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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0kla | Problem:
In a $3$ by $3$ grid of unit squares, an up-right path is a path from the bottom left corner to the top right corner that travels only up and right in steps of $1$ unit. For such a path $p$, let $A_{p}$ denote the number of unit squares under the path $p$. Compute the sum of $A_{p}$ over all up-right paths $p... | [
"Solution:\n\nEach path consists of $3$ steps up and $3$ steps to the right, so there are $\\binom{6}{3} = 20$ total paths. Consider the sum of the areas of the regions above all of these paths. By symmetry, this is the same as the answer to the problem. For any path, the sum of the areas of the regions above and b... | United States | HMMT Spring 2021 Guts Round | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 90 | |
0gkc | PA and PB be the tangents to circle $\omega$ from an external point $P$. Let $M$ and $N$ be the midpoints of $AP$ and $AB$, respectively. Extend $MN$ to meet $\omega$ at $C$, where $N$ is between $M$ and $C$. $PC$ meets $\omega$ at $D$ and extend $ND$ to intersect $PB$ at $Q$. Show that $MNQP$ is a rhombus. | [
"Observe that $AB \\perp NP$. Thus, $M$ is the circumcenter of $\\triangle ANP$ and hence $MN = MP$.\n\nIt can also be seen that $MN \\parallel PQ$.\n\nFrom the power of the point $M$, $PM^2 = MA^2 = ME \\cdot MC$.\n\nSo, $\\frac{PM}{ME} = \\frac{MC}{PM}$ and hence $\\triangle PME \\sim \\triangle CMP$.\n\nThus, $M... | Thailand | Thai Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centro... | English | proof only | null | |
0gnk | Let $k > 1$ be an integer and $p = 6k + 1$ be a prime number. Prove that for each $m = 2^p - 1$
$$
\frac{2^m - 1}{127m}
$$
is an integer. (Şahin Emrah). | [
"Let us show that both $m$ and $127$ divide $2^{m-1} - 1$. By Fermat's little theorem $2^p \\equiv 2 \\pmod{p}$, so $m = 2^p - 1 \\equiv 1 \\pmod{p}$, which implies $p \\mid m-1$. Therefore, $2^p - 1 \\mid 2^{m-1} - 1$, so $m \\mid 2^{m-1} - 1$.\n\nOn the other hand, $6 \\mid p-1$ so $63 = 2^6 - 1 \\mid 2^{p-1} - 1... | Turkey | Turkish Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0hgq | Find the smallest positive integer $n$ for which it's possible to cut the square into $2n$ squares of two different sizes: $n$ squares of one size and $n$ more of other size.
 | [
"The example of cutting into $2n = 18$ squares is provided in the fig. 10.\n\nSuppose that there exists an example for $n \\le 8$. Denote the sides of the squares by $a, b$, and of the large square as $N$. Let's first show that it's possible to represent $N$ as the sum $ka + bl$, where $k, l \\le n$, at least in tw... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residue... | English | proof and answer | 9 | |
0gir | Suppose $O$ is the circumcenter of $\triangle ABC$ and $E$, $F$ are points on segments $CA$ and $AB$ respectively with $E, F \ne A$. Let $P$ be a point such that $PB = PF$ and $PC = PE$. Let $OP$ intersect $CA$ and $AB$ at points $Q$ and $R$ respectively. Let the line passing through $P$ and perpendicular to $EF$ inter... | [
"**解法一：**令 $A^*$ 為 $A$ 關於 $\\odot(ABC)$ 的對徑點, $M, N$ 分別為 $\\overline{A^*E}$, $\\overline{A^*F}$ 的中點, 則 $\\overline{AC} \\perp \\overline{A^*C}$ 且 $\\overline{AB} \\perp \\overline{A^*B}$ 因此 $M, N$ 分別位於 $\\overline{CE}$, $\\overline{BF}$ 的中垂線上。由 $\\angle OMP = \\angle ONP = 90^\\circ$, 我們知道 $M, N, O, P$ 共圓。透過 $MN$ 平... | Taiwan | Taiwan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Adv... | Chinese; English | proof only | null | |
03rr | For each positive integer, define a function
$$
f(n) = \begin{cases} 0, & \text{if } n \text{ is the square of an integer,} \\ \left\lfloor \frac{1}{\sqrt{n}} \right\rfloor, & \text{if } n \text{ is not the square of an integer.} \end{cases}
$$
(Here $[x]$ denotes the maximum integer not exceeding $x$, and
$\{x\} = x -... | [
"For arbitrary $a, k \\in \\mathbb{N}^+$, if $k^2 < a < (k+1)^2$, we set\n$$\na = k^2 + m, \\quad m = 1, 2, \\dots, 2k,\n$$\n$$\n\\sqrt{a} = k + \\theta, \\quad 0 < \\theta < 1,\n$$\n$$\n\\left[ \\frac{1}{\\sqrt{a}} \\right] = \\left[ \\frac{1}{\\sqrt{a-k}} \\right] = \\left[ \\frac{\\sqrt{a+k}}{a-k^2} \\right] = \... | China | China Mathematical Competition (Extra Test) | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 768 | |
0eqd | Each interior angle of a regular 180 sided polygon is equal to
(A) 175° (B) 180° (C) 157.5° (D) 120° (E) 178° | [
"It is easier to work with exterior angles, since the sum of the exterior angles of a polygon is $360°$. There are $180$ exterior angles, which are all equal since the polygon is regular. Thus each exterior angle is equal to $360°/180 = 2°$, and each interior angle is equal to $180° - 2° = 178°$."
] | South Africa | South African Mathematics Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | MCQ | E | |
0khw | The graph of $f(x) = |\lfloor x \rfloor| - |\lfloor 1-x \rfloor|$ is symmetric about which of the following? (Here $\lfloor x \rfloor$ is the greatest integer not exceeding $x$.)
(A) the $y$-axis (B) the line $x = 1$ (C) the origin (D) the point $(\frac{1}{2}, 0)$ (E) the point $(1, 0)$ | [] | United States | AMC 10 A | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | MCQ | D | |
0i6j | Problem:
Determine the value of
$$
1 \cdot 2 - 2 \cdot 3 + 3 \cdot 4 - 4 \cdot 5 + \cdots + 2001 \cdot 2002.
$$ | [
"Solution:\n$2004002$. Rewrite the expression as\n$$\n2 + 3 \\cdot (4 - 2) + 5 \\cdot (6 - 4) + \\cdots + 2001 \\cdot (2002 - 2000)\n$$\n$$\n= 2 + 6 + 10 + \\cdots + 4002.\n$$\nThis is an arithmetic progression with $(4002 - 2)/4 + 1 = 1001$ terms and average $2002$, so its sum is $1001 \\cdot 2002 = 2004002$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2004002 | |
05hy | Problem:
Exercice 9. Dans tout l'exercice, on s'intéresse à des tableaux $5 \times 5$, dont les lignes (resp. les colonnes) sont notées $L_{0}, L_{1}, L_{2}, L_{3}, L_{4}$ du bas vers le haut (resp. $C_{0}, C_{1}, C_{2}, C_{3}, C_{4}$, de gauche à droite).
Dans chacune des 25 cases d'un tel tableau, on écrit un nombre... | [
"Solution:\n\nOn se donne un éventuel tableau cohérent. Commençons par dégager quelques unes de ses propriétés.\n\na) Pour tous $i, j$, on a $f(i, j) \\in \\{0,1,2,3,4,5\\}$.\nEn effet, dans un tableau cohérent, chaque $f(i, j)$ représente le nombre de fois où $i$ apparaît sur $L_{j}$, donc $f(i, j)$ est un entier ... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | [[[1, 0, 0, 0, 0], [2, 1, 0, 0, 0], [0, 1, 0, 1, 0], [1, 3, 2, 1, 1], [1, 0, 3, 3, 4]], [[2, 1, 0, 0, 0], [1, 0, 0, 0, 0], [0, 0, 0, 0, 1], [1, 4, 1, 1, 1], [1, 0, 4, 4, 3]]] | |
036q | Problem:
Tokens are placed in some of the cells of a table of size $8 \times 8$ such that:
(1) there is at least one token in any rectangle of size $2 \times 1$ and $1 \times 2$;
(2) there are two neighboring tokens in any rectangle of size $7 \times 1$ and $1 \times 7$.
Find the minimum possible number of tokens. | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof and answer | 42 | |
0jml | Problem:
The numbers $2^{0}, 2^{1}, \cdots, 2^{15}, 2^{16}=65536$ are written on a blackboard. You repeatedly take two numbers on the blackboard, subtract one from the other, erase them both, and write the result of the subtraction on the blackboard. What is the largest possible number that can remain on the blackboar... | [
"Solution:\n\nAnswer: $131069$\n\nIf we reverse the order of the numbers in the final subtraction we perform, then the final number will be negated. Thus, the possible final numbers come in pairs with opposite signs. Therefore, the largest possible number is the negative of the smallest possible number.\n\nTo get t... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 131069 | |
060m | Problem:
Soit $n \geqslant 2$ et $a_{1}, \ldots, a_{n} \in [0,1]$ des réels. Déterminer la valeur maximale que peut prendre le plus petit des nombres
$$
a_{1}-a_{1} a_{2},\ a_{2}-a_{2} a_{3},\ \ldots,\ a_{n}-a_{n} a_{1}
$$ | [
"Solution:\n\nSoit $i$ tel que $a_{i}$ soit minimal. Alors on a $a_{i+1} \\geqslant a_{i}$ (où $a_{n+1}=a_{1}$) et donc par inégalité arithmético-géométrique sur $a_{i}$ et $1-a_{i}$, on obtient\n$$\na_{i}-a_{i} a_{i+1} \\leqslant a_{i}-a_{i}^{2}=a_{i}(1-a_{i}) \\leqslant \\frac{1}{4}(a_{i}+1-a_{i})^{2}=\\frac{1}{4... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1/4 | |
0by4 | Let $E(x, y) = \frac{x}{y} + \frac{x+1}{y+1} + \frac{x+2}{y+2}$.
a) Solve in positive integers the equation $E(x, y) = 3$.
b) Prove that there exist infinitely many positive integers $n$ for which the equation $E(x, y) = n$ has solutions in positive integers. | [
"a) Rewrite the equation as\n$$\n\\left(\\frac{x}{y} - 1\\right) + \\left(\\frac{x+1}{y+1} - 1\\right) + \\left(\\frac{x+2}{y+2} - 1\\right) = 0,\n$$\nwhich is equivalent to $(x - y) \\left(\\frac{1}{y} + \\frac{1}{y+1} + \\frac{1}{y+2}\\right) = 0$, and the solutions are all pairs $(k, k)$, with $k \\in \\mathbb{N... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | a) All solutions are x = y with x, y positive integers. b) For every positive integer k, n = 11k + 3 has a solution with (x, y) = (6k + 1, 1), so infinitely many n of the form 11k + 3 work. | |
0g6c | 試證: 在等差數列 $11, 21, 31, 41, 51, 61, \ldots$ 中, 存在無窮多個質數。 | [
"考慮整數 $N = (11 \\cdot 31 \\cdot 41 \\cdot 61 \\cdots p_n)^5 - 1$。則 $N$ 與 $11, 41, \\cdots p_n$ 互質。令 $a = 11 \\cdot 31 \\cdot 41 \\cdot 61 \\cdots p_n$，則\n$$\nN = a^5 - 1 = (a-1)(a^4 + a^3 + a^2 + a + 1).\n$$\n顯然, $2 \\nmid (a^4 + a^3 + a^2 + a + 1)$, 但 $5 \\nmid (a^4 + a^3 + a^2 + a + 1)$。令 $p \\neq 5$ 是 $a^4 + a^3... | Taiwan | 二〇一二數學奧林匹亞競賽第二階段選訓營 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
0g9v | 設△ABC為銳角三角形，其中∠B≠∠C。設M為BC邊中點，E、F分別為過B、C點的高的垂足。令K、L分別為線段ME、MF中點。在直線KL上取一點T使得AT//BC。
證明：$TA = TM$。
Let $\triangle ABC$ be an acute-angled triangle, with $\angle B \neq \angle C$. Let $M$ be the midpoint of side $BC$, and $E$, $F$ be the feet of the altitudes from $B$, $C$, respectively. Denote by $K$, $L$ the midpoints ... | [
"不失一般性，設 $AB > AC$。作 $\\triangle AEF$ 的外接圓 $\\omega$。\n\n*Lemma 1.* 直線 $ME$、直線 $MF$ 以及直線 $AT$ 都是圓 $\\omega$ 的切線。\n\n\n\n*Proof.* 注意到 $E$, $F$ 兩點落在以線段 $BC$ 為直徑的圓上，且該圓圓心為 $M$。所以 $MC = ME = MF = MB$。於是\n$$\n\\begin{aligned}\n\\angle FEM &= \\angle FEB + \\angle BEM = \\angle FCB + \\angle MBE ... | Taiwan | 二〇一六數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05dw | Problem:
A permutation of the integers $1,2, \ldots, m$ is called fresh if there exists no positive integer $k < m$ such that the first $k$ numbers in the permutation are $1,2, \ldots, k$ in some order. Let $f_{m}$ be the number of fresh permutations of the integers $1,2, \ldots, m$.
Prove that $f_{n} \geq n \cdot f_... | [
"Solution:\n\nLet $\\sigma = (\\sigma_{1}, \\ldots, \\sigma_{n-1})$ be a fresh permutation of the integers $1,2, \\ldots, n-1$. We claim that for any $1 \\leq i \\leq n-1$ the permutation\n$$\n\\sigma^{(i)} = (\\sigma_{1}, \\ldots, \\sigma_{i-1}, n, \\sigma_{i}, \\ldots, \\sigma_{n-1})\n$$\nis a fresh permutation o... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2020 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
05x8 | Problem:
Trouver toutes les fonctions $f: \mathbb{N}^* \rightarrow \mathbb{N}^*$ telles que
$$
f^{f(n)}(m) + m n = f(m) f(n)
$$
pour tous $m, n \in \mathbb{N}^*$. | [
"Solution:\nEn $m=n$, on trouve que $f(m)^2 = f^{f(m)}(m) + m^2 > m^2$, ce qui donne $f(m) > m$ car $f(m)$ est positif, et donc comme $f(m)$ est entier, $f(m) \\geqslant m+1$. Par une récurrence simple, on trouve que pour tous les entiers $k$, $m$ strictement positifs, $f^{k}(m) \\geqslant m+k$.\n\nMais on remarque... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | f(n) = n + 1 for all positive integers n | |
0kr7 | Problem:
A random permutation of $\{1,2, \ldots, 100\}$ is given. It is then sorted to obtain the sequence $(1,2, \ldots, 100)$ as follows: at each step, two of the numbers which are not in their correct positions are selected at random, and the two numbers are swapped. If $s$ is the expected number of steps (i.e. swap... | [
"Solution:\nLet $f(n)$ be the expected number of steps if there are $n$ elements out of order. Let's consider one of these permutations and suppose that $a$ and $b$ are random elements that are out of order. The probability that swapping $a$ and $b$ sends $a$ to the proper place is $\\frac{1}{n-1}$, and the probabi... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 2425 | |
07k6 | In a scalene triangle $ABC$, points $X$ and $Y$ lie on $BC$ and $AC$, respectively, such that $BC \perp XY$. Points $T$ and $Z$ are the reflections of $Y$ and $X$ with respect to the midpoints of sides $AC$ and $BC$, respectively. Point $P$ lies on $ZT$ such that the circumcenter of triangle $XZP$ coincides with the ci... | [
"Let $O$ be the center of circumcircle of triangle $XPZ$ and let the line $XY$ intersect this circle for a second time at point $Q$. Suppose that $AA'$, $CC'$ are diameters of the circumcircle of triangle $ABC$.\n\nNote that points $C'$, $Q$ are the reflections of points $B$, $X$ with respect to the line passing th... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quad... | null | proof only | null | |
04em | There were $n$ students participating in a competition and each student solved exactly three problems. For any two students there is exactly one problem which they both solved, while each problem was solved by exactly $k$ students. For which positive integers $n$ and $k$ is that possible?
(Moscow olympiad 1947) | [
"For $k=1$ each problem was solved by only one student. That means that there are no other students, due to the condition that for any two students there is exactly one problem which they both solved.\n\nNow let $k \\ge 2$.\nLet $A$ be one of the students. He solved exactly three problems, but each of these three p... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | [(1, 1), (4, 2), (7, 3)] | |
0kj3 | Problem:
Let $f(x) = x^{3} - 3x$. Compute the number of positive divisors of
$$
\left\lfloor f\left(f\left(f\left(f\left(f\left(f\left(f\left(f\left(\frac{5}{2}\right)\right)\right)\right)\right)\right)\right)\right)\right\rfloor
$$
where $f$ is applied 8 times. | [
"Solution:\nNote that $f\\left(y + \\frac{1}{y}\\right) = \\left(y + \\frac{1}{y}\\right)^{3} - 3\\left(y + \\frac{1}{y}\\right) = y^{3} + \\frac{1}{y^{3}}$.\n\nThus, $f\\left(2 + \\frac{1}{2}\\right) = 2^{3} + \\frac{1}{2^{3}}$, and in general $f^{k}\\left(2 + \\frac{1}{2}\\right) = 2^{3^{k}} + \\frac{1}{2^{3^{k}}... | United States | HMMT Spring 2021 Guts Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 6562 | |
0er7 | Prove that there are infinitely many terms of the arithmetic sequence $1, 14, 27, 40, \dots, 1+13k, \dots$ which are of the form $222\ldots22$. In other words a number that is made up using only the digit $2$.
(Hint: $1001 = 7 \times 11 \times 13$) | [
"Note that $221 = 13 \\times 17$ and therefore $222 = 13 \\times 17 + 1$ is in the arithmetic sequence. Furthermore, $222\\,222\\,221$ is divisible by $13$ and hence $222\\,222\\,222$ is in the sequence, since\n$$\n222222 = 222 \\times 1001 = 222 \\times 7 \\times 11 \\times 13.\n$$\nIn fact, any number consisting ... | South Africa | South African Mathematics Olympiad Third Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0drc | In an acute-angled triangle $ABC$, $D$ is the point on $BC$ such that $AD$ bisects $\angle BAC$, $E$ and $F$ are the feet of the perpendiculars from $D$ onto $AB$ and $AC$ respectively. The segments $BF$ and $CE$ intersect at $K$. Prove that $AK$ is perpendicular to $BC$. | [
"\n\nLet the extension of $AK$ intersect $BC$ at $N$. Note that $AE = AF$ and $BD : DC = c : b$, where $b = AC$ and $c = AB$. The cevians $AN$, $BF$, $CE$ concur at $K$. By Ceva's theorem, we have $(BN/NC)(CF/FA)(AE/EB) = 1$. Thus $BN/NC = EB/CF$. On the other hand, $EB = BD \\cos B$ and $C... | Singapore | Singapur 2015 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0gyi | The sequence $(u_n)$ is defined by conditions: $u_0 = 0$, $u_1 = \frac{1}{3}$ and $\frac{1}{2} u_n = \frac{1}{2}(u_{n+1} + u_{n-1})$. Prove that for all positive integers $n$ it holds $|u_n| \le 1$. | [
"Let an angle $\\varphi \\in (0, \\frac{\\pi}{2})$ such that $\\sin \\varphi = \\frac{\\sqrt{5}}{3}$. Then $\\cos \\varphi = \\frac{2}{3}$ and from the recurrent relation we have: $u_{n+1} = 2u_n \\cos \\varphi - u_{n-1}$. Moreover, $u_1 = \\sin \\varphi$, $u_2 = \\frac{2}{\\sqrt{5}} \\sin \\varphi \\cos \\varphi =... | Ukraine | The Problems of Ukrainian Authors | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0d0g | A subset of $\{1, 2, 3, \dots, 30\}$ is called *delicious* if it does not contain elements $a$ and $b$ satisfying $a = 3b$. A delicious subset is called *super delicious* if it is delicious and no delicious set has more elements than it has. Determine the number of super delicious subsets. | [
"Partition the set $\\{1, 2, 3, \\dots, 30\\}$ into 20 subsets as follows:\n$$\n\\begin{aligned}\n& \\{1, 3, 9, 27\\}, \\\\\n& \\{2, 6, 18\\}, \\\\\n& \\{4, 12\\}, \\{5, 15\\}, \\{7, 21\\}, \\{8, 24\\}, \\{10, 30\\}, \\\\\n& \\{11\\}, \\{13\\}, \\{14\\}, \\{16\\}, \\{17\\}, \\{19\\}, \\{20\\}, \\\\\n& \\{22\\}, \\{... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 96 | |
0i7u | Problem:
Convex quadrilateral $M A T H$ is given with $H M / M T = 3 / 4$, and $\angle A T M = \angle M A T = \angle A H M = 60^{\circ}$. $N$ is the midpoint of $M A$, and $O$ is a point on $T H$ such that lines $M T$, $A H$, $N O$ are concurrent. Find the ratio $H O / O T$. | [
"Solution:\n\n$\\triangle M A T$ is equilateral, so $H M / A T = H M / M T = 3 / 4$. Also, $\\angle A H M = \\angle A T M$, so the quadrilateral is cyclic. Now, let $P$ be the intersection of $M T$, $A H$, $N O$. Extend $M H$ and $N O$ to intersect at point $Q$. Then by Menelaus's theorem, applied to triangle $A H ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 9/16 | |
0fy0 | Problem:
Sei $\pi=\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ eine Permutation der Zahlen $1,2, \ldots, n$. Die Verwuselung von $\pi$ ist die Anzahl Paare $(i, j)$ natürlicher Zahlen mit $1 \leq i<j \leq n$ und $a_{j}>a_{i}$. Beweise, dass es für jede ganze Zahl $k$ mit $0 \leq k \leq\binom{ n}{2}$ eine Permutation der ... | [
"Solution:\n\nWir verwenden Induktion nach $n$, der Fall $n=1$ ist trivial.\n\n- Sei $0 \\leq k \\leq\\binom{ n-1}{2}$. Nach Induktionsvoraussetzung existiert eine Permutation $\\left(a_{1}, \\ldots, a_{n-1}\\right)$ der Zahlen $1,2, \\ldots, n-1$ mit Verwuselung $k$. Setzt man nun $a_{n}= n$, dann besitzt die Perm... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
05j0 | Problem:
Soit $ABC$ un triangle rectangle isocèle en $B$. Soit $M$ un point de l'arc $AC$ du cercle de centre $B$ passant par $A$ et $C$, $H$ son projeté orthogonal sur $(AB)$. On note $I$ le centre du cercle inscrit à $BHM$ et $J$ le centre du cercle exinscrit dans l'angle $B$ ($J$ est donc l'intersection de la bisse... | [
"Solution:\n\n\n\nOn observe d'abord que $(MI) \\perp (MJ)$ puisque dans un triangle, les bissectrices intérieure et extérieure en un point sont perpendiculaires.\n\nD'autre part, $\\widehat{MIJ} = 180^\\circ - \\widehat{BIM} = \\widehat{IBM} + \\widehat{BMI} = \\frac{1}{2}(\\widehat{HBM} +... | France | null | [
"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 | |
0ieg | Problem:
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a smooth function such that $f'(x) = f(1-x)$ for all $x$ and $f(0) = 1$. Find $f(1)$. | [
"Solution:\nDifferentiating the given equation gives $f''(x) = -f(x)$. This has solutions of the form $A \\cos(x) + B \\sin(x)$. Since $f(0) = 1$, $A = 1$. Then $f'(x) = B \\cos(x) - \\sin(x)$ and\n\n$$\n\\begin{aligned}\nf(1-x) & = \\cos(1-x) + B \\sin(1-x) \\\\\n& = \\cos 1 \\cos x + \\sin 1 \\sin x + B \\sin 1 \... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Equations > ODEs",
"Precalculus > Trigonometric functions"
] | null | proof and answer | sec 1 + tan 1 | |
0fia | Problem:
Demostrar que en un cuadrilátero convexo de área unidad, la suma de las longitudes de todos los lados y diagonales no es menor que $2(2+\sqrt{2})$. | [
"Solution:\n\nSea el cuadrilátero de lados $a, b, c, d$ y diagonales $p$ y $q$. Trazando las paralelas por cada vértice a la diagonal que no pasa por él se forma un paralelogramo de área $2$ y lados $p$ y $q$.\nPor el teorema isoperimétrico, de todos los paralelogramos de área $2$, el cuadrado tiene perímetro mínim... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Equations and Inequalities > ... | null | proof only | null | |
0k9r | Problem:
Consider sequences $a$ of the form $a = (a_1, a_2, \ldots, a_{20})$ such that each term $a_i$ is either $0$ or $1$. For each such sequence $a$, we can produce a sequence $b = (b_1, b_2, \ldots, b_{20})$, where
$$
b_i = \begin{cases}
a_i + a_{i+1} & i = 1 \\
a_{i-1} + a_i + a_{i+1} & 1 < i < 20 \\
a_{i-1} + a_... | [
"Solution:\n\nLet the two sequences be $b$ and $\\hat{b}$. Then, observe that given $a$, if $b_1 = \\hat{b}_1$ and $b_2 = \\hat{b}_2$, then $b = \\hat{b}$ (since $a$ will uniquely determine the remaining elements in $b$ and $\\hat{b}$). Thus, $b$ and $\\hat{b}$ must start with $(1, 0, \\ldots)$ and $(0, 1, \\ldots)... | United States | HMMT November 2019 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 64 | |
0070 | Ana y Beto juegan en un tablero de $11$ filas y $9$ columnas. Primero Ana divide el tablero en $33$ zonas. Cada zona está formada por $3$ casillas contiguas alineadas vertical u horizontalmente, como muestra la figura.
Luego, Beto escribe en cada casilla uno de los n... | [] | Argentina | XX Olimpiada Matemática del Cono Sur | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Spanish | proof only | null | |
0fwk | Problem:
Sei $ABC$ ein Dreieck mit $\Varangle BAC \neq 45^{\circ}$ und $\Varangle ABC \neq 135^{\circ}$. Sei $P$ der Punkt auf der Geraden $AB$ mit $\Varangle CPB = 45^{\circ}$. Seien $O_1$ und $O_2$ die Umkreismittelpunkte der Dreiecke $ACP$ und $BCP$. Zeige, dass die Fläche des Vierecks $CO_1PO_2$ gleich gross ist w... | [
"Solution:\n\nDa $O_1$ und $O_2$ auf der Mittelsenkrechten der Strecke $AP$ liegen, ist das Dreieck $AO_1O_2$ kongruent zum Dreieck $PO_2O_1$. Es genügt also zu zeigen, dass die Fläche des Dreiecks $AO_1O_2$ halb so gross ist wie die Fläche des Dreiecks $ABC$. Wir zeigen zuerst, dass die beiden Dreiecke ähnlich sin... | Switzerland | SMO Finalrunde | [
"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 | |
0ejd | Problem:
V pravokotnem trikotniku $ABC$ s pravim kotom pri $C$, višina iz oglišča $C$ seka stranico $AB$ v točki $D$. Kolikšna je velikost kota $\angle BAC$, če velja $|CD| = \sqrt{12}\ \mathrm{cm}$ in $|AD| = 4\ \mathrm{cm}$?
(A) $\alpha \doteq 49,06^\circ$
(B) $\alpha \doteq 40,89^\circ$
(C) $\alpha \doteq 40,54^\c... | [
"Solution:\n\nUporabimo kotne funkcije v pravokotnem trikotniku $\\tan \\alpha = \\frac{v_c}{b_1} = \\frac{\\sqrt{12}}{4}$ in izračunamo kot $\\alpha \\doteq 40,89^\\circ$. Pravilen je odgovor B."
] | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | MCQ | B | |
00me | Let $ABCDE$ be a regular pentagon with center $M$. A point $P \neq M$ is chosen on the line segment $MD$. The circumcircle of $ABP$ intersects the line segment $AE$ in $A$ and $Q$ and the line through $P$ perpendicular to $CD$ in $P$ and $R$.
*Prove that $AR$ and $QR$ are of the same length.* | [
"Let $S$ denote the common point of $RP$ and $AE$, see Figure 1. Since we are given a regular pentagon, the angles in triangle $ABE$ are well known as $\\angle BAE = 108^\\circ$ and $\\angle ABE = \\angle AEB = 36^\\circ$. Since $BE$ and $CD$ are parallel, $RP$ is perpendicular to $BE$, and we therefore have $\\ang... | Austria | 48th Austrian Mathematical Olympiad National Competition (Final Round, part 1) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08uk | Both $m$ and $n$ are 3-digit positive integers, and $m$ and $n$ differ only in one of the 3 digits. Also, $n$ is an integral multiple of $m$. How many possible such pairs $(m, n)$ are there? | [
"From $m \\neq n$ and the fact that $n$ is a multiple of $m$, it follows that $n \\geq 2m$. Since $m$ is an integer greater than or equal to $100$, we have $n - m \\geq m \\geq 100$, and therefore, we see that $n$ and $m$ differ on the hundred's digit. This means that there exists an integer $k$ ($1 \\leq k \\leq 8... | Japan | Japan Junior Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 22 | |
0hw1 | Problem:
Equilateral triangles $\triangle ABD$, $\triangle ACE$, and $\triangle BCF$ are drawn outside $\triangle ABC$ on each of its sides, with centers $G$, $H$, and $I$, respectively.
a) Prove that $\triangle GHI$ is equilateral.
b) Let $G'$, $H'$, and $I'$ be the reflections of $G$, $H$, and $I$ across $AB$, $BC... | [
"Solution:\n\n\n\na) Let $O$ be the intersection of circles $ABD$ and $ACE$ (besides $A$). Since opposite angles in a cyclic quadrilateral sum to $180^\\circ$, $\\angle AOB = \\angle AOC = 120^\\circ$, so $\\angle BOC = 120^\\circ$ as well. Thus, quadrilateral $BOCF$ is also cyclic. Since $... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Advanced Configurations > Napoleon and Fermat points",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Misce... | null | proof only | null | |
0eg0 | Problem:
Polona v pekarni redno kupuje žemlje in rogljičke. Med tednom so pri nakupu vsaj šestih rogljičkov le-ti $15 \%$ cenejši, med vikendom pa na celoten nakup priznajo $10 \%$ popust. V torek je Polona kupila 6 rogljičkov in 5 žemelj ter plačala $2,27 \text{ €}$. V soboto pa je kupila 7 rogljičkov, 4 žemlje in vr... | [
"Solution:\n\nNastavimo sistem enačb na podlagi podatkov za torkov nakup $6 r \\cdot 0{,}85 + 5 z = 2{,}27$ € ter za sobotni nakup $(7 r + 4 z + 2 r) \\cdot 0{,}9 = 2{,}52$. Pri tem smo pozorni na $10 \\%$ popust med vikendom, $15 \\%$ popust na rogljičke med tednom in vrednost vrečke. Sistem enačb lahko rešujemo z... | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | weekend | |
0fwd | Problem:
Seien $n > 1$ und $m$ natürliche Zahlen. Ein Parlament besteht aus $m n$ Abgeordneten, die $2 n$ Kommissionen gebildet haben, sodass gilt:
(i) Jede Kommission besteht aus $m$ Abgeordneten.
(ii) Jeder Abgeordnete ist Mitglied in genau 2 Kommissionen.
(iii) Je zwei Kommissionen haben höchstens ein gemeinsames M... | [
"Solution:\n\nSei $K$ eine beliebige Kommission. Jedes der $m$ Mitglieder von $K$ ist nach (ii) noch in einer anderen Kommission, und diese sind nach (iii) alle verschieden. Insgesamt gibt es also mindestens $m+1$ Kommissionen, also gilt $m \\leq 2 n-1$.\n\nWir zeigen nun, dass $m=2 n-1$ möglich ist, indem wir reku... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | m = 2n − 1 | |
08mm | Problem:
Let $x$, $y$, $z$ be positive real numbers. Prove that:
$$
\frac{x+2y}{z+2x+3y} + \frac{y+2z}{x+2y+3z} + \frac{z+2x}{y+2z+3x} \leq \frac{3}{2}
$$ | [
"Solution:\nNotice that $\\sum_{cyc} \\frac{x+2y}{z+2x+3y} = \\sum_{cyc} \\left(1 - \\frac{x+y+z}{z+2x+3y}\\right) = 3 - (x+y+z) \\sum_{cyc} \\frac{1}{z+2x+3y}$.\n\nWe have to prove that $3 - (x+y+z) \\sum_{cyc} \\frac{1}{z+2x+3y} \\leq \\frac{3}{2}$ or $\\frac{3}{2(x+y+z)} \\leq \\sum_{cyc} \\frac{1}{z+2x+3y}$.\n\... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
09th | Problem:
Gegeven is een bord met $2 m$ rijen en $2 n$ kolommen, waarbij $m$ en $n$ positieve gehele getallen zijn. Je mag één pion plaatsen op een vakje van dit bord, maar niet het vakje linksonder of het vakje rechtsboven. Vervolgens begint een slak een wandeling te maken over het bord. De slak begint in het vakje li... | [
"Solution:\n\nNummer de rijen van beneden naar boven met $1$ tot en met $2 m$ en nummer de kolommen van links naar rechts met $1$ tot en met $2 n$. De slak begint dus in vakje $(1,1)$ en eindigt in vakje $(2 m, 2 n)$. Kleur nu de vakjes als een schaakbord, waarbij vakje $(i, j)$ zwart wordt als $i+j$ even is en wit... | Netherlands | IMO-selectietoets | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Exactly the squares (i, j) with i + j odd (the white squares in a chessboard coloring). | |
0il9 | Problem:
Four unit circles are centered at the vertices of a unit square, one circle at each vertex. What is the area of the region common to all four circles? | [
"Solution:\n\nThe desired region consists of a small square and four \"circle segments,\" i.e. regions of a circle bounded by a chord and an arc. The side of this small square is just the chord of a unit circle that cuts off an angle of $30^{\\circ}$, and the circle segments are bounded by that chord and the circle... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | pi/3 + 1 - sqrt(3) | |
0jw0 | Problem:
Let $V=\{1, \ldots, 8\}$. How many permutations $\sigma: V \rightarrow V$ are automorphisms of some tree?
(A graph consists of a some set of vertices and some edges between pairs of distinct vertices. It is connected if every two vertices in it are connected by some path of one or more edges. A tree $G$ on $V... | [
"Solution:\nWe decompose into cycle types of $\\sigma$. Note that within each cycle, all vertices have the same degree; also note that the tree has total degree $14$ across its vertices (by all its seven edges).\n\nFor any permutation that has a $1$ in its cycle type (i.e., it has a fixed point), let $1 \\leq a \\l... | United States | HMMT February 2016 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | 30212 | |
0l8m | A club has $42$ participants. In each group of $31$ participants of this club, there is always an acquainted couple man - woman. Prove that in this club, we can select $12$ acquainted couples. | [] | Vietnam | VIETNAMESE MATHEMATICAL COMPETITION FOR TEAM SELECTION | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof only | null | |
0cda | Consider a $4 \times 4$ array of pairwise distinct positive integers such that on each column, respectively row, one of the numbers is equal to the sum of the other three. Determine the least possible value of the largest number such an array may contain. | [
"<table>\n<tr><td>1</td><td>8</td><td>12</td><td><b>21</b></td></tr>\n<tr><td>7</td><td>9</td><td>20</td><td>4</td></tr>\n<tr><td>10</td><td>19</td><td>3</td><td>6</td></tr>\n<tr><td>18</td><td>2</td><td>5</td><td>11</td></tr>\n</table>\n\nThe lower bound is a consequence of the following slightly more general fact... | Romania | THE 73rd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS - FIRST SELECTION TEST | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 21 | |
0irs | Problem:
Find $\max \{\operatorname{Perimeter}(T)\}$ for $T$ a triangle contained in a regular septagon (7-sided figure) of unit edge length. Write your answer $N$ to 2 places after the decimal. If the correct answer rounded to 2 decimal places is $A$, you will receive 0 points if $N<A$ and $\lfloor\max \{0,25-50 \cdot... | [
"Solution:\nAnswer: 5.85086\n\nLet the septagon be $A_{0} A_{1} \\ldots A_{6}$.\n\nIf $x$ is a point that can move along the x-axis, the distance from $x$ to a fixed point $p$ is a convex function in the x-coordinate. Therefore, the sum of the distances from $x$ to two other points is convex too, so if $x$ is const... | United States | 1st Annual Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 5.85 | |
02yr | Problem:
Na figura a seguir, o quadrado maior possui área de $1\ \mathrm{cm}^2$ e o quadrado do meio área $M$. A área do quadrado menor, que possui um vértice sobre um lado do quadrado do meio, é $N$. Qual o valor de $N$ em função de $M$ ?
 | [
"Solution:\n\nSejam $AQ = x\\ \\mathrm{cm}$, $QR = a\\ \\mathrm{cm}$ e $AG = s\\ \\mathrm{cm}$. Como a área do quadrado maior é $1\\ \\mathrm{cm}^2$, segue que $AD = 1\\ \\mathrm{cm}$. Os triângulos retângulos $ARQ$ e $DQP$ possuem os mesmos ângulos, pois\n$$\n\\begin{aligned}\n\\angle AQR & = 180^\\circ - \\angle ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | N = ((1 - M)/2)^2 | |
09p8 | Find all pairs of natural numbers $(a, b)$ such that $a! \cdot (a+1)! = b!$. | [] | Mongolia | MMO2025 Round 2 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (a, b) = (1, 2) | |
05ip | Problem:
Soit $ABCD$ un quadrilatère convexe tel que les droites $(AD)$ et $(BC)$ ne sont pas parallèles. On suppose que les cercles de diamètre $[AB]$ et $[CD]$ se coupent en deux points $E$ et $F$ situés à l'intérieur de $ABCD$. Soit $\Gamma_{E}$ le cercle qui passe par les projetés orthogonaux de $E$ sur $(AB)$, $(... | [
"Solution:\n\n\n\nSoient $P, Q, R, S$ les projections de $E$ sur les droites $(DA)$, $(AB)$, $(BC)$, $(CD)$. Les points $P$ et $Q$ appartiennent au cercle de diamètre $[AE]$ de sorte que $\\widehat{QPE} = \\widehat{QAE}$. De même, $\\widehat{QRE} = \\widehat{QBE}$. Ainsi :\n$$\n\\widehat{QP... | France | Olympiades Françaises de Mathématiques - Test de Sélection | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0b70 | Consider a convex quadrilateral $ABCD$ having $\angle BCD = 120^\circ$, $\angle CBA = 45^\circ$, $\angle CBD = 15^\circ$ and $\angle CAB = 90^\circ$. Show that $AB = AD$. | [
"Consider the mirror image $E$ of point $C$ with respect to $A$. Since $\\angle CDB = \\angle CEB = 45^\\circ$, the quadrilateral $BCDE$ is cyclic, and $CE$ is a diameter of its circumcircle. Hence $A$ is its circumcenter, implying $AB = AD$."
] | Romania | NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0hfq | You are given some positive integer $n$. Prove that for any real numbers $a_1, a_2, ..., a_n$ there exists some number of form $k\sqrt{2}$, where $k$ is some positive integer, that all numbers $k\sqrt{2} + a_1, k\sqrt{2} + a_2, ..., k\sqrt{2} + a_n$ are irrational. | [
"Consider numbers $x_1 = \\sqrt{2}, x_2 = 2\\sqrt{2}, ..., x_{n+1} = (n + 1)\\sqrt{2}$. Suppose that for each $k = 1, n + 1$ at least one of the numbers $x_k + a_1, x_k + a_2, ..., x_k + a_n$ is rational. As we have $n$ numbers, and try $n + 1$ options, from the Dirichlet principle some two numbers of form $x_i + a... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
088q | Problem:
Un folletto sceglie due numeri dispari $x, y$ tali che $0 < y < x < 2012$, calcola $x^{2} - y^{2}$ e scrive il risultato su un foglio. Ogni mattina (a partire da quella del giorno successivo) si sveglia, legge il numero scritto sul foglio e, se questo numero è pari, lo sostituisce con la sua metà e va a fare ... | [
"Solution:\n\nLa risposta è $\\mathbf{(B)}$. $x^{2} - y^{2}$ può essere scomposto come $(x + y)(x - y)$. Non è possibile che $x + y$ e $x - y$ siano entrambi multipli di $4$, perché altrimenti lo sarebbe anche $(x + y) + (x - y) = 2x$ il quale tuttavia è il doppio di un numero dispari. Abbiamo che $\\max\\{x + y, x... | Italy | Olimpiadi della Matematica - Gara di Febbraio | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | B | |
0hnj | Problem:
Square $ABCD$ has side length $1$. A dilation is performed about point $A$, creating square $AB' C' D'$. If $BC' = 29$, determine the area of triangle $BDC'$. | [
"Solution:\n\n$29^{2} - 2 \\cdot \\frac{1}{2}(29)\\left(\\frac{29}{2}\\right) - \\frac{1}{2} = 420$."
] | United States | null | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 420 | |
0850 | Problem:
Sia $k \geq 1$ un numero naturale. Determinare in funzione di $k$ il numero di interi positivi $n$ con le seguenti proprietà:
a. in base dieci si scrivono con $k$ cifre, tutte dispari;
b. sono divisibili per $5$, e il quoziente $\frac{n}{5}$, scritto in base dieci, ha ancora $k$ cifre, tutte dispari. | [
"Solution:\n\nIl numero di interi positivi $n$ con le proprietà (a) e (b) è $3^{k-1}$.\n\nPer dimostrarlo si consideri il numero intero $\\frac{n}{5}$ e siano $a_{0}, a_{1}, \\ldots, a_{k-1}$ le sue cifre, ossia\n$$\n\\frac{n}{5}=a_{k-1} 10^{k-1}+\\ldots+a_{1} 10+a_{0}\n$$\nPer (b) le cifre $a_{i}$ sono tutte dispa... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 3^{k-1} | |
087l | Problem:
Sia $ABC$ un triangolo equilatero di centro $O$ e area $1$. Siano $D$, $E$, $F$ i punti simmetrici di $O$ rispetto ai tre lati del triangolo. Quanto vale l'area in comune ai triangoli $ABC$ e $DEF$?
(A) $\frac{1}{3}$
(B) $\frac{2 \sqrt{3}}{9}$
(C) $\frac{\sqrt{2}}{3}$
(D) $\frac{\sqrt{3}}{3}$
(E) $\frac{2}{3... | [
"Solution:\n\nLa risposta è $\\mathbf{(E)}$. L'area che stiamo cercando è uguale all'area del triangolo $ABC$ al quale abbiamo sottratto i tre triangoli più piccoli che partono dai vertici di $ABC$. Questi triangoli sono anche essi equilateri (sono omomorfi ad $ABC$) e tutti congruenti. Notiamo inoltre che $ABC$ e ... | Italy | UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | E | |
0akp | Let $t_k = a_1^k + a_2^k + \dots + a_n^k$, where $a_1, a_2, \dots, a_n$ are positive real numbers and $k \in \mathbb{N}$. Proof that
$$
\frac{t_5^2 t_1^6}{15} - \frac{t_4^4 t_2^2 t_1^2}{6} + \frac{t_2^3 t_4^5}{10} \ge 0.
$$ | [
"The inequality that we need to prove is equivalent with the inequality\n$$\n2t_5^2 t_1^6 - 5t_4^4 t_2^2 t_1^2 + 3t_2^3 t_4^5 \\ge 0,\n$$\nor with\n$$\n2t_5^2 t_1^6 + 3t_2^3 t_4^5 \\ge 5t_4^4 t_2^2 t_1^2.\n$$\nUsing the inequality between arithmetic and geometric mean we have\n$$\n2t_5^2 t_1^6 + 3t_2^3 t_4^5 \\ge 5... | North Macedonia | Macedonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
015p | Let $n$ be a positive integer. In a triangular grid of side length $n$, one object is located at each node. The following steps are permitted: choose three nodes that are pairwise neighbours to each other and cyclicly interchange the objects at these nodes. For which $n$ is it possible to rotate the whole set of object... | [
"We show that there exists an order of the nodes of the grid such that, choosing the nodes in this order, one can always bring the desired object to the chosen node so that the objects at the nodes chosen before do not move.\nIndeed, choose nodes row-by-row, starting from the shortest one. It is easy to see that $n... | Baltic Way | Baltic Way SHL | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | All positive integers n | |
01s4 | Find all pairs of polynomials $p(x), q(x) \in \mathbb{R}[x]$ satisfying the equality
$$
p(x^2) = p(x)q(1-x) + p(1-x)q(x)
$$
for all real $x$. | [
"(Solution by A. Asanau, D. Voynov.) Replace $x$ by $1-x$ in\n$$\np(x^2) = p(x)q(1-x) + p(1-x)q(x), \\quad (1)\n$$\nthen we have $p((x-1)^2) = p(1-x)q(x) + p(x)q(1-x) = p(x^2)$. It follows that the polynomial $p(x^2)$ is periodic, i.e., $p(x)$ is a constant polynomial, $p(x) = a$ for some $a \\in \\mathbb{R}$.\nIf ... | Belarus | SELECTION and TRAINING SESSION | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All solutions are given by constant p(x) = a. If a = 0, q(x) is arbitrary in R[x]. If a ≠ 0, then q must satisfy q(x) + q(1 − x) = 1 for all real x, equivalently q(x) = (x − 1/2) r((x − 1/2)^2) + 1/2 for some polynomial r(x) in R[x]. | |
01p6 | Find all polynomials $P(x)$ such that the equality
$$
(x - 1)P(x + 1) - (x + 1)P(x - 1) = 4P(x)
$$
holds for all real $x$. | [
"Answer: $P(x) = a x(x-1)(x+1)$, where $a \\in \\mathbb{R}$.\n\nSet $x = 1$ and $x = -1$ in the initial identity\n$$\n(x-1)P(x+1) - (x+1)P(x-1) = 4P(x). \\quad (1)\n$$\nThus we obtain $-2P(0) = 4P(1)$ and $-2P(0) = 4P(-1)$ respectively. Setting $x = 0$ in (1), we obtain $-P(1) - P(-1) = 4P(0)$, so, taking into acco... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | P(x) = a x(x-1)(x+1) for any real a | |
0hbw | Line $l$ is perpendicular to the side $AC$ of acute triangle $ABC$, and intersects $AC$ at $K$. $l$ intersects circumscribed circle of $\triangle ABC$ at $P$ and $T$ (point $P$ in the same half-plane w.r.t. $AC$ as vertex $B$). By $P_1$ and $T_1$ are denoted projections of points $P$ and $T$ to line $AB$. Furthermore, ... | [
"We denote by $B_1$ a projection of vertex $B$ to line $l$, then quadrilaterals $BB_1PP_1$ and $TKAT_1$ are inscribed, with diameters $BP$ and $AT$. Then, we have the following equalities of angles:\n\n$$\n\\angle P_1B_1K = \\pi - \\angle P_1B_1P = \\pi - \\angle P_1BP = \\\\\n\\angle ABP = \\pi - \\angle ATP = \\p... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"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 circle"
] | English | proof only | null | |
050x | We say that two real numbers $r$ and $s$ are *close* if $|r - s| = 10^u$ for some integer $u$. Let $y = ax + b$ be a linear function, for which there exist close numbers $x_1$ and $x_2$ so that the corresponding $y_1$ and $y_2$ are also close. Prove that for any close numbers $x'_1$ and $x'_2$, the corresponding $y'_1$... | [
"From the premises we get $|x_1 - x_2| = 10^u$ and $|y_1 - y_2| = |(ax_1 + b) - (ax_2 + b)| = 10^v$ for some integers $u, v$. So,\n$$\n10^v = |(ax_1 + b) - (ax_2 + b)| = |a(x_1 - x_2)| = |a| \\cdot |x_1 - x_2| = |a| \\cdot 10^u,\n$$\nwhich gives $|a| = \\frac{10^v}{10^u} = 10^{v-u}$. Let $x'_1, x'_2$ be any close r... | Estonia | Estonian Math Competitions | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof only | null | |
0779 | Problem:
Let $f$ be function defined from the set $\{(x, y): x, y$ reals, $x y \neq 0\}$ into the set of all positive real numbers such that
(i) $\quad f(x y, z)=f(x, z) f(y, z)$, for all $x, y \neq 0$;
(ii) $\quad f(x, y z)=f(x, y) f(x, z)$, for all $x, y \neq 0$;
(iii) $\quad f(x, 1-x)=1$, for all $x \neq 0,1$.
Prov... | [
"Solution:\n\nTaking $x=y=1$ in (ii), we get $f(1, z)^2 = f(1, z)$ so that $f(1, z) = 1$ for all $z \\neq 0$. Similarly, $x=y=-1$ gives $f(-1, z) = 1$ for all $z \\neq 0$. Using the second condition, we also get $f(z, 1) = f(z, -1) = 1$ for all $z \\neq 0$.\n\nObserve\n$$\nf\\left(\\frac{1}{x}, y\\right) f(x, y) = ... | India | Indian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
0egh | Problem:
Dana je kvadratna funkcija $f$ s predpisom $f(x) = (m-1)x^{2} + m x + m$, kjer $m \neq 1$.
a. Za $m=3$ izračunaj najmanjšo vrednost funkcije $f$.
b. Poišči vsa realna števila $m$, da bo funkcija $f$ strogo negativna za vsak $x$. | [
"Solution:\n\na.\nNajmanjša vrednost kvadratne funkcije je pri $q = \\frac{-D}{4a}$.\nIzračunamo diskriminanto $D = -15$.\nIzračunamo najmanjšo vrednost $q = \\frac{15}{8}$.\n\nb.\nFunkcija $f$ bo negativna za vsak $x$, ko bosta izpolnjena pogoja $(m-1 < 0)$ in $D < 0$.\nDiskriminanta kvadratne funkcije $f$ je $D =... | Slovenia | 18. tekmovanje v znanju matematike za dijake srednjih tehniških i strokovnih šol, Odbirno tekmovanje | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) 15/8; b) m < 0 | |
0i92 | Problem:
We are given triangle $ABC$, with $AB = 9$, $AC = 10$, and $BC = 12$, and a point $D$ on $BC$. $B$ and $C$ are reflected in $AD$ to $B'$ and $C'$, respectively. Suppose that lines $BC'$ and $B'C$ never meet (i.e., are parallel and distinct). Find $BD$. | [
"Solution:\n\nThe lengths of $AB$ and $AC$ are irrelevant. Because the figure is symmetric about $AD$, lines $BC'$ and $B'C$ meet if and only if they meet at a point on line $AD$. So, if they never meet, they must be parallel to $AD$. Because $AD$ and $BC'$ are parallel, triangles $ABD$ and $ADC'$ have the same are... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 6 | |
0djn | Prove that the inequality
$$
x^2\sqrt{1+2y^2} + y^2\sqrt{1+2x^2} \ge xy(x+y+\sqrt{2})
$$
holds for any two real numbers $x$ and $y$. | [
"By the quadratic-arithmetic mean inequality, we have\n$$\n1 + 2t^2 \\ge \\frac{1}{2}(1 + t\\sqrt{2})^2.\n$$\nIt follows that\n$$\n\\begin{align*}\nx^2 \\sqrt{1+2y^2} + y^2 \\sqrt{1+2x^2} &\\ge x^2 \\cdot \\frac{1+y\\sqrt{2}}{2} + y^2 \\cdot \\frac{1+x\\sqrt{2}}{2} \\\\\n&= \\frac{x^2 + y^2 + xy(x+y)\\sqrt{2}}{\\sq... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0f9r | Problem:
Does there exist a rectangle which can be dissected into 15 congruent polygons which are not rectangles? Can a square be dissected into 15 congruent polygons which are not rectangles? | [
"Solution:\n\nyes, yes\n\n\nBy stretching vertically we get a square."
] | Soviet Union | 24th ASU | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Yes; Yes | |
0afl | A positive integer is written in each $1 \times 1$ cell of a rectangular board. The following two steps are allowed:
i) the numbers in the cells of an arbitrarily chosen row are multiplied by $2$;
ii) the numbers in the cells of an arbitrarily chosen column are decreased by $1$.
Is it possible after finite repetition... | [
"We will show that it is possible. If there are $1$'s in the first column then we perform step i) on those rows that have $1$ in their first cell. Then we perform step ii) on the first column. We repeat this procedure until all the numbers in the first column are $1$'s. Then we perform ii) on the first column, so n... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms"
] | Macedonian, English | proof and answer | Yes | |
0ir1 | Problem:
A triangle (with non-zero area) is constructed with the lengths of the sides chosen from the set
$$
\{2, 3, 5, 8, 13, 21, 34, 55, 89, 144\}
$$
Show that this triangle must be isosceles (A triangle is isosceles if it has at least two sides the same length.) | [
"Solution:\nFirst note that the numbers in the list are an increasing sequence of Fibonacci numbers, i.e. starting with $5$, each number is the sum of the previous two numbers in the list. By the triangle inequality, we know that the longest side is always less than the sum of the other two sides. Suppose the trian... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
0706 | Problem:
$L$ and $L'$ are two perpendicular lines not in the same plane. $AA'$ is perpendicular to both lines, where $A$ belongs to $L$ and $A'$ belongs to $L'$. $S$ is the sphere with diameter $AA'$. For which points $P$ on $S$ can we find points $X$ on $L$ and $X'$ on $L'$ such that $XX'$ touches $S$ at $P$? | [] | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | Exactly the points on the two great circles obtained by intersecting the sphere with the two planes through AA′ that bisect the angle between the directions of L and L′; equivalently, the intersection of S with the planes making equal angles with L and L′. | |
0kdk | Problem:
Let $S=\{(x, y) \mid x>0, y>0, x+y<200$, and $x, y \in \mathbb{Z}\}$. Find the number of parabolas $\mathcal{P}$ with vertex $V$ that satisfy the following conditions:
- $\mathcal{P}$ goes through both $(100,100)$ and at least one point in $S$,
- $V$ has integer coordinates, and
- $\mathcal{P}$ is tangent to t... | [
"Solution:\nWe perform the linear transformation $(x, y) \\rightarrow (x-y, x+y)$, which has the reverse transformation $(a, b) \\rightarrow \\left(\\frac{a+b}{2}, \\frac{b-a}{2}\\right)$. Then the equivalent problem has a parabola has a vertical axis of symmetry, goes through $A=(0,200)$, a point $B=(u, v)$ in\n$$... | United States | HMMT February 2020 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Rotation",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 264 | |
0fnw | Sobre una circunferencia se marcan $2014$ puntos. Sobre cada uno de los segmentos cuyos extremos son dos de los $2014$ puntos, se escribe un número real no negativo. Se sabe que para cualquier polígono convexo cuyos vértices son algunos de los $2014$ puntos, la suma de los números escritos en sus lados es menor o igual... | [] | Spain | XXIX Olimpiada Iberoamericana de Matemáticas | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | 2013/2 | |
0kgq | Problem:
Squares $ABCD$ and $DEFG$ have side lengths $1$ and $\frac{1}{3}$, respectively, where $E$ is on $\overline{CD}$ and points $A$, $D$, $G$ lie on a line in that order. Line $CF$ meets line $AG$ at $X$. The length $AX$ can be written as $\frac{m}{n}$, where $m, n$ are positive integers and $\operatorname{gcd}(m... | [
"Solution:\n\n\n\nThere are a variety of solutions involving similar triangles. One fast way to solve the problem without hunting for many geometric relationships is to notice that, if one continues to add squares inscribed between $\\overline{AX}$ and $\\overline{XC}$ as shown in the diagr... | United States | HMMT November 2021 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 302 | |
0js2 | Problem:
Let $ABC$ be a triangle with $AB = 13$, $BC = 14$, $CA = 15$. Let $O$ be the circumcenter of $ABC$. Find the distance between the circumcenters of triangles $AOB$ and $AOC$. | [
"Solution:\n\nLet $S$, $T$ be the intersections of the tangents to the circumcircle of $ABC$ at $A$, $C$ and at $A$, $B$ respectively. Note that $ASCO$ is cyclic with diameter $SO$, so the circumcenter of $AOC$ is the midpoint of $OS$, and similarly for the other side. So the length we want is $\\frac{1}{2} ST$. Th... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle tr... | null | proof and answer | 91/6 | |
023q | Problem:
A sequência $x y z$ - Quais são os valores prováveis de $x, y$ e $z$ na sequência
$$
\frac{1}{2}, \frac{5}{8}, \frac{3}{4}, \frac{7}{8}, x, y, z ?
$$ | [
"Solution:\n\nIgualando os denominadores, verificamos que a sequência dada é igual a\n$$\n\\frac{4}{8}, \\frac{5}{8}, \\frac{6}{8}, \\frac{7}{8}, x, y, z\n$$\nAssim, o denominador é sempre $8$ e os numeradores são consecutivos. Logo, $x=\\frac{8}{8}=1$, $y=\\frac{9}{8}$ e $z=\\frac{10}{8}=\\frac{5}{4}$."
] | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | final answer only | x = 1, y = 9/8, z = 5/4 | |
06gv | Let $a$, $b$ and $c$ be positive real numbers such that $ab + bc + ca = 1$. Prove that
$$
\sqrt[4]{\frac{\sqrt{3}}{a} + 6\sqrt{3}b} + \sqrt[4]{\frac{\sqrt{3}}{b} + 6\sqrt{3}c} + \sqrt[4]{\frac{\sqrt{3}}{c} + 6\sqrt{3}a} \le \frac{1}{abc}.
$$
When does the equality hold? | [
"(IMO Shortlist 2004 A5 modified) By the power mean inequality, we have\n$$\n\\begin{aligned}\nS &:= \\sqrt[4]{\\frac{\\sqrt{3}}{a} + 6\\sqrt{3}b} + \\sqrt[4]{\\frac{\\sqrt{3}}{b} + 6\\sqrt{3}c} + \\sqrt[4]{\\frac{\\sqrt{3}}{c} + 6\\sqrt{3}a} \\\\\n&\\le 3\\sqrt[4]{\\frac{1}{3}\\left(\\frac{\\sqrt{3}}{a} + 6\\sqrt{... | Hong Kong | CHKMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | Equality holds when a = b = c = 1/√3. | |
0joe | Problem:
Let $ABCDE$ be a convex pentagon such that $\angle ABC = \angle ACD = \angle ADE = 90^{\circ}$ and $AB = BC = CD = DE = 1$. Compute $AE$. | [
"Solution:\nBy Pythagoras,\n$$\nAE^2 = AD^2 + 1 = AC^2 + 2 = AB^2 + 3 = 4\n$$\nso $AE = 2$."
] | United States | HMMT February | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2 | |
09se | Problem:
Laat cirkels $\Gamma_{1}$ en $\Gamma_{2}$, met middelpunten respectievelijk $O_{1}$ en $O_{2}$, elkaar snijden in twee verschillende punten $A$ en $B$. De lijn $O_{1} A$ snijdt $\Gamma_{2}$ nogmaals in $C$ en de lijn $O_{2} A$ snijdt $\Gamma_{1}$ nogmaals in $D$. De lijn door $B$ evenwijdig met $A D$ snijdt $... | [
"Solution:\n\nIn alle oplossingen bekijken we de configuratie waarin $A, B, E$ en $D$ in die volgorde op een cirkel liggen, $O_{1}, A$ en $C$ in die volgorde op een lijn en $O_{2}, A$ en $D$ in die volgorde op een lijn. De andere configuraties gaan analoog.\n\nVanwege koordenvierhoek $A B E D$ geldt $\\angle B E D=... | Netherlands | IMO-selectietoets | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dsx | In $\triangle ABC$, let $O$, $H$, and $N$ be its circumcenter, orthocenter, and nine-point center respectively. Let $AN$ meet the circumcircle of $\triangle ABC$ at $S$. Let the tangents to the circumcircle of $ABC$ at $B$ and $C$ meet at $D$. Show that $\angle DSH = \angle DOA$. | [] | Singapore | Singapore International Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Homothet... | null | proof only | null | |
05ns | Problem:
Soit $ABC$ un triangle non rectangle tel que $AB < AC$. On note $H$ le projeté de $A$ sur $(BC)$, et $E, F$ les projetés respectifs de $H$ sur $(AB)$ et $(AC)$. La droite $(EF)$ coupe $(BC)$ au point $D$. On considère le demi-cercle de diamètre $[CD]$ situé dans le même demi-plan délimité par $(CD)$ que $A$. ... | [
"Solution:\n\n\n\nComme $AEHF$ sont cocycliques (sur le cercle de diamètre $[AH]$), on a $(EF, EA) = (HF, HA) = (HF, CA) + (CA, CB) + (CB, HA) = (CA, CB) = (CF, CB)$, donc $(EF, EB) = (CF, CB)$. Par conséquent, $B, E, F, C$ sont cocycliques.\n\nD'après la puissance d'un point par rapport à ... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hk8 | Problem:
On a given street, there are $n$ houses numbered from $1$ to $n$. Let $a_{i}$ ($1 \leq i \leq n$) be the number of people living in the house numbered $i$, and let $b_{i}$ ($i \geq 1$) be the number of houses on the street in which at least $i$ people live. Prove that
$$
a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+b... | [
"Solution:\n\nLet us number the people in each house from $1$ up to the total number of people living there. Then for each $k \\geq 1$, the label $k$ is assigned as often as there is a house with $k$ or more people; thus there are $b_{k}$ people labeled $k$. The quantity\n$$\nb_{1}+b_{2}+b_{3}+\\cdots\n$$\nthus rep... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0dgy | Prove the identity
$$
\frac{n!}{x(x+1)(x+2)\dots(x+n)} = \frac{\binom{n}{0}}{x} - \frac{\binom{n}{1}}{x+1} + \frac{\binom{n}{2}}{x+2} - \dots + (-1)^n \frac{\binom{n}{n}}{x+n}.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0hkc | Problem:
If $a$ and $b$ are positive integers prove that
$$
a+b \leq 1+a b.
$$ | [
"Solution:\nSince $a$ and $b$ are positive integers we know that $a \\geq 1$ and $b \\geq 1$. This implies that $(a-1)(b-1) \\geq 0$ hence $a b - a - b + 1 \\geq 0$ which is equivalent to the given inequality."
] | United States | Berkeley Math Circle Monthly Contest 1 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
08cx | Problem:
Dato un numero reale $x$ compreso fra $0$ e $1$, consideriamo la sua scrittura decimale $0, c_{1} c_{2} c_{3} \ldots$ Chiamiamo $B(x)$ l'insieme delle diverse sottosequenze di sei cifre consecutive che compaiono nella sequenza $c_{1} c_{2} c_{3} \ldots$
Per esempio, $B(1 / 22)=\{045454,454545,545454\}$.
Det... | [] | Italy | Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | null | proof and answer | 7 | |
0kch | Problem:
In the future, MIT has attracted so many students that its buildings have become skyscrapers. Ben and Jerry decide to go ziplining together. Ben starts at the top of the Green Building, and ziplines to the bottom of the Stata Center. After waiting $a$ seconds, Jerry starts at the top of the Stata Center, and ... | [
"Solution:\n\nDefine the following lengths:\n\n\nNote that due to all the 3-4-5 triangles, we find $\\frac{x}{z}=\\frac{z}{y}=\\frac{4}{3}$, so $120=x+y=\\frac{25}{12} z$. Then,\n$$\nu=\\frac{5}{3} x=\\frac{20}{9} z=\\frac{16}{15} 120=128$$\nwhile\n$$v=\\frac{5}{4} y=\\frac{15}{16} z=\\frac... | United States | HMMO 2020 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 740 | |
0gcd | 一個長方形 $\mathcal{R}$, 其各邊邊長皆為奇整數。我們將之分割為若干個小的長方形, 使得每個小長方形的各邊邊長皆為正整數。證明其中至少有一個小長方形, 它和 $\mathcal{R}$ 各邊的距離的奇偶性相同。
(在此, 一個小長方形到 $\mathcal{R}$ 某一邊 $L$ 的距離, 定義為該小長方形平行 $L$ 的一邊中較靠近 $L$ 者與 $L$ 的垂直距離。) | [
"令 $R$ 的長寬分別為 $a, b$。將長方形切格為 $ab$ 個單位正方形, 並依據西洋棋盤方式將之交替塗為黑白。注意到由於 $a, b$ 皆為奇數, 四個角落的正方形必被塗成同色, 不失一般性設為黑色。\n\n對於每個長方形, 我們稱它為黑的 (白的) 若且唯若它的四個角落正方形都是黑的 (白的), 否則稱它為灰的。注意到以下事實：\n(i) 灰長方形包含相同數量的黑與白正方形；\n(ii) 黑長方形內的黑正方形比白正方形多一個；\n(iii) 白長方形內的白正方形比黑正方形多一個。\n\n現在, $R$ 是黑的, 所以其內的黑正方形比白正方形多一個; 搭配以上事實, 我們知道至少要有一個小長方形是黑的。現在令該長方... | Taiwan | 2018 數學奧林匹亞競賽第二階段選訓營, 模擬競賽(二) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0g55 | Problem:
Finde alle Tripel $(a, b, p)$ natürlicher Zahlen, sodass $p$ eine Primzahl ist und die Gleichung
$$
(a+b)^{p}=p^{a}+p^{b}
$$
erfüllt ist. | [
"Solution:\n$(a, b, p)=(1,1,2)$ ist die Einzige Lösung. Wir teilen das Problem in zwei Fälle.\n\n- Fall 1: $a=b$\nDie Gleichung wird zu $2^{p} \\cdot a^{p}=2 \\cdot p^{a}$, und da $4\\mid 2^{p}, 2\\mid p^{a}$ woraus folgt, dass $p=2$. Wenn wir diese neue Information in der ursprünglichen Gleichung einsetzen, ergibt... | Switzerland | Zweite Runde 2023 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (1, 1, 2) | |
09a7 | Find all the positive integers $n$ such that there exist $n$ sets $A_1, A_2, \dots, A_n$ such that each of them has exactly $5$ elements, any two of these $n$ sets have exactly one common element and the union of these sets consists of $n$ elements. | [
"Let us assume that the union of the sets consists of integers from $1$ to $n$. Let $S_i$ denote the number of sets that $i$ belongs to. Then the total number of elements of $n$ sets is $5n = S_1 + S_2 + \\dots + S_n$ ($\\bullet$).\n\nLet us assume $S_1 > 5$ and $1 \\in A_1, 1 \\in A_2, 1 \\in A_3, 1 \\in A_4, 1 \\... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 21 | |
0kcu | Problem:
Two points are chosen inside the square $\{(x, y) \mid 0 \leq x, y \leq 1\}$ uniformly at random, and a unit square is drawn centered at each point with edges parallel to the coordinate axes. The expected area of the union of the two squares can be expressed as $\frac{a}{b}$, where $a, b$ are relatively prime ... | [
"Solution:\n\nLet $\\Delta x$ and $\\Delta y$ be the positive differences between the $x$ coordinates and $y$ coordinates of the centers of the squares, respectively. Then, the length of the intersection of the squares along the $x$ dimension is $1-\\Delta x$, and likewise the length along ... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 1409 | |
0jxt | Problem:
Kelvin the Frog is hopping on a number line (extending to infinity in both directions). Kelvin starts at $0$. Every minute, he has a $\frac{1}{3}$ chance of moving $1$ unit left, a $\frac{1}{3}$ chance of moving $1$ unit right and $\frac{1}{3}$ chance of getting eaten. Find the expected number of times Kelvin... | [
"Solution:\n\nFirst we compute probability that the mouse returns to $0$ before being eaten. Then probability that it is at $0$ in $2n$ minutes without being eaten is given by $\\frac{1}{3^{2n}} \\binom{2n}{n}$. Therefore, the overall expectation is given by\n$$\n\\begin{aligned}\n& \\sum_{n \\geq 1} \\binom{2n}{n}... | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (3*sqrt(5) - 5)/5 | |
04n9 | Let $a$, $b$ and $c$ be positive real numbers such that
$$
a^2 + b^2 - c^2 = \sqrt{3}ab, \quad a^2 - b^2 + c^2 = \sqrt{2}ac.
$$
Determine the ratio $b : c$. (Hong Kong 2002) | [] | Croatia | Croatia_2018 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | sqrt(2):1 | |
0ela | Problem:
V koordinatni sistem nariši množico točk $(x, y)$, ki ustrezajo pogoju $(|x| \leq 2) \wedge (-2 \leq y \leq 4) \wedge (y \leq x)$. Izračunaj ploščino in obseg lika, ki nastane. | [
"Solution:\n\n1. V koordinatni sistem narišemo množico točk, ki ustreza pogoju $(|x| \\leq 2) \\wedge (-2 \\leq y \\leq 4) \\wedge (y \\leq x)$. Dobljena množica točk oblikuje pravokotni trikotnik s katetama $k_{1} = k_{2} = 4$. Izračunamo ploščino trikotnika $S = \\frac{k_{1} \\cdot k_{2}}{2} = 8$. S Pitagorovim i... | Slovenia | 23. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles"
] | null | final answer only | Area = 8; Perimeter = 8 + 4√2 | |
0dsk | In a convex quadrilateral $ABCD$, $\angle A < 90^\circ$, $\angle B < 90^\circ$ and $AB > CD$. Points $P$ and $Q$ are on the segments $BC$ and $AD$ respectively. Suppose the triangles $APD$ and $BQC$ are similar. Prove that $AB$ is parallel to $CD$. | [
"\n\nLet the lines $BC$ and $AD$ intersect at $M$. The conditions $\\angle A < 90^\\circ$, $\\angle B < 90^\\circ$ and $AB > CD$ imply that $C$ lies between $MB$ and $D$ lies between $MA$. Given the triangles $APD$ and $BQC$ are similar, we have $\\angle BCQ = \\angle ADP$ so that $\\angle ... | Singapore | Singapore Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0fgi | Problem:
Hallar todas las funciones $f$ definidas en el conjunto de los números reales no negativos y que toman valores reales no negativos, tales que
1) $\quad f(x f(y)) f(y)=f(x+y)$ para todo $x, y \geq 0$,
2) $\quad f(2)=0$,
3) $\quad f(x) \neq 0$ para $0 \leq x<2$. | [] | Spain | International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = 2/(2 - x) for 0 ≤ x < 2, and f(x) = 0 for x ≥ 2. | |
0497 | A bag contained $255$ balls numbered $1, 2, \ldots, 255$, and then $N$ students each took one ball from the bag. It turned out that none of the drawn numbers was exactly twice as big as any other drawn number. Determine the maximum possible $N$. (Ana Prlić) | [
"Group the observed numbers into sets:\n$$\nA_0 = \\{1\\}, A_1 = \\{2, 3\\}, A_2 = \\{4, 5, 6, 7\\}, A_3 = \\{8, 9, \\dots, 15\\}, A_4 = \\{16, 17, \\dots, 31\\}, \\\\\nA_5 = \\{32, 33, \\dots, 63\\}, A_6 = \\{64, 65, \\dots, 127\\}, A_7 = \\{128, 129, \\dots, 255\\}.\n$$\nThe number of elements of $A_k$ is $2^k$, ... | Croatia | CroatianCompetitions2011 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 170 | |
0iyn | Problem:
Let $ABC$ be a triangle with $AB = 16$ and $AC = 5$. Suppose the bisectors of angles $\angle ABC$ and $\angle BCA$ meet at point $P$ in the triangle's interior. Given that $AP = 4$, compute $BC$. | [
"Solution:\n\nAs the incenter of triangle $ABC$, point $P$ has many properties. Extend $AP$ past $P$ to its intersection with the circumcircle of triangle $ABC$, and call this intersection $M$. Now observe that\n\n$$\n\\angle PBM = \\angle PBC + \\angle CBM = \\angle PBC + \\angle CAM = \\beta + \\alpha = 90^\\circ... | United States | Harvard-MIT Mathematics Tournament | [
"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 and answer | 14 | |
0b1x | Problem:
What is the largest integer $k$ such that $k+1$ divides
$$
k^{2020}+2 k^{2019}+3 k^{2018}+\cdots+2020 k+2021 ?
$$ | [
"Solution:\nThe remainder when the polynomial $x^{2020}+2 x^{2019}+3 x^{2018}+\\cdots+2020 x+2021$ is divided by $x+1$ is\n$$\n(-1)^{2020}+2(-1)^{2019}+3(-1)^{2018}+\\cdots+2020(-1)+2021=1010(-1)+2021=1011\n$$\nTherefore, $k+1$ divides $k^{2020}+2 k^{2019}+3 k^{2018}+\\cdots+2020 k+2021$ precisely when $k+1$ divide... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1010 | |
0ek0 | Problem:
Izračunaj $x+y$, če je $\left(x+\sqrt{1+x^{2}}\right)\left(y+\sqrt{1+y^{2}}\right)=1$. | [
"Solution:\nDano enakost pomnožimo z $\\left(x-\\sqrt{1+x^{2}}\\right)$, da dobimo $\\left(x^{2}-\\left(1+x^{2}\\right)\\right)\\left(y+\\sqrt{1+y^{2}}\\right) = x-\\sqrt{1+x^{2}}$ oziroma $-y-\\sqrt{1+y^{2}} = x-\\sqrt{1+x^{2}}$. Slednje lahko preoblikujemo v $\\sqrt{1+x^{2}}-\\sqrt{1+y^{2}}=x+y$.\n\nČe dano enako... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 0 | |
07vr | Find all integer solutions $(x, y)$ to the equation
$$
x^2 - x y + y^2 = 2023.
$$ | [
"Note that $2023 = 17^2 \\cdot 7$ and let $(x, y)$ be a solution. If we assume that $y$ is not divisible by $17$, there exists a number $a$ such that $a y \\equiv x \\pmod{17}$. We then have\n$$\n0 = x^2 - x y + y^2 \\equiv y^2 (a^2 - a + 1) \\pmod{17},\n$$\nhence $a^2 - a + 1 \\equiv 0 \\pmod{17}$. This implies th... | Ireland | IRL_ABooklet_2023 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | The integer solutions are (17, 51), (-17, -51), (51, 17), (-51, -17), (34, 51), (-34, -51), (51, 34), (-51, -34), (-17, 34), (17, -34), (34, -17), and (-34, 17). | |
02zh | Problem:
O professor M. A. Luco escreveu no quadro a expressão:
$$
\frac{n^{2}-5 n+4}{n-4}
$$
Então, ele diz aos alunos que $n$ pode ser qualquer número natural, com exceção de $4$.
a) Qual o valor da expressão para $n=1$?
b) Marcos substituiu $n$ por um número natural e verificou que o valor da expressão é $5$. Marc... | [
"Solution:\na) Para $n=1$, temos:\n$$\n\\begin{aligned}\n\\frac{n^{2}-5 n+4}{n-4} & =\\frac{1^{2}-5 \\cdot 1+4}{1-4} \\\\\n& =\\frac{1-5+4}{-3} \\\\\n& =0\n\\end{aligned}\n$$\n\nb) Igualando a expressão a $5$, temos:\n$$\n\\begin{aligned}\n\\frac{n^{2}-5 n+4}{n-4} & =5 \\\\\nn^{2}-5 n+4 & =5(n-4) \\\\\nn^{2}-10 n+2... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a) 0; b) 6; c) all natural numbers except 3 |
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