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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0ifj | Problem:
Let $ABCD$ be a regular tetrahedron with side length $2$. The plane parallel to edges $AB$ and $CD$ and lying halfway between them cuts $ABCD$ into two pieces. Find the surface area of one of these pieces. | [
"Solution:\n\nThe plane intersects each face of the tetrahedron in a midline of the face; by symmetry it follows that the intersection of the plane with the tetrahedron is a square of side length $1$. The surface area of each piece is half the total surface area of the tetrahedron plus the area of the square, that ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 1 + 2*sqrt(3) | |
0aoq | Problem:
Let $x = \cos \theta$. Express $\cos 3\theta$ in terms of $x$. | [
"Solution:\n\n$4x^3 - 3x$\n\n$$\n\\begin{aligned}\n\\cos 3\\theta &= \\cos (2\\theta + \\theta) \\\\\n&= \\cos 2\\theta \\cos \\theta - \\sin 2\\theta \\sin \\theta \\\\\n&= (2 \\cos^2 \\theta - 1) \\cos \\theta - 2 \\sin^2 \\theta \\cos \\theta \\\\\n&= (2 \\cos^2 \\theta - 1) \\cos \\theta - 2(1 - \\cos^2 \\theta... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Precalculus > Trigonometric functions"
] | null | final answer only | 4x^3 - 3x | |
05xy | Problem:
Soit $ABC$ un triangle acutangle (dont tous les angles sont aigus) avec $BA \neq BC$. Soit $O$ le centre de son cercle circonscrit. La droite $(AB)$ intersecte le cercle circonscrit à $BOC$ une deuxième fois en $P \neq B$. Montrer que $PA = PC$. | [
"Solution:\n\n\n\nTraçons la figure dans le cas où $BC < BA$, le cas $BC > BA$ étant totalement analogue. Il s'agit de montrer que $PA = PC$, c'est-à-dire que $\\widehat{ACP} = \\widehat{PAC} \\ (= \\widehat{BAC})$.\n\nOr on a:\n$$\n\\begin{aligned}\n\\widehat{ACP} & = \\widehat{ACO} + \\wi... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
070n | Problem:
Representatives from $n > 1$ different countries sit around a table. If two people are from the same country then their respective right hand neighbors are from different countries. Find the maximum number of people who can sit at the table for each $n$. | [
"Solution:\n\nAnswer: $n^{2}$.\n\nObviously there cannot be more than $n^{2}$ people. For if there were, then at least one country would have more than $n$ representatives. But there are only $n$ different countries to choose their right-hand neighbours from. Contradiction.\n\nRepresent someone from country $i$ by ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n^2 | |
0fcq | Problem:
Por turno, en orden alfabético, tres amigos lanzan un dado. Quien saque un 6 en primer lugar gana lo apostado.
Por cada euro que apueste Carlos, ¿qué cantidad han de poner Ana y Blas para equilibrar el juego y lograr que sea equitativo, es decir, para que las expectativas de ganancia sean las mismas para los ... | [
"Solution:\n\nEl esquema en árbol nos ayudará a determinar las probabilidades que tienen cada uno de los amigos de ganar en este juego:\n\n\n\nPor cada $91\\,€$ en litigio, $36$ los debe poner Ana, $30$ Blas y $25$ Carlos.\n\nLuego, si Carlos apuesta $1\\,€$, Ana debe poner $1{,}44\\,€$ y B... | Spain | Fase Local | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | Ana 1.44 €, Blas 1.20 € (per 1 € by Carlos) | |
06l7 | There are $n \ge 3$ cities in a country and between any two cities $A$ and $B$, there is either a one way road from $A$ to $B$, or a one way road from $B$ to $A$ (but never both). Assume the roads are built such that it is possible to get from any city to any other city through these roads, and define $d(A, B)$ to be t... | [
"The answer is $\\frac{3}{2}$ for $n \\neq 4$ and $\\frac{19}{12}$ for $n = 4$.\n\nNote that for any distinct cities $A$ and $B$, exactly one of $d(A, B)$ and $d(B, A)$ is $1$, while the other is at least $2$. Thus it follows that $d(A, B) + d(B, A) \\geq 3$. The average for this pair is at least $\\frac{3}{2}$. Th... | Hong Kong | CHKMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Minimum average directed distance equals 3/2 for all n ≥ 3 with n ≠ 4, and equals 19/12 for n = 4. | |
0faw | Problem:
If $a > b > c > d > 0$ are integers such that $ad = bc$, show that $(a - d)^2 \geq 4d + 8$. | [
"Solution:\n\nWe need first that $a + d > b + c$. Put $a = m + h$, $d = m - h$, $b = m' + k$, $c = m' - k$. Then since $a - d > b - c$, we have $h > k$. But $m^2 - h^2 = ad = bc = {m'}^2 - k^2$, so $m > m'$ and hence $a + d > b + c$. Since $a$, $b$, $c$, $d$ are integers it follows that $(a + d - b - c) \\geq 1$.\n... | Soviet Union | 1st CIS | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebr... | null | proof only | null | |
03gq | Problem:
a) Find all positive integers with initial digit $6$ such that the integer formed by deleting this $6$ is $1/25$ of the original integer.
b) Show that there is no integer such that deletion of the first digit produces a result which is $1/35$ of the original integer. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) All integers of the form 625·10^m for m ≥ 0 (i.e., 625, 6250, 62500, …). b) No such integer exists. | |
0arz | Problem:
How many positive-integer pairs $(x, y)$ are solutions to the equation $\frac{x y}{x+y}=1000$. | [
"Solution:\n\n(ans. 49\n$(2 a_{1}+1)(2 a_{2}+1) \\cdots(2 a_{k}+1)$ where $1000=p_{1}^{a_{1}} p_{2}^{a_{2}} \\cdots p_{k}^{a^{k}}=2^{3} 5^{3} ;$ so 49. Let $\\frac{x y}{x+y}=n \\Rightarrow x y-n x-n y=0 \\Rightarrow(x-n)(y-n)=n^{2} \\Rightarrow x>n, y>n$. In the factorization $n^{2}=p_{1}^{2 a_{1}} p_{2}^{2 a_{2}} ... | Philippines | 13th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | final answer only | 49 | |
00ry | Let $n \ge 4$ points in the plane, no three of them are collinear. Prove that the number of parallelograms of area $1$, formed by these points, is at most $\frac{n^2-3n}{4}$. | [
"Fix a direction in the plane. We cannot have three points in the same line parallel to the direction so suppose that in that direction there are $k$ pairs of points, each pair belonging to a parallel line to the fixed direction. Then there are at most $k-1$ parallelograms of area $1$ formed by these $k$ pairs of p... | Balkan Mathematical Olympiad | BMO 2017 | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0b56 | Problem:
Fie $O$ un punct interior triunghiului ascuţitunghic $ABC$. Cercurile centrate în mijloacele laturilor triunghiului şi care trec prin $O$, se intersectează a doua oară în $K$, $L$ şi $M$.
Demonstraţi că $O$ este centrul cercului înscris în triunghiul $KLM$ dacă şi numai dacă $O$ este centrul cercului circumsc... | [] | Romania | BMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > In... | null | proof only | null | |
0g1c | Problem:
Das Hauptgebäude der ETH Zürich ist ein in Einheitsquadrate unterteiltes Rechteck. Jede Seite eines Quadrates ist eine Wand, wobei gewisse Wände Türen haben. Die Aussenwand des Hauptgebäudes hat keine Türen. Eine Anzahl von Teilnehmern der SMO hat sich im Hauptgebäude verirrt. Sie können sich nur durch Türen ... | [
"Solution:\n\nSobald zwei Teilnehmer nach einer Anweisung auf dem selben Quadrat sind, werden sie danach immer auf dem selben Quadrat sein, da sie immer in die gleiche Richtung gehen. Somit reduzieren wir das Problem auf dasselbe Problem mit einem Teilnehmer weniger. Dadurch genügt es zu zeigen, dass wir zwei versc... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
03w9 | Show that there are only finitely many triples $(a, b, c)$ of positive integers satisfying the equation $abc = 2009(a + b + c)$. | [
"There are at most six permutations for any three numbers $x, y, z$. It suffices to show that there are only finitely many triples $(a, b, c)$, with $a \\ge b \\ge c$, of positive integers satisfying the equation $abc = 2009(a + b + c)$. It follows that $abc \\le 2009 \\times (3a)$ or $bc \\le 2009 \\times 3 = 6027... | China | China Girls' Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
03nk | Problem:
A purse contains a finite number of coins, each with distinct positive integer values. Is it possible that there are exactly 2020 ways to use coins from the purse to make the value 2020? | [
"Solution:\nIt is possible.\nConsider a coin purse with coins of values $2, 4, 8, 2014, 2016, 2018, 2020$ and every odd number between $503$ and $1517$. Call such a coin big if its value is between $503$ and $1517$. Call a coin small if its value is $2, 4$ or $8$ and huge if its value is $2014, 2016, 2018$ or $2020... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
05fc | Problem:
Soit $ABC$ un triangle et $\Omega$ son cercle circonscrit. On note $X$ le point d'intersection des tangentes à $\Omega$ en $B$ et $C$. On note $\varphi$ l'angle $\widehat{B A X}$ et $\mu$ l'angle $\widehat{X A C}$. On note $Y$ le point de la droite $(A X)$ tel que $\widehat{A C Y}=\varphi$. Montrer que $\wide... | [
"Solution:\n\n\nOn redéfinit les points de l'énoncé de la manière qui nous arrange. On note $\\omega_{1}$ (resp. $\\omega_{2}$) le cercle passant par $A, B$ (resp. $A$ et $C$) et tangent en $A$ à $(A C)$ (resp. $(A B)$). On note $Y'$ l'intersection des cercles $\\omega_{1}$ et $\\omega_{2}$... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigo... | null | proof only | null | |
0j71 | Problem:
How many ways are there to color the vertices of a $2n$-gon with three colors such that no vertex has the same color as either of its two neighbors or the vertex directly across from it? | [
"Solution:\n\nAnswer: $3^{n} + (-2)^{n+1} - 1$\n\nLet the $2n$-gon have vertices $A_{1}, A_{2}, \\ldots, A_{2n}$, in that order. Consider the diagonals $d_{1} = (A_{1}, A_{n+1})$, $d_{2} = (A_{2}, A_{n+2})$, $\\cdots$, $d_{n} = (A_{n}, A_{2n})$. Suppose the three colors are red (R), green (G), and blue (B). Each di... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 3^n + (-2)^{n+1} - 1 | |
0l2f | Problem:
Find all polynomials $f$ that satisfy the equation
$$
\frac{f(3x)}{f(x)} = \frac{729(x-3)}{x-243}
$$
for infinitely many real values of $x$. | [
"Solution:\nThe above equation holds for infinitely many $x$ if and only if\n$$\n(x-243) f(3x) = 729(x-3) f(x)\n$$\nfor all $x \\in \\mathbb{C}$, because $(x-243) f(3x) - 729(x-3) f(x)$ is a polynomial, which has infinitely many zeroes if and only if it is identically $0$.\n\nWe now plug in different values to find... | United States | 25th Bay Area Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = a x^2 (x - 9)(x - 27)(x - 81)(x - 243) for arbitrary constant a | |
0br6 | Let $ABC$ be a non-equilateral triangle such that $m(\angle A) = 60^\circ$. Let $D$ and $E$ be the intersection points of the Euler line of triangle $ABC$ and the sides of the angle $\angle BAC$. Prove that the triangle $ADE$ is equilateral. | [
"Let $H$ and $O$ be the orthocenter and the circumcenter of $ABC$, $R$ the radius of the circumcenter; then $OH$ meets the line $AB$ at $D$ and the line $AC$ at $E$. Let $B'$ be the foot of the altitude from $B$ and $C'$ be the foot of the altitude from $C$.\n\nSince $BCC'B'$ is a cyclic quadrilateral we have $\\an... | Romania | 67th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Advanced... | English | proof only | null | |
0j6b | Problem:
Find the number of ordered triples $(a, b, c)$ of pairwise distinct integers such that $-31 \leq a, b, c \leq 31$ and $a+b+c>0$. | [
"Solution:\nAnswer: 117690\nWe will find the number of such triples with $a<b<c$. The answer to the original problem will then be six times what we will get. By symmetry, the number of triples $(a, b, c)$ with $a+b+c>0$ is equal to the number of those with $a+b+c<0$. Our main step is thus to find the number of trip... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 117690 | |
05gw | Problem:
Déterminer tous les entiers $n \geqslant 2$ vérifiant la propriété suivante : pour tous entiers $a_{1}, a_{2}, \ldots, a_{n}$ dont la somme n'est pas divisible par $n$, il existe un indice $i$ tel qu'aucun des nombres
$$
a_{i}, a_{i}+a_{i+1}, \ldots, a_{i}+\cdots+a_{i+n-1}
$$
n'est divisible par $n$ (pour $i>... | [
"Solution:\n\nCe sont exactement les nombres premiers!\n\nEn effet, si $n=ab$, on peut prendre $a_{1}=0$ et $a_{2}=\\cdots=a_{n}=a$. La somme des $a_{i}$ vaut $a(n-1)$ donc n'est pas divisible par $n$. Cependant, soit $1 \\leqslant i \\leqslant n$. Si $i+b-1 \\leqslant n$, alors le nombre $a_{i}+\\cdots+a_{i+b-1}=a... | France | Préparation Olympique Française de Mathématiques | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | All prime numbers | |
0crp | The numbers $1, 2, \dots, n^2$ are put in some order into the squares of a checkered $n \times n$ board, one number per square. Pete performs several moves according to the following rules. On the first move, he puts a token into some square. By any subsequent move, he may either put a new token into an arbitrary squar... | [
"Answer: $n$.\n\nLet us show that $n$ tokens are sufficient. Note that one token is enough for each row: you can put it in the square of the row with the minimal number, and then visit all the squares of the row in order of increasing numbers.\n\nOn the other hand, let us show that fewer than $n$ tokens may not be ... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n | |
00li | Let $a, b, c$ be integers such that
$$
\frac{ab}{c} + \frac{ac}{b} + \frac{bc}{a}
$$
is an integer.
Prove that each of the numbers
$$
\frac{ab}{c} \cdot \frac{ac}{b} \quad \text{and} \quad \frac{bc}{a}
$$
is an integer. | [
"Set $u := \\frac{ab}{c}$, $v := \\frac{ac}{b}$ and $w := \\frac{bc}{a}$. By assumption, $u + v + w$ is an integer. It is easily seen that $uv + uw + vw = a^2 + b^2 + c^2$ and $uvw = abc$ are integers, too.\n\nAccording to Vieta's formulae, the rational numbers $u, v, w$ are the roots of a cubic polynomial $x^3 + p... | Austria | National Competition | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0ith | Problem:
Alice rolls two octahedral dice with the numbers $2,3,4,5,6,7,8,9$. What's the probability the two dice sum to $11$? | [
"Solution:\n\nAnswer: $\\frac{1}{8}$\n\nNo matter what comes up on the first die, there is exactly one number that could appear on the second die to make the sum $11$, because $2$ can be paired with $9$, $3$ with $8$, and so on. So, there is a $\\frac{1}{8}$ chance of getting the correct number on the second die."
... | United States | 1st Annual Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 1/8 | |
059t | Mother wants to divide a cake of triangular shape between three kids. She makes a straight cut from one vertex to the midpoint of the opposite side and then another straight cut from another vertex to the midpoint of the opposite side. She gives the piece of quadrilateral shape to Anna, the triangular piece opposite to... | [
"*Answer:* all kids get the same amount.\n\nLet $S$ be the area of the initial triangle, $S_A$ be the area of the quadrilateral piece, $S_B$ be the area of Berta's triangle, and $S_1$ and $S_2$ be the areas of the remaining two triangles (Fig. 14). Then $S_1 + S_B = S_2 + S_B = \\frac{1}{2}S$ (a common altitude whi... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof and answer | All receive equal amounts. | |
01ev | Let $\mathbb{T} = \{1, 3, 6, 10, 15, \dots\}$ be the set of triangular numbers, i.e. numbers of the form $T_n = \frac{n(n+1)}{2}$. Let $f$ be a function defined on the set of positive integers such that
1) $f(n)$ is a positive integer for each $n$;
2) $f(uv) = f(u)f(v)$ for any pair $(u, v)$ of coprime numbers;
3) $f(a... | [
"It is not difficult to find $f(n)$ for small $n$:\n$$\nf(1 \\cdot 1) = f(1)f(1) \\text{ therefore } f(1) = 1;\n$$\n$$\nf(3) = f(1) + f(1) + f(1) = 3;\n$$\n$$\nf(5) = f(1 + 1 + 3) = 5;\n$$\n$$\nf(10) = f(1 + 3 + 6) = 4 + 3f(2) \\text{ and } f(10) = f(2 \\cdot 5) = f(2)f(5) = 5f(2) \\text{ therefore } f(2) = 2.\n$$\... | Baltic Way | Baltic Way shortlist | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
01d1 | Prove that
$$
\sum_{k=1}^{n} (-1)^k \binom{n}{k} \binom{kn}{n} = (-n)^n
$$ | [
"Consider an $n \\times n$ checkerboard and count the number $s$ of ways to color exactly one square in each column. On the one hand, $s = n^n$. On the other hand, if $a_i$ denotes the number of ways to color exactly $n$ squares such that some fixed $i$ columns do not contain a colored square, then $s = \\sum_{i=0}... | Baltic Way | Baltic Way 2016 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
0975 | Problem:
Pentru ce valori reale $\alpha$ ecuația $\sin 3x = \alpha \sin x + (4 - 2|\alpha|) \sin^2 x$ are aceeași mulțime de soluții reale ca și ecuația $\sin 3x + \cos 2x = 1 + 2 \sin x \cos 2x$? | [
"Solution:\nUtilizând identitățile $\\sin 3x = 3 \\sin x - 4 \\sin^3 x$ și $\\cos 2x = 1 - 2 \\sin^2 x$, ecuația a doua este echivalentă cu $\\sin x - 2 \\sin^2 x = 0$, adică $\\sin x = 0$ sau $\\sin x = \\frac{1}{2}$.\n\nUtilizând identitatea $\\sin 3x = 3 \\sin x - 4 \\sin^3 x$, prima ecuație este echivalentă cu ... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | [0, 1) ∪ {3, 4} ∪ (5, +∞) | |
0egp | Problem:
V kateri točki graf funkcije $f(x)=2 \log _{\sqrt{2}}(\sqrt{2} x-5)-4$ seka abscisno os?
(A) $\left(\frac{7 \sqrt{2}}{2}, 0\right)$
(B) $(7 \sqrt{2}, 0)$
(C) $(1-7 \sqrt{2}, 0)$
(D) $\left(\frac{\sqrt{2}}{7}, 0\right)$
(E) $\left(\frac{\sqrt{2}}{2}-7,0\right)$ | [
"Solution:\nPredpis funkcije $f$ enačimo z 0. Dobimo enačbo $2 \\log _{\\sqrt{2}}(\\sqrt{2} x-5)-4=0$. Rešitev enačbe je $x=\\frac{7 \\sqrt{2}}{2}$. Graf funkcije $f$ seka abscisno os v točki $\\left(\\frac{7 \\sqrt{2}}{2}, 0\\right)$."
] | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | MCQ | A | |
02vc | Problem:
A figura a seguir representa um triângulo $ABC$, retângulo em $C$, com uma circunferência no seu interior tangenciando os três lados $AB$, $BC$ e $CA$ nos pontos $C_1$, $A_1$ e $B_1$, respectivamente. Seja $H$ o pé da altura relativa ao lado $A_1C_1$ do triângulo $A_1B_1C_1$.
a) Cal... | [
"Solution:\n\nConsidere a figura a seguir.\n\n\n(a) Como $\\angle ACB=90^\\circ$, então $\\angle CBA=90^\\circ-\\angle BAC=90^\\circ-\\angle A$. Dado que $AB_1$ e $AC_1$ são tangentes à circunferência, segue que $AB_1=AC_1$. De modo semelhante, $BA_1=BC_1$. Assim, como $A_1BC_1$ e $AB_1C_1$... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 45° | |
0e0k | Let $ABC$ be an acute triangle and let $D$ be a point on the side $AB$. The circumcircle of the triangle $BCD$ intersects the side $AC$ at $E$. The circumcircle of the triangle $ADC$ intersects the side $BC$ at $F$. Let $O$ be the circumcentre of the triangle $CEF$. Prove that the points $D$ and $O$ and the circumcentr... | [
"Let $O_1, O_2, O_3$ and $O_4$ be the circumcentres of the triangles $ADE$, $ADC$, $BFD$ and $BCD$. The line $O_1O_2$ bisects the segment $AD$ and the two are perpendicular. Similarly, $O_3O_4$ bisects the segment $DB$ and these two are perpendicular as well.\n\n\n\nDenote the angles of the... | Slovenia | Selection Examinations for the IMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"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 | |
00g3 | Let $\mathbb{R}$ denote the set of all real numbers. Find all functions $f$ from $\mathbb{R}$ to $\mathbb{R}$ satisfying:
(i) there are only finitely many $s$ in $\mathbb{R}$ such that $f(s)=0$, and
(ii) $f\left(x^{4}+y\right)=x^{3} f(x)+f(f(y))$ for all $x, y$ in $\mathbb{R}$. | [
"The only such function is the identity function on $\\mathbb{R}$.\n\nSetting $(x, y)=(1,0)$ in the given functional equation (ii), we have $f(f(0))=0$. Setting $x=0$ in (ii), we find\n$$\n\\begin{equation*}\nf(y)=f(f(y)) \\tag{1}\n\\end{equation*}\n$$\n[1 mark.] and thus $f(0)=f(f(0))=0$ [1 mark.]. It follows from... | Asia Pacific Mathematics Olympiad (APMO) | XIV APMO | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x | |
0l0c | Problem:
Suppose that $a, b, c$, and $d$ are real numbers such that $a+b+c+d=8$. Compute the minimum possible value of
$$
20\left(a^{2}+b^{2}+c^{2}+d^{2}\right)-\sum_{\text{sym}} a^{3} b,
$$
where the sum is over all 12 symmetric terms. | [
"Solution:\nObserve that\n$$\n\\sum_{\\mathrm{sym}} a^{3} b=\\sum_{\\mathrm{cyc}} a \\cdot \\sum_{\\mathrm{cyc}} a^{3}-\\sum_{\\mathrm{cyc}} a^{4}=8 \\sum_{\\mathrm{cyc}} a^{3}-\\sum_{\\mathrm{cyc}} a^{4}\n$$\nso\n$$\n\\begin{aligned}\n20 \\sum_{\\mathrm{cyc}} a^{2}-\\sum_{\\mathrm{sym}} a^{3} b & =\\sum_{\\mathrm{... | United States | HMIC 2024 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 112 | |
0466 | Suppose $n \ge 2$ is a positive integer, and $A_1A_2 \cdots A_{2n}$ is a convex $2n$-gon inscribed in a circle. It is known that there exists a point $P$ inside this $2n$-gon such that
$$
\angle PA_1A_2 = \angle PA_2A_3 = \cdots = \angle PA_{2n-1}A_{2n} = \angle PA_{2n}A_1.
$$
Prove that the following equation holds
$$... | [
"All subscripts are to be understood modulo $2n$. Denote $\\angle PA_1A_2 = \\angle PA_2A_3 = \\cdots = \\alpha$. For $1 \\le i \\le 2n$, extend $A_iP$ to intersect the circumference at point $B_i$. Note that\n$$\n\\angle A_iB_iB_{i-1} = \\angle A_iA_{i-1}P = \\angle A_{i+1}A_iB_i = \\alpha.\n$$\nHence, $A_iA_{i+1}... | China | 2023 Chinese IMO National Team Selection Test | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | English | proof only | null | |
0e9f | Find all polynomials $p$ with real coefficients, such that
$$
p(p(x)) = (x^2 + x + 1)p(x)
$$
for all $x \in \mathbb{R}$. | [
"The zero polynomial is obviously a valid solution. Let $p$ be a non-zero polynomial and write $p(x) = a_nx^n + a_{n-1}x^{n-1} + \\dots + a_0$, where $a_n \\ne 0$. The leading term on the left-hand side of the equality is equal to $a_n(a_nx^n)^n = a_n^{n+1}x^{n^2}$, and the leading term on the right-hand side is $x... | Slovenia | National Math Olympiad in Slovenia | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | p(x) = 0 and p(x) = x^2 + x | |
09m1 | Let $a$, $b$, $c$ be real numbers such that the system of equations
$$
\begin{cases}
ax + by = 1 \\
x + cy = a \\
cx + y = b
\end{cases}
$$
has a real solution $(x, y)$.
Prove that $a^2 + b^2 + c^2 = 2abc + 1$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
0fde | Problem:
En el interior de un cuadrado $ABCD$ se construye el triángulo equilátero $ABE$. Sea $P$ el punto intersección de las rectas $AC$ y $BE$. Sea $F$ el punto simétrico del $P$ respecto de la recta $DC$. Se pide demostrar que:
a) el triángulo $CEF$ es equilátero.
b) el triángulo $DEF$ es rectángulo e isósceles.
c)... | [
"Solution:\n\n\na) Puesto que $\\angle EBC=90^{\\circ}-60^{\\circ}=30^{\\circ}$, los ángulos en la base del triángulo isósceles $BEC$ ( $BE=BC$ por construcción) son iguales a $75^{\\circ}$. Resulta que $\\angle ECP=\\angle ECB-\\angle PCB=75^{\\circ}-45^{\\circ}=30^{\\circ}$, de donde se s... | Spain | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0h6u | Does there a function $f: \mathbb{R} \rightarrow \mathbb{R}$ exist, that for any real numbers $x$, $y$ the following inequality is fulfilled:
$$
f(x - f(y)) \le x - y f(x)?
$$ | [
"**Answer:** it doesn't exist.\n\nLet us assume such function exists. Let us substitute $y = 0$ in the given inequality:\n$$\nf(x - f(0)) \\le x,\n$$\nThen, substitute $x = x + f(0)$ there:\n$$\nf(x) \\le x + f(0). \\quad (1)\n$$\nLet us substitute $x = f(y)$ in the initial inequality. We will get:\n$$\n\\begin{cas... | Ukraine | UkraineMO | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | it doesn't exist | |
009w | Evaluate the sum
$$
\frac{1 \cdot 4}{2 \cdot 5} + \frac{2 \cdot 7}{5 \cdot 8} + \dots + \frac{k(3k+1)}{(3k-1)(3k+2)} + \dots + \frac{99 \cdot 298}{296 \cdot 299}
$$ | [
"Denote $S_n = \\sum_{k=1}^{n} \\frac{k(3k+1)}{(3k-1)(3k+2)}$. Multiply all numerators by $12$ and all denominators by $4$ to obtain\n$$\n3S_n = \\sum_{k=1}^{n} \\frac{12k(3k+1)}{(6k-2)(6k+4)}.\n$$\nNow complete squares as follows:\n$$\n\\begin{aligned}\n12k(3k+1) &= 36k^2 + 12k = (6k+1)^2 - 1, \\\\\n(6k-2)(6k+4) &... | Argentina | Argentine National Olympiad 2015 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 9900/299 | |
0cps | Consider two cubic polynomials $F(x) = x^3 + a_1x^2 + a_2x + a_3$ and $G(x) = x^3 + b_1x^2 + b_2x + b_3$ with unit leading coefficients. All real roots of the equations $F(x) = 0$, $G(x) = 0$, and $F(x) = G(x)$ are written. It appears that eight distinct numbers are written. Prove that at least one of the minimal and t... | [
"Заметим, что у многочленов $F(x)$ и $G(x)$ не более, чем по три корня, а у многочлена $F(x) - G(x)$ (имеющего степень, не превосходящую 2) не больше двух корней. Поскольку у них в совокупности 8 корней, то у $F(x)$ и $G(x)$ ровно по три корня, а у $F(x) - G(x)$ ровно два, причём все они имеют кратность 1.\n\nПредп... | Russia | Russian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English, Russian | proof only | null | |
0dwe | Problem:
Za racionalno funkcijo $f(x)=\frac{a x+b}{c x+1}$ velja: $f(1)=\frac{3}{4}$, $f(2)=1$ in $f(-1)=-\frac{1}{2}$. Določi realne parametre $a, b$ in $c$ ter zapiši funkcijo $f(x)$. Zapis funkcije poenostavi. | [
"Solution:\n\nUpoštevamo zapisane pogoje in zapišemo enačbe $\\frac{a+b}{c+1}=\\frac{3}{4}$, $\\frac{2 a+b}{2 c+1}=1$ in $\\frac{-a+b}{-c+1}=-\\frac{1}{2}$. Odpravimo ulomke in rešimo sistem treh enačb s tremi neznankami. Dobimo rešitev $a=\\frac{2}{3}$, $b=c=\\frac{1}{3}$. Zapišemo funkcijo $f(x)=\\frac{\\frac{2}{... | Slovenia | 4. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | a=2/3, b=1/3, c=1/3; f(x) = (2x+1)/(x+3) | |
0chh | Let $ABCA'B'C'$ be a triangular regular prism, with lateral edges $AA'$, $BB'$, $CC'$. Consider the midpoint $D$ of the edge $BC$ and the parallelogram $ADB'E$. Let $F$ be the orthogonal projection of the point $A'$ on the line $AE$, $d$ be the intersection of the planes $(ADE)$ and $(A'CF)$ and $P$ be the intersection... | [
"AD is a median in $\\triangle ABC$, so the point $P$ is the baricenter of the triangle $ABC$ if and only if $AP = 2PD$. Since $PF \\parallel DE$, this is equivalent to $AF = 2FE$. Because $AA' \\parallel BB'$, $BD \\parallel EA'$ and $BB' \\perp BD$, the triangle $AEA'$ has a right angle at $A'$. Then $A'E$ and $A... | Romania | 74th Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
04qx | Solve the system of equations
$$
\frac{1}{xy} = \frac{x}{z} + 1, \quad \frac{1}{yz} = \frac{y}{x} + 1, \quad \frac{1}{zx} = \frac{z}{y} + 1
$$
in the domain of the real numbers. | [
"From the form of the equations it is immediate that $xyz \\neq 0$. Two of the numbers $x, y, z$ have to be of the same sign; then the right-hand side of the equation where the ratio of these two numbers occurs is positive, hence so must be the corresponding left-hand side, which implies that the third of the numbe... | Czech Republic | Czech-Slovak-Polish Match | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | x = y = z = ± sqrt(2)/2 | |
0d3p | Circles $\omega_{1}$ and $\omega_{2}$ meet at $P$ and $Q$. Segments $AC$ and $BD$ are chords of $\omega_{1}$ and $\omega_{2}$ respectively, such that segment $AB$ and ray $CD$ meet at $P$. Ray $BD$ and segment $AC$ meet at $X$. Point $Y$ lies on $\omega_{1}$ such that $PY \parallel BD$. Point $Z$ lies on $\omega_{2}$ s... | [
"Because quadrilateral $BPDQ$ is cyclic, we have $\\angle PQD = \\angle PBD$. Because quadrilateral $ACQP$ is cyclic, we have $\\angle PQC = 180^{\\circ} - \\angle CAP$. We deduce that\n$$\n\\begin{aligned}\n\\angle DQC & = \\angle PQC - \\angle PQD = 180^{\\circ} - \\angle CAP - \\angle PBD \\\\\n& = 180^{\\circ} ... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | English, Arabic | proof only | null | |
08d5 | Problem:
Sia $A B C D E F$ un esagono inscritto in una circonferenza e tale che $A B = B C$, $C D = D E$ ed $E F = A F$. Dimostrare che i segmenti $A D$, $B E$ e $C F$ concorrono (cioè hanno un punto in comune). | [
"Solution:\n\nPoiché $A B = B C$, usando il fatto che in una circonferenza a corde congruenti corrispondono angoli alla circonferenza congruenti, si ha $\\angle A E B = \\angle B E C$. Analogamente $\\angle C A D = \\angle D A E$ e $\\angle A C F = \\angle F C E$.\n\nDunque $A D$, $E B$ e $C F$ sono le tre bisettri... | Italy | XXXV Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07o9 | The points $E$ and $F$ are on the sides $AC$ and $AB$, respectively, of triangle $ABC$ such that $FE$ is parallel to $BC$. The lines $BE$ and $CF$ intersect at $G$. Prove that the line $AG$ passes through the midpoint of $BC$. | [
"Let $AG$ meet $BC$ at $D$. We need to show that $D$ is the midpoint of $BC$.\n\n\n\nAs $EF$ is parallel to $BC$, we have $\\frac{|CE|}{|EA|} = \\frac{|BF|}{|FA|}$. Ceva's Theorem tells us that\n$$\n\\frac{|BD|}{|DC|} \\cdot \\frac{|CE|}{|EA|} \\cdot \\frac{|FA|}{|BF|} = 1.\n$$\nTogether th... | Ireland | Ireland | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem"
] | null | proof only | null | |
0kc6 | Problem:
Let $\mathbb{N}_{>1}$ denote the set of positive integers greater than $1$. Let $f: \mathbb{N}_{>1} \rightarrow \mathbb{N}_{>1}$ be a function such that $f(m n) = f(m) f(n)$ for all $m, n \in \mathbb{N}_{>1}$. If $f(101!) = 101!$, compute the number of possible values of $f(2020 \cdot 2021)$. | [
"Solution:\n\nFor a prime $p$ and positive integer $n$, we let $v_{p}(n)$ denote the largest nonnegative integer $k$ such that $p^{k} \\mid n$. Note that $f$ is determined by its action on primes. Since $f(101!) = 101!$, by counting prime factors, $f$ must permute the set of prime factors of $101!$; moreover, if $p... | United States | HMMO 2020 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 66 | |
054n | Inside a circle of radius $1$ (or on the circumference), one marks $n$ points in such a way that the minimal distance between two marked points is as large as possible. Let $d_n$ be this distance between the two closest points. Is it true that $d_{n+1} < d_n$ for every natural number $n \ge 2$? | [
"We show that $d_6 \\le 1 \\le d_7$. For the first inequality, assume arbitrary six points $A_1, A_2, A_3, A_4, A_5, A_6$ being marked in the circle. Let the centre of the circle be $O$. If $A_i = O$ for some $i$, the distance between $A_i$ and any other marked points is at most $1$. Assume in the rest that $A_i = ... | Estonia | National Olympiad Final Round | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
... | English | proof and answer | No | |
0all | Problem:
Find the largest number $N$ so that
$$
\sum_{n=5}^{N} \frac{1}{n(n-2)} < \frac{1}{4}
$$ | [] | Philippines | 18th PMO Area Stage | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 24 | |
01yl | Find all positive integers $a$ for which there exists a polynomial $p(x)$ with integer coefficients such that
$$
p(\sqrt{2} + 1) = 2 - \sqrt{2} \quad \text{und} \quad p(\sqrt{2} + 2) = a.
$$ | [
"$a = 7k - 2$ where $k$ is an arbitrary positive integer.\n\nSuppose that the integer $a$ and the polynomial $p(x)$ satisfy the condition. For the polynomial $q(x) = p(x + 1)$, the equalities from the condition have the form $q(\\sqrt{2}) = 2 - \\sqrt{2}$ and $q(1 + \\sqrt{2}) = a$. Since the coefficients of the po... | Belarus | Belarus2022 | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Modular Arithmetic",
"Number Theory > Algebraic Number Theory > Algebraic numbers"
] | English | proof and answer | a = 7k - 2 for any positive integer k | |
0g6p | 試求
$$
\frac{1}{1 + \sqrt{3}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \cdots + \frac{1}{\sqrt{97} + \sqrt{99}}
$$
的整數部分。 | [
"利用\n$$\n\\frac{1}{\\sqrt{n} + \\sqrt{n+2}} \\le \\frac{1}{4} \\left( \\frac{1}{\\sqrt{n}} + \\frac{1}{\\sqrt{n+2}} \\right)\n$$\n可得\n$$\n\\begin{align*}\nS &= \\frac{1}{1+\\sqrt{3}} + \\frac{1}{\\sqrt{5}+\\sqrt{7}} + \\cdots + \\frac{1}{\\sqrt{97}+\\sqrt{99}} \\\\\n< & \\frac{1}{4} \\left( \\frac{1}{\\sqrt{1}} + \... | Taiwan | 二〇一二數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 2 | |
09oe | Let triangle $ABC$ be inscribed in a circle $\omega$, and let $M$ be the midpoint of arc $AB$ that does not contain point $C$. Let the tangent to $\omega$ at point $B$ intersect line $AC$ at point $P$. Let line $PM$ intersect $\omega$ again at point $G$. The tangent to $\omega$ at point $G$ intersects line $BC$ at poin... | [] | Mongolia | MMO2025 Round 3 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
040d | There is a stone at each vertex of a given regular $13$-gon, and the color of each stone is black or white. Prove that we may exchange the position of two stones such that the coloring of these stones are symmetric with respect to some symmetric axis of the $13$-gon. | [] | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0b0r | Problem:
Let $x = -\sqrt{2} + \sqrt{3} + \sqrt{5}$, $y = \sqrt{2} - \sqrt{3} + \sqrt{5}$, and $z = \sqrt{2} + \sqrt{3} - \sqrt{5}$. What is the value of the expression below?
$$
\frac{x^{4}}{(x-y)(x-z)} + \frac{y^{4}}{(y-z)(y-x)} + \frac{z^{4}}{(z-x)(z-y)}
$$ | [
"Solution:\nWriting the expression as a single fraction, we have\n$$\n\\frac{x^{4} y - x y^{4} + y^{4} z - y z^{4} - z x^{4} + x z^{4}}{(x-y)(x-z)(y-z)}\n$$\nNote that if $x = y$, $x = z$, or $y = z$, then the numerator of the expression above will be $0$. Thus, $(x-y)(x-z)(y-z)$ divides $x^{4} y - x y^{4} + y^{4} ... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 20 | |
0l8p | The positive real numbers $a$, $b$, $c$ satisfy the condition:
$$
21ab + 2bc + 8ca \le 12.
$$
Find the least value of the expression:
$$
P(a, b, c) = \frac{1}{a} + \frac{2}{b} + \frac{3}{c}.
$$ | [] | Vietnam | VIETNAMESE MATHEMATICAL COMPETITION FOR TEAM SELECTION | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | \frac{\left(7^{2/3} + 7^{1/3} + 2\right)^{3/2}}{\sqrt{2}\,\cdot 7^{1/3}} | |
0grk | Find all positive real numbers $c$ such that
$$
\frac{x^3y + y^3z + z^3x}{x + y + z} + \frac{4c}{xyz} \geq 2c + 2
$$
for all positive real numbers $x, y, z$. | [
"Answer: $c=1$.\nIf $x = y = z = \\sqrt[6]{4c}$ then we get\n$$\n\\frac{x^3 y + y^3 z + z^3 x}{x + y + z} + \\frac{4c}{xyz} = 4\\sqrt{c} \\ge 2c + 2\n$$\nand hence $(\\sqrt{c}-1)^2 \\le 0$. Therefore, all $c \\ne 1$ do not satisfy the inequality. Let us show that the inequality holds for $c = 1$. By AM-GM inequalit... | Turkey | 23rd Junior Turkish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 1 | |
01h0 | A triangle $ABC$ is given. Call its orthocenter $H$. Define $\omega$ as the circle through $B$, $C$, and $H$, and define $\Gamma$ as the circle with diameter $AH$. Let $X$ be the other intersection of $\omega$ and $\Gamma$, and let the reflection of $\Gamma$ over $AX$ be $\gamma$.
Suppose $\gamma$ and $\omega$ intersec... | [
"Let $M$ be the midpoint of $BC$. We first show that $X$ lies on $AM$. Consider $A'$, the reflection of $A$ across $M$. As $ABA'C$ is a parallelogram, we have that $\\angle BA'C = \\angle BAC = 180^\\circ - \\angle BHC$, which in turn gives us that $A'$ lies on $\\omega$. Now $\\angle HBA' = \\angle HBC + \\angle C... | Baltic Way | Baltic Way 2020 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0fce | Problem:
Tenemos una colección de esferas iguales que apilamos formando un tetraedro cuyas aristas tienen todas $n$ esferas. Calcula, en función de $n$, el número total de puntos de tangencia (contactos) que hay entre las esferas del montón. | [
"Solution:\n\nEl problema en el plano.\n\nAnalicemos primero el problema en el caso plano. Sea $A_{n}$ el número de contactos de $n$ esferas colocadas en un triángulo plano con $n$ esferas en cada uno de los lados (figura de la derecha). Fijémonos que el número total de esferas es, evidentemente, $T_{n}=\\frac{n(n+... | Spain | null | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n^3 - n | |
05nh | Problem:
Soit $ABC$ un triangle, et $\Gamma$ son cercle circonscrit. Soit $M$ le milieu de l'arc $BC$ ne contenant pas $A$. Un cercle $\mathscr{C}$ est tangent à $[AB), [AC)$ en $D$ et $E$ respectivement, et tangent intérieurement à $\Gamma$ en $F$. Montrer que $(DE), (BC)$ et $(FM)$ sont concourantes. | [
"Solution:\n\nSoit $I$ le centre du cercle inscrit dans $ABC$. La droite $(EF)$ recoupe $\\Gamma$ en un point $H$. Soit $J$ le point d'intersection de $(BC)$ avec $(FM)$.\n\nIl est facile de voir que $H$ est le milieu de l'arc $AC$ ne contenant pas $B$ : en effet, l'homothétie de centre $F$... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof only | null | |
02l3 | Problem:
Uma desigualdade - Os valores de $x$ que satisfazem $\frac{1}{x-1}>1$ são:
(a) $x<2$
(b) $x>1$
(c) $1<x<2$
(d) $x<1$
(e) $x>2$ | [
"Solution:\n\nNote que o inverso de um número $b$ só é maior do que 1 quando $b$ for positivo e menor do que 1. Portanto,\n$$\n\\frac{1}{x-1}>1 \\Longleftrightarrow 0<x-1<1 \\Longleftrightarrow 1<x<2\n$$\nA opção correta é (c)."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | c | |
0l1q | Problem:
Compute the number of even positive integers $n \leq 2024$ such that $1,2, \ldots, n$ can be split into $\frac{n}{2}$ pairs, and the sum of the numbers in each pair is a multiple of $3$. | [
"Solution:\n\nThere have to be an even number of multiples of $3$ at most $n$, so this means that $n \\equiv 0,2 \\pmod{6}$. (We can also say that there should be an equal number of $1 \\pmod{3}$ and $2 \\pmod{3}$ numbers, which gives the same restriction.)\n\nWe claim that all these work. We know there are an even... | United States | HMMT February 2024 Guts Round | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 675 | |
0d3d | Let $ABC$ be a triangle with $\angle A < \angle B \leq \angle C$, $M$ and $N$ the midpoints of sides $CA$ and $AB$, respectively, and $P$ and $Q$ the projections of $B$ and $C$ on the medians $CN$ and $BM$, respectively. Prove that the quadrilateral $MNPQ$ is cyclic. | [
"Because $\\angle BPC = \\angle BQC = 90^\\circ$, quadrilateral $BCQP$ is cyclic and therefore\n$$\n\\angle CPQ = \\angle CBQ.\n$$\nBecause $M$ and $N$ are midpoints of sides $AC$ and $AB$, segment $MN$ is parallel to side $BC$. Hence $\\angle NMQ = \\angle CBQ$.\n\nWe deduce that $\\angle NMQ = \\angle CPQ$. This ... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Arabic | proof only | null | |
07ef | Given two intersecting circles $\omega_1, \omega_2$. Points $A, C$ lie on $\omega_1$ and points $B, D$ lie on $\omega_2$ so that $AB, CD$ are common tangents of $\omega_1, \omega_2$. Let $M$ be the midpoint of $AB$. Tangent lines through $M$ to $\omega_1, \omega_2$ (other than $AB$) intersect $CD$ at $Y, X$. If $I$ be ... | [
"Suppose that $MX, MY$ touch $\\omega_2, \\omega_1$ at $E, F$ also let $N$ be the foot of perpendicular line through $I$ to $CD$.\n\n\n\nNotice that $MF = MA = MB = ME$. $N$ is the intersection of $XY$ and incircle of triangle $MXY$ so\n$$\n\\begin{align*}\nNY &= \\frac{YX + YM - MX}{2} \\\... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0db4 | Let $x$, $y$, $z$, $a$, $b$, $c$ be pairwise different integers from the set $\{1,2,3,4,5,6\}$. Find the smallest possible value for the expression $x y z + a b c - a x - b y - c z$. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof and answer | 10 | |
04o7 | Let $n$ be a positive integer. Points $A_1, A_2, \dots, A_n$ are located on the inside of a circle, and points $B_1, B_2, \dots, B_n$ are on the circle, so that the lines $\overline{A_1B_1}$, $\overline{A_2B_2}$, \dots, $\overline{A_nB_n}$ are mutually disjoint. A grasshopper can jump from point $A_i$ to point $A_j$ (f... | [
"Let a path be any line $A_iA_j$ and a wall be any line $A_kB_k$. We say that a path and a wall intersect if the path goes through an inner point of the wall. A path is good if there is no wall to intersect it, and a wall is irrelevant if it does not intersect any path.\n\n**Claim 1.** For any $i \\in \\{1,2,\\dots... | Croatia | Croatian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
02cp | Problem:
Um subconjunto - O conjunto $\{1,2,3, \ldots, 3000\}$ contém um subconjunto de 2000 elementos tal que nenhum elemento é o dobro do outro? | [
"Solution:\n\nVamos construir o subconjunto pedido da seguinte forma:\n- ele contém todos os números ímpares: $1,3,5, \\ldots, 2999$. Aqui já temos uma lista com 1500 números.\n- o conjunto não pode conter os números que são da forma $2 \\times$ (número ímpar),\n- o conjunto pode conter os números que são da forma ... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | No | |
0awz | Problem:
In cyclic pentagon $ABCDE$, $\angle ABD = 90^\circ$, $BC = CD$, and $AE$ is parallel to $BC$. If $AB = 8$ and $BD = 6$, find $AE^2$. | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 338/5 | |
0l5t | Problem:
Isabella has a bag with 20 blue diamonds and 25 purple diamonds. She repeats the following process 44 times: she removes a diamond from the bag uniformly at random, then puts one blue diamond and one purple diamond into the bag. Compute the expected number of blue diamonds in the bag after all 44 repetitions. | [
"Solution:\n\nLet $a = 20$ and $b = 25$ be the initial numbers of blue and purple diamonds, respectively, and let $c = 44$ be the number of times Isabella performs the operation. Suppose that at some point, the bag contains $x$ blue diamonds and $y$ purple diamonds, for $x + y = z$ total diamonds. After one step, t... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | final answer only | 173/4 | |
0kj6 | Let $ABCD$ be an isosceles trapezoid with $\overline{BC} \parallel \overline{AD}$ and $AB = CD$. Points $X$ and $Y$ lie on diagonal $\overline{AC}$ with $X$ between $A$ and $Y$, as shown in the figure. Suppose $\angle AXD = \angle BYC = 90^\circ$, $AX = 3$, $XY = 1$, and $YC = 2$. What is the area of $ABCD$?
![](attach... | [
"Let $A = (0, 0)$ and $C = (a, b)$.\n\nLet $X$ be on $AC$ such that $AX = 3$, $XY = 1$, $YC = 2$.\n\nLet $AC$ have length $AX + XY + YC = 3 + 1 + 2 = 6$.\n\nSo $AC = 6$.\n\nLet $X$ be at $\\left(\\frac{3a}{6}, \\frac{3b}{6}\\right) = \\left(\\frac{a}{2}, \\frac{b}{2}\\right)$.\n\nLet $Y$ be at $\\left(\\frac{4a}{6}... | United States | AMC 12 A | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | C | |
0azr | Problem:
Suppose that $\{a_n\}_{n \geq 1}$ is an increasing arithmetic sequence of integers such that $a_{a_{20}} = 17$ (where the subscript is $a_{20}$). Determine the value of $a_{2017}$. | [
"Solution:\n\nLet $a_1 = a$ be the first term of such arithmetic sequence and $d > 0$ be its common difference. Then the condition $a_{a_{20}} = 17$ is equivalent to\n$$\na_{a_{20}} = a + (a_{20} - 1)d = a + (a + 19d - 1)d = a(1 + d) + 19d^2 - d.\n$$\nSolving for $a$, we get\n$$\na = \\frac{-19d^2 + d + 17}{d + 1} ... | Philippines | 20th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 4013 | |
0egt | Problem:
Za realno število $a$ velja $a^{2}-\frac{1}{2} a=\frac{1}{4}$. Koliko je vrednost izraza $a^{3}-\frac{1}{2} a$?
(A) $-\frac{1}{4}$
(B) $\frac{1}{4}$
(C) $\frac{1}{2}$
(D) 4
(E) $\frac{1}{8}$ | [
"Solution:\n\nS pomočjo dane enakosti izračunamo\n$$\n\\begin{aligned}\na^{3}-\\frac{1}{2} a & =\\left(a^{3}-\\frac{1}{2} a^{2}\\right)+\\left(\\frac{1}{2} a^{2}-\\frac{1}{4} a\\right)-\\frac{1}{4} a=a\\left(a^{2}-\\frac{1}{2} a\\right)+\\frac{1}{2}\\left(a^{2}-\\frac{1}{2} a\\right)-\\frac{1}{4} a= \\\\\n& =\\frac... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | E | |
0f0l | Problem:
Prove that a 9 digit decimal number whose digits are all different, which does not end with 5 and or contain a 0, cannot be a square. | [
"Solution:\n\nLet $N$ be a 9-digit decimal number whose digits are all different, does not end with $5$, and does not contain a $0$.\n\nFirst, since $N$ has 9 digits, and all digits are different and nonzero, the digits must be $1,2,3,4,5,6,7,8,9$ in some order.\n\nLet us consider the sum of the digits:\n\n$$\n1 + ... | Soviet Union | ASU | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0bqd | a. Find all perfect squares of the form $aabcc$.
b. Let $n$ be a given positive integer. Prove that there exists a perfect square of the form $aab \underbrace{cc\dots c}_{2n \text{ times}}$. | [] | Romania | 67th NMO Shortlisted Problems | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English | proof and answer | Part a: 22500, 44100, 44944.
Part b: For any positive integer n, (10^n · 15)^2 = 225 followed by 2n zeros is a perfect square of the form aab with 2n copies of c, taking a = 2, b = 5, c = 0. | |
00w3 | Problem:
A segment $AB$ of unit length is marked on the straight line $t$. The segment is then moved on the plane so that it remains parallel to $t$ at all times, the traces of the points $A$ and $B$ do not intersect and finally the segment returns onto $t$. How far can the point $A$ now be from its initial position? | [
"Solution:\nThe point $A$ can move any distance from its initial position - see Figure 4 and note that we can make the height $h$ arbitrarily small.\n\n\nFigure 4"
] | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Unbounded; it can be arbitrarily large (any distance). | |
0hi3 | Does there exist a convex 2023-gon on the Cartesian plane with vertices at points whose coordinates are both integers, such that all its side lengths are equal? | [
"Suppose such a 2023-gon exists.\nLet its side be denoted by $a$, so $a^2$ is an integer, and its vertices as $(x_1, y_1)$, $(x_2, y_2)$, ..., $(x_{2023}, y_{2023})$. Consider the 2023-gon with the smallest value of $a^2$. We have $(x_i - x_{i+1})^2 + (y_i - y_{i+1})^2 = a^2$ for each $i$, where $x_{2024} = x_1$, $... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | No, such a polygon does not exist. | |
036b | Problem:
A card game is played by five persons. In a group of 25 persons all like to play that game. Find the maximum possible number of games which can be played if no two players are allowed to play simultaneously more than once. | [
"Solution:\nThe number of all pairs of players is $\\frac{25 \\cdot 24}{2} = 300$ and after each game 10 of them become impossible. Therefore at most $300 \\div 10 = 30$ games are possible.\n\nWe shall prove that 30 games are possible. We denote the pairs of players by $(m, n)$, where $1 \\leq m, n \\leq 5$ are int... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | 30 | |
0j1r | Let $ABC$ be a triangle. Point $M$ and $N$ lie on sides $AC$ and $BC$ respectively such that $MN \parallel AB$. Points $P$ and $Q$ lie on sides $AB$ and $CB$ respectively such that $PQ \parallel AC$. The incircle of triangle $CMN$ touches segment $AC$ at $E$. The incircle of triangle $BPQ$ touches segment $AB$ at $F$. ... | [
"\n**Solution** (By Gabriel Carroll). Let $\\omega_1, \\omega_C, \\omega_B$, and $\\omega$ denote the incircles of triangles $ABC, MNC, PBQ$, and $ARS$, respectively. Denote by $I$ and $I_1$ the incenters of triangles $ABC$ and $ARS$, respectively. Let $\\omega_1$ touch sides $AB$ and $AC$ ... | United States | Team Selection Test 2010 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0dsn | Find the largest positive integer $n$ such that there exist $n$ real polynomials where the sum of any two has no real roots but the sum of any three does. | [
"When $n = 3$, we can take the constant polynomials $f, g, h = -1, -2, 3$ which clearly satisfy the problem conditions.\n\nNow assume that $n = 4$ and let our polynomials be $f_1, f_2, f_3, f_4$.\nNote that for any $i, j$, we must have either $f_i(x) + f_j(x) > 0$ or $f_i(x) + f_j(x) < 0$ for all $x$ as otherwise i... | Singapore | Singapore Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | 3 | |
097q | Problem:
Numerele naturale $a, b, c, d$ şi $n$ verifică relaţiile $a^{2}-b^{2}=c^{2}-d^{2}=n$. Să se arate, că numărul $2(a+b)(c+d)(a c+b d-n)$ este un pătrat perfect. | [
"Solution:\n\nFolosind relaţiile, $a^{2}-b^{2}=c^{2}-d^{2}=n$, se obţine consecutiv:\n\n$$\n\\begin{gathered}\nE=2(a+b)(c+d)(a c+b d-n)=(a+b)(c+d)(2 a c+2 b d-2 n)=(a+b)(c+d)\\left[(b+d)^{2}-(a-c)^{2}\\right]= \\\\\n=(a+b)(c+d)(b+d-a+c)(b+d+a-c)=(a+b)(c+d+b-a) \\cdot(c+d)(a+b+d-c)= \\\\\n=[(a+b)(c+d)+(a+b)(b-a)] \\... | Moldova | Olimpiada Republicană la Matematică, Ziua a doua | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
05ar | a. Does there exist a positive integer $n$ such that the eight last digits of the number $n^2 + 1$ are the same as in the number $2n$, but the ninth digit from the end of these two numbers are different?
b. Does there exist a positive integer $n$ such that the nine last digits of the number $n^2 + 1$ are the same as i... | [
"The condition that the last $k$ digits of two numbers are the same is fulfilled if and only if the difference of these two numbers ends with exactly $k$ zeroes. Note that $n^2 + 1 - 2n = (n-1)^2$.\n\na. Let $n = 100010001$, then the number $n-1$ ends with 4 zeroes and the fifth digit from the end is 1. Hence the n... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | a) Yes; for example n = 100010001. b) No; such an integer does not exist. | |
05ab | There are 8 distinct points marked on a circle. Juku wants to draw as many triangles as possible in such a way that all vertices of each triangle he draws are at the marked points, and no two of these triangles share a side. Find the largest number of triangles that can be drawn under these conditions. | [
"Assume w.l.o.g. that the points marked on the circle are equally spaced and number the marked points counterclockwise with natural numbers $0$, $1$, $\\ldots$, $7$. Consider a triangle with vertices marked at points $0$, $1$, $3$ and its $7$ copies obtained by rotating the original triangle counterclockwise by $\\... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 8 | |
0kaa | Problem:
How many ways can one fill a $3 \times 3$ square grid with nonnegative integers such that no nonzero integer appears more than once in the same row or column and the sum of the numbers in every row and column equals $7$? | [
"Solution:\n\nIn what ways could we potentially fill a single row? The only possibilities are if it contains the numbers $(0,0,7)$ or $(0,1,6)$ or $(0,2,5)$ or $(0,3,4)$ or $(1,2,4)$. Notice that if we write these numbers in binary, in any choices for how to fill the row, there will be exactly one number with a $1$... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 216 | |
00as | Let $a$ and $b$ be rational numbers such that $a+b = a^2 + b^2$. Suppose that the common value $s = a+b = a^2 + b^2$ is not an integer, and write it as an irreducible fraction: $s = \frac{m}{n}$. Let $p$ be the least prime divisor of $n$. Find the minimum value of $p$. | [
"The minimum value of $p$ is $p = 5$. Write $a$ and $b$ as fractions with least common denominator $w$: $a = \\frac{u}{w}$, $b = \\frac{v}{w}$. In other words, if $a = \\frac{u'}{w'}$, $b = \\frac{v'}{w'}$ is another representation with common denominator $w'$, then $w' \\geq w$. The irreducible representation $s =... | Argentina | Argentine National Olympiad 2016 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | 5 | |
0al8 | Problem:
Find all complex numbers $z$ such that
$$
\frac{z^{4}+1}{z^{4}-1} = \frac{i}{\sqrt{3}}
$$ | [] | Philippines | AREA STAGE | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | {1/2 + i*sqrt(3)/2, -sqrt(3)/2 + i*1/2, -1/2 - i*sqrt(3)/2, sqrt(3)/2 - i*1/2} | |
0bcb | The measure of the angle $\hat{A}$ of the acute triangle $ABC$ is $60^\circ$, and $HI = HB$, where $I$ and $H$ are the incenter and the orthocenter of the triangle $ABC$. Find the measure of the angle $\hat{B}$. | [
"We have $m(\\angle BIC) \\equiv m(\\angle BHC) = 120^\\circ$, hence $B$, $H$, $I$, $C$ are situated on a circle.\n\nIf $m(\\angle B) > 60^\\circ$ then $m(\\angle HBI) = m(\\angle ABI) - m(\\angle ABH) = \\frac{1}{2}m(\\angle B) - 30^\\circ$.\n\nBut $m(\\angle HBI) = m(\\angle HIB) = m(\\angle HCB) = 90^\\circ - m(... | Romania | 62nd NMO SELECTION TESTS FOR THE JUNIOR BALKAN 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 > Miscellaneous > Angle chasing"
] | null | proof and answer | 80° | |
0cav | Let $ABCD$ be a rectangle, $GH \parallel BC$, with $G \in (AB)$ and $H \in (AD)$, and $EF \parallel DC$, with $E \in (AD)$ and $F \in (BC)$. Let $GH \cap EF = \{M\}$ and $AH \cap CE = \{K\}$. Prove that the point $K$ is on the circle passing through the feet of the altitudes of the triangle $DFG$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry... | null | proof only | null | |
01qq | Bob cuts an apple into either $20$ or $14$ pieces. Then he cuts one of these pieces into either $20$ or $14$ pieces. He repeats this procedure several times.
Can Bob obtain $1! + 2! + 3! + \dots + 1013! + 2014!$ small bits of the apple? | [
"Answer: yes, he can.\nIf Bob cuts a piece of the apple into $20$ pieces, then the total number of the pieces increases by $19$. If Bob cuts a piece of the apple into $14$ pieces, then the total number of the pieces increases by $13$. So, if Bob makes $x$ cuts into $20$ pieces and $y$ cuts into $14$ pieces, then th... | Belarus | Final Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | yes | |
0d6g | Let $f(x) = x^{2} + a x + b$ be a quadratic function with real coefficients $a, b$. It is given that the equation $f(f(x)) = 0$ has 4 distinct real roots and the sum of 2 roots among these roots is equal to $-1$. Prove that $b \leq \frac{-1}{4}$. | [
"Solution:\nFirst, we will prove that $f(x) = 0$ has some solutions (maybe not distinct).\nIndeed, if $f(x) = 0$ has no root, then it can be written as $f(x) = (x - c)^{2} + d$ with $d > 0$ and\n$$\nf(f(x)) = \\left((x - c)^{2} + d - c\\right)^{2} + d > 0.\n$$\nIt means $f(f(x)) = 0$ has no solution, which is a con... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0exi | Problem:
A circle is circumscribed about the triangle $ABC$. $X$ is the midpoint of the arc $BC$ (on the opposite side of $BC$ to $A$), $Y$ is the midpoint of the arc $AC$, and $Z$ is the midpoint of the arc $AB$. $YZ$ meets $AB$ at $D$ and $YX$ meets $BC$ at $E$. Prove that $DE$ is parallel to $AC$ and that $DE$ passe... | [
"Solution:\n$ZY$ bisects the angle $AYB$, so $AD/BD = AY/BY$. Similarly, $XY$ bisects angle $BYC$, so $CE/BE = CY/BY$. But $AY = CY$. Hence $AD/BD = CE/BE$. Hence triangles $BDE$ and $BAC$ are similar and $DE$ is parallel to $AC$.\n\nLet $BY$ intersect $AC$ at $W$ and $AX$ at $I$. $I$ is the incenter. $AI$ bisects ... | Soviet Union | 5th ASU | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0f8t | Problem:
Show that for each integer $n > 0$, there is a polygon with vertices at lattice points and all sides parallel to the axes, which can be dissected into $1 \times 2$ (and/or $2 \times 1$) rectangles in exactly $n$ ways. | [] | Soviet Union | 23rd ASU | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
02p1 | Emerald writes the integers from $1$ to $9$ in a $3 \times 3$ table, one number in each cell, each number appearing exactly once. Then she computes eight sums: the sums of three numbers on each row, the sums of the three numbers on each column and the sums of the three numbers on both diagonals.
a. Show a table such t... | [
"a.\nFor instance,\n\n| 1 | 2 | 3 |\n|---|---|---|\n| 4 | 5 | 6 |\n| 8 | 9 | 7 |\n\nThe trick is to only adjust the last row. The usual order $7$, $8$, $9$ yields all sums to be multiple of $3$, so it's just a matter of rearranging them.\n\nb.\nNo, it's not possible. First, notice that the sum of three numbers $x$,... | Brazil | Brazilian Math Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a. Example grid with rows: 1 2 3; 4 5 6; 8 9 7.
b. No, it is not possible. | |
0ekh | Problem:
Kristina je narisala 2 kvadrata, katerih dolžine stranice $v$ centimetrih so naravna števila, in osenčila del večjega kvadrata, ki leži zunaj manjšega kvadrata (glej sliko). Ploščina osenčenega območja je enaka $43~\mathrm{cm}^2$. Koliko kvadratnih centimetrov je vsota ploščin obeh Kristininih kvadratov?
![](... | [
"Solution:\n\nOznačimo dolžino stranice večjega kvadrata v centimetrih z $a$, manjšega pa z $b$. Tedaj je $43 = a^2 - b^2 = (a + b)(a - b)$. Ker pa sta $a + b$ in $a - b$ naravni števili in je $43$ praštevilo, sledi $a + b = 43$ in $a - b = 1$. Torej je $a = 22$ in $b = 21$. Vsota ploščin obeh Kristininih kvadratov... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | MCQ | B | |
08ye | A $3 \times 3$ grid made up of $9$ $1 \times 1$ squares is given. Suppose you want to distribute $9$ distinct positive integers chosen from the integers greater than or equal to $1$ and less than or equal to $9$ into $9$ square boxes of the grid. How many distinct ways of distributing the $9$ numbers are there if for a... | [
"$32$ ways\n\nFrom the grid of $9$ squares, we pick a $2 \\times 2$ four squares to fill in with numbers. Let as in the diagram (a) below $a$, $b$, $c$, $d$ be the numbers inserted into the $4$ squares. Then, we see that the difference between $a$ and $d$ is $5$ or less. In fact, since both $|a-b|$ and $|b-d|$ are ... | Japan | 2019 Japan Mathematical Olympiad First Stage | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 32 | |
07x6 | Let $n$ be a positive integer. Consider all arrangements of $n$ identical green coins, $n$ identical white coins and $n$ identical orange coins in a row. Each such arrangement of $3n$ coins can be considered as a sequence of *blocks*, where coins within a block have the same colour and any two adjacent blocks contain c... | [
"For any arrangement, we say that a position $k \\in \\{2, 3, \\dots, 3n\\}$ is a *change* if and only if the coin at position $k$ has a different colour to the coin at position $k-1$. Note that the number of blocks in any arrangement is one greater than the number of changes in that arrangement. For instance, the ... | Ireland | IRL_ABooklet_2024 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | A = 2, B = 1 | |
08rp | Let $n$ be a positive integer. Two people $P$, $Q$ play a game in which they call an integer $m$ ($1 \le m \le n$) alternately. $P$ calls the first number. They cannot call the numbers which are already called by themselves or by their opponent. The game is over when neither can call numbers. If the sum of the numbers ... | [
"Let the number called by a player in the $m$th turn be $N_m$. Then sequence $(N_1, \\dots, N_l)$ is called \"history up to the $l$th turn\". We call $j$ which satisfies $j \\neq N_1, \\dots, N_l$ \"free in the $l+1$th turn\". We are going to prove a proposition that if $n \\equiv 0, 4, 5 \\pmod 6$, $P$ can absolut... | Japan | Japanese Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | n ≡ 0, 4, 5 (mod 6) | |
0h24 | Find all pairs of prime numbers $(a, b)$, such that $a^b = b^a + 1$ is prime. | [
"**Answer:** $(2,3)$, $(3,2)$, $(2,2)$.\n\nClearly, either $a$ or $b$ is even. WLOG, $a = 2$. It is easy to see that $b = 2$ and $b = 3$ satisfy the condition. Suppose that $b > 3$. We have: $2^b = b^2 + 1$, $b > 3$. It is easy to see that the last expression is divisible by $3$ and cannot be prime."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | (2,3), (3,2), (2,2) | |
0byi | Determine the integers $x$ and $y$ for which $\sqrt{4^x + 5^y}$ is rational. | [
"We treat four cases:\n\n**I.** $x, y \\ge 0$\n\n$\\sqrt{4^x + 5^y}$ is rational if and only if $4^x + 5^y$ is a perfect square, i.e., there exists $n \\in \\mathbb{N}$ such that $4^x + 5^y = n^2$. Analyzing this equation modulo $3$, we have $4^x \\equiv 1 \\pmod{3}$, $5^y \\equiv (-1)^y \\pmod{3}$, and $n^2 \\equi... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (1, 1) and (-2, 1) | |
0enm | Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for all $x, y > 0$,
$$
f(yf(x))(x + y) = x^2(f(x) + f(y)).
$$ | [
"Setting $y = x$, we obtain $2x f(xf(x)) = 2x^2 f(x)$, and so $f(xf(x)) = x f(x)$ since $x > 0$.\n\nNow suppose $f(x) = f(y)$. Then\n$$\nx^2(f(x) + f(y)) = f(yf(x))(x + y) = f(yf(y))(x + y) = y f(y)(x + y)\n$$\nand so\n$$\n2x^2 f(x) = (x y + y^2) f(y) \\implies 2x^2 - x y - y^2 = (2x + y)(x - y) = 0 \\implies x = y... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1/x | |
0g2h | Problem:
Ein Hochhaus hat 7 Lifte, wobei aber jeder nur in 6 Stockwerken hält. Trotzdem gibt es für je zwei Stockwerke immer einen Lift, der die beiden Stockwerke direkt verbindet.
Zeige, dass das Hochhaus höchstens 14 Stockwerke haben kann, und dass ein solches Hochhaus mit 14 Stockwerken tatsächlich realisierbar is... | [
"Solution:\n\nOn construit d'abord un exemple d'une telle tour avec 14 étages comme suit:\n\n\n\nOù les $\\times$ désignent les étages où s'arrêtent chaque ascenseur. On vérifie facilement que pour chaque paire d'étages il existe un ascenseur qui les relie directement.\n\nPremière Solution:... | Switzerland | SMO - Vorrunde | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 14 | |
0gza | In a table $n \times n$ two players fill the lines one by one with numbers "+1" and "1". At first the first player fills the first line. Then second player -- second line, then first player fills third line. Then second -- forth line etc. In the end of filling lines, first player gets 1 point for every line or column, ... | [
"**Answer:** By even $k$ the first player collects $(3k + 2)$, the second $k$; by the odd $k$ the first $(3k + 1)$, second $(k + 1)$.\n\nLet $n = 2k$.\n\nIn the beginning we can look into such a strategy for every player. The first player fills numbers in arbitrary way. In this way he collects $k$ points. The secon... | Ukraine | The Problems of Ukrainian Authors | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Let n be the board size. The optimal scores are:
- If n ≡ 0 (mod 4): first player n/2, second player 3n/2.
- If n ≡ 1 (mod 4): first player (3n + 1)/2, second player (n − 1)/2.
- If n ≡ 2 (mod 4): first player n/2 + 1, second player 3n/2 − 1.
- If n ≡ 3 (mod 4): first player (3n − 1)/2, second player (n + 1)/2.
Equival... | |
0gnj | The sequence $\{x_n\}$ is defined by $x_1 = a$, $x_2 = b$ and $x_n = 2008x_{n-1} - x_{n-2}$ for all $n \ge 2$. Prove that there are positive integers $a$ and $b$ such that for all $n \ge 1$ the expression $1 + 2006x_nx_{n+1}$ is a perfect square. (Şahin Emrah). | [
"We prove that at $a = 1$, $b = 2008$ all terms of the sequence are perfect squares.\nLet us prove by induction that for all $n \\ge 1$\n$$\nx_n^2 + x_{n+1}^2 - 1 = 2008x_n x_{n+1}. \\quad (1)\n$$\n1. $n = 1 : 1^2 + 2008^2 - 1 = 2008 \\cdot 1 \\cdot 2008$.\n2. Suppose (1) is held for $n = k : x_k^2 + x_{k+1}^2 - 1 ... | Turkey | Team Selection Test for IMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null |
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