | If $k$ is an integer, let $\mathrm{c}(k)$ denote the largest cube that is less than or equal to $k$. Find all positive integers $p$ for which the following sequence is bounded:
$$
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
$$
# | Since $\mathrm{c}\left(a_{n}\right) \leqslant a_{n}$ for all $n \in \mathbb{N}, a_{n+1} \geqslant a_{n}$ with equality if and only if $\mathrm{c}\left(a_{n}\right)=a_{n}$. Hence the sequence is bounded if and only if it is eventually constant, which is if and only if $a_{n}$ is a perfect cube, for some $n \geqslant 0$.... | {
"problem_match": "# Problem 3",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2011-zz.jsonl",
"solution_match": "# Solution\n"
} | 101 | 637 |
2011 | T3 | 4 | null | Benelux_MO | Abby and Brian play the following game: They first choose a positive integer $N$. Then they write numbers on a blackboard in turn. Abby starts by writing a 1. Thereafter, when one of them has written the number $n$, the other writes down either $n+1$ or $2 n$, provided that the number is not greater than $N$. The playe... | (a) Abby has a winning strategy for odd $N$ : Observe that, whenever any player writes down an odd number, the other player has to write down an even number. By adding 1 to that number, the first player can write down another odd number. Since Abby starts the game by writing down an odd number, she can force Brian to w... | {
"problem_match": "# Problem 4",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2011-zz.jsonl",
"solution_match": "# Solution\n"
} | 139 | 534 |
2012 | T3 | 1 | null | Benelux_MO | A sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ of natural numbers is defined by the rule
$$
a_{n+1}=a_{n}+b_{n} \quad(n=1,2, \ldots)
$$
where $b_{n}$ is the last digit of $a_{n}$. Prove that such a sequence contains infinitely many powers of 2 if and only if $a_{1}$ is not divisible by 5 . | First we can observe that:
- If $a_{1}$ is divisible by 5 , then $a_{n}=a_{2}=0(\bmod 10) \forall n \geq 2$.
- If $a_{1}$ is not divisible by 5 , then for $n \geq 2: a_{n}$ is even, the sequence $b_{n}$ is periodic, its period is a cyclic permutation of $(2,4,8,6)$, and $a_{n+4}=a_{n}+20$.
(a) Let us suppose that $a_{... | {
"problem_match": "\nProblem 1.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2012-zz.jsonl",
"solution_match": "\nSolution."
} | 109 | 1,533 |
2012 | T3 | 2 | null | Benelux_MO | Find all quadruples $(a, b, c, d)$ of positive real numbers such that $a b c d=1$, $a^{2012}+2012 b=2012 c+d^{2012}$ and $2012 a+b^{2012}=c^{2012}+2012 d$. | Rewrite the last two equations into
$$
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, \... | {
"problem_match": "\nProblem 2.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2012-zz.jsonl",
"solution_match": "\nSolution."
} | 82 | 626 |
2012 | T3 | 3 | null | Benelux_MO | 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 $\Gamm... | 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 ... | {
"problem_match": "\nProblem 3.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2012-zz.jsonl",
"solution_match": "\nSolution."
} | 121 | 1,065 |
2012 | T3 | 4 | null | Benelux_MO | 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... | (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$ questi... | {
"problem_match": "\nProblem 4.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2012-zz.jsonl",
"solution_match": "# Solution."
} | 242 | 1,508 |
2013 | T3 | 1 | null | Benelux_MO | 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 ... | 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, \ldo... | {
"problem_match": "\nProblem 1.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2013-zz.jsonl",
"solution_match": "\nSolution."
} | 179 | 1,473 |
2013 | T3 | 2 | null | Benelux_MO | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(x+y)+y \leqslant f(f(f(x)))
$$
holds for all $x, y \in \mathbb{R}$. | Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function satisfying the given inequality (2). Writing $z$ for $x+y$, we find that $f(z)+(z-x) \leqslant f(f(f(x)))$, or equivalently
$$
f(z)+z \leqslant f(f(f(x)))+x
$$
for all $x, z \in \mathbb{R}$. Substituting $z=f(f(x))$ yields $f(f(f(x)))+f(f(x)) \leqslant f(f(f(x)... | {
"problem_match": "\nProblem 2.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2013-zz.jsonl",
"solution_match": "\nSolution."
} | 58 | 610 |
2013 | T3 | 3 | null | Benelux_MO | Let $\triangle A B C$ be a triangle with circumcircle $\Gamma$, and let $I$ be the center of the incircle of $\triangle A B C$. The lines $A I, B I$ and $C I$ intersect $\Gamma$ in $D \neq A, E \neq B$ and $F \neq C$. The tangent lines to $\Gamma$ in $F, D$ and $E$ intersect the lines $A I, B I$ and $C I$ in $R, S$ and... | We first prove that $|D B|=|D I|$. (This may also be claimed by referring to the lemma that $D$ is the centre of the circumcircle of $B I C I_{a}$.) By the constant angle theorem and the fact that $A D$ and $B E$ are angle bisectors of triangle $A B C$, we see that
$$
\angle D B I=\angle D B C+\angle C B I=\angle D A ... | {
"problem_match": "\nProblem 3.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2013-zz.jsonl",
"solution_match": "\nSolution."
} | 158 | 674 |
2013 | T3 | 4 | null | Benelux_MO | 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 $p$ the... | 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 tha... | {
"problem_match": "# Problem 4.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2013-zz.jsonl",
"solution_match": "# Solution."
} | 145 | 1,716 |
2014 | T3 | 2 | null | Benelux_MO | 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 c... | 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... | {
"problem_match": "# Problem 2",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2014-zz.jsonl",
"solution_match": "# Solution\n"
} | 209 | 543 |
2014 | T3 | 3 | null | Benelux_MO | 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.
# | 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 \... | {
"problem_match": "# Problem 3",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2014-zz.jsonl",
"solution_match": "# Solution\n"
} | 66 | 746 |
2014 | T3 | 4 | null | Benelux_MO | 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|... | 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 ... | {
"problem_match": "# Problem 4",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2014-zz.jsonl",
"solution_match": "# Solution\n"
} | 132 | 524 |
2015 | T3 | 2 | null | Benelux_MO | 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 th... | . 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_{... | {
"problem_match": "# Problem 2.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2015-zz.jsonl",
"solution_match": "\nSolution 1"
} | 122 | 853 |
2015 | T3 | 2 | null | Benelux_MO | 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 th... | . 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 $A C^{\prime}$,
$\Gam... | {
"problem_match": "# Problem 2.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2015-zz.jsonl",
"solution_match": "\nSolution 3"
} | 122 | 772 |
2015 | T3 | 4 | null | Benelux_MO | 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 pos... | 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 \leq... | {
"problem_match": "# Problem 4.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2015-zz.jsonl",
"solution_match": "\nSolution."
} | 165 | 691 |
2016 | T3 | 1 | null | Benelux_MO | 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 $i \neq j$. | 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 no... | {
"problem_match": "\nProblem 1.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2016-zz.jsonl",
"solution_match": "\nSolution."
} | 65 | 1,166 |
2016 | T3 | 2 | null | Benelux_MO | 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 $n$. | 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 $\operatorname{gcd}\left(d_{1}, d_{2}\right)=1$, whi... | {
"problem_match": "\nProblem 2.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2016-zz.jsonl",
"solution_match": "\nSolution."
} | 68 | 684 |
2016 | T3 | 3 | null | Benelux_MO | 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 $x, y \in \mathbb{R}$. | 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}... | {
"problem_match": "\nProblem 3.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2016-zz.jsonl",
"solution_match": "\nSolution I."
} | 78 | 548 |
2016 | T3 | 4 | null | Benelux_MO | 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 ... | 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... | {
"problem_match": "\nProblem 4.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2016-zz.jsonl",
"solution_match": "\nSolution II."
} | 137 | 610 |
2017 | T3 | null | null | Benelux_MO | . Find all functions $f: \mathbb{Q}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that
$$
f(x y) \cdot \operatorname{gcd}\left(f(x) f(y), f\left(\frac{1}{x}\right) f\left(\frac{1}{y}\right)\right)=x y f\left(\frac{1}{x}\right) f\left(\frac{1}{y}\right)
$$
for all $x, y \in \mathbb{Q}_{>0}$, where $\operatorname{gcd}(a, b)$ ... | Let $f: \mathbb{Q}_{>0} \rightarrow \mathbb{Z}_{>0}$ be a function satisfying
$$
f(x y) \cdot \operatorname{gcd}\left(f(x) f(y), f\left(\frac{1}{x}\right) f\left(\frac{1}{y}\right)\right)=x y f\left(\frac{1}{x}\right) f\left(\frac{1}{y}\right)
$$
for all $x, y \in \mathbb{Q}_{>0}$. Taking $y=\frac{1}{x}$ in (1), we o... | {
"problem_match": null,
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2017-zz.jsonl",
"solution_match": null
} | 147 | 1,068 |
2017 | T3 | null | null | Benelux_MO | . 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... | 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 th... | {
"problem_match": null,
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2017-zz.jsonl",
"solution_match": null
} | 242 | 1,214 |
2017 | T3 | null | null | Benelux_MO | . 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:
- the $n^{2}$ positive integers are pairwise distinct;
- if for each row and each column we compute the greatest common divisor of the $n$... | 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 \... | {
"problem_match": null,
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2017-zz.jsonl",
"solution_match": null
} | 210 | 1,916 |
2018 | T3 | 1 | null | Benelux_MO | (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
$$
\left(x+\frac{1}{y}\right)\left(x+\frac{1}{y}+2018\right)+\left(y+\frac{1}... | . By the inequality between arithmetic and quadratic means,
$$
\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+... | {
"problem_match": "# Problem 1",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2018-zz.jsonl",
"solution_match": "\nSolution 1"
} | 176 | 657 |
2018 | T3 | 1 | null | Benelux_MO | (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
$$
\left(x+\frac{1}{y}\right)\left(x+\frac{1}{y}+2018\right)+\left(y+\frac{1}... | . By the inequality between arithmetic and geometric means, $x / y+y / x \geqslant 2$, and hence
$$
\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}... | {
"problem_match": "# Problem 1",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2018-zz.jsonl",
"solution_match": "\nSolution 2"
} | 176 | 686 |
2018 | T3 | 2 | null | Benelux_MO | 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 den... | . 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... | {
"problem_match": "# Problem 2",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2018-zz.jsonl",
"solution_match": "\nSolution 2"
} | 223 | 1,469 |
2018 | T3 | 3 | null | Benelux_MO | 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... | . 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 ... | {
"problem_match": "# Problem 3",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2018-zz.jsonl",
"solution_match": "\nSolution 4"
} | 133 | 654 |
2018 | T3 | 4 | null | Benelux_MO | 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... | (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 mo... | {
"problem_match": "# Problem 4",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2018-zz.jsonl",
"solution_match": "# Solution\n"
} | 175 | 751 |
2018 | T3 | 4 | null | Benelux_MO | 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... | . 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\... | {
"problem_match": "# Problem 4",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2018-zz.jsonl",
"solution_match": "\nSolution 2"
} | 175 | 1,012 |
2019 | T3 | 1 | null | Benelux_MO | 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. | 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... | {
"problem_match": "# Problem 1.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2019-zz.jsonl",
"solution_match": "\nSolution."
} | 129 | 743 |
2019 | T3 | 2 | null | Benelux_MO | 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 ch... | 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 o... | {
"problem_match": "\nProblem 2.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2019-zz.jsonl",
"solution_match": "\nSolution."
} | 116 | 716 |
2019 | T3 | 4 | null | Benelux_MO | 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}-... | 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 $... | {
"problem_match": "\nProblem 4.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2019-zz.jsonl",
"solution_match": "\nSolution II."
} | 284 | 560 |
2020 | T3 | 1 | null | Benelux_MO | 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$.
# | 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... | {
"problem_match": "# Problem 1",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2020-zz.jsonl",
"solution_match": "# Solution\n"
} | 52 | 692 |
2020 | T3 | 2 | null | Benelux_MO | 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.
The segments on t... | 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 bo... | {
"problem_match": "# Problem 2",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2020-zz.jsonl",
"solution_match": "# Solution\n"
} | 144 | 774 |
2021 | T3 | 1 | null | Benelux_MO | (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
$$
(b) Find the largest non-negative integer $k$ such that it is possible to replace $k$ of the six maxima in this inequality by minima in such a way that the ine... | The left-hand-side of the inequality is invariant under permutations of $a, b, c, d$. We may therefore suppose that $a \geqslant b \geqslant c \geqslant d$, so that the inequality reduces to
$$
0 \leqslant 3 a+2 b+c=a+(a+b)+(a+b+c)
$$
We claim that each of the terms on the right-hand side is non-negative; this will p... | {
"problem_match": "# Problem 1",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2021-zz.jsonl",
"solution_match": "# Solution 1"
} | 157 | 734 |
2021 | T3 | 2 | null | Benelux_MO | Pebbles are placed on a $2021 \times 2021$ board in such a way that each square contains at most one pebble. The pebble set of a square of the board is the collection of all pebbles which are in the same row or column as this square. Determine the least number of pebbles that can be placed on the board in such a way th... | Let $N \geqslant 1$ be a positive integer. We claim that the least number of pebbles that can be placed on a $(2 N+1) \times(2 N+1)$ chessboard in such a way that no two squares of the board have the same pebble set is $3 N+1$. The problem has $N=1010$, so at least 3031 pebbles are needed.
We begin by placing $(2 N+1)... | {
"problem_match": "# Problem 2",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2021-zz.jsonl",
"solution_match": "# Solution\n"
} | 96 | 660 |
2022 | T3 | 1 | null | Benelux_MO | Let $n \geqslant 0$ be an integer, and let $a_{0}, a_{1}, \ldots, a_{n}$ be real numbers. Show that there exists $k \in\{0,1, \ldots, n\}$ such that
$$
a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n} \leqslant a_{0}+a_{1}+\cdots+a_{k}
$$
for all real numbers $x \in[0,1]$.
# | Define $s_{i}=a_{0}+a_{1}+\cdots+a_{i}$ for $i \in\{0,1, \ldots, n\}$. Thus $a_{0}=s_{0}$ and $a_{i}=s_{i}-s_{i-1}$ for all $i \in\{1,2, \ldots, n\}$. Hence
$$
\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}\righ... | {
"problem_match": "# Problem 1",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2022-zz.jsonl",
"solution_match": "# Solution 2"
} | 131 | 542 |
2022 | T3 | 2 | null | Benelux_MO | 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 o... | 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' direc... | {
"problem_match": "# Problem 2",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2022-zz.jsonl",
"solution_match": "# Solution 1"
} | 123 | 831 |
2022 | T3 | 4 | null | Benelux_MO | 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... | (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... | {
"problem_match": "# Problem 4",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2022-zz.jsonl",
"solution_match": "# Solution 2"
} | 113 | 640 |
2022 | T3 | 4 | null | Benelux_MO | 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... | (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 ther... | {
"problem_match": "# Problem 4",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2022-zz.jsonl",
"solution_match": "# Solution 3"
} | 113 | 806 |
2023 | T3 | 1 | null | Benelux_MO | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
(x-y)(f(x)+f(y)) \leqslant f\left(x^{2}-y^{2}\right) \quad \text { for all } x, y \in \mathbb{R}
$$
# | 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(... | {
"problem_match": "# Problem 1",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2023-zz.jsonl",
"solution_match": "# Solution\n"
} | 74 | 757 |
2023 | T3 | 2 | null | Benelux_MO | 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 al... | 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 prop... | {
"problem_match": "# Problem 2",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2023-zz.jsonl",
"solution_match": "# Solution\n"
} | 115 | 523 |
2023 | T3 | 3 | null | Benelux_MO | 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 tria... | 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... | {
"problem_match": "# Problem 3",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2023-zz.jsonl",
"solution_match": "# Solution 1"
} | 142 | 575 |
2023 | T3 | 3 | null | Benelux_MO | 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 tria... | 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^{\... | {
"problem_match": "# Problem 3",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2023-zz.jsonl",
"solution_match": "# Solution 2"
} | 142 | 645 |
2023 | T3 | 3 | null | Benelux_MO | 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 tria... | Since $A P E F$ and $A P B C$ are cyclic,
$$
\begin{aligned}
\angle C P F & =\angle B P A-\angle B P C-\angle F P A=\left(180^{\circ}-\angle B C A\right)-\angle B A C-\angle F E A \\
& =\left(180^{\circ}-\angle B C A-\angle B A C\right)-\angle B E D=\angle C B A-\angle B E D=\angle C B E-\angle B E D=\angle B D E=\ang... | {
"problem_match": "# Problem 3",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2023-zz.jsonl",
"solution_match": "# Solution 3"
} | 142 | 793 |
2023 | T3 | 4 | null | Benelux_MO | A positive integer $n$ is friendly if every pair of neighbouring digits of $n$, written in base 10, differs by exactly 1. For example, 6787 is friendly, but 211 and 901 are not.
Find all odd natural numbers $m$ for which there exists a friendly integer divisible by $64 m$.
# | Any friendly number divisible by 64 is divisible by 4 , and hence the number formed by its last two digits is a multiple of 4 , so ends in $00,04,08, \ldots$, or 96 . A friendly number divisible by 4 must therefore end in $12,32,56$, or 76 , so cannot be divisible by 5 . In particular, if $5 \mid \mathrm{m}$, then ther... | {
"problem_match": "# Problem 4",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2023-zz.jsonl",
"solution_match": "# Solution\n"
} | 79 | 1,122 |
2024 | T3 | 1 | null | Benelux_MO | (a) Let $a_{0}, a_{1}, \ldots, a_{2024}$ be real numbers such that $\left|a_{i+1}-a_{i}\right| \leqslant 1$ for $i=0,1, \ldots, 2023$.
Find the minimum possible value of
$$
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_{... | (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
$$
a_{0} ... | {
"problem_match": "# Problem 1",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2024-zz.jsonl",
"solution_match": "# Solution 1"
} | 273 | 681 |
2024 | T3 | 2 | null | Benelux_MO | Let $n$ be a positive integer. In a coordinate grid, a path from $(0,0)$ to $(2 n, 2 n)$ consists of $4 n$ consecutive unit steps $(1,0)$ or $(0,1)$. Prove that the number of paths that divide the square with vertices $(0,0)$, $(2 n, 0),(2 n, 2 n),(0,2 n)$ into two regions with even areas is
$$
\frac{\binom{4 n}{2 n}+... | 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... | {
"problem_match": "# Problem 2",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2024-zz.jsonl",
"solution_match": "# Solution 1"
} | 127 | 571 |
2024 | T3 | 2 | null | Benelux_MO | 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
$$
\frac{\binom{4 n}{2 n}+... | Define $Z_{m, n}$ to be the number of routes from $(0,0)$ to $(2 m, 2 n)$ that divide the rectangle with vertices $(0,0),(0,2 n),(2 m, 2 n)$ and $(2 m, 0)$ into two parts of even area. We call such paths good. We claim that
$$
2 Z_{m, n}=\binom{2 m+2 n}{2 m}+\binom{m+n}{m}
$$
for all $m, n$, which for $m=n$ establish... | {
"problem_match": "# Problem 2",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2024-zz.jsonl",
"solution_match": "# Solution 2"
} | 127 | 1,213 |
2024 | T3 | 3 | null | Benelux_MO | 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 inters... | 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=... | {
"problem_match": "# Problem 3",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2024-zz.jsonl",
"solution_match": "\nSolution 1"
} | 222 | 709 |
2024 | T3 | 3 | null | Benelux_MO | 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 inters... | 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 o... | {
"problem_match": "# Problem 3",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2024-zz.jsonl",
"solution_match": "# Solution 2"
} | 222 | 647 |
2024 | T3 | 3 | null | Benelux_MO | 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 inters... | 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}-\... | {
"problem_match": "# Problem 3",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2024-zz.jsonl",
"solution_match": "# Solution 3"
} | 222 | 1,004 |
2024 | T3 | 4 | null | Benelux_MO | 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... | 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... | {
"problem_match": "# Problem 4",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2024-zz.jsonl",
"solution_match": "# Solution 2"
} | 166 | 516 |
2025 | T3 | 3 | null | Benelux_MO | Let \(ABC\) be a triangle with incentre \(I\) and circumcircle \(\Omega\) . Let \(D, E, F\) be the midpoints of the arcs \(\overline{BC}, \overline{CA}, \overline{AB}\) of \(\Omega\) not containing \(A, B, C\) , respectively. Let \(D'\) be the point of \(\Omega\) diametrically opposite to \(D\) . Show that \(I, D'\) , ... | 
By definition of \(D\) , \(E\) and \(F\) , we know that \(ID, IE\) and \(IF\) are the angle bisectors of \(ABC\) . Using angles in \(\Omega\) , we can compute
\[\overline{EFI} = \overline{EFC} = \overline{EBC} = \frac{\overline{ABC}}{2} =... | {
"problem_match": "\nProblem 3.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl",
"solution_match": "\nSolution."
} | 121 | 530 |
2025 | T3 | 4 | null | Benelux_MO | Let \(a_{0},a_{1},\ldots ,a_{10}\) be integers such that, for each \(i\in \{0,1,\ldots ,2047\}\) , there exists a subset \(S\subseteq \{0,1,\ldots ,10\}\) with
\[\sum_{j\in S}a_{j}\equiv i\pmod {2048}.\]
Show that for each \(i\in \{0,1,\ldots ,10\}\) , there is exactly one \(j\in \{0,1,\ldots ,10\}\) such that \(... | We denote by \(\nu_{2}(a)\) the valuation 2- adic of the integer \(a\) . Let us prove by induction the more general statement that, for \(n\in \mathbb{N}_{>0}\) , if \(a_{0},a_{1},\ldots ,a_{n - 1}\) are integers such that, for each \(i\in \{0,1,\ldots ,2^{n} - 1\}\) , there exists a subset \(S\subseteq \{0,1,\ldots ,n... | {
"problem_match": "\nProblem 4.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl",
"solution_match": "\nSolution."
} | 264 | 1,671 |
2025 | T3 | 4 | null | Benelux_MO | Let \(a_{0},a_{1},\ldots ,a_{10}\) be integers such that, for each \(i\in \{0,1,\ldots ,2047\}\) , there exists a subset \(S\subseteq \{0,1,\ldots ,10\}\) with
\[\sum_{j\in S}a_{j}\equiv i\pmod {2048}.\]
Show that for each \(i\in \{0,1,\ldots ,10\}\) , there is exactly one \(j\in \{0,1,\ldots ,10\}\) such that \(... | We do the same induction as in the main Solution. If all \(a_{j}\) are even, then all the sums are even, contradicting the hypothesis. Therefore, without loss of generality, we can assume that \(a_{0}\) is odd. For each \(i\in \{0,1,2,\ldots ,2^{n} - 1\}\) , let \(S_{i}^{\prime}\subseteq \{0,1,\ldots ,n - 1\}\) be the ... | {
"problem_match": "\nProblem 4.",
"resource_path": "Benelux_MO/segmented/Benelux_en-olympiad_en-bxmo-problems-2025-zz.jsonl",
"solution_match": "\nAlternative Solution."
} | 264 | 967 |
2021 | T2 | 4 | null | Canada_MO | A function $f$ from the positive integers to the positive integers is called Canadian if it satisfies
$$
\operatorname{gcd}(f(f(x)), f(x+y))=\operatorname{gcd}(x, y)
$$
for all pairs of positive integers $x$ and $y$.
Find all positive integers $m$ such that $f(m)=m$ for all Canadian functions $f$. | Define an $m \in \mathbb{N}$ to be good if $f(m)=m$ for all such $f$. It will be shown that $m$ is good if and only if $m$ has two or more distinct prime divisors. Let $P(x, y)$ denote the assertion
$$
\operatorname{gcd}(f(f(x)), f(x+y))=\operatorname{gcd}(x, y)
$$
for a pair $x, y \in \mathbb{N}$. Let $x$ be a posit... | {
"problem_match": "\n## Problem No. 4.",
"resource_path": "Canada_MO/segmented/en-2021CMO_solutions_en-1.jsonl",
"solution_match": "\nSolution."
} | 84 | 1,125 |
2021 | T2 | 5 | null | Canada_MO | Nina and Tadashi play the following game. Initially, a triple $(a, b, c)$ of nonnegative integers with $a+b+c=$ 2021 is written on a blackboard. Nina and Tadashi then take moves in turn, with Nina first. A player making a move chooses a positive integer $k$ and one of the three entries on the board; then the player inc... | The answer is $3^{\text {number of 1's in binary expansion of } 2021}=3^{8}=6561$.
Throughout this solution, we say two nonnegative integers overlap in the $2^{\ell}$ position if their binary representations both have a 1 in that position. We say that two nonnegative integers overlap if they overlap in some position. O... | {
"problem_match": "\n## Problem No. 5.",
"resource_path": "Canada_MO/segmented/en-2021CMO_solutions_en-1.jsonl",
"solution_match": "\nSolution."
} | 142 | 2,031 |
2025 | T2 | 1 | null | Canada_MO | The \(n\) players of a hockey team gather to select their team captain. Initially, they stand in a circle, and each person votes for the person on their left.
The players will update their votes via a series of rounds. In one round, each player \(a\) updates their vote, one at a time, according to the following proc... | Initially, all players are in a cycle. Note that once a player leaves the cycle, they cannot rejoin. Furthermore, a new cycle cannot be created. Hence, at any point in time, the graph corresponding to the votes will be a functional graph with a single cycle.
We will first prove that after \(\lfloor \log_{2}n\rfloor\)... | {
"problem_match": "\nP1.",
"resource_path": "CANADA_MO/segmented/en-CMO2025-solutions.jsonl",
"solution_match": "\n## Solution 1"
} | 211 | 613 |
2025 | T2 | 1 | null | Canada_MO | The \(n\) players of a hockey team gather to select their team captain. Initially, they stand in a circle, and each person votes for the person on their left.
The players will update their votes via a series of rounds. In one round, each player \(a\) updates their vote, one at a time, according to the following proc... | We will use induction on \(n\) .
Inductive Hypothesis. Let \(G\) be any functional graph with \(n\) nodes and a single cycle. Then after \(n\) rounds of the given operation, \(G\) will become a self- loop with \(n - 1\) nodes pointing to it.
Base Case. The cases \(n \leq 2\) are clear.
Inductive Step. Assume that... | {
"problem_match": "\nP1.",
"resource_path": "CANADA_MO/segmented/en-CMO2025-solutions.jsonl",
"solution_match": "\n## Solution 2"
} | 211 | 569 |
2025 | T2 | 2 | null | Canada_MO | Determine all positive integers \(a, b, c, p\) where \(p\) and \(p + 2\) are odd primes and
\[2^{a}p^{b} = (p + 2)^{c} - 1.\] | The only solution is \((a, b, c, p) = (3, 1, 2, 3)\) . First, factor the right hand side. This gives us
\[2^{a}p^{b} = (p + 1)((p + 2)^{c - 1} + (p + 2)^{c - 2} + \dots +(p + 2) + 1).\]
Since \(\gcd (p, p + 1) = 1\) it must be the case that \(p + 1 = 2^{x}\) for some positive integer \(x \leq a\) and so \(p = 2^{x}... | {
"problem_match": "\nP2.",
"resource_path": "CANADA_MO/segmented/en-CMO2025-solutions.jsonl",
"solution_match": "\n## Solution \n\n"
} | 54 | 685 |
2025 | T2 | 3 | null | Canada_MO | A polynomial \(c_{d}x^{d} + c_{d - 1}x^{d - 1} + \dots + c_{1}x + c_{0}\) with degree \(d\) is reflexive if there is an integer \(n \geq d\) such that \(c_{i} = c_{n - i}\) for every \(0 \leq i \leq n\) , where \(c_{i} = 0\) for \(i > d\) . Let \(\ell \geq 2\) be an integer and \(p(x)\) be a polynomial with integer coe... | Let \(d\) be the degree of \(p\) and let \(k\) be any non- negative integer. We will choose
\[q(x) = \frac{x^{d + k + \ell}p\left(\frac{1}{x}\right) - p(x)}{x - 1},\] \[r(x) = \frac{p(x) - x^{d + k}p\left(\frac{1}{x}\right)}{x - 1}.\]
First, we must show that both \(q\) and \(r\) are integer polynomials. Consider t... | {
"problem_match": "\nP3.",
"resource_path": "CANADA_MO/segmented/en-CMO2025-solutions.jsonl",
"solution_match": "\n## Solution 1"
} | 194 | 920 |
2025 | T2 | 5 | null | Canada_MO | A rectangle \(R\) is divided into a set \(S\) of finitely many smaller rectangles with sides parallel to the sides of \(R\) such that no three rectangles in \(S\) share a common corner. An ant is initially located at the bottom-left corner of \(R\) . In one operation, we can choose a rectangle \(r \in S\) such that the... | Consider the following version of the problem:
A rectangle \(R\) is divided into a set \(S\) of finitely many smaller rectangles such that no three rectangles in \(S\) share a common corner. For each \(r \in S\) , draw two non- intersecting arcs inside \(r\) , connecting the pairs of adjacent corners of \(r\) (there ... | {
"problem_match": "\nP5.",
"resource_path": "CANADA_MO/segmented/en-CMO2025-solutions.jsonl",
"solution_match": "\n## Solution 1"
} | 160 | 564 |
2025 | T2 | 5 | null | Canada_MO | A rectangle \(R\) is divided into a set \(S\) of finitely many smaller rectangles with sides parallel to the sides of \(R\) such that no three rectangles in \(S\) share a common corner. An ant is initially located at the bottom-left corner of \(R\) . In one operation, we can choose a rectangle \(r \in S\) such that the... | Suppose that no rectangle was chosen in at least two operations. In particular, a rectangle cannot be selected in two consecutive operations.
At any point in the process, consider whether the last move by the ant was horizontal or vertical, and whether the most recently chosen rectangle was to the left or the right o... | {
"problem_match": "\nP5.",
"resource_path": "CANADA_MO/segmented/en-CMO2025-solutions.jsonl",
"solution_match": "\n## Solution 2"
} | 160 | 607 |
2022 | T2 | 2 | null | Canada_MO | Let $d(k)$ denote the number of positive integer divisors of $k$. For example, $d(6)=4$ since 6 has 4 positive divisors, namely, $1,2,3$, and 6 . Prove that for all positive integers $n$,
$$
d(1)+d(3)+d(5)+\cdots+d(2 n-1) \leq d(2)+d(4)+d(6)+\cdots+d(2 n)
$$ | For any integer $k$ and set of integers $S$, let $f_{S}(k)$ be the number of multiples of $k$ in $S$. We can count the number of pairs $(k, s)$ with $k \in \mathbb{N}$ dividing $s \in S$ in two different ways, as follows:
- For each $s \in S$, there are $d(s)$ pairs that include $s$, one for each divisor of $s$.
- For... | {
"problem_match": "\nP2.",
"resource_path": "Canada_MO/segmented/en-cmo2022-solutions-en.jsonl",
"solution_match": "\nSolution."
} | 109 | 630 |
2022 | T2 | 3 | null | Canada_MO | Let $n \geq 2$ be an integer. Initially, the number 1 is written $n$ times on a board. Every minute, Vishal picks two numbers written on the board, say $a$ and $b$, erases them, and writes either $a+b$ or $\min \left\{a^{2}, b^{2}\right\}$. After $n-1$ minutes there is one number left on the board. Let the largest poss... | Clearly $f(n)$ is a strictly increasing function, as we can form $f(n-1)$ with $n-1$ ones, and add the final one. However, we can do better; assume Vishal generates $f(n)$ on the board. After $n-2$ minutes, there are two numbers left, say they were formed by $x$ ones and $y$ ones, where $x+y=n$. Clearly the numbers are... | {
"problem_match": "\nP3:",
"resource_path": "Canada_MO/segmented/en-cmo2022-solutions-en.jsonl",
"solution_match": "\nSolution."
} | 139 | 2,292 |
2022 | T2 | 5 | null | Canada_MO | Let $A B C D E$ be a convex pentagon such that the five vertices lie on a circle and the five sides are tangent to another circle inside the pentagon. There are $\binom{5}{3}=10$ triangles which can be formed by choosing 3 of the 5 vertices. For each of these 10 triangles, mark its incenter. Prove that these 10 incente... | Let $I$ be the incenter of pentagon $A B C D E$. Let $I_{A}$ denote the incenter of triangle $E A B$ and $I_{a}$ the incenter $D A C$. Define $I_{B}, I_{b}, I_{C}, I_{c}, I_{D}, I_{d}, I_{E}, I_{e}$ similarly.
We will first show that $I_{A} I_{B} I_{C} I_{D} I_{E}$ are concyclic. Let $\omega_{A}$ be the circle with ce... | {
"problem_match": "\nP5.",
"resource_path": "Canada_MO/segmented/en-cmo2022-solutions-en.jsonl",
"solution_match": "\nSolution."
} | 94 | 576 |
2023 | T2 | 1 | null | Canada_MO | William is thinking of an integer between 1 and 50, inclusive. Victor can choose a positive integer $m$ and ask William: "does $m$ divide your number?", to which William must answer truthfully. Victor continues asking these questions until he determines William's number. What is the minimum number of questions that Vic... | The minimum number is 15 questions.
First, we show that 14 or fewer questions is not enough to guarantee success. Suppose Victor asks at most 14 questions, and William responds with "no" to each question unless $m=$ 1. Note that these responses are consistent with the secret number being 1. But since there are 15 prime... | {
"problem_match": "\nP1.",
"resource_path": "Canada_MO/segmented/en-cmo2023-solutions-en.jsonl",
"solution_match": "\nSolution."
} | 72 | 623 |
2023 | T2 | 3 | null | Canada_MO | An acute triangle is a triangle that has all angles less than $90^{\circ}$ ( $90^{\circ}$ is a Right Angle). Let $A B C$ be an acute triangle with altitudes $A D, B E$, and $C F$ meeting at $H$. The circle passing through points $D, E$, and $F$ meets $A D, B E$, and $C F$ again at $X, Y$, and $Z$ respectively. Prove th... | Let the circumcircle of $A B C$ meet the altitudes $A D, B E$, and $C F$ again at $I, J$, and $K$ respectively.
Lemma (9-point circle). $I, J, K$ are the reflections of $H$ across $B C, C A, A B$. Moreover, $D, E, F, X, Y, Z$ are the midpoints of $H I, H J, H K, H A, H B, H C$.
Proof. Since $A B D E$ and $A B I C$ ar... | {
"problem_match": "\nP3.",
"resource_path": "Canada_MO/segmented/en-cmo2023-solutions-en.jsonl",
"solution_match": "\nSolution."
} | 144 | 856 |
2023 | T2 | 4 | null | Canada_MO | Let $f(x)$ be a non-constant polynomial with integer coefficients such that $f(1) \neq 1$. For a positive integer $n$, define $\operatorname{divs}(n)$ to be the set of positive divisors of $n$.
A positive integer $m$ is $f$-cool if there exists a positive integer $n$ for which
$$
f[\operatorname{divs}(m)]=\operatorna... | Assume for the sake of contradiction that there are infinitely many $f$-cool integers.
If $f(x)$ has a negative leading coefficient, then a sufficiently large $f$-cool integer $m$ will have $f(m)<0$. But this implies $m$ is not $f$-cool, contradiction.
Thus $f(x)$ has a positive leading coefficient, so we can pick an ... | {
"problem_match": "\nP4.",
"resource_path": "Canada_MO/segmented/en-cmo2023-solutions-en.jsonl",
"solution_match": "\nSolution."
} | 218 | 788 |
2023 | T2 | 5 | null | Canada_MO | A country with $n$ cities has some two-way roads connecting certain pairs of cities. Someone notices that if the country is split into two parts in any way, then there would be at most $k n$ roads between the two parts (where $k$ is a fixed positive integer). What is the largest integer $m$ (in terms of $n$ and $k$ ) s... | The answer is $m=\left\lceil\frac{n}{4 k}\right\rceil$
Call a collection of cities independent if no two cities in the collection are joined by a road. Let $r$ and $k$ be integers such that $n=4 k q+r$ where $1 \leq r \leq 4 k$.
First we show that $m \leq\left\lceil\frac{n}{4 k}\right\rceil=q+1$. Let $K_{i}$ denote a ... | {
"problem_match": "\nP5.",
"resource_path": "Canada_MO/segmented/en-cmo2023-solutions-en.jsonl",
"solution_match": "\nSolution."
} | 105 | 786 |
2024 | T2 | 3 | null | Canada_MO | Let $N$ be the number of positive integers with 10 digits $\overline{d_{9} d_{8} \cdots d_{1} d_{0}}$ in base 10 (where $0 \leq d_{i} \leq 9$ for all $i$ and $d_{9}>0$ ) such that the polynomial
$$
d_{9} x^{9}+d_{8} x^{8}+\cdots+d_{1} x+d_{0}
$$
is irreducible in $\mathbb{Q}$. Prove that $N$ is even.
(A polynomial is... | Let $f(x)=d_{9} x^{9}+d_{8} x^{8}+\cdots+d_{1} x+d_{0}$. If $d_{0}=0$, then $f(x)$ is divisible by $x$ and thus reducible, so we may ignore all such polynomials. The remaining polynomials all have nonzero leading and constant coefficients.
For any polynomial $p(x)$ of degree $n$ with nonzero leading and constant coeff... | {
"problem_match": "\nP3.",
"resource_path": "Canada_MO/segmented/en-cmo2024-solutions-en.jsonl",
"solution_match": "\nSolution."
} | 162 | 853 |
2024 | T2 | 4 | null | Canada_MO | Centuries ago, the pirate Captain Blackboard buried a vast amount of treasure in a single cell of an $M \times N(2 \leq M, N)$ grid-structured island. You and your crew have reached the island and have brought special treasure detectors to find the cell with the treasure. For each detector, you can set it up to scan a ... | The following alternative approach from CMO competitor Marvin Mao of Bergen County Academies is another full solution.
Take the same construction as in Solution 1. For the bound, consider the following sets:
- $S_{\mathrm{CR}}:=\{\{(1,1),(1, N)\},\{(M, 1),(M, N)\}\}$, i.e. the pairs of corners on the same row;
- $S_{... | {
"problem_match": "\nP4.",
"resource_path": "Canada_MO/segmented/en-cmo2024-solutions-en.jsonl",
"solution_match": "\nSolution 2."
} | 208 | 939 |
2024 | T2 | 5 | null | Canada_MO | Initially, three non-collinear points, $A, B$, and $C$, are marked on the plane. You have a pencil and a double-edged ruler of width 1. Using them, you may perform the following operations:
- Mark an arbitrary point in the plane.
- Mark an arbitrary point on an already drawn line.
- If two points $P_{1}$ and $P_{2}$ a... | Claim 1. It is possible to draw internal/external angle bisectors.
Proof. Let $A, B, C$ be marked. To bisect $\angle A B C$, draw the parallel line to $A B$ unit 1 away from it on the opposite side as $C$, and draw the parallel line to $B C$ unit 1 away from it on the opposite side as $A$. Let these lines intersect at ... | {
"problem_match": "\nP5.",
"resource_path": "Canada_MO/segmented/en-cmo2024-solutions-en.jsonl",
"solution_match": "\n## Solution 1."
} | 201 | 925 |
2002 | T2 | 5 | null | Canada_MO | Let $\mathbb{N}=\{0,1,2, \ldots\}$. Determine all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that
$$
x f(y)+y f(x)=(x+y) f\left(x^{2}+y^{2}\right)
$$
for all $x$ and $y$ in $\mathbb{N}$. | We claim $f$ is a constant function. Define $g(x)=f(x)-f(0)$. Then $g(0)=0, g(x) \geq-f(0)$ and
$$
x g(y)+y g(x)=(x+y) g\left(x^{2}+y^{2}\right)
$$
for all $x, y$ in $\mathbb{N}$.
Letting $y=0$ shows $g\left(x^{2}\right)=0$ (in particular, $g(1)=g(4)=0$ ), and letting $x=y=1$ shows $g(2)=0$. Also, if $x, y$ and $z$ ... | {
"problem_match": "\n5.",
"resource_path": "Canada_MO/segmented/en-sol2002.jsonl",
"solution_match": "\n## Solution 2."
} | 90 | 598 |
2002 | T2 | 5 | null | Canada_MO | Let $\mathbb{N}=\{0,1,2, \ldots\}$. Determine all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that
$$
x f(y)+y f(x)=(x+y) f\left(x^{2}+y^{2}\right)
$$
for all $x$ and $y$ in $\mathbb{N}$. | Suppose that $W$ is the set of nonnegative integers and that $f: W \rightarrow W$ satisfies:
$$
x f(y)+y f(x)=(x+y) f\left(x^{2}+y^{2}\right) .
$$
We will show that $f$ is a constant function.
Let $f(0)=k$, and set $S=\{x \mid f(x)=k\}$.
Letting $y=0$ in $(*)$ shows that $f\left(x^{2}\right)=k \quad \forall x>0$, a... | {
"problem_match": "\n5.",
"resource_path": "Canada_MO/segmented/en-sol2002.jsonl",
"solution_match": "\nSolution 3."
} | 90 | 1,265 |
2003 | T2 | 5 | null | Canada_MO | Let $S$ be a set of $n$ points in the plane such that any two points of $S$ are at least 1 unit apart. Prove there is a subset $T$ of $S$ with at least $n / 7$ points such that any two points of $T$ are at least $\sqrt{3}$ units apart. | We will construct the set $T$ in the following way: Assume the points of $S$ are in the $x y$-plane and let $P$ be a point in $S$ with maximum $y$-coordinate. This point $P$ will be a member of the set $T$ and now, from $S$, we will remove $P$ and all points in $S$ which are less than $\sqrt{3}$ units from $P$. From th... | {
"problem_match": "\n5.",
"resource_path": "Canada_MO/segmented/en-sol2003.jsonl",
"solution_match": "\n## Solution\n\n"
} | 75 | 582 |
2004 | T2 | 4 | null | Canada_MO | Let $p$ be an odd prime. Prove that
$$
\sum_{k=1}^{p-1} k^{2 p-1} \equiv \frac{p(p+1)}{2} \quad\left(\bmod p^{2}\right)
$$
[Note that $a \equiv b(\bmod m)$ means that $a-b$ is divisible by $m$.] | Since $p-1$ is even, we can pair up the terms in the summation in the following way (first term with last, 2nd term with 2nd last, etc.):
$$
\sum_{k=1}^{p-1} k^{2 p-1}=\sum_{k=1}^{\frac{p-1}{2}}\left(k^{2 p-1}+(p-k)^{2 p-1}\right)
$$
Expanding $(p-k)^{2 p-1}$ with the binomial theorem, we get
$$
(p-k)^{2 p-1}=p^{2 p... | {
"problem_match": "\n4.",
"resource_path": "Canada_MO/segmented/en-sol2004.jsonl",
"solution_match": "\n## Solution\n\n"
} | 88 | 614 |
2004 | T2 | 5 | null | Canada_MO | Let $T$ be the set of all positive integer divisors of $2004^{100}$. What is the largest possible number of elements that a subset $S$ of $T$ can have if no element of $S$ is an integer multiple of any other element of $S$ ? | Assume throughout that $a, b, c$ are nonnegative integers. Since the prime factorization of 2004 is $2004=2^{2} \cdot 3 \cdot 167$,
$$
T=\left\{2^{a} 3^{b} 167^{c} \mid 0 \leq a \leq 200,0 \leq b, c \leq 100\right\}
$$
Let
$$
S=\left\{2^{200-b-c} 3^{b} 167^{c} \mid 0 \leq b, c \leq 100\right\}
$$
For any $0 \leq b,... | {
"problem_match": "\n5.",
"resource_path": "Canada_MO/segmented/en-sol2004.jsonl",
"solution_match": "\n## Solution\n\n"
} | 65 | 682 |
2005 | T2 | 5 | null | Canada_MO | Let's say that an ordered triple of positive integers $(a, b, c)$ is $n$-powerful if $a \leq b \leq c$, $\operatorname{gcd}(a, b, c)=1$, and $a^{n}+b^{n}+c^{n}$ is divisible by $a+b+c$. For example, $(1,2,2)$ is 5 -powerful.
a) Determine all ordered triples (if any) which are $n$-powerful for all $n \geq 1$.
b) Deter... | Let $T_{n}=a^{n}+b^{n}+c^{n}$ and consider the polynomial
$$
P(x)=(x-a)(x-b)(x-c)=x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c .
$$
Since $P(a)=0$, we get $a^{3}=(a+b+c) a^{2}-(a b+a c+b c) a+a b c$ and multiplying both sides by $a^{n-3}$ we obtain $a^{n}=(a+b+c) a^{n-1}-(a b+a c+b c) a^{n-2}+(a b c) a^{n-3}$. Applying ... | {
"problem_match": "\n5.",
"resource_path": "Canada_MO/segmented/en-sol2005.jsonl",
"solution_match": "\n## Solution 1"
} | 189 | 929 |
2005 | T2 | 5 | null | Canada_MO | Let's say that an ordered triple of positive integers $(a, b, c)$ is $n$-powerful if $a \leq b \leq c$, $\operatorname{gcd}(a, b, c)=1$, and $a^{n}+b^{n}+c^{n}$ is divisible by $a+b+c$. For example, $(1,2,2)$ is 5 -powerful.
a) Determine all ordered triples (if any) which are $n$-powerful for all $n \geq 1$.
b) Deter... | Let $p$ be a prime. By Fermat's Little Theorem,
$$
a^{p-1} \equiv \begin{cases}1(\bmod p), & \text { if } p \text { doesn't divide } a \\ 0(\bmod p), & \text { if } p \text { divides } a\end{cases}
$$
Since $\operatorname{gcd}(a, b, c)=1$, we have that $a^{p-1}+b^{p-1}+c^{p-1} \equiv 1,2$ or $3(\bmod p)$. Therefore i... | {
"problem_match": "\n5.",
"resource_path": "Canada_MO/segmented/en-sol2005.jsonl",
"solution_match": "\n## Solution 2"
} | 189 | 779 |
2006 | T2 | 1 | null | Canada_MO | Let $f(n, k)$ be the number of ways of distributing $k$ candies to $n$ children so that each child receives at most 2 candies. For example, if $n=3$, then $f(3,7)=0, f(3,6)=1$ and $f(3,4)=6$.
Determine the value of
$$
f(2006,1)+f(2006,4)+f(2006,7)+\cdots+f(2006,1000)+f(2006,1003) .
$$
Comment. Unfortunately, there w... | The number of ways of distributing $k$ candies to 2006 children is equal to the number of ways of distributing 0 to a particular child and $k$ to the rest, plus the number of ways of distributing 1 to the particular child and $k-1$ to the rest, plus the number of ways of distributing 2 to the particular child and $k-2$... | {
"problem_match": "\n1.",
"resource_path": "Canada_MO/segmented/en-sol2006.jsonl",
"solution_match": "\nSolution 1."
} | 175 | 557 |
2006 | T2 | 1 | null | Canada_MO | Let $f(n, k)$ be the number of ways of distributing $k$ candies to $n$ children so that each child receives at most 2 candies. For example, if $n=3$, then $f(3,7)=0, f(3,6)=1$ and $f(3,4)=6$.
Determine the value of
$$
f(2006,1)+f(2006,4)+f(2006,7)+\cdots+f(2006,1000)+f(2006,1003) .
$$
Comment. Unfortunately, there w... | The desired number is the sum of the coefficients of the terms of degree not exceeding 1003 in the expansion of $\left(1+x+x^{2}\right)^{2005}$, which is equal to the coefficient of $x^{1003}$ in the expansion of
$$
\begin{aligned}
\left(1+x+x^{2}\right)^{2005}\left(1+x+\cdots+x^{1003}\right) & =\left[\left(1-x^{3}\ri... | {
"problem_match": "\n1.",
"resource_path": "Canada_MO/segmented/en-sol2006.jsonl",
"solution_match": "\nSolution 2."
} | 175 | 749 |
2006 | T2 | 3 | null | Canada_MO | In a rectangular array of nonnegative real numbers with $m$ rows and $n$ columns, each row and each column contains at least one positive element. Moreover, if a row and a column intersect in a positive element, then the sums of their elements are the same. Prove that $m=n$. | [Y. Zhao] Let the term in the $i$ th row and the $j$ th column of the array be denoted by $a_{i j}$, and let $S=\{(i, j)$ : $\left.a_{i j}>0\right\}$. Suppose that $r_{i}$ is the sum of the $i$ th row and $c_{j}$ the sum of the $j$ th column. Then $r_{i}=c_{j}$ whenever $(i, j) \in S$. Then we have that
$$
\sum\left\{... | {
"problem_match": "\n3.",
"resource_path": "Canada_MO/segmented/en-sol2006.jsonl",
"solution_match": "\nSolution 2."
} | 63 | 833 |
2006 | T2 | 4 | null | Canada_MO | Consider a round-robin tournament with $2 n+1$ teams, where each team plays each other team exactly once. We say that three teams $X, Y$ and $Z$, form a cycle triplet if $X$ beats $Y, Y$ beats $Z$, and $Z$ beats $X$. There are no ties.
(a) Determine the minimum number of cycle triplets possible.
(b) Determine the max... | (a) The minimum is 0 , which is achieved by a tournament in which team $T_{i}$ beats $T_{j}$ if and only if $i>j$.
(b) Any set of three teams constitutes either a cycle triplet or a "dominated triplet" in which one team beats the other two; let there be $c$ of the former and $d$ of the latter. Then $c+d=\left(\begin{a... | {
"problem_match": "\n4.",
"resource_path": "Canada_MO/segmented/en-sol2006.jsonl",
"solution_match": "\nSolution 1."
} | 94 | 912 |
2008 | T2 | 4 | null | Canada_MO | Find all functions $f$ defined on the natural numbers that take values among the natural numbers for which
$$
(f(n))^{p} \equiv n \quad \bmod f(p)
$$
for all $n \in \mathbf{N}$ and all prime numbers $p$. | The substitution $n=p$, a prime, yields $p \equiv(f(p))^{p} \equiv 0(\bmod f(p))$, so that $p$ is divisible by $f(p)$. Hence, for each prime $p, f(p)=1$ or $f(p)=p$.
Let $S=\{p: p$ is prime and $f(p)=p\}$. If $S$ is infinite, then $f(n)^{p} \equiv n(\bmod p)$ for infinitely many primes $p$. By the little Fermat theore... | {
"problem_match": "\n4.",
"resource_path": "Canada_MO/segmented/en-sol2008.jsonl",
"solution_match": "\nSolution."
} | 61 | 586 |
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