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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
06wz | The Bank of Oslo issues coins made out of two types of metal: aluminium (denoted $A$) and copper (denoted $C$). Morgane has $n$ aluminium coins, and $n$ copper coins, and arranges her $2n$ coins in a row in some arbitrary initial order. Given a fixed positive integer $k \leqslant 2n$, she repeatedly performs the follow... | [
"Define a block to be a maximal subsequence of consecutive coins made out of the same metal, and let $M^{b}$ denote a block of $b$ coins of metal $M$. The property that there is at most one aluminium coin adjacent to a copper coin is clearly equivalent to the configuration having two blocks, one consisting of all $... | IMO | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Algorithms"
] | English | proof and answer | All pairs (n, k) with n ≤ k ≤ (3n+1)/2 | |
0612 | Problem:
Soit $c$ un entier positif ou nul. Trouver toutes les suites d'entiers strictement positifs $a_{1}, a_{2}, \ldots$ telles que pour tout entier strictement positif $n$, $a_{n}$ soit égal au nombre d'entiers $i$ vérifiant $a_{i} \leqslant a_{n+1}+c$. | [
"Solution:\n\nSoit $(a_{n})$ une suite vérifiant les conditions de l'énoncé pour la constante $c$. On commence par montrer que la suite est croissante. En effet, supposons par l'absurde qu'il existe un entier $n$ tel que $a_{n}>a_{n+1}$. Alors d'après l'hypothèse de l'énoncé, il y a strictement plus de $a_{i}$ infé... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a_n = n + c + 1 for all n | |
0g38 | Problem:
Wir betrachten ein weisses $5 \times 5$-Quadrat bestehend aus 25 Einheitsquadraten. Wie viele verschiedene Möglichkeiten gibt es, eines oder mehrere der Einheitsquadrate schwarz anzumalen, sodass die resultierende schwarze Fläche ein Rechteck bildet? | [
"Solution:\n\nLösung 1 (Tanish), Gegenüberliegende Ecken zählen:\nWir betrachten die Ecken eines beliebigen Rechtecks. Diese liegen in einem Raster bestehend aus 36 Punkten in einem $6 \\times 6$ Quadrat. Wir wählen einen dieser Punkte als den ersten Eckpunkt. Wenn wir nun den gegenüberliegenden Eckpunkt wählen, ha... | Switzerland | Vorrunde | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > S... | null | proof and answer | 225 | |
01sm | Let $\triangle ABC$ be an acute triangle with the orthocenter $H$. Let $D$ be the point such that $HABD$ is a parallelogram ($AB \parallel HD$, $AH \parallel BD$). Let $E$ be the point on the line $DH$ such that $AC$ bisects $HE$. The line $AC$ meets the circumcircle of the triangle $DCE$ at $C$ and $F$.
Prove that $EF... | [
"1. See IMO-2015 Shortlist, Problem G1."
] | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals ... | English | proof only | null | |
0i0b | Problem:
If $a$ and $b$ are randomly selected real numbers between $0$ and $1$, find the probability that the nearest integer to $\frac{a-b}{a+b}$ is odd. | [
"Solution:\n\nThe only reasonable way to do this problem is geometrically (yes, you can use integrals to find the areas of the triangles involved, but I don't consider that reasonable).\n\nFirst, let us find the points $(a, b)$ in the plane for which the nearest integer to $\\frac{a-b}{a+b}$ is $0$, i.e. $-\\frac{1... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1/3 | |
08ch | Problem:
Una bottiglia a forma di cono poggia sulla sua base. Viene riempita d'acqua finché il livello del liquido non raggiunge $8$ centimetri misurati in verticale sotto il vertice del cono. Se ora si capovolge la bottiglia, senza cambiare la quantità di acqua al suo interno, lo spazio che rimane vuoto nella parte s... | [] | Italy | Olimpiade Italiana di Matematica | [
"Geometry > Solid Geometry > Volume"
] | null | proof and answer | 1 + sqrt(85) centimeters | |
024h | Problem:
Número de retas - Sabemos que dois pontos distintos determinam uma única reta. Quantas retas são determinadas por dois quaisquer dos nove pontos marcados no quadriculado dado? | [
"Solution:\n\nSolução 1: Para contar o número de retas, dividiremos as retas de acordo com suas posições.\n- retas paralelas aos lados dos quadrados: três horizontais e três verticais;\n\n- retas paralelas às diagonais dos quadrados: $3+3=6$;\n\n- outras retas: temo... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 20 | |
00ze | Problem:
The jury of an olympiad has 30 members in the beginning. Each member of the jury thinks that some of his colleagues are competent, while all the others are not, and these opinions do not change. At the beginning of every session a voting takes place, and those members who are not competent in the opinion of m... | [
"Solution:\n\nFirst we note that if nobody is excluded in some session, then the situation becomes stable and nobody can be excluded in any later session.\n\nWe use induction to prove the slightly more general claim that if the jury has $2 n$ members, $n \\geq 2$, then after at most $n$ sessions nobody will be excl... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0k52 | Problem:
Equilateral triangle $A B C$ has circumcircle $\Omega$. Points $D$ and $E$ are chosen on minor arcs $A B$ and $A C$ of $\Omega$ respectively such that $B C = D E$. Given that triangle $A B E$ has area $3$ and triangle $A C D$ has area $4$, find the area of triangle $A B C$. | [
"Solution:\n\nA rotation by $120^{\\circ}$ about the center of the circle will take $A B E$ to $B C D$, so $B C D$ has area $3$. Let $A D = x$, $B D = y$, and observe that $\\angle A D C = \\angle C D B = 60^{\\circ}$. By Ptolemy's Theorem, $C D = x + y$. We have\n$$\n\\begin{aligned}\n& 4 = [A C D] = \\frac{1}{2} ... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 37/7 | |
0h53 | Numbers $a$, $b$, $c$, $d$ satisfy: $ab + cd > 0$, $ac + bd > 0$, and $a^2 + d^2 = c^2 + b^2$. Prove that $ad + bc > 0$. | [
"$$(ab + cd)(ac + bd) = a^2bc + ac^2d + ab^2d + bcd^2 = bc(a^2 + d^2) + ad(c^2 + b^2) = bc(a^2 + d^2) + ad(a^2 + d^2) = (a^2 + d^2)(bc + ad) > 0,$$\nwhich implies the required inequality."
] | Ukraine | Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0lde | Let $ABC$ be an acute, scalene triangle with the circumcircle ($O$) and angles $ACB$, $ABC$ are acute. Let $M$ be a point on arc $BC$ that does not contain $A$ and $AM$ is not perpendicular to $BC$. The line $AM$ meets the perpendicular bisector of $BC$ at $T$ and the circumcircle of triangle $AOT$ meets $O$ at $N$ ($N... | [
"\n\na) Since $AOTN$ is a cyclic quadrilateral, $\\angle TNO = \\angle TAO = \\angle TMO$, hence\n$$\n\\angle TMN = \\angle OMN - \\angle OMT = \\angle ONM - \\angle ONT = \\angle TNM\n$$\nwhich means $TM = TN$ or $OT$ is a perpendicular bisector of $MN$. On the other hand, $OT$ is a perpen... | Vietnam | VMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0577 | Some numbers are printed on a spool of paper, one below another. Starting from the third number, each number is the sum of the two preceding numbers. The number $2018$ occurs fourth, the number $2020$ occurs eighth. Which are the first and the second numbers? | [
"Let the first and the second number be $x$ and $y$, respectively. Computing the following numbers, we see that the fourth number is $x + 2y$ and the eighth one is $8x + 13y$. The conditions of the problem imply the system of equations\n$$\n\\begin{cases}\nx + 2y = 2018, \\\\\n8x + 13y = 2020,\n\\end{cases}\n$$\nso... | Estonia | Open Contests | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | first number = -7398, second number = 4708 | |
0cop | The base of a pyramid $SABCD$ is a parallelogram $ABCD$. Prove that for each point inside $SABCD$, the sum of volumes of tetrahedra $OSAB$ and $OSCD$ is equal to the sum of volumes of tetrahedra $OSBC$ and $OSDA$.
В основании четырёхугольной пирамиды $SABCD$ лежит параллелограмм $ABCD$. Докажите, что для любой точки $... | [
"Let $X$ be the intersection point of the ray $SO$ with the plane $ABCD$ (see Fig. 10). Since the point $O$ lies inside the pyramid, the point $X$ lies inside its base. At the same time, $S_{XAB} + S_{XCD} = S_{XBC} + S_{XDA}$ (one possible proof of this fact can be seen from Fig. 11—each sum equals half the area o... | Russia | Regional round | [
"Geometry > Solid Geometry > Volume"
] | English; Russian | proof only | null | |
0gb1 | 坐標平面上有兩定點 $B(-1, 0)$, $C(1, 0)$。平面上的有界非空子集 $S$ 如果滿足下列兩條件：
(i) $S$ 中存在一點 $T$, 使得對 $S$ 中的任意點 $Q$, 線段 $TQ$ 會完全落在 $S$ 裡;
(ii) 對任意的三角形 $P_1P_2P_3$, 都能夠在 $S$ 中找到唯一的一點 $A$ 及集合 $\{1, 2, 3\}$ 上的一個排列 $\sigma$, 使得三角形 $ABC$ 與 $P_{\sigma(1)}P_{\sigma(2)}P_{\sigma(3)}$ 相似。
則稱 $S$ 為一個「好集合」。
證明在集合 $\{(x, y): x \ge 0, y \ge 0\}$ 中, 存在兩... | [
"如果在 $\\triangle ABC$ 與 $\\triangle P_{\\sigma(1)}P_{\\sigma(2)}P_{\\sigma(3)}$ 的相似關係中, $BC$ 對應到 $\\triangle P_{\\sigma(1)}P_{\\sigma(2)}P_{\\sigma(3)}$ 的最長邊, 則有 $BC \\ge AB \\ge AC$。條件 $BC \\ge AB$ 等價於 $(x + 1)^2 + y^2 \\le 4$, 而 $AB \\ge AC$ 對第一象限的任意點均成立。故可首先定義:\n$$\nS := \\{(x, y) : (x + 1)^2 + y^2 \\le 4, x \\g... | Taiwan | 二〇一七數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
0jnu | Problem:
Determine the number of unordered triples of distinct points in the $4 \times 4 \times 4$ lattice grid $\{0,1,2,3\}^{3}$ that are collinear in $\mathbb{R}^{3}$ (i.e. there exists a line passing through the three points). | [
"Solution:\nAnswer: 376\n\nDefine a main plane to be one of the $xy$, $yz$, $zx$ planes. Define a space diagonal to be a set of collinear points not parallel to a main plane. We classify the lines as follows:\n\n(a) Lines parallel to two axes (i.e. orthogonal to a main plane). Notice that given a plane of the form ... | United States | HMMT February | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 376 | |
08ag | Problem:
Per ogni intero positivo $n$, sia $D_{n}$ il massimo comune divisore di tutti i numeri della forma $a^{n} + (a+1)^{n} + (a+2)^{n}$ al variare di $a$ fra tutti gli interi positivi.
a. Dimostrare che, per ogni $n$, $D_{n}$ è della forma $3^{k}$ per qualche intero $k \geq 0$.
b. Dimostrare che, per ogni $k \ge... | [
"Solution:\n\nPer definizione di $D_{n}$, prendendo $a = D_{n}$ vediamo che $D_{n}$ divide $D_{n}^{n} + (D_{n} + 1)^{n} + (D_{n} + 2)^{n}$, e prendendo $a = D_{n} + 1$ si ottiene anche che $D_{n}$ divide $(D_{n} + 1)^{n} + (D_{n} + 2)^{n} + (D_{n} + 3)^{n}$. Per differenza $D_{n}$ divide $((D_{n} + 1)^{n} + (D_{n} ... | Italy | Cesenatico | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Diophanti... | null | proof only | null | |
08uj | Prove that for positive real numbers $x, y, z$ the following inequality is satisfied:
$$
\frac{1+xy+xz}{(1+y+z)^2} + \frac{1+yz+yx}{(1+z+x)^2} + \frac{1+zx+zy}{(1+x+y)^2} \ge 1
$$ | [
"From the Cauchy-Schwarz Inequality we get\n$$\n\\left(1 + \\frac{y}{x} + \\frac{z}{x}\\right) \\left(1 + xy + xz\\right) \\geq \\left(1 + y + z\\right)^2\n$$\nMultiplying both sides of the inequality above by $\\frac{x}{(x+y+z)(1+y+z)^2}$, we obtain\n$$\n\\frac{1 + xy + xz}{(1 + y + z)^2} \\geq \\frac{x}{x + y + z... | Japan | Japan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0fio | Problem:
Dado un triángulo $ABC$, con baricentro $G$.
a) Prueba que para cualquier punto del plano $M$ se verifica:
$$
\overline{MA}^{2}+\overline{MB}^{2}+\overline{MC}^{2} \geq \overline{GA}^{2}+\overline{GB}^{2}+\overline{GC}^{2}
$$
obteniéndose la igualdad si y solamente si $M=G$.
b) Fijado un número $k>\overline{GA... | [
"Solution:\n\na) Del baricentro $G$ de $ABC$ sabemos que cumple:\n\nse tiene:\n$$\n\\overrightarrow{GA}+\\overrightarrow{GB}+\\overrightarrow{GC}=\\overrightarrow{0}\n$$\n$$\n\\begin{aligned}\n& \\overline{MA}^{2}=\\overrightarrow{MA}^{2}=(\\overrightarrow{MG}+\\overrightarrow{GA})^{2}=\\ov... | Spain | XXXV Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plan... | null | proof and answer | a) For all points M, MA^2 + MB^2 + MC^2 >= GA^2 + GB^2 + GC^2 with equality if and only if M = G. Equivalently, MA^2 + MB^2 + MC^2 = 3·MG^2 + (GA^2 + GB^2 + GC^2). b) The locus is the circle centered at G with radius r = sqrt((k − (GA^2 + GB^2 + GC^2))/3). | |
0jdv | Problem:
Let $S=\{1,2, \ldots, 2013\}$. Find the number of ordered triples $(A, B, C)$ of subsets of $S$ such that $A \subseteq B$ and $A \cup B \cup C=S$. | [
"Solution:\nAnswer: $5^{2013}$ OR $125^{671}$\nLet $n=2013$. Each of the $n$ elements can be independently placed in 5 spots: there are $2^{3}-1$ choices with element $x$ in at least one set, and we subtract the $2^{1}$ choices with element $x$ in set $A$ but not $B$. Specifying where the elements go uniquely deter... | United States | HMMT November 2013 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 5^2013 | |
0ahd | Let $n$ be a positive integer. Let $P_n = \{2^n, 2^{n-1} \cdot 3, 2^{n-2} \cdot 3^2, \dots, 3^n\}$. For each subset $X$ of $P_n$, we write $S_X$ for the sum of all elements of $X$, with the convention that $S_\emptyset = 0$ where $\emptyset$ is the empty set. Suppose that $y$ is a real number with $0 \le y \le 3^{n+1} ... | [
"Given $y$, we construct $Y$ algorithmically. Let $Y = \\emptyset$ and of course $S_\\emptyset = 0$. For $i=0$ to $m$, perform the following operation:\nIf $S_Y + 2^i 3^{m-i} \\le y$, then replace $Y$ by $Y \\cup \\{2^i 3^{m-i}\\}$.\nWhen this process is finished, we have a subset $Y$ of $P_m$ such that $S_Y \\le y... | North Macedonia | 29-th Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
08p6 | Problem:
Let $S_{n}$ be the sum of reciprocal values of non-zero digits of all positive integers up to (and including) $n$. For instance, $S_{13} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \fra... | [
"Solution:\nWe will first calculate $S_{999}$, then $S_{1999} - S_{999}$, and then $S_{2016} - S_{1999}$.\n\nWriting the integers from $1$ to $999$ as $001$ to $999$, adding eventually also $000$ (since $0$ digits actually do not matter), each digit appears exactly $100$ times in each position (as unit, ten, or hun... | JBMO | Junior Balkan Mathematics Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 7 | |
060i | Problem:
Montrer que pour tous les réels $x, y, z$, on a
$$
\frac{x^{2}-y^{2}}{2 x^{2}+1}+\frac{y^{2}-z^{2}}{2 y^{2}+1}+\frac{z^{2}-x^{2}}{2 z^{2}+1} \leqslant 0 .
$$ | [
"Solution:\n\nOn réécrit l'inégalité en\n$$\n\\frac{x^{2}}{2 x^{2}+1}+\\frac{y^{2}}{2 y^{2}+1}+\\frac{z^{2}}{2 z^{2}+1} \\leqslant \\frac{y^{2}}{2 x^{2}+1}+\\frac{z^{2}}{2 y^{2}+1}+\\frac{x^{2}}{2 z^{2}+1} .\n$$\nL'inégalité n'étant pas symétrique, on ne peut pas supposer sans perte de généralité que $x^{2} \\leqsl... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof only | null | |
0ftw | Problem:
Von $4n$ Punkten in einer Reihe sind $2n$ weiss und $2n$ schwarz gefärbt. Zeige, dass es $2n$ aufeinanderfolgende Punkte gibt, von denen genau $n$ weiss und $n$ schwarz sind. | [
"Solution:\n\nNummeriere die Punkte von links nach rechts mit $P_{1}, \\ldots, P_{4n}$. Für $1 \\leq k \\leq 2n+1$ sei $a_{k}$ die Anzahl weisser Punkte unter den $2n$ aufeinanderfolgenden $P_{k}, \\ldots, P_{k+2n-1}$. Wir müssen zeigen, dass $a_{k}=n$ ist für mindestens ein $k$.\n\nMan überlegt sich leicht, dass s... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0j1i | Problem:
There are 101 people participating in a Secret Santa gift exchange. As usual each person is randomly assigned another person for whom (s)he has to get a gift, such that each person gives and receives exactly one gift and no one gives a gift to themself. What is the probability that the first person neither gi... | [
"Solution:\n\n0.96039\n\nLet $D_{k}$ denote the number of derangements of $\\{1,2, \\ldots, k\\}$. (A derangement is a permutation in which no element appears in its original position.)\n\nCall the first three people $A$, $B$, and $C$. Let $X \\rightarrow Y$ denote that $X$ gives a gift to $Y$ and let $X \\nrightar... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 0.96039 | |
0aqv | Problem:
Let $s_{1} = 2^{2010}$. For $n > 2$, define
$$
s_{n+1} = \begin{cases}
\log_{\sqrt{2}} s_{n}, & s_{n} > 0 \\
0, & s_{n} \leq 0
\end{cases}
$$
Find the smallest $n$ such that $s_{n} \in [4,6]$. | [] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 6 | |
0eap | For a positive integer $n \ge 2$ let $\tau(n)$ denote the number of positive divisors of $n$, and let $\sigma(n)$ be the sum of these divisors. Prove that
$$
\frac{n+1}{2} \ge \frac{\sigma(n)}{\tau(n)} \ge \frac{3}{2\sqrt{2}}\sqrt{n}.
$$
For each of these inequalities also determine when the equality holds. | [
"First consider the right inequality. Let $n = p_1^{k_1} p_2^{k_2} \\cdots p_s^{k_s}$, (where $p_1, \\dots, p_s$ are distinct primes and $k_1, \\dots, k_s$ are positive integers). Then\n$$\n\\tau(n) = (k_1 + 1)(k_2 + 1) \\cdots (k_s + 1)\n$$\n$$\n\\begin{align*}\n\\sigma(n) &= \\left(1 + p_1 + \\cdots + p_1^{k_1}\\... | Slovenia | National Math Olympiad in Slovenia | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Right inequality: equality holds only for n equal to two. Left inequality: equality holds exactly when n is prime. | |
0idv | Problem:
Suppose $S$ is a finite set with a binary operation $\star$ - that is, for any elements $a, b$ of $S$, there is defined an element $a \star b$ of $S$. It is given that $(a \star b) \star(a \star b)=b \star a$ for all $a, b \in S$.
Prove that $a \star b=b \star a$ for all $a, b \in S$. | [
"Solution:\n\nWe have\n$$\n\\begin{aligned}\na \\star b & =(b \\star a) \\star(b \\star a) \\\\\n& =([a \\star b] \\star[a \\star b]) \\star([a \\star b] \\star[a \\star b]) \\\\\n& =[a \\star b] \\star[a \\star b] \\\\\n& =b \\star a .\n\\end{aligned}\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Abstract Algebra > Other"
] | null | proof only | null | |
0hl2 | Problem:
How many distinct sets of 8 positive odd integers sum to 20? | [
"Solution:\n\nThis is the same as the number of ways 8 nonnegative even integers sum to 12 (we subtract 1 from each integer in the above sum). All 11 possibilities are (leaving out 0s): $12$, $10+2$, $8+4$, $8+2+2$, $6+6$, $6+4+2$, $6+2+2+2+2$, $4+4+4$, $4+4+2+2$, $4+2+2+2+2$, $2+2+2+2+2+2$."
] | United States | null | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 11 | |
0dix | Let $ABC$ be an acute, non-isosceles triangle with circumcenter $O$, orthocenter $H$. Denote $d$ as any diameter of $(O)$ and $A'$ as the midpoint of $BC$. Let $A_1$ be the projection of $A$ on $d$, $A_2$ be the reflection of $A_1$ over the perpendicular bisector of $BC$ and $A_3$ be the midpoint of $A'A_2$. Define $B_... | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Translat... | English | proof and answer | 1/2 | |
0fyp | Problem:
Sei $ABC$ ein spitzwinkliges Dreieck mit Höhenschnittpunkt $H$. Eine Gerade durch $H$ schneide $AB$ bzw. $AC$ in den Punkten $D$ bzw. $E$, so dass $|AD| = |AE|$ gilt. Die Winkelhalbierende von $\angle BAC$ schneide den Umkreis von $ADE$ im Punkt $K \neq A$. Zeige, dass $HK$ die Strecke $BC$ halbiert. | [
"Solution:\n\nSeien $B_{1}$ und $B_{2}$ die Schnittpunkte der Geraden $AC$ und $DK$ mit $BH$. Seien $C_{1}$ und $C_{2}$ die Schnittpunkte der Geraden $AB$ und $EK$ mit $CH$. Wegen der Voraussetzung $|AD| = |EA|$ ist das Dreieck $EAD$ gleichschenklig und somit steht $AK$ senkrecht auf $DE$. Da $ADKE$ ein Sehnenviere... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous ... | null | proof only | null | |
01f5 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(x^3 + y^3) = f(x^3) + 3x^3f(x)f(y) + 3f(x)(f(y))^2 + y^6f(y)
$$
for every two real numbers $x$ and $y$. | [
"Answer: $f(x) = 0$ and $f(x) = x^3$.\n\nSubstituting $x = y = 0$ into the original equation, we obtain $3(f(0))^3 = 0$ which implies\n$$\nf(0) = 0. \\qquad (\\text{Eq-1})\n$$\nSubstituting $x = 0$ into the original equation and applying (Eq-1), we get\n$$\nf(y^3) = y^6 f(y). \\qquad (\\text{Eq-2})\n$$\nReplacing t... | Baltic Way | Baltic Way 2019 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = 0 for all real x; f(x) = x^3 for all real x | |
0iu6 | Problem:
A kite is a quadrilateral whose diagonals are perpendicular. Let kite $ABCD$ be such that $\angle B = \angle D = 90^{\circ}$. Let $M$ and $N$ be the points of tangency of the incircle of $ABCD$ to $AB$ and $BC$ respectively. Let $\omega$ be the circle centered at $C$ and tangent to $AB$ and $AD$. Construct ano... | [
"Solution:\nLet's focus on the right triangle $ABC$ and the semicircle inscribed in it since the situation is symmetric about $AC$. First we find the radius $a$ of circle $O$. Let $AB = x$ and $BC = y$. Drawing the radii $OM$ and $ON$, we see that $AM = x - a$ and $\\triangle AMO \\sim \\triangle ABC$. In other wor... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | (1 + sqrt(5)) / 2 | |
05t1 | Problem:
Dans le train, alors qu'elles rentrent de EGMOnd an Zee, Clara et Edwige jouent au jeu suivant. Initialement, l'entier $n=1 \times 2 \times \cdots \times 20$ est écrit sur une feuille de papier. Puis, chacune à son tour, et en commençant par Clara, les joueuses remplacent l'entier $n$ par un des nombres $k n ... | [
"Solution:\n\nIntéressons-nous au nombre $n$ écrit sur la feuille de papier au moment où la joueuse X s'apprête à jouer, et avant que la partie ne se termine. On factorise partiellement $n$ comme produit de nombres premiers : $n=2^{x} \\times 5^{y} \\times m$, où $x$ et $y$ sont des entiers naturels et $m$ est un e... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Edwige | |
0j5k | Problem:
Let $f:(0,1) \rightarrow (0,1)$ be a differentiable function with a continuous derivative such that for every positive integer $n$ and odd positive integer $a < 2^{n}$, there exists an odd positive integer $b < 2^{n}$ such that $f\left(\frac{a}{2^{n}}\right) = \frac{b}{2^{n}}$. Determine the set of possible va... | [
"Solution:\nAnswer: $\\{-1,1\\}$\n\nThe key step is to notice that for such a function $f$, $f^{\\prime}(x) \\neq 0$ for any $x$.\n\nAssume, for sake of contradiction, that there exists $0 < y < 1$ such that $f^{\\prime}(y) = 0$. Since $f^{\\prime}$ is a continuous function, there is some small interval $(c, d)$ co... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | {-1, 1} | |
0kkb | The expression $\frac{2021}{2020} - \frac{2020}{2021}$ is equal to the fraction $\frac{p}{q}$ in which $p$ and $q$ are positive integers whose greatest common divisor is 1. What is $p$?
(A) 1 (B) 9 (C) 2020 (D) 2021 (E) 4041 | [
"The given expression equals\n$$\n\\frac{2021}{2020} - \\frac{2020}{2021} = \\frac{2021^2 - 2020^2}{2020 \\cdot 2021}\n$$\n\n\\begin{aligned}\n&= \\frac{(2021 + 2020)(2021 - 2020)}{2020 \\cdot 2021} \\\\\n&= \\frac{4041}{2020 \\cdot 2021}.\n\\end{aligned}\n$$\nBecause $4041 - 2 \\cdot 2020 = 1$, it follows that $40... | United States | Fall 2021 AMC 10 B | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | MCQ | E | |
02i1 | Problem:
Num triângulo retângulo, definimos o cosseno de seus ângulos agudos $\alpha$ por: $\cos \alpha=\frac{\text{cateto adjacente}}{\text{hipotenusa}}=\frac{c}{a}$.
O triângulo retângulo da figura tem cateto $OA=1$. Escreva em ordem crescente os cossenos dos ângulos de $25^{\circ}, 41^{\... | [
"Solution:\n\nDe acordo com a definição de cosseno, temos: $\\cos 25^{\\circ}=\\frac{1}{OM}$, $\\cos 41^{\\circ}=\\frac{1}{ON}$ e $\\cos 58^{\\circ}=\\frac{1}{OB}$. Na figura, vemos que $OM<ON<OB$, logo $\\cos 58^{\\circ}<\\cos 41^{\\circ}<\\cos 25^{\\circ}$."
] | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | cos 58° < cos 41° < cos 25° | |
04lb | Let $\triangle ABC$ be an acute-angled triangle with $\angle BAC = 75^\circ$. Let $P$ be the midpoint of the side $\overline{BC}$, and let $M$ and $N$ be the feet of the altitudes from vertices $B$ and $C$ respectively. Determine the angle $\angle MPN$. | [] | Croatia | Mathematical competitions in Croatia | [
"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 > Trigonometry",
"Geometry > Plane Geometry > Analyt... | null | proof and answer | 30° | |
0e55 | For the real numbers $x$ and $\alpha$ we have
$$
x + \frac{1}{x} = 2 \cos \alpha .
$$
Prove that for all positive integers $n$
$$
x^n + \frac{1}{x^n} = 2 \cos(n \alpha) .
$$ | [
"Let us first consider the case $x > 0$. We have $(x - 1)^2 \\ge 0$ or $x^2 + 1 \\ge 2x$, and dividing by $x$ we get\n$$\nx + \\frac{1}{x} \\ge 2 \\ge 2 \\cos \\alpha = x + \\frac{1}{x},\n$$\nso $\\cos \\alpha = 1$ and $x+1/x = 2$. We conclude that $x = 1$. This implies $x^n+1/x^n = 1+1=2 = \\cos(n\\alpha)$.\n\nIf ... | Slovenia | National Math Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
094i | Problem:
Let $a$, $b$, $c$ and $d$ be positive real numbers with $a b c d = 1$. Prove that
$$
\frac{a b + 1}{a + 1} + \frac{b c + 1}{b + 1} + \frac{c d + 1}{c + 1} + \frac{d a + 1}{d + 1} \geq 4
$$
and determine all quadruples $(a, b, c, d)$ for which equality holds. | [
"Solution:\nBy assumption we have\n$$\n\\begin{aligned}\n\\frac{a b + 1}{a + 1} + \\frac{b c + 1}{b + 1} + \\frac{c d + 1}{c + 1} + \\frac{d a + 1}{d + 1} & = \\left(\\frac{a b + 1}{a + 1} + \\frac{c d + a b c d}{c + 1}\\right) + \\left(\\frac{b c + 1}{b + 1} + \\frac{d a + a b c d}{d + 1}\\right) = \\\\\n& = (a b ... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | The sum is at least 4, with equality if and only if (a, b, c, d) = (t, 1/t, t, 1/t) for some positive real t. | |
01hq | For which values of $n$ does there exist a circumscribed $n$-gon with side lengths $1, 2, \dots, n$ (in any order)? | [
"It exists if $n = 4k$ or $n = 4k + 1$ where $k$ is a positive integer.\n\nLet us consider $n$-gon $P_1P_2 \\dots P_n$. Tangent points of the inscribed circle divide each of its sides in two segments. Lengths of these segments that have a common vertex $P_i$ are equal. Denote the length of tangent segments that ori... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | n = 4k or n = 4k + 1 for positive integers k | |
0eki | Problem:
V posodi je $60 \%$ rdečih in $40 \%$ modrih bonbonov. $30 \%$ rdečih in $15 \%$ modrih bonbonov je čokoladnih. Koliko procentov bonbonov ni čokoladnih?
(A) $24 \%$
(B) $85 \%$
(C) $90 \%$
(D) $76 \%$
(E) $45 \%$ | [
"Solution:\n\nRdečih čokoladnih je $0,6 \\cdot 0,3 = 0,18$, kar je $18 \\%$. Modrih čokoladnih je $0,4 \\cdot 0,15 = 0,06$, kar je $6 \\%$. Vseh čokoladnih je vsota, torej $24 \\%$ in tistih, ki niso čokoladni $76 \\%$. Pravilen je odgovor $D$."
] | Slovenia | 22. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | D | |
00ky | Let $ABC$ be an isosceles triangle with $AC = BC$ and $\angle ACB < 60^\circ$. We denote the incenter and circumcenter by $I$ and $O$, respectively. The circumcircle of triangle $BIO$ intersects the leg $BC$ also at point $D \neq B$.
a. Prove that the lines $AC$ and $DI$ are parallel.
b. Prove that the lines $OD$ and... | [
"Note that the condition $\\angle ACB < 60^\\circ$ guarantees that $O$ lies between $I$ and $C$.\n\na.\nWe denote the angles of triangle $ABC$ by $\\alpha = \\angle BAC$, $\\beta = \\angle ABC$ and $\\gamma = \\angle ACB$. Let $K$ and $k$ be the circumcircles of $ABC$ and $BIO$, respectively. The inscribed angle th... | Austria | Austrian 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 only | null | |
09oh | Let $I$ be the incenter of scalene triangle $ABC$ inscribed in circle $\omega$. The line $BI$ intersects $\omega$ again at point $M$. Let $I_B$ be the reflection of point $I$ across line $AC$. Let $H$ be a point on the minor arc $AM$ of $\omega$ such that $\angle BHI_B = 90^\circ$. Let $T$ be the intersection point of ... | [
"Let $\\omega$ be a circle with center $O$, and let $MI_B \\cap \\omega = D$. Also, draw diameters $BP$ and $DE$ in the circle $\\omega$.\n\n**Lemma.** *Triangle IBD is isosceles.*\n\n*Proof.* Let $F$ be the intersection of lines $DM$ and $AC$, and let $G$ be the reflection of $M$ across $AC$. By the Incenter-Excen... | Mongolia | MMO2025 Round 4 | [
"Geometry > Plane Geometry > Transformations > Inversion",
"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 | |
0dll | 100 stones are placed next to each other in a row: black, white, black, white, ..., black, white. In each move, it is allowed to choose two black stones between which all stones are white and paint those white stones black, or choose two white stones with all stones between them black and paint those black stones white... | [] | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
0ifc | Problem:
An ant starts at one vertex of a tetrahedron. Each minute it walks along a random edge to an adjacent vertex. What is the probability that after one hour the ant winds up at the same vertex it started at? | [
"Solution:\n\nLet $p_n$ be the probability that the ant is at the original vertex after $n$ minutes; then $p_0 = 1$. The chance that the ant is at each of the other three vertices after $n$ minutes is $\\frac{1}{3}(1 - p_n)$. Since the ant can only walk to the original vertex from one of the three others, and at ea... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | (3^59 + 1)/(4 * 3^59) | |
0ddv | Find all positive integer $n \ge 3$ such that it is possible to mark the vertices of a regular $n$-gon with the numbers from $1$ to $n$ so that for any three vertices $A$, $B$ and $C$ with $AB = AC$, the number in $A$ is greater or smaller than both numbers in $B$, $C$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | n = 2^k for integers k ≥ 2 | |
0l3i | In the figure below $WXYZ$ is a rectangle with $WX = 4$ and $WZ = 8$. Point $M$ lies on $\overline{XY}$, point $A$ lies on $\overline{YZ}$, and $\angle WMA$ is a right angle. The areas of triangles $\triangle WXM$ and $\triangle WAZ$ are equal. What is the area of $\triangle WMA$?
(A) 13 (B) 1... | [
"**Answer (C):** Label the diagram as shown, where $MX = a$ and $ZA = b$.\n\nThe Pythagorean Theorem on $\\triangle WMA$ gives $WM^2 + MA^2 = WA^2$, which implies that\n$$\n4^2 + a^2 + (8-a)^2 + (4-b)^2 = 8^2 + b^2.\n$$\nExpanding and simplifying yields $a^2 - 8a - 4b + 16 = 0$. Because the... | United States | 2024 AMC 12 B | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | C | |
07q4 | A sequence $(u_n)_{n=0}^{\infty}$ is defined recursively as follows: $u_0 = 2$ and
$$
u_n = \frac{2u_{n-1} + 1}{u_{n-1} + 2}, \text{ for } n = 1, 2, 3, \dots$$
So, for example, $u_1 = \frac{5}{4}$, $u_2 = \frac{14}{13}$. Prove that for all positive integers $n$,
$$1 < u_n < 1 + \frac{2}{3^n}.$$ | [
"Introduce a new sequence $v_n = u_n - 1$. We have $v_0 = 1$ and the given recursion translates into\n$$\nv_n + 1 = \\frac{2v_{n-1} + 3}{v_{n-1} + 3}, \\text{ i.e. } v_n = \\frac{v_{n-1}}{v_{n-1} + 3} = 1 - \\frac{3}{v_{n-1} + 3}.\n$$\nWe have to prove that $0 < v_n < \\frac{2}{3^n}$. As $v_0 = 1 > 0$, the recursio... | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0jmb | Problem:
Find the maximum possible number of diagonals of equal length in a convex hexagon. | [
"Solution:\n\n7\n\nFirst, we will prove that 7 is possible. Consider the following hexagon $A B C D E F$ whose vertices are located at $A(0,0)$, $B\\left(\\frac{1}{2}, 1-\\frac{\\sqrt{3}}{2}\\right)$, $C\\left(\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right)$, $D(0,1)$, $E\\left(-\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 7 | |
0cpe | Nonzero real numbers $a$, $b$, $c$ are chosen so that each two of three equations $a x^{11} + b x^4 + c = 0$, $b x^{11} + c x^4 + a = 0$, $c x^{11} + a x^4 + b = 0$ share a common root. Prove that all three equations share a common root.
Ненулевые числа $a$, $b$, $c$ таковы, что любые два из трёх уравнений $a x^{11} +... | [
"Заметим сразу, что все корни наших уравнений — ненулевые, поскольку свободные члены не равны нулю.\n\nПусть $p$ — общий корень первых двух уравнений. Тогда имеем\n$$\n\\begin{aligned}\n0 &= b(a p^{11} + b p^4 + c) - a(b p^{11} + c p^4 + a) = p^4 (b^2 - a c) - (a^2 - b c), \\\\\n0 &= b(b p^{11} + c p^4 + a) - c(a p... | Russia | Russian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English, Russian | proof only | null | |
0h4j | Three cyclists start off at the same time and ride along the sides of a triangle $ABC$ along the route $AB \to BC \to CA$. Their speeds on each of the segments $AB$, $BC$, $CA$ are known: the first cyclist has speeds $12$, $10$ and $20$ mph respectively on the three sides, the second one rides $15$, $15$ and $10$ mph, ... | [
"Denote the sides of the triangle by $AB = x$, $BC = y$, $CA = z$. Then the following equality must hold:\n$$\n\\frac{x}{12} + \\frac{y}{10} + \\frac{z}{20} = \\frac{x}{15} + \\frac{y}{15} + \\frac{z}{10} = \\frac{x}{10} + \\frac{y}{20} + \\frac{z}{12} \\text{ or } 5x + 6y + 3z = 4x + 4y + 6z = 6x + 3y + 5z.\n$$\nH... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Geometry > Plane Geometry > Triangles",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 60 degrees | |
06fd | Let $x$ and $y$ be real numbers satisfying $x - y + 1 \neq 0$. If
$$
1 + \cos^2(2007x + 2008y - 1) = \frac{x^2 + y^2 + 2(1+x)(1-y)}{x - y + 1},
$$
find the minimum value of $xy$. | [
"The minimum value of $xy$ is $\\frac{1}{16120225}$.\n\nNote that the left-hand side is positive. Therefore, by rewriting the right-hand side as\n$$\n(x - y + 1) + \\frac{1}{x - y + 1},\n$$\nwe know that $x - y + 1 > 0$. Thus, we can apply the AM-GM inequality to obtain\n$$\n(x - y + 1) + \\frac{1}{x - y + 1} \\ge ... | Hong Kong | Year 2008 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 1/16120225 | |
0kjg | Problem:
Two distinct squares on a $4 \times 4$ chessboard are chosen, with each pair of squares equally likely to be chosen. A knight is placed on one of the squares. The expected value of the minimum number of moves it takes for the knight to reach the other square can be written as $\frac{m}{n}$, where $m, n$ are p... | [
"Solution:\n\nWe can do casework based on the position of the knight: corner, edge, or center. In each case, we can quickly compute all 15 distances by writing a 1 down in all squares reachable from the original square, then writing a 2 down in all blank squares reachable from a square with a 1, writing a 3 down in... | United States | HMMT November 2021 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 1205 | |
00zk | Problem:
Let $a, b, c, d$ be positive integers such that $a b = c d$. Prove that $a + b + c + d$ is not prime. | [
"Solution:\nAs $a b = c d$, we get $a(a + b + c + d) = (a + c)(a + d)$. If $a + b + c + d$ were a prime, then it would be a factor in either $a + c$ or $a + d$, which are both smaller than $a + b + c + d$.",
"Solution:\nLet $r = \\operatorname{gcd}(a, c)$ and $s = \\operatorname{gcd}(b, d)$. Let $a = a' r$, $b = ... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0ho2 | Problem:
Let $x$, $y$, and $z$ be real numbers.
a. Prove that
$$
x + y > |x - y|
$$
if and only if $x$ and $y$ are both positive.
b. Find an inequality involving the variables $x$, $y$, and $z$, using only the operations of addition, subtraction, multiplication and absolute value, that is true if and only if $x$, $y... | [
"Solution:\n\nNote that $|x-y|$ is the greater of $x-y$ and $-(x-y)$. So the above assertion is equivalent to\n$$\n\\begin{aligned}\n& x+y > x-y \\quad \\text{ and } \\quad x+y > -x+y \\\\\n& 2y > 0 \\quad \\text{ and } \\quad 2x > 0 \\\\\n& y > 0 \\quad \\text{ and } \\quad x > 0.\n\\end{aligned}\n$$\n\n\nSolution... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0c1g | Let $ABC$ be an acute triangle with $AB < AC$, let $G$ be its centroid and $D$ the foot of the altitude from $A$. The line $DG$ meets the small arc $BC$ of the circumcircle of triangle $ABC$ at point $E$. Prove that the line $AB$ is tangent to the circumcircle of triangle $BDE$. | [
"Let $F$ be the point in which the parallel through $A$ to $BC$ intersects again the circumcircle of triangle $ABC$. We prove that the points $D$, $G$ and $F$ are collinear. $AFCB$ is a cyclic trapezoid, hence a cyclic one. If $T$ is the orthogonal projection of point $F$ onto $BC$, then $AFTD$ is a rectangle. It i... | Romania | 69th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Colline... | null | proof only | null | |
0aot | Problem:
A circle is inscribed in $\triangle ABC$ with sides $AB = 4$, $BC = 6$, and $AC = 8$. If $P$ and $Q$ are the respective points of tangency of $\overline{AB}$ and $\overline{AC}$ with the circle, determine the length of chord $PQ$. | [
"Solution:\n$\\frac{3 \\sqrt{10}}{4}$\n\nApplying the Law of Cosines to $\\triangle ABC$, we get\n$$\n\\cos A = \\frac{4^{2} + 8^{2} - 6^{2}}{2 \\cdot 4 \\cdot 8} = \\frac{11}{16}\n$$\nLet $AP = AQ = x$, $PB = y$, and $QC = z$. Then we have the following system of equations:\n$$\n\\left\\{\\begin{array}{l}\nx + y =... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 3*sqrt(10)/4 | |
0gw4 | Mykolka and Andriyko are playing the following game. They write positive integers turn by turn thus forming a sequence $a_1, a_2, ..., a_{2006}$ obeying the following restrictions: $a_1 = 1$ (this first turn is fixed, and it is made by Mykolka), and $a_n \le a_{n+1} \le 3a_n$ for $1 \le n \le 2005$. If, after the last ... | [
"Андрійко може забезпечити собі перемогу. Для доведення досить показати, що Андрійко зможе записати число $a_{2006} = M - 1$, де $M = \\sum_{k=1}^{2005} a_k$. Очевидно, що $M - 1 \\ge a_{2005}$. Доведемо, що Андрійко може забезпечити й виконання нерівності $a_{2006} = M - 1 \\le 3a_{2005}$. Нехай Миколка своїм черг... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Andriyko | |
0agl | A total of $2010$ coins are arranged in $5$ boxes. In the beginning the numbers of coins in the boxes are consecutive numbers. Martha has to choose and take one of the boxes, but previously she is allowed to do the following transformation finitely many times: from a box in which there are at least four coins to move o... | [
"That number of coins can be achieved by the following procedure: In every move if the box that is second by number of coins there are at least four coins, we move one coin to every other box, and this transformation is not possible only in the moment when the arrangement in the boxes is $0, 1, 2, 3, 2004$ coins se... | North Macedonia | Macedonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 2004 | |
0jze | Problem:
Show that for positive real numbers $a$, $b$, and $c$,
$$
\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \geq \frac{(a+b+c)^{2}}{a b(a+b)+b c(b+c)+c a(c+a)}
$$ | [
"Solution:\nExpanding and rearranging the denominator gives\n$$\na^{2} b + a b^{2} + b^{2} c + b c^{2} + c^{2} a + a c = c^{2}(a+b) + a^{2}(b+c) + b^{2}(c+a).\n$$\nBy Cauchy-Schwarz,\n$$\n\\left(c^{2}(a+b) + a^{2}(b+c) + b^{2}(c+a)\\right)\\left(\\frac{1}{a+b} + \\frac{1}{b+c} + \\frac{1}{c+a}\\right) \\geq (a+b+c)... | United States | Berkeley Math Circle | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0cc2 | Let $\triangle ABC$ be a triangle and let $P$ and $Q$ be points on sides $AB$ and $AC$, respectively, such that $AP = AQ$ and line $PQ$ passes through the incenter $I$ of triangle $ABC$. Let $M$ be the second intersection point of the circumcircles of triangles $BPI$ and $CQI$. The lines $PM$ and $BI$ intersect at $D$ ... | [
"Since triangle $APQ$ is isosceles, we have $\\angle APQ = \\angle AQP$. $BMIP$ and $CMIQ$ are cyclic quadrilaterals, so $\\angle APQ = \\angle BMI$ and $\\angle AQP = \\angle CMI$. It follows that $\\angle BMI = \\angle CMI$. (1).\n\nLet $K$ be the point where lines $IM$ and $BC$ meet. Since $BI$ is the bisector o... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - FIRST SELECTION TEST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscel... | null | proof only | null | |
0boz | Is it possible to partition the set of the positive integers into two subsets such that none of them contains an infinitely long (non-constant) arithmetic progression? | [
"The answer is in the affirmative. For example, one can consider\n$$A = \\{1, 3, 4, 7, 8, 9, 13, 14, 15, 16, 21, \\dots\\},$$\n$$B = \\{2, 5, 6, 10, 11, 12, 17, 18, 19, 20, 26, \\dots\\},$$\ni.e., for the infinite word $W = w_1w_2\\cdots w_k \\cdots = aba^2b^2\\cdots a^nb^n\\cdots$, put $A$ the set of the indices $... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Algebraic Number Theory > Combinatorial number theory: van der Waerden's theorem",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Yes | |
04et | Sequence $(a_n)$ is defined recursively: $a_1 = 2$, $a_n = 2(n + a_{n-1})$ for $n \ge 2$. Prove that $a_n < 2^{n+2}$ for all $n \in \mathbb{N}$. | [
"First few elements of the sequence are\n$a_1 = 2$, $a_2 = 2(2+2) = 8$, $a_3 = 2(3+8) = 22$, $a_4 = 2(4+22) = 52$, $\\dots$\nLet us look at the differences $2^{n+2} - a_n$:\n$$\n\\begin{align*}\n2^3 - a_1 &= 8 - 2 = 6, \\\\\n2^4 - a_2 &= 16 - 8 = 8, \\\\\n2^5 - a_3 &= 32 - 22 = 10, \\\\\n2^6 - a_4 &= 64 - 52 = 1... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0c4f | Problem:
Fie $n$ şi $q$ două numere naturale, $n \geq 2$, $q \geq 2$ şi $q \not \equiv 1(\bmod 4)$, şi fie $K$ un corp finit care are exact $q$ elemente. Arătaţi că, oricare ar fi elementul $a$ din $K$, există $x$ şi $y$ în $K$, astfel încât $a = x^{2^{n}} + y^{2^{n}}$. (Orice corp finit este comutativ.) | [
"Solution:\nFie $p$ caracteristica lui $K$. Atunci $p$ este prim şi $q = p^{\\alpha}$, unde $\\alpha$ este un număr natural nenul. Cum $q \\not \\equiv 1(\\bmod 4)$, rezultă că şi $p \\not \\equiv 1(\\bmod 4)$, deci $p = 2$ sau $p \\equiv 3$ $(\\bmod 4)$ şi, în acest caz, $\\alpha$ este impar.\n\nDacă $p = 2$, iar ... | Romania | Olimpiada Naţională de Matematică Etapa Judeţeană şi a Municipiului Bucureşti | [
"Algebra > Abstract Algebra > Field Theory",
"Algebra > Abstract Algebra > Group Theory",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0gqe | Some cities of a country consisting of $n$ cities are connected by round trip flights so that there are at least $k$ flights from any city and any city is reachable from any city. Prove that for any such flight organization these flights can be distributed among $n-k$ air companies so that one can reach any city from a... | [
"The problem can be reformulated in terms of graph theory: Let $G$ be a connected graph with $n$ vertices. If the degree of each vertex is at least $k$, then the edges of $G$ can be colored into $n-k$ colors so that for any pair of vertices there is a path between them not containing identically colored edges. We w... | Turkey | Team Selection Test for IMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0jly | Problem:
Let $\mathcal{P}_1$, $\mathcal{P}_2$, $\mathcal{P}_3$ be pairwise distinct parabolas in the plane. Find the maximum possible number of intersections between two or more of the $\mathcal{P}_i$. In other words, find the maximum number of points that can lie on two or more of the parabolas $\mathcal{P}_1$, $\mat... | [
"Solution:\n\n12\n\nNote that two distinct parabolas intersect in at most 4 points, which is not difficult to see by drawing examples. Given three parabolas, each pair intersects in at most 4 points, for at most $4 \\cdot 3 = 12$ points of intersection in total. It is easy to draw an example achieving this maximum,... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 12 | |
05ev | Problem:
Soient $d_{1}$, $d_{2}$, $d_{3}$ des droites concourantes et $A, A^{\prime}$ des points sur la droite $d_{1}$, $B, B^{\prime}$ des points sur la droite $d_{2}$, $C, C^{\prime}$ des points sur la droite $d_{3}$ tels que les droites $(AB)$ et $(A^{\prime}B^{\prime})$ sont parallèles et les droites $(BC)$ et $(B... | [
"Solution:\n\nSoit $S$ le point de concours des trois droites. Puisque les droites $(AB)$ et $(A^{\\prime}B^{\\prime})$ sont parallèles, d'après le théorème de Thalès,\n$$\n\\frac{AS}{A^{\\prime}S} = \\frac{BS}{B^{\\prime}S}\n$$\nDe même on trouve\n$$\n\\frac{BS}{B^{\\prime}S} = \\frac{CS}{C^{\\prime}S}\n$$\nOn déd... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ket | Problem:
A polynomial $P(x)$ is a base-$n$ polynomial if it is of the form $a_{d} x^{d} + a_{d-1} x^{d-1} + \cdots + a_{1} x + a_{0}$, where each $a_{i}$ is an integer between $0$ and $n-1$ inclusive and $a_{d} > 0$. Find the largest positive integer $n$ such that for any real number $c$, there exists at most one base-... | [
"Solution:\nIt is equivalent to determine the largest $n$ such that we cannot find two distinct base-$n$ polynomials $P_{1}$ and $P_{2}$ such that $P_{1}(\\sqrt{2} + \\sqrt{3}) = P_{2}(\\sqrt{2} + \\sqrt{3})$. The difference of two base-$n$ polynomials is a polynomial with integer coefficients whose absolute values... | United States | HMMT February 2020 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Algebraic Number Theory > Algebraic numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 9 | |
0j6f | Problem:
Three not necessarily distinct positive integers between $1$ and $99$, inclusive, are written in a row on a blackboard. Then, the numbers, without including any leading zeros, are concatenated to form a new integer $N$. For example, if the integers written, in order, are $25$, $6$, and $12$, then $N=25612$ (a... | [
"Solution:\n\nAnswer: $825957$\n\nWe will divide this into cases based on the number of digits of $N$.\n\n- Case 1: $6$ digits. Then each of the three numbers must have two digits, so we have $90$ choices for each. So we have a total of $90^{3} = 729000$ possibilities.\n\n- Case 2: $5$ digits. Then, exactly one of ... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 825957 | |
0bs7 | Let $ABCD$ be a cyclic quadrilateral, and let diagonals $AC$ and $BD$ intersect at $X$. Let $C_1$, $D_1$ and $M$ be the midpoints of segments $CX$, $DX$ and $CD$, respectively. Lines $AD_1$ and $BC_1$ intersect at $Y$, and line $MY$ intersects diagonals $AC$ and $BD$ at different points $E$ and $F$, respectively. Prove... | [
"\n\nWe are to prove that $\\angle EXY = \\angle EFX$; alternatively, but equivalently, $\\angle AYX + \\angle XAY = \\angle BYF + \\angle XBY$.\n\nSince the quadrangle $ABCD$ is cyclic, the triangles $XAD$ and $XBC$ are similar, and since $AD_1$ and $BC_1$ are corresponding medians in thes... | Romania | 2016 European Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
00k5 | Determine all solutions of the Diophantine equation
$$
a^2 = b \cdot (b+7)
$$
in integers $a \geq 0$ and $b \geq 0$. | [
"We have the trivial estimate $a^2 = b \\cdot (b+7) \\ge b^2$ resulting in $a \\ge b$ due to the non-negativity of $a$ and $b$. On the other hand, the inequality between the arithmetic and the geometric mean implies that\n$$\na = \\sqrt{b(b+7)} \\le \\frac{b+(b+7)}{2} = b + \\frac{7}{2}.\n$$\nCombining these inequa... | Austria | Austria 2014 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | (a, b) = (0, 0) and (12, 9) | |
0ays | Problem:
Let $S=\{1,2, \ldots, 2018\}$. For each subset $T$ of $S$, take the product of all elements of $T$, with $1$ being the product corresponding to the empty set. The sum of all such resulting products (with repetition) is $N$. Two elements $m$ and $n$ of $S$, with $m<n$, are removed. The sum of all products over ... | [] | Philippines | 21st PMO Area Stage | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1008 | |
0fld | Problem:
En un polígono regular de $67$ lados trazamos todos los segmentos que unen dos vértices, incluidos los lados del polígono. Elegimos $n$ de estos segmentos y asignamos a cada uno de ellos un color entre $10$ colores posibles. Halla el valor mínimo de $n$ que garantiza, que independientemente de cuáles sean los... | [
"Solution:\n\nVeamos en primer lugar que con $n=2010$ no es suficiente.\nDiremos que un segmento es de tamaño $r$ si une dos vértices entre los que, por el camino más corto siguiendo los lados del polígono, hay otros $r-1$ vértices. Elegimos los $2010$ segmentos de tamaño mayor que $3$. Para cada $r \\in \\{1,2, \\... | Spain | XLVII Olimpiada Matemática Española, Fase nacional (Pamplona) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2011 | |
06xc | For each $1 \leqslant i \leqslant 9$ and $T \in \mathbb{N}$, define $d_{i}(T)$ to be the total number of times the digit $i$ appears when all the multiples of $1829$ between $1$ and $T$ inclusive are written out in base $10$.
Show that there are infinitely many $T \in \mathbb{N}$ such that there are precisely two disti... | [
"Let $n := 1829$. First, we choose some $k$ such that $n \\mid 10^{k} - 1$. For instance, any multiple of $\\varphi(n)$ would work since $n$ is coprime to $10$. We will show that either $T = 10^{k} - 1$ or $T = 10^{k} - 2$ has the desired property, which completes the proof since $k$ can be taken to be arbitrarily ... | IMO | International Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof only | null | |
0hiv | Problem:
Let $k$ and $n$ be positive integers such that $k < 2^{n+1} - 1$. Prove that there is a sum of exactly $n$ powers of $2$ that is divisible by $k$. (Example: if $k = 9$ and $n = 4$, then $2 + 4 + 16 + 32$ is divisible by $9$.) | [
"Solution:\n\nWrite $k$ in base $2$ and suppose there are $r$ $1$'s. Since $2^{n+1} - 1$ is the smallest number with at least $n+1$ $1$'s, we conclude $r \\leq n$. Also $k$ clearly divides a sum of $r$ powers of $2$, since $k$ is such a sum. Now we claim that if $k$ divides a sum of $i$ powers of $2$, it also divid... | United States | Berkeley Math Circle Take-Home Contest #1 | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
02ks | Problem:
Equação cúbica - Sobre a equação $2007 x^{3}+2006 x^{2}+2005 x=0$ é certo afirmar:
(a) Não possui raízes
(b) Tem 3 raízes reais distintas
(c) Tem 2 raízes iguais
(d) Tem apenas uma raiz real
(e) Tem 3 raízes positivas | [
"Solution:\n\nObservemos que $x=0$ é uma solução, logo as possibilidades (a) e (e) ficam descartadas. Agora só precisamos estudar as soluções de $2007 x^{2}+2006 x+2005=0$, que é uma equação do $2^{o}$ grau com discriminante\n$$\n\\begin{aligned}\n\\Delta & =2006^{2}-4 \\times 2007 \\times 2005=2006^{2}-4(2006+1)(2... | Brazil | Nível 3 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | d | |
0dw4 | Problem:
Dano je pozitivno realno število $p$. Med vsemi pari pozitivnih realnih števil $(x, y)$, ki ustrezajo enačbi $x y(x+y)=p$, poišči tistega, za katerega je vrednost izraza $x^{3}+y^{3}$ najmanjša. | [
"Solution:\n\nKer je $x^{3}+y^{3}-p=x^{3}+y^{3}-x^{2} y-x y^{2}=x^{2}(x-y)-y^{2}(x-y)=(x-y)^{2}(x+y) \\geq 0$, je vrednost izraza zagotovo večja ali enaka $p$. Enakost je dosežena le za $x=y=\\left(\\frac{p}{2}\\right)^{\\frac{1}{3}}$."
] | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | x = y = (p/2)^{1/3} | |
0dfl | Given is triangle $ABC$ with $AB > AC$. Circles $o_B$, $o_C$ are inscribed in angle $BAC$ with $o_B$ tangent to $AB$ at $B$ and $o_C$ tangent to $AC$ at $C$. Tangent to $o_B$ from $C$ different than $AC$ intersects $AB$ at $K$, and tangent to $o_C$ from $B$ different than $AB$ intersects $AC$ at $L$. Line $KL$ and the ... | [
"Note that the length of the segment of the common tangent to $o_B$ and $o_C$ joining the tangency points is equal to\n$$\nAB - AC = LB - LC = KC - KB,\n$$\nSAUDI ARABIAN IMO Booklet 2022\n\nwhich means that points $A$, $L$ lie on the one, and point $K$ on the other leg of some hyperbola $\\eta$ with foci $B$, $C$.... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
078i | In triangle $ABC$ with $CA = CB$, point $E$ lies on the circumcircle of $ABC$ such that $\angle ECB = 90^\circ$. The line through $E$ parallel to $CB$ intersects $CA$ in $F$ and $AB$ in $G$. Prove that the centre of the circumcircle of triangle $EGB$ lies on the circumcircle of triangle $ECF$. | [
"\nWe have $FG = FA$ since $FG$ is parallel to $BC$. But also $\\triangle GAE$ is a right angle triangle. Thus, if $F'$ is the midpoint of $GE$, then $\\angle GAF = \\angle FGA = \\angle F'GA = \\angle GAF'$ which implies $F \\equiv F'$. Thus, $F$ is the midpoint of $GE$.\nIf $O$ is the cir... | India | INMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic... | null | proof only | null | |
0fw6 | Problem:
Im Dreieck $A B C$ sei $J$ der Mittelpunkt des Ankreises, welcher die Seite $B C$ in $A_{1}$ und die Verlängerungen der Seiten $A C$ und $A B$ in $B_{1}$ bzw. $C_{1}$ berührt. Die Gerade $A_{1} B_{1}$ schneide die Gerade $A B$ rechtwinklig in $D$. Sei $E$ die Projektion von $C_{1}$ auf die Gerade $D J$. Besti... | [
"Solution:\n\nWir zeigen zuerst, dass die Punkte $C_{1}, E, C$ auf einer Geraden liegen, indem wir $C_{1} C \\perp D J$ zeigen. Sei $K$ der Schnittpunkt von $J C$ und $A_{1} B_{1}$. Weil $C B_{1}$ und $C A_{1}$ tangential zum Ankreis liegen, gilt $J C \\perp A_{1} B_{1}$ und wegen $\\angle K D C_{1}=\\angle D C_{1}... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | angle BEA1 = 90°, angle AEB1 = 90° | |
0973 | Problem:
Aflați aria maximală posibilă a patrulaterului lungimile laturilor căruia sunt egale $1~\mathrm{cm}$, $2 \sqrt{2}~\mathrm{cm}$, $3~\mathrm{cm}$ și $4~\mathrm{cm}$. | [
"Solution:\n\nEvident este faptul că patrulaterul de arie maximală este patrulater convex.\nFie $ABCD$ un patrulater convex. Dacă modificăm configurația patrulaterului doar schimbând cu locurile două laturi vecine, atunci aria patrulaterului nu se schimbă. Deci putem considera că laturile lui sunt aranjate în ordin... | Moldova | Olimpiada Republicană la Matematică, A doua zi | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 2 + 3√2 | |
0azs | Problem:
A triangle with side lengths $24, 70, 74$ is inscribed in a circle. Find the difference between the numerical values of the area and the circumference of the circle in terms of $\pi$. | [
"Solution:\nNotice that the given triangle is a right triangle with hypotenuse equal to $74$ because $12^{2} + 35^{2} = 37^{2}$ which implies that $24^{2} + 70^{2} = 74^{2}$. From this, we know that the area and circumference of the circle are $37^{2} (\\pi)$ or $1369 \\pi$ and $74 \\pi$, respectively. Thus, the di... | Philippines | 20th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Number Theory > Diophantine Equations > Pythagorean triples"
] | null | final answer only | 1295π | |
0kbj | Problem:
A palindrome is a positive integer that reads the same forward and backward, like $2552$ or $1991$. Find a positive integer greater than $1$ that divides all four-digit palindromes. | [
"Solution:\nWe claim that $11$ divides all four-digit palindromes. Note that any four-digit palindrome $abba$ is the sum\n$$\nabba = a00a + bb0 = a \\times 1001 + b \\times 110.\n$$\nNow, $110$ is a multiple of $11$, since it is $11 \\times 10$, and $1001$ is a multiple of $11$, since it is $11 \\times 91$. Thus, $... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 11 | |
0a5g | Problem:
Let $ABC$ be an isosceles triangle with $AB = AC$. Point $D$ lies on side $AC$ such that $BD$ is the angle bisector of $\angle ABC$. Point $E$ lies on side $BC$ between $B$ and $C$ such that $BE = CD$. Prove that $DE$ is parallel to $AB$. | [
"Solution:\n\nLet $E'$ be the point on line $BC$ such that $DE'$ is parallel to $AB$. We know that $E'$ lies between $B$ and $C$ because $D$ lies between $A$ and $C$. So it suffices for us to prove that $BE' = CD$.\n\n\n\n$\\angle ACB = \\angle CBA = \\angle CE'D$.\n\nTherefore triangle $CD... | New Zealand | NZMO Round One | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0hyr | Problem:
In the $O$-$E$ game, a round starts with player $A$ paying $c$ cents to player $B$. Then $A$ secretly arranges the numbers $1, 3, 5, 7, 9, 11, 13$ in some order as a sequence $a_{1}, a_{2}, \ldots, a_{7}$, and $B$ secretly arranges $2, 4, 6, 8, 10, 12, 14$ as a sequence $b_{1}, b_{2}, \ldots, b_{7}$. Finally,... | [
"Solution:\n\nLet $k$ be in $X$. There are $7!\\,6!\\,(8-k)$ choices of the sequences $a_{1}, a_{2}, \\ldots, a_{7}$ and $b_{1}, b_{2}, \\ldots, b_{7}$ for which $2k-1$ is an $a_{j}$ with $a_{j}<b_{j}$. Indeed, the $a$'s can be any of the $7!$ permutations of the 7 odd integers; then $j$ is the subscript such that ... | United States | BAMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 4 | |
0f4q | Problem:
A real number is assigned to each unit square in an infinite sheet of squared paper. Show that some cell contains a number that is less than or equal to at least four of its eight neighbors. | [] | Soviet Union | 16th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
05sj | Problem:
Soit $a_{1}, a_{2}, \ldots$ la suite d'entiers telle que $a_{1}=1$ et, pour tout entier $n \geqslant 1$,
$$
a_{n+1}=a_{n}^{2}+a_{n}+1
$$
Démontrer, pour tout entier $n \geqslant 1$, que $a_{n}^{2}+1$ divise $a_{n+1}^{2}+1$. | [
"Solution:\n\nPour tout entier $n \\geqslant 1$, posons $b_{n}=a_{n}^{2}+1$. On conclut en constatant directement que\n$$\nb_{n+1} \\equiv\\left(a_{n}^{2}+a_{n}+1\\right)^{2}+1 \\equiv\\left(b_{n}+a_{n}\\right)^{2}+1 \\equiv a_{n}^{2}+1 \\equiv b_{n} \\equiv 0 \\quad\\left(\\bmod b_{n}\\right)\n$$\n\n\nSolution alt... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0kls | Problem:
Let $n$ be the product of the first 10 primes, and let
$$
S=\sum_{x y \mid n} \varphi(x) \cdot y
$$
where $\varphi(x)$ denotes the number of positive integers less than or equal to $x$ that are relatively prime to $x$, and the sum is taken over ordered pairs $(x, y)$ of positive integers for which $x y$ divide... | [
"Solution:\nWe see that, for any positive integer $n$,\n$$\nS=\\sum_{x y \\mid n} \\varphi(x) \\cdot y=\\sum_{x \\mid n} \\varphi(x)\\left(\\sum_{y \\left\\lvert\\, \\frac{n}{x}\\right.} y\\right)=\\sum_{x \\mid n} \\varphi(x) \\sigma\\left(\\frac{n}{x}\\right) .\n$$\nSince $\\varphi$ and $\\sigma$ are both weakly ... | United States | HMMT Spring 2021 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1024 | |
0drz | A total of $731$ objects are put into $n$ nonempty bags where $n$ is a positive integer. These bags can be distributed into $17$ red boxes and also into $43$ blue boxes so that each red and each blue box contain $43$ and $17$ objects, respectively. Find the minimum value of $n$. | [
"We shall prove this general with $p$ red boxes and $q$ blue boxes and $pq$ objects, with $p, q$ coprime. Suppose the bags have been distributed as described, with each red box containing $q$ objects and each blue box containing $q$ objects. Suppose we have a collection of $r$ red and $b$ blue boxes. We say that a ... | Singapore | Singapur | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 59 | |
066e | Find all ordered pairs $(a, b)$ of positive integers for which the numbers $\frac{a^3b - 1}{a + 1}$ and $\frac{b^3a + 1}{b - 1}$ are both positive integers. | [] | Greece | Junior Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | [(1, 3), (2, 2), (3, 3)] | |
056o | Let $k$ be a positive integer. Find all positive integers $n$, such that it is possible to mark $n$ points on the sides of a triangle (different from its vertices) and connect some of them with a line in such a way that the following conditions are satisfied:
1) there is at least 1 marked point on each side;
2) for e... | [
"Let $A$, $B$, and $C$ be the sets of the points on different sides and $a$, $b$ and $c$ their cardinalities respectively. Let us count all the triplets $(p, q, r)$, where $p \\in A$, $q \\in B$, and $r \\in C$ and they are either pairwise connected or pairwise not connected. For all $p \\in A$, $q \\in B$ there ex... | Estonia | IMO Team Selection Contest | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 12k | |
03jo | Problem:
Let $ABC$ be a right angled triangle of area $1$. Let $A'B'C'$ be the points obtained by reflecting $A$, $B$, $C$ respectively, in their opposite sides. Find the area of $\triangle A'B'C'$. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | 3 | |
07vs | Suppose $a_1, a_2, \dots, a_n$ are positive numbers. Prove that the square root of the arithmetic mean of the numbers $a_i a_j$, $1 \le i < j \le n$ doesn't exceed the arithmetic mean of $a_k$, $k = 1, 2, \dots, n$, with equality iff $a_1 = a_2 = \dots = a_n$. | [
"The claim is that\n$$\n\\frac{\\sum_{1 \\le i < j \\le n} a_i a_j}{\\binom{n}{2}} \\le \\left( \\frac{\\sum_{k=1}^{n} a_k}{n} \\right)^2 . \\text{ Equivalently,}\n$$\n$$\n2n \\sum_{1 \\le i < j \\le n} a_i a_j \\le (n-1) \\left( \\sum_{k=1}^{n} a_k^2 + 2 \\sum_{1 \\le i < j \\le n} a_i a_j \\right), \\text{ or}\n$... | Ireland | IRL_ABooklet_2023 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0bfn | Denote $M$ the midpoint of a segment whose ends are the midpoints of the diagonals of quadrilateral $ABCD$. A circle $C$ with center $M$ and radius $r$ is drawn so that the vertices of the quadrilateral are outside $C$.
Prove that $PA + PB + PC + PD > 4r$ for all points $P$ which are inside $C$. | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
03w3 | The set of the permutation $X = (x_1, x_2, \dots, x_9)$ of $1, 2, \dots, 9$ is $A$. $\forall X \in A$, and let $f(X) = x_1 + 2x_2 + 3x_3 + \dots + 9x_9$, $M = \{f(X) \mid X \in A\}$. Find the value of $|M|$. (Posed by Xiong Bin) | [
"We prove for $n \\ge 4$. If the permutations $X_n = (x_1, x_2, \\dots, x_n)$ of $1, 2, \\dots, n$ consist of a set $A$, and $f(X_n) = x_1 + 2x_2 + 3x_3 + \\dots + nx_n$, $M_n = \\{f(X) \\mid X \\in A\\}$, then $|M_n| = \\frac{n^3 - n + 6}{6}$.\n\nUsing mathematical induction on $n$, we see that\n$$\nM_n = \\left\\... | China | China Southeastern Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof and answer | 121 | |
0cu0 | A class consists of 23 students. During a year, each student celebrated his birthday with some (at least one, but not all) of his classmates. For each pair of students, count the number of such parties in which they both participated. Is it possible that all such numbers for all pairs of students are equal? (I. Bogdano... | [
"Two possible examples are as follows.\n\n1) Arrange all the students in a circle and assume that each comes to all birthdays except for that of his clockwise neighbor. Then, for any two students $A$ and $B$, they miss only the birthdays where $A$ is absent and where $B$ is absent. Thus, any pair of students met at... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics"
] | English; Russian | proof and answer | Yes | |
0fls | Problem:
Hallar todas las funciones $f: \mathbb{R} \rightarrow \mathbb{R}$ de variable real con valores reales, tales que para todo $x, y \in \mathbb{R}$,
$$
(x-2) f(y) + f(y + 2 f(x)) = f(x + y f(x))
$$ | [
"Solution:\n\nSupongamos primeramente que $f(0) = 0$. Haciendo $x = 0$ en (1), $f(y) = 0$ para todo $y \\in \\mathbb{R}$. Esta función satisface la ecuación funcional dada (1).\n\nSea $f(0) \\neq 0$. Haciendo $y = 0$ en (1), se obtiene $(x-2) f(0) + f(2 f(x)) = f(x)$ para todo $x \\in \\mathbb{R}$. Claramente esto ... | Spain | 48 aME | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 and f(x) = x − 1 | |
0ayn | Problem:
Suppose there are 3 distinct green balls, 4 distinct red balls, and 5 distinct blue balls in an urn. The balls are to be grouped into pairs such that the balls in any pair have different colors. How many sets of six pairs can be formed? | [
"Solution:\n\nSince there are 7 green and red balls combined, and 5 blue balls, then there should be one pair with 1 green and 1 red, and all the other pairs must have a blue ball. There are $3 \\times 4$ ways of selecting a green-red pair, and $5!$ ways of selecting the partners for the blue balls. Thus, there are... | Philippines | 20th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | final answer only | 1440 | |
025c | Problem:
Quantos são os pares de números inteiros positivos $(x, y)$ tais que
$$
\frac{x y}{x+y}=144 ?
$$ | [
"Solution:\nA equação dada é equivalente a $x y = 144(x + y) = 144 x + 144 y$, portanto, isolando $x$, obtemos $x = \\frac{144 y}{y - 144}$. Como $x$ e $y$ devem ser inteiros positivos, o denominador $y - 144$ deve ser um número inteiro positivo, digamos, $y - 144 = n$. Substituindo essa expressão no valor de $x$, ... | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 45 | |
0jdt | Problem:
Given that $62^{2} + 122^{2} = 18728$, find positive integers $(n, m)$ such that $n^{2} + m^{2} = 9364$. | [
"Solution:\n$(30,92)$ OR $(92,30)$\n\nIf $a^{2} + b^{2} = 2c$, then\n$$\n\\left(\\frac{a+b}{2}\\right)^{2} + \\left(\\frac{a-b}{2}\\right)^{2} = \\frac{2a^{2} + 2b^{2}}{4} = \\frac{a^{2} + b^{2}}{2} = c.\n$$\nThus, $n = \\frac{62 + 122}{2} = 92$ and $m = \\frac{122 - 62}{2} = 30$ works."
] | United States | HMMT November 2013 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | (30, 92) or (92, 30) |
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