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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
02pu | Problem:
Tio Mané tem duas caixas, uma com sete bolas distintas numeradas de $1$ a $7$ e outra com oito bolas distintas numeradas com todos os números primos menores que $20$. Ele sorteia uma bola de cada caixa.
Sugestão: Calcule a probabilidade do produto ser ímpar. Qual é a probabilidade de que o produto dos número... | [
"Solution:\n\nO produto dos números sorteados é ímpar somente se as duas bolas sorteadas têm números ímpares.\n\nA probabilidade de sortearmos da primeira caixa uma bola com número ímpar é $\\frac{4}{7}$ e a probabilidade de sortearmos uma bola ímpar da segunda caixa é $\\frac{7}{8}$, porque esta contém bolas com o... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1/2 | |
06m1 | There is an $n \times n$ chessboard where $n \ge 4$ is a positive even number. The cells of the chessboard are coloured black and white such that adjacent cells sharing a common side have different colours. Let $A$ and $B$ be two interior cells (which means cells not lying on an edge of the chessboard) of distinct colo... | [
"For convenience, suppose all cells are unit squares. We first prove two claims.\n\n**Claim 1.** For any rectangular chessboard, a chess piece can move from one corner to the opposite corner by visiting every cell exactly once if at least one side has an odd length.\n\n**Claim 2.** For any rectangular chessboard, a... | Hong Kong | Year 2021 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
04wi | Let $ABC$ be an acute triangle, $O$ its circumcenter, $\omega$ its circumcircle, $H$ its orthocenter, and $M$ the midpoint of $BC$. Line $AH$ meets $\omega$ again at $D \neq A$ and line $DM$ meets $\omega$ again at $E \neq D$. Finally, line $AE$ meets the circumcircle of triangle $OME$ again at $F \neq A$. Prove that $... | [] | Czech Republic | National Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscell... | English | proof only | null | |
07sr | A pond has $2020$ lily pads arranged in a circle. At time zero, two frogs (Anthony and Clare) share the same lily pad. Every minute, Anthony jumps over $99$ lily pads in an anti-clockwise direction, to land on a pad $100$ removed from where the jump started. At the same time, Clare jumps over $100$ lily pads in a clock... | [
"**Solution 1.** As we are concerned with the relative position of Anthony and Clare, the answer is unchanged if Anthony remains stationary and Clare jumps $201$ lily pads to the right.\nWhen does Clare next come close to Anthony? She makes one circuit of the pond after roughly $2020/201 \\approx 10$ jumps, but in ... | Ireland | IRL_ABooklet_2020 | [
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | 201 | |
0j96 | Problem:
Triangle $ABC$ has $AB = 5$, $BC = 3\sqrt{2}$, and $AC = 1$. If the altitude from $B$ to $AC$ and the angle bisector of angle $A$ intersect at $D$, what is $BD$? | [
"Solution:\n\nLet $E$ be the foot of the perpendicular from $B$ to line $AC$. By the Law of Cosines, $\\cos \\angle BAC = \\frac{4}{5}$, and it follows that $BE = 3$ and $AE = 4$. Now, by the Angle Bisector Theorem, $\\frac{BD}{BE} = \\frac{AB}{AB + AE} = \\frac{5}{9}$, so $BD = \\frac{5}{3}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 5/3 | |
08oi | Problem:
Find the integer solutions of the equation
$$
x^{2}=y^{2}\left(x+y^{4}+2 y^{2}\right)
$$ | [
"Solution:\nIf $x=0$, then $y=0$ and conversely, if $y=0$, then $x=0$. It follows that $(x, y)=(0,0)$ is a solution of the problem. Assume $x \\neq 0$ and $y \\neq 0$ satisfy the equation. The equation can be transformed in the form $x^{2}-x y^{2}=y^{6}+2 y^{4}$. Then $4 x^{2}-4 x y^{2}+y^{4}=4 y^{6}+9 y^{4}$ and c... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | [(0, 0), (12, 2), (12, -2), (-8, 2), (-8, -2)] | |
0d29 | Let $S = \{0, 1, 2, 3, \ldots\}$ be the set of the non-negative integers. Find all strictly increasing functions $f: S \rightarrow S$ such that $n + f(f(n)) \leq 2 f(n)$ for every $n$ in $S$. | [
"Since $f$ is strictly increasing, we have $f(n) \\geq n$ for all $n$ in $S$.\n\nAssume that there exists an integer $n$ in $S$ such that $f(n) > n$. Let $n_0$ be the smallest such $n$ and write $f(n_0) = n_0 + k_0$, for some positive integer $k_0 \\geq 1$. Again, since $f$ is strictly increasing, we have $f(n) \\g... | Saudi Arabia | Selection tests for the International Mathematical Olympiad 2013 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | All functions of the form f(n) = n for n < n0 and f(n) = n + k0 for n ≥ n0, for some n0 ∈ S and some integer k0 ≥ 0. | |
0lff | Let $P(x) \in \mathbb{R}[x]$ be a monic, non-constant polynomial. Determine all continuous functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(f(P(x)) + y + 2023f(y)) = P(x) + 2024f(y),
$$
for all reals $x, y$. | [
"Because the polynomial $P(x)$ is non-constant and has the highest coefficient equal to $1$, there exists a constant $c$ such that $P(x)$ can take on all values above $[c, +\\infty)$.\nFrom equation (1), we get\n$$\nf(f(x) + y + 2023f(y)) = x + 2024f(y), \\quad \\forall x \\ge c, y \\in \\mathbb{R}. \\quad (2)\n$$\... | Vietnam | Team selection tests | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof and answer | f(x) = x for all real x | |
0g8s | 試證: 存在無窮多個正整數 $n$, 使得 $n^4 + n^2 + 1$ 的最大質因數, 和 $(n+1)^4 + (n+1)^2 + 1$ 的最大質因數相同。
Prove that there exist infinitely many positive integers $n$ such that the largest prime divisor of $n^4 + n^2 + 1$ is equal to the largest prime divisor of $(n + 1)^4 + (n + 1)^2 + 1$. | [
"令 $p_n$ 為 $n^4 + n^2 + 1$ 的最大質因數，並令 $q_n$ 為 $n^2 + n + 1$ 的最大質因數。\n易知 $p_n = q_{n^2}$。同時，由\n$$\nn^4+n^2+1 = (n^2+1)^2-n^2 = (n^2-n+1)(n^2+n+1) = ((n-1)^2+(n-1)+1)(n^2+n+1)\n$$\n知，對於所有 $n \\ge 2$，有 $p_n = \\max\\{q_n, q_{n-1}\\}$。又因 $n^2 - n + 1$ 是奇數，\n$$\n\\text{gcd}((n-1)^2+(n-1)+1, n^2+n+1) = \\text{gcd}(2n, n^2... | Taiwan | 二〇一四數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
07hl | In triangle $ABC$, with $AB < AC$, $I$ is the incenter, $E$ is the intersection of $A$-excircle, and $BC$. Point $F$ lies on the external angle bisector of $\angle BAC$ such that $E$ and $F$ lies on the same side of the line $AI$ and $\angle AIF = \angle AEB$. Point $Q$ lies on $BC$ such that $\angle AIQ = 90^\circ$. C... | [
"Let in triangle $ABC$, points $I_a$, $H$, and $D$ be the $A$-excenter, foot of the altitude from $A$, and foot of the angle bisector from $A$, respectively. Since $(AD, II_a) = -1$ and $\\angle AHD = 90^\\circ$, we have $\\angle IHD = \\angle DHI_a$. Further, $AIHQ$ is cyclic, So $\\angle QAI = \\angle IHD$, hence... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point... | English | proof only | null | |
0ab9 | The sequence $49$, $4489$, $444889$, ... is given so that the number $48$ is put in the middle of the preceding term. Prove that every term in the sequence is perfect square. | [
"Let us assume that $A = 444\\ldots488\\ldots89$, i.e. $A = \\overline{444\\ldots488\\ldots89}$.\n\n$$\n\\text{Then } A = 4 \\cdot \\overline{111\\ldots1} \\cdot 10^n + 8 \\cdot \\overline{111\\ldots1} + 1.\n$$\n$$\n\\text{Now } \\overline{111\\ldots1} = 10^{n-1} + 10^{n-2} + \\dots + 10^2 + 10 + 1 = \\frac{10^n - ... | North Macedonia | Macedonian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0hyv | Problem:
Pyramid $EARLY$ is placed in $(x, y, z)$ coordinates so that $E = (10, 10, 0)$, $A = (10, -10, 0)$, $R = (-10, -10, 0)$, $L = (-10, 10, 0)$, and $Y = (0, 0, 10)$. Tunnels are drilled through the pyramid in such a way that one can move from $(x, y, z)$ to any of the 9 points $(x, y, z-1)$, $(x \pm 1, y, z-1)$,... | [
"Solution:\n\nStart by figuring out the probabilities of ending up at each point on the way down the pyramid. Obviously we start at the top vertex with probability $1$, and each point on the next level down with probability $1/9$. Since each probability after $n$ steps will be some integer over $9^{n}$, we will loo... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 550/9^10 | |
03ff | Points $A$, $B$, $Y$ and $C$ lie in this order on circle $k$ with center $O$, such that $BC = 2$ cm, $\angle BAY = 42^\circ$ and $\angle CAY = 78^\circ$. It is known that the circle $\omega$ through the points $A$, $O$ and $B$ is tangent to the line $BY$.
The circle through the points $A$ and $C$, tangent to the line ... | [
"a) Clearly $\\angle BAC = \\angle BAY + \\angle CAY = 120^\\circ$, respectively $\\angle BOC = 360^\\circ - 2\\angle BAC = 120^\\circ$. Thus, if $M$ is the midpoint of $BC$, then $OM \\perp BC$ (because $BO = OC$), $\\angle BOM = 60^\\circ$ and $BM = \\frac{BC}{2} = 1$. Let $BO = x$ and from triangle $BOM$ we have... | Bulgaria | Bulgarian Spring Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | BO = 2√3/3 cm; angle YAN = 36° | |
0hop | Problem:
Given a triangle $ABC$ such that $\angle B = 90^{\circ}$, denote by $k$ the circle with center on $BC$ that is tangent to $AC$. Denote by $T$ a point of tangency of $k$ and the tangent from $A$ to $k$ (different from $AC$). If $B'$ is the midpoint of $AC$ and $M$ the intersection of $BB'$ and $AT$, prove that... | [
"Solution:\n\nLet $O$ be the center of $k$. The quadrilateral $ABTO$ can be inscribed in a circle, hence $\\angle CBB' = \\angle B'CB$ and $\\angle TAO = \\angle OAC$. Thus $\\angle MTB = \\angle ATB = \\angle AOB = \\angle ACO + \\angle OAC = \\angle B'BC + \\angle TAO = \\angle B'BC + \\angle TBO = \\angle TBM$, ... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ... | null | proof only | null | |
056t | A quadrilateral $ABCD$ has incenter $I$. Diagonals $AC$ and $BD$ intersect at point $P$. Given that $I$ lies inside the triangle $PAB$ (not on its side), prove that the area of the triangle $PAB$ is greater than the area of any of triangles $PBC$, $PCD$ and $PDA$. | [
"Denote the area of any triangle $\\triangle \\Delta$ by $S_{\\triangle \\Delta}$. Let $D'$ be the reflection of $D$ from $AC$ and $\\alpha = \\angle APB$ (Fig. 24). We have\n\n$$\nS_{PAB} = \\frac{1}{2} \\cdot PA \\cdot PB \\cdot \\sin \\alpha, \\qquad (4)\n$$\n$$\nS_{PBC} = \\frac{1}{2} \\cdot PB \\cdot PC \\cdot... | Estonia | Final Round of National Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ed2 | Find all real numbers $x$ which satisfy the equation
$$
\sqrt[3]{2x+13} - \sqrt[3]{2x-13} = 2.
$$ | [
"We raise the equation to the third power to get\n$$\n(2x + 13) - 3 \\left(\\sqrt[3]{2x + 13}\\right)^2 \\sqrt[3]{2x - 13} + 3 \\sqrt[3]{2x + 13} \\left(\\sqrt[3]{2x - 13}\\right)^2 - (2x - 13) = 8.\n$$\nWe then rearrange it\n$$\n26 - 3\\sqrt[3]{2x + 13}\\sqrt[3]{2x - 13}\\left(\\sqrt[3]{2x + 13} - \\sqrt[3]{2x - 1... | Slovenia | National Math Olympiad 2015 – First Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | {-7, 7} | |
0et8 | Let $ABC$ be a triangle, and let $T$ be a point on the extension of $AB$ beyond $B$, and $U$ a point on the extension of $AC$ beyond $C$, such that $BT = CU$. Moreover, let $R$ and $S$ be points on the extensions of $AB$ and $AC$ beyond $A$ such that $AS = AT$ and $AR = AU$. Prove that $R, S, T, U$ lie on a circle whos... | [
"\nFigure 2\n\nSince $AS = AT$, we have $\\angle AST = \\angle ATS = \\frac{1}{2}\\angle BAC$. Similarly, $\\angle ARU = \\angle AUR = \\frac{1}{2}\\angle BAC$, so that $\\angle S = \\angle R$. This implies that $R, S, T$ and $U$ are concylic.\n\nWe now show that the centre $O$ of the circl... | South Africa | The South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0coy | Does there exist a positive integer $k$ such that all positive integers from $1$ to $k$ can be partitioned into two groups, and all the numbers in each group can be written one after another (in some order without spaces between them) so that the two obtained numbers are identical?
Можно ли при каком-то натуральном $k... | [
"Предположим противное. Ясно, что $k \\ge 10$, так как в наборе цифр от $1$ до $9$ нет повторяющихся. Рассмотрим наибольшую степень десятки $10^n$, не превосходящую $k$. Последовательность цифр числа $10^n$ целиком войдет в одно из составленных чисел. Но тогда такая же последовательность из единицы и $n$ последующи... | Russia | Regional round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English; Russian | proof and answer | No such k exists | |
0h28 | Does there exist an increasing sequence of integers $0 = a_0 < a_1 < a_2 < \dots$, such that the following two conditions are satisfied:
1) every natural number can be written in the form $a_i + a_j$ for some (maybe equal) indices $i, j \ge 0$;
2) $a_n > \frac{n^2}{16}$ for all natural $n$? | [
"Let $(a_n)$ be the sequence of all natural numbers $k$ that in binary representation have $1$s only on even places or only on odd places. For instance, this sequence contains the numbers that have the following binary representations: $10000$, $10100$, $101$, $1000$. Evidently, the first condition is satisfied for... | Ukraine | Problems of Ukrainian Authors | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series",
"Discrete Mathematics > Combinatorics"
] | English | proof only | null | |
0242 | Problem:
Manuel é um matemático que gosta de jogos de cartas. Ele encontra os irmãos Jonas e Jonatan durante uma viagem de ônibus e propõe um jogo. Serão usados apenas os quatro ases do baralho, o de copas e o de ouros são vermelhos enquanto o de espadas e o de paus são pretos.
Manuel será o ... | [
"Solution:\n\n(a) Representaremos copas, espadas, ouros e paus pelas letras $C, E, O$ e $P$, respectivamente. A árvore de possibilidades é mostrada na figura a seguir:\n\n\n(b) Usando a árvore de possibilidades, há 12 resultados possíveis e em 8 deles Manuel vence. Portanto, a probabilidade... | Brazil | NÍVEL 3 | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | 2/3 | |
0a0i | Albert makes a sequence of numbers, the first $2023$ of which are exactly the numbers $1$ through $2023$, in some order. To determine each subsequent number, Albert takes the *median* of the $2023$ previous numbers. You find the median of $2023$ numbers by sorting them by size and taking exactly the middle number from ... | [
"1"
] | Netherlands | Dutch Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 1 | |
081u | Problem:
Determinare tutte le terne di interi positivi $ (x, y, z) $ che soddisfano il seguente sistema:
$$
\left\{\begin{array}{l}
45 x y^{2}=8 z^{3} \\
x y z<1000
\end{array}\right.
$$ | [
"Solution:\n\nL'unica soluzione è $(6,10,15)$. Osserviamo anzitutto che $45$ divide $z^{3}$, e quindi $z$ è divisibile per $3 \\cdot 5=15$. Analogamente $8$ divide $x y^{2}$, e quindi $x$ e $y$ non possono essere dispari entrambi. Poniamo $z=15 w$ e distinguiamo 3 casi:\n\n(1) $x$ dispari: allora $8$ divide $y^{2}$... | Italy | Progetto Olimpiadi di Matematica | [
"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 | (6, 10, 15) | |
02s7 | Problem:
Os ciclistas têm aversão ao número zero (porque é oval) e ao número oito (porque assim ficam as rodas após os acidentes). Quantos sócios podem se inscrever num clube de ciclistas se cada um deve possuir uma identificação de três dígitos, sem usar o dígito zero nem o dígito oito? | [
"Solution:\nJá que os ciclistas não usam o dígito $0$ e nem o $8$, restam os dígitos $1, 2, 3, 4, 5, 6, 7$ e $9$. Assim, há $8$ possibilidades para a escolha de cada dígito. Temos que escolher números de três dígitos. Logo, temos $8$ opções para o primeiro dígito, $8$ opções para o segundo dígito e $8$ opções para ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Other"
] | null | final answer only | 512 | |
0eun | When there is a link from a webpage $i$ to a webpage $j$, you can move from the webpage $i$ to the webpage $j$ directly. For $n \ge 2$, there are $n$ many webpages numbered from $1$ to $n$, and for all $i \in \{1, 2, \dots, n-1\}$, there is a link from the webpage $i$ to the webpage $(i+1)$.
Now you are allowed to add ... | [
"Let $T(n)$ be smallest number of new links so that the property holds. By using mathematical induction we will show that for all $n \\ge 2$,\n$$\nT(n) \\le 3(n-1) \\log_2 \\log_2 n.\n$$\n\nIt is trivial that $T(2) = 0 \\le 3(2-1)\\log_2\\log_2 2$, and $T(3) = 0 \\le 3(3-1)\\log_2\\log_2 3$.\nLet $n \\ge 4$. When $... | South Korea | 23rd Korean Mathematical Olympiad Final Round | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
07yh | Problem:
Determinare per quali $n \geq 3$ è possibile trovare $n$ numeri interi positivi tali che a due a due abbiano almeno un fattore in comune diverso da 1, ma a tre a tre siano primi tra loro.
Se si aggiunge la condizione che tutti i numeri interi siano minori di 5000, qual è il massimo valore di $n$ possibile? | [
"Solution:\n\nÈ possibile per tutti gli interi positivi $n \\geq 3$. Si associ ad ogni coppia di interi $\\{i, j\\}$ con $1 \\leq i < j \\leq n$ un numero primo $p_{ij} = p_{ji}$ in modo tale che a coppie distinte siano associati primi distinti.\nSi definisca poi, per $1 \\leq i \\leq n$, $a_{i}$ come il prodotto d... | Italy | null | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | All integers n at least three; with all numbers less than 5000, the maximum n is 4. | |
04a2 | Let $a, b > 1$ be real numbers such that $\log_b a$, $\log_{2b}(2a)$ and $\log_{4b}(4a)$, in that order, form an arithmetic progression. Prove that $a = b$. (Tonći Kokan) | [] | Croatia | CroatianCompetitions2011 | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof only | null | |
0krf | Problem:
Aerith picks five numbers and for every three of them, takes their product, producing ten products. She tells Bob that the nine smallest positive divisors of sixty are among her products. Can Bob figure out the last product? | [
"Solution:\n\nYes Bob can. Let the numbers be $v, w, x, y, z$ and let the missing product be $x y z$. The nine given products are $1,2,3,4,5,6,10,12,15$.\n\nPartition these products into the three sequences\n$$\ns_{1}=(v y z, v z x, v x y),\\quad s_{2}=(w y z, w z x, w x y),\\quad s_{3}=(v w x, v w y, v w z)\n$$\na... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 5/2 | |
02lr | Let $N = \{0, 1, 2, 3, \dots\}$. Given sets $A, B \subset N$, for each positive integer $n$ denote $r(A, B, n)$ as the number of solutions to the equation $a + b = n$, $a \in A, b \in B$.
Prove that there exists $n_0 \in N$ such that $r(A, B, n+1) > r(A, B, n)$ for all $n > n_0$ if and only if $N \setminus A$ and $N \s... | [
"First suppose that $r(A, B, n)$ is increasing for $n > n_0$. For the sake of simplicity, let $\\bar{A} = N \\setminus A$ and $\\bar{B} = N \\setminus B$. Then\n$$\nn+1 = r(N, N, n) = r(A, B, n) + r(\\bar{A}, B, n) + r(A, \\bar{B}, n) + r(\\bar{A}, \\bar{B}, n) \\\\\n\\iff r(\\bar{A}, B, n) + r(A, \\bar{B}, n) + r(... | Brazil | XXXI Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0kvq | Problem:
Let $A B C D$ be a rectangle with $A B=20$ and $A D=23$. Let $M$ be the midpoint of $C D$, and let $X$ be the reflection of $M$ across point $A$. Compute the area of triangle $X B D$. | [
"Solution:\n\n\nObserve that $[X B D]=[B A D]+[B A X]+[D A X]$. We will find the area of each of these triangles individually.\n- We have $[A B D]=\\frac{1}{2}[A B C D]$.\n- Because $A M=A X$, $[B A X]=[B A M]$ as the triangles have the same base and height. Thus, as $[B A M]$ have the same... | United States | HMMT November 2023 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | final answer only | 575 | |
04hf | Prove that for all $x \in [\frac{1}{111}, \frac{110}{111}]$ there exist numbers $a_i \in \{-1, 1\}, i = 1, 2, \dots, 101$ such that
$$
|x_{101} - x| \le \frac{1}{402},
$$
where
$x_0 = 1, \quad x_k = (x_{k-1} + 1)^{a_k}, \quad \text{for } k = 1, 2, \dots, 101.$ | [
"For a given positive integer $n$ let $S_n$ denote the set of all possible values that $x_n$ can obtain for different choices of numbers $a_i$, $1 \\le i \\le n$. For example:\n$$\nS_1 = \\{\\frac{1}{2}, 2\\}, \\quad S_2 = \\{\\frac{1}{3}, \\frac{2}{3}, \\frac{3}{2}, 3\\}, \\quad S_3 = \\{\\frac{1}{4}, \\frac{2}{5}... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0c3o | Problem:
Fie $p$ un număr prim mai mare ca $5$ şi $S=\{p-n^{2} \mid n \in \mathbb{N}, n^{2}<p\}$. Demonstraţi că $S$ conţine două elemente $a$ şi $b$ astfel încât $1<a<b$ şi $a$ divide $b$. | [
"Solution:\n\nVom arăta că cel mai mic element al lui $S$ care e mai mare ca $1$ divide un element mai mare al lui $S$.\n\nDacă $p$ este de forma $m^{2}+1$ cu $m \\in \\mathbb{N}$, arătăm că $p-(m-1)^{2}=2m$ divide $p^{2}-1=m^{2}$. Într-adevăr, cum $m$ este par, rezultă că $2m \\mid m^{2}$.\n\nDacă $p$ nu este de f... | Romania | Al cincilea test de selecţie pentru OBMJ | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0i6o | Problem:
Find all values of $x$ that satisfy $x = 1 - x + x^{2} - x^{3} + x^{4} - x^{5} + \cdots$ (be careful; this is tricky). | [
"Solution:\nMultiplying both sides by $1 + x$ gives $(1 + x)x = 1$, or $x = \\frac{-1 \\pm \\sqrt{5}}{2}$. However, the series only converges for $|x| < 1$, so only the answer $x = \\frac{-1 + \\sqrt{5}}{2}$ makes sense."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (-1 + sqrt(5))/2 | |
0jb8 | Problem:
A restaurant has some number of seats, arranged in a line. Its customers are in parties arranged in a queue. To seat its customers, the restaurant takes the next party in the queue and attempts to seat all of the party's member(s) in a contiguous block of unoccupied seats. If one or more such blocks exist, the... | [
"Solution:\nAnswer: 29 First, note that if there are only 28 seats, it is possible for the seating not to be possible, in the following way. The party of six could be seated in such a way that the remaining contiguous regions have sizes 10 and 12. Then, the party of 4 is seated in the middle of the region of size 1... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 29 | |
0b7s | Let $ABC$ be a triangle and let $D, E, F$ be the midpoints of the sides $BC, CA, AB$. Prove that $\angle DAC = \angle ABE$ if and only if $\angle AFC = \angle BDA$. | [
"Let $G$ be the centroid of the given triangle. Since $DF \\parallel AC$, one has $\\angle DAC = \\angle GDF$.\n\nIf $\\angle DAC = \\angle ABE$, then $\\angle GDF = \\angle FBG$, hence the quadrilateral $BFGD$ is cyclic, implying $\\angle AFC = \\angle BDA$. The converse holds by the same argument."
] | Romania | NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0kij | 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 | [
"**Answer (B):** The given condition means that either $x$ and $x^2$ represent the same angle or they represent supplementary angles. In the first case $x^2 = x + 360k$ for some integer $k$. By the quadratic formula,\n$$\nx = \\frac{1 \\pm \\sqrt{1 + 1440k}}{2}.\n$$\nThe least value of $x > 1$ occurs when $k = 1$ a... | United States | AMC 12 A | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | B | |
06ad | Let $Q_n$ be the set of $n$-tuples $x = (x_1, ..., x_n)$, $x_i \in \{0, 1, 2\}$, $i = 1, 2, ..., n$. A triad $(x, y, z)$, where $x = (x_1, x_2, ..., x_n)$, $y = (y_1, y_2, ..., y_n)$, $z = (z_1, z_2, ..., z_n)$, of distinguished elements of $Q_n$ is called good, if there exists at least one $i \in \{1, 2, ..., n\}$ for... | [
"We will use induction with respect to $n$. The case for $n = 1$ is obvious. We suppose that every good subset of $Q_{n-1}$ has at most $2\\left(\\frac{3}{2}\\right)^{n-1}$ elements.\n\nLet $A_0 = \\{(x_1, ..., x_n) \\in A : x_n \\neq 0\\}$. We define the subsets $A_1, A_2$ similarly, that is\n$$\nA_1 = \\{(x_1, ..... | Greece | 39th Hellenic Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0erc | $OABC$ is a rectangle of length $16$ cm. $M$ is the midpoint of $CB$ and $P$ lies on $AB$ so that $BP = 3$ cm. The length $OM$ is $17$ cm. The length of $OP$, in cm, is
(A) $18$
(B) $19$
(C) $20$
(D) $21$
(E) $22$ | [
"$CM = MB = 8$. Then by Pythagoras in $\\triangle OMC$, $OC = 15$. But then $AP = 15 - 3 = 12$, and Pythagoras in $\\triangle OAP$ shows $OP = 20$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | MCQ | C | |
05qg | Problem:
On dispose, dans le plan, 16 points deux à deux distincts, que l'on note $A_{i, j}$ pour $i, j \in \{1,2,3,4\}$. Ces points vérifient les relations d'alignement et de cocyclicité suivantes :
$\triangleright$ pour tout $i \in \{1,2,3,4\}$, les points $A_{i, 1}, A_{i, 2}, A_{i, 3}$ et $A_{i, 4}$ sont alignés;
... | [
"Solution:\n\n\nCi-dessous, on notera $\\Delta_{i}$ la droite $\\left(A_{i, 1} A_{i, 2}\\right)$ et $\\Delta^{j}$ la droite $\\left(A_{1, j} A_{2, j}\\right)$. Une figure et une chasse aux angles nous montrent que les droites $\\Delta_{1}$ et $\\Delta_{3}$ sont parallèles, puisque ( $\\Delt... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0grq | We say that a group of 25 students is a *team* if any two students in this group are friends. It is known that in the school any student belongs to at least one team but if any two students end their friendships at least one student does not belong to any team. We say that a team is *special* if at least one student of... | [
"Let us prove that any two friends $A$ and $B$ belong to some special team. Let us define a longest sequence $S_1, S_2, \\dots, S_m$ of students such that\n\n* for all $1 \\le i < j \\le m$ we have $S_i \\ne S_j$\n* for each $1 \\le i \\le m$ any team containing $S_i$ also contains $A, B, S_1, S_2, \\dots, S_{i-1}$... | Turkey | Team Selection Test | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
01k7 | 4 blue, 10 green and several (at least one) red points are marked on a plane. All points are distinct. It is known that the sum of the distances between the red points and the blue points is $21$, the sum of the distances between the red points and the green points is $2$.
Can the sum of the distances between the blue ... | [
"a) Let $n$ denote the number of red points. Then the following inequality holds (see (*) in the solution of Problem 2, Category C)\n$$\n10 \\cdot 21 - 4 \\cdot 2 \\le n \\cdot 20 \\le 10 \\cdot 21 + 4 \\cdot 2 \\iff 202 \\le 20n \\le 218.\n$$\nIt is easy to see that there are no integers $n$ satisfying this condit... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | a) no; b) yes | |
03ow | In a given trapezium $ABCD$, $AD \parallel BC$. Suppose $E$ is a variable point on $AB$, $O_1$ and $O_2$ are circumcenters of $\triangle AED$ and $\triangle BEC$ respectively. Prove that the length of $O_1O_2$ is a fixed value. (posed by Leng Gangsong) | [
"**Proof** As shown in the figure, we join $EO_1$ and $EO_2$, then $\\angle AEO_1 = 90^\\circ - \\angle ADE$, $\\angle BEO_2 = 90^\\circ - \\angle BCE$. Hence\n$\\angle O_1EO_2 = \\angle ADE + \\angle ECB$.\n\nSince $AD \\parallel BC$, through $E$ constructing a line parallel to $AD$, we ca... | China | China Western Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0j14 | Problem:
Let $p(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0$, where each $a_i$ is either $1$ or $-1$. Let $r$ be a root of $p$. If $|r| > \frac{15}{8}$, what is the minimum possible value of $n$? | [
"Solution:\n\nAnswer: $4$ We claim that $n = 4$ is the answer. First, we show that $n > 3$. Suppose that $n \\leq 3$. Let $r$ be the root of the polynomial with $|r| \\geq \\frac{15}{8}$. Then, by the Triangle Inequality, we have:\n$$\n\\begin{gathered}\n\\left|a_n r^n\\right| = \\left|a_{n-1} r^{n-1} + a_{n-2} r^{... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 4 | |
02e3 | A ball moves endlessly on a circular billiard table. When it hits the edge it is reflected. Show that if it passes through a point on the table three times, then it passes through it infinitely many times. | [
"Suppose $AB$ and $BC$ are two successive chords of the ball's path. Then by the reflection law $\\angle ABO = \\angle OBC$. But $OAB$ and $OBC$ are isosceles and so $\\angle AOB = \\angle BOC$. Hence $AB = BC$. So every chord of the path is the same length $d$.\n\nWe now claim that through any given point $P$ insi... | Brazil | VIII OBM | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
06xx | Let $a, b, c, d$ be positive integers satisfying
$$
\frac{a b}{a+b}+\frac{c d}{c+d}=\frac{(a+b)(c+d)}{a+b+c+d}
$$
Determine all possible values of $a+b+c+d$. | [
"Answer: The possible values are the positive integers that are not square-free.\n\nFirst, note that if we take $a=\\ell$, $b=k \\ell$, $c=k \\ell$, $d=k^{2} \\ell$ for some positive integers $k$ and $\\ell$, then we have\n$$\n\\frac{a b}{a+b}+\\frac{c d}{c+d}=\\frac{k \\ell^{2}}{\\ell+k \\ell}+\\frac{k^{3} \\ell^{... | IMO | International Mathematical Olympiad Shortlist | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All positive integers that are not square-free. | |
08lu | Problem:
Find all integers $n$ such that $n^{4}+8 n+11$ is a product of two or more consecutive integers. | [
"Solution:\nWe will prove that $n^{4}+8 n+11$ is never a multiple of $3$. This is clear if $n$ is a multiple of $3$. If $n$ is not a multiple of $3$, then\n$$\nn^{4}+8 n+11 = (n^{4}-1) + 12 + 8 n = (n-1)(n+1)(n^{2}+1) + 12 + 8 n,\n$$\nwhere $8 n$ is the only term not divisible by $3$. Thus $n^{4}+8 n+11$ is never t... | JBMO | 2008 Shortlist JBMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | n = 1 | |
00nt | Let $ABC$ be a triangle. Let $P$ be the point on the extension of $BC$ beyond $B$ such that $BP = BA$. Let $Q$ be the point on the extension of $BC$ beyond $C$ such that $CQ = CA$. Prove that the circumcenter $O$ of the triangle $APQ$ lies on the angle bisector of the angle $\angle BAC$.
Figu... | [
"Since $ACQ$ is an isosceles triangle, the perpendicular bisector of $AQ$ is the angle bisector of $\\angle QCA$. But the perpendicular bisector of $AQ$ also passes through the circumcenter $O$ of the triangle $APQ$.\n\nTherefore, $O$ lies on the angle bisector of $\\angle QCA$ which is the exterior angle bisector ... | Austria | AUT_ABooklet_2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03s6 | Express the polynomial in $x$ $f(x) = 1 - x + x^2 - x^3 + \cdots - x^{19} + x^{20}$ into a polynomial in $y$ $g(y) = a_0 + a_1 y + a_2 y^2 + \cdots + a_{19} y^{19} + a_{20} y^{20}$, where $y = x - 4$. Then $a_0 + a_1 + \cdots + a_{20} = \text{_________}$. | [
"The terms in the expression $f(x)$ form a geometric series with first term $1$ and common ratio $-x$. By the summation formula of geometric series,\n$$\nf(x) = \\frac{(-x)^{21} - 1}{-x - 1} = \\frac{x^{21} + 1}{x + 1}.\n$$\nSet $x = y+4$, $g(y) = \\frac{(y+4)^{21} + 1}{y+5}$. Let $y = 1$, we get\n$$\na_0 + a_1 + \... | China | China Mathematical Competition (Jiangxi) | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | final answer only | (5^21 + 1)/6 | |
0j56 | Problem:
Let $AMOL$ be a quadrilateral with $AM = 10$, $MO = 11$, and $OL = 12$. Given that the perpendicular bisectors of sides $AM$ and $OL$ intersect at the midpoint of segment $AO$, find the length of side $LA$. | [
"Solution:\nLet $D$ be the midpoint of $AM$ and $E$ be the midpoint of $AO$. Then, we note that $ADE \\sim AMO$, so $M$ is a right angle. Similarly, $L$ is a right angle. Consequently, we get that\n$$\nAO^{2} = OM^{2} + AM^{2} \\Rightarrow AL = \\sqrt{AO^{2} - OL^{2}} = \\sqrt{11^{2} + 10^{2} - 12^{2}} = \\sqrt{77}... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(77) | |
0e21 | Problem:
Izkopali so jamo v obliki kvadra. Njena globina je 12 krat večja od dolžine. Dolžina jame je $\frac{3}{2}$ njene širine. Vsota merskega števila prostornine jame in ploščine njenega dna je $\frac{7}{6}$. Izračunaj globino jame. | [
"Solution:\n\nIzberemo spremenljivke $a$ je dolžina, $b$ je širina in $c$ je globina. Nastavimo sistem enačb $12a = c$, $a = \\frac{3b}{2}$ in $abc + ab = \\frac{7}{6}$. Sistem uredimo in dobimo $48a^{3} + 4a^{2} - 7 = 0$. Rešimo enačbo tretje stopnje z uporabo Hornerjevega algoritma. Zapišemo rešitve $a = \\frac{1... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 6 | |
07qu | Prove for all complex numbers $z$ that
$$
|z|^2 + 2|z - 1| \ge 1,
$$
with equality iff $z = 1$. | [
"If $|z| \\ge 1$, then\n$$\n|z|^2 + 2|z - 1| \\ge 1 + 2|z - 1| \\ge 1,\n$$\nwith equality iff $z = 1$.\n\nIf $|z| < 1$, then\n$$\n|z - 1| \\ge |z| - 1 = 1 - |z|,\n$$\nand so\n$$\n|z|^2 + 2|z - 1| - 1 \\ge |z|^2 + 2(1 - |z|) - 1 = (|z| - 1)^2 > 0.\n$$"
] | Ireland | Ireland_2017 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
07hd | Incircle $\omega$ of triangle $ABC$ is tangent to sides $CB$ and $CA$ at $D$ and $E$, respectively. Point $X$ is the reflection of $D$ with respect to $B$. Suppose that the line $DE$ is tangent to the $A$-excircle at $Z$. Let the circumcircle of triangle $XZE$ intersects $\omega$ at $K$, for the second time. Prove that... | [
"Let $I$ be the incenter and $I_a$ the $A$-excenter of triangle $ABC$ and $J$ the $C$-excenter of triangle $CDE$. $J$ lies on $\\omega$, since\n$$\n\\angle EJD = 90^\\circ - \\frac{1}{2}\\angle ECD = \\angle CDE.\n$$\nWe claim that $A$, $J$, and $Z$ are collinear. Point $T$ lies on $AC$ such that $IT \\parallel DE$... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Transformations > Homothety",
"... | English | proof only | null | |
04wt | Prove, that for every integer $k \ge 1$ there is a positive integer $n$ with following property: in decimal representation of $2^n$ one can find a block of exactly $k$ consecutive zeros, i.e.
$$
2^n = \dots a \overline{00\dots0} b\dots, \quad \text{ $k$ zeros }
$$
where $a$, $b$ are nonzero digits. | [
"First we show there are arbitrarily long blocks of zeros in powers of $2$. To get at least $k$ zeros in $2^n$, this power must be of the form $y \\cdot 10^{m+k} + z$ with $y, z$ positive integers and $z$ having at most $m$ digits, i.e. $z < 10^m$. Thus it is sufficient to find $n, m$ with $2^n$ having residue less... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko 2006 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0imm | Problem:
A sequence $\{a_{n}\}_{n \geq 1}$ of positive reals is defined by the rule $a_{n+1} a_{n-1}^{5} = a_{n}^{4} a_{n-2}^{2}$ for integers $n > 2$ together with the initial values $a_{1} = 8$, $a_{2} = 64$, and $a_{3} = 1024$. Compute
$$
\sqrt{a_{1} + \sqrt{a_{2} + \sqrt{a_{3} + \cdots}}}
$$ | [
"Solution:\nAnswer: $3 \\sqrt{2}$.\n\nTaking the base-$2$ log of the sequence $\\{a_{n}\\}$ converts the multiplicative rule to a more familiar additive rule:\n$$\n\\log_{2}(a_{n+1}) - 4 \\log_{2}(a_{n}) + 5 \\log_{2}(a_{n-1}) - 2 \\log_{2}(a_{n-2}) = 0.\n$$\nThe characteristic equation is $0 = x^{3} - 4x^{2} + 5x ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 3 sqrt(2) | |
0f4b | Problem:
Find an example of $m$ and $n$ such that $m$ is the product of $n$ consecutive positive integers and also the product of $n + 2$ consecutive positive integers. Show that we cannot have $n = 2$. | [] | Soviet Union | 15th ASU | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | Example: m = 120, n = 3; there is no solution with n = 2. | |
0gpe | Graphistan has 2011 cities and Graph Air (GA) is running one-way flights between all pairs of these cities. Determine the maximum possible value of the integer $k$ such that no matter how these flights are arranged it is possible to travel between any two cities in Graphistan riding only GA flights so long as the absol... | [
"We want to find the largest integer $k$ such that a directed path exists from any vertex $v_1$ to any vertex $v_2$ in any directed complete graph $G$ with 2011 vertices satisfying $|\\text{indeg}(v) - \\text{outdeg}(v)| \\le k$ for all vertices $v$. The answer is 1005.\n\nObserve that $|\\text{indeg}(v) - \\text{o... | Turkey | Team Selection Test for IMO 2011 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1005 | |
0d9t | Let $ABC$ be a triangle with $M, N, P$ as midpoints of the segments $BC, CA, AB$ respectively. Suppose that $I$ is the intersection of angle bisectors of $\angle BPM$, $\angle MNP$ and $J$ is the intersection of angle bisectors of $\angle CNM$, $\angle MPN$. Denote ($\omega_{1}$) as the circle of center $I$ and tangent... | [
"1) Note that $\\angle MNC = \\angle MPB = \\angle A$ then by angle chasing, we have $IP \\parallel JN$. Denote $K = PJ \\cap IN$ then $K$ is the incenter of triangle $MNP$. Hence, $MK$ is the angle bisector of $\\angle NMP$, thus $MK \\parallel IP$. Denote $X = IN \\cap MP$ then\n$$\n\\frac{IP}{MK} = \\frac{XP}{XM... | Saudi Arabia | Team selection tests for BMO 2018 | [
"Geometry > Plane Geometry > Triangles",
"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 > Pl... | English | proof only | null | |
09bd | $ABC(\angle C = 90^\circ)$ тэгш өнцөгт гурвалжны $AA_1$ ба $BB_1$ биссектрисүүд $I$ цэгт огтлолцоно. $O$ нь $A_1B_1$-хэрчмийн дундаж цэг бол $OI \perp AB$ болохыг батал. | [
"\n\n$B_1$-цэгийг $AA_1$ биссектриссийн хувьд тэгш хэмтэй хувиргавал $AB$ гипотенуз дээр $B_2$ цэг гарна. Үүнтэй адил $A_1$ цэгийн $BB_1$ биссектриссийн хувьд тэгш хэмтэй хувиргавал $AB$ гипотенуз дээр $A_2$ цэг гарна.\n\n$\\angle B_1IA = \\angle IAB + \\angle IBA = \\frac{1}{2}(\\angle CAB... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | Mongolian | proof only | null | |
0btx | Find all non-negative integers $n$ so that $\sqrt{n+3} + \sqrt{\sqrt{n+3}}$ is an integer. | [
"Denote $m = \\sqrt{n+3} + \\sqrt{\\sqrt{n+3}}$. Then $n + \\sqrt{n+3} = (m - \\sqrt{n+3})^2$, whence $(2m+1)\\sqrt{n+3} = m^2 + 3$. Then there exists $p \\in \\mathbb{N}$ so that $n+3 = p^2$ and, since $p + \\sqrt{n+p} \\in \\mathbb{N}$, there exists $q \\in \\mathbb{N}$ so that $n+p = q^2$.\nEliminating $n$ yield... | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All n of the form k^4 − 3 with integers k ≥ 2. | |
04px | Find all triples $(x, y, z)$ of real numbers such that
$$
\begin{aligned}
(x^2 + 1)y &= z^2 + 1 \\
(y^2 + 1)z &= x^2 + 1 \\
(z^2 + 1)x &= y^2 + 1.
\end{aligned}
\qquad (\text{Ukraine 2017})
$$ | [] | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | (1, 1, 1) | |
0d2o | For a positive integer $n$, we consider all its divisors (including $1$ and itself). Suppose that $p\%$ of these divisors have their unit digit equal to $3$ (For example $n=117$, has six divisors, namely $1, 3, 9, 13, 39, 117$. Two of these divisors, namely $3$ and $13$, have unit digits equal to $3$. Hence for $n=117$... | [
"Let $n$ be a positive integer. Consider $D_{1}$, the set of all divisors of $n$ with unit digit $3$, and $D_{2}$ the set of all the other divisors of $n$.\n\nIf the unit digit of $n$ is different from $9$, consider the map $\\delta: D_{1} \\longrightarrow D_{2}$ defined by $\\delta(d) = \\frac{n}{d}$ for all $d \\... | Saudi Arabia | Selection tests for the International Mathematical Olympiad 2013 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 50 | |
00ct | Sea $n$ un número natural. Definimos $f(n)$ como la cantidad de maneras de escribir $n$ como suma de potencias de 2, donde se tiene en cuenta el orden en que aparece cada término. Por ejemplo, $f(4) = 6$ pues $4$ se puede escribir como $4$; $2+2$; $2+1+1$; $1+2+1$; $1+1+2$; $1+1+1+1$.
Hallar el menor $n$ mayor que $201... | [] | Argentina | Nacional OMA 2019 | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | Spanish | proof and answer | 2047 | |
0inx | Problem:
There are three video game systems: the Paystation, the WHAT, and the ZBoz2 $\pi$, and none of these systems will play games for the other systems. Uncle Riemann has three nephews: Bernoulli, Galois, and Dirac. Bernoulli owns a Paystation and a WHAT, Galois owns a WHAT and a ZBoz2 $\pi$, and Dirac owns a ZBoz... | [
"Solution:\n\nAnswer: $\\frac{7}{25}$. Since the games are not necessarily distinct, probabilities are independent. Multiplying the odds that each nephew receives a game he can play, we get $10/20 \\cdot 14/20 \\cdot 16/20 = 7/25$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 7/25 | |
0kly | As shown in the figure below, point $E$ lies in the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$. Point $F$ lies on $\overline{AD}$ so that $DE = DF$, and $ABCD$ is a square. What is the degree measure of $\angle AFE$?
(A) 160 (B) 164 (C) 166... | [] | United States | AMC 12 A | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | D | |
0kzv | Problem:
The numbers $1, 2, \ldots, 20$ are put into a hat. Claire draws two numbers from the hat uniformly at random, $a < b$, and then puts them back into the hat. Then, William draws two numbers from the hat uniformly at random, $c < d$.
Let $N$ denote the number of integers $n$ that satisfy exactly one of $a \leq n... | [
"Solution:\nThe number of integers that satisfy exactly one of the two inequalities is equal to the number of integers that satisfy the first one, plus the number of integers that satisfy the second one, minus twice the number of integers that satisfy both. Parity-wise, this is just the number of integers that sati... | United States | HMMT February 2024 Guts Round | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 181/361 | |
0f8s | Problem:
$ABCD$ has $AB = CD$, but $AB$ not parallel to $CD$, and $AD$ parallel to $BC$. The triangle $ABC$ is rotated about $C$ to $A'B'C$. Show that the midpoints of $BC$, $B'C$ and $A'D$ are collinear. | [] | Soviet Union | 23rd ASU | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0h0a | Let $x$, $y$, $z$ be real numbers such that $x \in (0, 1]$, $y \in (0, 1]$, $z \in (0, 1]$. Prove the inequality:
$$
\frac{x}{2+xy+yz} + \frac{y}{2+yz+zx} + \frac{z}{2+zx+xy} \le \frac{x+y+z}{x+y+z+xyz}
$$ | [
"Consider the obvious inequality:\n$$\n\\begin{aligned}\n0 \\le (1-x)(1-y)(1-z) &= 1 + xy + yz + zx - x - y - z - xyz \\\\\n&\\Rightarrow 1 + xy + yz + zx \\ge x + y + z + xyz \\\\\n&\\Rightarrow 2 + xy + yz \\ge x + y + z + xyz \\\\\n&\\Rightarrow \\frac{x}{2+xy+yz} \\le \\frac{x}{x+y+z+xyz}.\n\\end{aligned}\n$$\n... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0esm | It is estimated that there are $3\,000\,000$ km² of rain forest in Brazil, and it is being reduced at the rate of $7\,400$ km² per year. If this rate stays the same, then the number of years before there is no rain forest left is approximately
(A) $400$ (B) $300$ (C) $200$ (D) $150$ (E) $100$ | [
"The number of years is $\\frac{3\\,000\\,000}{7\\,400} = \\frac{30 \\times 100}{7.4} \\approx 400$"
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | English | MCQ | A | |
0ebb | Let $P$ and $Q$ be two points in the interior of the triangle $ABC$ such that $\angle PAC = \angle BAQ$ and $\angle CBP = \angle QBA$. Denote by $P_A, P_B$, and $P_C$ the orthogonal projections of the point $P$ onto the sides $BC, CA$, and $AB$, and by $Q_A, Q_B$, and $Q_C$ the orthogonal projections of the point $Q$ o... | [
"Since $\\angle PAC = \\angle BAQ$ the right-angle triangles $PAP_B$ and $QAQ_C$ have equal angles and hence are similar. It follows $\\frac{|AP|}{|AQ|} = \\frac{|AP_B|}{|AQ_C|}$. We also have $\\angle BAP = \\angle BAC - \\angle PAC = \\angle BAC - \\angle BAQ = \\angle QAC$. Hence the right-angle triangles $PAP_C... | Slovenia | Selection Examinations for the IMO 2015 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ddm | Let $ABC$ be an acute, non-isosceles triangle with the orthocenter $H$, circumcenter $O$ and $AD$ is the diameter of $(O)$. Suppose that the circle $(AHD)$ meets the lines $AB$, $AC$ at $F$, $E$, respectively. Denote $J$, $K$ as orthocenter and nine-point center of $AEF$. Prove that $HJ$ is parallel to $BC$ and $KO = K... | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous >... | null | proof only | null | |
0c6a | Let $n \in \mathbb{N}$, $(n \ge 2)$ be a positive integer, and let $A, B \in \mathcal{M}_n(\mathbb{R})$ be two matrices with real entries. Prove that there exists a complex number $z$, with $|z| = 1$, such that
$$
\operatorname{Re}(\det(A + zB)) \ge \det(A) + \det(B),
$$
where we denote by $\operatorname{Re}(w)$ the re... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Linear Algebra > Determinants",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
0k6o | Problem:
Complex numbers $a, b, c$ form an equilateral triangle with side length $18$ in the complex plane. If $|a+b+c|=36$, find $|bc+ca+ab|$. | [
"Solution:\n\nUsing basic properties of vectors, we see that the complex number $d=\\frac{a+b+c}{3}$ is the center of the triangle. From the given, $|a+b+c|=36 \\Longrightarrow |d|=12$. Then, let $a' = a-d$, $b' = b-d$, and $c' = c-d$. Due to symmetry, $|a'+b'+c'|=0$ and $|b'c'+c'a'+a'b'|=0$.\n\nFinally, we compute... | United States | HMMT February 2019 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | final answer only | 432 | |
022i | Problem:
No plano cartesiano, chama-se ponto inteiro a um ponto de coordenadas inteiras. Se $n$ é inteiro positivo, seja $f(n)$ o número de pontos inteiros que estão sobre o segmento que liga a origem ao ponto inteiro $(n, n+3)$, sem contar os extremos. Mostre que:
$$
f(n)=\left\{\begin{array}{lll}
2 & \text{ se } & n ... | [
"Solution:\nComecemos examinando alguns casos.\n- $f(1)$ é o número de pontos inteiros sobre o segmento que liga $(0,0)$ ao ponto $(1,4)$. Logo, $f(1)=0$.\n- $f(2)$ é o número de pontos inteiros sobre o segmento que liga $(0,0)$ ao ponto $(2,3)$. Logo, $f(2)=0$.\n- $f(3)$ é o número de pontos inteiros sobre o segme... | Brazil | Nível 3 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 2 if n is a multiple of 3; 0 otherwise | |
043x | Find the largest constant $C > 0$, such that for any integer $n \ge 2$, one can find real numbers $x_1, x_2, \dots, x_n \in [-1, 1]$ satisfying
$$
\prod_{1 \le i < j \le n} (x_i - x_j) \ge C^{\frac{n(n-1)}{2}}.
$$ | [
"For sufficiently large $n$, we guess the optimal choice of $(x_1, \\dots, x_n)$ on $[-1, 1]$ is close to the projection of points uniformly distributed on the unit circle onto the $x$ axis. To this end, we introduce the reference choice\n$$\na_k = \\cos \\vartheta_k = \\cos \\frac{2k-1}{2n} \\pi, \\quad k = 1, 2, ... | China | China National Team Selection Test | [
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Geometry > Plane Geometry > Analytic / Coordinate... | null | proof and answer | 1/2 | |
0l5n | Problem:
Compute the 100th smallest positive multiple of $7$ whose digits in base $10$ are all strictly less than $3$. | [
"Solution:\nWe construct an order-preserving bijection between positive multiples of $7$ in base $10$ whose digits are all less than $3$ and positive multiples of $7$. For any multiple of $7$ in base $10$ with digits all less than $3$, interpret it as a base $3$ number and convert it to a base $10$ decimal, which w... | United States | HMMT February | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 221221 | |
06va | You are given a set of $n$ blocks, each weighing at least $1$; their total weight is $2n$. Prove that for every real number $r$ with $0 \leqslant r \leqslant 2n-2$ you can choose a subset of the blocks whose total weight is at least $r$ but at most $r+2$. | [
"Claim. Suppose that you have $n$ blocks, each of weight at least $1$, and of total weight $s \\leqslant 2n$. Then for every $r$ with $-2 \\leqslant r \\leqslant s$, you can choose some of the blocks whose total weight is at least $r$ but at most $r+2$.\n\nProof. The base case $n=1$ is trivial. To prove the inducti... | IMO | IMO 2019 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0ewz | Problem:
What is the smallest number of tetrahedrons into which a cube can be partitioned? | [
"Solution:\nAnswer: 5.\n\nTetrahedral faces are triangular, so each cube face requires at least two tetrahedral faces. So at least 12 tetrahedral faces are needed in all. At most three faces of a tetrahedron can be mutually orthogonal (and no two faces can be parallel), so at most 3 faces from each tetrahedron can ... | Soviet Union | 4th ASU | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 5 | |
0ir2 | For a pair $A = (x_1, y_1)$ and $B = (x_2, y_2)$ of points on the coordinate plane, let $d(A, B) = |x_1 - x_2| + |y_1 - y_2|$. We call a pair $(A, B)$ of (unordered) points *harmonic* if $1 < d(A, B) \le 2$. Determine the maximum number of harmonic pairs among 100 points on the plane.
(This problem was suggested by Zu... | [
"We claim that there do not exist five points on the plane such that they are pairwise harmonic. Suppose such a set of five points exists. We say $(x_1, y_1)$ precedes $(x_2, y_2)$, denoted by $(x_1, y_1) \\prec (x_2, y_2)$, if and only if $x_1 \\le x_2$ and $y_1 \\le y_2$. This defines a partial order on the set. ... | United States | Team Selection Test | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 3750 | |
0kj1 | For each positive integer $n$, let $f_1(n)$ be twice the number of positive integer divisors of $n$, and for $j \ge 2$, let $f_j(n) = f_1(f_{j-1}(n))$. For how many values of $n \le 50$ is $f_{50}(n) = 12$?
(A) 7 (B) 8 (C) 9 (D) 10 (E) 11 | [] | United States | AMC 12 A | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | D | |
0emz | Show that for every natural number $n$, there exists a natural number $k$ such that the number $kn$ starts with the digits 2011. | [
"Suppose we wish to find a solution where $kn$ has $d+4$ digits. Then we must satisfy $2011 \\cdot 10^d \\le kn < 2012 \\cdot 10^d$, or equivalently\n$$\nk \\in \\left[ \\frac{2011}{n} \\cdot 10^d, \\frac{2012}{n} \\cdot 10^d \\right).\n$$\nThe size of this interval is $\\frac{10^d}{n}$. Clearly, we can choose $d$ ... | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0gnq | In a country between each pair of cities there is at most one direct road. There is a connection (using one or more roads) between any two cities even after the elimination of any given road. We say that the city $A$ can be $k$-directionally connected to the city $B$, if: we can orient at most $k$ roads such that after... | [
"The answer is: $k = 2n - 3$. If all pairs of cities are directly connected (complete graph) then all roads from $A$ and all roads to $B$ must be oriented: $k \\ge (n-1) + (n-1) - 1 = 2n - 3$ or if $\\deg(A) = \\deg(B) = n - 1$, all other vertices have degree $2$, then again we have to orient all $2(n-2) + 1 = 2n -... | Turkey | Turkish Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Menger's theorem / max-flow, min-cut"
] | English | proof and answer | 2n - 3 | |
0j9a | Problem:
$ABC$ is an isosceles triangle with $AB = 2$ and $\measuredangle ABC = 90^{\circ}$. $D$ is the midpoint of $BC$ and $E$ is on $AC$ such that the area of $AEDB$ is twice the area of $ECD$. Find the length of $DE$. | [
"Solution:\n\nAnswer: $\\frac{\\sqrt{17}}{3}$\n\nLet $F$ be the foot of the perpendicular from $E$ to $BC$. We have $[AEDB] + [EDC] = [ABC] = 2 \\Rightarrow [EDC] = \\frac{2}{3}$.\n\nSince we also have $[EDC] = \\frac{1}{2} (EF)(DC)$, we get $EF = FC = \\frac{4}{3}$.\n\nSo $FD = \\frac{1}{3}$, and $ED = \\frac{\\sq... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(17)/3 | |
01m0 | Point $K$ is the intersection point of medians $AL$, $BM$, and $CN$ of a triangle $ABC$.
Prove that $\angle ABM = \angle CAL$ if $\angle BLA = \angle CNA$. | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0k53 | Problem:
Circle $\omega_{1}$ of radius $1$ and circle $\omega_{2}$ of radius $2$ are concentric. Godzilla inscribes square $CASH$ in $\omega_{1}$ and regular pentagon $MONEY$ in $\omega_{2}$. It then writes down all $20$ (not necessarily distinct) distances between a vertex of $CASH$ and a vertex of $MONEY$ and multip... | [
"Solution:\n\nWe represent the vertices with complex numbers. Place the vertices of $CASH$ at $1$, $i$, $-1$, $-i$ and the vertices of $MONEY$ at $2\\alpha$, $2\\alpha\\omega$, $2\\alpha\\omega^{2}$, $2\\alpha\\omega^{3}$, $2\\alpha\\omega^{4}$ with $|\\alpha|=1$ and $\\omega=e^{\\frac{2\\pi i}{5}}$. We have that t... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof and answer | 1048577 | |
0d1g | Find all positive integers $n$ with the following property: there are two divisors $a$ and $b$ of the number $n$ such that $a^2 + b^2 + 1$ is a multiple of $n$. | [
"Since $a$ divides $n$ divides $a^2 + b^2 + 1$, it follows that $a \\mid b^2 + 1$. Similarly $b \\mid a^2 + 1$. Thus we have $\\gcd(a, b) = 1$.\nWe have $ab \\mid a^2 + b^2 + 1$. By symmetry, we can assume $a \\le b$. It follows that\n$$\na^2 + b^2 + 1 = kab \\quad (1)\n$$\nfor some positive integer $k$. In the cas... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | All n equal to 1 or 3, or of the form F_{2k-1} F_{2k+1} or 3 F_{2k-1} F_{2k+1} for k ≥ 1, where F_m denotes the mth Fibonacci number with F_0 = 0 and F_1 = 1. | |
0b0p | Problem:
A student writes the six complex roots of the equation $z^{6} + 2 = 0$ on the blackboard. At every step, he randomly chooses two numbers $a$ and $b$ from the board, erases them, and replaces them with $3ab - 3a - 3b + 4$. At the end of the fifth step, only one number is left. Find the largest possible value o... | [] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 730 | |
0jls | Problem:
Find the number of strictly increasing sequences of nonnegative integers with the following properties:
- The first term is $0$ and the last term is $12$. In particular, the sequence has at least two terms.
- Among any two consecutive terms, exactly one of them is even. | [
"Solution:\nAnswer: $144$\n\nFor a natural number $n$, let $A_{n}$ be a set containing all sequences which satisfy the problem conditions but with $12$ replaced by $n$. Also, let $a_{n}$ be the size of $A_{n}$.\n\nWe first consider $a_{1}$ and $a_{2}$. We get $a_{1}=1$, as the only sequence satisfying the problem c... | United States | HMMT November 2014 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 144 | |
032e | Problem:
Let $f(x) = x^{4} - x^{3} + 8a x^{2} - a x + a^{2}$ and $g(y) = y^{2} - y + 6a$.
a) Prove that $f(x) = \left(x^{2} - y_{1} x + a\right)\left(x^{2} - y_{2} x + a\right)$, where $y_{1}$ and $y_{2}$ are the roots of the equation $g(y) = 0$.
b) Find all values of $a$ such that the equation $f(x) = 0$ has four di... | [
"Solution:\nSince $y_{1} + y_{2} = 1$ and $y_{1} y_{2} = 6a$, we have\n$$\n\\begin{gathered}\n\\left(x^{2} - y_{1} x + a\\right)\\left(x^{2} - y_{2} x + a\\right) \\\\\n= x^{4} - \\left(y_{1} + y_{2}\\right) x^{3} + \\left(2a + y_{1} y_{2}\\right) x^{2} - a\\left(y_{1} + y_{2}\\right) x + a^{2} = f(x)\n\\end{gather... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (1/25, 1/24) | |
02gr | Given a circle and a point $A$ inside the circle, but not at its center. Find points $B, C, D$ on the circle which maximize the area of the quadrilateral $ABCD$. | [
"If we fix the length $BD$, then we obviously maximize area $BDA$ by taking the distance of $A$ from $BD$ as large as possible and hence by taking $BD$ perpendicular to $AO$ and on the opposite side of $O$ to $A$. We maximize $BCD$ by taking $C$ the midpoint of the arc $BD$. So suppose the radius of the circle is $... | Brazil | XXV OBM | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
067s | Every triangle in the plane $\mathbb{R}^2$ contains a Mediterranean point on its boundary or in its interior (even if the triangle degenerates into a line segment or a point). These Mediterranean points satisfy the following conditions:
(i) If a triangle is symmetric with respect to a line through the origin $(0, 0)$, ... | [
"Consider the auxiliary triangle $DEF$ with $D = (-12, 5)$, $E = (12, 5)$, and $F = (0, 13)$. Since the triangle $DEF$ is symmetric with respect to the $y$-axis, condition (i) yields that its Mediterranean point $M$ lies on the $y$-axis, and hence has coordinates $(0, m)$ with $5 \\le m \\le 13$. Define a non-negat... | Greece | Mediterranean Mathematical Competition PETER O' HALLORAN MEMORIAL | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | (0, 5) | |
0jko | Divide the plane into an infinite square grid by drawing all the lines $x = m$ and $y = n$ for integers $m$ and $n$. Next, if a square's upper-right corner has both coordinates even, color it black; otherwise, color it white (in this way, exactly $1/4$ of the squares are black and no two black squares are adjacent). Le... | [
"Classify the white squares into 3 groups in the following way: white squares of type A are vertically adjacent to black squares, white squares of type B are diagonally adjacent to black squares, and white squares of type C are horizontally adjacent to black squares. In addition to this, note that at any point in t... | United States | IMO Team Selection Team Selection Test | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
00kw | A *police emergency number* is a positive integer that ends with the digits $133$ in decimal representation. Prove that every police emergency number has a prime factor larger than $7$. | [
"Let $n = 1000k + 133$ be a police emergency number and assume that all its prime divisors are at most $7$. It is clear from the last digit that $n$ is odd and that $n$ is not divisible by $5$, so $1000k + 133 = 3^a 7^b$ for suitable integers $a, b \\ge 0$.\n\nThus $3^a 7^b \\equiv 133 \\pmod{1000}$.\n\nThis also i... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0l2a | Problem:
There are 1024 players, ranked from 1 (most skilled) to 1024 (least skilled), participating in a single elimination tournament. In each of the 10 rounds, the remaining players are paired uniformly at random. In each match, the player with a lower rank always wins, and the loser is eliminated from the tourname... | [
"Solution:\n\nThe probability that the rank $N$ player passes round $i$ (where $i=0$ is implied as 1) is\n$$\n\\frac{\\binom{1024-2^{i}}{N-1}}{\\binom{1023}{N-1}}\n$$\n\nSumming this from $i=0$ to $9$ (each represents the expectation of advancing one round), we must find the minimum $N$ for which the sum goes below... | United States | HMMT November 2024 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 350 | |
0bwt | Consider the positive integer $n = 7 + 7^2 + 7^3 + \ldots + 7^{2017}$.
a) Show that the remainders of the divisions of $7^{2018}$ by $6$ and by $48$ are equal.
b) Determine the last two decimal digits of the number $6n$. | [
"a) Since $7n = 7^2 + 7^3 + 7^4 + \\ldots + 7^{2017} + 7^{2018}$, we have $6n = 7n - n = 7^{2018} - 7$. It follows that $7^{2018} = 6n + 7 = 6(n + 1) + 1$, and the remainder of the division by $6$ of the number $7^{2018}$ is $1$.\n\nAlso, $n = 7 + 7^2(1 + 7) + 7^4(1 + 7) + \\ldots + 7^{2016}(1 + 7) = 7 + 8 \\cdot (... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | Both remainders are 1; the last two digits of 6n are 42. | |
0c34 | Consider a four-point configuration in the plane, every three points of which can be covered by a strip of unit width. Prove that:
a) the four points can be covered by a strip of width at most $\sqrt{2}$; and
b) if no strip of width less than $\sqrt{2}$ covers all four points, then they are the vertices of a square o... | [
"Both facts follow from the lemma below.\n\n**Lemma.** Consider a triangle some altitude of which has length at most $1$. If the triangle has an altitude of length at least $\\sqrt{2}$, say the $X$-altitude, then the internal angular span at $X$ does not exceed $45^\\circ$; equality holds if and only if the $X$-alt... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | null | proof only | null | |
035e | Problem:
Let $m$ be a positive integer, $A=\{-m,-m+1, \ldots, m-1, m\}$ and $f: A \rightarrow A$ be a function such that $f(f(n))=-n$ for every $n \in A$.
a) Prove that the number $m$ is even.
b) Find the number of all functions $f: A \rightarrow A$ with the required property. | [
"Solution:\n\na.\nLet $n \\in A$ and $O_{n}=\\{n, f(n),-n, f(-n)\\}$. Since $f(f(n))=-n$ and $f(f(-n))=n$, it follows easily that if $k \\in A$ then either $O_{k}=O_{n}$ or $O_{n} \\cap O_{k}=\\varnothing$. Moreover, we obtain $f(n) \\neq f(-n)$ for $n \\neq 0$.\n\nFurther, if $f( \\pm n)= \\pm n$, then $\\mp n=f(f... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | m is even; for m equal to two times k, the number of functions is m!/(m/2)! | |
0ce9 | Find all integers $n \ge 2$ so there exists a finite field $\mathbb{K}$ with $n$ elements, with the property: for every $x \in \mathbb{K}$ there exists $y \in \mathbb{K}$ such that $x^2 = y^3$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Algebra > Abstract Algebra > Field Theory",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof and answer | All prime powers n with 3 not dividing n − 1 (equivalently, n not congruent to 1 modulo 3). | |
02z3 | Problem:
Em uma escola, devem ser formados $n$ clubes, com $n \geq 3$ e cada um com 3 integrantes, de modo que para cada par de clubes haja exatamente um estudante que integra ambos.
a) Dê um exemplo de uma distribuição de 7 clubes que satisfaçam as condições mencionadas.
b) Verifique que se um estudante pertence a 4 ... | [
"Solution:\n\na) Nomeando os alunos por $A, B, C, D, E, F$ e $G$, os seguintes conjuntos representam 7 clubes que atendem às condições do enunciado:\n$$\n\\{A, B, C\\},\\{A, D, E\\},\\{A, F, G\\},\\{B, E, F\\},\\{B, D, G\\},\\{C, D, F\\},\\{C, E, G\\}\n$$\n\nb) Suponha que o estudante $A$ pertença aos seguintes 4 c... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 8 | |
082u | Problem:
a) Si dimostri che se in un triangolo vi sono due altezze di egual lunghezza, allora il triangolo è isoscele.
b) Si dimostri che se in un triangolo vi sono due mediane di egual lunghezza, allora il triangolo è isoscele.
c) Sui lati $A B$ e $A C$ di un triangolo $A B C$ si scelgano due punti $M, N$ in modo t... | [
"Solution:\n\na. Siano $B$ e $C$ i vertici da cui partono le due altezze uguali, rispettivamente relative ai lati $A C$ e $A B$. Calcolando l'area $S$ del triangolo rispetto ad $A B$ e ad $A C$ e chiamando $h$ la lunghezza comune delle due altezze, si ha\n$$\nS = \\frac{A B \\cdot h}{2} = \\frac{A C \\cdot h}{2}\n$... | Italy | Progetto Olimpiadi di Matematica 2003 GARA di SECONDO LIVELLO TRIENNIO | [
"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 | |
05el | Problem:
Soient $\omega$ et $\Omega$ deux cercles concentriques (c'est-à-dire qu'ils ont le même centre) de sorte que le cercle $\omega$ soit à l'intérieur du cercle $\Omega$. Soient $X$ et $Y$ deux points sur le cercle $\omega$. On note $P$ et $Q$ les points d'intersection respectifs du cercle $\Omega$ avec les tange... | [
"Solution:\n\n\n\nLe quadrilatère $XYQP$ semble être un trapèze isocèle sur la figure. On commence par le démontrer.\n\nSoit $d$ la médiatrice de $[XY]$. Puisque $O$ est le centre du cercle $\\omega$, $O$ appartient à $d$. Ainsi, les cercles centrés en $O$, et en particulier $\\omega$ et $\... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09iw | A point $D$ chosen on the side $BC$ of a triangle $ABC$ such that $AD = 2BD$. Let $E$ and $F$ be the foot of the perpendicular line from $D$ to the line $AB$ and the line $AC$, respectively. A line through $D$ perpendicular to the line $BC$ meets with $EF$ at $M$. A line through $M$ and parallel to $BC$ meets with $AB$... | [] | Mongolia | Mongolian Mathematical Olympiad Round 1 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null |
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