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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
03se | Let $\triangle ABC$ be a given triangle. If $|\vec{BA} - t \vec{BC}| \ge |\vec{AC}|$ for any $t \in \mathbb{R}$, then $\triangle ABC$ is ( ). | [
"Suppose $\\angle ABC = \\alpha$. Since $|\\vec{BA} - t \\vec{BC}| \\ge |\\vec{AC}|$, we have\n$$\n|\\vec{BA}|^2 - 2t \\vec{BA} \\cdot \\vec{BC} + t^2 |\\vec{BC}|^2 \\ge |\\vec{AC}|^2.\n$$\nLet\n$$\nt = \\frac{\\vec{BA} \\cdot \\vec{BC}}{|\\vec{BC}|^2},\n$$\nwe get\n$$\n|\\vec{BA}|^2 - 2|\\vec{BA}|^2 \\cos^2 \\alph... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | MCQ | C | |
0ccv | Find all positive integers $n$ so that $N = \frac{1}{n(n+1)}$ is a finite decimal fraction. | [
"The number $N$ is a finite decimal fraction if and only if its denominator is of the form $2^a \\cdot 5^b$, with $a, b \\in \\mathbb{N}$.\nSince $n$ and $n+1$ are coprime, the possible cases are:\nI) $n = 1$, $n+1 = 2^a \\cdot 5^b$ (with an obvious conclusion);\nII) $n = 5^b$, $n+1 = 2^a$;\nIII) $n = 2^a$, $n+1 = ... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | n = 1 and n = 4 | |
0ic9 | Problem:
If the system of equations
$$
\begin{aligned}
& |x+y|=99 \\
& |x-y|=c
\end{aligned}
$$
has exactly two real solutions $(x, y)$, find the value of $c$. | [
"Solution:\nIf $c<0$, there are no solutions. If $c>0$ then we have four possible systems of linear equations given by $x+y= \\pm 99$, $x-y= \\pm c$, giving four solutions $(x, y)$. So we must have $c=0$, and then we do get two solutions ($x=y$, so they must both equal $\\pm 99 / 2$)."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | c = 0 | |
026r | Problem:
A seguinte figura mostra um cubo.
Calcule o número de triângulos equiláteros que podem ser formados de modo que seus três vértices sejam vértices do cubo. | [
"Solution:\n\nSobre o cubo existem somente 3 distâncias possíveis entre os vértices:\n$$\nAB = \\ell, \\quad AR = \\ell \\sqrt{2} \\quad \\text{e} \\quad PD = \\ell \\sqrt{3}\n$$\nUm triângulo equilátero que usa os vértices do cubo devia então ter alguma dessas distâncias como a medida do seu lado.\n![](attached_im... | Brazil | null | [
"Geometry > Solid Geometry > 3D Shapes",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 8 | |
02pd | Problem:
Num torneio de tênis cada jogador passa para a rodada seguinte somente em caso de vitória. Se não for possível que sempre passe para a rodada seguinte um número par de jogadores, a organização do torneio decide quais rodadas determinados jogadores devem jogar. Por exemplo, um cabeça de chave pode, a critério d... | [
"Solution:\n\na. Na primeira rodada são realizadas $32$ partidas, das quais $32$ jogadores passam para a fase seguinte. Depois são realizadas $16$ partidas, classificando $16$ para a rodada seguinte e assim por diante. Assim, o número de partidas do torneio é\n$$\n32 + 16 + 8 + 4 + 2 + 1 = 63\n$$\n\nb. Como em cada... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) 63; b) 2010 | |
07rs | Suppose $p \ge 1$ and $a$, $b$, $c$ are the side lengths of a triangle. Prove that
$$
2(a^p + b^p + c^p) < (a+b+c)(a^{p-1} + b^{p-1} + c^{p-1}) \le 3(a^p + b^p + c^p).
$$ | [
"Notice that\n$$\n\\begin{align*}\n& (a + b + c)(a^{p-1} + b^{p-1} + c^{p-1}) - 2(a^p + b^p + c^p) \\\\\n&= (b + c)a^{p-1} + (c + a)b^{p-1} + (a + b)c^{p-1} - a^p - b^p - c^p \\\\\n&= (b + c - a)a^{p-1} + (c + a - b)b^{p-1} + (a + b - c)c^{p-1} > 0,\n\\end{align*}\n$$\nby the triangle inequality, since $a$, $b$, $c... | Ireland | Irish | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
06sj | Players $A$ and $B$ play a paintful game on the real line. Player $A$ has a pot of paint with four units of black ink. A quantity $p$ of this ink suffices to blacken a (closed) real interval of length $p$. In every round, player $A$ picks some positive integer $m$ and provides $1 / 2^{m}$ units of ink from the pot. Pla... | [
"No. Such a strategy for player $A$ does not exist.\n\nWe will present a strategy for player $B$ that guarantees that the interval $[0,1]$ is completely blackened, once the paint pot has become empty.\n\nAt the beginning of round $r$, let $x_{r}$ denote the largest real number for which the interval between $0$ and... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No; player A has no such winning strategy. | |
0ijg | Problem:
Four points are independently chosen uniformly at random from the interior of a regular dodecahedron. What is the probability that they form a tetrahedron whose interior contains the dodecahedron's center? | [
"Solution:\nTo any tetrahedron $P_{1} P_{2} P_{3} P_{4}$ we can associate a quadruple $\\left(\\epsilon_{(i j k)}\\right)$, where $(i j k)$ ranges over all conjugates of the cycle $(123)$ in the alternating group $A_{4}$: $\\epsilon_{i j k}$ is the sign of the directed volume $\\left[O P_{i} P_{j} P_{k}\\right]$. A... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Linear Algebra > Vectors",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | 1/8 | |
0c5t | Let $a$, $b$, $c$ be positive real numbers. Prove that
$$
\frac{1}{abc} + 1 \geq 3 \left( \frac{1}{a^2 + b^2 + c^2} + \frac{1}{a + b + c} \right).
$$ | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0har | Does the triangle, sides of which can be expressed as a positive integer in centimeters and 2 of its medians are perpendicular, exist? | [
"Let's find the condition on the sides of the triangle, under which the medians are perpendicular. Without loss of generality let's consider that medians $AM$ (from the edge $A$) and $BN$ (from the edge $B$) are perpendicular. Let's use vectors: $\\vec{c} = \\overrightarrow{AB}$, $\\vec{b} = \\overrightarrow{AC}$. ... | Ukraine | The Problems of Ukrainian Authors | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | Yes; for example, a triangle with sides 13 cm, 19 cm, and 22 cm has two perpendicular medians. | |
01p4 | A triangle $ABC$ is inscribed in the parabola $y = x^2$. Let $a$, $b$, $c$ be the abscissae of the midpoints of its sides.
Find the radius of the circumcircle of $\triangle ABC$. | [
"Answers: $R = 0.5\\sqrt{(1 + 4a^2)(1 + 4b^2)(1 + 4c^2)}$.\n\nLet $A(l; l^2)$, $B(m; m^2)$, $C(n; n^2)$ be coordinates of the vertices of the triangle $ABC$. Then\n$$\nAB = \\sqrt{(m-l)^2 + (m^2-l^2)^2} = |m-l|\\sqrt{1+(m+l)^2} = |m-l|\\sqrt{1+4c^2},\n$$\nsince $c = 0.5(m + l)$.\nSimilarly, $BC = |m - n|\\sqrt{1 + ... | Belarus | BelarusMO 2013_s | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | R = 0.5*sqrt((1 + 4a^2)(1 + 4b^2)(1 + 4c^2)) | |
0jzf | Problem:
Emily starts with an empty bucket. Every second, she either adds a stone to the bucket or removes a stone from the bucket, each with probability $\frac{1}{2}$. If she wants to remove a stone from the bucket and the bucket is currently empty, she merely does nothing for that second (still with probability $\fr... | [
"Solution:\n\nAnswer: $\\frac{\\binom{2017}{340}}{2^{2017}}$\n\nReplace 2017 with $n$ and 1337 with $k$ and denote the general answer by $f(n, k)$. I claim that $f(n, k) = \\frac{\\binom{n}{\\frac{n-k}{2}}}{2^n}$.\n\nWe proceed by induction on $n$.\n\nThe claim is obviously true for $n=0$ since $f(0,0)=1$. Moreover... | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | final answer only | binom(2017, 340) / 2^2017 | |
098o | Problem:
Găsiți toate funcțiile continue $f: \mathbf{R} \rightarrow \mathbf{R}$, care verifică relația
$$
3 \cdot f(2x+1) = f(x) + 5x, \quad \forall x \in \mathbf{R}
$$ | [
"Solution:\nMai întâi vom căuta soluțiile în clasa funcțiilor de gradul întâi, de forma $f(x) = a x + b$.\nPentru astfel de funcții obținem relațiile echivalente:\n$$\n\\begin{gathered}\n3 \\cdot (a \\cdot (2x+1) + b) = a x + b + 5x, \\quad \\forall x \\in \\mathbf{R}, \\quad \\Leftrightarrow \\\\\n6a x + 3a + 3b =... | Moldova | Moldova National Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x - 3/2 | |
0dp9 | Prove that for every prime number $p$ there exists infinitely many 4-tuples $(x, y, z, t)$ of pairwisely distinct positive integers such that the number
$$
(x^2 + pt^2)(y^2 + pt^2)(z^2 + pt^2)
$$
is a square of an integer. | [
"Firstly, note that the equation $x^2 - py^2 = 1$ has infinitely many solutions in positive integers. (Pell's equation)\nThen for every prime number $p$ there exist infinitely many positive integers $s$ and $t$ such that $s^2 - 1 = pt^2$.\nPutting $x = s^2 - 1$, $y = s + 1$, $z = s - 1$ we have\n$$\n(x^2 + s^2 - 1)... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
07jc | For a sequence of positive integers $x_1, x_2, \dots$, we perform the following operation: In the $i$-th step, we mark all rational numbers in the interval $[0, 1]$ with denominator $x_i$ (i.e., numbers of the form $j/x_i$ for $j \in \{0, 1, \dots, x_i\}$). Let $a_i$ be the length of the shortest interval whose two end... | [
"First, note that since $a_1 + a_2 + \\cdots + a_n$ is increasing according to the relation $a_1 + a_2 + \\cdots + a_n = \\frac{x_n}{n}$, the sequence $x_n$ must also be increasing. We prove by induction that $a_n = n$. First, notice that by the problem's condition, it holds up to $n = 5$. Assume the statement hold... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | x_n = n for all n | |
040h | Let $H$ and $O$ be the orthocenter and circumcenter of an acute triangle $ABC$ ($A$, $H$, $O$ are non-collinear). Suppose that $D$ is the projection of $A$ onto the line $BC$, and the perpendicular bisector of the segment $AO$ meets the line $BC$ at $E$. Prove that the midpoint of $OH$ is on the circumcircle of triangl... | [] | China | China Western Invitational Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
042p | Suppose positive real numbers $a$, $b$, and $c$ satisfy $a^2 + 4b^2 + 9c^2 = 4b + 12c - 2$. Find the minimum of $\frac{1}{a} + \frac{2}{b} + \frac{3}{c}$. | [
"By the given condition, we have\n$$\na^2 + (2b - 1)^2 + (3c - 2)^2 = 3.\n$$\nBy making use of the Cauchy inequality, we get\n$$\n3[a^2 + (2b - 1)^2 + (3c - 2)^2] \\geq (a + 2b - 1 + 3c - 2)^2,\n$$\nnamely, $(a + 2b + 3c - 3)^2 \\leq 9$. Therefore,\n$$\na + 2b + 3c \\leq 6.\n$$\nAgain, by the Cauchy inequality we g... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 6 | |
0f0a | Problem:
A sequence of finite sets of positive integers is defined as follows. $S_0 = \{m\}$, where $m > 1$. Then given $S_n$ you derive $S_{n + 1}$ by taking $k^2$ and $k + 1$ for each element $k$ of $S_n$. For example, if $S_0 = \{5\}$, then $S_2 = \{7, 26, 36, 625\}$. Show that $S_n$ always has $2^n$ distinct elemen... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
07t5 | Let $a$, $b$, $c > 0$. Prove that
$$
\sqrt[7]{\frac{a}{b+c} + \frac{b}{c+a}} + \sqrt[7]{\frac{b}{c+a} + \frac{c}{a+b}} + \sqrt[7]{\frac{c}{a+b} + \frac{a}{b+c}} \ge 3.
$$ | [
"Because $\\sqrt[3]{\\sqrt[7]{x}} = \\sqrt[21]{x}$, the three-term AM-GM inequality gives\n$$\n\\sqrt[7]{\\frac{a}{b+c} + \\frac{b}{c+a}} + \\sqrt[7]{\\frac{b}{c+a} + \\frac{c}{a+b}} + \\sqrt[7]{\\frac{c}{a+b} + \\frac{a}{b+c}} \n\\geq 3 \\sqrt[21]{\\left(\\frac{a}{b+c} + \\frac{b}{c+a}\\right) \\left(\\frac{b}{c+a... | Ireland | IRL_ABooklet_2020 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof only | null | |
0aou | Problem:
If $\sqrt[3]{x+5} - \sqrt[3]{x-5} = 1$, find $x^{2}$. | [
"Solution:\n$52$\nBy factoring with difference of two cubes, we have\n$$\n\\begin{aligned}\n& (\\sqrt[3]{x+5})^{3} - (\\sqrt[3]{x-5})^{3} \\\\\n& \\quad = (\\sqrt[3]{x+5} - \\sqrt[3]{x-5})\\left(\\sqrt[3]{(x+5)^{2}} + \\sqrt[3]{(x+5)(x-5)} + \\sqrt[3]{(x-5)^{2}}\\right)\n\\end{aligned}\n$$\nwhich can be simplified ... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 52 | |
0708 | Problem:
The incircle of the triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. $AD$ meets the circle again at $X$ and $AX = XD$. $BX$ meets the circle again at $Y$ and $CX$ meets the circle again at $Z$. Show that $EY = FZ$. | [] | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jne | Problem:
Call a string of letters $S$ an almost palindrome if $S$ and the reverse of $S$ differ in exactly two places.
Find the number of ways to order the letters in HMMTTHEMETEAM to get an almost palindrome. | [
"Solution:\nAnswer: $2160$\nNote that $T$, $E$, $A$ are used an odd number of times. Therefore, one must go in the middle spot and the other pair must match up. There are $3 \\cdot 2 \\left(\\frac{6!}{2!}\\right) = 2160$ ways to fill in the first six spots with the letters $T$, $H$, $E$, $M$, $M$ and a pair of diff... | United States | HMMT November 2015 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 2160 | |
05n7 | Problem:
Soit $a_{1}, a_{2}, \ldots, a_{n}$ des entiers strictement positifs. Pour tout $k=1,2, \ldots, n$, on note
$$
m_{k}=\max _{1 \leq \ell \leq k} \frac{a_{k-\ell+1}+a_{k-\ell+2}+\cdots+a_{k}}{\ell} .
$$
Montrer que pour tout $\alpha>0$, le nombre d'entiers $k$ tel que $m_{k}>\alpha$ est strictement plus petit qu... | [
"Solution:\n\nSoit $k_{1}$ le plus grand entier $k$ tel que $m_{k}>\\alpha$. Il existe $\\ell_{1}$ tel que $a_{k_{1}-\\ell_{1}+1}+\\cdots+a_{k_{1}}>\\ell_{1} \\alpha$.\nSoit $k_{2}$ le plus grand entier $\\leqslant k_{1}-\\ell_{1}$ tel que $m_{k}>\\alpha$. Il existe $\\ell_{2}$ tel que $a_{k_{2}-\\ell_{2}+1}+\\cdot... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0dq7 | In the triangle $ABC$ with $AC > AB$, $D$ is the foot of the perpendicular from $A$ onto $BC$ and $E$ is the foot of the perpendicular from $D$ onto $AC$. Let $F$ be the point on the line $DE$ such that $EF \cdot DC = BD \cdot DE$. Prove that $AF$ is perpendicular to $BF$. | [
"Since we are supposed to prove $\\angle AFB = 90^\\circ$, it means that the 4 points $A$, $B$, $D$, $F$ are concyclic. Note that $AC > AB$ implies that $\\angle B > \\angle C$. If $TD$ is the tangent to the circumcircle of the triangle $ABD$ with $B$ and $T$ lying on opposite sides of the line $AD$, then $\\angle ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06hw | If $x$ is a real number, find the minimum value of $|x + 1| + 2|x - 5| + |2x - 7| + \left|\frac{x - 11}{2}\right|$. | [
"It suffices to minimise $2|x+1| + 4|x-5| + 4\\left|x - \\frac{7}{2}\\right| + |x - 11|$ (which is 2 times the given expression), i.e. to find the minimum total distance from $x$ to the 11 numbers:\n$$\n-1, -1, \\frac{7}{2}, \\frac{7}{2}, \\frac{7}{2}, \\frac{7}{2}, 5, 5, 5, 5, 11.\n$$\nBy the triangle inequality, ... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 45/4 | |
08at | Problem:
Caboyara, famoso circense australiano, si esibisce anche quest'anno in un gran trucco. Predispone una scala spettacolare con $N = p_{1} \cdot p_{2} \cdot \ldots \cdot p_{2015}$ gradini, dove $p_{1}, p_{2}, \ldots, p_{2015}$ sono numeri primi distinti; i gradini che corrispondono a divisori di $N$ (compresi il... | [
"Solution:\n\nLa risposta è (B). Dato il gradino speciale corrispondente al divisore $d$ di $N$, l'$i$-esimo canguro vi salterà sopra se e solo se il primo $p_{i}$ è un fattore di $d$; la luce del gradino cambierà quindi colore tante volte quanti sono i fattori primi di $d$: sarà verde alla fine dell'esibizione se ... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | B | |
0hfa | Let $O$ be the circumcenter of triangle $ABC$ with $\angle A = 120^\circ$. Let $P$ and $Q$ denote the projections of $B$ onto $CO$ and $AO$, respectively. Let $M$ be the midpoint of $AO$. Prove that the circumcircle of $\triangle MPQ$ touches $AC$. | [
"Let $R$ be the projection of $B$ onto $AC$ (fig. 23). Then $P, Q, R$ are the projections of $B$ onto the sides of $\\triangle AOC$. Let $B_1$ be the isogonal conjugate of $B$ with respect to this triangle. Since $\\angle BAC = \\angle BOC = 120^\\circ$, we have $\\angle B_1AO = \\angle B_1OA$. Then the projection ... | Ukraine | Problems from Ukrainian Authors | [
"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 > Advanced Configurations... | English | proof only | null | |
0e87 | Problem:
Naj bo $E$ taka točka na stranici $CD$ pravokotnika $ABCD$, da je kot $\angle AEB$ pravi in velja $3|EA|=2|EC|$. Določi razmerje med dolžinama stranic pravokotnika $ABCD$. | [
"Solution:\n\nOznačimo $|AB|=|CD|=a$, $|BC|=|DA|=b$ in $|EC|=c$. Tedaj je $|EA|=\\frac{2}{3}c$ in $|ED|=a-c$. Po Pitagorovem izreku za trikotnik $AED$ velja\n$$\nb^2 + (a-c)^2 = \\frac{4}{9}c^2\n$$\n oziroma\n$$\nb^2 = -a^2 + 2ac - \\frac{5}{9}c^2.\n$$\nPo Pitagorovem izreku za trikotnika $BCE$ in $ABE$ velja\n$$\n... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 4*sqrt(3)/3 | |
0iel | Problem:
Let $B$ be a set of integers either bounded below or bounded above. Then show that if $S$ tiles all other integers $\mathbf{Z} \backslash B$, then $S$ tiles all integers $\mathbf{Z}$. | [
"Solution:\n\nAssume $B$ is bounded above; the other case is analogous. Let $a$ be the difference between the largest and smallest element of $S$. Denote the sets in the partition of $\\mathbf{Z} \\backslash B$ by $S_{k}$, $k \\in \\mathbf{Z}$, such that the minimum element of $S_{k}$, which we will denote $c_{k}$,... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0gmh | Find all prime numbers $p$ for which the number of ordered pairs of integers $(x, y)$ satisfying the conditions $y^2 \equiv x^3 - x \pmod p$ and $0 \le x, y < p$ is exactly $p$. | [] | Turkey | X. NATIONAL MATHEMATICAL OLYMPIAD | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | p = 2 and all odd primes congruent to 3 modulo 4 | |
022l | Problem:
Renato tem trinta melancias, Leandro tem dezoito melancias e Marcelo tem vinte e quatro jacas. Ao contrário de Leandro e Renato, Marcelo não gosta de jaca. Por outro lado, os três gostam de melancia. Os três fazem então um acordo: Marcelo dá as suas vinte e quatro jacas para Leandro e Renato, e as melancias d... | [
"Solution:\n\nA ideia é determinar o valor de melancias em termos de jacas. Como são $18+30=48$ melancias e $24$ jacas, temos a proporção\n$$\n\\begin{array}{ccc}\n48 \\text{ melancias} & - & 24 \\text{ jacas} \\\\\n1 \\text{ melancia} & - & x \\text{ jacas}\n\\end{array}\n$$\no que dá $x=1/2$. Ou seja, uma melanci... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | Renato receives 15 jackfruits; Leandro receives 9 jackfruits. | |
0ci5 | A natural number $n \ge 5$ will be called *special* if, no matter how we choose five distinct numbers from $1, 2, 3, \ldots, n$, we find among them four distinct numbers $a, b, c, d$ so that $a + b = c + d$.
a) Prove that $n = 6$ is special.
b) Find all the special numbers. | [
"a) Since $1+6 = 2+5 = 3+4$, every set of $5$ numbers from $1, 2, 3, 4, 5, 6$ contains $4$ distinct numbers $a, b, c, d$ so that $a + b = c + d = 7$.\n\nb) Indeed, no $4$ numbers out of $1, 2, 3, 5, 8$ provide equal sums: if we do not choose $8$, then $5+a > b+c$, for every $a, b, c \\in \\{1, 2, 3\\}$, and if we c... | Romania | 74th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 5, 6, 7 | |
0k0b | Problem:
New this year at HMNT: the exciting game of $R N G$ baseball! In RNG baseball, a team of infinitely many people play on a square field, with a base at each vertex; in particular, one of the bases is called the home base. Every turn, a new player stands at home base and chooses a number $n$ uniformly at random... | [
"Solution:\n\nFor $i=0,1,2,3$, let $P_{i}$ be the probability that a player on the $i$-th base scores a point before strikeout (with zeroth base being the home base). We have the following equations:\n$$\n\\begin{aligned}\nP_{0} & =\\frac{1}{5}\\left(P_{1}+P_{2}+P_{3}+1\\right) \\\\\nP_{1} & =\\frac{1}{5}\\left(P_{... | United States | HMMT November 2017 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 409/125 | |
0kif | Let $x$ be the least real number greater than $1$ such that $\sin x = \sin(x^2)$, where the arguments are in degrees. What is $x$ rounded up to the closest integer?
(A) 10 (B) 13 (C) 14 (D) 19 (E) 20 | [] | United States | AMC 12 A | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | B | |
01x9 | Let the sequence $(a_n)$ be constructed in the following way:
$$
a_1 = 1,\ a_2 = 1,\ a_{n+2} = a_{n+1} + \frac{1}{a_n},\ n = 1, 2, \dots
$$
Prove that $a_{180} > 19$. | [
"First five terms of the sequence are $a_1 = 1$, $a_2 = 1$, $a_3 = 2$, $a_4 = 3$ and $a_5 = 3.5$.\n$$\n\\text{Note that } a_{n+2}^2 = a_{n+1}^2 + \\frac{2a_{n+1}}{a_n} + \\frac{1}{a_n^2} = a_{n+1}^2 + 2 + \\frac{2}{a_{n-1}a_n} + \\frac{1}{a_n^2} > a_{n+1}^2 + 2.\n$$\nTherefore, $a_{180}^2 > a_5^2 + 350 = 362.25 > 1... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
02ol | 33 friends are collecting stickers for a 2011-sticker album. A distribution of stickers among the 33 friends is *incomplete* when there is a sticker that no friend has. Determine the least $m$ with the following property: every distribution of stickers among the 33 friends such that, for any two friends, there are at l... | [
"Since $2011 = 33 \\cdot 61 - 2$, consider the example where $S_i = \\{k \\in \\mathbb{Z} \\mid 61(i-1) < k \\le 61i\\}$ for $i = 1, 2, \\dots, 31$, $S_{32} = \\{k \\in \\mathbb{Z} \\mid 61 \\cdot 31 < k < 61 \\cdot 32\\}$ and $S_{33} = \\{k \\in \\mathbb{Z} \\mid 61 \\cdot 32 \\le k \\le 2011\\}$.\nNotice that $|S... | Brazil | Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1890 | |
03md | Problem:
Let $x$, $y$ and $z$ be positive real numbers. Show that $x^{2} + x y^{2} + x y z^{2} \geq 4 x y z - 4$. | [
"Solution:\nNote that\n$$\nx^{2} \\geq 4x - 4, \\quad y^{2} \\geq 4y - 4, \\quad \\text{and} \\quad z^{2} \\geq 4z - 4\n$$\nand therefore\n$$\nx^{2} + x y^{2} + x y z^{2} \\geq (4x - 4) + x(4y - 4) + x y (4z - 4) = 4 x y z - 4.\n$$"
] | Canada | Sun Life Financial Canadian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
05hs | Problem:
Déterminer tous les entiers $n \geqslant 1$ tels qu'il existe une permutation $(a_{1}, a_{2}, \ldots, a_{n})$ de $(1,2, \ldots, n)$ vérifiant la condition suivante :
$$
k \mid a_{1}+a_{2}+\ldots+a_{k}
$$
pour tout $k \in\{1,2, \ldots, n\}$. | [
"Solution:\n\nOn commence par regarder ce qu'il se passe pour $n=1,2,3$ : $n=1,3$ sont solutions mais $n=2$ ne l'est pas. Soit $n>3$ vérifiant la propriété de l'énoncé.\nOn a $n \\mid a_{1}+a_{2}+\\ldots+a_{n}=1+2+\\ldots+n=\\frac{n(n+1)}{2}$ et donc $\\frac{n+1}{2} \\in \\mathbb{Z}$ : $n$ est impair.\nEnsuite $n-1... | France | ENVOi 3 : ARITHMÉTIQUE | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n = 1 and n = 3 | |
0hz0 | Problem:
Carl and Bob can demolish a building in $6$ days, Anne and Bob can do it in $3$, Anne and Carl in $5$. How many days does it take all of them working together if Carl gets injured at the end of the first day and can't come back? Express your answer as a fraction in lowest terms. | [
"Solution:\n\nLet $a$ be the portion of the work that Anne does in one day, similarly $b$ for Bob and $c$ for Carl. Then what we are given is the system of equations $b + c = 1/6$, $a + b = 1/3$, and $a + c = 1/5$.\n\nThus in the first day they complete $a + b + c = \\frac{1}{2}(1/6 + 1/3 + 1/5) = 7/20$, leaving $1... | United States | Harvard-MIT Math Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 59/20 | |
0dly | Let $ABC$ be a non-isosceles triangle with centroid $G$ and inscribed in circle $(O)$. Let $M, N, P$ be the midpoints of $BC, CA, AB$ respectively. Let $D, E, F$ be the reflection points of the foot of the internal bisector of angles $A, B, C$ through $M, N, P$ respectively. Prove that the orthocenter of triangle $DEF$... | [
"Suppose $EF$ intersects $BC$ at $D'$. Let $(\\omega_A)$ be the circle with diameter $DD'$. By the angle bisector theorem, one can get\n$$\n\\frac{DB}{DC} = \\frac{AC}{AB}, \\quad \\frac{EC}{EA} = \\frac{AB}{BC}, \\quad \\frac{FA}{FB} = \\frac{BC}{CA}.\n$$\nBased on Ceva's theorem, $AD, BE, CF$ are concurrent and $... | Saudi Arabia | Saudi Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem"... | null | proof only | null | |
03sn | Let $ABCD$ be a convex quadrilateral. Let $O$ be the intersection of $AC$ and $BD$. Let $O$ and $M$ be the intersections of the circumcircle of $\triangle OAD$ with the circumcircle of $\triangle OBC$. Let $T$ and $S$ be the intersections of $OM$ with the circumcircle of $\triangle OAB$ and $\triangle OCD$ respectively... | [
"Since $\\angle BTO = \\angle BAO$ and $\\angle BCO = \\angle BMO$, $\\triangle BTM$ and $\\triangle BAC$ are similar. Hence,\n\n$$\n\\frac{TM}{AC} = \\frac{BM}{BC} \\qquad \\textcircled{1}\n$$\n\nSimilarly,\n$$\n\\triangle CMS \\sim \\triangle CBD.\n$$\nHence,\n$$\n\\frac{MS}{BD} = \\frac{CM}{BC} \\qquad \\textcir... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08ft | Problem:
Una complicata coreografia prevede una fila di 7 ballerine equispaziate su un palcoscenico piano; di fronte a quella delle ballerine vi è una fila identica, parallela, di altrettanti ballerini. La coreografa vuole assegnare a ciascuna ballerina alcuni ballerini in modo che, tracciando sul palcoscenico il segm... | [
"Solution:\n\nLa risposta è (D). Immaginiamo per comodità le ballerine ed i ballerini disposti in orizzontale, da sinistra a destra. La condizione del testo può essere tradotta nella maniera seguente. Chiamiamo $B_{i}$ l'insieme dei ballerini che vengono assegnati alla $i$-esima ballerina, con $i=1, \\ldots, 7$. Pe... | Italy | Italian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | MCQ | D | |
04m9 | In 2018 Matija will turn the age which is equal to three times the sum of digits of the year he was born in. The same statement is true for his grandfather. What age did his grandfather turn in the year Matija was born? | [] | Croatia | Croatia_2018 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 54 | |
0hkv | Problem:
Let $ABC$ be a triangle and suppose $AB = 3$, $BC = 4$, $CA = 5$. What is the distance from $B$ to line $AC$? | [
"Solution:\nFirst note that since $3^{2} + 4^{2} = 5^{2}$, we have $\\angle B = 90^{\\circ}$ by the (converse to the) Pythagorean theorem. We calculate the area of the triangle in two ways. Viewing $AB$ as the base and $BC$ as the height, we get that the area is $\\frac{1}{2} \\cdot 3 \\cdot 4 = 6$. But viewing $AC... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 12/5 | |
02oe | Problem:
Existe um número quadrado perfeito formado apenas por algarismos 0 e 6 ? | [] | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | Only zero; no positive perfect square is composed solely of the digits zero and six. | |
0l1h | Problem:
For any positive integer $n$, let $f(n)$ be the number of ordered triples $(a, b, c)$ of positive integers such that
- $\max (a, b, c)$ divides $n$ and
- $\operatorname{gcd}(a, b, c)=1$.
Compute $f(1)+f(2)+\cdots+f(100)$. | [
"Solution:\nWe will show that $\\sum_{m=1}^{n} f(m)=n^{3}$. Indeed, consider the map\n$$\n\\begin{aligned}\ng:\\{1, \\ldots, n\\}^{3} & \\rightarrow \\{1, \\ldots, n\\}^{4} \\\\\ng(a, b, c) & =\\left(\\frac{a}{\\operatorname{gcd}(a, b, c)}, \\frac{b}{\\operatorname{gcd}(a, b, c)}, \\frac{c}{\\operatorname{gcd}(a, b... | United States | HMMT November 2024 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 1000000 | |
01mc | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that for all real $x, y$
$$
f(x - f(x/y)) = x f(1 - f(1/y))
$$
and
a) $f(1 - f(1)) \neq 0$;
b*) $f(1 - f(1)) = 0$. | [
"$$\nf(x - f(x/y)) = x f(1 - f(1/y)), \\quad \\forall x, y \\in \\mathbb{R}, y \\neq 0. \\quad (*)\n$$\nSet $y = 1$ in $(*)$ then\n$$\nf(x - f(x)) = x f(1 - f(1)). \\tag{1}\n$$\nIf $f(t) = 0$ for some $t$ then (1) implies $f(t - f(t)) = f(t) = 0$, hence, in view of\n(*), $t = 0$ since $f(1 - f(1)) \\ne 0$. On the o... | Belarus | Selection and Training Session | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | All solutions are linear: f(x) = k x for some real k. Case a) f(1 − f(1)) ≠ 0 corresponds to k ∈ ℝ \ {0, 1}. Case b*) f(1 − f(1)) = 0 corresponds to k ∈ {0, 1}, i.e., f(x) ≡ 0 or f(x) = x. | |
0fxj | Problem:
Sei $n$ eine natürliche Zahl. Bestimme die Anzahl Permutationen $\left(a_{1}, \ldots, a_{n}\right)$ der Menge $\{1,2, \ldots, n\}$ mit der folgenden Eigenschaft:
$$
2\left(a_{1}+\ldots+a_{k}\right) \quad \text{ ist durch } k \text{ teilbar } \quad \forall k \in\{1,2, \ldots, n\}
$$ | [
"Solution:\nWir nennen eine Permutation mit dieser Eigenschaft gut. Für $n \\leq 3$ sind alle Permutationen gut. Weiter ist leicht zu sehen, dass für jede gute Permutation $\\left(a_{1}, \\ldots, a_{n}\\right)$ auch die beiden Permutationen\n$$\n\\left(a_{1}, \\ldots, a_{n}, n+1\\) \\quad \\text{ und } \\quad\\left... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | For n=1: 1; for n=2: 2; and for n ≥ 3: 3*2^(n-2). | |
006s | Juan tiene 11 pesas todas de distintos pesos y todas de pesos enteros. La suma de los pesos de las 11 pesas es 1810. Con estas pesas se pueden obtener todos los pesos enteros desde 1 hasta 1810.
Determinar los posibles valores de la sexta pesa, contando de menor a mayor. | [] | Argentina | Argentina 2009 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Spanish | proof and answer | 25, 26, 27, 28, 29, 30, 31, 32 | |
094q | Problem:
Let $A$ and $B$ be positive integers. Consider a sequence of positive integers $(x_{n})_{n \geq 1}$ such that
$$
x_{n+1} = A \cdot \operatorname{gcd}(x_{n}, x_{n-1}) + B \quad \text{ for every } n \geq 2
$$
Prove that the sequence attains only finitely many different values.
Remark. We denote by $\operatorna... | [
"Solution:\n\nLet $n \\geq 2$ be a positive integer such that $x_{n+1} > x_{n}$. Then\n$$\n\\frac{x_{n}}{\\operatorname{gcd}(x_{n}, x_{n-1})} < \\frac{x_{n+1}}{\\operatorname{gcd}(x_{n}, x_{n-1})} = A + \\frac{B}{\\operatorname{gcd}(x_{n}, x_{n-1})} \\leq A + B.\n$$\nFurthermore,\n$$\n\\begin{aligned}\n\\operatorna... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0g3i | Problem:
Find all finite sets $S$ of positive integers with at least two elements, such that if $m > n$ are two elements of $S$, then
$$
\frac{n^{2}}{m-n}
$$
is also an element of $S$. | [
"Solution:\nTrying to apply number theoretical methods to deduce something from the fact that $m-n$ divides $n^{2}$ does not seem to lead anywhere. Instead, we will try to find the extreme values that the quotient $n^{2} /(m-n)$ can achieve. This will give us some interesting bounds on the elements of $S$.\n\na. Fi... | Switzerland | Final round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Colori... | null | proof and answer | All such sets are exactly of the form S = {s, 2s} for any positive integer s. | |
0kwj | Problem:
Suppose $P(x)$ is a polynomial with real coefficients such that $P(t) = P(1) t^{2} + P(P(1)) t + P(P(P(1)))$ for all real numbers $t$. Compute the largest possible value of $P(P(P(1)))$. | [
"Solution:\n\nLet $(a, b, c) := (P(1), P(P(1)), P(P(P(1))))$, so $P(t) = a t^{2} + b t + c$ and we wish to maximize $P(c)$. Then we have that\n$$\n\\begin{aligned}\na & = P(1) \\\\\nb & = a + b + c \\\\\nc & = P(b) = a b^{2} + b b + c\n\\end{aligned}\n$$\nThe first equation implies $c = -b$. The third equation impl... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 1/9 | |
0ip4 | Problem:
A $3 \times 3 \times 3$ cube composed of 27 unit cubes rests on a horizontal plane. Determine the number of ways of selecting two distinct unit cubes from a $3 \times 3 \times 1$ block (the order is irrelevant) with the property that the line joining the centers of the two cubes makes a $45^{\circ}$ angle wit... | [
"Solution:\n\nThere are 6 such slices, and each slice gives 10 valid pairs (with no overcounting). Therefore, there are 60 such pairs."
] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 60 | |
0ah9 | Let the circles $k_1$ and $k_2$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_1$ and $k_2$, that touches $k_1$ and $k_2$ at $M$ and $N$, respectively. If $t \perp AM$ and $MN = 2AM$, evaluate $\angle NMB$. | [
"Let $P$ be the symmetric of $A$ with respect to $M$ (figure 1). Then $\\overline{AM} = \\overline{MP}$ and $t \\perp AP$, hence the triangle $APN$ is isosceles with $AP$ as its base, so $\\angle NAP = \\angle NPA$.\n\nWe have $\\angle BAP = \\angle BAM = \\angle BMN$ and $\\angle BAN = \\angle BNM$.\n\nThus we hav... | North Macedonia | XVI-th Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 45° | |
0bs4 | For each positive integer $n$ denote $x_n$ the number of the positive integers with $n$ digits, divisible with $4$, formed with digits $2$, $0$, $1$ or $6$.
a) Compute $x_1$, $x_2$, $x_3$ and $x_4$.
b) Find $n$ so that
$$
1 + \left\lfloor \frac{x_2}{x_1} \right\rfloor + \left\lfloor \frac{x_3}{x_2} \right\rfloor + \l... | [
"a) $x_1 = 1$ (0 is divisible with $4$), $x_2 = 4$ (the numbers $12$, $16$, $20$ and $60$ are divisible with $4$), $x_3 = 3 \\cdot 5$, (because the first digit cannot be $0$ and the last two can be $12$, $16$, $20$, $60$ and $00$), $x_4 = 3 \\cdot 4 \\cdot 5 = 60$ (because the first digit cannot be $0$, for the sec... | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | x1=1, x2=4, x3=15, x4=60; n=504 | |
0ks8 | Let $x_1 \le x_2 \le \dots \le x_{100}$ be real numbers such that $|x_1| + |x_2| + \dots + |x_{100}| = 1$ and $x_1 + x_2 + \dots + x_{100} = 0$. Among all such 100-tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m... | [
"Let $x_1 \\le x_2 \\le \\dots \\le x_{100}$, $S = x_1 + x_2 + \\dots + x_{100} = 0$, and $T = |x_1| + |x_2| + \\dots + |x_{100}| = 1$.\n\nWe want to maximize $x_{76} - x_{16}$.\n\nLet $k$ be the number of negative $x_i$'s, and $100-k$ the number of nonnegative $x_i$'s. Since the sum is $0$, the sum of the negative... | United States | AIME II | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | final answer only | 841 | |
0jjh | Problem:
Let $ABCD$ be a trapezoid with $AB \parallel CD$. The bisectors of $\angle CDA$ and $\angle DAB$ meet at $E$, the bisectors of $\angle ABC$ and $\angle BCD$ meet at $F$, the bisectors of $\angle BCD$ and $\angle CDA$ meet at $G$, and the bisectors of $\angle DAB$ and $\angle ABC$ meet at $H$. Quadrilaterals $... | [
"Solution:\n\nAnswer: $\\boxed{\\dfrac{256}{7}}$\n\nLet $M, N$ be the midpoints of $AD, BC$ respectively. Since $AE$ and $DE$ are bisectors of supplementary angles, the triangle $AED$ is right with right angle $E$. Then $EM$ is the median of a right triangle from the right angle, so triangles $EMA$ and $EMD$ are is... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | 256/7 | |
04w6 | Let $a$, $b$, $c$, $d$ be positive real numbers lying in the interval $[1, 2]$ that also satisfy the equation $(a + c)(b + d) = 8$. Prove that the inequality
$$
\frac{1}{a^2 + b^2 - 1} + \frac{1}{b^2 + c^2 - 1} + \frac{1}{c^2 + d^2 - 1} + \frac{1}{d^2 + a^2 - 1} \geq 1,
$$
is satisfied for all such quadruples and deter... | [
"To begin with, note that all the numbers are greater or equal to one, so the denominators are positive and the question is well-posed.\n\nAs a first step in our solution, let's observe that since $a$ and $b$ lie in a closed interval of length $1$, we must have $(a - b)^2 \\le 1$, which can be rearranged to\n$$\n\\... | Czech Republic | First Round of the 73rd Czech and Slovak Mathematical Olympiad (take-home part) | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (1, 2, 1, 2) and (2, 1, 2, 1) | |
046k | As shown below, let $ABCD$ be a cyclic quadrilateral such that the diagonals $AC$ and $BD$ are perpendicular with intersection point $E$. Point $F$ is on the side $AD$, the ray $FE$ meets the circumscribed circle of $ABCD$ at point $P$. Point $Q$ is on the segment $PE$ such that $PQ \cdot PF = PE^2$. The line through $... | [
"Proof: Let $EX \\parallel AF$, intersecting $AP$ at $X$, and $DP$ at $Y$. Extend $XQ$ and $YQ$ to intersect $BC$ at $S$ and $T$, respectively.\nFrom $\\frac{PQ}{PE} = \\frac{PE}{PF} = \\frac{PX}{PA}$, we get $XQ \\parallel AE$, similarly $YQ \\parallel DE$.\nSince $\\angle EXS = \\angle AEX = \\angle DAC = \\angle... | China | 22nd Chinese Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0gf8 | 三角形 $ABC$ 中，令點 $B'$, $C'$ 分別為邊 $AC$ 及 $AB$ 的中點，而點 $H$ 為通過頂點 $A$ 的高的垂足。證明：三角形 $AB'C'$，$BC'H$ 及 $B'CH$ 的外接圓共於一點 $I$，且直線 $HI$ 平分線段 $B'C'$。 | [
"\n\n設點 $F$ 為 $B'C'$ 的中點, 點 $A'$ 為 $BC$ 中點, 且設直線 $HF$ 與 $\\triangle BHC'$ 的外接圓再交於點 $I$, 又設直線 $AA'$ 與 $\\triangle ABC$ 的外接圓再交於點 $M$。三角形 $HB'C'$ 全等於 $\\triangle AB'C'$, 故相似於 $\\triangle ABC$。因為\n\n$$\n\\angle C'IF = \\angle ABC = \\angle A'MC,\n$$\n$$\n\\angle C'FI = \\angle AA'B = \\angle MA... | Taiwan | 2022 數學奧林匹亞競賽第一階段選訓營, 國際競賽實作(一) | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | Chinese; English | proof only | null | |
035x | Problem:
Prove that the equation
$$
x^{2} + 2y^{2} + 98z^{2} = \underbrace{77\ldots7}_{2005}
$$
has no integer solutions. | [
"Solution:\n\nAssume that the equation has a solution $(x_{0}, y_{0}, z_{0})$. Then $x_{0}^{2} + 2y_{0}^{2}$ is divisible by $7$. Since the remainders modulo $7$ of the perfect squares are $0, 1, 2$ and $4$, it follows that both $x_{0}$ and $y_{0}$ are divisible by $7$.\n\nThen the left hand side of the given equat... | Bulgaria | Bulgarian Mathematical Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
0jju | Let $a_1, a_2, a_3, \dots$ be a sequence of integers, with the property that every consecutive group of $a_i$'s averages to a perfect square. More precisely, for every positive integers $n$ and $k$, the quantity
$$
\frac{a_n + a_{n+1} + \dots + a_{n+k-1}}{k}
$$
is always the square of an integer. Prove that the sequenc... | [
"We prove the following equivalent statement: we show that if $f : \\mathbb{N} \\to \\mathbb{Z}$ is a function such that $(f(m) - f(n))(m - n)$ is always the square of an integer, then $f$ must be of the form $A^2x + B$ for integers $A, B$. First, since $p(f(n + p) - f(n))$ is a square for any prime $p$ and positiv... | United States | IMO Team Selection Test | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0kcf | Problem:
Let $ABC$ be a triangle with $AB = 5$, $AC = 8$, and $\angle BAC = 60^\circ$. Let $UVWXYZ$ be a regular hexagon that is inscribed inside $ABC$ such that $U$ and $V$ lie on side $BA$, $W$ and $X$ lie on side $AC$, and $Z$ lies on side $CB$. What is the side length of hexagon $UVWXYZ$? | [
"Solution:\n\nLet the side length of $UVWXYZ$ be $s$. We have $WZ = 2s$ and $WZ \\parallel AB$ by properties of regular hexagons. Thus, triangles $WCZ$ and $ACB$ are similar. $AWV$ is an equilateral triangle, so we have $AW = s$. Thus, using similar triangles, we have\n\n$$\n\\frac{WC}{WZ} = \\frac{AC}{AB} \\Longri... | United States | HMMT February | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 40/21 | |
0izv | Problem:
Triangle $A B C$ is given in the plane. Let $A D$ be the angle bisector of $\angle B A C$; let $B E$ be the altitude from $B$ to $A D$, and let $F$ be the midpoint of $A B$. Given that $A B=28$, $B C=33$, $C A=37$, what is the length of $E F$? | [
"Solution:\n\n$14$\n\n$\\triangle A B E$ is a right triangle, and $F$ is the midpoint of the hypotenuse (and therefore the circumcenter), so $E F = B F = A F = 14$."
] | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 14 | |
0hb7 | Find all natural numbers $a$ and $b$, for which the number $2^{a!} + 2^{b!}$ is a cube of a natural number.
*Recall that for a natural number $n$, $n! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot n$.*
(Arseniy Nicolaev) | [
"It is clear that if $n \\ge 3$, $n!$ is divisible by $3$, that is $n! = 3k$, for some natural number $k$. But then $2^{n!} = 2^{3k} = 8^k \\equiv 1 \\pmod{7}$. It's easy to see that the cubes of integer numbers are equal to $0$ or $\\pm 1$ modulo $7$. Without loss of generality, we may assume that $a \\ge b$.\n\nC... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | a = b = 2 | |
0d87 | Let $n = p_{1} p_{2} \ldots p_{2017}$ be the positive integer where $p_{1}, p_{2}, \ldots, p_{2017}$ are 2017 distinct odd primes. A triangle is called nice if it is a right triangle with integer side lengths and the inradius is $n$. Find the number of nice triangles (two triangles are considered different if their tup... | [
"Consider the right triangle $ABC$ with incircle ($I$) and $D, E$ are tangent points of $(I)$ on sides $AB, AC$. It is easy to see that $ADIE$ is a square of side $n$.\n\n\n\nDenote $BD = x$, $CE = y$ then we can see that from a pair $(x, y)$ with $x > n, y > n$, we can construct a right tr... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | English | proof and answer | 3^{2017} | |
0bhe | Let $A$ and $B$ be two matrices from $M_3(\mathbb{C})$, such that $(AB)^2 = A^2B^2$ and $(BA)^2 = B^2A^2$. Prove that $(AB - BA)^3 = O_3$. | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
0gsh | Let $x$, $y$, $z$ be real numbers satisfying $y > 2z > 4x$ and
$$
2(x^3 + y^3 + z^3) + 15(xy^2 + yz^2 + zx^2) > 16(x^2y + y^2z + z^2x) + 2xyz.
$$
Show that $4x + y > 4z$. | [
"Define $a = x - 2y$, $b = y - 2z$, $c = z - 2x$. Then $b, c > 0$ and the problem statement is $b > 2c$. Now since\n$$\nx = -\\frac{a + 2b + 4c}{7}, \\quad y = -\\frac{b + 2c + 4a}{7}, \\quad z = -\\frac{c + 2a + 4b}{7}\n$$\nwe get\n$$\n\\begin{aligned}\nS &= 16(x^2y + y^2z + z^2x) + 2xyz - 2(x^3 + y^3 + z^3) \\\\\... | Turkey | Team Selection Test for IMO 2019 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
00wx | Problem:
Let $[x]$ be the integer part of a number $x$, and $\{x\} = x - [x]$. Solve the equation
$$
[x] \cdot \{x\} = 1991 x.
$$ | [
"Solution:\nLet $f(x) = [x] \\cdot \\{x\\}$. Then we have to solve the equation $f(x) = 1991 x$.\n\nObviously, $x = 0$ is a solution.\n\nFor any $x > 0$ we have $0 \\leq [x] \\leq x$ and $0 \\leq \\{x\\} < 1$ which imply $f(x) < x < 1991 x$.\n\nFor $x \\leq -1$ we have $0 > [x] > x - 1$ and $0 \\leq \\{x\\} < 1$ wh... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | x = 0 and x = -1/1992 | |
013t | Problem:
Let the points $D$ and $E$ lie on the sides $B C$ and $A C$, respectively, of the triangle $A B C$, satisfying $B D = A E$. The line joining the circumcentres of the triangles $A D C$ and $B E C$ meets the lines $A C$ and $B C$ at $K$ and $L$, respectively. Prove that $K C = L C$. | [
"Solution:\n\nAssume that the circumcircles of triangles $A D C$ and $B E C$ meet at $C$ and $P$. The problem is to show that the line $K L$ makes equal angles with the lines $A C$ and $B C$. Since the line joining the circumcentres of triangles $A D C$ and $B E C$ is perpendicular to the line $C P$, it suffices to... | Baltic Way | Baltic Way 2005 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0cpv | For a positive integer $a$, by $P(a)$ we denote the maximal prime divisor of $a^2 + 1$. Prove that there exist infinitely many triples of distinct positive integers $a, b, c$ such that $P(a) = P(b) = P(c)$.
Для натурального $a$ обозначим через $P(a)$ наибольший простой делитель числа $a^2 + 1$. Докажите, что существуе... | [
"Сделаем сначала замечание, общее для всех трёх решений. Пусть $p$ — нечётное простое число, а $a < p$ — натуральное число такое, что $a^2 + 1$ делится на $p$; тогда числа $a$ и $p-a$ различны, и $P(a) = P(p-a) = p$. Действительно, числа $a^2 + 1$ и $(p-a)^2 + 1 = (a^2 + 1) + p(p - 2a)$ делятся на $p$ и меньше $p^2... | Russia | Russian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English, Russian | proof only | null | |
02le | Problem:
Em um táxi podem se sentar um passageiro na frente e três atrás. De quantas maneiras podem se sentar os quatro passageiros se um deles quer ficar na janela? | [
"Solution:\n\nO passageiro que quer ficar na janela tem 3 possíveis lugares para se sentar, o seguinte pode-se sentar em qualquer lugar livre, logo tem 3 possíveis lugares, o seguinte dois possíveis lugares, e o último não tem escolha. Concluímos que o número de formas de se sentar é $3 \\times 3 \\times 2 = 18$."
... | Brazil | Nível 3 | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | final answer only | 18 | |
09oq | Find the least number of digits in the multiple of $10^{2025} + 3$ such that the number of different digits in it is exactly two. | [
"Answer: 4050\nPut $n = 2025$ and $d = 10^n + 3$. Let $N = \\overline{a_m \\dots a_2 a_1}$ be a multiple of $d$ such that the number of different digits in $N$ is at most two. Then we can assume that $a_1 \\ne 0$ and clearly $m \\ge n+1$ since $N \\ge d$. Suppose that $2n > m$, and consider $A = \\overline{a_n \\do... | Mongolia | MMO2025 Round 4 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 4050 | |
0ejf | Problem:
Dolžina akvarija je $50~\mathrm{cm}$, širina $20~\mathrm{cm}$ in višina $25~\mathrm{cm}$. Koliko $\mathrm{cm}$ od zgornjega roba akvarija bo nivo vode, če vanj vlijemo $19$ litrov vode?
(A) $19~\mathrm{cm}$
(B) $1{,}9~\mathrm{cm}$
(C) $10{,}9~\mathrm{cm}$
(D) $6~\mathrm{cm}$
(E) $0{,}6~\mathrm{cm}$ | [
"Solution:\n\nUgotovimo, da je prostornina vode $19~\\mathrm{dm}^3$. Zapišemo podatke v obrazec za prostornino vode ($x$ je višina vode v akvariju) in dobimo enačbo $V = 5 \\cdot 2 \\cdot x = 19$. Izračunamo $x = 1{,}9~\\mathrm{dm} = 19~\\mathrm{cm}$. Nivo vode bo torej $6~\\mathrm{cm}$ nižje od zgornjega roba akva... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Solid Geometry > Volume",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
00s7 | Let $ABC$ be an acute triangle with $AB \neq AC$ and circumcircle $\Gamma$. The angle bisector of $BAC$ intersects $BC$ and $\Gamma$ at $D$ and $E$ respectively. Circle with diameter $DE$ intersects $\Gamma$ again at $F \neq E$. Point $P$ is on $AF$ such that $PB = PC$ and $X$ and $Y$ are feet of perpendiculars from $P... | [
"WLOG, assume $AB < AC$. Let $M$ be the midpoint of side $BC$ and let the circumcircle of $DFE$ intersect $AF$ again at $K$. Since\n$$\n90^\\circ + \\angle MED = 180^\\circ - \\angle MDE = \\angle ABC + \\frac{\\angle BAC}{2} = \\angle AFE = \\angle DFE + \\angle AFD = 90^\\circ + \\angle AFD\n$$\nit follows that\n... | Balkan Mathematical Olympiad | BMO 2017 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic q... | English | proof only | null | |
09y9 | Problem:
Bepaal alle drietallen $(x, y, z)$ van reële getallen waarvoor geldt
$$
\begin{aligned}
& x^{2}-y z=|y-z|+1, \\
& y^{2}-z x=|z-x|+1, \\
& z^{2}-x y=|x-y|+1
\end{aligned}
$$ | [
"Solution:\n\nHet stelsel vergelijkingen is symmetrisch: verwissel je bijvoorbeeld $x$ en $y$, dan blijft de derde vergelijking hetzelfde en wisselen de eerste twee vergelijkingen om. We kunnen dus zonder verlies van algemeenheid aannemen dat $x \\geq y \\geq z$. Dan wordt het stelsel:\n$$\n\\begin{aligned}\n& x^{2... | Netherlands | Selectietoets | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | The six solutions are (4/3, 4/3, -5/3), (4/3, -5/3, 4/3), (-5/3, 4/3, 4/3), (5/3, -4/3, -4/3), (-4/3, 5/3, -4/3), and (-4/3, -4/3, 5/3). | |
0284 | Problem:
Considere cinco números reais positivos ordenados por $0<a \leq b \leq c \leq d \leq e$. Sabe-se que sempre que tiramos um destes números, podemos separar os outros quatro em dois grupos tais que a soma dos números de um grupo é igual à soma dos números do outro grupo. Se uma sequência $(a, b, c, d, e)$ satis... | [
"Solution:\n\n(a) Considere os números $0<1 \\leq 1 \\leq 1 \\leq 3 \\leq 3$. Veja que se tirarmos um dos números iguais a $1$, podemos separar os restantes em dois grupos iguais a $(1,3)$. Se tirarmos um $3$, então podemos separar os restantes nos grupos $(1,1,1)$ e $(3)$.\n\n(b) Suponha que temos três números igu... | Brazil | NÍVEL 3 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) An example is 1, 1, 1, 3, 3.
b) If three terms are equal in a balanced sequence, then all five terms are equal.
c) Under the stated conditions, all five numbers are equal. | |
0lb9 | Let $ABCD$ be a convex quadrilateral inscribed in a circle $(O)$. Suppose that the line $AB$ meets the line $CD$ at $M$ and the line $AD$ meets the line $BC$ at $N$. Let $P, Q, S$, and $T$ be the intersection points of the bisectors of the angles $\widehat{MAN}$ and $\widehat{MBN}$, $\widehat{MBN}$ and $\widehat{MCN}$,... | [
"We consider the case that $B$ lies between $A$ and $M$, and $C$ lies between $B$ and $N$. In this case, the points $P$ and $T$ are in the same half plane of the line $SQ$. The proofs for other cases are similar.\n\n\n\n1. We have\n$$\n\\widehat{QPT} = \\widehat{BPA} = \\widehat{MBP} - \\wi... | Vietnam | Vietnam Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line,... | Vietnamese | proof only | null | |
06cm | Students have taken a test paper in each of $n$ ($n \ge 3$) subjects. It is known that for any subject exactly three students get the best score in the subject, and for any two subjects exactly one student gets the best score in every one of these two subjects. Determine the smallest $n$ so that the above conditions im... | [
"The smallest $n$ is $8$.\nWe use terminologies in set theory. Let $A_1, A_2, \\dots, A_n$ be sets corresponding to the $n$ subjects, while the elements correspond to the students getting the best score in that subject. It is given that $|A_j| = 3$ and $|A_i \\cap A_j| = 1$ for any $1 \\le i < j \\le n$. Suppose $n... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 8 | |
093c | Problem:
Let $n$ be a positive integer and $u_{1}, u_{2}, \ldots, u_{n}$ be positive integers not larger than $2^{k}$, for some integer $k \geq 3$. A representation of a non-negative integer $t$ is a sequence of non-negative integers $a_{1}, a_{2}, \ldots, a_{n}$ such that
$$
t = a_{1} u_{1} + a_{2} u_{2} + \cdots + a_... | [
"Solution:\nWe shall treat a representation of $t$ as a multiset with entries from $\\{u_{1}, \\ldots, u_{n}\\}$ whose sum is equal to $t$, where integers $a_{1}, \\ldots, a_{n}$ correspond to the multiplicities of numbers $u_{1}, \\ldots, u_{n}$ in the multiset. For a representation $M$, let the support of $M$ be ... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0c3a | Let $ABCD$ be a regular tetrahedron, with $AB = a$. Denote $M$ the midpoint of $[BC]$, $N$ the midpoint of $[AM]$, and take $P \in [BD]$ so that $BP = 3PD$. Compute the distance between the straight lines $AC$ and $NP$.
Gabriela Chişiu and Carmen Rusu | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | a * sqrt(6) / 12 | |
0fk1 | Problem:
Resolver, en el conjunto de los números reales, el sistema de ecuaciones
$$
\left.\begin{array}{l}
y^{3}-6 x^{2}+12 x-8=0 \\
z^{3}-6 y^{2}+12 y-8=0 \\
x^{3}-6 z^{2}+12 z-8=0
\end{array}\right\}
$$ | [
"Solution:\nDado que $3 t^{2}-6 t+4 \\geq 0$ para todo $t \\in \\mathbb{R}$, entonces cualquier solución $\\left(x_{0}, y_{0}, z_{0}\\right)$ del sistema verifica que $x_{0}^{3}>0, y_{0}^{3}>0, z_{0}^{3}>0$. Es decir, $x_{0}, y_{0}, z_{0}$ son números positivos.\n\nAdemás, sumando las tres ecuaciones resulta\n$$\n\... | Spain | Spanish Mathematical Olympiad - Local Stage | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (2, 2, 2) | |
0j3a | Problem:
Let $f(x, y) = x^{2} + 2x + y^{2} + 4y$. Let $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$, and $(x_{4}, y_{4})$ be the vertices of a square with side length one and sides parallel to the coordinate axes. What is the minimum value of $f(x_{1}, y_{1}) + f(x_{2}, y_{2}) + f(x_{3}, y_{3}) + f(x_{4}, y_{4})$? | [
"Solution:\nThe square's corners must be at $(x, y), (x+1, y), (x+1, y+1)$, and $(x, y+1)$ for some $x$ and $y$. So,\n$$\n\\begin{aligned}\nf(x_{1}, y_{1}) & + f(x_{2}, y_{2}) + f(x_{3}, y_{3}) + f(x_{4}, y_{4}) \\\\\n&= 2(x^{2} + 2x) + 2((x+1)^{2} + 2(x+1)) + 2(y^{2} + 4y) + 2((y+1)^{2} + 4(y+1)) \\\\\n&= 4x^{2} +... | United States | Harvard-MIT November Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | -18 | |
027a | Problem:
Um trapézio especial - A base $AD$ de um trapézio $ABCD$ mede $30~\mathrm{cm}$. Suponhamos que exista um ponto $E$ na base $AD$ tal que os triângulos $\triangle ABE$, $\triangle BCE$ e $\triangle CDE$ tenham perímetros iguais. Determine o comprimento de $BC$. | [
"Solution:\n\nQueremos provar que $AE$ é igual a $BC$. Para isso, suponhamos que $AE$ seja maior do que $BC$ e escolhamos o ponto $A'$ sobre $AE$ tal que $EA' = BC$. Por construção, $EA'$ e $BC$ são paralelos, de modo que $A'BCE$ é um paralelogramo e, em particular,\n$$\nA'B = CE\n$$\nPela desigualdade triangular, ... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Dist... | null | proof and answer | 15 cm | |
0510 | Find all triples of positive integers $(x, y, z)$, for which $x \cdot y! + 2y \cdot x! = z!$. | [
"Since the left-hand side is greater than both $x!$ and $y!$, obviously $z > x$ and $z > y$. So, both sides of the equation are divisible by both $x!$ and $y!$. Therefore, $x \\cdot y!$ is divisible by $x!$, which means that $y!$ is divisible by $(x-1)!$, giving $y \\ge x-1$. Analogously, $2y \\cdot x!$ is divisibl... | Estonia | Estonian Math Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (x, y, z) = (n, n+1, n+2) for all positive integers n, and (x, y, z) = (2, 1, 3) | |
0i7s | Problem:
Suppose that $A, B, C, D$ are four points in the plane, and let $Q, R, S, T, U, V$ be the respective midpoints of $A B, A C, A D, B C, B D, C D$. If $Q R = 2001$, $S U = 2002$, $T V = 2003$, find the distance between the midpoints of $Q U$ and $R V$. | [
"Solution:\nThis problem has far more information than necessary: $Q R$ and $U V$ are both parallel to $B C$, and $Q U$ and $R V$ are both parallel to $A D$. Hence, $Q U V R$ is a parallelogram, and the desired distance is simply the same as the side length $Q R$, namely $2001$. (See figure, next page.)"
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 2001 | |
086h | Problem:
Una pulce si trova inizialmente nel punto $(0,0)$ del piano cartesiano. Successivamente compie $n$ salti. Ogni salto viene effettuato in una a scelta delle quattro direzioni cardinali. Il primo salto è di lunghezza $1$, il secondo di lunghezza $2$, il terzo di lunghezza $4$, e così via, fino all'$n$-salto, ch... | [] | Italy | XXV OLIMPIADE ITALIANA DI MATEMATICA | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0j9u | Problem:
Let $x$ and $y$ be positive real numbers such that $x^{2} + y^{2} = 1$ and $(3x - 4x^{3})(3y - 4y^{3}) = -\frac{1}{2}$. Compute $x + y$. | [
"Solution:\n\nLet $x = \\cos(\\theta)$ and $y = \\sin(\\theta)$. Then, by the triple angle formulae, we have that $3x - 4x^{3} = -\\cos(3\\theta)$ and $3y - 4y^{3} = \\sin(3\\theta)$, so $-\\sin(3\\theta) \\cos(3\\theta) = -\\frac{1}{2}$. We can write this as $2 \\sin(3\\theta) \\cos(3\\theta) = \\sin(6\\theta) = 1... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | √6/2 | |
09zv | Problem:
Bepaal alle positieve gehele getallen $n \geq 2$ waarvoor er een positieve deler $m \mid n$ bestaat met
$$
n = d^{3} + m^{3},
$$
waarbij $d$ de kleinste deler van $n$ groter dan 1 is. | [
"Solution:\n\nDe kleinste deler van $n$ groter dan 1 is het kleinste priemgetal dat een deler is van $n$, dus $d$ is priem. Verder geldt dat $d \\mid n$, dus $d \\mid d^{3} + m^{3}$, dus $d \\mid m^{3}$. Hieruit volgt dat $m > 1$. Anderzijds is $m \\mid n$, dus $m \\mid d^{3} + m^{3}$, dus $m \\mid d^{3}$. Omdat $d... | Netherlands | MO-selectietoets | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 16, 72, 520 | |
0ce8 | Find the triples $(a, b, c)$ of positive integers so that $1 + 2^a + 3^b = 6^c$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (a,b,c) = (1,1,1), (5,1,2), (3,3,2) | |
0j1g | Problem:
What is the perimeter of the triangle formed by the points of tangency of the incircle of a $5$-$7$-$8$ triangle with its sides? | [
"Solution:\nLet $\\triangle ABC$ be a triangle with sides $a=7$, $b=5$, and $c=8$. Let the incircle of $\\triangle ABC$ be tangent to sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$. By the law of cosines (using the form $\\cos (A)=\\frac{b^{2}+c^{2}-a^{2}}{2 b c}$), we have\n$$\n\\begin{aligned}\n& \\cos (A... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Method... | null | final answer only | 3 + 9 sqrt(21) / 7 | |
0bsv | Let $p$ be an odd prime and let $G$ be a $(p+1)$-element group. If $p$ divides the number of automorphisms of $G$, prove that $p \equiv 3 \pmod 4$.
Bogdan Moldovan | [
"Since $p$ is a prime divisor of the number of automorphisms of $G$, some automorphism $f$ has order $p$. Since $f$ is a permutation of the set $G \\setminus \\{e\\}$, it follows that $f$ is a cycle of length $p$, so $G \\setminus \\{e\\} = \\{x, f(x), \\dots, f^{p-1}(x)\\}$, whatever $x$ in $G \\setminus \\{e\\}$.... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
00t4 | Given an integer $k \ge 2$, determine all functions $f$ from the positive integers into themselves such that $f(x_1)! + f(x_2)! + \dots + f(x_k)!$ is divisible by $x_1! + x_2! + \dots + x_k!$ for all positive integers $x_1, x_2, \dots, x_k$.
Albania | [
"The identity is the only function satisfying the condition in the statement. Begin by letting the $x$'s be all equal to $n$ to infer that $f(n)!$ is divisible by $n!$, so $f(n) \\ge n$ for all positive integers $n$.\n\n**Claim.** $f(p-1) = p-1$ for all but finitely many primes $p$.\n\nAssume the Claim for the mome... | Balkan Mathematical Olympiad | BMO Short List | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | f(n) = n for all positive integers n | |
0dor | Let $n \ge 2$ be an integer. Prove that
$$
(1^{n-1} + 2^{n-1} + \dots + (n-1)^{n-1}) + 1 \text{ is divisible by } n
$$
if and only if, for each prime factor $p$ of $n$
$$
\frac{n}{p}-1 \text{ is divisible by } p \text{ and } \frac{n}{p}-1 \text{ is divisible by } p-1.
$$ | [
"For simplicity let's enumerate the statements.\n(1) $(1^{n-1} + 2^{n-1} + \\dots + (n-1)^{n-1}) + 1$ is divisible by $n$\n(2) $\\frac{n}{p}-1$ divisible by $p$ and $\\frac{n}{p}-1$ divisible by $p-1$\nLet $n = Ap$. Firstly, observe that\n$$\n\\sum_{k=1}^{n-1} k^{n-1} \\equiv \\begin{cases} -A & (\\text{mod } p), &... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"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 > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0g4m | Problem:
The Tokyo Metro system is one of the most efficient in the world. There is some odd positive integer $k$ such that each metro line passes through exactly $k$ stations, and each station is serviced by exactly $k$ metro lines. One can get from any station to any other station using only one metro line - but thi... | [
"Solution:\n\nCall a metro line charming if it contains exactly one of the stations David wants to visit, and breathtaking if it contains two such stations. Every train station in $S$ has $k$ metro lines passing through it, of which $k-1$ link it to other members of $S$ and are breathtaking, and the last one is cha... | Switzerland | Switzerland Selection Solution | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0d5k | Let $ABC$ be a triangle and $G$ its centroid. Let $G_{a}$, $G_{b}$ and $G_{c}$ be the orthogonal projections of $G$ on sides $BC$, $CA$, respectively $AB$. If $S_{a}$, $S_{b}$ and $S_{c}$ are the symmetrical points of $G_{a}$, $G_{b}$, respectively $G_{c}$ with respect to $G$, prove that $AS_{a}$, $BS_{b}$ and $CS_{c}$... | [
"Let $H_{a}$ be the foot of altitude from $A$, $M_{a}$ the midpoint of side $BC$, $A'$ the intersection point of line $AS_{a}$ and side $BC$ and $A''$ the intersection point of the parallel line to $BC$ passing through $S_{a}$ with $AM_{a}$.\n\n\n\nBecause $GG_{a}$ and $AH_{a}$ are parallel... | Saudi Arabia | SAMC 2015 | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | English, Arabic | proof only | null | |
0hkm | Problem:
Find, with proof, all ways to write $1$ as a sum of three fractions, each with numerator $1$ and positive integer denominator. (The order of the fractions is irrelevant, so for instance $\frac{1}{2}+\frac{1}{4}+\frac{1}{4}$ is the same as $\frac{1}{4}+\frac{1}{4}+\frac{1}{2}$.) | [
"Solution:\nThere are three solutions:\n$$\n1 = \\frac{1}{3} + \\frac{1}{3} + \\frac{1}{3} = \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{4} = \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{6}\n$$\nNow we must prove that these are the only solutions. If the fraction $1/2$ appears in the expression, the remaining fractions must... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1 = 1/3 + 1/3 + 1/3 = 1/2 + 1/4 + 1/4 = 1/2 + 1/3 + 1/6 | |
0bx2 | If $a_1, a_2, \dots, a_{2017}$ are positive integers, show that the fraction
$$
\frac{9^{2017} - 7 \cdot 3^{2017} + 7}{9^{(a_1+a_2)(a_2+a_3)\dots(a_{2016}+a_{2017})(a_{2017}+a_1)} - 1}
$$
is reducible. | [
"The last digit of $3^{2017}$ is 3, so the last digit of the numerator is 5.\n$$\n(a_1+a_2)+(a_2+a_3)+\\dots+(a_{2016}+a_{2017})+(a_{2017}+a_1) = 2(a_1+a_2+\\dots+a_{2017}).\n$$\nThe sum of these 2017 nonnegative integers being even, one of the summands, and hence their product, is even, so the last digit of the nu... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0h2g | In a triangle $ABC$ the angle $A$ is twice as big as the angle $B$, and $CD$ is the bisector of the angle $C$. Prove that $BC = AC + AD$. | [
"Let $E$ be a point on $BC$ such that $AE$ is perpendicular to $CD$. Then $\\triangle ACE$ is isosceles, since the bisector of the angle $C$ is also an altitude of the triangle (fig. 23). Hence, $AC = CE$. $\\triangle ADE$ is isosceles, since the line $CD$ is perpendicular to $AE$ and divides $AE$ in half (altitude... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0cbm | Let $ABCD$ be a tetrahedron and $M$ be a variable point on the face $BCD$. The perpendicular from $M$ onto the plane $BCD$ meets the planes $ABC$, $ACD$ and $ADB$ in $M_1$, $M_2$ and $M_3$. Prove that the sum $MM_1 + MM_2 + MM_3$ is constant if and only if the tetrahedron's altitude from $A$ passes through the barycent... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter,... | null | proof only | null |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.