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 |
|---|---|---|---|---|---|---|---|---|---|
2024 | T2 | 2 | null | Nordic_MO | Let $\mathcal{Q}_{1}$ be a quadrilateral such that the midpoints of its sides lie on a circle. Prove that there exists a cyclic quadrilateral $\mathcal{Q}_{2}$ with the same sidelengths as $\mathcal{Q}_{1}$, such that two of the angles in $\mathcal{Q}_{2}$ are equal. | Let $A, B, C$ and $D$ be the vertices of $\mathcal{Q}_{1}$, and $K, L, M$ and $N$ be the midpoints of the sides $A B, B C, C D$ and $D A$, respectively.
We have $K L\|A C\| M N$ and $L M\|B D\| N K$, and thus $K L M N$ is a parallelogram. From the problem condition it is known that $K L M N$ is cyclic, and thus, from t... | {
"problem_match": "\n## Problem 2",
"resource_path": "Nordic_MO/segmented/en-2024-sol.jsonl",
"solution_match": "\n## Solution"
} | 77 | 545 |
2024 | T2 | 3 | null | Nordic_MO | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(f(x) f(y)+y)=f(x) y+f(y-x+1)
$$
for all $x, y \in \mathbb{R}$. | First note that $f$ satisfies the functional equation if and only if $-f$ does as well. We can therefore assume that $f(0) \geq 0$. We consider two cases, depending on whether $f$ is injective or not.
1. First assume that $f$ is not injective. Then there exist $a, t \in \mathbb{R}, t \neq 0$ such that $f(a+t)=$ $f(a)$... | {
"problem_match": "\n## Problem 3",
"resource_path": "Nordic_MO/segmented/en-2024-sol.jsonl",
"solution_match": "\n## Solution"
} | 61 | 802 |
2024 | T2 | 4 | null | Nordic_MO | Alice and Bob are playing a game. First, Alice chooses a partition $\mathcal{C}$ of the positive integers into a (not necessarily finite) set of sets, such that each positive integer is in exactly one of the sets in $\mathcal{C}$. Then Bob does the following operation a finite number of times.
Choose a set $S \in \math... | We will use the following notation. If $S \subseteq \mathbb{Z}^{+}$is a subset of the positive integers and $D$ is defined as in the problem statement, define $d(S)=D \backslash S$. In addition, the $d$-sequence of $S$, written $\left\{d^{S}\right\}_{n \in \mathbb{Z}^{+}}$, is defined recursively as $d_{0}^{S}=S$, and ... | {
"problem_match": "\n## Problem 4",
"resource_path": "Nordic_MO/segmented/en-2024-sol.jsonl",
"solution_match": "\n## Solution 1"
} | 174 | 1,125 |
2024 | T2 | 4 | null | Nordic_MO | Alice and Bob are playing a game. First, Alice chooses a partition $\mathcal{C}$ of the positive integers into a (not necessarily finite) set of sets, such that each positive integer is in exactly one of the sets in $\mathcal{C}$. Then Bob does the following operation a finite number of times.
Choose a set $S \in \math... | Use the same notation as in solution 1 . We wish to construct a partition $\bigcup_{i \in \mathbb{Z}^{+}} A_{i}$ of the positive integers, such that no two sets among all sets in $\left\{d^{A_{1}}\right\}_{n \in \mathbb{Z}^{+}},\left\{d^{A_{2}}\right\}_{n \in \mathbb{Z}^{+}}, \ldots$ are equal.
To begin with, consider ... | {
"problem_match": "\n## Problem 4",
"resource_path": "Nordic_MO/segmented/en-2024-sol.jsonl",
"solution_match": "\n## Solution 2"
} | 174 | 1,169 |
2011 | T1 | 2 | null | RMM | Determine all positive integers $n$ for which there exists a polynomial $f(x)$ with real coefficients, with the following properties:
(1) for each integer $k$, the number $f(k)$ is an integer if and only if $k$ is not divisible by $n$;
(2) the degree of $f$ is less than $n$.
(Hungary) GÉza Kós | We will show that such polynomial exists if and only if $n=1$ or $n$ is a power of a prime.
We will use two known facts stated in Lemmata 1 and 2.
Lemma 1. If $p^{a}$ is a power of a prime and $k$ is an integer, then $\frac{(k-1)(k-2) \ldots\left(k-p^{a}+1\right)}{\left(p^{a}-1\right)!}$ is divisible by $p$ if and onl... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2011-Sols2011D1.jsonl",
"solution_match": "\nSolution."
} | 85 | 2,690 |
2011 | T1 | 3 | null | RMM | A triangle $A B C$ is inscribed in a circle $\omega$. A variable line $\ell$ chosen parallel to $B C$ meets segments $A B$, $A C$ at points $D, E$ respectively, and meets $\omega$ at points $K, L$ (where $D$ lies between $K$ and $E$ ). Circle $\gamma_{1}$ is tangent to the segments $K D$ and $B D$ and also tangent to $... | Let $P$ be the meeting point of the common inner tangents to $\gamma_{1}$ and $\gamma_{2}$. Also, let $b$ be the angle bisector of $\angle B A C$. Since $K L \| B C, b$ is also the angle bisector of $\angle K A L$.
Let $\mathfrak{H}$ be the composition of the symmetry $\mathfrak{S}$ with respect to $b$ and the inversi... | {
"problem_match": "\nProblem 3.",
"resource_path": "RMM/segmented/en-2011-Sols2011D1.jsonl",
"solution_match": "\nSolution."
} | 178 | 769 |
2011 | T1 | 4 | null | RMM | Given a positive integer $n=\prod_{i=1}^{s} p_{i}^{\alpha_{i}}$, we write $\Omega(n)$ for the total number $\sum_{i=1}^{s} \alpha_{i}$ of prime factors of $n$, counted with multiplicity. Let $\lambda(n)=(-1)^{\Omega(n)}$ (so, for example, $\left.\lambda(12)=\lambda\left(2^{2} \cdot 3^{1}\right)=(-1)^{2+1}=-1\right)$.
... | Notice that we have $\Omega(m n)=\Omega(m)+\Omega(n)$ for all positive integers $m, n$ ( $\Omega$ is a completely additive arithmetic function), translating into $\lambda(m n)=\lambda(m) \cdot \lambda(n)$ (so $\lambda$ is a completely multiplicative arithmetic function), hence $\lambda(p)=-1$ for any prime $p$, and $\l... | {
"problem_match": "\nProblem 4.",
"resource_path": "RMM/segmented/en-2011-Sols2011D2.jsonl",
"solution_match": "\nSolution."
} | 193 | 1,499 |
2011 | T1 | 5 | null | RMM | For every $n \geq 3$, determine all the configurations of $n$ distinct points $X_{1}, X_{2}, \ldots, X_{n}$ in the plane, with the property that for any pair of distinct points $X_{i}, X_{j}$ there exists a permutation $\sigma$ of the integers $\{1, \ldots, n\}$, such that $\mathrm{d}\left(X_{i}, X_{k}\right)=\mathrm{d... | Let us first prove that the points must be concyclic. Assign to each point $X_{k}$ the vector $x_{k}$ in a system of orthogonal coordinates whose origin is the point of mass of the configuration, thus $\frac{1}{n} \sum_{k=1}^{n} x_{k}=0$.
Then $\mathrm{d}^{2}\left(X_{i}, X_{k}\right)=\left\|x_{i}-x_{k}\right\|^{2}=\le... | {
"problem_match": "\nProblem 5.",
"resource_path": "RMM/segmented/en-2011-Sols2011D2.jsonl",
"solution_match": "\nSolution."
} | 167 | 718 |
2011 | T1 | 6 | null | RMM | The cells of a square $2011 \times 2011$ array are labelled with the integers $1,2, \ldots, 2011^{2}$, in such a way that every label is used exactly once. We then identify the lefthand and right-hand edges, and then the top and bottom, in the normal way to form a torus (the surface of a doughnut).
Determine the large... | For the toroidal case, it is clear the statement of the problem is referring to the cells of a $\mathbb{Z}_{N} \times \mathbb{Z}_{N}$ lattice on the surface of the torus, labeled with the numbers $1,2, \ldots, N^{2}$, where one has to determine the least possible maximal absolute value $M$ of the difference of labels a... | {
"problem_match": "\nProblem 6.",
"resource_path": "RMM/segmented/en-2011-Sols2011D2.jsonl",
"solution_match": "\nSolution."
} | 746 | 2,671 |
2012 | T1 | 2 | null | RMM | Given a triangle $A B C$, let $D, E$, and $F$ respectively denote the midpoints of the sides $B C, C A$, and $A B$. The circle $B C F$ and the line $B E$ meet again at $P$, and the circle $A B E$ and the line $A D$ meet again at $Q$. Finally, the lines $D P$ and $F Q$ meet at $R$. Prove that the centroid $G$ of the tri... | . (Ilya Bogdanov and Marian Andronache) We also prove that $\angle(R P, P G)=\angle(R Q, Q G)$, or $\angle(D P, P G)=\angle(F Q, Q G)$.
Let $S$ be the point on ray $G D$ such that $A G \cdot G S=C G \cdot G F$ (so the points $A, S, C, F$ are concyclic). Then $G P \cdot G E=G P \cdot \frac{1}{2} G B=\frac{1}{2} C G \cd... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2012-Solutions2012-1.jsonl",
"solution_match": "\nSolution 2"
} | 122 | 876 |
2012 | T1 | 3 | null | RMM | Each positive integer number is coloured red or blue. A function $f$ from the set of positive integer numbers into itself has the following two properties:
(a) if $x \leq y$, then $f(x) \leq f(y)$; and
(b) if $x, y$ and $z$ are all (not necessarily distinct) positive integer numbers of the same colour and $x+y=z$, then... | For integer $x, y$, by a segment $[x, y]$ we always mean the set of all integers $t$ such that $x \leq t \leq y$; the length of this segment is $y-x$.
If for every two positive integers $x, y$ sharing the same colour we have $f(x) / x=f(y) / y$, then one can choose $a=\max \{f(r) / r, f(b) / b\}$, where $r$ and $b$ ar... | {
"problem_match": "\nProblem 3.",
"resource_path": "RMM/segmented/en-2012-Solutions2012-1.jsonl",
"solution_match": "\nSolution."
} | 129 | 782 |
2012 | T1 | 5 | null | RMM | Given a positive integer number $n \geq 3$, colour each cell of an $n \times n$ square array one of $\left[(n+2)^{2} / 3\right]$ colours, each colour being used at least once. Prove that the cells of some $1 \times 3$ or $3 \times 1$ rectangular subarray have pairwise distinct colours. | For more convenience, say that a subarray of the $n \times n$ square array bears a colour if at least two of its cells share that colour.
We shall prove that the number of $1 \times 3$ and $3 \times 1$ rectangular subarrays, which is $2 n(n-2)$, exceeds the number of such subarrays, each of which bears some colour. Th... | {
"problem_match": "\nProblem 5.",
"resource_path": "RMM/segmented/en-2012-Solutions2012-2.jsonl",
"solution_match": "\nSolution."
} | 84 | 1,107 |
2012 | T1 | 6 | null | RMM | Let $A B C$ be a triangle and let $I$ and $O$ respectively denote its incentre and circumcentre. Let $\omega_{A}$ be the circle through $B$ and $C$ and tangent to the incircle of the triangle $A B C$; the circles $\omega_{B}$ and $\omega_{C}$ are defined similarly. The circles $\omega_{B}$ and $\omega_{C}$ through $A$ ... | Let $\gamma$ be the incircle of the triangle $A B C$ and let $A_{1}, B_{1}, C_{1}$ be its contact points with the sides $B C, C A, A B$, respectively. Let further $X_{A}$ be the point of contact of the circles $\gamma$ and $\omega_{A}$. The latter circle is the image of the former under a homothety centred at $X_{A}$. ... | {
"problem_match": "\nProblem 6.",
"resource_path": "RMM/segmented/en-2012-Solutions2012-2.jsonl",
"solution_match": "\nSolution."
} | 160 | 1,769 |
2013 | T1 | 1 | null | RMM | For a positive integer $a$, define a sequence of integers $x_{1}, x_{2}, \ldots$ by letting $x_{1}=a$ and $x_{n+1}=2 x_{n}+1$ for $n \geq 1$. Let $y_{n}=2^{x_{n}}-1$. Determine the largest possible $k$ such that, for some positive integer $a$, the numbers $y_{1}, \ldots, y_{k}$ are all prime.
(Russia) Valery Senderov | The largest such is $k=2$. Notice first that if $y_{i}$ is prime, then $x_{i}$ is prime as well. Actually, if $x_{i}=1$ then $y_{i}=1$ which is not prime, and if $x_{i}=m n$ for integer $m, n>1$ then $2^{m}-1 \mid 2^{x_{i}}-1=y_{i}$, so $y_{i}$ is composite. In particular, if $y_{1}, y_{2}, \ldots, y_{k}$ are primes fo... | {
"problem_match": "\nProblem 1.",
"resource_path": "RMM/segmented/en-2013-Solutions2013-1.jsonl",
"solution_match": "\nSolution."
} | 117 | 550 |
2013 | T1 | 2 | null | RMM | Does there exist a pair $(g, h)$ of functions $g, h: \mathbb{R} \rightarrow \mathbb{R}$ such that the only function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \in \mathbb{R}$ is the identity function $f(x) \equiv x$ ?
(United Kingdom) Alexander Betts | . Such a tester pair exists. We may biject $\mathbb{R}$ with the closed unit interval, so it suffices to find a tester pair for that instead. We give an explicit example: take some positive real numbers $\alpha, \beta$ (which we will specify further later). Take
$$
g(x)=\max (x-\alpha, 0) \quad \text { and } \quad h(x... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2013-Solutions2013-1.jsonl",
"solution_match": "\nSolution 1"
} | 109 | 797 |
2013 | T1 | 2 | null | RMM | Does there exist a pair $(g, h)$ of functions $g, h: \mathbb{R} \rightarrow \mathbb{R}$ such that the only function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \in \mathbb{R}$ is the identity function $f(x) \equiv x$ ?
(United Kingdom) Alexander Betts | . Let us agree that a sequence $\mathbf{x}=\left(x_{n}\right)_{n=1,2, \ldots}$ is cofinally non-constant if for every index $m$ there exists an index $n>m$ such that $x_{m} \neq x_{n}$.
Biject $\mathbb{R}$ with the set of cofinally non-constant sequences of 0 's and 1 's, and define $g$ and $h$ by
$$
g(\epsilon, \mat... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2013-Solutions2013-1.jsonl",
"solution_match": "\nSolution 2"
} | 109 | 634 |
2013 | T1 | 2 | null | RMM | Does there exist a pair $(g, h)$ of functions $g, h: \mathbb{R} \rightarrow \mathbb{R}$ such that the only function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \in \mathbb{R}$ is the identity function $f(x) \equiv x$ ?
(United Kingdom) Alexander Betts | . (Ilya Bogdanov) We will show that there exists a tester pair of bijective functions $g$ and $h$.
First of all, let us find out when a pair of functions is a tester pair. Let $g, h: \mathbb{R} \rightarrow \mathbb{R}$ be arbitrary functions. We construct a directed graph $G_{g, h}$ with $\mathbb{R}$ as the set of vert... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2013-Solutions2013-1.jsonl",
"solution_match": "\nSolution 3"
} | 109 | 1,058 |
2013 | T1 | 4 | null | RMM | Let $P$ and $P^{\prime}$ be two convex quadrilateral regions in the plane (regions contain their boundary). Let them intersect, with $O$ a point in the intersection. Suppose that for every line $\ell$ through $O$ the segment $\ell \cap P$ is strictly longer than the segment $\ell \cap P^{\prime}$. Is it possible that t... | The answer is in the affirmative: Given a positive $\epsilon<2$, the ratio in question may indeed be greater than $2-\epsilon$.
To show this, consider a square $A B C D$ centred at $O$, and let $A^{\prime}, B^{\prime}$, and $C^{\prime}$ be the reflections of $O$ in $A, B$, and $C$, respectively. Notice that, if $\ell$... | {
"problem_match": "\nProblem 4.",
"resource_path": "RMM/segmented/en-2013-Solutions2013-2.jsonl",
"solution_match": "\nSolution."
} | 116 | 2,444 |
2013 | T1 | 5 | null | RMM | Given an integer $k \geq 2$, set $a_{1}=1$ and, for every integer $n \geq 2$, let $a_{n}$ be the smallest $x>a_{n-1}$ such that:
$$
x=1+\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{x}{a_{i}}}\right\rfloor .
$$
Prove that every prime occurs in the sequence $a_{1}, a_{2}, \ldots$.
(Bulgaria) Alexander Ivanov | . We prove that the $a_{n}$ are precisely the $k$ th-power-free positive integers, that is, those divisible by the $k$ th power of no prime. The conclusion then follows.
Let $B$ denote the set of all $k$ th-power-free positive integers. We first show that, given a positive integer $c$,
$$
\sum_{b \in B, b \leq c}\lef... | {
"problem_match": "\nProblem 5.",
"resource_path": "RMM/segmented/en-2013-Solutions2013-2.jsonl",
"solution_match": "\nSolution 1"
} | 123 | 642 |
2013 | T1 | 5 | null | RMM | Given an integer $k \geq 2$, set $a_{1}=1$ and, for every integer $n \geq 2$, let $a_{n}$ be the smallest $x>a_{n-1}$ such that:
$$
x=1+\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{x}{a_{i}}}\right\rfloor .
$$
Prove that every prime occurs in the sequence $a_{1}, a_{2}, \ldots$.
(Bulgaria) Alexander Ivanov | . (Ilya Bogdanov) For every $n=1,2,3, \ldots$, introduce the function
$$
f_{n}(x)=x-1-\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{x}{a_{i}}}\right\rfloor
$$
Denote also by $g_{n}(x)$ the number of the indices $i \leq n$ such that $x / a_{i}$ is the $k$ th power of an integer. Then $f_{n}(x+1)-f_{n}(x)=1-g_{n}(x)$ for ... | {
"problem_match": "\nProblem 5.",
"resource_path": "RMM/segmented/en-2013-Solutions2013-2.jsonl",
"solution_match": "\nSolution 2"
} | 123 | 920 |
2013 | T1 | 6 | null | RMM | $2 n$ distinct tokens are placed at the vertices of a regular $2 n$-gon, with one token placed at each vertex. A move consists of choosing an edge of the $2 n$-gon and interchanging the two tokens at the endpoints of that edge. Suppose that after a finite number of moves, every pair of tokens have been interchanged exa... | Step 1. Enumerate all the tokens in the initial arrangement in clockwise circular order; also enumerate the vertices of the $2 n$-gon accordingly. Consider any three tokens $i<j<k$. At each moment, their cyclic order may be either $i, j, k$ or $i, k, j$, counted clockwise. This order changes exactly when two of these t... | {
"problem_match": "\nProblem 6.",
"resource_path": "RMM/segmented/en-2013-Solutions2013-2.jsonl",
"solution_match": "\nSolution."
} | 95 | 892 |
2015 | T1 | 1 | null | RMM | Does there exist an infinite sequence of positive integers $a_{1}, a_{2}, a_{3}, \ldots$ such that $a_{m}$ and $a_{n}$ are coprime if and only if $|m-n|=1$ ?
(Peru) Jorge Tipe | The answer is in the affirmative.
The idea is to consider a sequence of pairwise distinct primes $p_{1}, p_{2}, p_{3}, \ldots$, cover the positive integers by a sequence of finite non-empty sets $I_{n}$ such that $I_{m}$ and $I_{n}$ are disjoint if and only if $m$ and $n$ are one unit apart, and set $a_{n}=\prod_{i \in... | {
"problem_match": "\nProblem 1.",
"resource_path": "RMM/segmented/en-2015-Solutions_RMM2015-1.jsonl",
"solution_match": "\nSolution."
} | 62 | 688 |
2015 | T1 | 2 | null | RMM | For an integer $n \geq 5$, two players play the following game on a regular $n$-gon. Initially, three consecutive vertices are chosen, and one counter is placed on each. A move consists of one player sliding one counter along any number of edges to another vertex of the $n$-gon without jumping over another counter. A m... | We shall prove that the first player wins if and only the exponent of 2 in the prime decomposition of $n-3$ is odd.
Since the game is identical for both players, has finitely many possible states and always terminates, we can label the possible states Wins od Losses according as whether a player faced with that positi... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2015-Solutions_RMM2015-1.jsonl",
"solution_match": "\nSolution."
} | 143 | 1,205 |
2015 | T1 | 3 | null | RMM | A finite list of rational numbers is written on a blackboard. In an operation, we choose any two numbers $a, b$, erase them, and write down one of the numbers
$$
a+b, a-b, b-a, a \times b, a / b(\text { if } b \neq 0), b / a(\text { if } a \neq 0) .
$$
Prove that, for every integer $n>100$, there are only finitely ma... | We prove the problem statement even for all positive integer $n$.
There are only finitely many ways of constructing a number from $n$ pairwise distinct numbers $x_{1}, \ldots, x_{n}$ only using the four elementary arithmetic operations, and each $x_{k}$ exactly once. Each such formula for $k>1$ is obtained by an elemen... | {
"problem_match": "\nProblem 3.",
"resource_path": "RMM/segmented/en-2015-Solutions_RMM2015-1.jsonl",
"solution_match": "\nSolution."
} | 172 | 1,184 |
2015 | T1 | 5 | null | RMM | Let $p \geq 5$ be a prime number. For a positive integer $k$ we denote by $R(k)$ the remainder of $k$ when divided by $p$. Determine all positive integers $a<p$ such that
$$
m+R(m a)>a
$$
for every $m=1,2, \ldots, p-1$.
(Bulgaria) Alexander Ivanov | The required integers are $p-1$ along with all the numbers of the form $\lfloor p / q\rfloor, q=$ $2, \ldots, p-1$. In other words, these are $p-1$, along with the numbers $1,2, \ldots,\lfloor\sqrt{p}\rfloor$, and also the (distinct) numbers $\lfloor p / q\rfloor, q=2, \ldots,\left\lfloor\sqrt{p}-\frac{1}{2}\right\rflo... | {
"problem_match": "\nProblem 5.",
"resource_path": "RMM/segmented/en-2015-Solutions_RMM2015-2.jsonl",
"solution_match": "\nSolution."
} | 88 | 689 |
2015 | T1 | 6 | null | RMM | Given a positive integer $n$, determine the largest real number $\mu$ satisfying the following condition: for every $4 n$-point configuration $C$ in an open unit square $U$, there exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater ... | The required maximum is $\frac{1}{2 n+2}$. To show that the condition in the statement is not met if $\mu>\frac{1}{2 n+2}$, let $U=(0,1) \times(0,1)$, choose a small enough positive $\epsilon$, and consider the configuration $C$ consisting of the $n$ four-element clusters of points $\left(\frac{i}{n+1} \pm \epsilon\rig... | {
"problem_match": "\nProblem 6.",
"resource_path": "RMM/segmented/en-2015-Solutions_RMM2015-2.jsonl",
"solution_match": "\nSolution."
} | 90 | 2,145 |
2016 | T1 | 2 | null | RMM | Given positive integers $m$ and $n \geq m$, determine the largest number of dominoes ( $1 \times 2$ or $2 \times 1$ rectangles) that can be placed on a rectangular board with $m$ rows and $2 n$ columns consisting of cells ( $1 \times 1$ squares) so that:
(i) each domino covers exactly two adjacent cells of the board;
(... | . The required maximum is $m n-\lfloor m / 2\rfloor$ and is achieved by the brick-like vertically symmetric arrangement of blocks of $n$ and $n-1$ horizontal dominoes placed on alternate rows, so that the bottom row of the board is completely covered by $n$ dominoes.
To show that the number of dominoes in an arrangeme... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2016-Solutions_RMM2016-1.jsonl",
"solution_match": "\nSolution 1"
} | 135 | 807 |
2016 | T1 | 2 | null | RMM | Given positive integers $m$ and $n \geq m$, determine the largest number of dominoes ( $1 \times 2$ or $2 \times 1$ rectangles) that can be placed on a rectangular board with $m$ rows and $2 n$ columns consisting of cells ( $1 \times 1$ squares) so that:
(i) each domino covers exactly two adjacent cells of the board;
(... | . (sketch by Ilya Bogdanov) We present an alternative proof of the bound.
Label the rows upwards $0,1, \ldots, m-1$, and the columns from the left to the right by $0,1, \ldots, 2 n-1$; label each cell by the pair of its column's and row's numbers, so that $(1,0)$ is the second left cell in the bottom row. Colour the ce... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2016-Solutions_RMM2016-1.jsonl",
"solution_match": "\nSolution 2"
} | 135 | 922 |
2016 | T1 | 3 | null | RMM | A cubic sequence is a sequence of integers given by $a_{n}=n^{3}+b n^{2}+c n+d$, where $b, c$ and $d$ are integer constants and $n$ ranges over all integers, including negative integers.
(a) Show that there exists a cubic sequence such that the only terms of the sequence which are squares of integers are $a_{2015}$ and... | The only possible value of $a_{2015} \cdot a_{2016}$ is 0 . For simplicity, by performing a translation of the sequence (which may change the defining constants $b, c$ and $d$ ), we may instead concern ourselves with the values $a_{0}$ and $a_{1}$, rather than $a_{2015}$ and $a_{2016}$.
Suppose now that we have a cubi... | {
"problem_match": "\nProblem 3.",
"resource_path": "RMM/segmented/en-2016-Solutions_RMM2016-1.jsonl",
"solution_match": "\nSolution."
} | 133 | 866 |
2016 | T1 | 4 | null | RMM | Let $x$ and $y$ be positive real numbers such that $x+y^{2016} \geq 1$. Prove that $x^{2016}+y>$ 1 - 1/100. | If $x \geq 1-1 /(100 \cdot 2016)$, then
$$
x^{2016} \geq\left(1-\frac{1}{100 \cdot 2016}\right)^{2016}>1-2016 \cdot \frac{1}{100 \cdot 2016}=1-\frac{1}{100}
$$
by Bernoulli's inequality, whence the conclusion.
If $x<1-1 /(100 \cdot 2016)$, then $y \geq(1-x)^{1 / 2016}>(100 \cdot 2016)^{-1 / 2016}$, and it is sufficie... | {
"problem_match": "\nProblem 4.",
"resource_path": "RMM/segmented/en-2016-Solutions_RMM2016-2.jsonl",
"solution_match": "\nSolution."
} | 53 | 558 |
2016 | T1 | 5 | null | RMM | A convex hexagon $A_{1} B_{1} A_{2} B_{2} A_{3} B_{3}$ is inscribed in a circle $\Omega$ of radius $R$. The diagonals $A_{1} B_{2}, A_{2} B_{3}$, and $A_{3} B_{1}$ concur at $X$. For $i=1,2,3$, let $\omega_{i}$ be the circle tangent to the segments $X A_{i}$ and $X B_{i}$, and to the arc $A_{i} B_{i}$ of $\Omega$ not c... | (a) Let $\ell_{1}$ be the tangent to $\Omega$ parallel to $A_{2} B_{3}$, lying on the same side of $A_{2} B_{3}$ as $\omega_{1}$. The tangents $\ell_{2}$ and $\ell_{3}$ are defined similarly. The lines $\ell_{1}$ and $\ell_{2}, \ell_{2}$ and $\ell_{3}, \ell_{3}$ and $\ell_{1}$ meet at $C_{3}, C_{1}, C_{2}$, respectivel... | {
"problem_match": "\nProblem 5.",
"resource_path": "RMM/segmented/en-2016-Solutions_RMM2016-2.jsonl",
"solution_match": "\nSolution."
} | 250 | 950 |
2016 | T1 | 6 | null | RMM | A set of $n$ points in Euclidean 3-dimensional space, no four of which are coplanar, is partitioned into two subsets $\mathcal{A}$ and $\mathcal{B}$. An $\mathcal{A B}$-tree is a configuration of $n-1$ segments, each of which has an endpoint in $\mathcal{A}$ and the other in $\mathcal{B}$, and such that no segments for... | The configurations of segments under consideration are all bipartite geometric trees on the points $n$ whose vertex-parts are $\mathcal{A}$ and $\mathcal{B}$, and transforming one into another preserves the degree of any vertex in $\mathcal{A}$, but not necessarily that of a vertex in $\mathcal{B}$.
The idea is to dev... | {
"problem_match": "\nProblem 6.",
"resource_path": "RMM/segmented/en-2016-Solutions_RMM2016-2.jsonl",
"solution_match": "\nSolution."
} | 276 | 1,976 |
2017 | T1 | 1 | null | RMM | (a) Prove that every positive integer $n$ can be written uniquely in the form
$$
n=\sum_{j=1}^{2 k+1}(-1)^{j-1} 2^{m_{j}}
$$
where $k \geq 0$ and $0 \leq m_{1}<m_{2}<\cdots<m_{2 k+1}$ are integers.
This number $k$ is called the weight of $n$.
(b) Find (in closed form) the difference between the number of positive int... | (a) We show by induction on the integer $M \geq 0$ that every integer $n$ in the range $-2^{M}+1$ through $2^{M}$ can uniquely be written in the form $n=\sum_{j=1}^{\ell}(-1)^{j-1} 2^{m_{j}}$ for some integers $\ell \geq 0$ and $0 \leq m_{1}<m_{2}<\cdots<m_{\ell} \leq M$ (empty sums are 0 ); moreover, in this unique re... | {
"problem_match": "\nProblem 1.",
"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-1.jsonl",
"solution_match": "\nSolution."
} | 162 | 1,477 |
2017 | T1 | 2 | null | RMM | Determine all positive integers $n$ satisfying the following condition: for every monic polynomial $P$ of degree at most $n$ with integer coefficients, there exists a positive integer $k \leq n$, and $k+1$ distinct integers $x_{1}, x_{2}, \ldots, x_{k+1}$ such that
$$
P\left(x_{1}\right)+P\left(x_{2}\right)+\cdots+P\l... | There is only one such integer, namely, $n=2$. In this case, if $P$ is a constant polynomial, the required condition is clearly satisfied; if $P=X+c$, then $P(c-1)+P(c+1)=$ $P(3 c)$; and if $P=X^{2}+q X+r$, then $P(X)=P(-X-q)$.
To rule out all other values of $n$, it is sufficient to exhibit a monic polynomial $P$ of ... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-1.jsonl",
"solution_match": "\nSolution."
} | 144 | 965 |
2017 | T1 | 3 | null | RMM | Let $n$ be an integer greater than 1 and let $X$ be an $n$-element set. A non-empty collection of subsets $A_{1}, \ldots, A_{k}$ of $X$ is tight if the union $A_{1} \cup \cdots \cup A_{k}$ is a proper subset of $X$ and no element of $X$ lies in exactly one of the $A_{i}$ s. Find the largest cardinality of a collection ... | . (Ilya Bogdanov) The required maximum is $2 n-2$. To describe a ( $2 n-2$ )-element collection satisfying the required conditions, write $X=\{1,2, \ldots, n\}$ and set $B_{k}=\{1,2, \ldots, k\}$, $k=1,2, \ldots, n-1$, and $B_{k}=\{k-n+2, k-n+3, \ldots, n\}, k=n, n+1, \ldots, 2 n-2$. To show that no subcollection of th... | {
"problem_match": "\nProblem 3.",
"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-1.jsonl",
"solution_match": "\nSolution 1"
} | 174 | 1,363 |
2017 | T1 | 3 | null | RMM | Let $n$ be an integer greater than 1 and let $X$ be an $n$-element set. A non-empty collection of subsets $A_{1}, \ldots, A_{k}$ of $X$ is tight if the union $A_{1} \cup \cdots \cup A_{k}$ is a proper subset of $X$ and no element of $X$ lies in exactly one of the $A_{i}$ s. Find the largest cardinality of a collection ... | . Proceed again by induction on $n$ to show that the cardinality of a collection of proper non-empty subsets of $X$, no subcollection of which is tight, does not exceed $2 n-2$.
Consider any collection $\mathcal{B}$ of proper non-empty subsets of $X$ with no tight subcollection (we call such collection good). Assume t... | {
"problem_match": "\nProblem 3.",
"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-1.jsonl",
"solution_match": "\nSolution 2"
} | 174 | 1,195 |
2017 | T1 | 3 | null | RMM | Let $n$ be an integer greater than 1 and let $X$ be an $n$-element set. A non-empty collection of subsets $A_{1}, \ldots, A_{k}$ of $X$ is tight if the union $A_{1} \cup \cdots \cup A_{k}$ is a proper subset of $X$ and no element of $X$ lies in exactly one of the $A_{i}$ s. Find the largest cardinality of a collection ... | . We provide yet another proof of the estimate $|\mathcal{B}| \leq 2 n-2$, using the notion of a good collection from Solution 2. Arguing indirectly, we assume that there exists a good collection $\mathcal{B}$ with $|\mathcal{B}| \geq 2 n-1$, and choose one such for the minimal possible value of $n$. Clearly, $n>2$.
F... | {
"problem_match": "\nProblem 3.",
"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-1.jsonl",
"solution_match": "\nSolution 3"
} | 174 | 1,065 |
2017 | T1 | 4 | null | RMM | In the Cartesian plane, let $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ be the graphs of the quadratic functions $f_{1}(x)=p_{1} x^{2}+q_{1} x+r_{1}$ and $f_{2}(x)=p_{2} x^{2}+q_{2} x+r_{2}$, where $p_{1}>0>p_{2}$. The graphs $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ cross at distinct points $A$ and $B$. The four tangents to... | . Let $\mathcal{A}_{i}$ and $\mathcal{B}_{i}$ be the tangents to $\mathcal{G}_{i}$ at $A$ and $B$, respectively, and let $C_{i}=\mathcal{A}_{i} \cap \mathcal{B}_{i}$. Since $f_{1}(x)$ is convex and $f_{2}(x)$ is concave, the convex quadrangle formed by the four tangents is exactly $A C_{1} B C_{2}$.
Lemma. If $C A$ and... | {
"problem_match": "\nProblem 4.",
"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-2.jsonl",
"solution_match": "\nSolution 1"
} | 204 | 737 |
2017 | T1 | 4 | null | RMM | In the Cartesian plane, let $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ be the graphs of the quadratic functions $f_{1}(x)=p_{1} x^{2}+q_{1} x+r_{1}$ and $f_{2}(x)=p_{2} x^{2}+q_{2} x+r_{2}$, where $p_{1}>0>p_{2}$. The graphs $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ cross at distinct points $A$ and $B$. The four tangents to... | . Use the standard equation of a tangent to a smooth curve in the plane, to deduce that the tangents at two distinct points $A$ and $B$ on the parabola of equation $y=p x^{2}+q x+r$,
$p \neq 0$, meet at some point $C$ whose coordinates are
$$
x_{C}=\frac{1}{2}\left(x_{A}+x_{B}\right) \quad \text { and } \quad y_{C}=p ... | {
"problem_match": "\nProblem 4.",
"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-2.jsonl",
"solution_match": "\nSolution 2"
} | 204 | 1,013 |
2017 | T1 | 5 | null | RMM | Fix an integer $n \geq 2$. An $n \times n$ sieve is an $n \times n$ array with $n$ cells removed so that exactly one cell is removed from every row and every column. A stick is a $1 \times k$ or $k \times 1$ array for any positive integer $k$. For any sieve $A$, let $m(A)$ be the minimal number of sticks required to pa... | . Given $A, m(A)=2 n-2$, and it is achieved, for instance, by dissecting $A$ along all horizontal (or vertical) grid lines. It remains to prove that $m(A) \geq 2 n-2$ for every $A$.
By holes we mean the cells which are cut out from the board. The cross of a hole in $A$ is the union of the row and the column through th... | {
"problem_match": "\nProblem 5.",
"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-2.jsonl",
"solution_match": "\nSolution 1"
} | 134 | 772 |
2017 | T1 | 5 | null | RMM | Fix an integer $n \geq 2$. An $n \times n$ sieve is an $n \times n$ array with $n$ cells removed so that exactly one cell is removed from every row and every column. A stick is a $1 \times k$ or $k \times 1$ array for any positive integer $k$. For any sieve $A$, let $m(A)$ be the minimal number of sticks required to pa... | . In order to prove $m(A) \geq 2 n-2$, it suffices to show that there are $2 n-2$ cells in $A$, no two of which may be contained in the same stick.
To this end, consider the bipartite graph $G$ with parts $G_{h}$ and $G_{v}$, where the vertices in $G_{h}$ (respectively, $G_{v}$ ) are the $2 n-2$ maximal sticks $A$ is ... | {
"problem_match": "\nProblem 5.",
"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-2.jsonl",
"solution_match": "\nSolution 3"
} | 134 | 784 |
2017 | T1 | 6 | null | RMM | Let $A B C D$ be any convex quadrilateral and let $P, Q, R, S$ be points on the segments $A B, B C, C D$, and $D A$, respectively. It is given that the segments $P R$ and $Q S$ dissect $A B C D$ into four quadrilaterals, each of which has perpendicular diagonals. Show that the points $P, Q, R, S$ are concyclic.
NikOla... | . We start with a lemma which holds even in a more general setup.
Lemma 1. Let $P Q R S$ be a convex quadrangle whose diagonals meet at $O$. Let $\omega_{1}$ and $\omega_{2}$ be the circles on diameters $P Q$ and $R S$, respectively, and let $\ell$ be their radical axis. Finally, choose the points $A, B$, and $C$ outsi... | {
"problem_match": "\nProblem 6.",
"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-2.jsonl",
"solution_match": "\nSolution 1"
} | 105 | 1,072 |
2017 | T1 | 6 | null | RMM | Let $A B C D$ be any convex quadrilateral and let $P, Q, R, S$ be points on the segments $A B, B C, C D$, and $D A$, respectively. It is given that the segments $P R$ and $Q S$ dissect $A B C D$ into four quadrilaterals, each of which has perpendicular diagonals. Show that the points $P, Q, R, S$ are concyclic.
NikOla... | . (Ilya Bogdanov) To begin, we establish a useful lemma.
Lemma 2. If $P$ is a point on the side $A B$ of a triangle $O A B$, then
$$
\frac{\sin A O P}{O B}+\frac{\sin P O B}{O A}=\frac{\sin A O B}{O P}
$$
Proof. Let $[X Y Z]$ denote the area of a triangle $X Y Z$, to write
$0=2([A O B]-[P O B]-[P O C])=O A \cdot O B ... | {
"problem_match": "\nProblem 6.",
"resource_path": "RMM/segmented/en-2017-Solutions_RMM2017-2.jsonl",
"solution_match": "\nSolution 2"
} | 105 | 1,496 |
2018 | T1 | 2 | null | RMM | Determine whether there exist non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying
$$
P(x)^{10}+P(x)^{9}=Q(x)^{21}+Q(x)^{20} .
$$
## Ilya Bogdanov, Russia | . Letting $r$ and $s$ be integers such that $r \geq 2$ and $s \geq 2 r$, we show that if $P^{r}+P^{r-1}=$ $Q^{s}+Q^{s-1}$, then $Q$ is constant.
Let $m=\operatorname{deg} P$ and $n=\operatorname{deg} Q$. A degree inspection in the given relation shows that $m \geq 2 n$.
We will prove that $P(P+1)$ has at least $m+1$ d... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2018-RMM2018-Day1-English.jsonl",
"solution_match": "\nSolution 2"
} | 66 | 629 |
2018 | T1 | 3 | null | RMM | Ann and Bob play a game on an infinite checkered plane making moves in turn; Ann makes the first move. A move consists in orienting any unit grid-segment that has not been oriented before. If at some stage some oriented segments form an oriented cycle, Bob wins. Does Bob have a strategy that guarantees him to win?
Max... | The answer is in the negative: Ann has a strategy allowing her to prevent Bob's victory.
We say that two unit grid-segments form a low-left corner (or LL-corner) if they share an endpoint which is the lowest point of one and the leftmost point of the other. An up-right corner (or $U R$-corner) is defined similarly. Th... | {
"problem_match": "\nProblem 3.",
"resource_path": "RMM/segmented/en-2018-RMM2018-Day1-English.jsonl",
"solution_match": "\nSolution."
} | 73 | 743 |
2018 | T1 | 4 | null | RMM | Let $a, b, c, d$ be positive integers such that $a d \neq b c$ and $\operatorname{gcd}(a, b, c, d)=1$. Prove that, as $n$ runs through the positive integers, the values $\operatorname{gcd}(a n+b, c n+d)$ may achieve form the set of all positive divisors of some integer.
Raul Alcantara, Peru | . We extend the problem statement by allowing $a$ and $c$ take non-negative integer values, and allowing $b$ and $d$ to take arbitrary integer values. (As usual, the greatest common divisor of two integers is non-negative.) Without loss of generality, we assume $0 \leq a \leq c$. Let $S(a, b, c, d)=\left\{\operatorname... | {
"problem_match": "\nProblem 4.",
"resource_path": "RMM/segmented/en-2018-RMM2018-Day2-English.jsonl",
"solution_match": "\nSolution 1"
} | 92 | 916 |
2018 | T1 | 4 | null | RMM | Let $a, b, c, d$ be positive integers such that $a d \neq b c$ and $\operatorname{gcd}(a, b, c, d)=1$. Prove that, as $n$ runs through the positive integers, the values $\operatorname{gcd}(a n+b, c n+d)$ may achieve form the set of all positive divisors of some integer.
Raul Alcantara, Peru | . (Alexander Betts) For positive integers $s$ and $t$ and prime $p$, we will denote by $\operatorname{gcd}_{p}(s, t)$ the greatest common $p$-power divisor of $s$ and $t$.
Claim 1. For any positive integer $n, \operatorname{gcd}(a n+b, c n+d) \mid a d-b c$.
Proof. This is clear from the identity
$$
a(c n+d)-c(a n+b)=a... | {
"problem_match": "\nProblem 4.",
"resource_path": "RMM/segmented/en-2018-RMM2018-Day2-English.jsonl",
"solution_match": "\nSolution 2"
} | 92 | 1,029 |
2018 | T1 | 5 | null | RMM | Let $n$ be a positive integer and fix $2 n$ distinct points on a circumference. Split these points into $n$ pairs and join the points in each pair by an arrow (i.e., an oriented line segment). The resulting configuration is good if no two arrows cross, and there are no arrows $\overrightarrow{A B}$ and $\overrightarrow... | . The required number is $\binom{2 n}{n}$. To prove this, trace the circumference counterclockwise to label the points $a_{1}, a_{2}, \ldots, a_{2 n}$.
Let $\mathcal{C}$ be any good configuration and let $O(\mathcal{C})$ be the set of all points from which arrows emerge. We claim that every $n$-element subset $S$ of $... | {
"problem_match": "\nProblem 5.",
"resource_path": "RMM/segmented/en-2018-RMM2018-Day2-English.jsonl",
"solution_match": "\nSolution 1"
} | 107 | 600 |
2018 | T1 | 5 | null | RMM | Let $n$ be a positive integer and fix $2 n$ distinct points on a circumference. Split these points into $n$ pairs and join the points in each pair by an arrow (i.e., an oriented line segment). The resulting configuration is good if no two arrows cross, and there are no arrows $\overrightarrow{A B}$ and $\overrightarrow... | . Use the counterclockwise labelling $a_{1}, a_{2}, \ldots, a_{2 n}$ in the solution above.
Letting $D_{n}$ be the number of good configurations on $2 n$ points, we establish a recurrence relation for the $D_{n}$. To this end, let $C_{n}=\frac{(2 n)!}{n!(n+1)!}$ the $n$th Catalan number; it is well-known that $C_{n}$ i... | {
"problem_match": "\nProblem 5.",
"resource_path": "RMM/segmented/en-2018-RMM2018-Day2-English.jsonl",
"solution_match": "\nSolution 2"
} | 107 | 792 |
2018 | T1 | 6 | null | RMM | Fix a circle $\Gamma$, a line $\ell$ tangent to $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ cross $\ell$ at $Y$ and $Z$. Prove that, as $X$ traces $\Omega$, the circle $X Y Z$ i... | Assume $\Gamma$ of unit radius and invert with respect to $\Gamma$. No reference will be made to the original configuration, so images will be denoted by the same letters. Letting $\Gamma$ be centred at $G$, notice that inversion in $\Gamma$ maps tangents to $\Gamma$ to circles of unit diameter through $G$ (hence inter... | {
"problem_match": "\nProblem 6.",
"resource_path": "RMM/segmented/en-2018-RMM2018-Day2-English.jsonl",
"solution_match": "\nSolution."
} | 105 | 844 |
2019 | T1 | 1 | null | RMM | Amy and Bob play the game. At the beginning, Amy writes down a positive integer on the board. Then the players take moves in turn, Bob moves first. On any move of his, Bob replaces the number $n$ on the blackboard with a number of the form $n-a^{2}$, where $a$ is a positive integer. On any move of hers, Amy replaces th... | The answer is in the negative. For a positive integer $n$, we define its square-free part $S(n)$ to be the smallest positive integer $a$ such that $n / a$ is a square of an integer. In other words, $S(n)$ is the product of all primes having odd exponents in the prime expansion of $n$. We also agree that $S(0)=0$.
Now ... | {
"problem_match": "\nProblem 1.",
"resource_path": "RMM/segmented/en-2019-RMM2019-Day1-English.jsonl",
"solution_match": "\nSolution."
} | 133 | 630 |
2019 | T1 | 2 | null | RMM | Let $A B C D$ be an isosceles trapezoid with $A B \| C D$. Let $E$ be the midpoint of $A C$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $A B E$ and $C D E$, respectively. Let $P$ be the crossing poitn of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $P E$ is tang... | . If $A B C D$ is a rectangle, the statement is trivial due to symmetry. Hence, in what follows we assume $A D \nVdash B C$.
Let $F$ be the midpoint of $B D$; by symmetry, both $\omega$ and $\Omega$ pass through $F$. Let $P^{\prime}$ be the meeting point of tangents to $\omega$ at $F$ and to $\Omega$ at $E$. We aim to... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2019-RMM2019-Day1-English.jsonl",
"solution_match": "\nSolution 1"
} | 124 | 582 |
2019 | T1 | 2 | null | RMM | Let $A B C D$ be an isosceles trapezoid with $A B \| C D$. Let $E$ be the midpoint of $A C$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $A B E$ and $C D E$, respectively. Let $P$ be the crossing poitn of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $P E$ is tang... | . Let $Q$ be the isogonal conjugate of $P$ with respect to $\triangle A E D$, so $\angle(Q A, A D)=$ $\angle(E A, A P)=\angle(E B, B A)$ and $\angle(Q D, D A)=\angle(E D, D P)=\angle(E C, C D)$. Now our aim is to prove that $Q E \| C D$; this will yield that $\angle(E C, C D)=\angle(A E, E Q)=\angle(P E, E D)$, whence ... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2019-RMM2019-Day1-English.jsonl",
"solution_match": "\nSolution 2"
} | 124 | 557 |
2019 | T1 | 2 | null | RMM | Let $A B C D$ be an isosceles trapezoid with $A B \| C D$. Let $E$ be the midpoint of $A C$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $A B E$ and $C D E$, respectively. Let $P$ be the crossing poitn of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $P E$ is tang... | . (Dan Carmon) Let $O$ be the intersection of the diagonals $A C$ and $B D$. Let $F$ be the midpoint of $B D$. Let $S$ be the second intersection point of the circumcircles of triangles $A O F$ and $D O E$. We will prove that $S D$ and $S E$ are tangent to $\Omega$; the symmetric argument would then imply also that $S ... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2019-RMM2019-Day1-English.jsonl",
"solution_match": "\nSolution 3"
} | 124 | 961 |
2019 | T1 | 3 | null | RMM | Given any positive real number $\varepsilon$, prove that, for all but finitely many positive integers $v$, any graph on $v$ vertices with at least $(1+\varepsilon) v$ edges has two distinct simple cycles of equal lengths.
(Recall that the notion of a simple cycle does not allow repetition of vertices in a cycle.)
Russ... | Fix a positive real number $\varepsilon$, and let $G$ be a graph on $v$ vertices with at least $(1+\varepsilon) v$ edges, all of whose simple cycles have pairwise distinct lengths.
Assuming $\varepsilon^{2} v \geq 1$, we exhibit an upper bound linear in $v$ and a lower bound quadratic in $v$ for the total number of si... | {
"problem_match": "\nProblem 3.",
"resource_path": "RMM/segmented/en-2019-RMM2019-Day1-English.jsonl",
"solution_match": "\nSolution."
} | 80 | 1,991 |
2021 | T1 | 1 | null | RMM | Let $T_{1}, T_{2}, T_{3}, T_{4}$ be pairwise distinct collinear points such that $T_{2}$ lies between $T_{1}$ and $T_{3}$, and $T_{3}$ lies between $T_{2}$ and $T_{4}$. Let $\omega_{1}$ be a circle through $T_{1}$ and $T_{4}$; let $\omega_{2}$ be the circle through $T_{2}$ and internally tangent to $\omega_{1}$ at $T_{... | Let $O_{i}$ be the centre of $\omega_{i}, i=1,2,3,4$. Notice that the isosceles triangles $O_{i} T_{i} T_{i-1}$ are similar (indices are reduced modulo 4 ), to infer that $\omega_{4}$ is internally tangent to $\omega_{1}$ at $T_{4}$, and $O_{1} O_{2} O_{3} O_{4}$ is a (possibly degenerate) parallelogram.
Let $F_{i}$ b... | {
"problem_match": "\nProblem 1.",
"resource_path": "RMM/segmented/en-2021-RMM2021-Day1-English_Solutions.jsonl",
"solution_match": "\nSolution."
} | 285 | 965 |
2021 | T1 | 2 | null | RMM | Xenia and Sergey play the following game. Xenia thinks of a positive integer $N$ not exceeding 5000. Then she fixes 20 distinct positive integers $a_{1}, a_{2}, \ldots, a_{20}$ such that, for each $k=1,2, \ldots, 20$, the numbers $N$ and $a_{k}$ are congruent modulo $k$. By a move, Sergey tells Xenia a set $S$ of posit... | Sergey can determine Xenia's number in 2 but not fewer moves.
We first show that 2 moves are sufficient. Let Sergey provide the set $\{17,18\}$ on his first move, and the set $\{18,19\}$ on the second move. In Xenia's two responses, exactly one number occurs twice, namely, $a_{18}$. Thus, Sergey is able to identify $a_... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2021-RMM2021-Day1-English_Solutions.jsonl",
"solution_match": "\nSolution."
} | 171 | 583 |
2021 | T1 | 3 | null | RMM | A number of 17 workers stand in a row. Every contiguous group of at least 2 workers is a brigade. The chief wants to assign each brigade a leader (which is a member of the brigade) so that each worker's number of assignments is divisible by 4. Prove that the number of such ways to assign the leaders is divisible by 17 ... | Assume that every single worker also forms a brigade (with a unique possible leader). In this modified setting, we are interested in the number $N$ of ways to assign leadership so that each worker's number of assignments is congruent to 1 modulo 4.
Consider the variables $x_{1}, x_{2}, \ldots, x_{17}$ corresponding to... | {
"problem_match": "\nProblem 3.",
"resource_path": "RMM/segmented/en-2021-RMM2021-Day1-English_Solutions.jsonl",
"solution_match": "\nSolution."
} | 82 | 603 |
2021 | T1 | 4 | null | RMM | Consider an integer $n \geq 2$ and write the numbers $1,2, \ldots, n$ down on a board. A move consists in erasing any two numbers $a$ and $b$, and, for each $c$ in $\{a+b,|a-b|\}$, writing $c$ down on the board, unless $c$ is already there; if $c$ is already on the board, do nothing. For all integers $n \geq 2$, determ... | The answer is in the affirmative for all $n \geq 2$. Induct on $n$. Leaving aside the trivial case $n=2$, deal first with particular cases $n=5$ and $n=6$.
If $n=5$, remove first the pair $(2,5)$, notice that $3=|2-5|$ is already on the board, so $7=2+5$ alone is written down. Removal of the pair $(3,4)$ then leaves e... | {
"problem_match": "\nProblem 4.",
"resource_path": "RMM/segmented/en-2021-RMM2021-Day2-English_Solutions.jsonl",
"solution_match": "\nSolution."
} | 131 | 878 |
2021 | T1 | 4 | null | RMM | Consider an integer $n \geq 2$ and write the numbers $1,2, \ldots, n$ down on a board. A move consists in erasing any two numbers $a$ and $b$, and, for each $c$ in $\{a+b,|a-b|\}$, writing $c$ down on the board, unless $c$ is already there; if $c$ is already on the board, do nothing. For all integers $n \geq 2$, determ... | . We will prove the following, more general statement:
Claim. Write down a finite number (at least two) of pairwise distinct positive integers on a board. A move consists in erasing any two numbers $a$ and $b$, and, for each $c$ in $\{a+b,|a-b|\}$, writing $c$ down on the board, unless $c$ is already there; if $c$ is a... | {
"problem_match": "\nProblem 4.",
"resource_path": "RMM/segmented/en-2021-RMM2021-Day2-English_Solutions.jsonl",
"solution_match": "\nSolution 2"
} | 131 | 642 |
2021 | T1 | 5 | null | RMM | Let $n$ be a positive integer. The kingdom of Zoomtopia is a convex polygon with integer sides, perimeter $6 n$, and $60^{\circ}$ rotational symmetry (that is, there is a point $O$ such that a $60^{\circ}$ rotation about $O$ maps the polygon to itself). In light of the pandemic, the government of Zoomtopia would like t... | Let $P$ denote the given polygon, i.e., the kingdom of Zoomtopia. Throughout the solution, we interpret polygons with integer sides and perimeter $6 k$ as $6 k$-gons with unit sides (some of their angles may equal $180^{\circ}$ ). The argument hinges on the claim below:
Claim. Let $P$ be a convex polygon satisfying the... | {
"problem_match": "\nProblem 5.",
"resource_path": "RMM/segmented/en-2021-RMM2021-Day2-English_Solutions.jsonl",
"solution_match": "\nSolution."
} | 159 | 733 |
2021 | T1 | 6 | null | RMM | Initially, a non-constant polynomial $S(x)$ with real coefficients is written down on a board. Whenever the board contains a polynomial $P(x)$, not necessarily alone, one can write down on the board any polynomial of the form $P(C+x)$ or $C+P(x)$, where $C$ is a real constant. Moreover, if the board contains two (not n... | The required polynomials are all polynomials of an even degree $d \geq 2$, and all polynomials of odd degree $d \geq 3$ with negative leading coefficient.
Part I. We begin by showing that any (non-constant) polynomial $S(x)$ not listed above is not $(A, B)$-nice for some pair $(A, B)$ with either $|A|=|B|=2$, or $|A|=... | {
"problem_match": "\nProblem 6.",
"resource_path": "RMM/segmented/en-2021-RMM2021-Day2-English_Solutions.jsonl",
"solution_match": "\nSolution."
} | 315 | 2,667 |
2023 | T1 | 2 | null | RMM | Fix an integer $n \geqslant 3$. Let $\mathcal{S}$ be a set of $n$ points in the plane, no three of which are collinear. Given different points $A, B, C$ in $\mathcal{S}$, the triangle $A B C$ is nice for $A B$ if Area $(A B C) \leqslant \operatorname{Area}(A B X)$ for all $X$ in $\mathcal{S}$ different from $A$ and $B$... | For convenience, a triangle whose vertices all lie in $\mathcal{S}$ will be referred to as a triangle in $\mathcal{S}$. The argument hinges on the following observation:
Given any partition of $\mathcal{S}$, amongst all triangles in $\mathcal{S}$ with at least one vertex in each part, those of minimal area are all ade... | {
"problem_match": "\nProblem 2.",
"resource_path": "RMM/segmented/en-2023-RMM2023-Day1-English_Solutions.jsonl",
"solution_match": "\nSolution."
} | 187 | 1,178 |
2023 | T1 | 3 | null | RMM | Let $n \geqslant 2$ be an integer, and let $f$ be a $4 n$-variable polynomial with real coefficients. Assume that, for any $2 n$ points $\left(x_{1}, y_{1}\right), \ldots,\left(x_{2 n}, y_{2 n}\right)$ in the plane, $f\left(x_{1}, y_{1}, \ldots, x_{2 n}, y_{2 n}\right)=0$ if and only if the points form the vertices of ... | The smallest possible degree is $2 n$. In what follows, we will frequently write $A_{i}=$ $\left(x_{i}, y_{i}\right)$, and abbreviate $P\left(x_{1}, y_{1}, \ldots, x_{2 n}, y_{2 n}\right)$ to $P\left(A_{1}, \ldots, A_{2 n}\right)$ or as a function of any $2 n$ points.
Suppose that $f$ is valid. First, we note a key pr... | {
"problem_match": "\nProblem 3.",
"resource_path": "RMM/segmented/en-2023-RMM2023-Day1-English_Solutions.jsonl",
"solution_match": "\nSolution."
} | 143 | 3,017 |
2023 | T1 | 4 | null | RMM | Given a triangle $A B C$, let $H$ and $O$ be its orthocentre and circumcentre, respectively. Let $K$ be the midpoint of the line segment $A H$. Let further $\ell$ be a line through $O$, and let $P$ and $Q$ be the orthogonal projections of $B$ and $C$ onto $\ell$, respectively. Prove that $K P+K Q \geq B C$.
# Russia, ... | . Denote by $R$ and $O$ the circumradius and the circumcentre of triangle $A B C$, respectively. As in Solution 1, we see that $M K=R$.
Assume now that $\ell$ is fixed, while $A$ moves along the fixed circle $(A B C)$. Then $K$ will move along a cricle centred at $M$ with radius $R$. We must show that for each point $... | {
"problem_match": "\nProblem 4.",
"resource_path": "RMM/segmented/en-2023-RMM2023-Day2-English_Solutions.jsonl",
"solution_match": "\nSolution 4"
} | 104 | 811 |
2023 | T1 | 5 | null | RMM | Let $P(x), Q(x), R(x)$ and $S(x)$ be non-constant polynomials with real coefficients such that $P(Q(x))=R(S(x))$. Suppose that the degree of $P(x)$ is divisible by the degree of $R(x)$.
Prove that there is a polynomial $T(x)$ with real coefficients such that $P(x)=R(T(x))$.
## Iran, NaVid Safaei | . All polynomials in the solution have real coefficients. As usual, the degree of a polynomial $f(x)$ is denoted $\operatorname{deg} f(x)$.
Of all pairs of polynomials $P(x), R(x)$, satisfying the conditions in the statement, choose one, say, $P_{0}(x), R_{0}(x)$, so that $P_{0}(Q(x))=R_{0}(S(x))$ has a minimal (posit... | {
"problem_match": "\nProblem 5.",
"resource_path": "RMM/segmented/en-2023-RMM2023-Day2-English_Solutions.jsonl",
"solution_match": "\nSolution 2"
} | 93 | 1,451 |
2023 | T1 | 6 | null | RMM | Let $r, g, b$ be non-negative integers. Let $\Gamma$ be a connected graph on $r+g+b+1$ vertices. The edges of $\Gamma$ are each coloured red, green or blue. It turns out that $\Gamma$ has
- a spanning tree in which exactly $r$ of the edges are red,
- a spanning tree in which exactly $g$ of the edges are green and
- a ... | . Induct on $n=r+g+b$. The base case, $n=1$, is clear.
Let now $n>1$. Let $V$ denote the vertex set of $\Gamma$, and let $T_{r}, T_{g}$, and $T_{b}$ be the trees with exactly $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. Consider two cases.
Case 1: There exists a partition $V=A \sqcup B$ of the ver... | {
"problem_match": "\nProblem 6.",
"resource_path": "RMM/segmented/en-2023-RMM2023-Day2-English_Solutions.jsonl",
"solution_match": "\nSolution 1"
} | 153 | 837 |
2023 | T1 | 6 | null | RMM | Let $r, g, b$ be non-negative integers. Let $\Gamma$ be a connected graph on $r+g+b+1$ vertices. The edges of $\Gamma$ are each coloured red, green or blue. It turns out that $\Gamma$ has
- a spanning tree in which exactly $r$ of the edges are red,
- a spanning tree in which exactly $g$ of the edges are green and
- a ... | . For a spanning tree $T$ in $\Gamma$, denote by $r(T), g(T)$, and $b(T)$ the number of red, green, and blue edges in $T$, respectively.
Assume that $\mathcal{C}$ is some collection of spanning trees in $\Gamma$. Write
$$
\begin{array}{rlrl}
r(\mathcal{C}) & =\min _{T \in \mathcal{C}} r(T), & & g(\mathcal{C})=\min _{... | {
"problem_match": "\nProblem 6.",
"resource_path": "RMM/segmented/en-2023-RMM2023-Day2-English_Solutions.jsonl",
"solution_match": "\nSolution 2"
} | 147 | 1,733 |
2007 | T2 | 3 | null | Serbia_MO | Одредити све парове природних бројева $(x, n)$ који су решења једначине
$$
x^{3}+2 x+1=2^{n}
$$
Време за рад 270 минута.
Сваки задатак вреди 7 поена.
## СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА
такмичење ученика средњих школа из математике
Београд, 03.04.2007.
## Други дан | Провером се добија да је за $n \leqslant 2$ једино решење пар $(1,2)$. Докажимо да за $n \geqslant 3$ нема решења.
Број $x$ мора бити непаран, па је $x^{2}+2 \equiv 3(\bmod 8)$. Сада из $x\left(x^{2}+2\right) \equiv-1$ $(\bmod 8)$ следи да је $x \equiv 5(\bmod 8)$. Шта више, како $3 \mid x\left(x^{2}+2\right)$ (уколик... | {
"problem_match": "\n3. ",
"resource_path": "Serbia_MO/segmented/sr-2007_smo_resenja.jsonl",
"solution_match": "\n3. "
} | 149 | 620 |
2007 | T2 | 4 | null | Serbia_MO | Нека је $k$ природан број. За сваку функцију $f: \mathbb{N} \rightarrow \mathbb{N}$, нека је низ функција $\left(f_{m}\right)_{m \geqslant 1}$ дефинисан са $f_{1}=f$ и $f_{m+1}=f \circ f_{m}$ за $m \geqslant 1$. Функција $f$ је $k$-фина уколико за све $n \in \mathbb{N}$ важи
$$
f_{k}(n)=f(n)^{k}
$$
(a) За које $k$ по... | Свака функција је 1-фина, па је одговор на оба дела задатка потврдан. Нека је надаље $k \geqslant 2$. Свака $k$-фина функција је 1 -1 јер из $f(m)=f(n)$ следи $m^{k}=f_{k}(m)=f_{k}(n)=n^{k}$, тј. $m=n$.
(a) Одговор: ДА. Конструишимо функцију $f$ индуктивно на следећи начин. Нека је $n$ најмањи природан број чија слика... | {
"problem_match": "\n4. ",
"resource_path": "Serbia_MO/segmented/sr-2007_smo_resenja.jsonl",
"solution_match": "\n4. "
} | 219 | 861 |
2007 | T2 | 5 | null | Serbia_MO | Дат је неједнакокраки троугао $A B C$. Нека су $A D, B E, C F$ симетрале углова овог троугла $(D \in B C, E \in A C, F \in A B)$. Нека су $K_{a}, K_{b}, K_{c}$ тачке на уписаном кругу троугла $A B C$ такве да су $D K_{a}, E K_{b}, F K_{c}$ тангенте на уписани круг и да $K_{a} \notin B C, K_{b} \notin A C, K_{c} \notin ... | Докажимо да су троуглови $K_{a} K_{b} K_{c}$ и $A_{1} B_{1} C_{1}$ хомотетични. Да бисмо то доказали, довољно је да докажемо да је $K_{a} K_{b} \| A_{1} B_{1}$, односно $K_{a} K_{b} \| A B$ (аналогно ће следити и за друге парове страница).
Означимо $M=K_{a} K_{b} \cap B C$, са $S$ означимо центар уписаног круга, и са ... | {
"problem_match": "\n5. ",
"resource_path": "Serbia_MO/segmented/sr-2007_smo_resenja.jsonl",
"solution_match": "\n5. "
} | 249 | 2,103 |
2007 | T2 | 6 | null | Serbia_MO | Нека је $k$ природан број. Доказати да за позитивне реалне бројеве $x, y, z$ чији је збир једнак 1 , важи неједнакост
$$
\frac{x^{k+2}}{x^{k+1}+y^{k}+z^{k}}+\frac{y^{k+2}}{y^{k+1}+z^{k}+x^{k}}+\frac{z^{k+2}}{z^{k+1}+x^{k}+y^{k}} \geqslant \frac{1}{7}
$$
Када важи једнакост?
Време за рад 270 минута.
Сваки задатак вр... | Дати израз је симетричан, па се без губљења општости може претпоставити да је $x \geqslant y \geqslant z$. Тада је
$$
x^{k+1}+y^{k}+z^{k} \leqslant y^{k+1}+z^{k}+x^{k} \leqslant z^{k+1}+x^{k}+y^{k}
$$
Заиста, довољно је доказати прву неједнакост, тј. да је $x^{k+1}+y^{k} \leqslant y^{k+1}+$ $x^{k}$. Ова неједнакост ј... | {
"problem_match": "\n6. ",
"resource_path": "Serbia_MO/segmented/sr-2007_smo_resenja.jsonl",
"solution_match": "\n6. "
} | 204 | 1,070 |
2008 | T2 | 1 | null | Serbia_MO | У скупу целих бројева решити једначину
$$
12^{x}+y^{4}=2008^{z} . \quad \text { (Милош Милосављевић) }
$$ | За $x<0$ или $z \leq 0$ једино решење је тривијално $(0,0,0)$. Надаље је $z>0$. Како је $2008=2^{3} \cdot 251$, обе стране једначине су дељиве са 251. Претпоставимо да је $x=2 x_{1}$ парно. Тада $\left(2^{x_{1}}\right)^{2} \equiv-\left(y^{2}\right)^{2}(\bmod 251)$, што дизанјем на 125 -ти степен даје $1 \equiv\left(2^{... | {
"problem_match": "\n1. ",
"resource_path": "Serbia_MO/segmented/sr-2008_smo_resenja.jsonl",
"solution_match": "\n1. "
} | 61 | 1,039 |
2008 | T2 | 2 | null | Serbia_MO | Дат је троугао $A B C$. Нека су тачке $D$ и $E$ на правој $A B$ такве да је $D-A-B-E, A D=A C$ и $B E=B C$. Симетрале унутрашњих углова код темена $A$ и $B$ секу наспрамне странице у тачкама $P$ и $Q$, редом, а описану кружницу око троугла $A B C$ у тачкама $M$ и $N$, редом. Права која спаја тачку $A$ са центром кружни... | Означимо са $U$ центар описаног круга $\triangle B M E$. Применимо инверзију са центром $A$ и квадратом полупречника $A B \cdot A C$. Тачке $B$ и $C$ се сликају у тачке $B^{\prime}$ и $C^{\prime}$ симетричне тачкама $C$ и $B$ у односу на $A P$, тачке $P$ и $M$ се сликају једна у другу, а $E$ се слика у тачку $E^{\prime... | {
"problem_match": "\n2. ",
"resource_path": "Serbia_MO/segmented/sr-2008_smo_resenja.jsonl",
"solution_match": "\n2. "
} | 265 | 589 |
2008 | T2 | 3 | null | Serbia_MO | Доказати да за позитивне реалне бројеве $a, b$ и $c$, такве да је $a+b+c=1$, важи неједнакост
$$
\frac{1}{b c+a+\frac{1}{a}}+\frac{1}{c a+b+\frac{1}{b}}+\frac{1}{a b+c+\frac{1}{c}} \leqslant \frac{27}{31}
$$
(Марко Радовановић са сараунииима)
Време за рад 270 минута.
Сваки задатак вреди 7 поена.
## СРПСКА МАТЕМАТИ... | Тражена неједнакост је очигледно еквивалентна неједнакости
$$
\frac{a}{p+a^{2}}+\frac{b}{p+b^{2}}+\frac{c}{p+c^{2}} \leq \frac{27}{31}
$$
где је $a+b+c=1$ и $p=a b c+1$. Посматраћемо функцију
$$
f(x)=\frac{3(a+b+c)}{3 x+a^{2}+b^{2}+c^{2}}-\frac{a}{x+a^{2}}-\frac{b}{x+b^{2}}-\frac{c}{x+c^{2}}
$$
Доказаћемо да важи $... | {
"problem_match": "\n3. ",
"resource_path": "Serbia_MO/segmented/sr-2008_smo_resenja.jsonl",
"solution_match": "\n3. "
} | 225 | 1,599 |
2008 | T2 | 4 | null | Serbia_MO | Свака тачка равни је обојена са једном од 3 боје. Доказати да постоји троугао за који важи:
$1^{\circ}$ сва 3 темена тог троугла су обојена истом бојом;
$2^{\circ}$ полупречник описане кружнице тог троугла је 2008;
$3^{\circ}$ један угао троугла је два или три пута већи од неког од друга два угла.
. Разликујемо два случаја.
(i) Међу пет црвених темена не постоје два суседна. Сваком положају црвених тачака (до на ротацију) одговара композиција броја ... | {
"problem_match": "\n4. ",
"resource_path": "Serbia_MO/segmented/sr-2008_smo_resenja.jsonl",
"solution_match": "\n4. "
} | 224 | 584 |
2008 | T2 | 5 | null | Serbia_MO | Нека је низ $\left(a_{n}\right)_{n \geqslant 1}$ дефинисан са $a_{1}=3, a_{2}=11$ и $a_{n}=4 a_{n-1}-a_{n-2}$, за $n \geqslant 3$. Доказати да је сваки члан овог низа облика $a^{2}+2 b^{2}$ за неке природне $a$ и $b$.
(Ђорђе Баралић) | Имамо $a_{1}=1+2 \cdot 1^{2}, a_{2}=3^{2}+2 \cdot 1^{2}, a_{3}=3^{2}+2 \cdot 4^{2}, a_{4}=11^{2}+2 \cdot 4^{2}$, итд. Доказаћемо индукцијом по $n$ да важи
$$
a_{2 n-1}=a_{n-1}^{2}+2\left(\frac{a_{n}-a_{n-1}}{2}\right)^{2} \quad \text { и } \quad a_{2 n}=a_{n}^{2}+2\left(\frac{a_{n}-a_{n-1}}{2}\right)^{2},
$$
при чему... | {
"problem_match": "\n5. ",
"resource_path": "Serbia_MO/segmented/sr-2008_smo_resenja.jsonl",
"solution_match": "\n5. "
} | 134 | 1,252 |
2008 | T2 | 6 | null | Serbia_MO | Нека је $A B C D E$ конвескан петоугао у коме је $A B=1, \varangle B A E=\varangle A B C=$ $120^{\circ}, \varangle C D E=60^{\circ}$ и $\varangle A D B=30^{\circ}$. Доказати да је површина петоугла $A B C D E$ мања од $\sqrt{3}$.
(Милош Милосављевић)
## РЕШЕЊА | Нека је $k$ круг описан око троугла $A B D$, и $l$ права кроз $D$ паралелна са
$A B$. Полупречник круга $k$ је 1 . Полуправе $B C$ и $A E$ секу $k$ у тачкама $H$ и $I$, а праву $l$ у $F$ и $G$, редом. Троуглови $F C D$ и $G D E$ су слични јер је $\angle C F D=\angle D G E=60^{\circ}$ и $\angle F C D=120^{\circ}-\angle ... | {
"problem_match": "\n6. ",
"resource_path": "Serbia_MO/segmented/sr-2008_smo_resenja.jsonl",
"solution_match": "\n6. "
} | 137 | 1,848 |
2009 | T2 | 1 | null | Serbia_MO | Нека су $\alpha$ и $\beta$ углови неједнакокраког троугла $A B C$ код темена $A$ и $B$, редом. Нека симетрале ових углова секу наспрамне странице троугла у $D$ и $E$, редом. Доказати да оштар угао између правих $D E$ и $A B$ није већи од $\frac{|\alpha-\beta|}{3}$.
(Душан Ђукић) | Као и обично, означимо $\varangle A C B=\gamma$ и $B C=a, C A=b, A B=c$, при чему је без смањења општости $a>b$ и $\alpha>\beta$. Нека је $F$ тачка пресека правих $D E$ и $A B$, а $\varphi$ угао између ових правих. Из односа $\frac{B D}{D C}=\frac{c}{b}$ и $\frac{C E}{E A}=\frac{a}{c}$ лако налазимо $B D=\frac{a c}{b+c... | {
"problem_match": "\n1. ",
"resource_path": "Serbia_MO/segmented/sr-2009_smo_resenja.jsonl",
"solution_match": "\n1. "
} | 139 | 563 |
2009 | T2 | 2 | null | Serbia_MO | Одредити најмањи природан број који је дељив са 2009 и коме је збир цифара једнак 2009. | Пошто је $2009=223 \cdot 9+2$, тражени број има бар 224 цифре. Посматраћемо 224 -цифрене бројеве $x=\overline{c_{223} c_{222} \ldots c_{1} c_{0}}$. Јасно је да је $c_{223} \geq 2$. Притом, ако је $c_{223}=2$ онда је $c_{222}=\cdots=c_{0}=9$ и $x=3 \cdot 10^{223}-1 \equiv 3 \cdot 10-1$, а то није дељиво са $2009=7^{2} \... | {
"problem_match": "\n2. ",
"resource_path": "Serbia_MO/segmented/sr-2009_smo_resenja.jsonl",
"solution_match": "\n2. "
} | 52 | 964 |
2009 | T2 | 3 | null | Serbia_MO | Одредити највећи природан број $n$ за који постоје различити скупови $S_{1}, S_{2}, \ldots, S_{n}$ такви да је:
$1^{\circ}\left|S_{i} \cup S_{j}\right| \leqslant 2004$ за свака два цела броја $1 \leqslant i, j \leqslant n$, и
$2^{\circ} S_{i} \cup S_{j} \cup S_{k}=\{1,2, \ldots, 2008\}$ за свака три цела броја $1 \le... | Сваки скуп $S_{i}$ има највише 2003 елемената. Заиста, ако је $\left|S_{i}\right|=2004$, из услова $1^{\circ}$ следи да је $S_{j} \subset S_{i}$ за све $j$, противно услову $2^{\circ}$. Посматрајмо скупове
$$
G_{\{i, j\}}=\{1,2, \ldots, 2008\} \backslash\left(S_{i} \cup S_{j}\right) \quad \text { за } 1 \leq i, j \leq... | {
"problem_match": "\n3. ",
"resource_path": "Serbia_MO/segmented/sr-2009_smo_resenja.jsonl",
"solution_match": "\n3. "
} | 263 | 756 |
2009 | T2 | 4 | null | Serbia_MO | Нека је $n \in \mathbb{N}$ и $A_{n}$ скуп свих пермутација $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ скупа $\{1,2, \ldots, n\}$ таквих да важи
$$
k \mid 2\left(a_{1}+a_{2}+\cdots+a_{k}\right) \quad \text { за свако } 1 \leqslant k \leqslant n
$$
Одредити број елемената скупа $A_{n}$.
(Видан Говедарииа) | Означимо са $F_{n}$ број елемената скупа $A_{n}$. Имамо $F_{1}=1, F_{2}=2$ и $F_{3}=6$. За $n>3$, посматрајмо било коју пермутацију $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ у $A_{n}$. Како $n-1$ дели $2\left(a_{1}+\cdots+a_{n-1}\right)=n(n+1)-2 a_{n} \equiv 2-2 a_{n}(\bmod n-1)$, следи да је $a_{n}$ једнако $1, \frac... | {
"problem_match": "\n4. ",
"resource_path": "Serbia_MO/segmented/sr-2009_smo_resenja.jsonl",
"solution_match": "\n4. "
} | 161 | 697 |
2009 | T2 | 6 | null | Serbia_MO | Нека је $k$ уписана кружница неједнакокраког $\triangle A B C$, чији је центар $S$. Кружница $k$ додирује странице $B C, C A, A B$ у тачкама $P, Q, R$, редом. Права $Q R$ сече праву $B C$ у тачки $M$. Нека кружница која садржи тачке $B$ и $C$ додирује $k$ у тачки $N$. Описана кружница $\triangle M N P$ сече праву $A P$... | Посматрајмо хомотетију са центром $N$ која слика круг $k$ у круг $B C N$; нека она слика тачку $P$ у $P_{1}$. Тангента на круг $B C N$ у $P_{1}$ је паралелна тангенти на $k$ у $P$, тј. правој $B C$, што значи да је $P_{1}$ средиште лука $B C$ круга $B C N$. Дакле, $N P$ је симетрала угла $C N B$, па је $\frac{B N}{C N}... | {
"problem_match": "\n6. ",
"resource_path": "Serbia_MO/segmented/sr-2009_smo_resenja.jsonl",
"solution_match": "\n6. "
} | 218 | 665 |
2010 | T2 | 1 | null | Serbia_MO | Неки од $n$ градова су повезани авионским линијама (све линије су двосмерне). Постоји тачно $m$ линија. Нека је $d_{i}$ број линија које полазе из града $i$, за $i=1,2, \ldots, n$. Ако је $1 \leqslant d_{i} \leqslant 2010$, за свако $i=1,2, \ldots, n$, доказати да важи
$$
\sum_{i=1}^{n} d_{i}^{2} \leqslant 4022 m-2010... | Услов задатка нам даје $0 \leqslant\left(d_{i}-1\right)\left(2010-d_{i}\right)$ за све $i$, тј. $d_{i}^{2} \leqslant 2011 d_{i}-2010$. Користећи услов $\sum_{i=1}^{n} d_{i}=2 m$, сабирањем ових неједнакости добијамо
$$
\sum_{i=1}^{n} d_{i}^{2} \leqslant 2011 \cdot \sum_{i=1}^{n} d_{i}-2010 n=4022 m-2010 n
$$
а једнак... | {
"problem_match": "\n1. ",
"resource_path": "Serbia_MO/segmented/sr-2010_smo_resenja.jsonl",
"solution_match": "\n1. "
} | 227 | 614 |
2010 | T2 | 2 | null | Serbia_MO | У оштроуглом $\triangle A B C$ тачка $M$ је средиште странице $B C$, а тачке $D, E$ и $F$ су подножја висина из темена $A, B$ и $C$, редом. Нека је $H$ ортоцентар $\triangle A B C, S$ средиште дужи $A H$, а $G$ пресек дужи $F E$ и $A H$. Ако је $N$ тачка пресека тежишне дужи $A M$ и описане кружнице $\triangle B C H$, ... | Нека је $A^{\prime}$ тачка таква да је $A B A^{\prime} C$ паралелограм. Тада важи $\varangle B A^{\prime} C=$ $\varangle B A C=180^{\circ}-\varangle B H C=180^{\circ}-$ $\varangle B N C$, па су тачке $A^{\prime}, B, C, H, N$ на истом кругу, тј. кругу над пречником $H A^{\prime}$. Одавде је $\varangle A N H=90^{\circ}$,... | {
"problem_match": "\n2. ",
"resource_path": "Serbia_MO/segmented/sr-2010_smo_resenja.jsonl",
"solution_match": "\n2. "
} | 191 | 816 |
2010 | T2 | 3 | null | Serbia_MO | Нека је $A$ бесконачан подскуп скупа природних бројева. Одредити све природне бројеве $n$ такве да за свако $a \in A$ важи
$$
a^{n}+a^{n-1}+\ldots+a^{1}+1 \mid a^{n!}+a^{(n-1)!}+\ldots+a^{1!}+1
$$
(Милош Милосавлевић)
## СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА
такмичење ученика средњих школа из математике
Ниш, 07.04.2010.
... | Означимо $P(x)=x^{n}+x^{n-1}+\cdots+1$ и $Q(x)=x^{n!}+\cdots+x^{1!}+1$; нека је $Q(x)=C(x) P(x)+R(x)$, где су $C$ и $R$ полиноми са целим коефицијентима и $\operatorname{deg} R<\operatorname{deg} P$. По услову задатка $P(a) \mid Q(a)$, и самим тим $P(a) \mid R(a)$, за бесконачно много целих бројева $a$. Како за довољно... | {
"problem_match": "\n3. ",
"resource_path": "Serbia_MO/segmented/sr-2010_smo_resenja.jsonl",
"solution_match": "\n3. "
} | 188 | 1,471 |
2010 | T2 | 4 | null | Serbia_MO | Нека је $O$ центар описане кружнице $\triangle A B C$. Права кроз $O$ сече странице $C A$ и $C B$ у тачкама $D$ и $E$, редом, и описану кружницу $\triangle A B O$ у тачки $P$ унутар троугла (различитој од $O$ ). Тачка $Q$ на страници $A B$ је таква да је $\frac{A Q}{Q B}=\frac{D P}{P E}$. Доказати да је $\varangle A P ... | Нека је $X$ тачка на полуправој $A P$ таква да је $E X \| A C$. По Талесовој теореми је $A P: P X=D P: P E=$ $A Q: Q B$, одакле следи $B X \| Q P$.
Права $P E$ је спољашња симетрала угла $A P B$ и полови угао $B P X$. Такође, пошто је $\varangle B E X=180-$ $\varangle A C B$ и $\varangle B P X=180^{\circ}-\varangle A ... | {
"problem_match": "\n4. ",
"resource_path": "Serbia_MO/segmented/sr-2010_smo_resenja.jsonl",
"solution_match": "\n4. "
} | 179 | 844 |
2010 | T2 | 5 | null | Serbia_MO | Таблица димензија $n \times n$, на чијим пољима су бројеви $1,2, \ldots, n^{2}$ (на сваком пољу тачно један број и сваки број на тачно једном пољу) назива се нишка ако сви производи од по $n$ бројева који се налазе на $n$ „разбацаних" поља дају исти остатак при дељењу са $n^{2}+1$. Да ли постоји нишка таблица за:
(a) ... | (a) Претпоставимо да постоји нишка таблица $8 \times 8$ и да производ ма којих 8 разбацаних бројева даје остатак $r$ по модулу $8^{2}+1=65=5 \cdot 13$. Сви бројеви у таблици се могу поделити на 8 дисјунктних осморки разбацаних бројева. Међу овим осморкама постоји једна која садржи умножак броја 13 и једна која не садрж... | {
"problem_match": "\n5. ",
"resource_path": "Serbia_MO/segmented/sr-2010_smo_resenja.jsonl",
"solution_match": "\n5. "
} | 227 | 542 |
2010 | T2 | 6 | null | Serbia_MO | Нека су $a_{0}$ и $a_{n}$ различити делиоци природног броја $m>1$, а низ природних бројева $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$ такав да задовољава
$$
a_{i+1}=\left|a_{i} \pm a_{i-1}\right| \quad \text { за } 0<i<n
$$
Ако је НЗД $\left(a_{0}, \ldots, a_{n}\right)=1$, доказати да у низу постоји члан који је мањи од $\... | Посматрајмо два најмања (различита) члана низа, $p$ и $q$. Ако је $\min \{p, q\}=$ 1, тврђење тривијално важи; зато надаље претпостављамо да је $p, q>1$.
Лема 1. Постоје индекси $k$ и $l$ за које је $a_{k}=p, a_{l}=q$ и $|k-l| \leq 2$.
Доказ. Нека је $a_{k}=p$ и $a_{l}=q(k<l)$. Претпоставимо да је $r=l-k>2$. Доказаће... | {
"problem_match": "\n6. ",
"resource_path": "Serbia_MO/segmented/sr-2010_smo_resenja.jsonl",
"solution_match": "\n6. "
} | 188 | 1,791 |
2011 | T2 | 2 | null | Serbia_MO | Нека је $n$ непаран природан број такав да су бројеви $\varphi(n)$ и $\varphi(n+1)$ степени броја два ( $\varphi(n)$ је број природних бројева не већих од $n$ и узајамно простих са $n$ ). Доказати да је $n+1$ степен броја два или је $n=5$.
(Марко Радовановић) | Ако је $n=\prod_{i=1}^{k} p_{i}^{r_{k}}$ канонска факторизација $n$, важи $\varphi(n)=\prod_{i=1}^{k} p_{i}^{r_{k}-1}\left(p_{i}-\right.$ 1), па пошто $n$ нема других простих чинилаца осим двојке, мора бити $a_{i}=1$ и $p_{i}-1=2^{b_{i}}$ за свако $i$ и неке $b_{i}$. Како $2^{b_{i}}+1$ може бити прост само ако је $b_{i... | {
"problem_match": "\n2. ",
"resource_path": "Serbia_MO/segmented/sr-2011_smo_resenja.jsonl",
"solution_match": "\n2. "
} | 130 | 1,193 |
2011 | T2 | 3 | null | Serbia_MO | Нека је $H$ ортоцентар, а $O$ центар описане кружнице оштроуглог троугла $A B C$. Тачке $D$ и $E$ су подножја висина из $A$ и $B$, редом. Обележимо са $K$ пресечну тачку правих $O D$ и $B E$, а са $L$ пресечну тачку правих $O E$ и $A D$. Нека је $X$ друга пресечна тачка кружница описаних око троуглова $H K D$ и $H L E$... | Ако је $X$ центар описаног круга $\triangle O D E$, онда је $90^{\circ}-\angle K D E=90^{\circ}-$ $\angle O D E=\angle X E O=\angle X E L=\angle X H D=\angle X K D$ (сви углови су оријентисани), одакле следи да је $X K \perp D E$; аналогно $X L \perp D E$, тј. $K$ и $L$ леже на симетрали дужи $D E$, па је $D E H O$ јед... | {
"problem_match": "\n3. ",
"resource_path": "Serbia_MO/segmented/sr-2011_smo_resenja.jsonl",
"solution_match": "\n3. "
} | 313 | 797 |
2011 | T2 | 6 | null | Serbia_MO | Скуп $T$ садржи 66 тачака, а скуп $P$ садржи 16 правих у равни. За тачку $A \in T$ и праву $l \in P$ кажемо да су иниидентни пар ако $A \in l$. Доказати да број инцидентних парова не може бити већи од 159 , као и да постоји
![](https://cdn.mathpix.com/cropped/2024_06_04_ecf1fe71dc588d40a3adg-2.jpg?height=51&width=1501... | Означимо са $A_{1}, \ldots, A_{66}$ тачке скупа $T$ и са $a_{i}$ број правих из $P$ које садрже $A_{i}$. Тада је број парова правих које се секу у $A_{i}$ једнак $\binom{a_{i}}{2}$, а број инцидентних парова $I=\sum a_{i}$. Како се сваке две праве секу у највише једној тачки, важи $\sum_{i=1}^{66}\binom{a_{i}}{2} \leqs... | {
"problem_match": "\n6. ",
"resource_path": "Serbia_MO/segmented/sr-2011_smo_resenja.jsonl",
"solution_match": "\n6. "
} | 218 | 938 |
2012 | T2 | 2 | null | Serbia_MO | Наћи све природне бројеве $a$ и $b$ такве да
$$
a\left|b^{2}, \quad b\right| a^{2} \quad \text { и } \quad a+1 \mid b^{2}+1 . \quad \text { (Душан Ђукић) }
$$ | Нека је $b^{2}=c a$. Услови задатка дају $b^{2}=c a \mid a^{4}$ и $a+1 \mid c a+1$, а то је еквивалентно са
$$
c \mid a^{3} \quad \text { и } \quad a+1 \mid c-1 .
$$
Нека је $c=d(a+1)+1, d \in \mathbb{N}_{0}$. Како је $a^{3} \equiv-1(\bmod (a+1))$, имамо $\frac{a^{3}}{c} \equiv-1(\bmod (a+1))$, тј. $\frac{a^{3}}{c}=e... | {
"problem_match": "\n2. ",
"resource_path": "Serbia_MO/segmented/sr-2012_smo_resenja.jsonl",
"solution_match": "\n2. "
} | 82 | 758 |
2012 | T2 | 3 | null | Serbia_MO | У неким чворовима квадратне решетке $2012 \times 2012$ налази се мува и $k$ паукова. Један потез састоји се у следећем: мува се помера на суседан чвор или остаје на истом месту, а након тога се сваки од $k$ паукова помера на неки суседан чвор или остаје на истом месту (у једном чвору може бити више паукова). У сваком т... | Један паук не може да ухвати муву. Довољно је да мува не мрда ако паук није на суседном пољу, односно да се помери на поље дијагонално супротно пауковом ако јесте.
Доказаћемо да су два паука довољна у оба дела задатка - означимо их са $P$ и $Q$, муву са $M$, а $x$ - и $y$-координату тачке $A$ са $A_{x}$ и $A_{y}$.
(a... | {
"problem_match": "\n3. ",
"resource_path": "Serbia_MO/segmented/sr-2012_smo_resenja.jsonl",
"solution_match": "\n3. "
} | 442 | 1,055 |
2012 | T2 | 4 | null | Serbia_MO | Наћи све природне бројеве $n$ за које постоји пермутација ( $p_{1}, p_{2}, \ldots, p_{n}$ ) бројева $(1,2, \ldots, n)$ таква да скупови $\left\{p_{i}+i \mid 1 \leqslant i \leqslant n\right\}$ и $\left\{p_{i}-i \mid 1 \leqslant i \leqslant n\right\}$ чине потпуне системе остатака по модулу $n$. (Марко Ђикић) | Претпоставимо да таква пермутација постоји. Како је $\left\{p_{i}+i \mid 1 \leq i \leq\right.$ $n\}$ потпун систем остатака по модулу $n$, важи $\sum_{k=1}^{n} k \equiv \sum_{i=1}^{n}\left(p_{i}+i\right) \equiv$ $\sum_{i=1}^{n} i+\sum_{i=1}^{n} p_{i} \equiv 2 \sum_{k=1}^{n} k(\bmod n)$, дакле $\sum_{k=1}^{n} k=\frac{n(... | {
"problem_match": "\n4. ",
"resource_path": "Serbia_MO/segmented/sr-2012_smo_resenja.jsonl",
"solution_match": "\n4. "
} | 160 | 618 |
2012 | T2 | 5 | null | Serbia_MO | Нека је $\mathcal{K}$ целобројна решетка. Да ли постоји бијекција $f: \mathbb{N} \rightarrow \mathcal{K}$ таква да за све међусобно различите $a, b, c \in \mathbb{N}$ важи
$$
\text { НЗД }(a, b, c)>1 \quad \Longrightarrow \quad f(a), f(b), f(c) \text { нису колинеарне? }
$$
(Целобројна решетка је скуп тачака у равни ... | Поређајмо све тачке решетке у низ $A_{1}, A_{2}, \ldots$. Ово се може урадити нпр. спирално: $(0,0),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1), \ldots$ Конструисаћемо индуктивно пример бијекције са траженим својством.
Ставимо $f(1)=A_{1}$. Претпоставимо да су $f(1), \ldots, f(n-1)$ одређене и узмимо за $f(n)$ тачку $A_{m... | {
"problem_match": "\n5. ",
"resource_path": "Serbia_MO/segmented/sr-2012_smo_resenja.jsonl",
"solution_match": "\n5. "
} | 185 | 895 |
2012 | T2 | 6 | null | Serbia_MO | Нека композиција садржи $n>1$ вагона са златницима. Постоје две врсте наизглед истих златника: прави и лажни. У сваком вагону се налазе златници само једне врсте. Златници исте врсте су исте масе, док златници различитих врста немају исту масу. Маса правог златника је позната.
Одредити минималан број мерења на дигитал... | Доказаћемо да је минималан број мерења једнак 2. Означимо тежине правог и лажног златника са $x$ и $y$ редом, и нека је $a_{i}=1$ ако су златници у $i$-том вагону лажни, а $a_{i}=0$ у супротном.
Узмимо у првом мерењу по један златник из сваког вагона. Тада је $a_{1}+$ $a_{2}+\cdots+a_{n}=\frac{n x-m_{1}}{x-y}$, где је... | {
"problem_match": "\n6. ",
"resource_path": "Serbia_MO/segmented/sr-2012_smo_resenja.jsonl",
"solution_match": "\n6. "
} | 277 | 1,315 |
2013 | T2 | 2 | null | Serbia_MO | Нека је
$$
S_{n}=\left\{\binom{n}{n},\binom{2 n}{n},\binom{3 n}{n}, \ldots,\binom{n^{2}}{n}\right\}, \quad \text { за } n \in \mathbb{N}
$$
a) Доказати да постоји бесконачно много сложених природних бројева $n$ таквих да $S_{n}$ није потпун систем остатака по модулу $n$.
б) Доказати да постоји бесконачно много сложе... | (а) Доказаћемо да $n=2 p$ задовољава услове, где је $p>2$ прост број. Имамо
$$
\binom{2 k p}{2 p}=k \prod_{i=1}^{p-1} \frac{2 k p-i}{2 p-i} \cdot(2 k-1) \prod_{i=1}^{p-1} \frac{2 k p-p-i}{p-i} \equiv k(2 k-1) \quad(\bmod p)
$$
Конкретно, одавде је $\binom{2 k p}{2 p}$ дељиво са $p$ за $k \in\left\{\frac{p+1}{2}, p, 2... | {
"problem_match": "\n2. ",
"resource_path": "Serbia_MO/segmented/sr-2013_smo_resenja.jsonl",
"solution_match": "\n2. "
} | 204 | 665 |
2013 | T2 | 3 | null | Serbia_MO | Нека су $M, N$ и $P$ средишта страница $B C, A C$ и $A B$, редом, а $O$ центар описане кружнице оштроуглог троугла $A B C$. Кружнице описане око троуглова $B O C$ и $M N P$ секу се у различитим тачкама $X$ и $Y$ унутар троугла $A B C$. Доказати да је
$$
\varangle B A X=\varangle C A Y
$$
(Марко Ђикић)
## Министарств... | Обележимо са $k_{1}$ и $k_{2}$ редом кругове $M N P$ и $B O C$. Круг $k_{1}$ је Ојлеров круг у $\triangle A B C$ и пролази кроз подножја висина $D, E$ из $B, C$ и средиште $O_{1}$ дужи $A H$, где је $H$ ортоцентар $\triangle A B C$.
Покажимо да друга пресечна тачка $Z$ праве $A Y$ и круга $k_{1}$ лежи на Ојлеровом кру... | {
"problem_match": "\n3. ",
"resource_path": "Serbia_MO/segmented/sr-2013_smo_resenja.jsonl",
"solution_match": "\n3. "
} | 252 | 701 |
2013 | T2 | 4 | null | Serbia_MO | Одредити све $n \in \mathbb{N}$ за које је могуће поделити скуп $\{1,2, \ldots, 3 n\}$ на $n$ дисјунктних трочланих подскупова облика $\{a, b, c\}$ у којима су $b-a$ и $c-b$ различити бројеви из скупа $\{n-1, n, n+1\}$.
(Душан Ђукић) | Тражена партиција скупа $\{1,2, \ldots, 3 n\}$ одговара партицији темена правилног $3 n$-угла $P_{1} P_{2} \ldots P_{3 n}$ на тројке $\left\{A_{i}, B_{i}, C_{i}\right\}$ такве да су углови сваког од троуглова $A_{i} B_{i} C_{i}$ једнаки $\frac{n-1}{3 n} \pi, \frac{n}{3 n} \pi$ и $\frac{n+1}{3 n} \pi$. Погодним обележав... | {
"problem_match": "\n4. ",
"resource_path": "Serbia_MO/segmented/sr-2013_smo_resenja.jsonl",
"solution_match": "\n4. "
} | 130 | 721 |
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