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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
08yx | In triangle $ABC$, let $P$ and $Q$ be points on side $BC$ and suppose that the orthocenter of triangle $ACP$ and the orthocenter of triangle $ABQ$ coincide. Given that $AB = 10$, $AC = 11$, $BP = 5$, $CQ = 6$, find the length of $BC$. | [
"Let $H$ be the orthocenter of triangle $ABC$ and $K$ be the common orthocenter of triangle $ABQ$ and triangle $ACP$. Let $D$ be the foot of the perpendicular from $A$ to line $BC$. By definition, $A, D, H, K$ are collinear.\n\nSince $AB = 10$, $AC = 11$ and $BC > 5$, both $\\angle B$ and $\\angle C$ are less than ... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | sqrt(231) | |
0fo8 | Los puntos $P$ y $Q$ están en el lado $BC$ del triángulo acutángulo $ABC$ de modo que $\angle PAB = \angle BCA$ y $\angle CAQ = \angle ABC$. Los puntos $M$ y $N$ están en las rectas $AP$ y $AQ$, respectivamente, de modo que $P$ es el punto medio de $AM$, y $Q$ es el punto medio de $AN$. Demostrar que las rectas $BM$ y ... | [
"**Solución por Daniel Lasaosa Medarde, Pamplona, España.** Sean $E, F$ los puntos medios respectivos de $CA, AB$. Claramente, $\\angle MBA = \\angle PFA$ y $\\angle NCA = \\angle QEA$. Ahora bien, como por construcción $ABC, PBA$ y $QAC$ son semejantes, llamando $D$ al punto medio de $BC$, se tiene que $\\angle PF... | Spain | LV Olimpiada Internacional de Matemáticas | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Spanish | proof only | null | |
0kea | Problem:
The integer $202020$ is a multiple of $91$. For every positive integer $n$, show how $n$ additional $2$'s may be inserted into the digits of $202020$ so that the resulting $(n+6)$-digit integer is also a multiple of $91$. For example, a possible way to do this when $n=5$ is $22020220222$ (the inserted $2$'s a... | [
"Solution:\n\nEvery integer of the form $202\\ldots2020$ (where the dots represent any number of $2$'s) is divisible by $91$. There are a variety of ways to discover and prove this fact, some of which we outline here:\n\nMethod 1.\nFor $n=1$, there are four options: $2202020$, $2022020$, $2020220$, and $2020202$. L... | United States | Bay Area Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0h4u | Let $a, b, c, d$ be positive integers satisfying $ab = cd$. Is it possible that $a+b+c+d$ is a prime number? | [
"It follows from the problem statement that $\\frac{ab}{c}$ is a positive integer. Then there should exist positive integers $m, n, x, y$ such that $c = mn$, $a = mx$, $b = ny$. This implies that $d = \\frac{ab}{c} = xy$, and so $a+b+c+d = mx+ny+mn+xy = (n+x)(m+y)$, which is, obviously, not prime."
] | Ukraine | Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | No | |
02we | Problem:
Quando Paulo fez 15 anos, convidou 43 amigos para uma festa. O bolo tinha a forma de um polígono regular de 15 lados e havia 15 velas sobre ele. As velas foram colocadas de tal maneira que não havia três velas em linha reta. Paulo dividiu o bolo em pedaços triangulares onde cada corte ligava duas velas ou lig... | [
"Solution:\n\nSeja $n$ o número de triângulos em que se pode dividir o bolo com as condições dadas. Somaremos os ângulos interiores destes triângulos de duas formas:\n\n- Por um lado, como cada triângulo possui soma dos ângulos internos igual a $180^{\\circ}$, a soma de todos os ângulos internos deles é $180^{\\cir... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 43 | |
0dx4 | Problem:
Dana je funkcija $f(x)=m x-(3 m-20)$, kjer je $m$ rešitev enačbe
$$
\frac{(\sqrt[3]{2})^{6} \cdot 2^{-6}}{0,5^{2} \cdot 8^{-\frac{2}{3}}} = m \cdot \frac{1}{(\sqrt[3]{2})^{9}}
$$
Nariši graf funkcije $f$. | [
"Solution:\n\nNajprej rešimo enačbo\n$$\n\\frac{(\\sqrt[3]{2})^{6} \\cdot 2^{-6}}{0,5^{2} \\cdot 8^{-\\frac{2}{3}}} = m \\cdot \\frac{1}{(\\sqrt[3]{2})^{9}}\n$$\nLeva stran enačbe je enaka\n$$\n\\frac{2^{2} \\cdot 2^{-6}}{2^{-2} \\cdot 2^{-2}} = 1\n$$\ndesna stran pa $m \\cdot 2^{-3}$. Iz enakosti $1 = m \\cdot 2^{... | Slovenia | 6. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | y = 8x - 4 | |
0kjz | Problem:
Let $n$ be a positive integer. Claudio has $n$ cards, each labeled with a different number from $1$ to $n$. He takes a subset of these cards, and multiplies together the numbers on the cards. He remarks that, given any positive integer $m$, it is possible to select some subset of the cards so that the differen... | [
"Solution:\nWe require that $n \\geq 15$ so that the product can be divisible by $25$ without being even. In addition, for any $n > 15$, if we can acquire all residues relatively prime to $100$, we may multiply them by some product of $\\{1,2,4,5,15\\}$ to achieve all residues modulo $100$, so it suffices to acquir... | United States | HMMT Spring 2021 Guts Round | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 17 | |
0en6 | Find all triples $(p, q, r)$ of prime numbers which satisfy
$$
(p + 1)(q + 2)(r + 3) = 4pqr.
$$ | [
"Dividing both sides of the equation by $pqr$, we obtain\n$$\n\\left(1 + \\frac{1}{p}\\right) \\left(1 + \\frac{2}{q}\\right) \\left(1 + \\frac{3}{r}\\right) = 4.\n$$\nIf $p, q, r \\ge 5$, then\n$$\n\\left(1 + \\frac{1}{p}\\right) \\left(1 + \\frac{2}{q}\\right) \\left(1 + \\frac{3}{r}\\right) \\le \\frac{6}{5} \\c... | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (2, 3, 5), (5, 3, 3), (7, 5, 2) | |
0eil | Problem:
Dan je trikotnik $A B C$. Naj bosta $D$ in $E$ taki točki, ki ležita zaporedoma na poltrakih $C A$ in $C B$, a ne na stranicah trikotnika $A B C$, da velja $|A D|=|B E|=|A B|$. Presečišče premic $A E$ in $B D$ označimo z $G$, središče trikotniku $A B C$ včrtane krožnice pa z $I$. Dokaži, da premica GI poteka ... | [
"Solution:\n\nKer je trikotnik $A B E$ enakokrak z vrhom pri $B$, velja\n$$\n\\Varangle E A B=\\frac{1}{2}(\\pi-\\Varangle A B E)=\\frac{1}{2} \\Varangle C B A .\n$$\nPremica $B I$ je simetrala kota $\\Varangle C B A$, zato je $\\frac{1}{2} \\Varangle C B A=\\Varangle I B A$. Od tod sledi $\\Varangle E A B=\\Varang... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"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 | |
0c4b | Problem:
Fie $ABC$ un triunghi şi $E, F$ două puncte arbitrare pe laturile $(AB)$, respectiv $(AC)$. Cercul circumscris triunghiului $AEF$ intersectează a doua oară cercul circumscris triunghiului $ABC$ în punctul $M$. Fie $D$ simetricul lui $M$ faţă de dreapta $EF$ şi $O$ centrul cercului circumscris triunghiului $AB... | [
"Solution:\n\nDacă $M = A$, adică cercurile sunt tangente în $A$, atunci omotetia de centru $A$ care transformă cercul circumscris triunghiului $AEF$ în cercul circumscris triunghiului $ABC$ transformă segmentul $[EF]$ într-un segment paralel cu acesta, $[BC]$. Atunci $D \\in BC$ dacă şi numai dacă $[EF]$ este lini... | Romania | Al patrulea test de selecţie pentru OBMJ | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents"... | null | proof only | null | |
01gn | Prove that there exists a polygon on a square grid that can be tiled with dominoes ($1 \times 2$ or $2 \times 1$ figures) in exactly 2020 ways. | [
"Let $A_n$ be a figure that consists of a $2 \\times 2$ square combined with $n$ L-shapes ($A_3$ is shown in Figure ?? (a)). We will prove by induction that $A_n$ can be tiled with dominoes in exactly $2n + 2$ ways.\n\nThe base case for $A_0$ is evident. Now assume that we have proved it for $A_k$ and consider the ... | Baltic Way | Baltic Way 2020 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ity | Problem:
Compute
$$
1 \cdot 2^{2} + 2 \cdot 3^{2} + 3 \cdot 4^{2} + \cdots + 19 \cdot 20^{2}
$$ | [
"Solution:\nWe can write this as\n$$\n(1^{3} + 2^{3} + \\cdots + 20^{3}) - (1^{2} + 2^{2} + \\cdots + 20^{2})\n$$\nwhich is equal to\n$$\n44100 - 2870 = 41230\n$$"
] | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 41230 | |
0ayk | Problem:
In the figure below, five circles are tangent to line $\ell$. Each circle is externally tangent to two other circles. Suppose that circles $A$ and $B$ have radii $4$ and $225$, respectively, and that $C_1$, $C_2$, $C_3$ are congruent circles. Find their common radius.
 | [] | Philippines | 21st PMO Area Stage | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 9/4 | |
0ary | Problem:
Find the positive integers $n$ so that $2^{8} + 2^{11} + 2^{n}$ is a perfect square. | [
"Solution:\n(ans. 12)\n$$\n\\begin{aligned}\n& m^{2} = 2^{8} + 2^{11} + 2^{n} \\Rightarrow 2^{n} = m^{2} - 2^{8} - 2^{11} = m^{2} - 2^{8}(1 + 2^{3}) = \\\\\n& m^{2} - (3 \\cdot 2^{4})^{2} = (m - 3 \\cdot 2^{4})(m + 3 \\cdot 2^{4}) = (m - 48)(m + 48) \\Rightarrow m - 48 = \\\\\n& 2^{k},\\ m + 48 = 2^{l},\\ k + l = n... | Philippines | 13th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 12 | |
0g1y | Problem:
David und Linus spielen folgendes Spiel: David wählt eine Teilmenge $Q$ der Menge $\{1, \ldots, 2018\}$. Dann wählt Linus eine natürliche Zahl $a_{1}$ und berechnet die Zahlen $a_{2}, \ldots, a_{2018}$ rekursiv, wobei $a_{n+1}$ das Produkt der positiven Teiler von $a_{n}$ ist.
Sei $P$ die Menge der natürlich... | [
"Solution:\n\nWir beweisen, dass Linus eine Gewinnstrategie hat. Sei $a_{1}=p^{k_{1}}$ für eine Primzahl $p$ und eine nichtnegative ganze Zahl $k_{1}$. Dann ist $a_{n}$ von der Form $p^{k_{n}}$ für jedes $n$ und es gilt aufgrund der Rekursionsvorschrift\n$$\na_{n+1}=1 \\cdot p \\cdot \\ldots \\cdot p^{k_{n}}=p^{0+1... | Switzerland | SMO-Selektion | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | null | proof and answer | Linus | |
02dd | Show that there are at least 3 and at most 4 powers of $2$ with $m$ digits. For which $m$ are there $4$? | [
"Take $n$ to be the smallest integer such that $2^n \\ge 10^{m-1}$. Then $2^{n-1} < 10^{m-1}$, so $2^{n+2} < 8 \\cdot 10^{m-1} < 10^m$. So $2^n$, $2^{n+1}$ and $2^{n+2}$ all have $m$ digits. Thus there are at least $3$ powers of $2$ with $m$ digits.\n\n$2^{n-1} \\ge 5 \\cdot 10^{m-2}$ (otherwise $2^n < 10^{m-1}$). ... | Brazil | III OBM | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | There are at least 3 and at most 4 powers of two with m digits. There are 4 precisely for those m such that there exists an integer n with (m−1)/log10 2 ≤ n < m/log10 2 − 3 (equivalently, an integer lies between (m−1)/log10 2 and m/log10 2 − 3). | |
0ijt | Problem:
Let $ABC$ be a triangle such that $AB=2$, $CA=3$, and $BC=4$. A semicircle with its diameter on $\overline{BC}$ is tangent to $\overline{AB}$ and $\overline{AC}$. Compute the area of the semicircle. | [
"Solution:\nLet $O$, $D$, and $E$ be the midpoint of the diameter and the points of tangency with $\\overline{AB}$ and $\\overline{AC}$ respectively. Then $[ABC]=[AOB]+[AOC]=\\frac{1}{2}(AB+AC) r$, where $r$ is the radius of the semicircle. Now by Heron's formula, $[ABC]=\\sqrt{\\frac{9}{2} \\cdot \\frac{1}{2} \\cd... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 27π/40 | |
0eq1 | If $x = \sqrt[3]{900}$ then
(A) $7 < x < 8$ (B) $9 < x < 10$ (C) $11 < x < 12$ (D) $10 < x < 11$ (E) $12 < x < 13$ | [
"Answer B.\nSince $9^3 = 729 < 900$ and $10^3 = 1000 > 900$, it follows that $9 < \\sqrt[3]{900} < 10$."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | MCQ | B | |
0gpy | In a non-isosceles triangle $ABC$ let $O$ and $I$ be the circumcenter and the incenter, respectively. Let $D, E, F$ be the midpoints of the sides $[BC], [AC], [AB]$, respectively. Let $T$ be the foot of the perpendicular from $I$ to $[AB]$, $P$ be the circumcenter of the triangle $DEF$ and $Q$ be the midpoint of the li... | [
"Let $H$ be the orthocenter of the triangle $ABC$. Let the points $K, M, L$ be the intersection of the lines $AI$ and $OD$, $AI$ and $OH$, $AH$ and $OI$. We know that $AH = 2OD$. By a simple angle chasing we see that $AI$ is an angle bisector of $HAO$. Therefore\n$$\n\\frac{AO}{OD} = 2\\frac{AO}{AH} = 2\\frac{OM}{M... | Turkey | Team Selection Test for IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry ... | English | proof and answer | 4 | |
0bgy | Let $M$ be a nonempty set of positive reals so that, for every $a$, $b$, $c$ in $M$, the number $ab + bc + ca$ is rational. Prove that $\frac{a}{b}$ is rational for every $a$, $b$ in $M$. | [
"Let $a$, $b$, $c$ be arbitrary elements of $M$. By hypothesis, $ab + bc + ca$ is rational for all $a$, $b$, $c$ in $M$.\n\nLet us fix $a$ and $b$ in $M$, and let $c$ vary over $M$.\n\nFor $c = a$, we have:\n\n$$\nab + bc + ca = ab + ba + aa = 2ab + a^2\n$$\nwhich is rational.\n\nFor $c = b$, we have:\n\n$$\nab + b... | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof only | null | |
03xu | Let $n_1, n_2, \dots, n_{26}$ be pairwise distinct positive integers, satisfying:
(1) In the decimal representation of each $n_i$, each digit belongs to the set $\{1, 2\}$;
(2) For any $i, j$, $n_j$ cannot be obtained from $n_i$ by adding some digits on the right.
Find the least possible value of $\sum_{i=1}^{26} S(... | [
"Given two positive integers $a, b$ in decimal representation, we say $a$ contains $b$ if $a$ can be obtained from $b$ by adding some digits on the right. We first prove a lemma.\n\n**Lemma** Let $n_1, n_2, ..., n_r$ be pairwise distinct positive integers with digit 1 or 2. If none contains another, then the number... | China | China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 179 | |
0c5u | Determine the whole numbers $a$, $b$, $c$ for which
$$
\frac{a+1}{3} = \frac{b+2}{4} = \frac{5}{c+3}.
$$ | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a=2, b=2, c=2 | |
03vi | The number of rational solutions to the system of equations
$$
\begin{cases} x + y + z = 0, \\ xyz + z = 0, \\ xy + yz + xz + y = 0 \end{cases}, \text{ is } (\quad).
$$ | [
"If $z = 0$, then\n$$\n\\begin{cases} x + y = 0, \\\\ xy + y = 0 \\end{cases}.\n$$\nIt follows that\n$$\n\\begin{cases} x = 0, \\\\ y = 0 \\end{cases} \\text{ or } \\begin{cases} x = -1, \\\\ y = 1. \\end{cases}\n$$\nIf $z \\neq 0$, from $xyz + z = 0$ we get\n$$\nxy = -1. \\qquad \\textcircled{1}\n$$\nFrom $x + y +... | China | China Mathematical Competition | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | English | proof and answer | 2 | |
0d7x | Let $n$ be a given positive integer. Prove that there are infinitely many pairs of positive integers $(a, b)$ with $a, b > n$ such that
$$
\prod_{i=1}^{2015}(a+i) \mid b(b+2016) ; \quad \prod_{i=1}^{2015}(a+i) \nmid b ; \quad \prod_{i=1}^{2015}(a+i) \nmid(b+2016) .
$$ | [
"The given problem can be generalized as follows:\nGiven three positive integers $k, m, n$. Let $k_{1}, k_{2}, \\ldots, k_{m}$ be any positive integer. Prove that there are infinitely many pair of positive integers $(a, b)$ such that\n$$\n\\prod_{k=1}^{m}\\left(a+k_{i}\\right) \\mid b(b+k) \\text{ but } \\prod_{k=1... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0fkb | Problem:
Halla dos enteros positivos $a$ y $b$ conociendo su suma y su mínimo común múltiplo. Aplícalo en el caso de que la suma sea $3972$ y el mínimo común múltiplo $985928$. | [
"Solution:\n\nSea $p$ un número primo que divide a la suma $a+b$ y a su mínimo común múltiplo $[a, b]$. Como $p \\mid [a, b]$ al menos divide a uno de los dos enteros $a$ ó $b$. Si $p \\mid a$, al dividir $p$ a la suma $a+b$, también $p \\mid b$. (Obviamente el mismo razonamiento vale si hubiéramos supuesto que $p ... | Spain | XLIV Olimpiada Matemática Española | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Inter... | null | proof and answer | a = 1964, b = 2008 | |
0l4l | Problem:
Let $S_{n}$ be the sum of the first $n$ prime numbers. For example,
$$
S_{5}=2+3+5+7+11=28 .
$$
Does there exist an integer $k$ such that $S_{2023}<k^{2}<S_{2024}$ ? | [
"Solution:\nClaim: There exists an integer $k$ such that $S_{2023}<k^{2}<S_{2024}$.\n\nProof: Let $k$ be the smallest integer such that $S_{2023}<k^{2}$. Note that\n$$\nk^{2}=1+3+5+\\cdots+O_{k}\n$$\nwhere $O_{k}=2k-1$, the $k^{\\text{th}}$ odd number. Furthermore, observe that $O_{k} \\leq p_{2023}$, the $2023^{\\... | United States | 25th Bay Area Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
03b0 | The cells of $2010 \times 2010$ table are filled with integers. Adding 1 to all numbers in a particular row or column is called a *move*. We say that a table is *good*, if after finitely many moves it is transformed in a table of equal numbers.
a) Find the largest natural number $n$ for which there exists good table c... | [] | Bulgaria | Selection test for 51. International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a) n = 4018; b) the largest number is 2^4019 − 1 | |
0c6c | Let $n \ge 3$ be a positive integer.
a) Prove that there exist $z_1, z_2, \dots, z_n \in \mathbb{C}$ such that
$$
\frac{z_1}{z_2} + \frac{z_2}{z_3} + \dots + \frac{z_{n-1}}{z_n} + \frac{z_n}{z_1} = n i.
$$
b) For which $n$ there exist the complex numbers $z_1, z_2, \dots, z_n$, having the same absolute value, such th... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | n divisible by 4 | |
0cot | The numbers $1, 2, \ldots, 10000$ are placed into the cells of a square grid $100 \times 100$ (each number appears exactly once) so that every two numbers which differ by $1$ are placed into two cells sharing a common side. Consider the $5000$ pairs of cells containing the numbers that differ by $5000$. For each pair, ... | [
"Ответ. $50\\sqrt{2}$.\n\nПронумеруем в квадрате строки (снизу вверх) и столбцы (слева направо) числами от $1$ до $100$; будем обозначать клетку парой номеров ее строки и столбца. Назовем расстоянием между клетками расстояние между их центрами. Клетки назовем парными, если числа в них различаются на $5000$.\n\nЗаме... | Russia | Regional round | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Ge... | English; Russian | proof and answer | 50√2 | |
0iam | Problem:
For $x$ a real number, let $f(x)=0$ if $x<1$ and $f(x)=2x-2$ if $x \geq 1$. How many solutions are there to the equation
$$
f(f(f(f(x))))=x ?
$$ | [
"Solution:\n\nCertainly $0, 2$ are fixed points of $f$ and therefore solutions. On the other hand, there can be no solutions for $x<0$, since $f$ is nonnegative-valued; for $0 < x < 2$, we have $0 \\leq f(x) < x < 2$ (and $f(0) = 0$), so iteration only produces values below $x$, and for $x > 2$, $f(x) > x$, and ite... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2 | |
05hq | Problem:
Soient $a$ et $b$ deux entiers tels que $a + b$ n'est pas divisible par $3$. Montrer que l'on ne peut pas colorier les entiers relatifs en trois couleurs de sorte que pour tout entier $n$, les trois nombres $n$, $n+a$ et $n+b$ soient de couleurs différentes. | [
"Solution:\n\nSupposons par l'absurde qu'il existe un tel coloriage. On notera $x \\sim y$ si $x$ et $y$ sont de la même couleur.\n\nPour tout $n$, les entiers $(n+a)+b$ et $n+a$ ne sont pas de la même couleur, et de même $n+a+b$ n'est pas de la même couleur que $n+b$, donc $n+a+b \\sim n$. Il vient immédiatement $... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0cio | We say that a natural number is *special* if it can be written as the sum of two or more consecutive natural numbers. Determine how many natural numbers less than $1000$ are special. | [
"Let $n$ be a natural number. We want to count the number of $n < 1000$ that can be written as the sum of two or more consecutive natural numbers.\n\nLet the consecutive numbers be $a, a+1, \\ldots, a+k-1$ for $k \\geq 2$. Their sum is:\n$$\nS = a + (a+1) + \\cdots + (a+k-1) = k a + \\frac{k(k-1)}{2}\n$$\nSo $n = k... | Romania | 75th NMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 989 | |
0hkw | Problem:
A $6 \times 6$ square is covered by nonoverlapping dominos ($2 \times 1$ rectangles, placed horizontally or vertically). Prove that there must be a horizontal line or a vertical line that passes through the interior of the big square, but which does not cut the interior of any domino. | [
"Solution:\n\nConsider all the grid lines of the big square. If some such line is intersected by $d$ dominos, we claim $d$ is even. Proof: Looking at the portion of the board on one side of the line, we find that the number of squares there is divisible by $6$ and so is even; on the other hand, each of the $d$ domi... | United States | Berkeley Math Circle Take-Home Contest #2 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0eib | Problem:
Poišči vsa naravna števila $n$, katerih kub je enak vsoti kvadratov treh ne nujno različnih deliteljev števila $n$. | [
"Solution:\nNaj bodo $a, b$ in $c$ delitelji števila $n$, za katere je $n^{3}=a^{2}+b^{2}+c^{2}$. Tedaj velja $a, b, c \\leq n$, zato je $n^{3}=a^{2}+b^{2}+c^{2} \\leq 3 n^{2}$ oziroma $n \\leq 3$. Primer $n=1$ odpade, ker 1 ni vsota treh naravnih števil. Če je $n=2$, sta edina delitelja 1 in 2. Toda nobena od vsot... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 3 | |
0izr | Problem:
In how many ways can you fill a $3 \times 3$ table with the numbers $1$ through $9$ (each used once) such that all pairs of adjacent numbers (sharing one side) are relatively prime? | [
"Solution:\n\n2016\n\nThe numbers can be separated into four sets. Numbers in the set $A=\\{1,5,7\\}$ can be placed next to anything. The next two sets are $B=\\{2,4,8\\}$ and $C=\\{3,9\\}$. The number $6$, which forms the final set $D$, can only be placed next to elements of $A$. The elements of each group can be ... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 2016 | |
063d | Problem:
Eine Menge $A$ von ganzen Zahlen heißt zulässig, wenn sie folgende Eigenschaft hat:
Für $x, y \in A$ ($x = y$ ist erlaubt) gilt $x^{2} + k x y + y^{2} \in A$ für jede ganze Zahl $k$.
Man bestimme alle Paare $m, n$ von Null verschiedener ganzer Zahlen, für welche die einzige zulässige Menge, die sowohl $m$ als... | [
"Solution:\n\nFür ein Paar $m, n$ mit $\\operatorname{ggT}(|m|,|n|) = d > 1$ ist $m^{2} + k m n + n^{2}$ durch $d$ teilbar, so dass als Menge $A$ auch die Menge aller ganzzahligen Vielfachen von $d$ in Frage kommt, die das Element $1$ nicht enthält und daher von $\\mathbb{Z}$ verschieden ist.\n\nNun betrachten wir ... | Germany | 1. Auswahlklausur | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | All pairs (m, n) of nonzero integers with gcd(|m|, |n|) = 1 | |
0eby | Let $D$ and $E$ be points on the sides $BC$ and $CA$ of the triangle $ABC$ respectively. The circumcircle of the triangle $CDE$ and the line through $C$, which is parallel to $AB$, intersect again in a point $L$. The line $DL$ intersects the side $AB$ in a point $M$. Denote by $N$ the point on the line $AB$ such that $... | [
"Denote the intersection of the lines $CT$ and $AB$ by $U$. We want to prove that\n$$\n\\frac{|AU|}{|UB|} = \\frac{|AC|}{|CB|}, \\qquad (2)\n$$\nsince it will follow that $CT$ is the angle bisector of $\\angle ACB$.\n\nSince the points $C, E, D$, and $L$ are concyclic, and the lines $AB$ and $CL$ are parallel, we h... | Slovenia | Selection Examinations for the IMO 2015 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocent... | null | proof only | null | |
0f6r | Problem:
Find all pairs $(x, y)$ such that $|\sin x - \sin y| + \sin x \sin y \leq 0$. | [] | Soviet Union | 19th ASU | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | All pairs where sin x = 0 and sin y = 0, i.e., x = k*pi and y = l*pi for integers k, l. | |
06bv | Let $\mathbb{Z}$ denote the set of integers. Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that $f(-1) = f(1)$ and $f(x) + f(y) = f(x + 2xy) + f(y - 2xy)$ for all integers $x, y$. | [
"$f$ can be any function such that $f(2^k m) = f(2^k)$ for any $k \\ge 0$ and odd $m$, where $f(0), f(1), f(2), f(2^2), \\dots$ are arbitrary integers.\n\nWe label the equation as follows.\n$$\nf(x) + f(y) = f(x + 2xy) + f(y - 2xy) \\quad (1)\n$$\nPutting $x = 1$ and $y = n$ in (1), we obtain\n$$\nf(1) + f(n) = f(2... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Other"
] | null | proof and answer | All functions f: Z → Z such that f(0) is arbitrary and, for each k ≥ 0, f(2^k m) = c_k for all odd m, where the values c_k = f(2^k) are arbitrary integers. Equivalently, for every nonzero n written as n = 2^k m with m odd, f(n) depends only on k (the 2-adic valuation), and f is even. | |
0hkd | Problem:
The five-digit number $9A65B$ is divisible by $36$, where $A$ and $B$ are digits. Find all possible values of $A$ and $B$. | [
"Solution:\n\nA number is divisible by $36$ if and only if it is divisible by $9$ and $4$.\n\nFor it to be divisible by $4$, the last two digits must make a two-digit number divisible by $4$, and the only possibilities for this are $B = 2$ or $B = 6$.\n\nFor it to be divisible by $9$, the sum of the digits must be ... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | A=5, B=2 or A=1, B=6 | |
0195 | Determine all positive integers $d$ such that whenever $d$ divides an integer $n$, $d$ will also divide any integer $m$ having the same digits as $n$. | [
"Answer: $d = 1$, $d = 3$ or $d = 9$. It is known that $1$, $3$ and $9$ have the given property. Assume that $d$ is a $k$ digit number such that whenever $d$ divides an integer $n$, $d$ will also divide any integer $m$ having the same digits as $n$. Then there exists a $k+2$ digit number $10a_1a_2\\dots a_k$ which ... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 1, 3, 9 | |
02x6 | Problem:
Na figura a seguir, todos os retângulos são iguais e possuem o perímetro de $8~\mathrm{cm}$. Qual o perímetro total da figura?
 | [
"Solution:\n\nPodemos deslizar os blocos e formar uma nova figura com o mesmo perímetro da anterior. Se o lado menor do bloco mede $a$ e o maior mede $b$ então $2a + 2b = 8~\\mathrm{cm}$. No perímetro da nova figura, temos 8 segmentos de tamanho $a$ e 8 de tamanho $b$. Assim, o seu perímetro é $8a + 8b = 4(2a + 2b)... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 32 cm | |
0enn | Find all primes $p$ and $q$ such that
$$
2q^p - p^q = 7.
$$ | [
"It is easy to see that $p$ is odd and $p \\ne q$, so $p \\ge 3$ and $(p,q) = 1$.\n\nIf $q = 2$, then $2^{p+1} = 7 + p^2$. The only solution is $p = 3$, as $2^{n+1} > 7 + n^2$ for $n \\ge 4$.\n\nFor $q \\ge 3$, by Fermat's Little Theorem we get that $q^p \\equiv_p q$, so $p \\mid 2q^p - 7 \\equiv 2q - 7$ and simila... | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | [(3, 2), (3, 5)] | |
05m8 | Problem:
Soit $A$ un point extérieur à un cercle $\mathscr{C}$ de centre $O$. Un point $P$ se déplace sur $\mathscr{C}$. Soit $M$ le point d'intersection entre $(A P)$ et la bissectrice de $\widehat{P O A}$. Montrer que $M$ se déplace sur un cercle que l'on décrira. | [
"Solution:\n\n\n\nNotons $d = O A$ et $R$ le rayon du cercle.\n\nD'après le théorème de la bissectrice, on a $M P / M A = O P / O A$, donc\n$$\n\\frac{M A}{A P} = \\frac{M A}{M A + M P} = \\frac{1}{1 + \\frac{M P}{M A}} = \\frac{1}{1 + \\frac{O P}{O A}} = \\frac{O A}{O A + O P} = \\frac{d}{... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | Let d = OA and R be the radius of the given circle, and set k = d / (d + R). The locus of M is the circle with center O′ on the line AO such that AO′ = k · AO = d^2 / (d + R), and radius kR = dR / (d + R). Equivalently, M lies on the circle centered at O′ with radius kR, which also passes through O. | |
03g7 | Let $n$ be a natural number. King Arthur has invited $2^n - 1$ knights to an audience in Camelot. Merlin the Magician arranged the knights in a list numbered from $1$ to $2^n - 1$. It turned out that any two knights with numbers $a, b, a < b$ are friends if and only if $0 \le b - 2a \le 1$. The king chose a natural num... | [
"Let us construct a graph $T$ with vertex-set which is the set of all knights, numbered as in the first list. Two vertices are adjacent if the corresponding knights are friends. It can be seen that $T$ is a fully balanced binary tree - see fig. 1. We label each vertex with the knight's number in the first list. Let... | Bulgaria | TST for BMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Algorithms"
] | English | proof only | null | |
05xa | Problem:
Pour tout entier $n \geqslant 1$, on pose $u_{n}=1!+2!+\ldots+n!$. Montrer qu'il existe une infinité de nombres premiers divisant au moins l'un des termes de la suite $\left(u_{n}\right)$. | [
"Solution:\n\nSupposons l'inverse : alors il existe des nombres premiers $p_{1}<\\ldots<p_{r}$ tels que pour tout $n \\geqslant 1$, $u_{n}=1!+2!+\\ldots+n!$ soit le produit des $p_{i}^{a_{i}(n)}$, où les $a_{i}(n)$ sont des entiers positifs.\n\nSi $n \\geqslant 1$ est tel que $a_{i}(n)<v_{p_{i}}((n+1)!)$, alors\n$$... | France | Préparation Olympique Française de Mathématiques - Envoi 3: Arithmétique | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
04h7 | Let $a$, $b$ and $c$ be the side-lengths of a triangle with perimeter $1$. Prove that
$$
\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2} < 1 + \frac{\sqrt{2}}{2}.
$$
(APMO 2003) | [
"Without loss of generality, we may assume $a \\ge b \\ge c$.\nThe triangle inequality and $a + b + c = 1$ imply that $a < b + c = 1 - a$, i.e. $a < \\frac{1}{2}$.\nSince $b \\le a$, it follows that $\\sqrt{a^2 + b^2} \\le \\sqrt{2a^2} = a\\sqrt{2} < \\frac{\\sqrt{2}}{2}$.\nSince $c \\le b$, it follows that $b^2 + ... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
064e | Problem:
Beweisen Sie, dass es für jede beliebige nichtnegative ganze Zahl $z$ genau ein geordnetes Paar $(m, n)$ positiver ganzer Zahlen $m, n$ gibt, so dass $2 z = (m+n)^2 - m - 3 n$ gilt. | [
"Solution:\n\nUmformen von (1) ergibt $2 z = (m+n-1)^2 + m - n - 1 = 2 m - 2 + (m+n-1)(m+n-2)$, woraus $z+1 = m + \\frac{(m+n-1)(m+n-2)}{2}$ folgt. Wir setzen $m+n-1 = k$ und erhalten $z+1 = m + \\frac{k(k-1)}{2}$, wobei $0 < m \\leq k$ und $k \\in \\mathbb{Z}, k \\geq 1$ gilt. Außerdem ist $\\frac{k(k-1)}{2} \\in ... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
06tw | Let $ABC$ be a triangle with circumcircle $\Gamma$ and incentre $I$. Let $M$ be the midpoint of side $BC$. Denote by $D$ the foot of perpendicular from $I$ to side $BC$. The line through $I$ perpendicular to $AI$ meets sides $AB$ and $AC$ at $F$ and $E$ respectively. Suppose the circumcircle of triangle $AEF$ intersect... | [
"Let $AM$ meet $\\Gamma$ again at $Y$ and $XY$ meet $BC$ at $D'$. It suffices to show $D' = D$. We shall apply the following fact.\n\n- Claim. For any cyclic quadrilateral $PQRS$ whose diagonals meet at $T$, we have\n$$\n\\frac{QT}{TS} = \\frac{PQ \\cdot QR}{PS \\cdot SR}\n$$\nProof. We use $[W_1 W_2 W_3]$ to denot... | IMO | IMO 2016 Shortlisted Problems | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Polar triang... | English | proof only | null | |
00cd | Los tres enteros $2000$, $19$ y $n$ están escritos en el pizarrón. Ana y Beto juegan el siguiente juego:
Comienza Ana y luego juegan por turnos. Cada jugada consiste en borrar uno de los números del pizarrón y reemplazarlo por la diferencia de los otros dos (el mayor menos el menor). Solo están permitidas las jugadas e... | [] | Argentina | Nacional OMA 2019 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Spanish | proof and answer | Ana | |
0kzr | Two teams are in a best-two-out-of-three playoff: the teams will play at most $3$ games, and the winner of the playoff is the first team to win $2$ games. The first game is played on Team $A$'s home field, and the remaining games are played on Team $B$'s home field. Team $A$ has a $\frac{2}{3}$ chance of winning at hom... | [
"There are three ways for Team $A$ to win the playoff: win the first two games; win the first game, lose the second game, and win the third game; or lose the first game and win the second and third games. The probability that it wins in one of these ways is\n$$\n\\frac{2}{3} \\cdot p + \\frac{2}{3} \\cdot (1-p) \\c... | United States | AMC 10 A | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | E | |
0bnf | Problem:
Határozd meg azokat a folytonos és növekvő $f:[0, \infty) \rightarrow \mathbb{R}$ függvényeket, amelyekre
$$
\int_{0}^{x+y} f(t) \mathrm{d} t \leq \int_{0}^{x} f(t) \mathrm{d} t+\int_{0}^{y} f(t) \mathrm{d} t
$$
bármely $x, y \in[0, \infty)$ esetén!
Problem:
Determinaţi funcţiile continue şi crescătoare $f:... | [
"Solution:\n\nInegalitatea din enunţ este echivalentă cu\n$$\n\\int_{x}^{x+y} f(t) \\mathrm{d} t \\leq \\int_{0}^{y} f(t) \\mathrm{d} t,\n$$\nde unde\n$$\n\\int_{0}^{y} f(t+x) \\mathrm{d} t \\leq \\int_{0}^{y} f(t) \\mathrm{d} t,\n$$\noricare ar fi $x \\geq 0$ şi $y \\geq 0$.\n\nCum $f(t+x) \\geq f(t)$, oricare ar ... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All constant functions: f(x) = c for all x ≥ 0, where c is a real constant. | |
0eio | Problem:
8 prijateljev, od tega 3 dekleta, se bo v zabaviščnem parku peljalo $z$ velikim razglednim kolesom. Vseh 8 prijateljev se bo naključno razporedilo v 4 proste kabine razglednega kolesa, v vsako kabino po 2 prijatelja. Kolikšna je verjetnost, da nobeni 2 dekleti ne bosta sedeli v isti kabini?
(20 točk) | [
"Solution:\n\nNaj bo $A$ dogodek, da nobeni 2 dekleti ne sedita v isti kabini. Verjetnost dogodka $A$ izračunamo po formuli\n$$\nP(A)=\\frac{\\text{ugodne razporeditve}}{\\text{vse razporeditve}}\n$$\nVseh možnih razporeditev je $8!$, saj moramo 8 prijateljev na poljuben način razporediti na 8 sedišč, kar pomeni, d... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Odbirno tekmovanje | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 4/7 | |
00hw | Let $ABCD$ be a parallelogram. Let $W$, $X$, $Y$, and $Z$ be points on sides $AB$, $BC$, $CD$, and $DA$, respectively, such that the incenters of triangles $AWZ$, $BXW$, $CYX$ and $DZY$ form a parallelogram. Prove that $WXYZ$ is a parallelogram. | [
"Let the four incenters be $I_{1}$, $I_{2}$, $I_{3}$, and $I_{4}$ with inradii $r_{1}$, $r_{2}$, $r_{3}$, and $r_{4}$ respectively (in the order given in the question). Without loss of generality, let $I_{1}$ be closer to $AB$ than $I_{2}$. Let the acute angle between $I_{1}I_{2}$ and $AB$ (and hence also the angle... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance... | English | proof only | null | |
0elb | Problem:
Izračunaj koordinate presečišč grafov funkcije $f(x)=x^{4}-2 x^{3}-7 x+2$ in funkcije $g(x)=3 x^{3}-8 x^{2}-1$. Zapiši smerni koeficient premice skozi ti dve presečišči. Izračunaj tangens manjšega od kotov med to premico in premico z enačbo $3 x+2 y-11=0$. | [
"Solution:\n\n1. Presečišče grafov funkcij določimo tako, da enačimo funkcijska predpisa $f(x)=g(x)$. Dobimo enačbo višje stopnje $x^{4}-2 x^{3}-7 x+2=3 x^{3}-8 x^{2}-1$. Enačbo preoblikujemo in dobimo $x^{4}-5 x^{3}+8 x^{2}-7 x+3=0$.\n\nUgotovimo, da je $x_{1}=1$ rešitev te enačbe, naredimo Hornerjev algoritem in ... | Slovenia | 23. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | Intersections: (1, -6) and (3, 8); slope: 7; tangent of the smaller angle: 17/19. | |
03vv | Let $f(x) = ax + b$, with $a, b$ real numbers; $f_1(x) = f(x)$, $f_{n+1}(x) = f(f_n(x))$, $n = 1, 2, \dots$. If $f_7(x) = 128x + 381$, then $a + b = \underline{\hspace{2cm}}$. | [
"$$\n\\begin{aligned}\nf_n(x) &= a^n x + (a^{n-1} + a^{n-2} + \\dots + a + 1)b \\\\\n&= a^n x + \\frac{a^n - 1}{a - 1} \\times b.\n\\end{aligned}\n$$\nAs $f_7(x) = 128x + 381$, we have $a^7 = 128$ and $\\frac{a^7 - 1}{a - 1} \\times b = 381$. Then $a = 2$, $b = 3$. The answer is $a + b = 5$."
] | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | final answer only | 5 | |
0c67 | Determine the largest value the expression $\sum_{1 \le i < j \le 4} (x_i + x_j) \sqrt{x_i x_j}$ may achieve, as $x_1, x_2, x_3, x_4$ run through the non-negative real numbers that add up to $1$. Determine also the $x_i$ at which the maximum is achieved. | [
"The required maximum is $3/4$ and is achieved if and only if the $x_i$ are all equal to $1/4$. To prove this, use the binomial expansion of $(\\sqrt{x_i} - \\sqrt{x_j})^4$ to write\n$$\n4(x_i + x_j)\\sqrt{x_i x_j} = x_i^2 + 6x_i x_j + x_j^2 - (\\sqrt{x_i} - \\sqrt{x_j})^4,\n$$\n\n\\begin{aligned}\n4 \\sum_{1 \\le ... | Romania | SELECTION TESTS FOR THE 2019 BMO AND IMO | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | Maximum value 3/4, achieved uniquely when x1 = x2 = x3 = x4 = 1/4. | |
0dyf | Problem:
Členi $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ geometrijskega zaporedja so naravna števila, manjša od 2008. Število $a_{2}$ je deljivo s $5$, $a_{3}$ je deljivo s $4$, $a_{4}$ je deljivo s $3$, število $a_{1}$ pa ni deljivo s $6$. Nobeno praštevilo ne deli vseh $5$ členov zaporedja. Izračunaj člene tega zaporedja. | [
"Solution:\n\nKer so $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ členi geometrijskega zaporedja, jih lahko zapišemo v obliki $a_{i}=a_{1} \\cdot q^{i-1}$ za $i=2,3,4,5$, kjer je $q$ neko realno število. Toda $q=\\frac{a_{2}}{a_{1}}$ je kvocient dveh naravnih števil, torej je racionalno število. Zapišimo $q=\\frac{m}{n}$ ko... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof and answer | 625, 750, 900, 1080, 1296 | |
09ds | Given $n$ different points in a plane, prove that among these points it is always possible to find 3 points forming an angle not exceeding $\frac{\pi}{n}$. | [
"If 3 of points lie simultaneously on a line there is nothing to prove. Therefore suppose that, no two of them don't lie on a straight line and . We claim there is a convex polygon with $k \\le n$ vertices such that all $n$ points lie inside of the polygon. It is obvious that there exists an angle of the polygon no... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0evu | Show that there exists a positive integer $K$ satisfying the following: for any prime $p > K$, the number of integers $1 \le a \le p$ such that $a^{p-1} - 1$ is divisible by $p^2$ is less than or equal to $\frac{p}{2^{2024}}$. | [
"Define\n$$\nS = \\{1 \\le a \\le p^2 : p^2|a^{p-1} - 1\\}, \\quad S_p = S \\cap \\{1, 2, \\dots, p\\}.\n$$\nWe observe that $S$ is closed in multiplication mod $p^2$: if $a, b \\in S$ and $ab \\equiv c \\pmod{p^2}$ then $c \\in S$.\n\n**Lemma 1.** We have $|S| \\le p-1$.\n\n*Proof.* (Note: actually we have $|S| = ... | South Korea | The 37th Korean Mathematical Olympiad Final Round | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0h6s | Prove that for all natural $n \ge 2$ the following number is composite:
$$
\frac{n^{1003} + n^{1002} + n^{1001} + 1}{n+1}
$$ | [
"Since\n$$\n\\begin{aligned}\nn^{1003} + n^{1002} + n^{1001} + 1 &= n^{1001}(n^2 + 1) + n^{1002} + 1 \\\\\n&= n^{1001}(n^2 + 1) + (n^2 + 1)(n^{1000} - n^{998} + n^{996} - \\dots - n^2 + 1)\n\\end{aligned}\n$$\nhence, the numerator is divisible by $n^2 + 1$. We can see that the numerator is also divisible by $n+1$, ... | Ukraine | UkraineMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
02rt | Problem:
Um quadrado mágico é uma tabela quadrada na qual a soma dos números em qualquer linha ou coluna é constante. Por exemplo,
| 1 | 5 | 9 |
| :--- | :--- | :--- |
| 8 | 3 | 4 |
| 6 | 7 | 2 |
é um quadrado mágico, o qual usa os números de 1 a 9. Como o leitor pode verificar, a soma em qualquer linha ou coluna é ... | [
"Solution:\n\na) A soma de todos os ímpares de 1 a 17 é 81. Como são três colunas no quadrado, e todas as colunas (e linhas) têm a mesma soma, a soma em cada coluna deve ser $81 / 3 = 27$. Daí, deduzimos que o número que falta na terceira coluna (a dos números 13 e 3) é o número 11, pois aí teremos $11 + 13 + 3 = 2... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Other"
] | null | proof and answer | a) X = 7; b) One valid hypermagic square: rows 8 1 6; 3 5 7; 4 9 2. | |
0e4t | Problem:
Točka $O_{1}$ je središče krožnice $\mathcal{K}_{1}$ in leži na krožnici $\mathcal{K}_{2}$ s središčem $O_{2}$. Krožnici $\mathcal{K}_{1}$ in $\mathcal{K}_{2}$ se sekata v točkah $A$ in $B$. Krožnica $\mathcal{K}_{1}$ seka daljico $O_{1} O_{2}$ v točki $C$. Premica $B C$ seka krožnico $\mathcal{K}_{2}$ v točk... | [
"Solution:\n\nOznačimo $\\angle A B D=\\alpha$. Zaradi tetivnosti štirikotnika $A O_{1} B D$ je $\\angle A O_{1} D=\\angle A B D=\\alpha$. Središčni kot $\\angle A O_{1} C$ v krožnici $\\mathcal{K}_{1}$ je dvakrat večji od obodnega kota $\\angle A B C$, zato je $\\angle A O_{1} C=2 \\alpha$. Od tod sledi\n$$\n\\beg... | Slovenia | 55. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0k8o | Problem:
In $\triangle ABC$, $AB = 2019$, $BC = 2020$, and $CA = 2021$. Yannick draws three regular $n$-gons in the plane of $\triangle ABC$ so that each $n$-gon shares a side with a distinct side of $\triangle ABC$ and no two of the $n$-gons overlap. What is the maximum possible value of $n$? | [
"Solution:\n\nIf any $n$-gon is drawn on the same side of one side of $\\triangle ABC$ as $\\triangle ABC$ itself, it will necessarily overlap with another triangle whenever $n > 3$. Thus either $n = 3$ or the triangles are all outside $ABC$. The interior angle of a regular $n$-gon is $180^\\circ \\cdot \\frac{n-2}... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 11 | |
0aad | Problem:
Let $n$ be a positive integer. Alice and Bob play the following game. First, Alice picks $n+1$ subsets $A_{1}, \ldots, A_{n+1}$ of $\{1, \ldots, 2^{n}\}$ each of size $2^{n-1}$. Second, Bob picks $n+1$ arbitrary integers $a_{1}, \ldots, a_{n+1}$. Finally, Alice picks an integer $t$. Bob wins if there exists a... | [
"Solution:\n\nBob has a winning strategy for every $n \\in \\mathbb{N}$. Initially, note that Bob wins if and only if he can \"shift\" the sets $A_{1}, \\ldots, A_{n+1}$ modulo $2^{n}$ such that they together cover every residue class. For a set of integers $C \\subset \\mathbb{Z}$ and $r \\in \\mathbb{N}$, let $C+... | Nordic Mathematical Olympiad | The 35th Nordic Mathematical Contest | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Alice has no winning strategy for any positive integer; Bob wins for all n. | |
0gmw | Find the largest area of a heptagon two of whose diagonals are perpendicular and whose vertices lie on a unit circle. | [] | Turkey | Team Selection Examination for the International Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | null | |
08gu | Problem:
Let $a$ and $b$ be positive real numbers such that $3 a^{2}+2 b^{2}=3 a+2 b$. Find the minimum value of
$$
A=\sqrt{\frac{a}{b(3 a+2)}}+\sqrt{\frac{b}{a(2 b+3)}}
$$ | [
"Solution:\n\nBy the Cauchy-Schwarz inequality we have that\n$$\n5\\left(3 a^{2}+2 b^{2}\\right)=5\\left(a^{2}+a^{2}+a^{2}+b^{2}+b^{2}\\right) \\geq (3 a+2 b)^{2}\n$$\n(or use that the last inequality is equivalent to $(a-b)^{2} \\geq 0$).\nSo, with the help of the given condition we get that $3 a+2 b \\leq 5$. Now... | JBMO | null | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 2/√5 | |
07uq | The equation $AB \times CD = EFGH$, where each of the letters $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$ represents a different digit and the values of $A$, $C$ and $E$ are all non-zero, has many solutions, e.g., $46 \times 85 = 3910$. Find the smallest value of the four-digit number $EFGH$ for which there is a solution. | [
"**Solution 1.** Consider factorisations of numbers $EFGH$ with all digits different, and identify all factorisations consisting of two 2-digit numbers. Start with the smallest possible number $1023$ and stop when a solution is found.\n$$\n\\begin{align*}\n1023 &= 3 \\cdot 11 \\cdot 31 = 11 \\cdot 93 = 31 \\cdot 33... | Ireland | IRL_ABooklet | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1058 | |
0brt | The altitudes $AA_1, BB_1, CC_1$ of the acute triangle $ABC$ intersect at $H$. Let $A_2$ be the reflection of point $A$ in the line $B_1C_1$ and let $O$ be the circumcenter of triangle $ABC$.
a) Prove that the points $O, A_2, B_1, C$ are cocyclic.
b) Prove that the points $O, H, A_1, A_2$ are cocyclic. | [
"a) The angles $\\angle ABC$ and $\\angle AB_1C_1$ are equal, therefore so are their complementary angles, $\\angle BAA_1$ and $\\angle A_2AC$.\nIt follows that the rays ($AH$ and ($AA_2$ are isogonal, hence $A_2 \\in (AO$). As $AO = CO$, we have $\\angle ACO \\equiv \\angle OAC \\equiv \\angle AA_2B_1$, therefore ... | 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 > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneo... | English | proof only | null | |
03sb | Suppose an ellipse with points $B_0$ and $B_1$ as the foci intercepts side $AB_i$ of $\triangle AB_0B_1$ at $C_i$ ($i = 0, 1$). Taking an arbitrary point $P_0$ on the extending line of $AB_0$, draw arc $\overarc{P_0Q_0}$ with $B_0$, $B_0P_0$ as the center and radius respectively, intercepting the extending line of $C_1... | [
"(1) From the properties of an ellipse we know\n$$\nB_1C_0 + C_0B_0 = B_1C_1 + C_1B_0.\n$$\nAlso, it is obvious that\n$$\nB_0P_0 = B_0Q_0, \\quad C_1B_0 + B_0Q_0 = C_1P_1,\n$$\n$$\nB_1C_1 + C_1P_1 = B_1C_0 + C_0Q_1, \\quad C_0Q_1 = C_0B_0 + B_0P'_0.\n$$\nAdding these equations, we get $B_0P_0 = B_0P'_0$.\nTherefore... | China | China Mathematical Competition (Extra Test) | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0dkn | Find all positive integers $n$ such that there exist $n$ consecutive positive integers whose sum is a perfect square. | [] | Saudi Arabia | Saudi Booklet | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | All positive integers n such that, writing n = 2^e r with r odd, one has e = 0 or e is odd. Equivalently, all odd n and those even n with v2(n) odd; the excluded n are exactly those with v2(n) an even positive integer (n = 2^{2k} r with k ≥ 1 and r odd). | |
09r5 | Problem:
Zij $n$ een positief geheel getal deelbaar door 4. We bekijken permutaties $(a_{1}, a_{2}, \ldots, a_{n})$ van $(1,2, \ldots, n)$ met de volgende eigenschap: voor elke $j$ geldt dat als we $i=a_{j}$ nemen, dan $a_{i}+j=n+1$. Bewijs dat er precies $\frac{\left(\frac{1}{2} n\right)!}{\left(\frac{1}{4} n\right)!... | [
"Solution:\n\nZij $t \\in \\{1,2, \\ldots, n\\}$. Stel dat $a_{t}=t$, dan kunnen we $i=j=t$ kiezen en geldt dus $a_{t}+t=n+1$, dus $2 t=n+1$. Maar $n$ is deelbaar door 4, dus $n+1$ is oneven. Tegenspraak. Stel nu dat $a_{t}=n+1-t$. Dan kunnen we $i=n+1-t$ en $j=t$ kiezen en geldt dus $a_{n+1-t}+t=n+1$, dus $a_{n+1-... | Netherlands | Toets 6 juni 2012 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | ((n/2)!)/((n/4)!) | |
03lh | Problem:
Let $p$ be an odd prime. Prove that
$$
\sum_{k=1}^{p-1} k^{2p-1} \equiv \frac{p(p+1)}{2} \quad\left(\bmod p^{2}\right)
$$
[Note that $a \equiv b(\bmod m)$ means that $a-b$ is divisible by $m$.] | [
"Solution:\nSince $p-1$ is even, we can pair up the terms in the summation in the following way (first term with last, 2nd term with 2nd last, etc.):\n$$\n\\sum_{k=1}^{p-1} k^{2p-1} = \\sum_{k=1}^{\\frac{p-1}{2}} \\left(k^{2p-1} + (p-k)^{2p-1}\\right)\n$$\nExpanding $(p-k)^{2p-1}$ with the binomial theorem, we get\... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
0efg | Problem:
Za linearno funkcijo $f$ velja $f(-1)+f(3)=-8$ in $f(1)+f(5)=4$. Koliko je $f(2)+f(7)$ ?
(A) 12
(B) 13
(C) 14
(D) 15
(E) 16 | [
"Solution:\nLinearna funkcija je oblike $f(x)=k \\cdot x+n$. Iz prve zveze dobimo $2k+2n=-8$, iz druge pa $6k+2n=4$. Iz tega sledi $k=3$ in $n=-7$, $f(x)=3x-7$. Tako velja $f(2)+f(7)=13$. Pravilen odgovor je (B)."
] | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | B | |
03k3 | Problem:
Solve the equation
$$
x^{2} + \frac{x^{2}}{(x+1)^{2}} = 3
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | (1 + sqrt(5))/2, (1 - sqrt(5))/2 | |
06u3 | Let $n$ be an odd positive integer. In the Cartesian plane, a cyclic polygon $P$ with area $S$ is chosen. All its vertices have integral coordinates, and the squares of its side lengths are all divisible by $n$. Prove that $2 S$ is an integer divisible by $n$. | [
"Let $P = A_{1} A_{2} \\ldots A_{k}$ and let $A_{k+i} = A_{i}$ for $i \\geqslant 1$. By the Shoelace Formula, the area of any convex polygon with integral coordinates is half an integer. Therefore, $2 S$ is an integer. We shall prove by induction on $k \\geqslant 3$ that $2 S$ is divisible by $n$. Clearly, it suffi... | IMO | IMO 2016 Shortlisted Problems | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
07ra | Determine, with proof, the smallest positive multiple of $99$ all of whose digits are either $1$ or $2$. | [
"We call a number *eligible* if its digits are all either $1$ or $2$. A number is divisible by $99 = 9 \\cdot 11$ if and only if it is divisible by both $9$ and $11$. Suppose $N \\in \\mathbb{N}$ has base-10 expansion $a_n a_{n-1} \\dots a_2 a_1$. We define three digit-sums (full, odd, even):\n$$\nS(N) := \\sum_{1 ... | Ireland | Ireland_2017 | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 1122222222 | |
0jvy | Problem:
Let the sequence $a_{i}$ be defined as $a_{i+1} = 2^{a_{i}}$. Find the number of integers $1 \leq n \leq 1000$ such that if $a_{0} = n$, then $100$ divides $a_{1000} - a_{1}$. | [
"Solution:\nWe claim that $a_{1000}$ is constant mod $100$.\n\n$a_{997}$ is divisible by $2$. This means that $a_{998}$ is divisible by $4$. Thus $a_{999}$ is constant mod $5$. Since it is also divisible by $4$, it is constant $\\bmod\\ 20$. Thus $a_{1000}$ is constant $\\bmod\\ 25$, since $\\phi(25) = 20$. Since $... | United States | HMMT November 2016 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic... | null | proof and answer | 50 | |
01ik | Find all (not necessarily strictly) monotonic functions $f: \mathbb{R} \to \mathbb{R}$ with
$$
f(x+y)^3 = f(x^3) + f(y^3) \text{ for all } x, y \in \mathbb{R}.
$$ | [
"By substituting $x = y = 0$ we see that $f(0)^3 = 2f(0)$ which implies $f(0)^3 - 2f(0) = 0$. This means that $f(0) = 0$, $f(0) = -\\sqrt{2}$ or $f(0) = \\sqrt{2}$.\n\nBy substituting $y = 0$ we get $f(x)^3 = f(x^3) + f(0^3) = f(x^3) + f(0)$.\n\n1) Let's first investigate the case $f(0) = 0$: in this case, $f(x)^3 ... | Baltic Way | Baltic Way 2023 Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All such functions are: f(x) = 0 for all x; f(x) = cube_root(x); f(x) = -cube_root(x); f(x) = sqrt(2) for all x; f(x) = -sqrt(2) for all x. | |
0i09 | Problem:
How many non-empty subsets of $\{1,2,3,4,5,6,7,8\}$ have exactly $k$ elements and do not contain the element $k$ for some $k=1,2, \ldots, 8$. | [
"Solution:\nProbably the easiest way to do this problem is to count how many non-empty subsets of $\\{1,2, \\ldots, n\\}$ have $k$ elements and do contain the element $k$ for some $k$. The element $k$ must have $k-1$ other elements with it to be in a subset of $k$ elements, so there are $\\binom{n-1}{k-1}$ such sub... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 127 | |
0ldh | For every positive integer $n$, let $x_n = C_{2n}^n$.
1. Show that if $\frac{2017^k}{2} < n < 2017^k$ for some positive integer $k$ then $x_n$ is a multiple of $2017$.
2. Find all positive integer $h > 1$ such that there exist positive integers $N, T$ such that for all $n > N$ then $(x_n)$ is a periodic sequence mod ... | [
"1) We prove that the statement is true for all odd prime $p$ instead of $2017$. Suppose there exists a positive integer $k$ such that $\\frac{p^k}{2} < n < p^k$. We have\n$$\nv_p(x_n) = v_p(C_{2n}^n) = v_p((2n)!) - 2v_p(n!).\n$$\nBecause $\\frac{p^k}{2} < n < p^k$ so $p^k < 2n < 2p^k < p^{k+1}$, hence\n$$\nv_p((2n... | Vietnam | Vietnamese Team Selection Test for IMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | h = 2 | |
00rh | Find all integers $n \ge 2$ for which there exist the real numbers $a_k$, $1 \le k \le n$, which are satisfying the following conditions:
$$
\sum_{k=1}^{n} a_k = 0, \quad \sum_{k=1}^{n} a_k^2 = 1 \text{ and } \sqrt{n} \cdot \left( \sum_{k=1}^{n} a_k^3 \right) = 2(b\sqrt{n} - 1), \text{ where } b = \max_{1 \le k \le n} ... | [
"We have: $\\left(a_k + \\frac{1}{\\sqrt{n}}\\right)^2 (a_k - b) \\le 0 \\Rightarrow \\left(a_k^2 + \\frac{2}{\\sqrt{n}} \\cdot a_k + \\frac{1}{n}\\right) (a_k - b) \\le 0 \\Rightarrow a_k^3 \\le \\left(b - \\frac{2}{\\sqrt{n}}\\right) a_k^2 + \\left(\\frac{2b}{\\sqrt{n}} - \\frac{1}{n}\\right) a_k + \\frac{b}{n} \... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | All even integers n ≥ 2 | |
08x1 | Let $a_1, a_2, \dots$ be an infinite sequence of distinct non-zero real numbers for which $\frac{a_{i+1}}{a_i} + \frac{a_{i+1}}{a_i}$ takes the same value lying in between 0 and 2 for each $i \ge 1$. Express in terms of $a_1, a_2, a_3$ the smallest number $c$ satisfying the following condition:
Condition: For any pair ... | [
"$$\n\\boxed{\\frac{2}{\\sqrt{4 - \\left(\\frac{a_2}{a_1} + \\frac{a_2}{a_3}\\right)^2}}}\n$$\n\nFix a positive integer $x$. Let for a positive integer $y$ greater than $x$,\n$$\nb_y = \\frac{a_x a_{x+1} + a_{x+1} a_{x+2} + \\dots + a_{y-1} a_y}{a_x a_y}.\n$$\nThen, we get\n$$\n\\begin{aligned}\nb_y + b_{y+2} &= \\... | Japan | Japan 2013 Initial Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 2 / sqrt(4 - (a_2/a_1 + a_2/a_3)^2) | |
0i1a | Problem:
A permutation of the numbers $1,2, \ldots, n$ is called "bad" if it contains a subsequence of 10 numbers in decreasing order, and "good" otherwise. For example, for $n=15$,
$$
15,13,1,12,7,11,9,8,10,6,5,4,3,2,14
$$
is a bad permutation, because it contains the subsequence
$$
15,13,12,11,10,6,5,4,3,2
$$
Prove ... | [
"Solution:\n\nConsider any permutation of $1,2, \\ldots, n$. Let the \"height\" of a number $i$ in the permutation be the length of the longest decreasing subsequence ending in $i$. Then, the permutation is bad if and only if some number has height at least 10. Also note that all numbers with the same height must b... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0fih | Problem:
Probar que existe una sucesión de enteros positivos $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ tal que
$$
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}
$$
es un cuadrado perfecto para todo entero positivo $n$. | [
"Solution:\n\nLo haremos por inducción sobre $n$. Para $n=2$ basta tomar $a_{1}=3, a_{2}=4$ con $3^{2}+4^{2}=5^{2}$. Supongamos por hipótesis de inducción que\n$$\na_{1}^{2}+a_{2}^{2}+\\cdots+a_{n}^{2}=k^{2}\n$$\nVeamos que podemos encontrar un entero positivo $a_{n+1}$ tal que $k^{2}+a_{n+1}^{2}=p^{2}$.\nEn efecto... | Spain | Olimpiada Matemática Española | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0iut | Problem:
Let $f$ be a function that takes in a triple of integers and outputs a real number. Suppose that $f$ satisfies the equations
$$
\begin{aligned}
f(a, b, c) & = \frac{f(a+1, b, c) + f(a-1, b, c)}{2} \\
f(a, b, c) & = \frac{f(a, b+1, c) + f(a, b-1, c)}{2} \\
f(a, b, c) & = \frac{f(a, b, c+1) + f(a, b, c-1)}{2}
... | [
"Solution:\n\nAnswer: 8\n\nNote that if we have the value of $f$ at the 8 points: $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(0,0,1)$, $(0,1,1)$, $(1,0,1)$, $(1,1,0)$, $(1,1,1)$, we can calculate the value for any triple of points because we have that $f(a+1, b, c) - f(a, b, c)$ is constant for any $a$, if $b$ and $c$ are f... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Functional equations"
] | null | proof and answer | 8 | |
01hu | During one test a service dog smells a pile of coins and barks if there is a fake coin in it. If a dog is sick, whether it barks or not does not depend on the presence of fake coin, it happens randomly. Suppose $k \le 2^s$ and that we have $2^k$ coins, exactly one of which is fake. Assume that we have an excess number ... | [
"Number the coins by $k$-digit binary numbers from $00\\ldots0$ to $11\\ldots1$. Let $A_i$ be the set of coins which have $0$ in $i$-th position of the binary number. The first $k$ tests we perform with the help of $k$ different dogs. In the $i$-th test we determine whether the set $A_i$ contains the fake coin. Wit... | Baltic Way | Baltic Way 2021 Shortlist | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Logic"
] | null | proof only | null | |
08ij | Problem:
The side lengths of the triangle $ABC$ satisfy the relations $a > b \geq 2c$. Prove that the altitudes of the triangle $ABC$ can not be the sides of any triangle. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
04ko | Determine all real numbers $a$ such that the equation
$$
x^2 - (5-a)x + a^2 - 11a - 46 = 0
$$
has two real solutions, one of which is less than $2$, and the other greater than $2$. | [
"Let the roots of the quadratic be $x_1$ and $x_2$, with $x_1 < 2 < x_2$.\n\nThe quadratic equation is $x^2 - (5-a)x + a^2 - 11a - 46 = 0$.\n\nBy Vieta's formulas:\n- $x_1 + x_2 = 5 - a$\n- $x_1 x_2 = a^2 - 11a - 46$\n\nSince the equation must have two real solutions, the discriminant must be non-negative:\n$$\nD =... | Croatia | Mathematical competitions in Croatia | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | -4 < a < 13 | |
0ff3 | Problem:
Encontrar todas la soluciones $(x, y)$ reales del sistema de ecuaciones
$$
\left.\begin{array}{c}
x^{2}-x y+y^{2}=7 \\
x^{2} y+x y^{2}=-2
\end{array}\right\}
$$ | [
"Solution:\nComo la segunda ecuación se puede escribir en la forma\n$$\nx y(x+y)=-2\n$$\nvamos a escribir la primera de manera relativamente parecida:\n$$\n(x+y)^{2}-3 x y=7\n$$\nHaciendo el cambio de variables $x+y=s, \\quad x y=p$ obtenemos el sistema equivalente\n$$\n\\left.\\begin{array}{c}\ns^{2}-3 p=7 \\\\\ns... | Spain | TANDA III | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | {(-1, 2), (2, -1), (1 + sqrt(2), 1 - sqrt(2)), (1 - sqrt(2), 1 + sqrt(2)), ((-9 + sqrt(57))/6, (-9 - sqrt(57))/6), ((-9 - sqrt(57))/6, (-9 + sqrt(57))/6)} | |
05z5 | Problem:
Soit $x, y, z$ trois nombres réels vérifiant $x+y+z=2$ et $xy+yz+zx=1$. Déterminer la valeur maximale que peut prendre $x-y$. | [
"Solution:\n\nSoit $(x, y, z)$ un triplet vérifiant l'énoncé. Quitte à échanger $x$ et le maximum du triplet, puis $z$ et le minimum du triplet, on peut supposer $x \\geqslant z \\geqslant y$, tout en augmentant $x-y$.\n\nComme $x+y+z=2$, on a $4=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=x^2+y^2+z^2+2$, donc $x^2+y^2+z^2=2... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 2/sqrt(3) | |
0ld8 | There are $m$ boys and $n$ girls took part in a festival of singing couples. Each day, there are couples sang songs in a presentation. Each couple had a boy and a girl. In a presentation, every one sang at least one song, and a boy sang a duet with a girl not more than one time. Two presentations were different if ther... | [
"We denote the boys by the integers from $1$ to $m$ and girls by the integers from $1$ to $n$. Each presentation is presented by a table of $m$ rows (corresponding to $m$ girls) and $n$ columns (corresponding to $n$ boys). Each cell takes a label $0$ or $1$ as shown in the following:\n\n* Cell in $i$-th row and $j$... | Vietnam | Vietnam Mathematical Olympiad 2015 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
03gg | Problem:
$ABCD$ is a quadrilateral with $AD = BC$. If $\angle ADC$ is greater than $\angle BCD$, prove that $AC > BD$. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
05m6 | Problem:
Dans un pays, se trouvent 100 villes. Chacune de ces villes est reliée à exactement trois autres villes par des routes directes dans les deux sens. Prouver qu'il existe une ville $A$ à partir de laquelle on peut aller de ville en ville et revenir en $A$, sans jamais passer deux fois par une même route, et en ... | [
"Solution:\n\nPuisqu'il n'y a qu'un nombre fini de villes, on peut considérer un chemin $C$ de longueur maximale. Soit $v_{0}$ une des villes extrémités de $C$ et parcourons $C$ en partant de $v_{0}$ en numérotant les villes au fur et à mesure. La maximalité de $C$ assure que les trois villes reliées à $v_{0}$ par ... | France | TEST DU GROUPE A ET DES CANDIDATES À L'ÉPREUVE EGMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
04i9 | Determine all pairs $(a, b)$ of integers such that the intersections of the parabola $y = x^2 + a x + b$ and the coordinate axes form a triangle whose area is equal to $3$. | [] | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (-1, -2), (1, -2), (-4, 3), (4, 3), (-5, 6), (5, 6) | |
0k5n | Problem:
Dylan has a $100 \times 100$ square, and wants to cut it into pieces of area at least $1$. Each cut must be a straight line (not a line segment) and must intersect the interior of the square. What is the largest number of cuts he can make? | [
"Solution:\n\nSince each piece has area at least $1$ and the original square has area $10000$, Dylan can end up with at most $10000$ pieces. There is initially $1$ piece, so the number of pieces can increase by at most $9999$. Each cut increases the number of pieces by at least $1$, so Dylan can make at most $9999$... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 9999 | |
0c0n | Let $n$ be a positive integer and let $Q$ be an $n \times n \times n$ cube. A box is a subset of $Q$ of one of the forms $1 \times n \times n$, $n \times 1 \times n$ or $n \times n \times 1$. Each $1 \times 1 \times 1$ cell of $Q$ is coloured one of several many colours. Consider the colour set of each box (each colour... | [
"The required maximum is $n(n+1)(2n+1)/6$ and is achieved, for instance, by colouring cells as described below. For each colour, we list the set of all cells bearing that colour:\n* $n$ singletons of the form $\\{(i, i, i)\\}$, where $1 \\le i \\le n$;\n* $3\\binom{n}{2}$ doubletons of the form $\\{(i, j, j), (j, i... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n(n+1)(2n+1)/6 | |
0hnb | Problem:
On an infinite chessboard, two squares are said to touch if they share at least one vertex and they are not the same square. Suppose that the squares are colored black and white such that
- there is at least one square of each color;
- each black square touches exactly $m$ black squares;
- each white square t... | [
"Solution:\n\nThe answer is no. There are many tilings to demonstrate this; one of the simplest is to divide the board into horizontal stripes and color every third stripe black. In this tiling, $m=2$ and $n=5$."
] | United States | Berkeley Math Circle Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No | |
0htr | Problem:
Let $0 < a_{0} \leq a_{1} \leq \cdots \leq a_{n}$. If $z$ is a complex number such that $a_{0} z^{n} + a_{1} z^{n-1} + \cdots + a_{n} = 0$ prove that $|z| \geq 1$. | [
"Solution:\n\nAssume that $|z| < 1$. If $a_{0} z^{n} + a_{1} z^{n-1} + \\cdots + a_{n} = 0$ then $a_{0} z^{n+1} + a_{1} z^{n} + \\cdots + a_{n} z = 0$ and subtracting these two equations leads to $a_{0} z^{n+1} + (a_{1} - a_{0}) z^{n} + \\cdots + (a_{n} - a_{n-1}) z - a_{n} = 0$, or equivalently $a_{n} = a_{0} z^{n... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0gjm | 令 $n \ge 3$ 為一正整數。有來自 $k$ 間學校的共 $n$ 位咒術師，編號為 1 到 $n$。已知當兩個咒術師對決時，編號小的咒術師會獲勝。此外，對於任何 $\{1, 2, \dots, n\}$ 的重排 $\{x_1, x_2, \dots, x_n\}$，$x_1$ 對 $x_2$ 號、$x_2$ 號對 $x_3$ 號一直到 $x_{n-1}$ 號對 $x_n$ 號咒術師的 $n-1$ 場對決的獲勝者中，包含 $k$ 間學校的人各至少一位。試證 $n \ge 2^k$。
Let $n \ge 3$ be a positive integer. There are $n$ Jujutsushis in total f... | [
"將 $n$ 名咒術師當成 $n$ 個點 $A_1, \\dots, A_n$，以學校為顏色對各點塗色，並對所有 $1 \\le i < j \\le n$，將 $A_iA_j$ 連線並塗上 $A_i$ 的顏色。我們先證明以下關鍵引理。\n\n**引理：** 對於第 $i$ 種顏色，存在 $1 \\le p \\le n$，使得 $A_1, A_2, \\dots, A_p$ 中的 $i$ 色點數量多於 $p/2$。\n\n證明：若否，則存在 $i$ 使得對於所有 $1 \\le p \\le n$，前 $p$ 個點中都至多只有 $\\lfloor p/2 \\rfloor$ 個 $i$ 色點。令 $A_{x_1}, A_{... | Taiwan | IMO 3J, Independent Study 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Chinese; English | proof only | null |
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