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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0jqg | Problem:
Let $H, I, O, \Omega$ denote the orthocenter, incenter, circumcenter and circumcircle of a scalene acute triangle $A B C$. Prove that if $\angle B A C=60^{\circ}$ then the circumcenter of $\triangle I H O$ lies on $\Omega$. | [
"Solution:\nFirst, we show that the five points $B, O, H, I, C$ all lie on a circle. To see this, note that\n$$\n\\begin{aligned}\n& \\angle B I C=90^{\\circ}+\\frac{1}{2} \\angle B A C=120^{\\circ} \\\\\n& \\angle B O C=2 \\angle B A C=120^{\\circ} \\\\\n& \\angle B H C=180^{\\circ}-\\angle B A C=120^{\\circ} .\n\... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dzx | Let $a_1, a_2, \dots, a_n$ be positive integers. Assume that for some positive integer $k$, $k < n$, the following is true: if we choose any $k$ of the given $n$ numbers their sum is divisible by $n$. Prove that $a_1 + a_2 + \dots + a_n$ is also divisible by $n$. | [
"If $k = 1$, each of the numbers $a_1, a_2, \\dots, a_n$ is divisible by $n$, so $n$ also divides their sum.\n\nNow, let $1 < k < n$ and let $i \\neq j$. Since the set $\\{a_1, a_2, \\dots, a_n\\} \\setminus \\{a_i, a_j\\}$ has $n-2 \\ge k-1$ elements, one can choose an arbitrary subset $S$ with $k-1$ elements. The... | Slovenia | Slovenija 2008 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
0jpo | Problem:
A prime number $p$ is twin if at least one of $p+2$ or $p-2$ is prime and sexy if at least one of $p+6$ and $p-6$ is prime.
How many sexy twin primes (i.e. primes that are both twin and sexy) are there less than $10^{9}$? Express your answer as a positive integer $N$ in decimal notation; for example, 52149522... | [
"Solution:\n\nAnswer: 1462105\n\nThe Hardy-Littlewood conjecture states that given a set $A$ of integers, the number of integers $x$ such that $x+a$ is a prime for all $a \\in A$ is\n$$\n\\frac{x}{(\\ln x)^{|A|}} \\prod_{p} \\frac{1-\\frac{w(p ; A)}{p}}{\\left(1-\\frac{1}{p}\\right)^{k}}(1+o(1))\n$$\nwhere $w(p ; A... | United States | HMMT February | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Algorithms"
] | null | final answer only | 1462105 | |
0i2d | Problem:
What is the 18th digit after the decimal point of $\frac{10000}{9899}$? | [
"Solution:\n$\\frac{10000}{9899}$ satisfies $100(x-1) = 1.01 x$, so each pair of adjacent digits is generated by adding the previous two pairs of digits. So the decimal is $1.01020305081321345590\\ldots$, and the 18th digit is $5$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 5 | |
07t1 | Define $f(r) = 2\sqrt{r} + \sqrt{r-1} + \sqrt{r+1}$, $g(r) = 2\sqrt{r} + \sqrt{r-1} - \sqrt{r+1}$ and $h(r) = 2\sqrt{r} - \sqrt{r-1} + \sqrt{r+1}$. Calculate to two decimal places
$$
\sum_{r=1}^{2020} \frac{1}{f(r) \cdot g(r) \cdot h(r)}.
$$ | [
"The answer is 0.25 to 2DP (or more accurately, 0.247219).\nThe key observation is that\n$$\n\\frac{1}{f(r)g(r)h(r)} = \\frac{1}{4} \\left( 2\\sqrt{r} - \\sqrt{r-1} - \\sqrt{r+1} \\right).\n$$\nTwo see why, we apply the difference of two squares twice:\n$$\n\\begin{aligned} & \\left(2\\sqrt{r} - \\sqrt{r-1} - \\sqr... | Ireland | IRL_ABooklet_2020 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 0.25 | |
040a | Prove that for any real number $M > 2$, there exists a strictly increasing infinite sequence of positive integers $a_1, a_2, \dots$ satisfying both the following two conditions:
(1) $a_i > M^i$ for any positive integer $i$.
(2) An integer $n$ is non-zero if and only if there exists a positive integer $m$ and $b_1, b_2,... | [
"For given $M > 2$, we construct by induction a sequence $\\{a_n\\}$ that satisfies the requirements. Take $a_1, a_2$ that satisfy $a_2 - a_1 = 1$ and $a_1 > M^2$. Now suppose $a_1, a_2, \\dots, a_{2k}$ are already chosen, such that $a_i > M^i$, $i = 1, 2, \\dots, 2k$ and such that the set $A_k = \\{b_1a_1 + \\dots... | China | China Mathematical Olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
01nt | Two squares with the centers $C_1$ and $B_1$ are constructed on the sides $AB$ and $AC$ outside of the acute-angled triangle $ABC$, respectively. The square $C_1B_1DE$ is constructed on the segment $C_1B_1$ so that $A$ and $D$ lie in the different half-planes with respect to $C_1B_1$.
Prove that the center of the squar... | [
"Let $H$ be a foot of perpendicular from $A$ onto $BC$. Since $\\angle AB_1C = 90^\\circ$, we have $\\angle AHC + \\angle AB_1C = 90^\\circ + 90^\\circ = 180^\\circ$. Thus, points $A, B_1, C, H$ lie on the same circle. Since $AB_1 = B_1C$, we see that $HB_1$ is a bisector of $\\angle AHC$, so $\\angle AHB_1 = \\ang... | Belarus | Belorusija 2012 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0c75 | A positive integer $m$ is perfect if the sum of all its positive divisors, $1$ and $m$ inclusive, is equal to $2m$. Determine the positive integers $n$ such that $n^n + 1$ is a perfect number. | [
"There is only one such integer, namely, $n = 3$; it is readily checked that $3^3 + 1 = 28$ is perfect.\n\nIf $n$ is odd, then $n^n + 1$ is even, so it is of the form $2^{p-1}(2^p - 1)$, where $p$ and $2^p - 1$ are both prime (Euler's theorem on the structure of perfect even integers). Rule out the trivial case $n ... | Romania | SELECTION TESTS FOR THE 2019 BMO AND IMO | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | n = 3 | |
02nl | Let $P(n)$ be a polynomial with real coefficients. Prove that there exist integers $n$ and $k$ such that $k$ has at most $n$ digits and at least $P(n)$ divisors. | [
"Let $d$ be the degree of $P$, and consider $d+1$ distinct primes $p_1, p_2, \\dots, p_{d+1}$. Let $A = p_1p_2\\dots p_{d+1}$ and $k_N = A^N$. If $A$ has $r$ digits, then $k_N$ has at most $rN$ digits. On the other hand, $k_N$ has $(N+1)^{d+1}$ positive divisors. Since $P$ has degree $d$, $P(rN) < (N+1)^{d+1}$ for ... | Brazil | Brazilian Math Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
0ge8 | 給定正整數 $n$ 與 $k$，其中 $n \ge k+1$。某星球上有 $n$ 個國家，其中一些國家之間有建交，並且每個國家都至少有 $k$ 個邦交國。邪惡反派杜蘭莎想要分化這些國家，因此進行以下操作：
(1) 他首先選兩個國家 $A$ 與 $B$，由其各自發起陣營 $\mathcal{A}$ 與 $\mathcal{B}$ (故 $A \in \mathcal{A}$ 且 $B \in \mathcal{B}$)。
(2) 所有其他國家各自選擇要加入陣營 $\mathcal{A}$ 還是 $\mathcal{B}$。
(3) 在所有國家選擇完陣營後，對於任何兩個有建交的國家 $X$ 與 $Y$，若 $X \in \mathcal... | [
"讓我們以國家為點集 $S$ 做完全圖, 並對於每條邊 $AB$, 賦予其一個非負整數值 $d(A, B)$。對於每個點 $A$, 令 $d(A) = \\sum_{B \\neq A} d(A, B)$。對於任兩個點集 $\\mathcal{A}$ 與 $\\mathcal{B}$, 令 $d(\\mathcal{A}, \\mathcal{B}) = \\sum_{A \\in \\mathcal{A}, B \\in \\mathcal{B}} d(A, B)$。令 $T$ 為有最小 $d(T)$ 的其中一個點。我們將證明以下強化命題：\n若 $\\min_{A \\neq T} d(A) \\ge k$, 則存在 $... | Taiwan | 2021 數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
08tc | A plane lying in the 3-dimensional space splits the space into 2 parts. We call one such part (excluding the plane) a **half space**. Let $S$ be a set consisting of 10 points in the space, no 4 among them lying on a same plane. Determine the number of subsets of $S$ which can be obtained as an intersection of $S$ with ... | [
"[260].\nFirst, let us explain some terminologies used in the sequel.\nFor sets $A$ and $B$, we denote by $A \\setminus B$ the set of those elements in $A$ not in $B$.\nFor a finite set $A$, we denote by $|A|$ the number of elements in $A$. For a mapping $F: A \\to B$ and for $b \\in B$, we denote by $F^{-1}(b)$ th... | Japan | Japan Mathematical Olympiad First Round | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 260 | |
04de | For a given positive integer $k$ let $S(k)$ denote the sum of all numbers from the set $\{1, 2, \dots, k\}$ relatively prime to $k$. Let $m$ be a positive integer and $n$ an odd positive integer. Prove that there exist positive integers $x$ and $y$ such that $m$ divides $x$ and $2S(x) = y^n$. | [
"For $k = 1$, $S(k) = 1$. For $k > 1$, note that for a positive integer $a < k$ relatively prime to $k$ the number $k - a$ is also relatively prime to $k$. Hence the summands in the sum $S(k)$ can be paired and each pair has sum $k$. The number of positive integers relatively prime to $k$ (and less than $k$) is $\\... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0h15 | In a trapezoid $ABCD$ with the bases $AD$ and $BC$, a point $F$ is chosen on the side $CD$. Let $E$ be the point of intersection of the lines $AF$ and $BD$. A point $G$ is chosen on the side $AB$ so that $EG \perp AD$. Let $H$ be the point of intersection of the lines $CG$ and $BD$, and let $I$ be the point of intersec... | [
"Without loss of generality, assume that $BC < AD$. Let $S$ be the point of any intersection of the lines $AB$ and $CD$, and $T$ the point of intersection of the lines $AF$ and $DG$ (fig. 34).\n\nFirst we prove that the points $S$, $H$, and $T$ are collinear. To this end, we use Menelaus' theorem for the triangle $... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0k68 | Problem:
In triangle $ABC$, $AB = 13$, $BC = 14$, $CA = 15$. Squares $ABB_1A_2$, $BCC_1B_2$, $CAA_1C_2$ are constructed outside the triangle. Squares $A_1A_2A_3A_4$, $B_1B_2B_3B_4$, $C_1C_2C_3C_4$ are constructed outside the hexagon $A_1A_2B_1B_2C_1C_2$. Squares $A_3B_4B_5A_6$, $B_3C_4C_5B_6$, $C_3A_4A_5C_6$ are const... | [
"Solution:\n\nWe can use complex numbers to find synthetic observations. Let $A = a$, $B = b$, $C = c$. Notice that $B_2$ is a rotation by $-90^{\\circ}$ (counter-clockwise) of $C$ about $B$, and similarly $C_1$ is a rotation by $90^{\\circ}$ of $B$ about $C$. Since rotation by $90^{\\circ}$ corresponds to multipli... | United States | HMMT February 2019 February 16, 2019 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic /... | null | proof and answer | 19444 | |
0f3x | Problem:
The sequence $\{a_n\}$ of positive integers is such that
(1) $a_n \leq n^{3/2}$ for all $n$, and
(2) $m - n$ divides $a_m - a_n$ (for all $m > n$).
Find $a_n$. | [] | Soviet Union | 15th ASU | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | a_n = 1 for all n, or a_n = n for all n | |
07sh | In the usual notation for a triangle $ABC$, prove that
$$
2 \sin \frac{A}{2} \le \frac{a}{\sqrt{bc}},
$$
with equality iff $c = b$. Deduce, or prove otherwise, that
$$
8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le 1,
$$
and
$$
\sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2} \le \frac{a^3 + b^3 +... | [
"The half-angle formula, $2\\sin^2(A/2) = 1 - \\cos A$, and the Cosine-Rule give\n$$\n4 \\sin^2 \\frac{A}{2} = 2(1 - \\cos A) = 2 - \\frac{b^2 + c^2 - a^2}{bc} = \\frac{a^2 - (b-c)^2}{bc} \\le \\frac{a^2}{bc},\n$$\nwith equality iff $b = c$. Because $\\sin(A/2) > 0$, we can take square roots to obtain the first ine... | Ireland | IRL_ABooklet_2020 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0hgz | On a rectangular board $100 \times 300$ two people take turns coloring the uncolored cells. The first one paints yellow, the second one paints blue. The coloring is completed when every cell on the board is colored. A *sequence of* cells is a set of cells in which two consecutive cells share a common side (all cells in... | [
"Let's divide the entire board into vertical dominoes, whose larger sides are parallel to the larger side of the board. Let's consider the second player's strategy: he paints the second square of the domino whose first square was painted by the first player in the previous move. Then it is clear that no sequence of... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0fb0 | Problem:
$A$ and $B$ lie on a circle. $P$ lies on the minor arc $AB$. $Q$ and $R$ (distinct from $P$) also lie on the circle, so that $P$ and $Q$ are equidistant from $A$, and $P$ and $R$ are equidistant from $B$. Show that the intersection of $AR$ and $BQ$ is the reflection of $P$ in $AB$. | [
"Solution:\n\n\n\nLet $AR$ and $BQ$ meet at $X$. Since arcs $QA$ and $AP$ are equal, we have $\\angle ABX = \\angle ABP$. Similarly, $\\angle BAX = \\angle BAP$. Side $AB$ is common, so triangles $ABX$ and $ABP$ are congruent. Hence $X$ is the reflection of $P$."
] | Soviet Union | 1st CIS | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0do5 | Problem:
Одредити све природне бројеве $n$ ($n>1$) који имају следеће својство: ако су $a_{1}, a_{2}, a_{3}, \ldots, a_{k}$ сви природни бројеви мањи од $n$ и узајамно прости са $n$ и важи поредак $a_{1}<a_{2}<a_{3}<\cdots<a_{k}$, онда ниједан од збирова $a_{i}+a_{i+1}$ за $i=1,2, \ldots, k-1$ није дељив са 3. | [
"Solution:\n\nЗа $n \\leqslant 28$ услов задатка је задовољен само за $n \\in\\{2,4,10\\}$. Нека је $n>28$. Приметимо да је низ $a_{i}$ симетричан у односу на $\\frac{n}{2}$. Дакле, $a_{i}+a_{k+1-i}=n$. Ако $2 \\nmid n$, онда је $a_{1}=1, a_{2}=2$ и $3 \\mid a_{1}+a_{2}$. С друге стране, ако $3 \\mid n$, одаберимо ... | Serbia | 13. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 2, 4, 10 | |
02lf | Problem:
Múltiplos de 3 e quadrados perfeitos - Escreve-se em ordem crescente cada múltiplo de $3$ cuja soma com o número $1$ é um quadrado perfeito:
$$
3 ; \quad 15 ; \quad 24 ; 48 ; \quad \ldots
$$
Qual é o múltiplo na posição $2006^{\circ}$ ? | [
"Solution:\n\nChamemos $a$ um número qualquer da lista, então sabemos que:\n- $a$ é múltiplo de $3$\n- $a+1$ é um quadrado: $a+1=k^{2}$, sendo $k$ um número natural.\nAssim $a=k^{2}-1$, e logo\n$$\na=(k-1)(k+1)\n$$\nComo $a$ é divisível por $3$, então ou $k+1$ ou $k-1$ é divisível por $3$. Logo, $k$ não é divisível... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 9060099 | |
0ehh | Problem:
a) Reši enačbi $9^{x} - 4 \cdot 5^{2x-3} = 25^{x-1} + 2 \cdot 3^{2x-3}$ in $\sqrt{x+1+\sqrt{x+1+\sqrt{x+3}}} = \sqrt{x+3}$.
b) Zapiši splošno obliko polinoma z vodilnim koeficientom 2, katerega ničle so enake rešitvam enačb iz a). | [
"Solution:\nPreoblikovanje enačbe $3^{2x} - 2 \\cdot 3^{2x-3} = 5^{2x-2} + 4 \\cdot 5^{2x-3}$\n\nIzpostavitev skupnega faktorja\n\nUreditev do oblike $3^{2x-5} = 5^{2x-5}$\n\nRešitev enačbe $x = \\frac{5}{2}$\n\nReševanje korenske enačbe\n\nUreditev do oblike $x^{2} - 7x + 6 = 0$\n\nRešitev enačbe je $x = 1$, druga... | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a) First equation: x = 5/2; second equation: x = 1. b) p(x) = 2x^2 − 7x + 5. | |
00z8 | Problem:
Let $S$ be a set of integers containing the numbers $0$ and $1996$. Suppose further that any integer root of any non-zero polynomial with coefficients in $S$ also belongs to $S$. Prove that $-2$ belongs to $S$. | [
"Solution:\n\nConsider the polynomial $W(x) = 1996x + 1996$. As $W(-1) = 0$ we conclude that $-1 \\in S$.\n\nNow consider the polynomial $U(x) = -x^{1996} - x^{1995} - \\cdots - x^{2} - x + 1996$. As $U(1) = 0$ we have $1 \\in S$.\n\nFinally, let $T(x) = -x^{10} + x^{9} - x^{8} + x^{7} - x^{6} + x^{3} - x^{2} + 199... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0bao | Let $a$, $b$, $c$ be distinct positive integers.
a) Prove that $a^2b^2 + a^2c^2 + b^2c^2 \ge 9$.
b) If, moreover, $ab + ac + bc + 3 = abc > 0$, show that
$$
(a - 1)(b - 1) + (a - 1)(c - 1) + (b - 1)(c - 1) \ge 6.
$$ | [
"a) At least one of the numbers $a$, $b$, $c$ has modulus at least $2$, whence $a^2b^2 + a^2c^2 + b^2c^2 \\ge 1 \\cdot 1 + 1 \\cdot 4 + 1 \\cdot 4 = 9$.\n\nb) The required inequality can be successively written\n$$\n\\begin{aligned}\n& ab + ac + bc - 2(a + b + c) + 3 \\ge 6 \\\\\n& ab + ac + bc - 3 \\ge 2(a + b + c... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0i6v | Problem:
Let $T_{L} = \sum_{n=1}^{L} \left\lfloor n^{3} / 9 \right\rfloor$ for positive integers $L$. Determine all $L$ for which $T_{L}$ is a square number. | [
"Solution:\nSince $T_{L}$ is square if and only if $9 T_{L}$ is square, we may consider $9 T_{L}$ instead of $T_{L}$.\nIt is well known that $n^{3}$ is congruent to $0, 1$, or $8$ modulo $9$ according as $n$ is congruent to $0, 1$, or $2$ modulo $3$. (Proof: $(3m + k)^{3} = 27m^{3} + 3(9m^{2})k + 3(3m)k^{2} + k^{3}... | United States | HMMT 2002 | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | L in {1, 2} | |
00fb | Let $n$ be an integer such that $n>3$. Suppose that we choose three numbers from the set $\{1,2, \ldots, n\}$. Using each of these three numbers only once and using addition, multiplication, and parenthesis, let us form all possible combinations.
a. Show that if we choose all three numbers greater than $n / 2$, then t... | [
"In both items, the smallest chosen number is at least $2$: in part (a), $n / 2 > 1$ and in part (b), $p$ is a prime. So let $1 < x < y < z$ be the chosen numbers. Then all possible combinations are\n$$\nx + y + z, \\quad x + y z, \\quad x y + z, \\quad y + z x, \\quad (x + y) z, \\quad (z + x) y, \\quad (x + y) z,... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
02iw | Problem:
Sinal de um produto e sinal de um quociente: $a$, $b$, $c$ e $d$ são quatro números não nulos tais que os quocientes $\frac{a}{5}$, $\frac{-b}{7 a}$, $\frac{11}{a b c}$, $\frac{-18}{a b c d}$ são positivos. Determine os sinais de $a$, $b$, $c$ e $d$. | [
"Solution:\n\n- $\\frac{a}{5} > 0 \\Rightarrow a > 0$\n\n- Temos $a > 0 \\Rightarrow 7a > 0$, logo: $\\frac{-b}{7a} > 0 \\Rightarrow -b > 0 \\Rightarrow b < 0$\n\n- $\\frac{11}{a b c} > 0 \\Rightarrow a b c > 0$\n\n- $\\frac{-18}{a b c d} > 0 \\Rightarrow a b c d < 0$, como $a b c > 0$ segue que $d < 0$."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | a > 0, b < 0, c < 0, d < 0 | |
0krm | Problem:
You start with a single piece of chalk of length $1$. Every second, you choose a piece of chalk that you have uniformly at random and break it in half. You continue this until you have $8$ pieces of chalk. What is the probability that they all have length $\frac{1}{8}$? | [
"Solution:\n\nThere are $7!$ total ways to break the chalks. How many of these result in all having length $\\frac{1}{8}$? The first move gives you no choice. Then, among the remaining $6$ moves, you must apply $3$ breaks on the left side and $3$ breaks on the right side, so there are $\\binom{6}{3} = 20$ ways to o... | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1/63 | |
0csw | Let $ABCD$ be an inscribed quadrilateral for which the rays $AB$ and $DC$ meet at $K$. It happened that the points $B$, $D$, and the midpoints of the segments $AC$ and $KC$ are concyclic. Find all possible values the angle $ADC$ may get. | [
"Обозначим через $N$ и $M$ середины отрезков $KC$ и $AC$ соответственно. Тогда $MN$ — средняя линия в треугольнике $AKC$, поэтому $\\angle BAC = \\angle NMC$. Кроме того, $\\angle BAC = \\angle BDC$, так как четырехугольник $ABCD$ — вписанный.\n\nПусть точки $M$ и $N$ лежат с одной стороны от прямой $BD$. Тогда $M$... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 90° | |
0b6g | Let $f : [0, 1] \to \mathbb{R}$ be a continuous function and $n \ge 3$ be an integer. Show that one can find $n$ numbers $a_1, a_2, \dots, a_n$ in the interval $[0, 1]$, in arithmetic progression, such that
$$
\int_{0}^{1} f(x) \, dx = \frac{1}{n} \sum_{k=1}^{n} f(a_k).
$$ | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | English | proof only | null | |
0g5b | 設 $k$ 為一給定實數。試找出所有從實數映至實數的函數 $f(x)$ 滿足對任意實數 $x, y$, 均有
$$
f(x) + (f(y))^2 = k f(x + y^2).
$$ | [
"若 $k \\neq 1$, 則將 $y = 0$ 代入原式得到\n$$\n(f(0))^2 = (k - 1)f(x).\n$$\n因此 $f(x)$ 為常數函數, 代入原式後可解出 $f(x) = 0$ 或 $f(x) = \\frac{1}{k-1}$。\n\n若 $k = 1$, 則將 $x = 0$ 代入原式得到\n$$\nf(y)^2 = f(y^2).\n$$\n藉由上式我們可以知道對任意 $y > 0$ 均有 $f(y) > 0$ 並且 $f(1) = 0$ 或 $1 > f(x)$。將上式代回原題中：\n$$\n\\begin{align*}\n& f(x) + (f(y))^2 = f(x + y^2)... | Taiwan | 二〇一一數學奧林匹亞競賽第一階段選訓營,模擬競賽(一) | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | If k ≠ 1: f(x) ≡ 0 or f(x) ≡ k − 1. If k = 1: f(x) ≡ 0 or f(x) ≡ x. | |
0dny | Problem:
Нека је $k$ кружница описана око $\triangle A B C$, а $k_{a}$ приписана кружница наспрам темена $A$. Две заједничке тангенте кружница $k$ и $k_{a}$ секу праву $B C$ у тачкама $P$ и $Q$. Доказати да важи $\varangle P A B=\varangle Q A C$.
(Дуиан Ђукић) | [
"Solution:\n\nНека унутрашња и спољна симетрала угла $B A C$ секу праву $B C$ редом у тачкама $D$ и (можда бесконачној) $D_{1}$. Заједничке тангенте се секу у центру $T$ позитивне хомотетије $\\mathscr{H}$ која слика приписани круг $\\omega_{a}$ у описани круг $\\Omega$. Ако је $T$ бесконачна тачка, $\\mathscr{H}$ ... | Serbia | 11. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analyt... | null | proof only | null | |
0jw9 | Problem:
A positive integer $k$ is called powerful if there are distinct positive integers $p, q, r, s, t$ such that $p^{2}$, $q^{3}$, $r^{5}$, $s^{7}$, $t^{11}$ all divide $k$. Find the smallest powerful integer. | [
"Solution:\nFirst of all, $1024$ is powerful because it can be divided evenly by $16^{2} = 256$, $8^{3} = 512$, $4^{5} = 1024$, $2^{7} = 128$, and $1^{11} = 1$.\n\nNow we show that $1024$ is the smallest powerful number. Since $s \\neq t$, at least one of them is at least $2$. If $t \\geq 2$ or $s \\geq 3$, then we... | United States | HMMT November 2017 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1024 | |
0gmx | Let $E$ and $F$ be two points on the side $CD$ of a convex quadrilateral $ABCD$ satisfying $0 < DE = FC < CD$. Let $K$ be the second point of intersection of the circumcircles of the triangles $ADE$ and $ACF$, and let $L$ be the second point intersection of the circumcircles of the triangles $BDE$ and $BCF$. Show that ... | [
"Let $\\{M\\} = DC \\cap AK$ and $\\{N\\} = DC \\cap BL$. Considering the powers of the point $M$ with respect to the circles $ADE$ and $AFC$ we obtain\n$$\nME \\cdot MD = MK \\cdot MA = MF \\cdot MC.\n$$\nSince $DE = FC$, we conclude that $M$ is the midpoint of the line segment $CD$. Similarly,\n$$\nNE \\cdot ND =... | Turkey | 14th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0gb8 | 令 $n, k$ 為正整數且 $n > k$。有甲、乙兩人:
1. 甲先私自在紙條上寫下一個 $n$ 位數的 01 序列, 並在黑板上寫下所有跟這個 01 序列恰有 $k$ 個位數不同的所有長度為 $n$ 的 01 序列。舉例來說, 如果 $n=3, k=1$, 且紙條上的 01 序列 101, 則甲必須在黑板上寫下 001, 111 和 100.
2. 接著, 乙看著黑板上的所有序列, 試圖猜測紙條上的序列是什麼。他每次可以猜一個 $n$ 位數的 01 序列, 而甲必須誠實回答他是否猜對了。
對於每組 $(n, k)$, 試求最小的正整數 $m$, 使得乙存在一個猜測策略, 能保證在 $m$ 次猜測內猜到正解。 | [
"令紙條上的序列為 $X$.\n先考慮 $n \\neq 2k$. 若 $X$ 的首項為 1, 則在黑板上將有 $C(n-1, k)$ 條首項為 1, $C(n-1, k-1)$ 條首項為 0, 注意到 $C(n-1, k) \\neq C(n-1, k-1)$. 由此可知, 乙只需計算首項為 1 與為 0 的序列數量, 便可確知 $X$ 的首項為何。依據同樣的方法, 乙可以確知 $X$ 的每一項為何。故乙第一次便可猜中。\n\n現在考慮 $n = 2k$. 當 $k = 1$ 時, 易知需猜兩次。當 $k \\ge 2$ 時, 注意到若我們將紙條上的 $X$ 的每一項都換掉 (0 換成 1, 1 換成 0), 則黑板上仍會... | Taiwan | 二〇一七數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | m = 1 if n ≠ 2k, and m = 2 if n = 2k. | |
0jqx | Problem:
Let $n$ be the smallest positive integer with exactly $2015$ positive factors. What is the sum of the (not necessarily distinct) prime factors of $n$? For example, the sum of the prime factors of $72$ is $2+2+2+3+3=14$. | [
"Solution:\nAnswer: $116$\n\nNote that $2015 = 5 \\times 13 \\times 31$ and that $N = 2^{30} \\cdot 3^{12} \\cdot 5^{4}$ has exactly $2015$ positive factors. We claim this is the smallest such integer. Note that $N < 2^{66}$.\n\nIf $n$ has $3$ distinct prime factors, it must be of the form $p^{30} q^{12} r^{4}$ for... | United States | HMMT November 2015 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | 116 | |
0eob | Points $E$ and $F$ lie inside a square $ABCD$ such that the two triangles $ABF$ and $BCE$ are equilateral. Show that $DEF$ is an equilateral triangle. | [
"We have $\\angle FAD = \\angle BAD - \\angle BAF = 90^\\circ - 60^\\circ = 30^\\circ$. Since $AF = AB = AD$, triangle $AFD$ is isosceles, which means that $\\angle ADF = \\angle AFD = \\frac{180^\\circ - \\angle FAD}{2} = 75^\\circ$ and $\\angle CDF = \\angle CDA - \\angle ADF = 90^\\circ - 75^\\circ = 15^\\circ$.... | South Africa | South African Mathematics Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0k92 | Problem:
Show that $x^{2}+y^{2}=z^{5}+z$ has infinitely many relatively prime integer solutions. | [
"Solution:\n\nWe use that there are infinitely many primes $p=4k+1$, and that all these primes can be written as a sum of two squares. Now, let $z=p$ be such a prime. Then $p=a^{2}+b^{2}$ for some integers $a$ and $b$, so\n$$\nz^{5}+z = p(p^{4}+1) = (a^{2}+b^{2})(p^{4}+1) = (a p^{2}+b)^{2} + (b p^{2}-a)^{2}.\n$$\nT... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Other"
] | null | proof only | null | |
02jg | Problem:
Em um quente dia de verão, 64 crianças comeram, cada uma, um sorvete pela manhã e outro à tarde. Os sorvetes eram de 4 sabores: abacaxi, banana, chocolate e doce de leite. A tabela abaixo mostra quantas crianças consumiram um destes sabores pela manhã e outro à tarde; por exemplo, o número 7 na tabela indica ... | [
"Solution:\n\nVamos primeiro analisar a informação contida na diagonal da tabela indicada pelos números dentro dos quadradinhos.\n\n| | TARDE | | | | |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| | | Ab... | Brazil | Brazilian Mathematical Olympiad | [
"Math Word Problems"
] | null | MCQ | C | |
0kfv | For a finite simple graph $G$, we define $G'$ to be the graph on the same vertex set as $G$, where for any two vertices $u \neq v$, the pair $\{u, v\}$ is an edge of $G'$ if and only if $u$ and $v$ have a common neighbor in $G$. Prove that if $G$ is a finite simple graph which is isomorphic to $(G')'$, then $G$ is also... | [
"**Claim** — The graph $G'$ has at least as many triangles as $G$, and has strictly more if $G$ has any fatal vertices.\n*Proof*. Obviously any triangle in $G$ persists in $G'$. Moreover, suppose $v$ is a fatal vertex of $G$. Then the neighbors of $G$ will form a clique in $G'$ which was not there already, so there... | United States | USA IMO TST | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0723 | Problem:
Show that for every real number $a$ the equation
$$
8 x^{4}-16 x^{3}+16 x^{2}-8 x+a=0
$$
has at least one non-real root and find the sum of all the non-real roots of the equation. | [
"Solution:\nSubstituting $x = y + \\frac{1}{2}$ in the equation, we obtain the equation in $y$:\n$$\n8 y^{4} + 4 y^{2} + a - \\frac{3}{2} = 0\n$$\nUsing the transformation $z = y^{2}$, we get a quadratic equation in $z$:\n$$\n8 z^{2} + 4 z + a - \\frac{3}{2} = 0\n$$\nThe discriminant of this equation is $32(2 - a)$... | India | INMO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Sum of non-real roots = 1 if a ≤ 3/2; and = 2 if a > 3/2. | |
0gt4 | $ABCD$ is a convex quadrilateral with $\angle ABC = \angle BCD$. The lines $AD$ and $BC$ intersect at $P$, and the line passing through $P$ and parallel to $AB$ intersects the line $BD$ at $T$. Show that $\angle ACB = \angle PCT$. | [
"Let the line passing through $A$ and parallel to $BC$ intersect the line $CD$ at $E$. Since\n$$\n\\frac{EA}{CP} = \\frac{AD}{PD} = \\frac{AB}{PT} \\quad \\text{and} \\quad \\angle EAB = \\angle CPT,\n$$\none has $EAB \\sim CPT$, hence $\\angle PCT = \\angle AEB$. On the other hand, one has $\\angle ABC = \\angle B... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof only | null | |
0b19 | Problem:
Find the smallest positive integer that is $20\%$ larger than one integer and $19\%$ smaller than another. | [
"Solution:\nSuppose $N$ is our integer. Then we have $N = \\frac{6}{5} x = \\frac{81}{100} y$ for some integers $x$ and $y$. In particular, $x$ is divisible by $5$ and $y$ is divisible by $100$. By multiplying by $100$, we have $120 x = 81 y$, and the smallest integers that satisfy this as well as the previous cond... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 162 | |
05mf | Problem:
Au club théâtre d'un lycée, on a formé 14 groupes de 4 élèves afin de travailler les scènes d'une pièce. Deux groupes différents ont toujours un et un seul élève en commun.
a) Prouver qu'il existe un élève qui appartient à au moins 5 groupes.
b) Chaque élève du club théâtre est membre d'au moins un groupe. Co... | [
"Solution:\n\na) Soit $G$ un des groupes. Chacun des 13 autres groupes a un élève en commun avec $G$. Or $13 = 3 \\times 4 + 1$ donc, d'après le principe des tiroirs, un des quatre élèves de $G$, disons $x$, appartient à au moins 4 des treize autres groupes. Avec $G$, cela signifie que $x$ appartient à au moins 5 g... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 43 | |
082v | Problem:
Sono dati tre interi positivi $a, b, c$. Posto $x = a b$, $y = a c$, $z = b c$, quale delle seguenti affermazioni è vera?
(A) Se $x, y, z$ sono dei quadrati, allora $a, b, c$ sono dei quadrati
(B) se $x, y, z$ sono pari, allora $a, b, c$ sono pari
(C) se $x, y, z$ sono dei cubi, allora $a, b, c$ sono dei cub... | [
"Solution:\n\nLa risposta è (C). Dato un primo $p$ indichiamo con $x_{p}, y_{p}$ e $z_{p}$ l'esponente (eventualmente nullo) con cui $p$ compare nella fattorizzazione rispettivamente di $x$, di $y$ e di $z$; in modo analogo fissiamo $a_{p}, b_{p}$ e $c_{p}$. L'ipotesi di (C) è equivalente a dire che $x_{p}, y_{p}, ... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | MCQ | C | |
0ifn | Problem:
Let $x$, $y$, and $z$ be distinct real numbers that sum to $0$. Find the maximum possible value of
$$
\frac{x y + y z + z x}{x^{2} + y^{2} + z^{2}}.
$$ | [
"Solution:\n$-1/2$\n\nNote that $0 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2 x y + 2 y z + 2 z x$. Rearranging, we get that $x y + y z + z x = -\\frac{1}{2}(x^2 + y^2 + z^2)$, so that in fact the quantity is always equal to $-1/2$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | -1/2 | |
0cy8 | Let $f: \mathbb{N} \rightarrow [0, \infty)$ be a function satisfying the following conditions:
a) $f(4) = 2$;
b) $\frac{1}{f(0) + f(1)} + \frac{1}{f(1) + f(2)} + \ldots + \frac{1}{f(n) + f(n+1)} = f(n+1)$, for all integers $n \geq 0$.
Find $f(n)$ in closed form. | [
"For $k \\geq 0$ we get\n$$\nf(k+2) - f(k+1) = \\frac{1}{f(k+1) + f(k+2)}\n$$\nhence $f^{2}(k+2) = f^{2}(k+1) + 1$. It follows\n$$\n4 = f^{2}(4) = f^{2}(3) + 1\n$$\nhence $f(3) = \\sqrt{3}$. Also, $3 = f^{2}(3) = f^{2}(2) + 1$ implies\n$$\nf(2) = \\sqrt{2} \\quad \\text{and} \\quad 2 = f^{2}(2) = f^{2}(1) + 1\n$$\n... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | f(n) = sqrt(n) | |
0ffs | Problem:
Sea $d$ la suma de las longitudes de todas las diagonales de un polígono convexo plano de $n$ vértices $(n>3)$, y sea $p$ su perímetro. Demostrar que
$$
n-3<\frac{2d}{p}<\left[\frac{n}{2}\right]\left[\frac{n+1}{2}\right]-2
$$
siendo $[x]$ la parte entera de $x$. | [] | Spain | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0ah4 | Solve the equation $p^{2q} + q^{2p} = r$ in the set of prime numbers. | [
"It is clear that $r > 2$, from where $r$ must be an odd prime number. One of the numbers $p$ or $q$ must be $2$, and the other must be an odd prime number. Without loss of generality, let $q = 2$ and $p$ be odd. But then the equation is of the form $p^4 + 2^{2p} = r$, i.e. $p^4 + 4 \\cdot 2^{4k} = r$ where $p = 2k... | North Macedonia | Macedonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | no solution | |
0jbq | Problem:
Purineqa is making a pizza for Arno. There are five toppings that she can put on the pizza. However, Arno is very picky and only likes some subset of the five toppings. Purineqa makes five pizzas, each with some subset of the five toppings. For each pizza, Arno states (with either a "yes" or a "no") if the pi... | [
"Solution:\n\nWe claim the only way for Purineqa to deduce Arno's preferences is for each pizza to contain exactly one topping, with no topping be repeated. It is obvious that she can deduce the toppings in this case.\n\nWe now claim that this is not possible with any other combination. Suppose that Arno tells Puri... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Exactly the five pizzas each having a single distinct topping; that is, one pizza with each topping and no other combinations. | |
0hsh | Problem:
Solve $x = \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}$ for $x$. | [
"Solution:\n\n$\\frac{1 + \\sqrt{5}}{2}$."
] | United States | null | [
"Algebra > Intermediate Algebra > Other",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | (1+sqrt(5))/2 | |
04t0 | We are given a positive integer $n > 2$. Find the greatest of all the numbers $d$, satisfying the following condition: For any set of $n$ integers one can choose its three different subsets so that the sum of elements each of which is an integer multiple of $d$. (The selected subsets need not be disjoint.) (Jaromír Šim... | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | n − 1 | |
0b3t | Problem:
For a real number $x$, let $\lfloor x\rfloor$ denote the greatest integer not exceeding $x$. Consider the function
$$
f(x, y)=\sqrt{M(M+1)}(|x-m|+|y-m|)
$$
where $M=\max (\lfloor x\rfloor,\lfloor y\rfloor)$ and $m=\min (\lfloor x\rfloor,\lfloor y\rfloor)$. The set of all real numbers $(x, y)$ such that $2 \l... | [
"Solution:\n\nFix $m \\geq 2$. First note that $x, y \\geq m$, and that $M \\geq 2$, implying $\\sqrt{M(M+1)} \\geq 2$. Thus, $f(x, y) \\leq 2$ necessarily implies $|x-m|+|y-m| \\leq 1$, and so $|x-m|,|y-m| \\leq 1$. This implies $x \\in [m-1, m+1]$, and so $x \\in [m, m+1]$. Similarly, $y \\in [m, m+1]$. Now if $x... | Philippines | 24th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities... | null | proof and answer | 2021 | |
0h1k | Sequence $(a_n)$ is defined as follows:
$$
a_1 = 1,\ a_2 = 2,\ a_{n+2} = (n+1)(a_n + a_{n+1})
$$
for each natural $n$. How many zeros does $a_{2011}$ end with? | [
"We first prove that $a_n = n!$ by PMI:\n$$\na_{n+2} = (n+1)(n! + (n+1)!) = n!(n+1)(n+2) = (n+2)!\n$$\nOne can now find the number of zeros using the well-known formula:\n$$\n\\left[ \\frac{2011}{5} \\right] + \\left[ \\frac{2011}{5^2} \\right] + \\left[ \\frac{2011}{5^3} \\right] + \\dots = 402 + 80 + 16 + 3 = 501... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 501 | |
0ios | There is a set of $n$ coins with distinct integer weights $w_1, w_2, \dots, w_n$. It is known that if any coin with weight $w_k$, $1 \le k \le n$, is removed from the set, the remaining coins can be split into two groups of the same weight. (The number of coins in the two groups can be different.) Find all $n$ for whic... | [
"The only such $n$ are $1$ and odd $n$ at least $7$. We divide into cases: $n$ even, $n = 1$, $n = 3$, $n = 5$, and $n \\ge 7$ odd.\n\n**Case 1:** $n$ even. Suppose for contradiction that such a set exists, and choose one with minimal total weight. Let $s$ be the sum of the weights, and note that\n$$\ns - w_1 \\equ... | United States | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | n = 1 or n is odd and at least 7 | |
0b2i | Problem:
In quadrilateral $ABCD$, $\angle CBA = 90^\circ$, $\angle BAD = 45^\circ$, and $\angle ADC = 105^\circ$. Suppose that $BC = 1 + \sqrt{2}$ and $AD = 2 + \sqrt{6}$. What is the length of $AB$?
(a) $2\sqrt{3}$
(b) $2 + \sqrt{3}$
(c) $3 + \sqrt{2}$
(d) $3 + \sqrt{3}$ | [] | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | a | |
02fp | Let $p(x)$ be the polynomial $x^3 + 14x^2 - 2x + 1$. Let $p^{(n)}(x)$ denote $p(p^{(n-1)}(x))$. Show that there is an integer $N$ such that $p^{(N)}(x) - x$ is divisible by $101$ for all integers $x$. | [
"We have $p(x) - p(y) = (x^3 + 14x^2 - 2x + 1) - (y^3 + 14y^2 - 2y + 1) = (x - y)(x^2 + xy + y^2 + 14x + 14y - 2)$. Since $101$ is prime, $p(x) \\equiv p(y) \\pmod{101} \\Leftrightarrow x \\equiv y \\pmod{101}$ or $x^2 + xy + y^2 + 14x + 14y - 2 \\equiv 0 \\pmod{101}$.\n\nCompleting squares, we have\n$$\n\\begin{al... | Brazil | XVIII OBM | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
07fe | Consider a triangle $\triangle ABC$ with circumcenter $O$ and incenter $I$. The incircle touches sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. Let $K$ be a point such that $KF$ is tangent to circumcircle of $\triangle BFD$ and $KE$ is tangent to circumcircle of $\triangle CED$. Prove that $BC$, $OI$ and ... | [
"Note that\n$$\n\\frac{XB}{XC} = \\frac{\\sin \\angle XOB}{\\sin \\angle XOC} = \\frac{\\sin \\angle IOB}{\\sin \\angle IOC}.\n$$\nWhich is equivalent to\n$$\n\\frac{\\sin\\left(\\frac{\\angle C}{2}\\right)}{\\sin\\left(\\frac{|\\angle B - \\angle A|}{2}\\right)} \\cdot \\frac{\\sin\\left(\\frac{|\\angle B - \\angl... | Iran | 37th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Tria... | English | proof only | null | |
0iq1 | Problem:
Consider a $7 \times 7$ chessboard that starts out with all the squares white. We start painting squares black, one at a time, according to the rule that after painting the first square, each newly painted square must be adjacent along a side to only the square just previously painted. The final figure painte... | [
"Solution:\n\na.\n\n\nb.\n\nFigure 1: Solutions to (a), (b) and (c)"
] | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | null | proof only | null | |
036u | Problem:
Find all sequences of positive integers $\{a_{n}\}_{n=1}^{\infty}$, such that $a_{4}=4$ and the identity
$$
\frac{1}{a_{1} a_{2} a_{3}}+\frac{1}{a_{2} a_{3} a_{4}}+\cdots+\frac{1}{a_{n} a_{n+1} a_{n+2}}=\frac{(n+3) a_{n}}{4 a_{n+1} a_{n+2}}
$$
holds true for every positive integer $n \geq 2$. | [
"Solution:\nWe rewrite the recurrence relation as\n$$\n\\frac{(n+2) a_{n-1}}{4 a_{n} a_{n+1}}+\\frac{1}{a_{n} a_{n+1} a_{n+2}}=\\frac{(n+3) a_{n}}{4 a_{n+1} a_{n+2}} \\Longleftrightarrow (n+2) a_{n+2}=\\frac{(n+3) a_{n}^{2}-4}{a_{n-1}}\n$$\nfor $n \\geq 3$. Setting $n=2$ in the initial relation we obtain $4\\left(a... | Bulgaria | Team selection test for 47. IMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | a_n = n for all n | |
0ebf | Problem:
Marko se je odločil, da bo privarčeval 600 evrov, ki jih bo porabil za nakup novega računalnika. Po štirih mesecih varčevanja je ugotovil, da so njegovi mesečni prihranki glede na vrstni red mesecev varčevanja v razmerju $4: 3: 2: 2$. Za nakup novega računalnika mu tako manjka le še polovica najmanjšega meseč... | [] | Slovenia | 15. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 200 | |
0fab | Problem:
$X$ is a set with $100$ members. What is the smallest number of subsets of $X$ such that every pair of elements belongs to at least one subset and no subset has more than $50$ members? What is the smallest number if we also require that the union of any two subsets has at most $80$ members? | [] | Soviet Union | 25th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | First case: 6; with the union constraint: 6 | |
0e1m | Jure has drawn a regular enneagon (a 9-sided polygon). He wants to arrange the numbers $1$ to $9$ at its vertices so that the sum of the numbers at any three consecutive vertices does not exceed some positive integer $n$. What is the least possible $n$ with which he can succeed? | [
"We will show that $n = 16$.\n\nAs one can see in the figure, it is possible to arrange the numbers $1$ to $9$ at the vertices so that the sum of any three consecutive numbers is at most $16$.\n\nLet us show that for $n < 16$ this cannot be achieved. Assume, to the contrary, that it is possible to do so. If we add ... | Slovenia | National Math Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 16 | |
0e19 | Let $ABC$ be an acute triangle such that $|AB| > |BC| > |AC|$. Let $D$ be a point different from $C$ on the segment $BC$, such that $|AC| = |AD|$. Let $H$ denote the orthocentre of the triangle $ABC$, and let $A_1, B_1$ be the feet of the altitudes from $A$ and $B$, respectively. Denote the intersection of the lines $A... | [
"Let $C_1$ be the foot of the altitude to the side $AB$. The triangle $CAD$ is isosceles since $|AC| = |AD|$.\n\n\n\nThe line $AA_1$ is the altitude in this isosceles triangle, so $\\angle CDH = \\angle HCD = \\angle C_1CB = \\frac{\\pi}{2} - \\angle CBA$. This implies that\n$$\n\\begin{ali... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03i6 | Problem:
Determine the largest real number $z$ such that
$$
\begin{gathered}
x + y + z = 5 \\
x y + y z + x z = 3
\end{gathered}
$$
and $x$, $y$ are also real. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 13/3 | |
06ge | Let $a$, $b$, $c$ and $d$ be positive real numbers, all larger than or equal to $1$. Prove that
a. $abcd(a+b+c+d-4)^4 \ge (a-1)(b-1)(c-1)(d-1)(a+b+c+d)^4$,
b. $\dfrac{d}{a+b+c} + \dfrac{a}{b+c+d} + \dfrac{b}{c+d+a} + \dfrac{c}{d+a+b} \ge \dfrac{4}{3}$. | [
"a.\nWLOG assume $a \\ge b \\ge c \\ge d$. Then $\\dfrac{a-1}{a} \\ge \\dfrac{b-1}{b} \\ge \\dfrac{c-1}{c} \\ge \\dfrac{d-1}{d}$. By Chebyshev's inequality, we have\n$$\n\\sum_{\\text{cyc}} \\left( a \\cdot \\frac{a-1}{a} \\right) \\ge \\frac{1}{4} \\left( \\sum_{\\text{cyc}} a \\right) \\left( \\sum_{\\text{cyc}} ... | Hong Kong | IMO HK TST | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
07p2 | Find with proof, all triples of non-negative integers $(x, y, n)$ satisfying
$$
(x^4 + 1)^3 + (y^4 + 1)^3 = 2014^n.
$$ | [
"Consider the given equation modulo $13$. The squares modulo $13$ are $0$, $\\pm 1$, $\\pm 3$, $\\pm 4$, so the fourth powers modulo $13$ are $0$, $1$, $3$, $9$. Therefore $x^4 + 1$ must be one of $1$, $2$, $4$, $10$ modulo $13$. For $z$ congruent to $1$, $2$, $4$, $10$ modulo $13$, the value of $z^3$ can only be $... | Ireland | Irska 2014 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | No solutions | |
0l76 | Problem:
Determine, with proof, whether a square can be dissected into finitely many (not necessarily congruent) triangles, each of which has interior angles $30^{\circ}$, $75^{\circ}$, and $75^{\circ}$. | [
"Solution:\nAssume for sake of contradiction that such a dissection exists. It has exactly half as many $30^{\\circ}$ angles as $75^{\\circ}$ angles.\n\nAround any intersection point except the square's vertices, the only angles that can appear are $30^{\\circ}$, $75^{\\circ}$, and $180^{\\circ}$. The only combinat... | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | No; such a dissection does not exist. | |
0kax | We say a function $f: \mathbb{Z}_{\ge 0} \times \mathbb{Z}_{\ge 0} \to \mathbb{Z}$ is *great* if for any nonnegative integers $m$ and $n$,
$$
f(m+1, n+1)f(m, n) - f(m+1, n)f(m, n+1) = 1.
$$
If $A = (a_0, a_1, \dots)$ and $B = (b_0, b_1, \dots)$ are two sequences of integers, we write $A \sim B$ if there exists a great ... | [
"**First solution (Nikolai Beluhov)** Let $k = a_0 = b_0 = c_0 = d_0$. We let $f, g, h$ be great functions for $(A, B), (B, C), (C, D)$ and write the following infinite array:\n$$\n\\left[ \\begin{array}{ccccccccc} \\vdots & \\vdots & b_3 & \\vdots & \\vdots & & & \\\\ \\cdots & g(2,2) & g(2,1) & b_2 & f(1,2) & f(2... | United States | USA IMO TST | [
"Algebra > Linear Algebra > Determinants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
080z | Problem:
Qual è il numero minimo di carte che bisogna pescare da un ordinario mazzo di 52 per avere almeno il $50\%$ di probabilità di estrarre una o più carte di cuori?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6 . | [
"Solution:\n\nLa risposta è (B). Infatti la probabilità di non aver estratto una carta di cuori dopo aver pescato due carte è\n$$\n\\frac{39}{52} \\cdot \\frac{38}{51} = \\frac{19}{34} > \\frac{1}{2}\n$$\nmentre dopo 3 carte è\n$$\n\\frac{39}{52} \\cdot \\frac{38}{51} \\cdot \\frac{37}{50} < \\left(\\frac{3}{4}\\ri... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | B | |
05jb | Problem:
Prouver que, pour tout réel $a \geqslant 0$, on a
$$
a^{3}+2 \geqslant a^{2}+2 \sqrt{a}
$$ | [
"Solution:\nPour tout réel $a \\geqslant 0$, on a\n$$\n\\begin{aligned}\na^{3}-a^{2}-2 \\sqrt{a}+2 & =a^{2}(a-1)-2(\\sqrt{a}-1) \\\\\n& =a^{2}(\\sqrt{a}-1)(\\sqrt{a}+1)-2(\\sqrt{a}-1) \\\\\n& =(\\sqrt{a}-1)\\left(a^{2}(\\sqrt{a}+1)-2\\right) \\quad (1)\n\\end{aligned}\n$$\nOr:\n- Si $a \\geqslant 1$ alors $\\sqrt{a... | France | Olympiades Françaises de Mathématiques, Envoi Numéro 3 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0epe | If $t$ toffees cost $c$ cents, the number of toffees that can be bought for $r$ rands is
(A) $\frac{100rc}{t}$ (B) $\frac{100rt}{c}$ (C) $\frac{100r}{ct}$ (D) $\frac{rt}{100c}$ (E) $\frac{100c}{rt}$ | [
"Since $t$ toffees cost $c$ cents, each toffee costs $\\frac{c}{t}$ cents.\n\n$r$ rands equals $100r$ cents.\n\nThe number of toffees that can be bought for $100r$ cents is thus $100r \\div \\frac{c}{t} = \\frac{100rt}{c}$"
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | MCQ | B | |
0f2s | Problem:
Show that there is no positive integer $n$ for which $1000^{n} - 1$ divides $1978^{n} - 1$. | [] | Soviet Union | ASU | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0knq | A finite set $S$ of positive integers has the property that, for each $s \in S$, and each positive integer divisor $d$ of $s$, there exists a unique element $t \in S$ satisfying $\gcd(s, t) = d$. (The elements $s$ and $t$ could be equal.)
Given this information, find all possible values for the number of elements of $S... | [] | United States | USAMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 2^n for any nonnegative integer n | |
0762 | Problem:
In an acute triangle $ABC$, $O$ is the circumcenter, $H$ is the orthocenter and $G$ is the centroid. Let $OD$ be perpendicular to $BC$ and $HE$ be perpendicular to $CA$, with $D$ on $BC$ and $E$ on $CA$. Let $F$ be the midpoint of $AB$. Suppose the areas of triangles $ODC$, $HEA$ and $GFB$ are equal. Find all... | [
"Solution:\n\nLet $R$ be the circumradius of $\\triangle ABC$ and $\\Delta$ its area. We have $OD = R \\cos A$ and $DC = \\frac{a}{2}$, so\n$$\n[ODC] = \\frac{1}{2} \\cdot OD \\cdot DC = \\frac{1}{2} \\cdot R \\cos A \\cdot R \\sin A = \\frac{1}{2} R^2 \\sin A \\cos A\n$$\nAgain $HE = 2R \\cos C \\cos A$ and $EA = ... | India | INMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | π/3 or π/4 | |
0kt2 | Problem:
Let $(x_{1}, y_{1}),(x_{2}, y_{2}),(x_{3}, y_{3}),(x_{4}, y_{4})$, and $(x_{5}, y_{5})$ be the vertices of a regular pentagon centered at $(0,0)$. Compute the product of all positive integers $k$ such that the equality
$$
x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k}+x_{5}^{k}=y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}... | [
"Solution:\nWithout loss of generality let the vertices of the pentagon lie on the unit circle. Then, if $f(\\theta)=\\cos (\\theta)^{k}$ and $g(\\theta)=\\sum_{j=0}^{4} f(\\theta+2 j \\pi / 5)$, the condition becomes $g(\\theta)=g(\\pi / 2-\\theta)$, or $g(\\theta)=g(\\theta+\\pi / 2)$, since $g$ is an odd functio... | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 1152 | |
032z | Problem:
A table with $m$ rows and $n$ columns is given. At any move one chooses some empty cells such that any two of them lie in different rows and columns, puts a white piece in any of these cells and then puts a black piece in the cells whose lines and columns contain white pieces. The game is over if it is not pos... | [
"Solution:\nWe put $m+n-1$ white pieces by $m+n-1$ moves consecutively in the cells $(1,1), (2,1), \\ldots, (m,1), (1,2), (1,3), \\ldots, (1,n)$.\n\nWe shall call a closed chain of cells such that at least two numbers of rows and columns change alternatively a zig-zag cycle.\n\nNote that there is at least one black... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | m + n - 1 | |
08v1 | A hexagon $ABCDEF$ is inscribed in a circle. If the sides $AB$ and $DE$ are parallel and so are the sides $BC$ and $EF$, prove that the sides $CD$ and $FA$ are also parallel. | [
"Since the sides $BC$ and $FA$ are not parallel, the lines $BC$ and $FA$ intersect. Call the point of their intersection $X$. Similarly, let $Y$ and $Z$ be the points of intersections of the lines $BC$, $DE$ and of the lines $DE$, $FA$, respectively.\nSince the quadrilateral $ABCD$ is inscribed in the circle, we ha... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
070l | Problem:
There are $98$ points on a circle. Two players play alternately as follows. Each player joins two points which are not already joined. The game ends when every point has been joined to at least one other. The winner is the last player to play. Does the first or second player have a winning strategy? | [
"Solution:\n\nAssume there are $n$ points. The first to play so that $n-2$ points each have at least one segment loses, because the other player simply joins the last two points and the game ends. But there are $N = (n-3)(n-4)/2$ possible plays amongst the first $n-3$ points to get a segment. For $n = 1$ or $2 \\bm... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | First player | |
06pd | A unit square is dissected into $n>1$ rectangles such that their sides are parallel to the sides of the square. Any line, parallel to a side of the square and intersecting its interior, also intersects the interior of some rectangle. Prove that in this dissection, there exists a rectangle having no point on the boundar... | [
"Call the directions of the sides of the square horizontal and vertical. A horizontal or vertical line, which intersects the interior of the square but does not intersect the interior of any rectangle, will be called a splitting line. A rectangle having no point on the boundary of the square will be called an inter... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof only | null | |
02ej | Let $Z$ be the set of integers. $f: Z \rightarrow Z$ is defined by $f(n) = n - 10$ for $n > 100$ and $f(n) = f(f(n+11))$ for $n = 100$. Find the set of possible values of $f$. | [
"We show that $f(n) = 91$ for $n \\le 100$. For $n = 100, 99, \\dots, 90$, we have $f(n) = f(f(n+11)) = f(n+11-10) = f(n+1)$. But $f(101) = 91$, so $f(n) = 91$ for $n = 100, 99, \\dots, 90$. Now for $n = 89, 88, \\dots, 1, 0, -1, \\dots$ we use induction. We have $f(n) = f(f(n+11)) = f(91)$ which, by induction, is ... | Brazil | XI OBM | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | all integers greater than or equal to 91 | |
0eii | Problem:
Nad stranicami trikotnika $ABC$ konstruiramo pozitivno orientirane rombe $BAA_{1}B_{2}$, $CBB_{1}C_{2}$ in $ACC_{1}A_{2}$. Dokaži, da lahko iz daljic $A_{1}A_{2}$, $B_{1}B_{2}$ in $C_{1}C_{2}$ z vzporednimi premiki sestavimo trikotnik, če te daljice niso vzporedne.
(20 točk) | [
"Solution:\n\nVpeljimo vektorje $\\vec{x} = A A_{1}$, $\\vec{y} = B B_{1}$ in $\\vec{z} = C C_{1}$. Tedaj je $A A_{2} = \\vec{z}$, $B B_{2} = \\vec{x}$ in $C C_{2} = \\vec{y}$, saj sta nasprotni stranici vsakega romba vzporedni in enako dolgi. Sledi\n$$\n\\overrightarrow{A_{1}A_{2}} = \\vec{z} - \\vec{x}, \\quad \\... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Odbirno tekmovanje | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
08zc | Find the number of tuples of positive integers $(a_1, a_2, \dots, a_{2022})$ which satisfy $a_1 < a_2 < \dots < a_{2022}$ and
$$
a_1^2 - 6^2 \ge a_2^2 - 7^2 \ge \dots \ge a_{2022}^2 - 2027^2.
$$ | [
"Since $1 \\le a_1 < a_2 < \\dots < a_{2022}$, we have $a_i \\ge i$ for all $1 \\le i \\le 2022$. Also, the condition $a_1^2 - 6^2 \\ge a_2^2 - 7^2 \\ge \\dots \\ge a_{2022}^2 - 2027^2$ is equivalent to $a_{i+1}^2 - a_i^2 \\le (i+6)^2 - (i+5)^2 = 2i + 11$ ($1 \\le i \\le 2021$).\n\nWhen $a_{i+1} \\ge a_i + 3$ holds... | Japan | Japan 2022 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 10 | |
0imu | Problem:
$A$, $B$, $C$, and $D$ are points on a circle, and segments $\overline{AC}$ and $\overline{BD}$ intersect at $P$, such that $AP = 8$, $PC = 1$, and $BD = 6$. Find $BP$, given that $BP < DP$. | [
"Solution:\n\nLet $BP = x$ and $DP = 6 - x$ (since $BP < DP$ and $BD = 6$).\n\nBy the Power of a Point theorem (intersecting chords),\n$$\nAP \\cdot PC = BP \\cdot PD\n$$\nSubstitute the given values:\n$$\n8 \\cdot 1 = x \\cdot (6 - x)\n$$\n$$\n8 = 6x - x^2\n$$\n$$\nx^2 - 6x + 8 = 0\n$$\n$$\n(x - 4)(x - 2) = 0\n$$\... | United States | $10^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles"
] | null | final answer only | 2 | |
0513 | Prove that for any positive integer $n$ the sum of the first $n$ primes is greater than $n^2$. | [
"First notice that the $n$-th prime $p_n$ satisfies the inequality $p_n \\ge 2n - 1$. Indeed, the claim holds for the first prime $p_1 = 2$. Since all other primes are odd and there is exactly $n-1$ odd numbers between $2$ and $2n$, there are at most $n$ prime numbers less or equal to $2n-1$, hence $p_n \\ge 2n - 1... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
05bu | Ats and Pets take turns to write representations of the number $15$ as the sum of three distinct single-digit positive integers. On every move, each player must write a sum that has exactly one common addend with the previous sum, no common addends with the second previous sum, and less than three common addends with a... | [
"There are $8$ possible sums:\n$$\n\\begin{array}{ccc}\n1+5+9, & 2+6+7, & 3+4+8, \\\\\n1+6+8, & 2+4+9, & 3+5+7, \\\\\n2+5+8, & 4+5+6.\n\\end{array}\n$$\nNote that no two of the three sums in the first row have any common addends, and the same applies to the three sums in the second row. Moreover, each sum in the fi... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Pets (the second player) | |
0cvq | Define the *distance* between two cells of a checkered board as the smallest number of chess king's moves needed to come from one cell to the other. Determine the largest $n$ for which one can mark $n$ cells of a $100 \times 100$ board so that none of the distances between two marked cells equals $15$. | [
"Разобьём доску на $9$ квадратов $30 \\times 30$, $6$ прямоугольников $10 \\times 30$ и один квадрат $10 \\times 10$ (см. рис. 10).\n\nВ каждом квадрате $30 \\times 30$ клетки разбиваются на $15^2$ четвёрок так, что расстояние между любыми клетками в одной четвёрке равно $15$ (каждая четвёрка состоит из клеток с ко... | Russia | Regional round | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 3025 | |
0bzs | Let $A$ be a finite ring and let $a, b \in A$ be two elements with the property $(ab - 1)b = 0$. Prove that $b(ab - 1) = 0$. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Abstract Algebra > Ring Theory"
] | null | proof only | null | |
0ig2 | Problem:
Find all real solutions to $x^{3} + (x+1)^{3} + (x+2)^{3} = (x+3)^{3}$. | [
"Solution:\nThe equation simplifies to $3x^{3} + 9x^{2} + 15x + 9 = x^{3} + 9x^{2} + 27x + 27$, or equivalently, $2x^{3} - 12x - 18 = 2(x-3)(x^{2} + 3x + 3) = 0$. The discriminant of $x^{2} + 3x + 3$ is $-3 < 0$, so the only real solution is $x = 3$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 3 | |
0at3 | Problem:
Determine, with proof, the smallest positive integer $n$ with the following property: For every choice of $n$ integers, there exist at least two whose sum or difference is divisible by $2009$. | [
"Solution:\nWe show that the least integer with the desired property is $1006$. We write $2009 = 2 \\cdot 1004 + 1$.\n\nConsider the set $\\{1005, 1006, \\ldots, 2009\\}$, which contains $1005$ integers. The sum of every pair of distinct numbers from this set lies between $2011$ and $4017$, none of which is divisib... | Philippines | Philippines Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 1006 | |
03x2 | The circle $\Gamma_1$, with radius $r$, is internally tangent to the circle $\Gamma_2$ at $S$. The chord $AB$ of $\Gamma_2$ is tangent to $\Gamma_2$ at $C$. Let $M$ be the midpoint of the arc $\widehat{AB}$ (not containing $S$), and let $N$ be the foot of the perpendicular from $M$ to the line $AB$. Prove that $AC \tim... | [
"\n\n**Solution** It is well known that $S, C, M$ are collinear. Indeed, consider the dilation centered at $S$ that sends $\\Gamma_1$ to $\\Gamma_2$. Then the line $AB$ is sent to the line $l$ parallel to $AB$ and tangent to $\\Gamma_2$, i.e. the line tangent to $\\Gamma_2$ at $M$ (the midp... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0dc0 | Given $10$ quadratic equations
$$
x^{2} + a_{1} x + b_{1} = 0,\quad x^{2} + a_{2} x + b_{2} = 0,\quad \ldots,\quad x^{2} + a_{10} x + b_{10} = 0
$$
each of them has two distinct real roots and the set of all roots is $S = \{ \pm 1, \pm 2, \ldots, \pm 10 \}$. Find the minimum value of the sum $T = b_{1} + b_{2} + \cdots... | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | -385 | |
03z2 | Suppose $a$ and $b$ are positive real numbers satisfying $\frac{1}{a} + \frac{1}{b} \le 2\sqrt{2}$ and $(a-b)^2 = 4(ab)^3$. Then $\log_a b = \underline{\hspace{2cm}}$. | [
"From $\\frac{1}{a} + \\frac{1}{b} \\le 2\\sqrt{2}$, we have $a + b \\le 2\\sqrt{2}ab$. On the other hand,\n$$\n(a + b)^2 = 4ab + (a - b)^2 = 4ab + 4(ab)^3 \\\\\n\\ge 4 \\cdot 2\\sqrt{ab} \\cdot (ab)^3 = 8(ab)^2,\n$$\nand that means\n$$\na + b \\ge 2\\sqrt{2}ab. \\qquad \\textcircled{1}\n$$\nTherefore,\n$$\na + b =... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | -1 | |
0abq | One student was multiplying two numbers. During the multiplication he switched the last digit of the first number, which was $4$, with $1$. So he obtained $525$ as a result instead of $600$. Which numbers did the student multiply? | [
"From the condition in the problem we get that when the first number is reduced by $4-1=3$, then their product is reduced by $600-525=75$. So the second number is $75 \\div 3 = 25$. The first number is $600 \\div 25 = 24$. The numbers are $24$ and $25$."
] | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 24 and 25 | |
0hf8 | Do positive real numbers $x, y, z$ have to be equal, if they satisfy
$$
\frac{xy + 1}{x + 1} = \frac{yz + 1}{y + 1} = \frac{zx + 1}{z + 1}
$$ | [
"From the statement, we have $(xy + 1)(y + 1) = (yz + 1)(x + 1)$, so $xy^2 + xy + y = xyz + x + yz$. Similarly $yz^2 + yz + z = xyz + y + zx$, and also $zx^2 + zx + x = xyz + z + xy$. After adding these three equations, we get $xy^2 + yz^2 + zx^2 = 3xyz$, or $\\frac{y}{z} + \\frac{z}{x} + \\frac{x}{y} = 3$. But for... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
08tz | 10 language interpreters are invited to participate in an international mathematical contest. Each of the 10 interpreters is proficient in exactly 2 among the 5 following languages: Greek, Slovanian, Vietnamese, Spanish and German. Furthermore, the combinations of the 2 proficient languages among the 5 are all differen... | [
"Let us call the interpreter proficient in the pair of different languages $X$, $Y$ $I_{XY}$. Suppose the interpreters are assigned to 5 rooms satisfying the condition of the problem. For each room call the language which the 2 occupants of the room can speak the **common** language of the room. By assumptions made... | Japan | Japan Mathematical Olympiad First Round | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 144 | |
042e | Suppose sequence $\{a_n\}$ consists of nine terms, which satisfy $a_1 = a_9 = 1$ and $\frac{a_{i+1}}{a_i} \in \{2, 1, -\frac{1}{2}\}$ for any $i \in \{1, 2, \dots, 8\}$. Then the number of sequences like this is ______. | [
"Let $b_i = \\frac{a_{i+1}}{a_i}$ ($1 \\le i \\le 8$). Then for each $\\{a_n\\}$ satisfying the given condition, we have\n$$\n\\prod_{i=1}^{8} b_i = \\prod_{i=1}^{8} \\frac{a_{i+1}}{a_i} = \\frac{a_9}{a_1} = 1, \\text{ with } b_i \\in \\{2, 1, -\\frac{1}{2}\\} (1 \\le i \\le 8). \\qquad \\textcircled{1}\n$$\nConver... | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics"
] | English | final answer only | 491 | |
05by | Let $n$ be any natural number. Find the least natural number $k$ such that it is possible to write a natural number from $1$ through $k$ into every cell of an $n \times n$ table in such a way that the sum of every two cells with a common side differs from all other such sums. The numbers in different cells do not have ... | [
"The least and the largest among the sums of two natural numbers from $1$ through $k$ are $1+1=2$ and $k+k=2k$, respectively. Thus there can be at most $2k-1$ distinct sums. The number of distinct pairs of cells with a common side is $n-1$ in every row and column. As the total number of rows and columns is $2n$, th... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n^2 - n + 1 | |
0a79 | Problem:
Let $a_{1}, a_{2}, \ldots, a_{n}$ be real numbers. Prove
$$
\sqrt[3]{a_{1}^{3}+a_{2}^{3}+\ldots+a_{n}^{3}} \leq \sqrt{a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}}
$$
When does equality hold in (1)? | [
"Solution:\nIf $0 \\leq x \\leq 1$, then $x^{3 / 2} \\leq x$, and equality holds if and only if $x=0$ or $x=1$.\n\nThe inequality is true as an equality if all the $a_{k}$'s are zero. Assume that at least one of the numbers $a_{k}$ is non-zero. Set\n$$\nx_{k}=\\frac{a_{k}^{2}}{\\sum_{j=1}^{n} a_{j}^{2}}\n$$\nThen $... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 4 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds if and only if either all the numbers are zero, or exactly one number is positive and all the others are zero. | |
0ia4 | Problem:
Take a clay sphere of radius $13$, and drill a circular hole of radius $5$ through its center. Take the remaining "bead" and mold it into a new sphere. What is this sphere's radius? | [
"Solution:\nLet $r$ be the radius of the sphere. We take cross sections of the bead perpendicular to the line of the drill and compare them to cross sections of the sphere at the same distance from its center. At a height $h$, the cross section of the sphere is a circle with radius $\\sqrt{r^{2}-h^{2}}$ and thus ar... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Volume"
] | null | proof and answer | 12 | |
00of | Let $ABC$ be an acute triangle with $AB > AC$. Let $D$, $E$ and $F$ denote the feet of its altitudes on $BC$, $AC$ and $AB$, respectively. Let $S$ denote the intersection of lines $EF$ and $BC$.
Prove that the circumcircles $k_1$ and $k_2$ of the two triangles $AEF$ and $DES$ touch in $E$. | [
"\nFigure 3: Problem 14\n\nLet $t_1$ be the tangent line to $k_1$ in point $E$ and let $t_2$ be the tangent line to $k_2$ in point $E$. The tangent-secant theorem applied to circle $k_1$ gives\n$$\n\\angle(EF, t_1) = \\angle FAE = \\alpha\n$$\nwith the usual notation for the angles in trian... | Austria | Austrian Mathematical Olympiad | [
"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 ... | English | proof only | null |
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