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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0is5 | Problem:
What is the smallest prime divisor of $5^{7^{10^{7^{10}}}} + 1$? | [
"Solution:\nNotice that $5$ to any power is odd, so this number is even. Then $2$ is a prime divisor. It also happens to be the smallest prime."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 2 | |
0d25 | Let $ABC$ be an acute triangle, and let $AA_{1}$, $BB_{1}$, and $CC_{1}$ be its altitudes. Segments $AA_{1}$ and $B_{1}C_{1}$ meet at point $K$. The perpendicular bisector of segment $A_{1}K$ intersects sides $AB$ and $AC$ at $L$ and $M$, respectively. Prove that points $A$, $A_{1}$, $L$, and $M$ lie on a circle. | [
"Because $LM$ is parallel to $BC$, the problem is equivalent to proving that $\\angle AA_{1}L = \\angle AML = \\angle ACB$. We present two solutions:\n\n\n\nFirst solution. Let $L_{1}$ be the point on $AB$ such that $\\angle AA_{1}L_{1} = \\angle ACB$ and let us prove that $L_{1} = L$.\n\nQ... | Saudi Arabia | Selection tests for the International Mathematical Olympiad 2013 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry... | English | proof only | null | |
0jhu | Problem:
There are $n$ children and $n$ toys such that each child has a strict preference ordering on the toys. We want to distribute the toys: say a distribution $A$ dominates a distribution $B \neq A$ if in $A$, each child receives at least as preferable of a toy as in $B$. Prove that if some distribution is not dom... | [
"Solution:\n\nSuppose we have a distribution $A$ assigning each child $C_{i}$, $i=1,2, \\ldots, n$, toy $T_{i}$, such that no child $C_{i}$ gets their top preference $T_{i}^{\\prime} \\neq T_{i}$. Then, pick an arbitrary child $C_{1}$ and construct the sequence of children $C_{i_{1}}, C_{i_{2}}, C_{i_{3}}, \\ldots$... | United States | HMMT 2013 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof only | null | |
0kg6 | What is the value of $\frac{(2112-2021)^2}{169}$?
(A) 7 (B) 21 (C) 49 (D) 64 (E) 91 | [
"Solution:\nObserve that $2112 - 2021 = 91 = 7 \\cdot 13$ and that $169 = 13^2$. Thus\n$$\n\\frac{(2112 - 2021)^2}{169} = \\frac{(7 \\cdot 13)^2}{13^2} = 7^2 = 49.\n$$"
] | United States | AMC 10 A | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | C | |
05a1 | A field is a $2020 \times 2021$ grid with a positive integer written into each cell, such that no number repeats in any row or column. An onion consists of $4$ consecutive cells in a row or column, whose numbers add up to exactly $4 \cdot 2021$. Find the largest possible number of onions on the field. | [
"*Answer:* $1009 \\cdot 4041$.\n\nIn a column of length $2020$ there are $2017$ potential onions. However there cannot be two consecutive onions, as they share $3$ cells, meaning their fourth numbers would have to be equal. So a column can have at most $1009$ onions. Analogously a row of length $2021$ can also have... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1009 * 4041 | |
04yi | Maryam and Artur play a game on a board, taking turns. At the beginning, the polynomial $XY - 1$ is written on the board. Artur is the first to make a move. In each move, the player replaces the polynomial $P(X, Y)$ on the board with one of the following polynomials of their choice:
a) $X \cdot P(X, Y)$
b) $Y \cdot P... | [
"We claim that Maryam can always achieve that the polynomial on the board at the end of her turn has the form $P(X,Y) = f(XY)$ where $f \\in \\mathbb{Z}[T]$ can be written as\n$$\nT^n - \\sum_{i=0}^{n-1} a_i T^i \\quad \\text{for integers } n > 0 \\text{ and } a_i \\ge 0, \\text{ not all of them zero.} \\quad (1)\n... | Czech-Polish-Slovak Mathematical Match | CAPS Match 2025 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof only | null | |
0k2z | Problem:
A square in the $x y$-plane has area $A$, and three of its vertices have $x$-coordinates $2$, $0$, and $18$ in some order. Find the sum of all possible values of $A$. | [
"Solution:\nMore generally, suppose three vertices of the square lie on lines $y = y_{1}$, $y = y_{2}$, $y = y_{3}$. One of these vertices must be adjacent to two others. If that vertex is on $y = y_{1}$ and the other two are on $y = y_{2}$ and $y = y_{3}$, then we can use the Pythagorean theorem to get that the sq... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 1168 | |
0c0l | a) Let $n$ be a composite positive integer. Show that there exist integers $a_1, a_2, \dots, a_n$ whose sum is not divisible by $n$, but $n$ divides (at least) one of the numbers
$$
a_k, \quad a_k + a_{k+1}, \quad \dots, \quad a_k + a_{k+1} + \dots + a_{k+n-1},
$$
for all positive integers $k \le n$; $a_k = a_{k-n}$ if... | [
"a) Write $n = ab$, where $a$ and $b$ both are integers greater than $1$, and let $a_1 = a_2 = \\dots = a_{n-1} = a$ and $a_n = 0$. The sum $a_1 + a_2 + \\dots + a_n = a \\cdot (n-1)$ is clearly not divisible by $n$. Consider any positive integer $k \\le n$: If $k \\le n-b$, then $a_k + a_{k+1} + \\dots + a_{k+b-1}... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0dv3 | Problem:
V živalski vrt naselijo družino risov. Število risov $N$ po $t$ letih $(t \geq 0)$ določa funkcija $N=10 \cdot e^{\frac{2}{5} t}$.
a) Koliko risov šteje družina ob naselitvi?
b) Koliko let bi potrebovali v živalskem vrtu, da bi družina risov štela 100 članov? Rezultat zaokroži na celo število.
Zapiši odgov... | [
"Solution:\n\na)\nZa $t=0$:\n$$\nN = 10 \\cdot e^{\\frac{2}{5} \\cdot 0} = 10 \\cdot e^{0} = 10 \\cdot 1 = 10\n$$\nOdgovor: 10 risov\n\nb)\nPoiščemo $t$, da bo $N=100$:\n$$\n100 = 10 \\cdot e^{\\frac{2}{5} t}\n$$\n$$\n10 = e^{\\frac{2}{5} t}\n$$\n$$\n\\ln 10 = \\ln e^{\\frac{2}{5} t}\n$$\n$$\n\\ln 10 = \\frac{2}{5}... | Slovenia | 2. matematično tekmovanje dijakov srednjih tehniških in strokovnih šol | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | a) 10; b) 6 | |
0i5u | Problem:
For any integer $n$, define $\lfloor n\rfloor$ as the greatest integer less than or equal to $n$. For any positive integer $n$, let
$$
f(n) = \lfloor n \rfloor + \left\lfloor \frac{n}{2} \right\rfloor + \left\lfloor \frac{n}{3} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor.
$$
For how many v... | [
"Solution:\n\nNotice that, for fixed $a$, $\\lfloor n / a \\rfloor$ counts the number of integers $b \\in \\{1, 2, \\ldots, n\\}$ which are divisible by $a$; hence, $f(n)$ counts the number of pairs $(a, b)$, $a, b \\in \\{1, 2, \\ldots, n\\}$ with $b$ divisible by $a$. For any fixed $b$, the number of such pairs i... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | final answer only | 55 | |
0fjf | Problem:
Sea $n$ un número natural, y $m$ el que resulta al escribir en orden inverso las cifras de $n$. Determinar, si existen, los números de tres cifras que cumplen $2 m+S=n$, siendo $S$ la suma de las cifras de $n$. | [
"Solution:\n\nTenemos las expresiones (en base 10)\n$$\n\\begin{aligned}\n& n = a b c = c + 10 b + 100 a \\\\\n& m = c b a = 100 c + 10 b + a\n\\end{aligned}\n$$\nque, sustituidas en $2 m + S = n$ nos da\n$$\n200 c + 20 b + 2 a + (a + b + c) = 100 a + 10 b + c\n$$\nes decir\n$$\n200 c + 11 b - 97 a = 0\n$$\nPor lo ... | Spain | Olimpiada Matemática Española | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | No three-digit numbers satisfy the condition. | |
01xi | The sequence $a_1, a_2, a_3, \dots$ of positive integers is defined in the following way: $a_1$ is given, and for each $n \ge 2$ the number $a_n$ is the smallest positive integer divisible by $n$, which is not less than $a_{n-1}$. (For example, if $a_5 = 115$, then $a_6 = 120, a_7 = 126, a_8 = 128$.)
Prove that if $a_... | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0i3x | Problem:
Boris was given a Connect Four game set for his birthday, but his color-blindness makes it hard to play the game. Still, he enjoys the shapes he can make by dropping checkers into the set. If the number of shapes possible modulo (horizontal) flips about the vertical axis of symmetry is expressed as $9(1+2+\cd... | [
"Solution:\n\nThere are $9^{7}$ total shapes possible, since each of the 7 columns can contain anywhere from 0 to 8 checkers. The number of shapes symmetric with respect to a horizontal flip is the number of shapes of the leftmost four columns, since the configuration of these four columns uniquely determines the c... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 729 | |
01w4 | a) Find all real numbers $a$ such that the parabola $y = x^2 - a$ and the hyperbola $y = 1/x$ intersect each other in three different points.
b) Find the locus of the centers of circumcircles of such triples of intersection points when $a$ takes all possible values. | [
"a) First we find the values of $a$ for which the parabola is tangent to the hyperbola (see fig.). Let $\\alpha$ be the abscissa of the tangency point. At this point the derivatives of functions $x^2-a$ and $1/x$ are equal, i.e. $2\\alpha = -1/\\alpha^2$, whence $\\alpha = -\\frac{1}{\\sqrt[3]{2}}$ and the ordinate... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | a) All real a such that a > (3/2)·∛2.
b) Locus of circumcenters: the vertical ray x = 1/2 with y < 1/2 − (3/4)·∛2. | |
0kfz | We say a nondegenerate triangle whose angles have measures $\theta_1, \theta_2, \theta_3$ is *quirky* if there exists integers $r_1, r_2, r_3$, not all zero, such that
$$
r_1\theta_1 + r_2\theta_2 + r_3\theta_3 = 0.
$$
Find all integers $n \ge 3$ for which a triangle with side lengths $n-1, n, n+1$ is quirky. | [
"The answer is $n = 3, 4, 5, 7$.\nWe first introduce a variant of the $k$th Chebyshev polynomials in the following lemma (which is standard, and easily shown by induction).\n\n**Lemma**\nFor each $k \\ge 0$ there exists $P_k(X) \\in \\mathbb{Z}[X]$, monic for $k \\ge 1$ and with degree $k$, such that\n$$\nP_k(X + X... | United States | USA TSTST | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's ... | English | proof and answer | n = 3, 4, 5, 7 | |
05df | Problem:
A word is a finite sequence of letters from some alphabet. A word is repetitive if it is a concatenation of at least two identical subwords (for example, $a b a b a b$ and $a b c a b c$ are repetitive, but $a b a b a$ and $a a b b$ are not). Prove that if a word has the property that swapping any two adjacent ... | [
"Solution:\nIn this and the subsequent solutions we refer to a word with all letters identical as constant.\nLet us consider a nonconstant word $W$, of length $|W|=w$, and reach a contradiction. Since the word $W$ must contain two distinct adjacent letters, be it $W=A a b B$ with $a \\neq b$, we may assume $B=c C$ ... | European Girls' Mathematical Olympiad (EGMO) | European Girls' Mathematical Olympiad | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
08ax | Problem:
Una pulce si trova inizialmente su un vertice di un poligono regolare di 2015 lati; compie una sequenza di salti in senso antiorario: al primo salto si sposta di un vertice (da quello iniziale al vicino), al secondo di tre, al terzo di cinque, e così via, di modo che all' $n$-esimo parte da un vertice e atter... | [
"Solution:\n\nLa risposta è 48. Osserviamo che dopo $n$ passi la pulce ha coperto una distanza totale di $n^{2}$ vertici dal punto di partenza: questo è chiaramente vero per $n=0$ e $n=1$, e d'altro canto, se dopo $n$ passi la pulce ha coperto una distanza di $n^{2}$ vertici, allora al passo successivo la distanza ... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 48 | |
09a2 | Let $A$ be a nonempty subset of the positive integers. If $x \in A$, then $[\sqrt[3]{x}] \in A$ and $[9x] \in A$ holds for any $x$. Prove that $A$ is the set of all positive integers. ($[x]$ denotes the integer part of $x$) | [
"Since $A$ is a nonempty subset of the positive integers, $A$ has a minimum element $m$. If $m > 1$ then $m > \\sqrt[3]{m} \\ge [\\sqrt[3]{m}]$ and $[\\sqrt[3]{m}] \\in A$. It is contrary to that $m$ is the minimum element. So $m = 1$.\n\nSince $1 \\in A$, $9^k \\in A$. From this $[\\sqrt[3]{81}] = 4 \\in A$ and $4... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | Mongolian | proof only | null | |
0gj9 | 德克斯特的實驗室裡有 $2024$ 台機器人，每台有德克斯特各自設定好的程式。某天，他調皮搗蛋的姊姊蒂蒂會闖進實驗室，在每個機器人的額頭上寫下一個在 $\{1, 2, \dots, 113\}$ 內的整數。每台機器人此時偵測到除了它自己以外的所有機器人額頭上的數字，並立即依據其程式，各別且同時猜測自己的數字。
試求最大正整數 $k$，讓德克斯特存在設定程式的方法，使得不論數字如何分布，都至少有 $k$ 台機器人猜對自己的數字。
Dexter's Laboratory has $2024$ robots, each with a program setup by Dexter. One day, his naughty sister... | [
"$k = \\lfloor 2024/113 \\rfloor = 17$。一般性地，對於 $n$ 台機器人與 $m$ 個數字，$k = \\lfloor n/m \\rfloor$。\n\n估計：將機器人編號 $1$ 到 $n$，數字的集合為 $C = \\{0, 1, \\dots, m-1\\}$，第 $i$ 台機器人戴的數字為 $x_i \\in C$，程式則為\n$$\nf_i(x_1, x_2, \\dots, x_{i-1}, x_{i+1}, \\dots, x_n) : C^n \\to C.\n$$\n假設蒂蒂以隨機的方式讓在 $i$ 台機器人寫上 $X_i$，其中 $X_i$ 服從 $C$ 上的均勻分... | Taiwan | IMO 3J, Independent Study 1 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Chinese; English | proof and answer | 17 | |
07w4 | Suppose that $a$, $b$, $c > 0$ and $a^2 + b^2 + c^2 = 3$. Prove that
$$
\frac{a}{(2a + 3)^2} + \frac{b}{(2b + 3)^2} + \frac{c}{(2c + 3)^2} \le \frac{3}{25}.
$$ | [
"**Solution 1.** Observe that\n$$\n\\begin{aligned}\n\\frac{a}{(2a + 3)^2} &= \\frac{a}{4a^2 + 12a + 9} = \\frac{a}{4(a - 1)^2 + 20a + 5} \\\\\n&\\leq \\frac{a}{20a + 5} = \\frac{1}{20} \\cdot \\frac{4a}{4a + 1} = \\frac{1}{20} - \\frac{1}{20} \\cdot \\frac{1}{4a + 1}.\n\\end{aligned}\n$$\n\nSo, the LHS of the orig... | Ireland | IRL_ABooklet_2023 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0at9 | Problem:
Two circles of radius $12$ have their centers on each other. As shown in the figure, $A$ is the center of the left circle, and $AB$ is a diameter of the right circle. A smaller circle is constructed tangent to $AB$ and the two given circles, internally to the right circle and externally to the left circle, as ... | [
"Solution:\n\nLet $R$ be the common radius of the larger circles, and $r$ that of the small circle. Let $C$ and $D$ be the centers of the right large circle and the small circle, respectively. Let $E$, $F$ and $G$ be the points of tangency of the small circle with $AB$, the left large circl... | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3√3 | |
09w0 | For a positive integer $n$, we consider an $n \times n$-board and tiles with sizes $1 \times 1$, $1 \times 2$, ..., $1 \times n$. In how many ways can exactly $\frac{1}{2}n(n+1)$ squares of the board be coloured red, so that the red squares can be covered by placing the $n$ tiles horizontally on the board, as well as b... | [
"The number of red squares must equal the total number of squares covered by the $n$ tiles, hence the tiles are only put on top of red squares. Consider a colouring of the board and the corresponding *horizontal covering* by the tiles (where all tiles are placed horizontally) and the *vertical covering*. We will de... | Netherlands | IMO Team Selection Test 1, June 2020 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2^{2n-2} | |
0g7c | 設四邊形 $ABCD$ 外接一圓, 兩對角線 $AC$ 與 $BD$ 交於 $E$ 點。令射線 $DA$ 與射線 $CB$ 交於 $F$ 點; $G$ 為平面上一點使得 $ECGD$ 為平行四邊形; $H$ 點為 $E$ 點對直線 $AD$ 的反射點。試證: $D, H, F, G$ 四點共圓。 | [
"我們先證明三角形 $FDG$ 與 $FBE$ 相似。因為 $ABCD$ 共圓, 三角形 $EAB$ 與 $EDC$ 相似, 同樣地 $FAB$ 與 $FCD$ 也相似。由平行四邊形 $ECGD$ 可得 $GD = EC$ 及 $\\angle CDG = \\angle DCE$。由圓周角性質可得 $\\angle DCE = \\angle DCA = \\angle DBA$。因此\n$$\n\\angle FDG = \\angle FDC + \\angle CDG = \\angle FBA + \\angle ABD = \\angle FBE, \\\\\n\\frac{GD}{EB} = \\frac{CE... | Taiwan | 二〇一三數學奧林匹亞競賽第三階段選訓營,模擬競賽(一) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ds2 | A square is cut into several rectangles, none of which is a square, so that the sides of each rectangle are parallel to the sides of the square. For each rectangle with sides $a$, $b$, $a < b$, compute the ratio $a/b$. Prove that sum of these ratios is at least $1$. | [
"Without loss of generality, we may assume the square has area $1$. Let the sides of the rectangles be $a_i$, $b_i$, $a_i \\leq b_i$ and $S_i = a_i b_i$. We have $\\sum S_i = 1$ and\n$$\n\\sum \\frac{a_i}{b_i} = \\sum \\frac{S_i}{b_i^2} \\geq \\sum S_i = 1.\n$$"
] | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0j43 | Problem:
Five of James' friends are sitting around a circular table to play a game of Fish. James chooses a place between two of his friends to pull up a chair and sit. Then, the six friends divide themselves into two disjoint teams, with each team consisting of three consecutive players at the table. If the order in ... | [
"Solution:\n\nAnswer: $5$\n\nNote that the team not containing James must consist of three consecutive players who are already seated. We have $5$ choices for the player sitting furthest clockwise on the team of which James is not a part. The choice of this player uniquely determines the teams, so we have a total o... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 5 | |
0jft | Problem:
Chords $\overline{AB}$ and $\overline{CD}$ of circle $\omega$ intersect at $E$ such that $AE=8$, $BE=2$, $CD=10$, and $\angle AEC=90^{\circ}$. Let $R$ be a rectangle inside $\omega$ with sides parallel to $\overline{AB}$ and $\overline{CD}$, such that no point in the interior of $R$ lies on $\overline{AB}$, $... | [
"Solution:\n\nAnswer: $26+6 \\sqrt{17}$\n\nBy power of a point, $(CE)(ED) = (AE)(EB) = 16$, and $CE + ED = CD = 10$. Thus $CE$, $ED$ are $2$, $8$. Without loss of generality, assume $CE = 8$ and $DE = 2$.\n\nAssume our circle is centered at the origin, with points $A = (-3,5)$, $B = (-3,-5)$, $C = (5,-3)$, $D = (-5... | United States | HMMT November 2013 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations an... | null | proof and answer | 26 + 6 sqrt(17) | |
070t | Problem:
Show that any integer greater than $10$ whose digits are all members of $\{1,3,7,9\}$ has a prime factor $\geq 11$. | [
"Solution:\nSuch a number cannot be divisible by $2$ (or its last digit would be even) or by $5$ (or its last digit would be $0$ or $5$). So if the result is false then the number must be of the form $3^{m} 7^{n}$ for non-negative integers $m, n$. But we claim that a number of this form must have even $10$s digit.\... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0992 | Problem:
Fie pătratul $ABCD$ cu latura de lungime $3~\mathrm{cm}$. Punctele $M$, $N$ și $P$ sunt situate pe latura $(AB)$, $(BC)$ și, respectiv, $(CD)$ astfel încât patrulaterul $AMNP$ este un trapez cu bazele $MN$ și $AP$, având diagonale perpendiculare și $BN=1~\mathrm{cm}$. Calculați perimetrul trapezului $AMNP$. | [
"Solution:\n\nCum $BN \\parallel AD$ și $MN \\parallel AP$, atunci unghiurile $BNM$ și $DAP$ sunt congruente, fapt care implică asemănarea triunghiurilor $BNM$ și $DAP$. Deoarece $BN=1~\\mathrm{cm}$, iar $AD=3~\\mathrm{cm}$, rezultă că $AP=3 \\cdot MN$. Dacă $MP \\cap AN=\\{O\\}$, atunci din congruențele unghiurilo... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 5 + 2√5 cm | |
0f8x | Problem:
Can 77 blocks each $3 \times 3 \times 1$ be assembled to form a $7 \times 9 \times 11$ block? | [] | Soviet Union | 23rd ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | No | |
0g33 | Problem:
Sei $n$ eine natürliche Zahl. Ein Volleyballteam bestehend aus $n$ Frauen und $n$ Männern stellt sich für ein Spiel auf. Dabei besetzt jedes Teammitglied eine der Positionen $1,2, \ldots, 2 n$, wobei sich genau die Positionen $1$ und $n+1$ ausserhalb des Spielfelds befinden. Während des Spiels rotieren alle T... | [
"Solution:\n\nWir visualisieren die Aufgabe wie folgt: Die Personen stehen in einem Kreis. Die Teammitglieder auf den Positionen $k$ und $n+k$ für $1 \\leq k \\leq n$ stehen sich dabei gegenüber.\n\nBeobachtung: Eine Frau muss immer gegenüber von einem Mann stehen, sonst gibt es eine Rotation, bei der sich zwei Fra... | Switzerland | Vorrunde 2019 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | (n!)^2 2^n | |
0gxx | Find all pairs of positive integers $(x, y)$ satisfying the equation:
$$x^y + y^x = 2008.$$ | [
"Let $x \\le y$. $x=1$ gives us an obvious solution $y=2007$.\n\nIn what follows, suppose that $2 \\le x \\le y$. We can check for small values of $x$:\n\n$2^{10} = 1024 < 2008 < 2^{11} = 2048$, $3^6 = 729 < 2008 < 3^7 = 2187$, \n$4^5 = 1024 < 2008 < 4^6 = 4096$, $5^4 = 625 < 2008 < 5^5 = 3125$, $6^4 = 1296 <... | Ukraine | The Problems of Ukrainian Authors | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (1, 2007) and (2007, 1) | |
0cqy | A function $f: \mathbb{R} \to \mathbb{R}$ is given. Assume that $(f(x))^2 \le f(y)$ for every $x > y$. Prove that each value of $f$ lies in $[0, 1]$. (A. Khrabrov) | [
"По условию $f(y) \\ge (f(y+1))^2 \\ge 0$ для любого $y$, поэтому все значения функции неотрицательны.\n\nПусть теперь $f(x_0) = 1 + a > 1$ для некоторого $x_0$. Доказем индукцией по $n$, что для любого $y < x_0$ верно неравенство $f(y) > 1 + 2^n a$.\n\nПри $n = 1$ имеем $f(y) \\ge (f(x_0))^2 = 1 + 2a + a^2 > 1 + 2... | Russia | XL Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
01wc | Is it possible to represent the polynomial of seven variables
$$
Q(x_1, x_2, \dots, x_7) = (x_1 + x_2 + \dots + x_7)^2 + 2(x_1^2 + x_2^2 + \dots + x_7^2)
$$
as a sum of squares of seven polynomials with non-negative integer coefficients:
$$
Q(x_1, \dots, x_7) = P_1(x_1, \dots, x_7)^2 + P_2(x_1, \dots, x_7)^2 + \dots + ... | [
"Answer: yes, it is possible.\nFor example, consider the following representation:\n$$\n(x_1 + x_2 + \\dots + x_7)^2 + 2(x_1^2 + x_2^2 + \\dots + x_7^2) = (x_1 + x_2 + x_4)^2 + \\\\\n\\qquad +(x_2 + x_3 + x_5)^2 + (x_3 + x_4 + x_6)^2 + (x_4 + x_5 + x_7)^2 + \\\\\n\\qquad +(x_5 + x_6 + x_1)^2 + (x_6 + x_7 + x_2)^2 +... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | Yes. For example:
(x1 + x2 + x3 + x4 + x5 + x6 + x7)^2 + 2(x1^2 + x2^2 + x3^2 + x4^2 + x5^2 + x6^2 + x7^2)
=
(x1 + x2 + x4)^2 + (x2 + x3 + x5)^2 + (x3 + x4 + x6)^2 + (x4 + x5 + x7)^2 + (x5 + x6 + x1)^2 + (x6 + x7 + x2)^2 + (x7 + x1 + x3)^2. | |
07e3 | Let $1 < t < 2$ be a real number. Prove that for all sufficiently large positive integers like $d$, there is a monic polynomial $P(x)$ of degree $d$, such that all of its coefficients are either $+1$ or $-1$ and
$$
|P(t) - 2019| < 1.
$$ | [
"At first we shall prove following lemma:\n**Lemma.** Let $b_n$ be a sequence of positive real numbers satisfying\n$$\nb_n \\leq 2b_0 + b_1 + \\cdots + b_{n-1},\n$$\nthen for each real number $z$ where\n$$\n|z| \\leq 2b_0 + b_1 + \\cdots + b_n,\n$$\nthere are $a_0, a_1, \\dots, a_n \\in \\{1, -1\\}$ such that\n$$\n... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof only | null | |
094v | Problem:
Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $D$ be a point on the line $AC$ such that $AB = AD$ and $A$ lies between $C$ and $D$. Suppose that there are two points $E \neq F$ on the circumcircle of the triangle $DBC$ such that $AE = AF = BC$. Prove that the line $EF$ passes through the circumc... | [
"Solution:\n\nLet $N$ be the midpoint of arc $BAC$. Then triangle $NBC$ is equilateral as $\\angle BNC = 60^{\\circ}$ and $N$ lies on the perpendicular bisector of $BC$. Moreover, $N$ lies on the angle bisector of the angle $DAB$, which is the perpendicular bisector of segment $BD$ considering the isosceles triangl... | Middle European Mathematical Olympiad (MEMO) | MEMO Szeged | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing... | null | proof only | null | |
02in | Problem:
Um atleta corre $5000~\mathrm{m}$ por semana em uma quadra de esportes, que tem uma pista curta e outra longa. Em uma semana ele treinou seis dias, sendo que a cada dia correu uma vez na pista longa e duas na pista curta. Na semana seguinte ele treinou sete dias, sendo que a cada dia correu uma vez em cada pi... | [
"Solution:\n\nDenotemos por $x$ e $y$ os comprimentos das pistas longa e curta, respectivamente.\n\nNuma semana, ele corre $6(x+2y)$ e na outra $7(x+y)$. Como, em cada semana, ele corre os mesmos $5000$ metros, temos:\n$$\n6(x+2y) = 7(x+y)\n$$\nSegue que $6x + 12y = 7x + 7y$, e portanto, $5y = x$. Assim, o comprime... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
02nm | The cells of a $3 \times 3$ table were numbered from $1$ to $9$, each number appearing exactly once. For each row the cell with the greatest number is colored red and the cell with the smallest number is colored green. Let $A$ be the smallest of the numbers in the red cells and $B$ the greatest of the numbers in the gr... | [
"a. For example,\n\n| 7 | 4 | 1 |\n|---|---|---|\n| 8 | 5 | 2 |\n| 9 | 6 | 3 |\n\nNotice that $A = \\min\\{7, 8, 9\\} = 7$ and $B = \\max\\{1, 2, 3\\} = 3$, so $A - B = 4$.\n\nb. For example,\n\n| 1 | 2 | 3 |\n|---|---|---|\n| 4 | 5 | 7 |\n| 6 | 8 | 9 |\n\nNotice that $A = \\min\\{3, 7, 9\\} = 3$ and $B = \\max\\{1... | Brazil | Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a: Example arrangement with rows 7 4 1 / 8 5 2 / 9 6 3 gives A = 7 and B = 3, so A − B = 4.
b: Example arrangement with rows 1 2 3 / 4 5 7 / 6 8 9 gives A = 3 and B = 6, so A − B = −3.
c: Not possible (A = 4 and B = 3 cannot occur). | |
0ft6 | Problem:
Sei $a_{1}<a_{2}<\ldots<a_{n}$ eine Folge positiver ganzer Zahlen mit der Eigenschaft, dass für $i<j$ die Dezimaldarstellung von $a_{j}$ nicht mit jener von $a_{i}$ beginnt (zum Beispiel können die Zahlen 137 und 13729 nicht beide in der Folge vorkommen). Beweise, dass gilt
$$
\sum_{i=1}^{n} \frac{1}{a_{i}} \... | [
"Solution:\n\nFür eine Folge $a=\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ wie in der Aufgabenstellung setzen wir\n$$\ns(a)=\\sum_{i=1}^{n} \\frac{1}{a_{i}}\n$$\nEnthält die Folge eine $k+1$-stellige Zahl $b=\\left(d_{k} \\ldots d_{1} d_{0}\\right)_{(10)},\\ k \\geq 1$, dann kann sie nach Voraussetzung die $k$-st... | Switzerland | IMO - Selektion | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | proof only | null | |
0d2p | The points of the plane have been colored by $2013$ different colors. We say that a triangle $\triangle ABC$ has the color $X$ if its three vertices $A$, $B$, $C$ have the color $X$. Prove that there are infinitely many triangles with the same color and the same area. | [
"Consider $2014$ parallel lines. Each line contains infinitely many points. Since the number of the colors is finite, by the pigeonhole principle, there exist on each line infinitely many points of the same color. Choose for each line one color for which there exist infinitely many points. Since there are $2013$ co... | Saudi Arabia | Preselection tests for the full-time training | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0ew9 | Problem:
Given a fixed circle $C$ and a line $L$ through the center $O$ of $C$. Take a variable point $P$ on $L$ and let $K$ be the circle center $P$ through $O$. Let $T$ be the point where a common tangent to $C$ and $K$ meets $K$. What is the locus of $T$? | [
"Solution:\n\nLet the common tangent meet $C$ at $S$. Let $X$ be the intersection of $C$ and $OP$ lying between $O$ and $P$. $PT = PO$, hence $\\angle POT = \\angle PTO$, so $\\angle OPT = 180^{\\circ} - 2\\angle POT$. But $PT$ and $OS$ are parallel, because both are perpendicular to the common tangent. Hence $\\an... | Soviet Union | 2nd ASU | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | The pair of fixed tangents to the circle that are perpendicular to the given line. | |
0e6e | Problem:
Natančno izračunaj dolžini stranic pravokotnika, katerega obseg je $4~\mathrm{cm}$, kot med diagonalama pa $60^\circ$. | [
"Solution:\n\nNarišemo skico in zapišemo zvezi med dolžinami stranic: $\\tan 30^\\circ = \\frac{b}{a}$ in $2a + 2b = 4$.\n\nIzrazimo npr. $b = \\frac{\\sqrt{3}}{3} a$ in vstavimo v $2a + 2b = 4$. Dobimo $2a + 2 \\frac{\\sqrt{3}}{3} a = 4$.\n\nIzračunamo\n\n$$\na = 3 - \\sqrt{3}~\\mathrm{cm}\n$$\n\nin\n\n$$\nb = \\s... | Slovenia | Državno tekmovanje | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | Side lengths: 3 − √3 cm and √3 − 1 cm | |
00ks | Anton chooses as starting number an integer $n \ge 0$ which is not a square. Berta adds to this number its successor $n + 1$. If this sum is a perfect square, she has won. Otherwise, Anton adds to this sum, the subsequent number $n + 2$. If this sum is a perfect square, he has won. Otherwise, it is again Berta's turn a... | [
"We will prove that Anton wins for the infinity of starting numbers $3x^2-1$ with $x \\ge 1$.\nSince $3x^2 - 1 \\equiv 2 \\pmod 3$, it cannot be a perfect square. After Berta adds the subsequent integer $3x^2$, the sum $6x^2 - 1$ is also $\\equiv 2 \\pmod 3$ and consequently not a perfect square. Now Anton adds the... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
08o3 | Problem:
Find all ordered triples $(x, y, z)$ of integers satisfying $20^{x} + 13^{y} = 2013^{z}$. | [
"Solution:\nAs $20 \\cdot 13 = 2^{2} \\cdot 5 \\cdot 13$ and $2013 = 3 \\cdot 11 \\cdot 61$ are relatively prime, $x$, $y$ and $z$ must be nonnegative.\n\nConsidering the equation modulo $3$, we observe that $x$ must be odd. Now considering the equation modulo $7$, we obtain $(-1) + (-1)^{y} \\equiv 4^{z} \\pmod{7}... | JBMO | 17th Junior Balkan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | no solutions | |
01jx | Find all possible non-zero integers $a, b, c$, so that two distinct roots of the equation $ax^2+bx+c = 0$ are also the roots of the equation $x^3 + bx^2 + ax + c = 0$. | [
"Answer: $a = 2$, $b = 4$, $c = -4$.\n\nLet $x_1, x_2$ be distinct roots of the equation $ax^2+bx+c=0$, i.e. be the zeroes of the function $f(x) = ax^2+bx+c$ and $g(x) = x^3+bx^2+ax+c$. By condition, $f(0) = g(0) = c \\ne 0$. Let $F(x) = g(x) - f(x)$. Then $0, x_1, x_2$ are the distinct zeroes of the polynomial $F(... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a = 2, b = 4, c = -4 | |
05o3 | Problem:
On considère un échiquier $3 \times 3$. Au début, on écrit le chiffre $0$ dans chacune des $9$ cases. Ensuite, à chaque étape, on effectue l'opération suivante : on choisit deux cases ayant un côté commun, puis on rajoute $1$ au nombre écrit dans ces deux cases, ou bien on retranche $1$ au nombre écrit dans c... | [
"Solution:\n\nCe n'est pas possible. On colorie l'échiquier $3 \\times 3$ en noir et blanc de manière usuelle (de sorte qu'une case noire n'ait que des cases blanches comme voisins et qu'une case blanche n'ait que des cases noires comme voisins). On vérifie que la somme des nombres sur les cases noires est toujours... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Not possible | |
0jd6 | A social club has $2k + 1$ members, each of whom is fluent in the same $k$ languages. Any pair of members always talk to each other in only one language. Suppose that there were no three members such that they use only one language among them. Let $A$ be the number of three-member subsets such that the three distinct p... | [
"The answer is $\\binom{2k+1}{3} - k(2k+1)$, or $\\frac{2k(k-2)(2k+1)}{3}$.\n\nWe will treat the social club as a complete graph on $2k+1$ vertices, where each language corresponds to one color of edge between pairs of vertices. Let $V = \\{v_1, \\dots, v_{2k+1}\\}$ be the set of vertices, $L = \\{l_1, \\dots, l_k\... | United States | IMO Team Selection Test | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | 2k(k-2)(2k+1)/3 | |
0dmv | Problem:
Нека је $n \geqslant 2$ природан број и нека позитивни реални бројеви $a_{0}, a_{1}, \ldots, a_{n}$ задовољавају једнакост
$$
\left(a_{k-1}+a_{k}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1} \quad \text{ за свако } k=1,2, \ldots, n-1 \text{. }
$$
Доказати да је $a_{n}<\frac{1}{n-1}$. | [
"Solution:\n\nДата једнакост је еквивалентна са\n$$\n\\frac{1}{a_{k}+a_{k+1}}=1+\\frac{1}{a_{k-1}+a_{k}}\n$$\nза свако $k>0$. Индукцијом следи $\\frac{1}{a_{k}+a_{k+1}}=k+\\frac{1}{a_{0}+a_{1}}$ за $k>0$, одакле добијамо да је $\\frac{1}{a_{n-1}+a_{n}} > n-1$ и према томе $a_{n}<\\frac{1}{n-1}$."
] | Serbia | СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
0h9p | Can one draw $6$ circles on a plane, such that each one passes through the centers of exactly three other circles? | [
"Possible example is shown on Fig. 18 where each segment has length $1$.\n\n"
] | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | Yes | |
0evm | Let $\mathbb{R}^+$ be the set of positive real numbers. Let $f : \mathbb{R}^+ \to \mathbb{R}^+$ be a function satisfying the following.
For each positive real number $x$, there exists $y \in \mathbb{R}$ such that
$$
(x + f(y))(y + f(x)) \leq 4,
$$
and the number of such $y$'s is finite.
Prove that $f(x) > f(y)$ for eve... | [
"For each positive real number $x$, let $A_x$ be the set of positive real numbers $y$ satisfying $(x + f(y))(y + f(x)) \\le 4$. The following are easy consequences by the definition.\n(1) If $x \\in A_y$, then $y \\in A_x$.\n(2) For $x < y$, if $f(x) \\le f(y)$ then $A_y \\subseteq A_x$.\nWe prove the following lem... | South Korea | The 36th KOREAN MATHEMATICAL OLYMPIAD Final Round | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
0lgi | Problem:
A convex hexagon $A B C D E F$ is inscribed in a circle. Prove the inequality
$$
A C \cdot B D \cdot C E \cdot D F \cdot A E \cdot B F \geq 27 A B \cdot B C \cdot C D \cdot D E \cdot E F \cdot F A
$$ | [
"Solution:\nLet\n$$\nd_{1} = A B \\cdot B C \\cdot C D \\cdot D E \\cdot E F \\cdot F A, \\quad d_{2} = A C \\cdot B D \\cdot C E \\cdot D F \\cdot A E \\cdot B F, \\quad d_{3} = A D \\cdot B E \\cdot C F\n$$\nApplying Ptolemy's theorem to quadrilaterals $A B C D$, $B C D E$, $C D E F$, $D E F A$, $E F A B$, $F A B... | Zhautykov Olympiad | XIV International Zhautykov Olympiad in Mathematics | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
093l | Problem:
Let $ABC$ be an acute-angled triangle such that $AB < AC$. Let $D$ be the point of intersection of the perpendicular bisector of the side $BC$ with the side $AC$. Let $P$ be a point on the shorter arc $AC$ of the circumcircle of the triangle $ABC$ such that $DP \parallel BC$. Finally, let $M$ be the midpoint ... | [
"Solution:\n\nLet the line $DP$ intersect the circumcircle of the triangle $ABC$ again at a point $Q$. We can see that\n$$\n\\angle ADQ = \\angle ACB = \\angle APB \\quad \\text{and} \\quad \\angle AQD = \\angle AQP = \\angle ABP\n$$\nThis implies that the triangles $AQD$ and $ABP$ are similar and therefore the equ... | Middle European Mathematical Olympiad (MEMO) | MEMO Team Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
01cl | Let $m$ and $n$ be positive integers and let the integer $X \ge \max(m, n)$. Show that there exist integers $u$ and $v$, not both equal to $0$, such that
$$
\max(|u|, |v|) \le \sqrt{X} \quad \text{and} \quad 0 \le m u + n v \le 2\sqrt{X}.
$$ | [
"There are $[\\sqrt{X} + 1]^2 \\ge X + 1$ pairs $(a, b)$ such that $0 \\le a, b \\le \\sqrt{X}$, and for these\n$$\n0 \\le m a + n b \\le 2 X \\sqrt{X}.\n$$\nTwo linear combinations $m a + n b \\ge m a' + n b'$ differ by at most $2\\sqrt{X}$, and so\n$$\n\\max(|a - a'|, |b - b'|) \\le \\sqrt{X} \\quad \\text{and} \... | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
01rz | Let the incircle of the triangle $ABC$ touch the side $AB$ at point $Q$; the incircles of the triangles $QAC$ and $QBC$ touch $AQ$, $AC$ and $BQ$, $BC$ at points $P$, $T$ and $D$, $F$, respectively.
Prove that $PDFT$ is a cyclic quadrilateral. | [
"(Solution by A. Gaponenko, D. Voynov.) First, note that incircles of the triangles $QAC$ and $QBC$ touch $CQ$ at the same point $X$ (well-known fact). Hence $CF = CX = CT$. Also $AP = AT$, $BF = BD$. Now we have\n$$\n\\angle TFD = 180^\\circ - \\angle TFC - \\angle BFD =\n$$\n$$\n= 180^\\circ - (90^\\circ - \\frac... | Belarus | SELECTION and TRAINING SESSION | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0bf7 | Problem:
Legyen $f: \mathbb{R} \rightarrow \mathbb{R}$ egy monoton függvény.
a) Igazold, hogy $f$-nek minden $x_{0} \in \mathbb{R}$ pontban van jobb- és baloldali határértéke!
b) Értelmezzük a $g: \mathbb{R} \rightarrow \mathbb{R}$, $g(x)=\lim _{t \nearrow x} f(t)$ függvényt, vagyis $g(x)$ az $f$ függvény baloldali ... | [] | Romania | Matematika tantárgyverseny Megyei szakasz | [
"Precalculus > Limits",
"Precalculus > Functions"
] | null | proof only | null | |
0l4n | Let $n$ be a positive integer. Ana and Banana play a game. Banana thinks of a function $f: \mathbb{Z} \to \mathbb{Z}$ and a prime number $p$. He tells Ana that $f$ is nonconstant, $p < 100$, and $f(x+p) = f(x)$ for all integers $x$. Ana's goal is to determine the value of $p$. She writes down $n$ integers $x_1, \dots, ... | [] | United States | TST2025 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 186 | |
0b2k | Problem:
For a real number $t$, $\lfloor t \rfloor$ is the greatest integer less than or equal to $t$ and $\{ t \} = t - \lfloor t \rfloor$ is the fractional part of $t$. How many real numbers $x$ between $1$ and $23$ satisfy $\lfloor x \rfloor \{ x \} = 2 \sqrt{x}$?
(a) $18$
(b) $19$
(c) $20$
(d) $21$ | [] | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | a | |
0jqr | Problem:
In triangle $A B C$, a point $M$ is selected in its interior so that $\angle M A B = 10^{\circ}$, $\angle M B A = 20^{\circ}$, $\angle M C A = 30^{\circ}$ and $\angle M A C = 40^{\circ}$. Determine the value of $\angle M B C$. | [
"Solution:\n\nLet $X$ be on side $B C$ so that $A M$ bisects $\\angle A B X$. Let $Y$ be on side $A C$ so that $B M$ bisects $\\angle A B Y$. Denote by $Z$ the intersection of these two lines; thus $M$ is the incenter of $\\triangle A B Z$. Then, $\\angle B M Z = 90^{\\circ} + \\frac{1}{2} \\angle B A Z = 100^{\\ci... | 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 and answer | 60° | |
06ry | In an acute triangle $A B C$ the points $D$, $E$ and $F$ are the feet of the altitudes through $A$, $B$ and $C$ respectively. The incenters of the triangles $A E F$ and $B D F$ are $I_{1}$ and $I_{2}$ respectively; the circumcenters of the triangles $A C I_{1}$ and $B C I_{2}$ are $O_{1}$ and $O_{2}$ respectively. Prov... | [
"Let $\\angle C A B=\\alpha$, $\\angle A B C=\\beta$, $\\angle B C A=\\gamma$. We start by showing that $A$, $B$, $I_{1}$ and $I_{2}$ are concyclic. Since $A I_{1}$ and $B I_{2}$ bisect $\\angle C A B$ and $\\angle A B C$, their extensions beyond $I_{1}$ and $I_{2}$ meet at the incenter $I$ of the triangle. The poi... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > ... | null | proof only | null | |
0dqw | Find all functions $f: \mathbb{R} \to \mathbb{R}$ so that
$$
(x+y)(f(x)-f(y)) = (x-y)f(x+y)
$$
for all $x, y \in \mathbb{R}$. | [
"Suppose that $f$ is a solution. Let\n$$\na = \\frac{1}{2}(f(1) - f(-1)), \\quad b = \\frac{1}{2}(f(1) + f(-1))\n$$\nand $g(x) = f(x) - a x - b x^2$. Then\n$$\n(x+y)(g(x)-g(y)) = (x-y)g(x+y)\n$$\nand $g(1) = g(-1) = 0$. Letting $y = 1$ and $y = -1$ above give\n$$\n\\begin{aligned}\n(x+1)g(x) &= (x-1)g(x+1) \\\\\nxg... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | f(x) = a x + b x^2 for arbitrary real constants a and b | |
0lek | Let $a$, $b$ and $c$ be non-negative real numbers such that
$$
2(a^2 + b^2 + c^2) + 3(ab + bc + ca) = 5(a + b + c).
$$
Prove that $4(a^2 + b^2 + c^2) + 2(ab + bc + ca) + 7abc \le 25$. | [
"Let $p = a + b + c$, $q = ab + bc + ca$ and $r = abc$, we have\n$$\n2(p^2 - 2q) + 3q = 5p \\text{ or } 2p^2 = 5p + q. \\quad (1)\n$$\n\nWe need to prove that $4(p^2 - 2q) + 2q + 7r \\le 25$ or $4p^2 + 7r \\le 25 + 6q$.\nSince $q = p^2 - 5p$, the inequality is equivalent to:\n$$\n7r + 30p \\le 8p^2 + 25.\n$$\nNotic... | Vietnam | TST | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0fgh | Problem:
Sean $A$, $B$ vértices adyacentes de un $n$-ágono regular $(n \geq 5)$ del plano que tiene centro en $O$. Un triángulo $XYZ$ que es congruente con $OAB$ e inicialmente coincide con él, se mueve en el plano de forma que $Y$ y $Z$ describan la frontera del polígono, dejando $X$ en el interior. Hallar el lugar ge... | [] | Spain | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | The locus of X is the circle centered at O with radius equal to the apothem, that is R cos(pi/n). | |
0afy | Банкнота од 100 денари треба да се раситни на монети од 2 и 5 денари, при што нивниот број е 32. Ако такво раситнување постои, колку монети од 2 и колку монети од 5 денари се употребени? | [
"І начин: Ако избереме сите 32 монети да се 2 денари, тогаш би имале 64 денари, па банкнотата од 100 денари не е раситнета. Ако една монета од 32-те монети од 2 денари се замени со монета од 5 денари, сумата се зголемува за $5-2=3$ денари. Значи, сумата од 64 денари треба да ја зголемиме за $100-64=36$ денари. Спор... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | Macedonian, English | proof and answer | 20 coins of 2 denari and 12 coins of 5 denari | |
00ah | Decide if there is an arithmetic progression of $2016$ natural numbers that are not perfect powers but their product is a perfect power.
(A perfect power is a number of the form $n^k$ where $n$ and $k$ are natural numbers with $n \ge 2$, $k \ge 2$.) | [
"For each $n \\ge 3$ there exists an arithmetic progression of length $n$ with these properties. The solution uses the remark that if $l \\in \\mathbb{N}$ is divisible by a prime $p$ but not by $p^2$—in which case we say that $l$ is exactly divisible by $p$—then $l$ is not a perfect power.\n\nStart a construction b... | Argentina | Argentine National Olympiad 2016 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | Yes; such an arithmetic progression exists (indeed for any length at least three, including 2016). | |
06gy | For any positive integer $a$, define $M(a)$ to be the number of positive integers $b$ for which $a+b$ divides $ab$. Find all integer(s) $a$ with $1 \le a \le 2013$, so that $M(a)$ is largest possible in the range of $a$. | [
"The answer is $a = 1680$.\nSince $ab \\equiv a(-a) = -a^2 \\pmod{a+b}$, we have $a+b \\mid ab$ if and only if $a+b \\mid a^2$. This shows $a+b$ can be any divisor of $a^2$ greater than $a$. Also, any such divisor corresponds to a unique positive integer $b$. Therefore, $M(a)$ is the number of divisors of $a^2$ gre... | Hong Kong | The Sixteenth Hong Kong (China) Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | English; Chinese | proof and answer | 1680 | |
0l8e | In a group of people, some are friends (friendship is mutual) and each person $p$ has a list $f_1(p), f_2(p), \dots, f_{d(p)}(p)$ of their friends, where $d(p)$ is the number of friends $p$ has. Additionally, any two people are connected by a series of friendships. Each person also has a *water balloon*. The following ... | [
"Given a person $p$, let $F(p)$ be the set of friends of $p$. Choose a person $p$ with the most friends. Note that for each friend $q$ of $p$, $p$ receives a water balloon from $q$ once out of every $d(q)$ turns. Since $p$ always receives 1 water balloon, we must have\n$$\n\\sum_{q \\in F(p)} \\frac{1}{d(q)} = 1.\n... | United States | United States of America — TST Selection Test | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
07hy | Let $\frac{1}{2} < s < 1$. An insect jumps on points in the interval $[0, 1]$. If the insect is on point $a$, it can jump to either $s \times a$ or $(a - 1) \times s + 1$. Prove that for any arbitrary point $c$ in the interval $[0, 1]$, the insect can jump in such a way that after a few steps, it will be at a distance ... | [
"We call the number $x$ an $n$-digit number if it can be written as $x = \\sum_{i=0}^{n} a_i s^i$ where $\\forall 0 \\le i \\le n$, $a_i$ is either $0$ or $1-s$ (note that $a_n$ can be $0$ too). Using the operations presented in the problem, starting from $a$, we can obtain numbers of the form $a s^N + b$ where $b$... | Iran | 40th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0job | Problem:
Let $a$ and $b$ be real numbers randomly (and independently) chosen from the range $[0,1]$. Find the probability that $a$, $b$ and $1$ form the side lengths of an obtuse triangle. | [
"Solution:\n\nAnswer: $\\frac{\\pi-2}{4}$\n\nWe require $a+b>1$ and $a^{2}+b^{2}<1$. Geometrically, this is the area enclosed in the quarter-circle centered at the origin with radius $1$, not including the area enclosed by $a+b<1$ (an isosceles right triangle with side length $1$). As a result, our desired probabil... | United States | HMMT November 2015 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | (π - 2)/4 | |
040w | Given an $n \times n$ grid, we call two cells in it adjacent if they have a common side. At the beginning, each cell is assigned number $+1$. An operation on the grid is defined as follows: one chooses a cell, and then changes the signs of every number in its adjacent cells (but does not change the sign of the number i... | [
"We denote the cell in the $i$-row and $j$-column of the grid as $A_{ij}$ ($i, j \\in \\{1, 2, \\dots, n\\}$).\n\nWhen $n = 2k$, $k \\in \\mathbb{N}^*$, we mark each $A_{ij}$ satisfying $i + j \\equiv 0 \\pmod{2}$ with color red (presented by shaded areas), and that satisfying $j - i \\equiv 3 \\pmod{4}$ and $j - i... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | all even integers n ≥ 2 | |
02wa | Problem:
Considere a coleção de todos os números de 5 dígitos cuja soma dos dígitos é 43. Um desses números é escolhido ao acaso. Qual a probabilidade dele ser múltiplo de 11? | [
"Solution:\nA soma máxima dos dígitos de um número de cinco dígitos é 45, que corresponde à soma dos dígitos de $99999$. Para que um número possua soma de seus dígitos $43$ podem ocorrer dois casos: ou ele terá três dígitos iguais a $9$ e dois iguais a $8$ ou ele terá quatro dígitos iguais a $9$ e um igual a $7$.\n... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 1/5 | |
0b5j | Let $\omega$ be a circle in the plane and $A,B$ two points lying on it. We denote by $M$ the midpoint of $AB$ and let $P \neq M$ be a new point on $AB$. Build circles $\gamma$ and $\delta$ tangent to $AB$ at $P$ and to $\omega$ at $C$, respectively $D$. Consider $E$ to be the point diametrically opposed to $D$ in $\ome... | [
"Let us begin by noticing that since $DE$ is a diameter, we have $\\angle DBE = 90^\\circ$, and if the circumcenter of $\\triangle BMC$ would lie on $BE$, we could conclude that $DB$ is tangent to the circumcircle of $\\triangle BMC$. Thus $\\angle DBA \\equiv \\angle BCM$. Now since $A$, $B$, $C$, $D$ are on $\\om... | Romania | Local Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Advanced Configurations > Is... | English | proof only | null | |
0771 | Let $n \ge 1$ be an integer and consider the sum
$$
x = \sum_{k \ge 0} \binom{n}{2k} 2^{n-2k} 3^k = \binom{n}{0} 2^n + \binom{n}{2} 2^{n-2} \cdot 3 + \binom{n}{4} 2^{n-4} \cdot 3^2 + \dots
$$
Show that $2x - 1$, $2x$, $2x + 1$ form the sides of a triangle whose area and inradius are also integers. | [] | India | INMO-2017 | [
"Number Theory > Diophantine Equations > Pell's equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | English | proof only | null | |
065o | Determine positive integers $x$, $y$, $z$ which satisfy the system
$$
\begin{aligned}
x + y + z &= xy + yz + zx \\
xyz &= 1.
\end{aligned}
$$
and have the least possible sum. | [
"**First solution**\nWe write the system in the form\n$$\nxy + yz + zx = x + y + z \\quad (1)\n$$\n$$\nxyz = 1. \\quad (2)\n$$\nSubtracting the two equations by parts we find\n$$\n\\begin{aligned}\n& xyz - (xy + yz + zx) = 1 - (x + y + z) \\\\\n\\Leftrightarrow \\quad & xyz - xy - yz - zx + x + y + z - 1 = 0 \\\\\n... | Greece | SELECTION EXAMINATION | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (1, 1, 1) | |
0glg | Let $p_1, p_2, \dots$ be the sequence of positive integers defined by $p_1 = 2$, and for all positive integers $n$, $p_{n+1}$ is defined to be the least prime number dividing $n p_1^{1!} p_2^{2!} \cdots p_n^{n!} + 1$.
Prove that all prime numbers appear in this sequence. | [
"Let $q_1, q_2, \\dots$ be the prime numbers in ascending order.\nAssume to the contrary that there exists a prime number not appearing in the sequence. Let $q_s$ be the least such prime.\nNote that by Fermat's little theorem we have $p_k^{k!} \\equiv 1 \\pmod{q_s}$ for all $k \\ge q_s - 1$.\nLet $r$ be the remaind... | Thailand | The 13th Thailand Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
040t | The sum of all the positive integers $n$ satisfying $\frac{1}{4} < \sin \frac{\pi}{n} < \frac{1}{3}$ is ______. | [
"As $\\sin x$ is a convex function for $x \\in (0, \\frac{\\pi}{6})$, we have $\\frac{3}{\\pi}x < \\sin x < x$. Then\n$$\n\\sin \\frac{\\pi}{13} < \\frac{\\pi}{13} < \\frac{1}{4}, \\sin \\frac{\\pi}{12} > \\frac{3}{\\pi} \\times \\frac{\\pi}{12} = \\frac{1}{4},\n$$\n$$\n\\sin \\frac{\\pi}{10} < \\frac{\\pi}{10} < \... | China | China Mathematical Competition | [
"Precalculus > Trigonometric functions",
"Calculus > Differential Calculus > Derivatives"
] | English | proof and answer | 33 | |
09ed | Every cell of the board $3 \times 3$ is coloured either by red or blue. Find the number of all colourings in which there are no $2 \times 2$ square in which all cells are red. | [
"| | | |\n|---|---|---|\n| 1 | 2 | 3 |\n| 4 | 5 | 6 |\n| 7 | 8 | 9 |\n\nDenote $A_i$, $i = 1, 2, 4, 5$ colourings such that there is at least one red square left up corner of which at one of $1, 2, 4, 5$. Therefore by inclusion-exclusion principle number of such colourings equals to\n$$\n|A| = |A_1| + |A_2| +... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof and answer | 417 | |
0dpv | Let $M$ be the midpoint of side $AB$ in triangle $ABC$. $B_1$ is a point on segment $AC$ such that $CB = CB_1$. The circumcircles of triangles $ABC$ and $BMB_1$, $\omega$ and $\omega_1$, intersect for the second time at point $K$. Let $Q$ be the midpoint of arc $ACB$ of $\omega$. Lines $B_1Q$ and $BC$ intersect at poin... | [
"Let the bisector of $\\angle ACB$ intersect $\\omega$ for the second time at point $N$. Note that $N$ is the midpoint of arc $AB$ (that does not contain $C$) of $\\omega$. Also, it is easy to see that line $CN$ is the perpendicular bisector of segment $BB_1$. Thus, $NA = NB = NB_1$, i.e. points $A, B$ and $B_1$ li... | Silk Road Mathematics Competition | SILK ROAD MATHEMATICS COMPETITION XX | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0cxk | Points $M$ and $N$ are considered in the interior of triangle $ABC$ such that $\widehat{MAB} = \widehat{NAC}$ and $\widehat{MBA} = \widehat{NBC}$. Prove that
$$
\frac{AM \cdot AN}{AB \cdot AC} + \frac{BM \cdot BN}{BA \cdot BC} + \frac{CM \cdot CN}{CA \cdot CB} = 1.
$$ | [
"\n\nLet $K$ be a point on the ray $BN$ such that $\\widehat{BMA} = \\widehat{BCK}$. It is clear that $K$ is outside of triangle because $\\widehat{BMA} > \\widehat{ACB}$. We have $\\triangle ABM \\sim \\triangle KBC$, hence\n$$\n\\frac{AB}{BK} = \\frac{BM}{BC} = \\frac{AM}{CK} . \\tag{1}\n... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry... | English | proof only | null | |
06it | Let $f(x) = x^6 - x^5 - x^3 - x^2 - x$ and $g(x) = x^4 - x^3 - x^2 - 1$. If $a, b, c, d$ are the four roots of the equation $g(x) = 0$, find the value of $f(a) + f(b) + f(c) + f(d)$. | [
"Since $a$ is a root of $g(x) = 0$, we have $a^4 - a^3 - a^2 - 1 = 0$. Using this relation, we get\n$$\nf(a) = a^6 - a^5 - a^3 - a^2 - a = (a^2+1)(a^4 - a^3 - a^2 - 1) + a^2 - a + 1 = a^2 - a + 1.\n$$\nOf course, the same also applies to $b, c$ and $d$. Therefore, we have\n$$\nf(a) + f(b) + f(c) + f(d) = (a^2 + b^2... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | 6 | |
0109 | Problem:
In a convex pentagon $A B C D E$, the sides $A E$ and $B C$ are parallel and $\angle A D E=\angle B D C$. The diagonals $A C$ and $B E$ intersect at $P$. Prove that $\angle E A D=\angle B D P$ and $\angle C B D=\angle A D P$. | [
"Solution:\n\n\nFigure 4\n\nLet $\\mathcal{C}_{1}$ and $\\mathcal{C}_{2}$ be the circumcircles of triangles $A E D$ and $B C D$, respectively. Let $D P$ meet $\\mathcal{C}_{2}$ for the second time at $F$ (see Figure 4). Since $\\angle A D E=\\angle B D C$, the ratio of the lengths of the se... | Baltic Way | Baltic Way 1998 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09q9 | Problem:
Voor een niet-negatief geheel getal $n$ noemen we een permutatie $\left(a_{0}, a_{1}, \ldots, a_{n}\right)$ van $\{0,1, \ldots, n\}$ kwadratisch als $k+a_{k}$ een kwadraat is voor $k=0,1, \ldots, n$. Bewijs dat er voor elke niet-negatieve gehele $n$ een kwadratische permutatie van $\{0,1, \ldots, n\}$ bestaat... | [
"Solution:\n\nWe bewijzen dit met inductie naar $n$. Voor $n=0$ werkt de permutatie $(0)$, want $0+0$ is een kwadraat.\n\nZij nu $l \\geq 0$ en neem aan dat er voor elke $n \\leq l$ een kwadratische permutatie bestaat (de inductiehypothese). We bekijken $n=l+1$. Zij $m$ zodat $m^{2}$ het kleinste kwadraat groter da... | Netherlands | Dutch TST | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Other"
] | null | proof only | null | |
0hyh | Problem:
Finitely many cards are placed in two stacks, with more cards in the left stack than the right. Each card has one or more distinct names written on it, although different cards may share some names. For each name, we define a "shuffle" by moving every card that has this name written on it to the opposite stac... | [
"Solution:\n\nLet the number of cards be $c$ and let the number of distinct names be $n$. Each card contains a set of names; denote these sets by $S_{1}, S_{2}, \\ldots, S_{c}$ (some of these sets may share elements). Now let $E$ be a subset of the set of $n$ names, and denote by $D(E)$ the difference of the number... | United States | 1st Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0k3l | Problem:
A row of fifty coins with integer denominations is given, such that the sum of the denominations is odd. Alice and Bob alternate taking either coin at the left end of the row or the right end of the row, with Alice playing first. Prove that Alice can always ensure she gets more than half the money. | [
"Solution:\n\nColor the coins alternatively black and white. Since $50$ is even, on Alice's turn, the coins at either end of the row are different colors.\n\nThus Alice could guarantee getting all of the black coins, she could also guarantee getting all of the white coins. Since either the sum of the black coins is... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
09ri | Problem:
Gegeven is een onbekende rij $a_{1}, a_{2}, a_{3}, \ldots$ van gehele getallen die voldoet aan de volgende eigenschap: voor elk priemgetal $p$ en elk positief geheel getal $k$ geldt
$$
a_{p k+1}=p a_{k}-3 a_{p}+13
$$
Bepaal alle mogelijke waarden van $a_{2013}$. | [
"Solution:\n\nLaat $q$ en $t$ priemgetallen zijn. Vul in $k=q, p=t$ :\n$$\na_{q t+1}=t a_{q}-3 a_{t}+13\n$$\nVul ook in $k=t, p=q$ :\n$$\na_{q t+1}=q a_{t}-3 a_{q}+13\n$$\nBeide uitdrukkingen rechts zijn dus gelijk aan elkaar, waaruit volgt\n$$\nt a_{q}-3 a_{t}=q a_{t}-3 a_{q}\n$$\noftewel\n$$\n(t+3) a_{q}=(q+3) a_... | Netherlands | MO-selectietoets | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 2016 | |
039o | Some of the squares of an $n \times n$ table are mined. In each square the number of the mined squares amongst this square and its neighbors (i.e. those which have common side or vertex with it) is written. Is it always possible to determine which squares are mined if:
a) $n = 2000$;
b) $n = 2007$? | [
"We denote the rows by $i = 1, \\dots, n$ and the columns by $j = 1, \\dots, n$ and let $a(i; j)$ be the number written in the square $(i; j)$.\n\na) No! Consider the table $A$ where the squares $(i; j)$ are mined if and only if $i \\equiv j \\equiv 1 \\pmod 3$ and the table $B$ where the squares $(i; j)$ are mined... | Bulgaria | Winter Mathematical Competition | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) No; b) Yes | |
0cik | Find the continuous and surjective functions $f : \mathbb{R} \to \mathbb{R}$ with the following two properties:
(i) $2f(1) = f(0) + f(2)$;
(ii) for any real numbers $a, b, c$, if $f(a), f(b), f(c)$ are terms of an arithmetic progression then $a, b, c$ are also terms of an arithmetic progression. | [] | Romania | 75th NMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | All functions f(x) = ax + b with a ≠ 0. | |
0bto | Find all positive integers $a$ and $b$ so that $\frac{a+1}{b}$ and $\frac{b+2}{a}$ are simultaneously positive integers. | [
"Since the fractions $f_1 = \\frac{a+1}{b}$ and $f_2 = \\frac{b+2}{a}$ are positive integers, $a+1 \\ge b$ and $b+2 \\ge a$, hence $a+1 \\ge b \\ge a-2$. This leaves the cases:\n\n1. $b = a + 1$: then $f_2 = 1 + \\frac{3}{a} \\in \\mathbb{N}$, whence $a = 1, b = 2$ or $a = 3, b = 4$, which are convenient values.\n\... | 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 | (a,b) ∈ {(1,1), (1,2), (3,1), (3,4), (5,3)} | |
0czn | Find all positive integers $n$ such that $27^{n}-2^{n}$ is a perfect square. | [
"For $n=1$ we have $27-2=25=5^{2}$. We will prove that there are no other positive integers with this property.\nIf $n$ is odd, $n=2k+1$, then\n$$\n\\begin{aligned}\n27^{n}-2^{n} &= 27^{2k+1}-2^{2k+1} = (28-1)^{2k+1}-2 \\cdot 4^{k} \\\\\n&= 4m+(-1)^{2k+1} = 4m-1\n\\end{aligned}\n$$\nIt is easy to see that integers ... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 1 | |
02x5 | Problem:
a) Um campeonato terá 7 competidores. Cada um deles jogará exatamente um jogo contra todos os outros. Qual o total de jogos do campeonato?
b) Um campeonato terá $n$ competidores e cada um deles jogará exatamente um jogo contra todos os outros. Verifique que o total de jogos é $\frac{n \cdot (n-1)}{2}$.
c) Um... | [
"Solution:\na) Sejam $\\{a, b, c, d, e, f, g\\}$ os jogadores. O jogador $a$ participará dos jogos associados aos 6 pares: $(a, b), (a, c), (a, d), (a, e), (a, f)$ e $(a, g)$. A princípio, como já contamos o jogo associado a $(a, b)$, para contar as partidas do jogador $b$, basta contabilizarmos as partidas associa... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) 21. b) n(n−1)/2. c) Every such tournament has a master; for example, the player with the most wins is a master. d) Yes, João’s only game was against Maria; in fact there were 10 other competitors. | |
07fz | Alice and Bob take turns alternatively on a $2020 \times 2020$ board with Alice starting the game. In each move every person colors a cell that has not been colored yet and will be rewarded with as many points as the colored cells in the same row and column. When the table is colored completely, the points determine th... | [
"We claim that Bob has a winning strategy and the maximum point difference he can make sure will happen is $\\frac{2020^2}{2}$.\n\nFirst we show Bob's strategy. Let $\\ell$ be the vertical line that dissects the table into two equal tables. After Alice colors a cell, Bob can easily color the cell symmetric to that ... | Iran | 37th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Bob; maximum guaranteed point difference 2020^2/2 | |
0g9r | Let $O$ be the circumcenter of triangle $ABC$, and $\omega$ be the circumcircle of triangle $BOC$. Line $AO$ intersects with circle $\omega$ again at the point $G$. Let $M$ be the midpoint of side $BC$, and the perpendicular bisector of $BC$ meets circle $\omega$ at the points $O$ and $N$.
Prove that the midpoint of t... | [
"令 $H$ 為 $A$ 到 $BC$ 的垂足。易見 $AN$、$AO$ 直徑圓與圓 $BOC$ 共點於一點 $S$。設 $AM$ 交圓 $OMG$ 於另一點 $V$, $GV$ 交圓 $BOC$ 於另一點 $U$, 且交 $AN$ 於點 $T$。因為 $\\angle NUG = \\angle MOG = \\angle MVU$, 也就是 $UN \\parallel AV$, 所以\n$$\n\\angle TGA = \\angle AMO = \\angle HAM = \\angle SAO = \\angle TAG\n$$\n(因為 $AN$ 是共軛中線, 故 $AM$, $AN$ 等角共軛, $AH$, ... | Taiwan | 二〇一六數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incente... | null | proof only | null | |
039z | Find all values of the real parameter $a$ such that the equation $\sin 2x \sin 4x - \sin x \sin 3x = a$ has a unique solution in the interval $[0, \pi)$. | [
"Let us analyze the equation:\n$$\n\\sin 2x \\sin 4x - \\sin x \\sin 3x = a\n$$\nfor $x \\in [0, \\pi)$.\n\nFirst, use the product-to-sum formulas:\n$$\n\\sin A \\sin B = \\frac{1}{2}[\\cos(A-B) - \\cos(A+B)]\n$$\nSo,\n$$\n\\sin 2x \\sin 4x = \\frac{1}{2}[\\cos(2x - 4x) - \\cos(2x + 4x)] = \\frac{1}{2}[\\cos(-2x) -... | Bulgaria | Fall Mathematical Competition | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | English | proof and answer | 1 | |
0byx | Consider $A \in M_n(\mathbb{C})$ with $n \ge 2$ such that $\det A = 0$ and denote by $A^*$ its adjutant. Show that $(A^*)^2 = (\text{tr} A^*) A^*$. | [
"Since $\\det A = 0$, we have $\\text{rang}(A) \\le n-1$. If $\\text{rang}(A) \\le n-2$, then $A^* = O_n$ and $(A^*)^2 = O_n$ (*).\n\nIf $\\text{rang}(A) = n-1$, from $AA^* = (\\det A)I_n = O_n$, results $0 \\ge \\text{rang}(A) + \\text{rang}(A^*) - n$, therefore $\\text{rang}(A^*) \\le 1$.\n\nIf $\\text{rang}(A^*)... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
0esl | How many integers between $100$ and $1000$ are multiples of $7$?
(A) $120$ (B) $125$ (C) $128$ (D) $132$ (E) $140$ | [
"The multiples of $7$ include $98$ ($= 7 \\times 14$), $105$, $112$, $\\ldots$, $994$ ($= 7 \\times 142$), $1001$, $\\ldots$ Of these, $142 - 14 = 128$ are between $100$ and $1000$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | C | |
04ng | In a quadrilateral $ABCD$ it holds that $\angle DBC = \angle DCB = 50^\circ$ and $\angle DAB = \angle ABC = \angle BDC$. Prove that $AC \perp BD$. (Ratko Višak) | [] | Croatia | Croatia_2018 | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
0d9f | A convex polygon is divided into some triangles. Let $V$ and $E$ be respectively the set of vertices and the set of edges of all triangles (each vertex in $V$ may be some vertex of the polygon or some point inside the polygon). The polygon is said to be good if the following conditions hold:
i. There are no 3 vertices... | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof and answer | Exactly the convex polygons whose number of sides is divisible by three. | |
05ad | Does there exist a positive integer $n$ such that
$$
1950^n + 1934^n = 2024^n?
$$ | [
"Answer: No.\n\nThe numbers $1950$, $1934$, and $2024$ give remainders $4$, $2$, and $1$, respectively, when divided by $7$. Raising $4$ to powers $n = 1, 2, 3, 4, 5, 6, \\dots$ results in remainders $4, 2, 1, 4, 2, 1, \\dots$, and raising $2$ to the same powers results in remainders $2, 4, 1, 2, 4, 1, \\dots$. Thu... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | No | |
0c9b | A *lattice point* in the Cartesian plane is a point whose coordinates are both integral. A *lattice polygon* is a polygon whose vertices are lattice points. Let $\Gamma$ be a convex lattice polygon. Prove that $\Gamma$ is contained in a convex lattice polygon $\Delta$ exactly one vertex of which is not a vertex of $\Ga... | [
"Let $T$ be the extra vertex of a desired polygon $\\Delta$; then $\\Delta$ is the convex hull of $T$ and $\\Gamma$. Thus, a point $T$ fits the bill if and only if this convex hull contains no vertices of $\\Gamma$ in its interior.\n\nEach segment $AB$ joining two lattice points is partitioned by lattice points int... | Romania | Romanian Master of Mathematics | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
0aw7 | Problem:
Find the last two digits of $2^{100}$. | [
"Solution:\n\nNote that $2^{12} \\equiv 96 \\pmod{100} \\equiv -4 \\pmod{100}$. Thus, $2^{100} \\equiv (2^{12})^{8} (2^{4}) \\pmod{100} \\equiv (-4)^{8} 2^{4} \\pmod{100} \\equiv 2^{20} \\pmod{100} \\equiv (-4) 2^{8} \\pmod{100} \\equiv 76 \\pmod{100}$"
] | Philippines | 18th PMO National Stage Oral Phase | [
"Number Theory > Modular Arithmetic"
] | null | final answer only | 76 | |
0186 | The non-negative real numbers $a$, $b$, $c$ satisfy $a + b + c = 1$. What is the largest possible value of
$$
a^2b + ab^2 + b^2c + bc^2 + a^2c + ac^2?
$$ | [
"The largest possible value is $\\frac{1}{4}$, it is obtained (for example) when $a = b = \\frac{1}{2}$ and $c = 0$.\nFirst rewrite the expression:\n$$\n\\begin{aligned}\na^2b + ab^2 + b^2c + bc^2 + a^2c + ac^2 &= ab(1-c) + bc(1-a) + ac(1-b) \\\\\n&= ab + bc + ac - 3abc \\\\\n&= ab(1-3c) + c(1-c) .\n\\end{aligned}\... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 1/4 | |
0e5k | Prove that for any positive real numbers $a$, $b$ and $c$ the following holds:
$$
a + \sqrt{ab} + \sqrt[3]{abc} \le \frac{4}{3}(a + b + c).
$$ | [
"Recall that the inequality of arithmetic and geometric means states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list of numbers. From this we get\n$$\na + \\sqrt{ab} + \\sqrt[3]{abc} = a + \\sqrt{\\frac{a}{2} \\cdot 2b} + \\sqrt[3]{\\fr... | Slovenia | Selection Examinations for the IMO 2012 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null |
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