id stringlengths 4 4 | problem_markdown stringlengths 36 3.59k | solutions_markdown listlengths 0 10 | images images listlengths 0 15 | country stringclasses 58
values | competition stringlengths 3 108 ⌀ | topics_flat listlengths 0 12 | language stringclasses 18
values | problem_type stringclasses 4
values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0f0g | Problem:
A king moves on an $8 \times 8$ chessboard. He can move one square at a time, diagonally or orthogonally (so away from the borders he can move to any of eight squares). He makes a complete circuit of the board, starting and finishing on the same square and visiting every other square just once. His trajectory... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | At least 28 moves parallel to the sides are required, and a circuit with exactly 28 such moves exists. With board side length 8, the maximum trajectory length is 28 + 36√2 and the minimum is 64. | |
02qm | Problem:
Considere uma pilha de cartas numeradas de $1$ a $104$. Um embaralhamento dessa pilha consiste em intercalar as $52$ cartas de cima com as $52$ de baixo, de modo que a carta que estava no topo fique em segundo lugar de cima para baixo. A figura mostra dois embaralhamentos seguidos a partir da situação inicial... | [
"Solution:\n\na) Vamos calcular a posição ocupada, após um embaralhamento, pela $n$-ésima carta da pilha. Há dois casos a considerar:\n\n1. Primeiro caso: $n \\leq 52$ (ou seja, a carta está na metade superior da pilha)\n\nNeste caso, após um embaralhamento, ficarão acima dela as primeiras $n$ cartas da metade infe... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Other"
] | null | proof and answer | a) Positions of card 5 after successive shuffles: 1→10th, 2→20th, 3→40th, 4→80th, 5→55th, 6→5th.
b) From the initial state, after one shuffle the card originally in position n moves to 2n if n ≤ 52, and to 2n − 105 if n > 52.
c) The two cards that swap positions at every shuffle are those initially at positions 35 and ... | |
05fe | Problem:
Soit $ABC$ un triangle, $H$ son orthocentre et $M$ le milieu du segment $[BC]$. Soit $d$ une droite passant par le point $M$. On suppose que $d$ coupe le cercle de diamètre $[AH]$ en $P$ et $Q$. Montrer que l'orthocentre du triangle $APQ$ est sur le cercle circonscrit du triangle $ABC$.
![](attached_image_1.... | [
"Solution:\n\nSoit $N$ le milieu de $[PQ]$ et $K$ l'orthocentre du triangle $APQ$. Si on appelle $K'$ le symétrique de $K$ par rapport au milieu $N$ du segment $[PQ]$, alors $[PQ]$ et $[KK']$ ont le même milieu $N$ donc $PKQK'$ est un parallélogramme donc $\\widehat{K'PA} = \\widehat{K'PQ} + \\widehat{QPA'} = \\wid... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fsf | Problem:
Bestimme alle Funktionen $f: \mathbb{R} \backslash\{0\} \rightarrow \mathbb{R}$, für die gilt
$$
\frac{1}{x} f(-x)+f\left(\frac{1}{x}\right)=x \quad \forall x \in \mathbb{R} \backslash\{0\}
$$ | [
"Solution:\nFür $x \\neq 0$ können wir in der Gleichung $x$ durch $-1 / x$ ersetzen und erhalten\n$$\n-x f\\left(\\frac{1}{x}\\right)+f(-x)=-\\frac{1}{x}\n$$\nMultipliziert man die ursprüngliche Gleichung mit $x$ und addiert sie zu (1), dann folgt $2 f(-x)=x^{2}-1 / x$. Ersetzt man hier schliesslich noch $x$ durch ... | Switzerland | IMO - Selektion | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = (x^2 + 1/x)/2 for all x ≠ 0 | |
06ul | Let $p \geqslant 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1, \ldots, p-1\}$ that was not chosen before by either of the two players and then chooses an element $a_{i}$ of the set $\{0,1,2,3,4,5,6,7,... | [
"If $p=2$ or $p=5$ then Eduardo chooses $i=0$ and $a_{0}=0$ in the first move, and wins, since, independently of the next moves, $M$ will be a multiple of $10$.\n\nNow assume that the prime number $p$ does not belong to $\\{2,5\\}$. Eduardo chooses $i=p-1$ and $a_{p-1}=0$ in the first move. By Fermat's Little Theor... | IMO | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
036v | Problem:
Consider the equations
$$
3^{2x+3} - 2^{x+2} = 2^{x+5} - 9^{x+1}
$$
and
$$
a \cdot 5^{2x} + |a-1| 5^{x} = 1
$$
where $a$ is a real number.
a) Solve the equation (1).
b) Find the values of $a$ such that the equations (1) and (2) are equivalent. | [
"Solution:\n\na) The equation (1) can be written as $9^{x} = 2^{x}$, i.e. $\\left(\\frac{9}{2}\\right)^{x} = 1$ and $x = 0$.\n\nb) We plug the only solution $x = 0$ of (1) in (2) and obtain $|a-1| = 1-a \\Longleftrightarrow a \\leq 1$. In this case $0$ is a solution of (2) and we have to decide when (2) has no othe... | Bulgaria | Spring Mathematical Competition | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a) x = 0; b) a ∈ [0, 1] ∪ {−1} | |
02v0 | Problem:
(a) Encontre todos os pares $(m, n)$ de inteiros não negativos que satisfazem a equação
$$
5 m + 8 n = 120
$$
(b) Sejam $a$, $b$ e $c$ números inteiros positivos com $c > 1$ tais que
$$
\frac{1}{a} + \frac{1}{b} = \frac{1}{c}
$$
Prove que pelo menos um dos números $a + c$ ou $b + c$ é um número composto, ou ... | [
"Solution:\n\na) Podemos reescrever a equação como\n$$\n\\begin{aligned}\n8 n & = 120 - 5 m \\\\\n& = 5(24 - m)\n\\end{aligned}\n$$\nTemos que $8$ divide o lado direito e não possui fator comum com $5$. Consequentemente $24 - m$ deve ser um múltiplo de $8$. Sabendo que $24$ é um múltiplo de $8$, temos que $m$ é um ... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | Part (a): (m, n) = (0, 15), (8, 10), (16, 5), (24, 0). | |
0hj3 | Problem:
Let $g_{1}(x), g_{2}(x), \ldots, g_{5}(x)$ be polynomials with integer coefficients. Suppose that their product $f(x) = g_{1}(x) g_{2}(x) \cdots g_{5}(x)$ satisfies $f(1999) = 2000$. Prove that for some $i \in \{1,2,3,4,5\}$, the sum of the coefficients of $g_{i}(x)$ is odd. | [
"Solution:\n\nFor some $i$, $g_{i}(1999)$ is odd. Indeed, if this were false, each $g_{i}(1999)$ would be divisible by $2$, so their product, $f(1999) = 2000$, would be divisible by $2^{5}$, which is not the case.\n\nNow write $g_{i}(x) = c_{n} x^{n} + c_{n-1} x^{n-1} + \\cdots + c_{0}$. Then\n$$\ng_{i}(1999) - (c_... | United States | Berkeley Math Circle Take-Home Contest #6 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof only | null | |
04xr | Let $ABC$ be a right triangle with hypotenuse $AB$ and $P$ be a point inside the shorter arc $AC$ of the circumcircle of the triangle $ABC$. The line perpendicular to $CP$ which passes through $C$ intersects the lines $AP$ and $BP$ in points $K$ and $L$ respectively. Prove that the ratio of the areas of the triangles $... | [
"Throughout the solution, we shall denote by $S_{XYZ}$ the area of the triangle $XYZ$.\n\n\nFig. 3\n\nLet $PR$ be the diameter of the circumcircle of $ABC$ (Fig. 3). Then $ARBP$ is a rectangle. Since $BR$ is parallel to $PA$ we have $S_{PBK} = S_{PRK}$, which implies\n$$\nS_{BKL} = S_{LPR}.... | Czech-Polish-Slovak Mathematical Match | 12th Czech-Polish-Slovak Mathematics Competition | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
08ib | Problem:
Through the arbitrary point of the triangle $ABC$ construct (explain the steps of the construction) a straight line which divides the triangle $ABC$ in two parts so that the ratio of their areas is equal to $3/4$. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0evf | For a given positive integer $k$, define two sequences $\{a_n\}$ and $\{b_n\}$ as follows:
$$
\begin{aligned}
a_1 &= k, & a_2 &= k, & a_{n+2} &= a_n a_{n+1} \quad (n \ge 1) \\
b_1 &= 1, & b_2 &= k, & b_{n+2} &= \frac{b_{n+1}^3 + 1}{b_n} \quad (n \ge 1)
\end{aligned}
$$
For any positive integer $n$, show that $a_{2n}b_{... | [
"**Lemma 2.** For an odd prime $p$ such that $p|k$, $(c_m, p) = 1$ ($-1 \\le m \\le n+1$).\n\n*Proof.* Suppose there is an odd prime $p$ such that $p|(c_m, c_{m+1})$. Then Lemma 1 implies $p|k$, which contradicts Lemma 2. $\\square$\n\n**Lemma 4.** For an odd prime $p$ such that $p^m||c_n$ with $m \\ge 1$, $p^m|(c_... | South Korea | Korean Mathematical Olympiad Final Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0ik3 | A mathematical frog jumps along the number line. The frog starts at $1$, and jumps according to the following rule: if the frog is at integer $n$, then it can jump either to $n + 1$ or to $n + 2^{m_n+1}$ where $2^{m_n}$ is the largest power of $2$ that is a factor of $n$. Show that if $k \ge 2$ is a positive integer an... | [
"For $i \\ge 0$ and $k \\ge 1$, let $x_{i,k}$ denote the minimum number of jumps needed to reach the integer $n_{i,k} = 2^i k$. We must prove that\n$$\nx_{i,k} > x_{i,1} \\qquad (*)\n$$\nfor all $i \\ge 0$ and $k \\ge 2$. We prove this using the method of descent.\n\nFirst note that $(*)$ holds for $i = 0$ and all ... | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
04kw | Let $n$ be a positive integer. Vertices of a regular $2n$-gon are painted red and blue in an alternating fashion. All the sides and diagonals of the $2n$-gon are drawn. If the number of segments with ends of the same colour is 3192, find the number of segments with ends of different colours. (Mea Bombardelli) | [] | Croatia | Mathematical competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 3249 | |
07cr | $ABCD$ is a square that is partitioned into rectangles such that no point is a corner of 4 rectangles. All corner points of rectangles are colored with 2 colors such that any two diagonal corners in a rectangle (of the partition) have different colors. If $A$ and $C$ have the same color, prove that $B$ and $D$ also hav... | [
"Consider a graph $G$ as follows, vertices of rectangles of the partition are vertices of $G$, two vertices are connected if they are opposite corners of a rectangle of the partition. Obviously, there are $2n$ edges in $G$ where $n$ is the number of rectangles used in the partition.\n\nNote that if some point is a ... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
002v | Consideramos todas las sucesiones finitas de términos positivos menores o iguales que $3$ y suma mayor que $100$. Para una tal sucesión $\alpha$ consideramos una subsucesión cuya suma $S$ difiera lo menos posible de $100$, y definimos el defecto de $\alpha$ por $|S-100|$. Hallar el máximo valor del defecto cuando $\alp... | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Español | proof and answer | 100/67 | |
0iqd | Problem:
Let $ABC$ be an equilateral triangle. Let $\Omega$ be a circle inscribed in $ABC$ and let $\omega$ be a circle tangent externally to $\Omega$ as well as to sides $AB$ and $AC$. Determine the ratio of the radius of $\Omega$ to the radius of $\omega$. | [
"Solution:\n\nAnswer: $3$"
] | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | 3 | |
0h80 | Olesya chose 5 numbers from the set $\{1; 2; 3; 4; 5; 6; 7\}$. She told Pavlik the product of these numbers and asked whether the sum of these numbers is odd or even. Pavlik replied that he could not determine it for sure. What product might have had Olesya? | [
"If Pavlik knows the product, he could determine the product of the last two numbers that were not chosen. Since he could not determine the pairing of the product, then he could not determine the numbers that were not chosen although he knows their product. Consider all products of two numbers in the given set:\n$$... | Ukraine | UkraineMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 420 | |
03c8 | Let $\triangle ABC$ be acute triangle. Point $M$ is arbitrary point on the side $AB$, and $N$ is the midpoint of $AC$. Denote by $P$ and $Q$ the feet of the perpendiculars from $A$ to the lines $MC$ and $MN$, respectively. Prove that when $M$ vary then the circumcenter of $\triangle PQN$ lies on a fixed line. | [] | Bulgaria | First Team Selection Test for 56th IMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry ... | English | proof only | null | |
0aal | Let $ABCD$ be a cyclic quadrilateral inscribed in a circle $\omega$ having center $O$. The lines $AD$ and $BC$ intersect at $E$, and the lines $AB$ and $CD$ intersect at $F$. Let $P$ be a point on the segment $EF$ such that $OP \perp EF$. The circle $\Gamma_1$ passes through the points $A, E$ and is tangent to $\omega$... | [] | North Macedonia | Fourth Memorial Mathematical Contest | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geome... | English | proof only | null | |
00u1 | Let $ABC$ be an acute triangle such that $CA \neq CB$ with circumcircle $\omega$ and circumcentre $O$. Let $\tau_A, \tau_B$ be the tangents to $\omega$ at $A$ and $B$, which meet at $X$. Now, let $Y$ be the foot of the perpendicular from $O$ onto $CX$, and let the line through $C$ parallel to $AB$ meet $\tau_A$ at $Z$.... | [
"Firstly observe that $OAXB$ is cyclic, with diameter $OX$, and $Y$ also lies on this circle since $OY \\perp XC$. Hence:\n$$\n\\angle AZC = \\angle XAB = \\angle ABX = \\angle AYX\n$$\nand so $CYAZ$ is cyclic.\n\n\n\nLet $M$ be the intersection of $YZ$ and $AC$ and let $CY$ intersect $\\om... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations... | English | proof only | null | |
0e1q | Find all positive integers $n$ for which there exists a polynomial $P(x)$ with integer coefficients such that $P(d) = \left(\frac{n}{d}\right)^2$ for each positive divisor $d$ of $n$. | [
"Obviously, such a polynomial exists for $n = 1$. In this case the only condition is that $P(1) = 1$ and the polynomial $P(x) = x$ has this property.\n\nIf $n$ is a prime, then its only two divisors are $1$ and $n$. The polynomial $P$ must satisfy the conditions $P(1) = n^2$ and $P(n) = 1$. Let us write $P(x) = ax ... | Slovenia | Selection Examinations for the IMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 1, all primes, and 6 | |
0c7r | Let $p$ be a prime number. For any permutation $\sigma \in S_p$, we define the matrix $A_\sigma = (a_{ij})_{i,j=1,p} \in \mathcal{M}_p(\mathbb{Z})$, whose entries are $a_{ij} = \sigma^{i-1}(j)$, for any $i, j = 1, p$, where $\sigma^0$ is the identity permutation, and $\sigma^k = \underbrace{\sigma \circ \sigma \circ \d... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Linear Algebra > Determinants",
"Algebra > Linear Algebra > Matrices",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof only | null | |
06tj | Let $A B C$ be an acute triangle, and let $M$ be the midpoint of $A C$. A circle $\omega$ passing through $B$ and $M$ meets the sides $A B$ and $B C$ again at $P$ and $Q$, respectively. Let $T$ be the point such that the quadrilateral $B P T Q$ is a parallelogram. Suppose that $T$ lies on the circumcircle of the triang... | [
"Let $S$ be the center of the parallelogram $B P T Q$, and let $B' \\neq B$ be the point on the ray $B M$ such that $B M = M B'$ (see Figure 1).\nIt follows that $A B C B'$ is a parallelogram. Then, $\\angle A B B' = \\angle P Q M$ and $\\angle B B' A = \\angle B' B C = \\angle M P Q$, and so the triangles $A B B'$... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > A... | English | proof and answer | sqrt(2) | |
09om | The integer sequences $\{a_n\}$, $\{b_n\}$, $\{c_n\}$ are defined as follows: $a_1 = 3$, $b_1 = c_1 = 1$ and
$$
\begin{align*}
a_{n+1} &= 3b_n + c_n, \\
b_{n+1} &= a_n + b_n, \\
c_{n+1} &= b_n + c_n
\end{align*}
$$
for $n \ge 2$. Show that for any odd integer $N \ge 3$
i) there exists an index $m$ such that $a_m - 1, ... | [] | Mongolia | MMO2025 Round 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
045o | Find the minimal integer $n \ge 3$ such that there exist $n$ points $A_1, A_2, \dots, A_n$ on a plane with no three being colinear, such that for every $1 \le i \le n$, the midpoint of the segment $A_iA_{i+1}$ is contained in the segment $A_jA_{j+1}$ for some $j \ne i$. Here, $A_{n+1} = A_1$. | [
"**Solution:** First, we prove that $n = 6$ satisfies the condition. Let $P_1, P_2, \\dots, P_6$ be the vertices of a regular hexagon in consecutive order. Let $A_1 = P_1, A_2 = P_3, A_3 = P_5, A_4 = P_2, A_5 = P_6, A_6 = P_4$. Then, these 6 points satisfy the conditions. See the figure below.\n![](attached_image_1... | China | Chinese Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | 6 | |
0gbm | 平面上有一 $\triangle ABC$ 及一點 $O$, $X$ 在射線 $AC$ 上。取點 $X_1$ 在射線 $BA$ 上，使得 $AX = AX_1$，且 $A$ 位於 $B$ 與 $X_1$ 之間。再取點 $X_2$ 在射線 $BC$ 上使得 $\overline{XX_2} \parallel \overline{OC}$。
試證: 當 $X$ 在 $AC$ 射線上移動時, $\triangle BX_1X_2$ 的外接圓圓心的軌跡為直線的一部分。 | [
"在 $AC$ 上取定固定的兩點 $X$ 與 $Y$，並作相應的 $X_1, X_2, Y_1, Y_2$。設 $BX_1X_2$ 與 $BY_1Y_2$ 的外接圓再交於點 $T$。則只要證明對 $AC$ 射線上任意一點 $Z$，相應的 $Z_1, Z_2$ 會與 $B, T$ 共圓即可，此時 $BZ_1Z_2$ 的外接圓圓心會在 $BT$ 線段的中垂線上。\n\n(註：若 $B, T$ 兩點重合，則 $\\triangle BX_1X_2$ 的外接圓與 $\\triangle BY_1Y_2$ 的外接圓切於 $B$ 點。此時變成證明 $\\triangle BZ_1Z_2$ 的外接圓亦與 $\\triangle BX_1X... | Taiwan | 二〇一八數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0he1 | Find all tuples of positive integers $(a, b, c)$, which satisfy the equation:
$$
a + b + c^2 = abc.
$$ | [
"Consider this equation as a square equation with respect to $c$. Then,\n$$\nc^2 - abc + (a + b) = 0 \\Rightarrow D = (ab)^2 - 4(a + b) = m^2.\n$$\nIf $b = 1$, $D_1 = a^2 - 4a - 4 = (a - 2)^2 - 8 = x^2 - 8$, where $x = a - 2$.\nIf $x = 3$, $D_1 = 1$ is a perfect square, so $a = 5$, $c^2 - 5c + 6 = 0$, which yields ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (2,2,2), (5,1,2), (1,5,2), (5,1,3), (1,5,3), (3,2,1), (2,3,1), (3,2,5), (2,3,5) | |
05ry | Problem:
Soit $f: \mathbb{Z} \to \mathbb{R}$ une fonction telle que
$$
f(n+2) = (f(n+1) + f(n)) / 2
$$
pour tout entier $n$. On suppose que $f$ est bornée, c'est-à-dire qu'il existe un réel $F$ tel que $-F \leqslant f(n) \leqslant F$ pour tout $n$.
Démontrer que $f$ est une fonction constante. | [
"Solution:\n\nSupposons que $f$ n'est pas constante. Il existe donc un entier $n$ et un réel $\\varepsilon \\neq 0$ tels que $f(n+1)-f(n)=\\varepsilon$.\n\nPour tout entier $k$, posons $\\Delta_{k}=f(k+1)-f(k)$. Alors\n$$\n\\Delta_{k+1}=f(k+2)-f(k+1)=(f(k+1)+f(k)) / 2-f(k+1)=-\\Delta_{k} / 2\n$$\net, réciproquement... | France | Préparation Olympique Française de Mathématiques - Test du 15 Mai 2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0dk3 | With positive integer $n > 1$, let $4n - 3$ positive numbers are written on the board (not necessarily all different). It is known that any 4 pairwise different numbers from that list form an arithmetic progression. Prove that some number is written on the board at least $n$ times. | [
"Let $d$ be the number of distinct values appear on the table, let them be\n$$\na_1 < a_2 < \\dots < a_d.\n$$\nAssuming by contradiction that no value appears at least $n$ times, then each value appears at most $n-1$ times. So we can count the amount of numbers on the board to get\n$$\n4n - 3 \\le d(n - 1) \\implie... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
028s | Problem:
Um número sortudo é aquele cuja soma de seus algarismos é divisível por $7$. Por exemplo, $7$, $25$ e $849$ são números sortudos. O menor par de números sortudos é $7$ e $16$.
a. Encontre oito números consecutivos, dos quais dois são números sortudos.
b. Encontre $12$ números consecutivos, tal que nenhum se... | [
"Solution:\n\na. A sequência de oito números consecutivos de $52$ a $59$ tem exatamente dois números sortudos: $52$ e $59$. Outro exemplo é qualquer sequência de $8$ números que contenha $59$ e $61$, por exemplo: $55, 56, 57, 58, 59, 60, 61, 62$.\n\nb. Dois exemplos: $994, \\ldots, 1005$ e $7994, \\ldots, 8005$. Ex... | Brazil | Nível 2 | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) 52–59 (lucky: 52 and 59); another example: 55–62 (lucky: 59 and 61).
b) 994–1005 or 7994–8005 (none lucky).
c) Any sequence of 13 consecutive integers contains at least one lucky number. | |
02la | Problem:
Um teste vocacional - Foi feito um teste vocacional em 1000 estudantes de uma escola. A tabela a seguir apresenta os resultados por área de estudo e sexo.
| | Exatas | Humanas | Biológicas |
|:-----------:|:------:|:-------:|:----------:|
| Masculino | 232 | 116 | 207 |
| Feminino... | [
"Solution:\n\na) De exatas temos $232 + 112 = 344$ estudantes, logo a probabilidade de escolher ao acaso um aluno de exatas é $\\frac{344}{1000} = 0,344$.\n\nb) Como o número de estudantes do sexo masculino é $555$, temos que a probabilidade de ser da área de humanas é $\\frac{116}{555} = 0,209$.\n\nc) O número de ... | Brazil | Brazilian Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Other",
"Math Word Problems"
] | null | final answer only | a) 0.344; b) 116/555 ≈ 0.209; c) 180/387 ≈ 0.465 | |
0b30 | Problem:
Denote by $\mathbb{Q}^{+}$ the set of positive rational numbers. A function $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}$ satisfies
- $f(p)=1$ for all primes $p$, and
- $f(a b)=a f(b)+b f(a)$ for all $a, b \in \mathbb{Q}^{+}$.
For which positive integers $n$ does the equation $n f(c)=c$ have at least one solution... | [
"Solution:\nWe claim that either $n$ is the product of distinct primes, or $n=1$.\n\nDefine $g(x)=\\frac{f}{x}$. The equation we are trying to solve becomes $g(c)=\\frac{1}{n}$.\n\nThe definition of the function becomes $g(p)=\\frac{1}{p}$ for all primes $p$, and\n$$\ng(a b)=g(a)+g(b)\n$$\nSubstituting $a=1$ in the... | Philippines | 23rd Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All squarefree positive integers (equivalently, n = 1 or n is a product of distinct primes). | |
01q5 | The graph of the hyperbola $y = \frac{1}{x}$ is drawn on the Cartesian plane $Oxy$. Three snails start simultaneously from the origin $O$ and move along the abscissa axis $Ox$ (each snail has its constant speed). Let $A(t)$, $B(t)$ and $C(t)$ be points on the graph of the hyperbola such that their abscissae are equal t... | [
"Let $a(t)$, $b(t)$, $c(t)$ be the abscissae of the first, the second and the third snails at the moment $t$, respectively, and $u_a, u_b, u_c$ be their speeds respectively.\nFirst, suppose that all snails move in the same direction. Without loss of generality we assume that the snails move along the positive part ... | Belarus | Final Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
0ew4 | Problem:
$ABCD$ is any convex quadrilateral. Construct a new quadrilateral as follows. Take $A'$ so that $A$ is the midpoint of $DA'$; similarly, $B'$ so that $B$ is the midpoint of $AB'$; $C'$ so that $C$ is the midpoint of $BC'$; and $D'$ so that $D$ is the midpoint of $CD'$. Show that the area of $A'B'C'D'$ is five... | [
"Solution:\n\nCompare the triangles $A'B'A$ and $ADB$. The base of $A'B'A$ can be taken as $A'A$, which is the same length as $AD$. The height of $A'B'A$ is $AB'$ times $\\sin B'A'A'$, which is twice $AB$ times $\\sin BAD$. So $\\text{area } A'B'A = 2 \\times \\text{area } ADB$.\n\nSimilarly, $\\text{area } B'C'B =... | Soviet Union | 2nd ASU | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a3b | Problem:
Op een $2023 \times 2023$ bord staan er op sommige vakjes een kever, met hoogstens één kever per vakje. Na een minuut gaat elke kever een vakje naar links of rechts, of een vakje naar boven of beneden. Na elke volgende minuut gaat elke kever weer een vakje verder, maar ze maken altijd een bocht van 90 graden.... | [
"Solution:\n\nHet antwoord is $2022^{2}+1$.\n\nEerst laten we zien dat $2022^{2}$ inderdaad onvoldoende is. Hiervoor delen we een $2022 \\times 2022$ deelbord op in $2 \\times 2$ bordjes. In elk zo'n bordje kunnen we vier kevers plaatsen zodanig dat ze in een met de klok mee een cirkeltje (vierkantje) lopen.\n\nSte... | Netherlands | IMO-selectietoets III | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2022^2 + 1 | |
0la4 | Let $k$ be a positive real number and $ABC$ is an acute triangle which is not isosceles. Denote by $O$ the circumcenter and let $AD$, $BE$ and $CF$ be its inner angle-bisectors. Choose the points $L$, $M$ and $N$ on $AD$, $BE$ and $CF$ respectively such that $\frac{AL}{AD} = \frac{BM}{BE} = \frac{CN}{CF} = k$. Denote b... | [] | Vietnam | Vijetnam 2008 | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Ge... | null | proof and answer | null | |
0erx | The inscribed circle of triangle $ABC$, with centre $I$, touches sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. Let $P$ be a point, on the same side of $FE$ as $A$, for which $\angle PFE = \angle BCA$ and $\angle PEF = \angle ABC$. Prove that $P$, $I$ and $D$ lie on a straight line. | [
"\nSince $\\angle AEI = \\angle AFI = 90^\\circ$, the points $A$, $E$, $F$, $I$ lie on a circle with diameter $AI$. Moreover, since $\\angle PFE = \\angle BCA$ and $\\angle PEF = \\angle ABC$ by our assumptions on $P$, triangles $ABC$ and $PEF$ are similar, so $\\angle EPF = \\angle BAC = \... | South Africa | South African Mathematics Olympiad Third Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
08re | $m, n, a, a', b$ and $b'$ are positive integers with $2 \le m \le n$, $a, a' \le m$, $b, b' \le n$ and $(a, b) \ne (a', b')$.
There is a rectangular town like a chessboard with $m$ avenues and $n$ streets. $\langle x, y \rangle$ represents the intersection of the $x$-th avenue from the west and the $y$-th street from t... | [
"By walking one block from one intersection to another, the sum of the coordinates changes its parity. Since there are totally $mn - 1$ blocks in the way, $(m, n, a, a', b, b')$ must satisfy one of the followings:\n(1) $mn$ is odd, and both $a+b$ and $a'+b'$ are even, or\n(2) $mn$ is even, and one of $a+b$ and $a'+... | Japan | The 16th Japanese Mathematical Olympiad - The Final Round | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | There exists such a path if and only if the parity condition holds and none of the small obstructions occur. Precisely: (i) If the total number of intersections mn is odd, then both a+b and a'+b' are even; or (ii) if mn is even, then exactly one of a+b and a'+b' is odd. In addition, the following exceptional cases admi... | |
08dp | Problem:
Alberto, Barbara e Carlo fanno un gioco. Questo gioco si compone di $k$ turni, al termine di ciascuno dei quali il primo classificato riceve $a_{1}$ punti, il secondo $a_{2}$ e il terzo $a_{3}$, con $a_{1}>a_{2}>a_{3}>0$ tutti interi. I punteggi finali di Alberto, Barbara e Carlo sono rispettivamente 22, 9, 9... | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Notiamo che durante il gioco vengono assegnati $k\\left(a_{1}+a_{2}+a_{3}\\right)$ punti in totale, quindi possiamo scrivere $k\\left(a_{1}+a_{2}+a_{3}\\right)=22+9+9=40$, da cui deduciamo che $k$ e $a_{1}+a_{2}+a_{3}$ sono entrambi divisori di 40. Notiamo inoltre che si... | Italy | Progetto Olimpiadi della Matematica | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | C | |
0e8c | A table tennis tournament is conducted in the following way. In each round, if the number of contestants is odd, one contestant is drawn to proceed automatically to the next round. After that, pairs are drawn from the other contestants. The contestants from each pair compete against each other, and the winner proceeds ... | [
"Starting with $2013$ contestants $1 + \\frac{2012}{2} = 1007$ of them go to the second round and $1 + \\frac{1006}{2} = 504$ make it to the third round. For the fourth and the fifth round we get $\\frac{504}{2} = 252$ and $\\frac{252}{2} = 126$ contestants, respectively. For the sixth and the seventh round we have... | Slovenia | National Math Olympiad 2013 - Final Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | f(2013) = 11; least n with f(n) = 11 is 1025 | |
0hee | Given $n$ pairwise distinct positive numbers. Vasia wrote on the board all possible numbers of the form $a + \frac{b}{c}$, where $a, b, c$ are distinct numbers from the given set. Can one always find two numbers on the board such that the larger one is not more than twice the other one, if
a) $n=4$?
b) $n=3$? | [
"a) We will show that for four given numbers one can always find two numbers on the board such that one is not more than twice the other one. Suppose by contradiction, it is not true. Let the given numbers be $a < b < c < d$. Then $d + \\frac{b}{a} > c + \\frac{b}{a} \\Rightarrow d + \\frac{b}{a} > 2(c + \\frac{b}{... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) Yes. b) No; for example, the set {ten to the minus one, ten to the minus three, ten to the minus nine}. | |
0gcq | 在三角形 $ABC$ 的內部選一個點 $T$。令 $A_1, B_1, C_1$ 分別為 $T$ 對直線 $BC, CA, AB$ 的反射點。記三角形 $A_1B_1C_1$ 的外接圓為 $\Omega$。設直線 $A_1T, B_1T, C_1T$ 分別與圓 $\Omega$ 再交於點 $A_2, B_2, C_2$。證明: 直線 $AA_2, BB_2, CC_2$ 共點, 且其交點在 $\Omega$ 上。 | [
"By $\\angle (\\boldsymbol{l}, n)$ we always mean the directed angle of the lines $l$ and $n$, taken modulo $180^\\circ$.\n\n\n\nLet $CC_2$ meet $\\Omega$ again at $K$ (as usual, if $CC_2$ is tangent to $\\Omega$, we set $K = C_2$). We show that the line $BB_2$ contains $K$; similarly, $AA_... | Taiwan | 二〇一九數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
02yu | Problem:
A figura abaixo representa uma estante com duas prateleiras com cinco pilhas de livros, sendo três delas com dois livros e duas delas com apenas um livro. Alice e Luiz inventaram um jogo no qual cada um deles, alternadamente, retira um ou dois livros de uma das pilhas de livros. Vence aquele que tirar o últim... | [
"Solution:\n\nAquele que deixar duas pilhas com 1 livro cada para o adversário jogar, vence o jogo. Para conseguir isso, basta Alice tirar, em seu primeiro lance, uma pilha com 2 livros. Após isso, Luiz terá 3 opções, que são:\n\nI) Tirar uma pilha com 2 livros: basta Alice tirar a outra pilha com 2 livros que cheg... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Alice | |
08wg | Let $A$ be a positive integer which is a multiple of $3$ but not of $9$. When we add to $A$ the product of all the digits of $A$, then we obtain a number which is a multiple of $9$. Find the smallest possible value for such a number $A$. | [
"$138$\n\n$138$ is a multiple of $3$, but not of $9$, and since $138 + 1 \\times 3 \\times 8 = 162$ is a multiple of $9$, the number $138$ satisfies the conditions of the problem. We will show in the sequel that if $A < 138$, then $A$ does not satisfy the conditions of the problem.\n\nNow, if a number $A$ satisfies... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 138 | |
0e1u | Find all real $x$ in $[0, 2\pi)$ for which all terms of the sequence
$$
a_n = \frac{1}{\cos(nx)}
$$
are integers. | [
"The numbers $a_1$ and $a_2$ are integers. Since $a_1 = \\frac{1}{\\cos x}$, $a_2 = \\frac{1}{\\cos(2x)}$ and $\\cos(2x) = 2(\\cos x)^2 - 1$, we have $a_2 = \\frac{1}{2\\cos^2 x - 1} = \\frac{a_1^2}{2 - a_1^2}$. Now, $a_2$ is an integer, so $2 - a_1^2$ is a divisor of $a_1^2$. So, $2 - a_1^2$ divides $2 - a_1^2$ an... | Slovenia | National Math Olympiad | [
"Precalculus > Trigonometric functions"
] | English | proof and answer | x in {0, π/3, 2π/3, π, 4π/3, 5π/3} | |
0dw1 | Problem:
V razredu je 25 dijakov. Rok je računal, koliko točk je v povprečju dosegel posamezen dijak pri šolski nalogi. Najprej je izračunal povprečje 74,5 točk, a se je spomnil, da je pozabil upoštevati svoj dosežek. Ko ga je upošteval, je izračunal povprečje 75 točk. Koliko točk je dosegel Rok pri šolski nalogi? Zap... | [
"Solution:\n\nKer je vsak izmed 24 dijakov dosegel v povprečju $74,5$ točke, so vsi skupaj dosegli $24 \\cdot 74,5 = 1788$ točk. Ko je Rok upošteval tudi svoje točke, je izračunal povprečje $75$ točk, zato je $25$ dijakov doseglo skupaj $25 \\cdot 75 = 1875$ točk. Razlika $1875 - 1788$ predstavlja število točk, ki ... | Slovenia | 4. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 87 | |
0brw | Determine the positive integers expressible in the form $\dfrac{x^2 + y}{xy + 1}$, for at least two pairs $(x, y)$ of positive integers. | [
"We show that $1$ is the only positive integer satisfying the condition in the statement. Clearly, $1 = \\dfrac{1^2 + y}{1 \\cdot y + 1}$ for any positive integer $y$, so $1$ is expressible in the required form for infinitely many pairs of positive integers.\n\nNext, we prove that any integer $n \\ge 2$ is uniquely... | Romania | 67th NMO Selection Tests for BMO and IMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 1 | |
0jx5 | Problem:
Find (in terms of $n \geq 1$) the number of terms with odd coefficients after expanding the product:
$$
\prod_{1 \leq i<j \leq n}\left(x_{i}+x_{j}\right)
$$
e.g., for $n=3$ the expanded product is given by $x_{1}^{2} x_{2}+x_{1}^{2} x_{3}+x_{2}^{2} x_{3}+x_{2}^{2} x_{1}+x_{3}^{2} x_{1}+x_{3}^{2} x_{2}+2 x_{1} ... | [
"Solution:\nNote that if we take $(\\bmod\\ 2)$, we get that\n$$\n\\prod_{1 \\leq i<j \\leq n}\\left(x_{i}+x_{j}\\right) \\equiv \\prod_{1 \\leq i<j \\leq n}\\left(x_{j}-x_{i}\\right)=\\operatorname{det}(M)\n$$\nwhere $M$ is the matrix with $M_{ij}=x_{i}^{j-1}$. This is called a Vandermonde determinant. Expanding t... | United States | February 2017 | [
"Algebra > Linear Algebra > Determinants",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | n! | |
0hle | Problem:
Ken is the best sugar cube retailer in the nation. Trevor, who loves sugar, is coming over to make an order. Ken knows Trevor cannot afford more than $127$ sugar cubes, but might ask for any number of cubes less than or equal to that. Ken prepares seven cups of cubes, with which he can satisfy any order Trevo... | [
"Solution:\n\nThe only way to fill seven cups to satisfy the above condition is to use a binary scheme, so the cups must contain $1, 2, 4, 8, 16, 32$, and $64$ cubes of sugar."
] | United States | null | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 64 | |
0613 | Problem:
On dispose de $a + b$ bols alignés sur une rangée. Les $a$ premiers bols contiennent une pomme tandis que les $b$ derniers bols contiennent une poire.
Une opération consiste à déplacer une pomme du bol $i$ au bol $i+1$ et une poire du bol $j$ au bol $j-1$, où $i$ et $j$ sont deux entiers tels que $i-j$ est p... | [
"Solution:\n\nCommençons par montrer qu'il n'est pas possible d'y arriver lorsque $a$ et $b$ sont impairs tous les deux. Soit $I$ la somme alternée des nombres de fruits dans les bols. Dans les positions initiales et finales décrites, on a alors $I=0$ car $a+b$ est pair. En outre, lors de chaque opération, $I$ est ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Possible if and only if ab is even | |
0ctf | Let $ABC$ be an acute-angled triangle with $AC < BC$; let $M$ be the midpoint of $AB$. Let $CC'$ be a diameter in the circumcircle $\Omega$ of the triangle $ABC$. The line $CM$ meets the lines $AC'$ and $BC'$ at $K$ and $L$, respectively. Denote by $\Delta$ the triangle formed by the line through $K$ perpendicular to $... | [
"Let $E$, $U$, and $V$ be the vertices of $\\Delta$, with $EU \\parallel AC$ and $EV \\parallel BC$. Let $C'E$ meet $\\Omega$ again at $X$. Then $X$ is the required tangency point. There are various methods to prove this.\n\n*Hint 1.* Let $CX$ meet $AC'$ and $BC'$ at $Q$ and $P$, respectively (see Fig. 19). One may... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > A... | English; Russian | proof only | null | |
0dwz | Problem:
Vaške obrekljivke si svoje čenče vsak dan izmenjujejo po telefonu, in sicer tako, da vsaki dve med seboj govorita natanko enkrat. Nekega dne je vsaka obrekljivka poklicala vsaj eno izmed ostalih. Dokaži, da obstajajo tri, za katere velja, da je prva poklicala drugo, druga tretjo in tretja prvo. | [
"Solution:\n\nNaj bo $n$ obrekljivk. Potem je vsaka poklicala med $1$ in $n-1$ ostalih, zato po Dirichletovem načelu obstajata dve, ki sta poklicali enako število obrekljivk, recimo $k$. Brez izgube splošnosti lahko predpostavimo, da je prva poklicala drugo. Prva je poklicala še $k-1$ ostalih, druga pa $k$. Med pre... | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
06vv | Prove that there is a constant $c>0$ and infinitely many positive integers $n$ with the following property: there are infinitely many positive integers that cannot be expressed as the sum of fewer than $c n \log (n)$ pairwise coprime $n^{\text{th}}$ powers. | [
"$n$ is divisible by $\\varphi\\left(p^{e}\\right)$ for every prime power $p^{e}$ exactly dividing $N$.\nThis property ensures that all $n^{\\text{th}}$ powers are congruent to $0$ or $1$ modulo each such prime power $p^{e}$, and hence that any sum of $m$ pairwise coprime $n^{\\text{th}}$ powers is congruent to $m$... | IMO | IMO 2019 Shortlisted Problems | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divis... | English | proof only | null | |
0685 | Let $AB\Gamma\Delta$ ($A\Delta\parallel B\Gamma$) be a trapezium with $\hat{A} = \hat{B} = 90^\circ$ and $A\Delta < B\Gamma$. Denote by $E$ the point of intersection of the two non parallel sides $AB$ and $\Gamma\Delta$, $Z$ the symmetric point of $A$ with respect to the line $B\Gamma$ and $M$ the midpoint of $EZ$. It ... | [
"Let the line $\\Delta Z$ meets $\\Gamma M, B\\Gamma$ at points $K, N$, respectively. Then in the triangle $A\\Delta Z$, $B$ is the midpoint of $AZ$ and $BN\\parallel A\\Delta$. Hence $N$ is the midpoint of $Z\\Delta$. Therefore in the triangle $ZE\\Delta$, $MN$ connects the midpoints of two sides, and hence:\n\n$$... | Greece | 33rd Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ac7 | $2n$ ($n > 1$) points are given in the plane. Line $p$ lies in the plane and does not cross the given points. Prove that this line cuts no more than $n^2$ segments with end in the given points. | [
"Let the given points and the line be lying in the plane $\\pi$. The line $p$ is dividing the plane into the two half-planes $\\pi_1$ and $\\pi_2$. If in one of the half-planes lies $m$ points, then in the other lies $2n - m$ points. A segment with endpoints in the same half-plane does not cut with the line. The li... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0k4u | Problem:
Ben "One Hunna Dolla" Franklin is flying a kite $KITE$ such that $IE$ is the perpendicular bisector of $KT$. Let $IE$ meet $KT$ at $R$. The midpoints of $KI$, $IT$, $TE$, $EK$ are $A$, $N$, $M$, $D$, respectively. Given that $[MAKE]=18$, $IT=10$, $[RAIN]=4$, find $[DIME]$.
Note: $[X]$ denotes the area of the... | [
"Solution:\n\nLet $[KIR]=[RIT]=a$ and $[KER]=[TER]=b$. We will relate all areas to $a$ and $b$. First,\n$$\n[RAIN]=[RAI]+[INR]=\\frac{1}{2} a+\\frac{1}{2} a=a.\n$$\nNext, we break up $[MAKE]=[MAD]+[AKD]+[DEM]$. We have\n$$\n\\begin{aligned}\n{[MAD]} & =\\frac{AD \\cdot DM}{2}=\\frac{1}{2} \\cdot \\frac{IE}{2} \\cdo... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals"
] | null | final answer only | 16 | |
0i1o | Problem:
Find the real solutions of $(2x+1)(3x+1)(5x+1)(30x+1) = 10$. | [
"Solution:\n$(2x+1)(3x+1)(5x+1)(30x+1) = [(2x+1)(30x+1)][(3x+1)(5x+1)] = (60x^2 + 32x + 1)(15x^2 + 8x + 1) = (4y+1)(y+1) = 10$, where $y = 15x^2 + 8x$.\n\nThe quadratic equation in $y$ yields $y = 1$ and $y = -\\frac{9}{4}$.\n\nFor $y = 1$, we have $15x^2 + 8x - 1 = 0$, so $x = \\frac{-4 \\pm \\sqrt{31}}{15}$.\n\nF... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | x = (-4 + sqrt(31)) / 15, x = (-4 - sqrt(31)) / 15 | |
07fm | Let $A_1, A_2, \dots, A_k$ be points on the unit circle. Prove that
$$
\sum_{1 \le i < j \le k} d(A_i, A_j)^2 \le k^2,
$$
where $d(A_i, A_j)$ denotes the distance between $A_i, A_j$. | [
"Assume that the circle mentioned in the problem is the unit circle on the complex plane. Then we can say every vertex $A_i$ is equivalent to a complex number $z_i$ such that $|z_i| = 1$. On the other hand we have $d(A_i, A_j) = |z_i - z_j|$. So we should prove that\n$$\n\\sum_{1 \\le i < j \\le k} |z_i - z_j|^2 \\... | Iran | 37th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Circles",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
09q1 | Problem:
Zij $A B C D$ een trapezium met $A B \| C D$, $2|A B|=|C D|$ en $B D \perp B C$. Zij $M$ het midden van $C D$ en zij $E$ het snijpunt van $B C$ en $A D$. Zij $O$ het snijpunt van $A M$ en $B D$. Zij $N$ het snijpunt van $O E$ en $A B$.
a. Bewijs dat $A B M D$ een ruit is.
b. Bewijs dat de lijn $D N$ door he... | [
"Solution:\n\nOplossing I. Uit $2|A B|=|C D|$ en $A B \\| C D$ volgt dat $A B$ een middenparallel in driehoek $C D E$ is. Dus $A$ is het midden van $D E$. Omdat $\\angle D B E=90^{\\circ}$, geldt volgens Thales dat $A$ het middelpunt is van de cirkel door $D, B$ en $E$. Dus $|A D|=|A E|=|A B|$ en we wisten al dat $... | Netherlands | Dutch TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane G... | null | proof only | null | |
05cw | In an isosceles triangle $ABC$ with $AB = AC$, the bisector of the angle $BAC$ intersects $BC$ at $D$. The bisector of the angle $ABC$ intersects the perpendicular bisector of $AD$ at $E$. Prove that the bisector of the angle $ACB$ is perpendicular to $DE$. | [
"As the $A$-bisector is also the altitude, we have $BD \\perp AD$. Let $I$ be the incenter of $ABC$ and $E' \\neq B$ the intersection of $BI$ and the circumcircle of $ABD$ (Fig. 2). As the angles $\\angle ABE' = \\angle DBE'$ correspond to the arcs $E'A$ and $E'D$ of this circle, we have $E'A = E'D$. Thus $E'$ lies... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
00jc | Solve the equation
$$
x^4 y^3 (y - x) = x^3 y^4 - 216
$$
in integers. | [
"The given equation is equivalent to\n$$\nx^3 y^4 + x^4 y^3 (x - y) = 216 \\iff (xy)^3 (x^2 - xy + y) = 6^3.\n$$\nBoth $x$ and $y$ must therefore be divisors of $6$, and therefore equal to $\\pm 1, \\pm 2, \\pm 3$ or $\\pm 6$. Also, $xy|6$ must hold. This means that $|x|$ and $|y|$ can only both equal $1$ or the se... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (-3, -2), (2, 3), (1, 6) | |
0c2h | a. Consider the positive integers $a$, $b$, $c$ so that $a < b < c$ and $a^2 + b^2 = c^2$. If $a_1 = a^2$, $a_2 = ab$, $a_3 = bc$, $a_4 = c^2$, prove that $a_1^2 + a_2^2 + a_3^2 = a_4^2$ and $a_1 < a_2 < a_3 < a_4$.
b. Show that for any $n \in \mathbb{N}$, $n \ge 3$, there exist the positive integers $a_1, a_2, \dots,... | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pythagorean triples"
] | null | proof only | null | |
0f2h | Problem:
Seven elves sit at a table. Each elf has a cup. In total the cups contain $3$ liters of milk. Each elf in turn gives all his milk to the others in equal shares. At the end of the process each elf has the same amount of milk as at the start. What was that? | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | final answer only | 3/7 liters | |
08d4 | Problem:
Luigi ha disegnato sul proprio quaderno un triangolo isoscele $ABC$ in cui i lati uscenti da $A$ sono uguali e, dopo aver tracciato la bisettrice interna all'angolo $\widehat{ABC}$ che interseca il lato $AC$ in $P$, si è accorto che la circonferenza per $B, P, C$ passava anche per il punto medio di $AB$. Si è... | [
"Solution:\n\nLa risposta è (C). Sia $M$ il punto medio di $AB$. Gli angoli $\\widehat{PBM}$ e $\\widehat{PCM}$ sono congruenti perché insistono sullo stesso arco. Dal momento che il triangolo è isoscele, gli angoli $\\widehat{ABC}$ e $\\widehat{ACB}$ sono uguali e quindi, per differenza di angoli uguali, $\\wideha... | Italy | Gara di Febbraio | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | C | |
02l4 | Problem:
O caminho da pulga - Para percorrer um caminho reto de 10 metros de comprimento, uma pulga usa a seguinte estratégia: a cada dia ela percorre a metade do caminho que faltava no dia anterior. Portanto, no primeiro dia ela percorre 5 metros, no segundo 2,5 metros e assim por diante (o tamanho da pulga é desprez... | [
"Solution:\n\nNo $1^{\\mathrm{o}}$ pulo a pulga percorre $10 \\times \\frac{1}{2}$, no $2^{\\mathrm{o}}$, $10 \\times \\frac{1}{2^{2}}$, e assim por diante.\n\nOs pulos da pulga\n\n\n\nDepois de 7 dias a pulga terá percorrido\n$$\n\\begin{aligned}\n& 10\\left(\\frac{1}{2}\\right)+10\\left(\... | Brazil | Nível 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | a) After 7 days: 10*(127/128) = 9.921875 m; after 10 days: 10*(1023/1024) = 9.990234375 m. b) From day 14 onward the flea is within 0.001 m of the end. | |
06mq | Four non-overlapping families dine at a restaurant together. Each family consists of one father, one mother and three children. There are five tables, which are red, blue, yellow, green and brown respectively. Every table can accommodate four people, and every person must sit at some table. How many ways are there to a... | [
"There are 278400 ways to assign the people.\n\nNote that each family can be partitioned in either a (4, 1) fashion (where exactly one parent sits with three children and the other is alone) or a (3, 2) fashion (where one parent sits with two children and the other sits with one).\n\n**Case 1.** All families are pa... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 278400 | |
0fxw | Problem:
Finde alle Funktionen $f: \mathbb{R} \rightarrow \mathbb{R}$, sodass für alle reellen $x, y$ gilt
$$
f(f(x)-f(y))=(x-y)^2 f(x+y)
$$ | [
"Solution:\nMit $x=y$ folgt $f(0)=0$. Mit $x=0$ respektive $y=0$ erhält man die beiden Gleichungen $f(-f(y))=y^2 f(y)$ und $f(f(x))=x^2 f(x)$ und durch Kombination weiter\n$$\nf(f(x))=f(-f(x)) \\quad \\forall x \\in \\mathbb{R}\n$$\nOffenbar ist die Nulfunktion $f \\equiv 0$ eine Lösung, wir nehmen daher im Folgend... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0, f(x) = x^2, f(x) = -x^2 | |
01ls | Is it possible to arrange the numbers $1, 2, \ldots, 2011$ over the circle in some order so that among any $25$ successive numbers at least $8$ numbers are multiples of $5$ or $7$ (or both $5$ and $7$)? | [
"Answer: it is not possible."
] | Belarus | Selection and Training Session | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof and answer | not possible | |
03h5 | Problem:
During a certain election campaign, $p$ different kinds of promises are made by the various political parties ($p>0$). While several parties may make the same promise, any two parties have at least one promise in common; no two parties have exactly the same set of promises. Prove that there are no more than $... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
02dw | Given a regular dodecahedron of side $a$. Take two pairs of opposite faces: $E, E'$ and $F, F'$. For the pair $E, E'$ take the line joining the centers of the faces and take points $A$ and $C$ on the line each a distance $m$ outside one of the faces. Similarly, take $B$ and $D$ on the line joining the centers of $F, F'... | [
"The centers of the faces of a regular dodecahedron form a regular icosahedron. Let $P, P'$ be two opposite vertices of a regular icosahedron. Then 5 of the remaining vertices are adjacent to $P$ and the other 5 are adjacent to $P'$. So we may label the other pair of opposite vertices $Q$ and $Q'$, where $Q$ is adj... | Brazil | VI OBM | [
"Geometry > Solid Geometry > 3D Shapes"
] | English | proof and answer | (1+sqrt(5))/2 | |
08mq | Problem:
Let $x_{i} > 1$, for all $i \in \{1,2,3, \ldots, 2011\}$. Prove the inequality
$$
\sum_{i=1}^{2011} \frac{x_{i}^{2}}{x_{i+1}-1} \geq 8044
$$
where $x_{2012} = x_{1}$. When does equality hold? | [
"Solution:\nRealize that $(x_{i}-2)^{2} \\geq 0 \\Leftrightarrow x_{i}^{2} \\geq 4(x_{i}-1)$. So we get:\n$$\n\\frac{x_{1}^{2}}{x_{2}-1} + \\frac{x_{2}^{2}}{x_{3}-1} + \\ldots + \\frac{x_{2011}^{2}}{x_{1}-1} \\geq 4\\left(\\frac{x_{1}-1}{x_{2}-1} + \\frac{x_{2}-1}{x_{3}-1} + \\ldots + \\frac{x_{2011}-1}{x_{1}-1}\\r... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Minimum value 8044, with equality if and only if all the numbers are equal to 2. | |
0jdp | Problem:
Chim Tu has a large rectangular table. On it, there are finitely many pieces of paper with nonoverlapping interiors, each one in the shape of a convex polygon. At each step, Chim Tu is allowed to slide one piece of paper in a straight line such that its interior does not touch any other piece of paper during ... | [
"Solution:\n\nLet the pieces of paper be $P_{1}, P_{2}, \\ldots, P_{n}$ in the Cartesian plane. It suffices to show that for any constant distance $D$, they can be slid so that each pairwise distance is at least $D$. Then, we can apply this using $D$ equal to the diameter of the rectangle, sliding all but at most o... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0bo6 | Problem:
Adott az $\left(x_{n}\right)_{n \geq 1}$ valós számsorozat, amelynek tagjai az $[1, \infty)$ halmazból vannak. Ha az $\left(y_{n}^{(k)}\right)_{n \geq 1}, y_{n}^{(k)}=\left[x_{n}^{k}\right], n \geq 1$, sorozat konvergens minden $k \in \mathbb{N}^{*}$ szám esetén, bizonyítsd be, hogy az $\left(x_{n}\right)_{n ... | [
"Solution:\n\nPentru $k \\in \\mathbb{N}^{*}$, $\\left(y_{n}^{(k)}\\right)_{n \\geq 1}$ este un şir convergent de numere naturale nenule. Atunci există $n_{k}, a_{k} \\in \\mathbb{N}^{*}$ astfel ca $y_{n}^{(k)}=a_{k}, \\forall n \\geq n_{k}$. Ca urmare, $x_{n}^{k} \\in\\left[a_{k}, a_{k}+1\\right), \\forall n \\geq... | Romania | Olimpiada Naţională de Matematică, Etapa Judeţeană şi a Municipiului Bucureşti | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0eoz | On a line segment with the points $R$, $A$, $T$ and $E$, the ratio $RA : AT = 1 : 3$ and $AT : TE = 5 : 2$. The ratio $RA : TE$ is
(A) $1 : 2$ (B) $1 : 3$ (C) $3 : 5$ (D) $5 : 6$ (E) $3 : 2$ | [
"We are given $\\frac{RA}{AT} = \\frac{1}{3}$ and $\\frac{AT}{TE} = \\frac{5}{2}$, so\n$$\n\\frac{RA}{TE} = \\frac{RA}{AT} \\times \\frac{AT}{TE} = \\frac{1}{3} \\times \\frac{5}{2} = \\frac{5}{6}.\n$$"
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | MCQ | D | |
02nb | Problem:
Qual é o maior número possível de elementos de um subconjunto de $\{1,2, \ldots, 100\}$ tal que nenhum de seus elementos seja um múltiplo de algum outro? | [
"Solution:\nInicialmente, observemos que o conjunto $\\{51,52,53, \\ldots, 100\\}$ tem 50 elementos e nenhum de seus elementos é múltiplo de outro. Assim, o subconjunto com o maior número de elementos e que satisfaz a propriedade exigida tem, no mínimo, 50 elementos.\n\nPara concluir que 50 é o maior número possíve... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 50 | |
0b4l | Problem:
Determine the sum of all positive integers $n$ for which $2[\tau(n)]^{2}=3 n$, where $\tau(n)$ is the number of positive divisors of $n$. | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 96 | |
04nd | Determine all pairs $(a, b)$ of positive integers such that $a^2 b$ divides $b^2 + 3a$. | [
"The given condition implies that\n$$\na^2 b k = b^2 + 3a\n$$\nfor some positive integer $k$. The equation $a^2 b k - 3a = b^2$ implies $a \\mid b^2$ and $b \\mid 3a$. Therefore, the numbers\n$$\n\\frac{b^2}{a}, \\frac{3a}{b} \\quad \\text{and} \\quad \\frac{b^2 + 3a}{a^2 b} = \\frac{b}{a^2} + \\frac{3}{a b}\n$$\na... | Croatia | Croatia_2018 | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | [(1, 1), (1, 3)] | |
0gsn | Let $ABC$ be an acute triangle with circumcircle $\omega$ centered at $O$. A point $P$ is chosen on the extension of the segment $BC$. The line $AP$ meets $\omega$ again at $L$. Let $K$ be the reflection of $L$ over the line $OP$ and $M$ be the point of intersection of lines $AK$ and $OP$. Prove that $\overline{OMB} + ... | [
"Note that $K$ lies on $\\omega$. Therefore,\n$$\n\\widehat{OKM} = \\widehat{OKA} = \\widehat{OAK} = \\widehat{OAM}.\n$$\nOn the other hand, since $OP$ is the perpendicular bisector of $[KL]$ we get $\\widehat{OLM} = \\widehat{OKM}$. Therefore, $\\widehat{OAM} = \\widehat{OLM}$, which in turn implies that $A, L, M,... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line,... | null | proof only | null | |
0h1q | There are $2012$ piles of stones. The first pile contains $2^0$ stones, the second pile contains $2^1$ stones, the third pile contains $2^2$ stones, and so on. The $2012$-th pile contains $2^{2011}$ stones. At one step, one can pick three piles and add $2$ stones to the first pile, $3$ stones to the second pile, and $4... | [
"After each operation the total number of stones changes by a number which is divisible by $9$. At the end, the total number is $2012 \\cdot 3^{1005}$, which is divisible by $9$. However, at the starting moment the total number is $2^0 + 2^1 + 2^2 + \\ldots + 2^{2011} = 2^{2012} - 1$, which is not divisible by $9$.... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | No | |
09un | Problem:
In een land zijn 2018 steden, waarvan sommige met elkaar verbonden zijn door wegen. Elke stad is verbonden met ten minste drie andere steden. Het is mogelijk om van elke willekeurige stad naar elke andere willekeurige stad te reizen via één of meer wegen. Bekijk voor elk tweetal steden de kortste route tussen... | [
"Solution:\n\nHet grootste aantal wegen dat in een kortste route kan voorkomen, is 1511. We geven eerst een land waarvoor dit aantal aangenomen wordt. Deel de steden op in 504 groepen: twee groepen van vijf steden $\\left(A_{0}, B_{0}, C_{0}, D_{0}, E_{0}\\right)$ en $\\left(A_{503}, B_{503}, C_{503}, D_{503}, E_{5... | Netherlands | Selectietoets | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1511 | |
01hr | Let $t$ be a non-zero real number and $n$ be a positive integer. Let $a_1, a_2, \dots, a_{2n-1}$ be real numbers (not necessarily distinct). Prove that there exist indices $i_1 < i_2 < \dots < i_n$ such that $a_{i_k} - a_{i_{l}} \neq t$ for any $1 \le k, l \le n$. | [
"Let $G = (V, E)$ be a graph with vertex set $V = \\{1, 2, \\dots, 2n-1\\}$ and edge set $E = \\{\\{i, j\\} : |a_i - a_j| = t\\}$. Note that $G$ has no odd cycles. Indeed, if $j_1, \\dots, j_{2k+1}$ is a cycle, then for all $l = 1, 3, 5, \\dots, 2k-1$ the number $a_{jl}$ differs from $a_{j,l}$ by $2t$ or $0$. Hence... | Baltic Way | Baltic Way 2021 Shortlist | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0i3y | Problem:
On the planet Lemniscate, the people use the elliptic table of elements, a far more advanced version of our periodic table. They're not very good at calculus, though, so they've asked for your help. They know that Kr is somewhat radioactive and deteriorates into Pl, a very unstable element that deteriorates t... | [
"Solution:\n\nThis problem is long-winded since it's giving an autonomous linear system of differential equations without using any such language (and it includes a number of subtle references). The system we have is $x' = -x$, $y' = x - 2y$. It's not hard to see that $x = 10 e^{-t}$ satisfies the first equation an... | United States | Harvard-MIT Math Tournament | [
"Calculus > Differential Equations > ODEs",
"Calculus > Differential Calculus > Applications"
] | null | proof and answer | ln 2 | |
0kuk | Problem:
Compute the number of tuples $\left(a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right)$ of (not necessarily positive) integers such that $a_{i} \leq i$ for all $0 \leq i \leq 5$ and
$$
a_{0}+a_{1}+\cdots+a_{5}=6
$$ | [
"Solution:\nLet $b_{i}=i-a_{i}$, so $b_{i} \\geq 0$. Then\n$$\n15-\\left(b_{0}+b_{1}+\\cdots+b_{5}\\right)=6 \\Longrightarrow b_{0}+b_{1}+\\cdots+b_{5}=9\n$$\nBy stars and bars, the answer is $\\binom{14}{5}=2002$."
] | United States | HMMT November 2023 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 2002 | |
000c | Sean $a$, $b$ y $c$ números reales positivos. Demostrar que
$$
\frac{a+b}{c^2} + \frac{c+a}{b^2} + \frac{b+c}{a^2} \ge \frac{9}{a+b+c} + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}
$$ | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | español | proof only | null | |
0l4m | Six points $A$, $B$, $C$, $D$, $E$, and $F$ lie in a straight line in that order. Suppose that $G$ is a point not on the line and that $AC = 26$, $BD = 22$, $CE = 31$, $DF = 33$, $AF = 73$, $CG = 40$, and $DG = 30$. Find the area of $\triangle BGE$. | [
"Because $CD = AF - AC - DF = 14$, the side lengths of $\\triangle CDG$ are $14$, $30$, and $40$. By Heron's Formula,\n$$\n\\text{Area}(\\triangle CDG) = \\sqrt{42(42 - 14)(42 - 30)(42 - 40)} = 168,\n$$\nimplying that the distance from $G$ to line $CD$ is $\\frac{2\\cdot168}{CD} = 24$. Then because $BE = BD + CE - ... | United States | AIME II | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 468 | |
089c | Problem:
Una ed una sola delle seguenti affermazioni è falsa. Quale?
(A) "La (B) è falsa"
(B) "La (C) è falsa"
(C) "La (E) è vera"
(D) "La (A) è vera"
(E) "Tre delle precedenti sono vere". | [
"Solution:\n\nLa risposta è (B). Se supponiamo che la frase \"La (B) è falsa\" sia falsa, allora anche la (D) è falsa, contro l'ipotesi del problema. Quindi la frase \"La (B) è falsa\" è vera, quindi l'affermazione falsa è la (B) e tutte le altre risultano vere."
] | Italy | Olimpiadi di Matematica | [
"Discrete Mathematics > Logic"
] | null | MCQ | B | |
07ar | Does there exist a non-identity function $f: \mathbb{N} \to \mathbb{N}$ that:
The number of divisors of $m$ is $f(n)$, if and only if the number of divisors of $f(m)$ is $n$, for each two natural numbers $m$ and $n$. | [
"The answer is Yes!\nLet $d(n)$ be the number of divisors of natural number $n$. We want to construct $f$ such that for any positive integer $m$, $f(d(f(m))) = d(m)$. Let $A_k = \\{n \\in \\mathbb{N} \\mid d(n) = k\\}$. For example, $A_1 = \\{1\\}$ and $A_2$ is the set of prime numbers. Note that $A_k$ has an infin... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof only | null | |
0a6b | Problem:
Let $A$, $B$, $C$, $D$, $E$ be five different points on the circumference of a circle in that (cyclic) order. Let $F$ be the intersection of chords $BD$ and $CE$. Show that if $AB = AE = AF$ then lines $AF$ and $CD$ are perpendicular. | [
"Solution:\nLet $x = \\angle ABF$ and let $y = \\angle BCA$. Since $\\triangle ABF$ is isosceles, we get $\\angle BFA = x$. Since $AB = AE$, it follows that arcs $BA$ and $AE$ are equal. Since equal arcs subtend equal angles, every angle subtended by either arc $AB$ or $AE$ must be equal to $\\angle BCA = y$.\n\n$$... | New Zealand | NZMO Round Two | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ixr | Problem:
Let $p_{0}(x), p_{1}(x), p_{2}(x), \ldots$ be polynomials such that $p_{0}(x) = x$ and for all positive integers $n$, $\frac{d}{dx} p_{n}(x) = p_{n-1}(x)$. Define the function $p(x): [0, \infty) \to \mathbb{R}$ by $p(x) = p_{n}(x)$ for all $x \in [n, n+1]$. Given that $p(x)$ is continuous on $[0, \infty)$, com... | [
"Solution:\nBy writing out the first few polynomials, one can guess and then show by induction that $p_{n}(x) = \\frac{1}{(n+1)!}(x+1)^{n+1} - \\frac{1}{n!} x^{n}$. Thus the sum evaluates to $e^{2010} - e^{2009} - 1$ by the series expansion of $e^{x}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | e^{2010} - e^{2009} - 1 | |
0fr2 | Problem:
Sean $a_{1}, a_{2}, \ldots, a_{2020}$ 2020 números reales de manera que la suma de 1009 de ellos cualesquiera es positiva. Demostrar que la suma de los 2020 números también es positiva. | [
"Solution:\nSea $S$ la suma de los 2020 números dados. Podemos escribir el número $1009 S$ como la suma de 1009 veces $S$. Esto da como resultado\n$$\n\\begin{aligned}\n1009 S & = a_{1} + a_{2} + a_{3} + \\ldots + a_{1008} + a_{1009} + a_{1010} + \\ldots + a_{2018} + a_{2019} + a_{2020} \\\\\n& + a_{1} + a_{2} + a_... | Spain | FASE LOCAL DE LA OLIMPIADA MATEMÁTICA ESPAÑOLA. | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
09l2 | Let $a$, $b$, and $c$ be real numbers.
(1)
Show that if $a^2 + b^2 = c^2$ and $a^3 + b^3 = c^3$, then either $a = 0$ or $b = 0$.
(2)
Show that if $a^3 + b^3 = c^3$ and $a^4 + b^4 = c^4$, then either $a = 0$ or $b = 0$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0gom | Let $n \ge 2$ be an integer and $E = \{1, 2, \dots, n\}$. If $A_1, A_2, \dots, A_k$ are subsets of $E$ and exactly one of $A_i \cap A_j$, $A'_i \cap A_j$, $A_i \cap A'_j$ and $A'_i \cap A'_j$ is empty for all $1 \le i < j \le k$, then determine the maximum possible value of $k$.
[For $A \subset E$, $A'$ denotes the el... | [
"The answer is $2n-3$ and an example is $\\{1\\}, \\{2\\}, \\dots, \\{n\\}, \\{1,2\\}, \\{1,2,3\\}, \\dots, \\{1,2,3,\\dots, n-2\\}$.\n\nWe will prove it by induction on $n$. For $n=2$, it is clear that $k$ is at most $1$. For $n=3$, it is easy to check that $k \\le 3$. Let us assume that the answer is $2n-5$ for $... | Turkey | 19th Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2n-3 | |
0jel | Problem:
Define a multiplication table to be a rectangular array in which every row is labeled with a different positive integer, every column is labeled with a different positive integer, and every cell is labeled with the product of its row and column numbers, for instance:
| $\times$ | 2 | 6 | 4 | 3 |
| :---: | :--... | [
"Solution:\nThe answer is $2 \\cdot 2012 - 1 = 4013$.\n\nTo get a multiplication table with only 4013 products, it suffices to fill the rows and columns with successive powers of two: $2^{0}, 2^{1}, 2^{2}, \\ldots, 2^{2012}$ for both the rows and the columns. Then the products within the table will be $2^{0}, 2^{1}... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4013 | |
017v | The $n \times n$ board is colored in $n$ colors such that the main diagonal (from top-left to right-bottom) is colored in the first color; the two adjacent diagonals are colored in the second color; the two next diagonals (one from above and one from below) are colored in the third color, etc.; the two corners (top-rig... | [
"Use the usual coordinate system for which the cells of the main diagonal have coordinates $(k, k)$, where $k = 1, \\dots, n$. Let $(k, f(k))$ be the coordinates of the $k$-th rook. Then by color restrictions for rooks we have\n$$\n\\sum_{k=1}^{n} (f(k) - k)^2 = \\sum_{i=0}^{n-1} i^2 = \\frac{n(n-1)(2n-1)}{6}.\n$$\... | Baltic Way | BALTIC WAY | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | proof only | null | |
0j25 | Define the sequence $a_1, a_2, a_3, \dots$ by $a_1 = 1$ and
$$
a_n = a_{\lfloor n/2 \rfloor} + a_{\lfloor n/3 \rfloor} + \dots + a_{\lfloor n/n \rfloor} + 1
$$
for $n > 1$. Prove that there are infinitely many $n$ such that
$$
a_n \equiv n \pmod{2^{2010}}.
$$
(This problem was suggested by Gabriel Carroll.) | [
"Our solution will be based upon the following key observation, for which we provide two different proofs.\n\n**Lemma 1.** Let $p$ be a prime. If $p^s$ divides $n$, then $2^{s-1}$ divides $a_n - a_{n-1}$, where $a_0 = 0$.\n\n*First proof of Lemma 1.* We argue by induction on $s$ and then $n$. First, the base case $... | United States | Team Selection Test 2010 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
06qk | Five identical empty buckets of 2-liter capacity stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. Then Ci... | [
"No, the Stepmother cannot enforce a bucket overflow and Cinderella can keep playing forever. Throughout we denote the five buckets by $B_{0}, B_{1}, B_{2}, B_{3}$, and $B_{4}$, where $B_{k}$ is adjacent to bucket $B_{k-1}$ and $B_{k+1}$ ($k=0,1,2,3,4$) and all indices are taken modulo $5$. Cinderella enforces that... | IMO | IMO Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | No; the Stepmother cannot enforce an overflow, and Cinderella can prevent any bucket from overflowing indefinitely. | |
0kfx | Let $a, b, c$ be fixed positive integers. There are $a+b+c$ ducks sitting in a circle, one behind the other. Each duck picks either *rock*, *paper*, or *scissors*, with $a$ ducks picking rock, $b$ ducks picking paper, and $c$ ducks picking scissors.
A *move* consists of an operation of one of the following three forms:... | [
"The maximum possible number of moves is $\\max(ab, ac, bc)$.\nFirst, we prove this is best possible. We define a *feisty triplet* to be an unordered triple of ducks, one of each of rock, paper, scissors, such that the paper duck is between the rock and scissors duck and facing the rock duck, as shown. (There may b... | United States | USA TSTST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | max(ab, ac, bc) | |
043u | As illustrated in Fig. 2.1, in the acute $\triangle ABC$, $AB < AC$, $I$ is the incentre, and $\odot O$ is the circumcentre. Let $M$, $N$ be the midpoints of $\overarc{BAC}$ and $\overarc{BC}$, respectively. Let $D$ be a point on $\overarc{AC}$ such that $AD // BC$. The inscribed circle of $\triangle ABC$ against $\ang... | [
"Fig. 2.1\n\nFirst, we need a lemma.\n\n**Lemma** Let $R'$ be the midpoint of $BC$. Then $\\angle AMI = \\angle IR'B$, $IR' // AE$.\n\n**Proof of lemma** As illustrated in Fig. 2.2, let $I_b$ and $I_c$ be the escentres of $\\triangle ABC$ relative to the vertices $B$ and $C$, respectively. Then $A, M, I_b$, and $I_... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
... | null | proof only | null | |
02ct | Problem:
Dividir um cubo - Se dividirmos um cubo de $1~\mathrm{m}$ de aresta em cubinhos de $1~\mathrm{mm}$ de aresta, que altura terá uma coluna formada por todos os cubinhos, dispostos sucessivamente um em cima do outro?
(a) $1~\mathrm{m}$
(b) $1~\mathrm{km}$
(c) $10~\mathrm{km}$
(d) $100~\mathrm{km}$
(e) $1000~\ma... | [
"Solution:\n\nConvertendo metros em milímetros temos: $1~\\mathrm{m} = 1000~\\mathrm{mm}$. Assim, o cubo ficou dividido em $1000 \\times 1000 = 10^{6}$ cubinhos de lado $1~\\mathrm{mm}$ cada um. Colocando-se lado a lado os $10^{6}$ cubinhos, teremos uma coluna de comprimento\n$$\n1000 \\times 1000 = 10^{6}~\\mathrm... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Solid Geometry > Volume"
] | null | MCQ | e | |
0hgc | Let $P$ be a point of the intersection of diagonals of the cyclic quadrilateral $ABCD$. The circumscribed circles of $\triangle APD$ and $\triangle BPC$ intersect the line $AB$ at points $E$ and $F$ correspondingly. $Q$ and $R$ are projections of the point $P$ onto the lines $FC$ and $DE$. Prove that $AB \parallel QR$. | [
"Denote by $d(Z, XY)$ the distance from the point $Z$ to the line $XY$. From the equality of inscribed angles it follows that (fig. 6)\n$$\n\\angle PEB = \\angle ADB = \\angle ACB = \\angle PFA \\Rightarrow PE = PF.\n$$\nFurthermore,\n$$\n\\angle EDP = \\angle BAP = \\angle BDC, \\quad \\angle FCP = \\angle PBA = \... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null |
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