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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0ier | Problem:
Let $S$ be the set of lattice points inside the circle $x^{2}+y^{2}=11$. Let $M$ be the greatest area of any triangle with vertices in $S$. How many triangles with vertices in $S$ have area $M$? | [
"Solution:\nThe boundary of the convex hull of $S$ consists of points with $(x, y)$ or $(y, x) = (0, \\pm 3)$, $(\\pm 1, \\pm 3)$, and $(\\pm 2, \\pm 2)$. For any triangle $T$ with vertices in $S$, we can increase its area by moving a vertex not on the boundary to some point on the boundary. Thus, if $T$ has area $... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | final answer only | 16 | |
0b47 | Problem:
In parallelogram $WXYZ$, the length of diagonal $WY$ is $15$, and the perpendicular distances from $W$ to lines $YZ$ and $XY$ are $9$ and $12$, respectively. Find the least possible area of the parallelogram. | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 108 | |
00od | A positive integer is called *powerful* if all exponents in its prime factorization are $\geq 2$.
Prove that there are infinitely many pairs of powerful consecutive positive integers. | [
"The numbers $8 = 2^3$ and $9 = 3^2$ form a pair of consecutive powerful numbers.\n\nWe now show that for each pair $(k, k+1)$ of powerful positive integers we can find a new pair, namely the pair $(4k(k+1), (2k+1)^2)$. Obviously, $4k(k+1)+1 = (2k+1)^2$. Since $k$ and $k+1$ are powerful, the product $4k(k+1) = 2^2k... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | English | proof only | null | |
005t | Problem:
Sea $ABC$ un triángulo con $BC = 1$ y ángulo $BAC$ agudo. Sean $D$ la intersección de la bisectriz interior del ángulo $BAC$ y el lado $BC$, $H$ el ortocentro y $O$ el circuncentro de $ABC$. Encuentre el valor de $AB:AC$ en el caso de que $HOCB$ y $AHDO$ sean cíclicos. | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry >... | Spanish | proof and answer | 2:1 | |
0hq5 | Problem:
What is the remainder when $2^{2001}$ is divided by $2^{7}-1$? | [
"Solution:\n$2^{2001 \\bmod 7} = 2^{6} = 64$."
] | United States | null | [
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | final answer only | 64 | |
0dhg | Let $u_n$ be the least common multiple of the first $n$ terms of a strictly increasing sequence of positive integers $a_1, a_2, a_3, \dots, a_{1000}$. Prove that
$$
\sum_{k=1}^{1000} \frac{1}{u_k} \le 2.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0c02 | For any positive integer $n$ and for any column matrix
$$
X = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} \in \mathcal{M}_{n,1}(\mathbb{Z}),
$$
we denote by $\delta(X)$ the greatest common divisor of the numbers $x_1, x_2, \dots, x_n$. Let $n \in \mathbb{N}$, $n \ge 2$, and $A \in \mathcal{M}_n(\mathbb{Z}... | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Linear Algebra > Vectors",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0bxm | Show that for every integer $n \ge 3$ there exist positive integers $x_1, x_2, \dots, x_n$, pairwise different, so that $\{2, n\} \subset \{x_1, x_2, \dots, x_n\}$ and
$$
\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n} = 1.
$$ | [
"It is known that\n$$\n\\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\dots + \\frac{1}{(n-1)n} + \\frac{1}{n} = \\left(\\frac{1}{1} - \\frac{1}{2}\\right) + \\left(\\frac{1}{2} - \\frac{1}{3}\\right) + \\dots + \\left(\\frac{1}{n-1} - \\frac{1}{n}\\right) + \\frac{1}{n} = 1.\n$$\nIf $n$ is not of the form $k(k+1... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Number Theory > Other"
] | English | proof only | null | |
0dmp | Problem:
Неки од $n$ градова су повезани авионским линијама (све линије су двосмерне). Постоји тачно $m$ линија. Нека је $d_{i}$ број линија које полазе из града $i$, за $i=1,2, \ldots, n$. Ако је $1 \leqslant d_{i} \leqslant 2010$, за свако $i=1,2, \ldots, n$, доказати да важи
$$
\sum_{i=1}^{n} d_{i}^{2} \leqslant 40... | [
"Solution:\n\nУслов задатка нам даје $0 \\leqslant (d_{i}-1)(2010-d_{i})$ за све $i$, тј. $d_{i}^{2} \\leqslant 2011 d_{i}-2010$. Користећи услов $\\sum_{i=1}^{n} d_{i}=2 m$, сабирањем ових неједнакости добијамо\n$$\n\\sum_{i=1}^{n} d_{i}^{2} \\leqslant 2011 \\cdot \\sum_{i=1}^{n} d_{i}-2010 n=4022 m-2010 n\n$$\nа ... | Serbia | Serbian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Equality holds if and only if n is even, or n is odd and at least 2011. | |
0kxr | Problem:
A random permutation $a = (a_{1}, a_{2}, \ldots, a_{40})$ of $(1, 2, \ldots, 40)$ is chosen, with all permutations being equally likely. William writes down a $20 \times 20$ grid of numbers $b_{ij}$ such that $b_{ij} = \max(a_{i}, a_{j+20})$ for all $1 \leq i, j \leq 20$, but then forgets the original permuta... | [
"Solution:\n\nWe can deduce information about $a$ from the grid $b$ by looking at the largest element of it, say $m$. If $m$ fills an entire row, then the value of $a$ corresponding to this row must be equal to $m$. Otherwise, $m$ must fill an entire column, and the value of $a$ corresponding to this column must be... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 10/13 | |
0hlt | Problem:
Prove that for $n \geq 1$, the last $n+2$ digits of $11^{10^{n}}$ are $6000 \ldots 0001$, with $n$ zeros between the 6 and the final 1. | [
"Solution:\nThe statement is that $11^{10^{n}} = k \\cdot 10^{n+2} + 6 \\cdot 10^{n+1} + 1$ for some positive integer $k$; we prove it by induction on $n$.\n\nFor the base case $n=1$, we expand $(10+1)^{10}$ by the binomial theorem and obtain\n$$\n\\left(\\text{ terms divisible by } 10^{3}\\right) + \\binom{10}{2} ... | United States | Berkeley Math Circle | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
0g10 | Problem:
Soit $ABC$ un triangle avec $AB \neq AC$ et soit $M$ le milieu de $BC$. La bissectrice de $\angle BAC$ coupe la droite $BC$ en $Q$. Soit $H$ le pied de la hauteur en $A$ sur $BC$. La perpendiculaire à $AQ$ passant par $A$ coupe la droite $BC$ en $S$. Montrer que $MH \cdot QS = AB \cdot AC$. | [
"Solution:\n\nSoit $I$ le point d'intersection de la médiatrice de $BC$ et de la bissectrice de $\\angle BAC$. Celui-ci est bien défini car $AB \\neq AC$. Il est connu que la bissectrice d'un angle d'un triangle et la médiatrice du côté opposé s'intersectent sur le cercle circonscrit du triangle. De là s'ensuit que... | Switzerland | IMO-Selektion | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"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 | |
09nc | Let $\alpha$ be a real number. Suppose that there exists a $2024 \times 2024$ real matrix $(a_{ij})$ such that $\sum_{k=1}^{2024} a_{ik}^2 = 1$ for all $i$ and $\sum_{k=1}^{2024} a_{ik}a_{jk} = \alpha$ for all $i \neq j$.
Find the maximum possible value of $\alpha$.
(Nyamdavaa Amar) | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Vectors",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | 1 | |
062w | Problem:
Es sei $n$ eine positive ganze Zahl und $b$ die größte ganze Zahl, die kleiner als $(\sqrt[3]{28}-3)^{-n}$ ist. Man beweise, dass $b$ nicht durch 6 teilbar sein kann. | [
"Solution:\n\nDie komplexe Zahl $\\omega=\\frac{-1+\\sqrt{3} i}{2}$ ist bekanntlich eine dritte Einheitswurzel und erfüllt $\\omega^{2}=\\frac{-1-\\sqrt{3} i}{2}$, $\\omega^{3}=1$ und $\\omega^{2}+\\omega+1=0$, insbesondere gilt\n$$\n1+\\omega^{j}+\\omega^{2 j}= \\begin{cases}3, & \\text{ falls } j \\text{ durch dr... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0fa2 | Problem:
A polygon can be transformed into a new polygon by making a straight cut, which creates two new pieces each with a new edge. One piece is then turned over and the two new edges are reattached. Can repeated transformations of this type turn a square into a triangle? | [] | Soviet Union | 25th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
08cq | Problem:
Siano $ABC$ un triangolo e $P$ un suo punto interno. Sia $H$ il punto sul lato $BC$ tale che la bisettrice dell'angolo $\widehat{AHP}$ è perpendicolare alla retta $BC$. Sapendo che $\widehat{ABC}=\widehat{HPC}$ e $\widehat{BPC}=130^\circ$, determinare la misura dell'angolo $\widehat{BAC}$. | [
"Solution:\n\nSia $P'$ il simmetrico del punto $P$ rispetto alla retta $BC$. Si noti che i punti $A$, $H$, $P'$ sono allineati, dato che $\\widehat{AHP'}=\\widehat{AHP}+2\\widehat{PHC}=\\widehat{AHP}+2\\left(90^\\circ-\\widehat{AHP}/2\\right)=180^\\circ$.\n\nOra, $\\widehat{AP'C}=\\widehat{HP'C}=\\widehat{HPC}$ per... | Italy | GARA di FEBBRAIO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | null | proof and answer | 50° | |
09vt | For an integer $n \ge 3$ we consider a circle containing $n$ vertices. To each vertex we assign a positive integer, and these integers do not necessarily have to be distinct. Such an assignment of integers is called *stable* if the product of any three adjacent integers is $n$. For how many values of $n$ with $3 \le n ... | [
"Suppose $n$ is not a multiple of 3 and that we have a stable assignment of the numbers $a_1, a_2, \\dots, a_n$, in that order on the circle. Then we have $a_i a_{i+1} a_{i+2} = n$ for all $i$, where the indices are considered modulo $n$. Hence,\n$$\na_{i+1} a_{i+2} a_{i+3} = n = a_i a_{i+1} a_{i+2},\n$$\nwhich yie... | Netherlands | BxMO Team Selection Test, March 2020 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 680 | |
04vn | Do there exist mutually distinct real numbers $a$, $b$, $c$ such that the numbers $a^2 + b$, $b^2 + c$, $c^2 + a$ are equal, in some order, to the numbers $a + b^2$, $b + c^2$, $c + a^2$? | [] | Czech Republic | School Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | No | |
08b6 | Problem:
Siano $\Gamma$ una circonferenza, $AB$ una sua corda, $C$ un punto interno ad $AB$, $r$ una retta per $C$ tale che, dette $D$ ed $E$ le intersezioni di $r$ con $\Gamma$, esse si trovino in parti opposte rispetto all'asse di $AB$. Siano poi $\Gamma_{D}$ la circonferenza tangente esternamente a $\Gamma$ in $D$ ... | [] | Italy | XXXI Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0iyc | Problem:
Let $\triangle ABC$ be an equilateral triangle with height $13$, and let $O$ be its center. Point $X$ is chosen at random from all points inside $\triangle ABC$. Given that the circle of radius $1$ centered at $X$ lies entirely inside $\triangle ABC$, what is the probability that this circle contains $O$? | [
"Solution:\n\nThe set of points $X$ such that the circle of radius $1$ centered at $X$ lies entirely inside $\\triangle ABC$ is itself a triangle, $A'B'C'$, such that $AB$ is parallel to $A'B'$, $BC$ is parallel to $B'C'$, and $CA$ is parallel to $C'A'$, and furthermore $AB$ and $A'B'$, $BC$ and $B'C'$, and $CA$ an... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | sqrt(3)*pi/100 | |
041p | Find all positive real numbers $t$ with the following property: there exists an infinite set $X$ of real numbers such that the inequality
$$
\max\{|x-(a-d)|, |y-a|, |z-(a+d)|\} > td
$$
holds for all (not necessarily distinct) $x, y, z \in X$, all real numbers $a$ and all positive real numbers $d$. | [
"The answer is $0 < t < \\frac{1}{2}$.\n\nFirstly, for $0 < t < \\frac{1}{2}$, choose $\\lambda \\in \\left(0, \\frac{1-2t}{2(1+t)}\\right)$, let $x_i = \\lambda^i$, $X = \\{x_1, x_2, \\dots\\}$. We claim that for all (not necessarily distinct) $x, y, z \\in X$, all real numbers $a$ and all positive real numbers $d... | China | China Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | (0, 1/2) | |
055l | Let $B = (-1,0)$ and $C = (1,0)$ be fixed points on the coordinate plane. A nonempty, bounded subset $S$ of the plane is said to be *nice* if
(i) there is a point $T \in S$ such that for every point $Q \in S$, the segment $TQ$ lies entirely in $S$; and
(ii) for any triangle $P_1P_2P_3$, there exists a unique point $A \... | [
"See IMO 2016 shortlist, problem G3."
] | Estonia | IMO Team Selection Contest II | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate... | English | proof only | null | |
0iwa | Problem:
Let $G$ be a finite graph in which every vertex has degree $k$. Prove that the chromatic number of $G$ is at most $k+1$. | [
"Solution:\n\nWe find a good coloring with $k+1$ colors. Order the vertices and color them one by one. Since each vertex has at most $k$ neighbors, one of the $k+1$ colors has not been used on a neighbor, so there is always a good color for that vertex. In fact, we have shown that any graph in which every vertex ha... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0602 | Problem:
Soient $a, b$ deux entiers tels que $\mathrm{pgcd}(a, b)$ a au moins deux facteurs premiers distincts. Soit $S=\{x \in \mathbb{N} \mid x \equiv a[b]\}$. Un élément de $S$ est dit irréductible s'il ne peut pas s'écrire comme un produit d'au moins deux éléments de $S$ (pas forcément distincts).
Montrer qu'il e... | [
"Solution:\n\nSoit $d=\\operatorname{pgcd}(a, b)$, et écrivons $a=d a'$, $b=d b'$. Commençons par traiter le cas où $\\operatorname{pgcd}(d, b')>1$. Si c'est le cas, alors on a $\\operatorname{pgcd}(a^2, b)=d\\operatorname{pgcd}(d a'^2, b')>d$. Donc, pour tout $k \\geqslant 2$, on a $a^{k} \\not \\equiv a[b]$. En p... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > ... | null | proof only | null | |
06mb | Someone obtained a positive integer $n$ by concatenating the positive integers from 1 to 100000 in order, i.e. $n = 123456789101112...9999899999100000$. How many times does '2022' appears as consecutive digits of $n$? | [
"For clearer illustration, we add a comma between two integers as we form $n$ by concatenating the integers from $1$ to $100000$, i.e. we write\n$$\nn = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, \\dots, 99998, 99999, 100000.\n$$\n\nWe count the number of occurrences of $2022$ according to the positions of the commas w... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Discrete Mathematics > Other"
] | English | proof and answer | 44 | |
04bf | Eleonora has many cubes with all white sides. First she takes one cube and puts it in an empty box. Then she takes one cube at a time and paints some of its sides green, in such a way that the cube is different from all the other cubes that are already in the box, and then she puts that cube also in the box. At most ho... | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 10 | |
00dx | Ana writes lists of numbers, according to the following rules. The first number of the list must be an integer greater than 1. To compute the next number, Ana finds the least prime number that divides the last written number, and divides this number by that prime. Ana repeats this procedure many times until she writes ... | [
"We claim that, if the exponent of the greatest prime divisor $p$ of $n$ is greater than or equal to $3$, then $n$ is cuboso. In fact, at some step of the process, we will get $p^3$ on the board, after we divide by every smaller prime divisor and the remaining factors $p$.\n\nConversely, if a number is cuboso, then... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 44 | |
0lgb | Problem:
Find the largest real $C$ such that for all pairwise distinct positive real $a_{1}, a_{2}, \ldots, a_{2019}$ the following inequality holds
$$
\frac{a_{1}}{\left|a_{2}-a_{3}\right|}+\frac{a_{2}}{\left|a_{3}-a_{4}\right|}+\ldots+\frac{a_{2018}}{\left|a_{2019}-a_{1}\right|}+\frac{a_{2019}}{\left|a_{1}-a_{2}\righ... | [
"Solution:\nWithout loss of generality we assume that $\\min \\left(a_{1}, a_{2}, \\ldots, a_{2019}\\right)=a_{1}$. Note that if $a, b, c$ $(b \\neq c)$ are positive, then $\\frac{a}{|b-c|}>\\min \\left(\\frac{a}{b}, \\frac{a}{c}\\right)$. Hence\n$$\nS=\\frac{a_{1}}{\\left|a_{2}-a_{3}\\right|}+\\cdots+\\frac{a_{201... | Zhautykov Olympiad | XV International Zhautykov Olympiad in Mathematics | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 1010 | |
026f | Problem:
Nas igualdades abaixo, cada letra representa um algarismo:
$$
AB + BC = CD \quad \text{ e } \quad AB - BC = BA
$$
quanto vale $A+B+C+D$ ? | [
"Solution:\n\n23"
] | Brazil | Desafios | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 23 | |
05im | Problem:
Ruby a effectué une série de mouvements avec son Rubik's cube. (Par exemple, elle peut tourner la face du haut dans le sens des aiguilles d'une montre, puis la face de fond de 180 degrés, puis la face de droite dans le sens contraire des aiguilles d'une montre. Ou n'importe quelle autre série de rotations de ... | [
"Solution:\n\nNotons $x_{0}$ la configuration de départ et $x_{n}$ la configuration après $n$ répétitions de la série de Ruby. Comme le nombre total de configurations d'un Rubik's cube est fini, à un moment une configuration va se répéter : $x_{n}=x_{m}$ avec $n<m$. Demandons maintenant à Ruby d'inverser sa série d... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Abstract Algebra > Group Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0gbd | 設三角形 $ABC$ 的外接圓為 $\Gamma$, 內心為 $I$。令 $M$ 為 $BC$ 邊的中點。由 $I$ 向 $BC$ 引垂線, 設垂足為 $D$。通過 $I$ 且與 $AI$ 垂直的直線分別與 $AB$, $AC$ 交於 $F$, $E$ 點。設三角形 $AEF$ 的外接圓與 $\Gamma$ 的另一交點為 $X$。證明直線 $XD$ 與 $AM$ 的交點落在 $\Gamma$ 上。 | [
"設 $AM$ 與 $\\Gamma$ 再交於 $Y$ 點，而 $XY$ 與 $BC$ 交於 $D'$ 點。由同一法，只需證明 $D = D'$，我們首先證明下面的引理。\n\n引理. 設圓內接四邊形 $PQRS$ 的兩條對角線交於 $T$ 點, 則有\n$$\n\\frac{QT}{TS} = \\frac{PQ \\cdot QR}{PS \\cdot SR}.\n$$\n\n引理證明. 將三角形 $W_1W_2W_3$ 的 (有向) 面積記為 $[W_1W_2W_3]$, 則有\n$$\n\\frac{QT}{TS} = \\frac{[PQR]}{[PSR]} = \\frac{\\frac{1}{2}PQ \\cd... | Taiwan | 二〇一七數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homot... | null | proof only | null | |
07im | In the triangle $ABC$, $I$ is the incenter and $M$ is the midpoint of the part of arc $BC$ on the circumcircle of $ABC$ that doesn't contain $A$. $X$ is a point on the external angle bisector of $\angle BAC$. The line $BX$ meets the circumcircle of $BIC$ at $T$, for the second time. The point $Y$ is on the circumcircle... | [
"Suppose that $O$ is the circumcenter of $AXC$. Since $M$ is the circumcenter of $BIC$:\n$$\n\\angle XTC = \\angle BTC = \\frac{1}{2}\\angle BMC = \\frac{1}{2}(\\angle B + \\angle C) = 180^{\\circ} - \\angle CAX\n$$\nYielding $XACT$ is cyclic. Thus, $CT$ would be the common perpendicular to the circumcircles of $XA... | Iran | 41th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
06qu | Let $f$ be a non-constant function from the set of positive integers into the set of positive integers, such that $a-b$ divides $f(a)-f(b)$ for all distinct positive integers $a, b$. Prove that there exist infinitely many primes $p$ such that $p$ divides $f(c)$ for some positive integer $c$. | [
"Assume that there are only finitely many primes $p_{1}, p_{2}, \\ldots, p_{m}$ that divide some function value produced of $f$.\nThere are infinitely many positive integers $a$ such that $v_{p_{i}}(a)>v_{p_{i}}(f(1))$ for all $i=1,2, \\ldots, m$, e.g. $a=\\left(p_{1} p_{2} \\ldots p_{m}\\right)^{\\alpha}$ with $\\... | IMO | IMO Problem Shortlist | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof only | null | |
0daq | Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove that
$$
\frac{a}{b} + \frac{b}{a} + \frac{b}{c} + \frac{c}{b} + \frac{c}{a} + \frac{a}{c} \geq 2 \sqrt{2} \left( \sqrt{\frac{1-a}{a}} + \sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}} \right).
$$ | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0hfc | Is it possible for some positive integers $a$ and $d$ to satisfy:
$$
a) [a, a + d] = [a, a + 2d];
$$
$$
b) [a, a + d] = [a, a + 4d];
$$
where by $[x, y]$ we denote the least common multiple of integer $x, y$? | [
"a) As $a + 2d > a$, there exists some power of a prime $p^k$, that $a + 2d$ is divisible by $p^k$ and $a$ isn't divisible by $p^k$. From the given equality it follows that $a + d$ must be divisible by $p^k$. But then $2(a + d) - (a + 2d) = a$ is divisible by $p^k$, contradicting the choice of $p^k$. This contradic... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | a) No, impossible. b) Yes; for example a = 4 and d = 2. | |
0b35 | Problem:
In rolling three fair twelve-sided dice simultaneously, what is the probability that the resulting numbers can be arranged to form a geometric sequence?
(a) $\frac{1}{72}$
(b) $\frac{5}{288}$
(c) $\frac{1}{48}$
(d) $\frac{7}{288}$ | [] | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | d | |
0h9f | Determine the greatest positive integer that has pairwise distinct digits and is divisible by each of its digits. | [
"Clearly, $0$ is not one of the digits. In order to be divisible by $5$ last digit has to be $5$, but the number will not be divisible by $2$, $4$, $6$ and $8$, so it will consist of more digits if $5$ is not one of them. But then if all other digits are present, it is not divisible by $3$, $6$ and $9$. Therefore, ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 9867312 | |
0e2r | Let $ABC$ be an isosceles triangle with the apex at $C$. Let $D$ and $E$ be two points on the sides $AC$ and $BC$, such that the angle bisectors $\angle DEB$ and $\angle ADE$ meet at $F$, which lies on the segment $AB$. Prove that $F$ is the midpoint of $AB$. | [
"Denote $\\angle BAC = \\alpha$, $\\angle ADF = \\varphi$ and $\\angle FEB = \\psi$. The triangle $ABC$ is isosceles with the apex at $C$, so $\\angle CBA = \\angle BAC = \\alpha$. The segments $DF$ and $EF$ bisect the angles $\\angle ADE$ and $\\angle DEB$, so $\\angle FDE = \\varphi$ and $\\angle DEF = \\psi$.\n\... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0g5q | 將數字 $1, 2, \dots, n$ 任意排成數列 $a_1, a_2, \dots, a_n$, 然後執行下面的操作:
選擇兩個連續對 $(a_j, a_{j+1})$ 與 $(a_k, a_{k+1})$, 其中 $j \le k - 2$, 然後交換這兩對數字的位置; 也就是新的排列為:
$$
(a'_1, \dots, a'_{j-1}, a'_j, a'_{j+1}, a'_{j+2}, \dots, a'_{k-1}, a'_k, a'_{k+1}, \dots) = (a_1, \dots, a_{j-1}, a_k, a_{k+1}, a_{j+2}, \dots, a_{k-1}, a_j, a_{j+1}, ... | [
"(a)\n考慮排列的 inversion 數 ($inv$), 例如排列 $23514$ 的 inversion 有 $(2, 1)$, $(3, 1)$, $(5, 1)$, $(5, 4)$, 所以 $inv = 4$。\n經過一次操作, 改變 $inv$ 值的部分只需要考慮到:\n$$\n\\begin{aligned}\n& (a_j, a_{j+1}, *, a_k, a_{k+1}) \\rightarrow (*, a_k, a_{k+1}, a_j, a_{j+1}) \\\\\n& \\rightarrow (a_k, a_{k+1}, *, a_j, a_{j+1}).\n\\end{aligned}\... | Taiwan | 二〇一一數學奧林匹亞競賽第一階段選訓營,獨立研究(三) | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | (a) No. (b) Yes. | |
06rg | Let $n$ be a positive integer and let $W = \ldots x_{-1} x_{0} x_{1} x_{2} \ldots$ be an infinite periodic word consisting of the letters $a$ and $b$. Suppose that the minimal period $N$ of $W$ is greater than $2^{n}$.
A finite nonempty word $U$ is said to appear in $W$ if there exist indices $k \leq \ell$ such that $... | [
"Throughout the solution, all the words are nonempty. For any word $R$ of length $m$, we call the number of indices $i \\in \\{1,2, \\ldots, N\\}$ for which $R$ coincides with the subword $x_{i+1} x_{i+2} \\ldots x_{i+m}$ of $W$ the multiplicity of $R$ and denote it by $\\mu(R)$. Thus a word $R$ appears in $W$ if a... | IMO | 52nd International Mathematical Olympiad 2011 Shortlist | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
01c8 | Let $q$ be a fixed positive rational number. Call number $x$ *charismatic* if there exist a positive integer $n$ and integers $\alpha_1, \alpha_2, \dots, \alpha_n$ such that
$$
x = (q+1)^{\alpha_1} \cdot (q+2)^{\alpha_2} \cdots (q+n)^{\alpha_n}.
$$
a) Prove that $q$ can be chosen in such a way that every positive rati... | [
"a) Take $q = 1$ and let $x$ be any positive rational number. Let $n = p-1$ where $p$ is the largest prime number that divides either the numerator or the denominator of $x$. Then all prime numbers occurring in the canonical representation of $x$ with non-zero exponent are in the form $1+i$ with $1 \\le i \\le n$. ... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | null | proof and answer | a) q = 1 works; every positive rational number is charismatic.
b) No. For q = 1/3, x = 1 is charismatic but x + 1 = 2 is not charismatic. | |
0i2z | Problem:
For each positive integer $n$, let $a_{n}$ be the number of permutations $\tau$ of $\{1,2, \ldots, n\}$ such that $\tau(\tau(\tau(x)))=x$ for $x=1,2, \ldots, n$. The first few values are
$$
a_{1}=1,\ a_{2}=1,\ a_{3}=3,\ a_{4}=9.
$$
Prove that $3^{334}$ divides $a_{2001}$.
(A permutation of $\{1,2, \ldots, n\... | [
"Solution:\n\nConsider the permutations $\\tau$ of $\\{1,2, \\ldots, n\\}$ such that $\\tau(\\tau(\\tau(x)))=x$ for $x=1,2, \\ldots, n$. Then for each $x \\in\\{1,2, \\ldots, n\\}$, there are only two possibilities. Either $x$ is a fixed point; i.e., $\\tau(x)=x$, or else $x$ is a member of a 3-cycle $(xyz)$; i.e. ... | United States | 3rd Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
06p0 | For $x \in (0,1)$ let $y \in (0,1)$ be the number whose $n$th digit after the decimal point is the $\left(2^{n}\right)$th digit after the decimal point of $x$. Show that if $x$ is rational then so is $y$.
(Canada) | [
"Since $x$ is rational, its digits repeat periodically starting at some point. We wish to show that this is also true for the digits of $y$, implying that $y$ is rational.\n\nLet $d$ be the length of the period of $x$ and let $d = 2^{u} \\cdot v$, where $v$ is odd. There is a positive integer $w$ such that\n$$\n2^{... | IMO | IMO 2006 Shortlisted Problems | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof only | null | |
0hnv | Problem:
Given three squares of dimensions $2 \times 2$, $3 \times 3$, and $6 \times 6$, choose two of them and cut each into 2 figures, such that it is possible to make another square from the obtained 5 figures. | [
"Solution:\nThe solution is shown in the picture below\n\n"
] | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | A 7×7 square (e.g., cut the 3×3 into 3×2 and 3×1 and the 2×2 into two 2×1 pieces, then assemble with the 6×6) | |
0knx | Problem:
Emily's broken clock runs backwards at five times the speed of a regular clock. Right now, it is displaying the wrong time. How many times will it display the correct time in the next 24 hours? It is an analog clock (i.e. a clock with hands), so it only displays the numerical time, not AM or PM. Emily's clock... | [
"Solution:\n\nWhen comparing Emily's clock with a normal clock, the difference between the two times decreases by 6 seconds for every 1 second that passes. Since this difference is treated as 0 whenever it is a multiple of 12 hours, the two clocks must agree once every $\\frac{12}{6}=2$ hours. Thus, in a 24 hour pe... | United States | HMMT November | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 12 | |
08j2 | Problem:
Let $n \geq 1$ be a positive integer. A permutation $(a_{1}, a_{2}, \ldots, a_{n})$ of the numbers $(1,2, \ldots, n)$ is called quadratique if among the numbers $a_{1}, a_{1}+a_{2}, \ldots, a_{1}+a_{2}+\ldots+a_{n}$ there exists at least one perfect square. Find the greatest number $n$, which is less than $20... | [] | JBMO | The third selection test for IMO 2003 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Pell's equations"
] | null | proof and answer | 1681 | |
034v | Problem:
In an acute $\triangle ABC$ with $CA \neq CB$ and incenter $O$ denote by $A_1$ and $B_1$ the tangent points of its excircles to the sides $CB$ and $CA$, respectively. The line $CO$ meets the circumcircle of $\triangle ABC$ at point $P$ and the line through $P$ which is perpendicular to $CP$ meets the line $AB... | [
"Solution:\n\nLet $K = CO \\cap AB$ and $\\Varangle AKC = \\varphi$. Denote by $M$ and $N$ the intersection points of $QO$ with $CB$ and $CA$, respectively. The Sine theorem for $\\triangle APQ$ gives\n$$\n\\frac{AQ}{PQ} = \\frac{\\sin \\left(90^\\circ + \\beta\\right)}{\\sin \\frac{\\gamma}{2}} = \\frac{\\cos \\be... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry ... | null | proof only | null | |
00zo | Problem:
Determine all functions $f$ from the real numbers to the real numbers, different from the zero function, such that $f(x) f(y) = f(x-y)$ for all real numbers $x$ and $y$. | [
"Solution:\n\nAnswer: $f(x) \\equiv 1$ is the only such function.\n\nSince $f$ is not the zero function, there is an $x_{0}$ such that $f\\left(x_{0}\\right) \\neq 0$. From $f\\left(x_{0}\\right) f(0) = f\\left(x_{0} - 0\\right) = f\\left(x_{0}\\right)$ we then get $f(0) = 1$.\n\nThen by $f(x)^{2} = f(x) f(x) = f(x... | Baltic Way | Baltic Way 1997 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) ≡ 1 | |
0gd7 | 令 $m, n \ge 2$ 為整數且令 $f(x_1, \dots, x_n)$ 為一實係數多項式使得對每一個 $x_1, x_2, \dots, x_n \in \{0, 1, \dots, m-1\}$,
$$
f(x_1, \dots, x_n) = \left[ \frac{x_1 + \dots + x_n}{m} \right]
$$
均成立。試證: $f$ 的次數至少為 $n$。 | [
"We transform the problem to a single variable question by the following.\n\n*Lemma.* Let $a_1, \\dots, a_n$ be nonnegative integers and let $G(x)$ be a nonzero polynomial with $\\deg G \\le a_1 + \\dots + a_n$. Suppose that some polynomial $F(x_1, \\dots, x_n)$ satisfies\n$$\nF(x_1, \\dots, x_n) = G(x_1 + \\dots +... | Taiwan | 二〇一九數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | proof only | null | |
05wl | Problem:
Soit $k \geqslant 2$ un entier. Trouver le plus petit entier $n \geqslant k+1$ pour lequel il existe un ensemble $E$ de $n$ réels, deux à deux distincts, dont chaque élément peut s'écrire comme la somme de $k$ autres éléments de $E$, qui sont eux-mêmes deux à deux distincts. | [
"Solution:\n\nSoit $n$ un entier et $E$ un ensemble tels que décrits dans l'énoncé. On trie les éléments de $E$ dans l'ordre croissant : ce sont $x_{1}<x_{2}<\\cdots<x_{k}$. Enfin, soit $s$ la somme de tous les éléments de $S$.\nPuisque $x_{1} \\geqslant x_{2}+x_{3}+\\cdots+x_{k+1}$ et $x_{n} \\leqslant x_{n-1-k}+x... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | k+4 | |
06na | Find the remainder when $1 \times 2 \times 3 \times 4 + 2 \times 3 \times 4 \times 5 + \dots + 2023 \times 2024 \times 2025 \times 2026$ is divided by $1000$. | [
"Answer: $120$\nLet $f(n) = n(n+1)(n+2)(n+3)(n+4)$. Note that\n$$\n\\begin{aligned}\nf(n+1) - f(n) &= (n+1)(n+2)(n+3)(n+4)(n+5) - n(n+1)(n+2)(n+3)(n+4) \\\\\n&= 5(n+1)(n+2)(n+3)(n+4).\n\\end{aligned}\n$$\nHence we have\n$$\n1 \\times 2 \\times 3 \\times 4 = \\frac{f(1) - f(0)}{5}, \\quad 2 \\times 3 \\times 4 \\tim... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 120 | |
02oh | We call a number *pal* if it doesn't have a zero digit and the sum of the squares of the digits is a perfect square. For example, $122$ and $34$ are *pal* but $304$ and $12$ are not *pal*. Prove that there exists a *pal* number with $n$ digits, $n > 1$. | [
"Consider the number $\\underbrace{55\\dots5}_{n \\text{ times}}$. The sum of the squares of its digits is $n \\cdot 5^2 = 25n$. We can exchange any two fives by one three and one four, so the sum of the squares decreases by $5^2$, until we run out of fives. So we can get any sum from $25 \\cdot \\lfloor n/2 \\rflo... | Brazil | Brazilian Math Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
05qr | Problem:
Soit $ABCD$ un quadrilatère convexe d'aire $S$. On note $a = AB$, $b = BC$, $c = CD$ et $d = DA$. Pour toute permutation $x, y, z, t$ de $a, b, c, d$, montrer que
$$
S \leqslant \frac{1}{2}(xy + zt)
$$ | [
"Solution:\n\nSi $x$ et $y$ sont adjacents, sans perte de généralité on se ramène à montrer que $S \\leqslant \\frac{1}{2}(ab + cd)$. Cela découle de $S_{ABC} = \\frac{1}{2} AB \\cdot BC \\cdot \\sin \\widehat{ABC} \\leqslant \\frac{1}{2} ab$, et de même $S_{CDA} \\leqslant \\frac{1}{2} cd$.\n\nSi $x$ et $y$ sont d... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
04gj | Let $a$, $b$, $c$ and $d$ be real numbers such that
$$
2 \cos a + 6 \cos b + 7 \cos c + 9 \cos d = 0, \\
2 \sin a - 6 \sin b + 7 \sin c - 9 \sin d = 0.
$$
If $\cos(b+c) \neq 0$, determine the value of $\frac{\cos(a+d)}{\cos(b+c)}$. | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Linear Algebra > Vectors"
] | English | proof and answer | 7/3 | |
0bi4 | Find all primes $p$ and $q$, with $p \le q$, so that
$$
p(2q + 1) + q(2p + 1) = 2(p^2 + q^2).
$$ | [
"The equality can be written $p+q = 2(p-q)^2$, which shows that $p$ is odd.\nIf $p \\ge 5$, then $p$ and $q$ leave remainder 1 or 2 when divided by 3.\nWe will show that in this case the equality is impossible. Indeed, if $p$ and $q$ leave the same remainder mod 3, then $3 \\mid 2(p-q)^2$ and $3 \\nmid p+q$; if $p$... | Romania | 65th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | p = 3, q = 5 | |
0gnr | In a computer network consisting of $2008$ computers no two cycles intersect. At time $t = 0$, a hacker hacks into a computer in this network; and at time $t = 1$, the network administrator installs a protective software to an unhacked computer. For each positive integer $k$, at time $t = 2k$, the hacker hacks into ano... | [
"The answer is $671$.\n\nFirst consider a network that has a single cycle $\\{C_1, C_2, C_3, C_4, C_5, C_6\\}$, and has chains of $668$, $667$ and $667$ computers starting at $C_1$, $C_3$ and $C_5$, respectively. Hacker takes $C_1$ in the first move. Then if the administrator takes $C_4$, the moves $C_2$, $C_3$, $C... | Turkey | 16th Turkish Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 671 | |
0iuc | Problem:
Given a rearrangement of the numbers from $1$ to $n$, each pair of consecutive elements $a$ and $b$ of the sequence can be either increasing (if $a < b$) or decreasing (if $b < a$). How many rearrangements of the numbers from $1$ to $n$ have exactly two increasing pairs of consecutive elements? | [
"Solution:\n\nNotice that each such permutation consists of $3$ disjoint subsets of $\\{1, \\ldots, n\\}$ whose union is $\\{1, \\ldots, n\\}$, each arranged in decreasing order. For instance, if $n=6$, in the permutation $415326$ (which has the two increasing pairs $15$ and $26$), the three sets are $\\{4,1\\}$, $... | United States | $12^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 3^n - (n+1)·2^n + n(n+1)/2 | |
021v | Problem:
Quem é menor? - Sem usar calculadora, decida qual dos números $33^{12}$, $63^{10}$ e $127^{8}$ é o menor. | [
"Solution:\nObservemos que:\n$$\n\\begin{aligned}\n& 33^{12} > 32^{12} = \\left(2^{5}\\right)^{12} = 2^{60} \\\\\n& 63^{10} < 64^{10} = \\left(2^{6}\\right)^{10} = 2^{60} \\\\\n& 127^{8} < 128^{8} = \\left(2^{7}\\right)^{8} = 2^{56}\n\\end{aligned}\n$$\nLogo, o maior dos números é $33^{12}$.\n\nPor outro lado, $\\f... | Brazil | Nível 2 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Equations and Inequalities"
] | null | proof and answer | 127^8 | |
0btr | Let $ABC$ be a right triangle with legs $AB$ and $AC$. The bisector of the angle $ACB$ intersects $AB$ in $D$ and the perpendicular in $B$ on $BC$ in $E$. Denote $F$ the reflection of $E$ across $B$ and $P$ the intersection of the lines $DF$ and $BC$. Prove that $EP \perp CF$.
Cătălin Cristea | [
"In $\\triangle BEC$, $m(\\angle CEB) = 180^\\circ - m(\\angle EBC) - m(\\angle ECB) = 90^\\circ - m(\\angle ECB)$. In $\\triangle ADC$, $m(\\angle ADC) = 180^\\circ - m(\\angle DAC) - m(\\angle ACD) = 90^\\circ - m(\\angle ACD)$. Since $m(\\angle ACD) = m(\\angle ECB)$, it follows $\\angle ADC \\equiv \\angle CEB$... | Romania | 67th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0ixh | Problem:
A $k$-clique of a graph is a set of $k$ vertices such that all pairs of vertices in the clique are adjacent. The clique number of a graph is the size of the largest clique in the graph. Does there exist a graph which has a clique number smaller than its chromatic number? | [
"Solution:\n\nConsider a graph with 5 vertices arranged in a circle, with each vertex connected to its two neighbors. If only two colors are used, it is impossible to alternate colors to avoid using the same color on two adjacent vertices, so the chromatic number is 3. Its clique number is 2, so we have found such ... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | Yes; for example, the five-cycle has clique number 2 and chromatic number 3. | |
0b3p | Problem:
You roll a fair 12-sided die repeatedly. The probability that all the primes show up at least once before seeing any of the other numbers can be expressed as a fraction $p / q$ in lowest terms. What is $p+q$? | [
"Solution:\nThere are 5 primes which are at most 12 - namely $2, 3, 5, 7$ and $11$. Notice that if a number has already been seen, we can effectively ignore all future occurrences of the number. Thus, the desired probability is the fraction of the permutations of $(1, 2, \\ldots, 12)$ such that the primes all occur... | Philippines | 24th Philippine Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | final answer only | 793 | |
09jg | Prove that the number of $4 \times 4$ Latin squares is $576$. Here a $4 \times 4$ Latin square is a $4 \times 4$ array filled with numbers from $1$ to $4$, each occurring exactly once in each row and exactly once in each column. | [] | Mongolia | Mongolian Mathematical Olympiad Round 1 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 576 | |
005s | Sea $ABC$ un triángulo obtusángulo en $C$ tal que $2B\hat\{A\}C = A\hat\{B\}C$. Sea $P$ un punto sobre el lado $AB$ tal que $BP = 2BC$. Sea $M$ el punto medio de $AB$ ($M$ está entre $P$ y $B$). Probar que la perpendicular al lado $AC$, trazada por $M$, corta a $PC$ en su punto medio. | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | Spanish | proof only | null | |
0dol | Let $a_1, a_2, \dots, a_{2003}$ be a sequence of real numbers. A term $a_k$, $1 \le k \le 2003$, is said to be a *leading term*, if at least one of the expressions $a_k, a_k+a_{k+1}, \dots, a_k+a_{k+1}+\dots+a_{2003}$ is positive. Prove that the sum of all leading terms is positive provided that the sequence has at lea... | [
"We solve this problem for any sequence having $n$ terms applying induction with respect to $n$.\n\nThe case $n = 1$ is clear.\n\nSuppose that the statement is true for all sequences of length less than $n$.\n\nNow consider a sequence $a_1, a_2, \\dots, a_n$.\n\n*Case 1. $a_1$ is not a leading term.*\n\nThen the se... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0i6d | Problem:
Let $ABCD$ be a quadrilateral, and let $E, F, G, H$ be the respective midpoints of $AB, BC, CD, DA$. If $EG = 12$ and $FH = 15$, what is the maximum possible area of $ABCD$? | [
"Solution:\nThe area of $EFGH$ is $EG \\cdot FH \\sin \\theta / 2$, where $\\theta$ is the angle between $EG$ and $FH$. This is at most $90$. However, we claim the area of $ABCD$ is twice that of $EFGH$. To see this, notice that $EF = AC / 2 = GH$, $FG = BD / 2 = HE$, so $EFGH$ is a parallelogram. The half of this ... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 180 | |
0c5b | Let $\triangle ABC$ be an equilateral triangle and let $A_1, A_2$ be points on the side $BC$, $B_1, B_2$ be points on the side $CA$, $C_1, C_2$ be points on the side $AB$, such that $BA_1 < BA_2$, $CB_1 < CB_2$, $AC_1 < AC_2$ and $A_1A_2 = B_1B_2 = C_1C_2$. Prove that there exists a triangle whose side lengths are $A_2... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
0jvv | Problem:
For any positive integer $n$, $S_{n}$ be the set of all permutations of $\{1,2,3, \ldots, n\}$. For each permutation $\pi \in S_{n}$, let $f(\pi)$ be the number of ordered pairs $(j, k)$ for which $\pi(j)>\pi(k)$ and $1 \leq j<k \leq n$. Further define $g(\pi)$ to be the number of positive integers $k \leq n$ ... | [
"Solution:\nDefine an $n \\times n$ matrix $A_{n}(x)$ with entries $a_{i, j}=x$ if $i \\equiv j \\pm 1(\\bmod n)$ and 1 otherwise. Let $F(x)=\\sum_{\\pi \\in S_{n}}(-1)^{f(\\pi)} x^{g(\\pi)}$ (here $(-1)^{f(\\pi)}$ gives the sign $\\prod \\frac{\\pi(u)-\\pi(v)}{u-v}$ of the permutation $\\pi$ ). Note by constructio... | United States | HMMT February 2016 | [
"Algebra > Linear Algebra > Determinants",
"Algebra > Linear Algebra > Matrices",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof and answer | 995*2^998 | |
0ieh | Problem:
Let $P(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + \cdots + a_{0}$ be a polynomial with real coefficients, $a_{n} \neq 0$. Suppose every root of $P$ is a root of unity, but $P(1) \neq 0$. Show that the coefficients of $P$ are symmetric; that is, show that $a_{n} = a_{0}, a_{n-1} = a_{1}, \ldots$ | [
"Solution:\n\nSince the coefficients of $P$ are real, the complex conjugates of the roots of $P$ are also roots of $P$. Now, if $x$ is a root of unity, then $x^{-1} = \\bar{x}$. But the roots of\n$$\nx^{n} P\\left(x^{-1}\\right) = a_{0} x^{n} + a_{1} x^{n-1} + \\cdots + a_{n}\n$$\nare then just the complex conjugat... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
018j | Let $T$ denote the 15-element set $T = \{10a+b : 1 \le a < b \le 6\}$. Let $S \subseteq T$ be a subset of $T$ in which all 6 digits appear and in which no 3 members exist which contain together all 6 digits 1, 2, ..., 6. Determine the largest possible size $n$ of $S$. | [
"Consider the numbers of $T$, which contain $1$ or $2$. Certainly, no $3$ of them can contain all $6$ digits and all $6$ digits appear. Hence $n \\ge 9$.\n\nConsider the partitions:\n12, 36, 45,\n13, 24, 56,\n14, 26, 35,\n15, 23, 46,\n16, 25, 34.\n\nSince every row is a partition of $\\{1, 2, ..., 6\\}$, it contain... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 9 | |
0fwp | Problem:
Seien $x$, $y$, $n$ natürliche Zahlen mit $x \geq 3$, $n \geq 2$ und
$$
x^{2}+5=y^{n}
$$
Zeige, dass jeder Primteiler $p$ von $n$ die Kongruenz $p \equiv 1(\bmod 4)$ erfüllt. | [
"Solution:\n\nIst $n=2k$ gerade, dann ist $5=(y^{k})^{2}-x^{2}$ Differenz von zwei Quadraten. Dies ist nur für $y^{k}=3$ und $x=2$ möglich, im Widerspruch zu $x \\geq 3$. Folglich ist jeder Primteiler von $n$ ungerade.\n\nIst $x$ ungerade, dann ist die linke Seite durch $2$ aber nicht durch $4$ teilbar, im Widerspr... | Switzerland | IMO Selektion 2008 | [
"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 > Quadratic residues"
] | null | proof only | null | |
07mv | Prove that the point of intersection of the line $L = \{(x, y) \mid y = mx\}$ and the line joining the reflection of $(1, 1)$ in $L$ with the point $(-1, 1)$ lies on the circle of unit radius centred at $(0, 1)$. | [
"First of all, the reflection $(x', y')$ of $(x_0, y_0)$ in $L$ is given by\n$$\nx' = \\frac{(1 - m^2)x_0 + 2my_0}{1 + m^2}, \\quad y' = \\frac{2mx_0 + (m^2 - 1)y_0}{1 + m^2}.\n$$\nThis is so because\n$$\nx' + x_0 = \\frac{2(x_0 + my_0)}{1 + m^2}, \\quad y' + y_0 = \\frac{2m(x_0 + my_0)}{1 + m^2}.\n$$\nThus the ref... | Ireland | Ireland | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01mh | Prove that the number $9 \cdot 2^m$ can be presented as the sum of the squares of three positive integers for any positive integer $m$. | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | English | proof only | null | |
07v7 | Let $\triangle ABC$ be a triangle with circumcentre $O$. The perpendicular line from $O$ to $BC$ intersects line $BC$ at $M$ and line $AC$ at $P$, and the perpendicular line from $O$ to $AC$ intersects line $AC$ at $N$ and line $BC$ at $Q$. Let $D$ be the intersection point of lines $PQ$ and $MN$. Construct the paralle... | [
"(a) Because $OP$ is the perpendicular bisector of $BC$, $\\triangle PBC$ is isosceles with symmetry axis $PO$. Thus $|PB| = |PC|$ and $\\angle PBC = \\angle PCB = \\angle C$. Hence $\\angle APB = 2\\angle C$ (external angle).\nSimilarly, $OQ$ is the perpendicular bisector of $AC$ hence $\\triangle QAC$ is isoscele... | Ireland | IRL_ABooklet | [
"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 > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneo... | English | proof only | null | |
0cxc | Find all pairs ($m, n$) of integers, $m, n \geq 2$ such that $m n - 1$ divides $n^{3} - 1$. | [
"The solutions are $\\left(k, k^{2}\\right)$ and $\\left(k^{2}, k\\right)$, with $k \\geq 2$.\nWe have $m n - 1 \\mid \\left(n^{3} - 1\\right) m - n^{2}(m n - 1) = n^{2} - m$, hence\n$$\nm n - 1 \\mid m\\left(n^{2} - m\\right) - (m n - 1) n = n - m^{2}\n$$\nIf $n > m^{2}$, then $m n - 1 \\leq n - m^{2} \\leq n - 1$... | Saudi Arabia | SAMC | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All pairs are (k, k^2) and (k^2, k) with k ≥ 2. | |
04fu | Let $a$ and $b$ be different real numbers and define $s = a - b$ and $t = a^3 - b^3$. Express $(a+b)^2$ in terms of $s$ and $t$. (Santos J. Prob. Seminar) | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (4t/s - s^2)/3 | |
0fl0 | Problem:
Sean $a$, $b$, $c$ números reales positivos tales que $a b c = 1$. Prueba la desigualdad siguiente
$$
\left(\frac{a}{1+a b}\right)^{2}+\left(\frac{b}{1+b c}\right)^{2}+\left(\frac{c}{1+c a}\right)^{2} \geq \frac{3}{4}
$$ | [
"Solution:\nComo $a b c = 1$, entonces\n$$\n\\left(\\frac{a}{1+a b}\\right)^{2} = \\left(\\frac{c a}{a b c + c}\\right)^{2} = \\left(\\frac{c a}{1 + c}\\right)^{2}.\n$$\nAnálogamente se obtienen\n$$\n\\left(\\frac{b}{1+b c}\\right)^{2} = \\left(\\frac{a b}{1+a}\\right)^{2} \\text{ y } \\left(\\frac{c}{1+c a}\\right... | Spain | Fase Nacional de la XLV Olimpiada Matemática Española | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0ak4 | Find all the pairs $(m,n)$ of integers which satisfy the equation
$$m^5 - n^5 = 16mn.$$ | [
"If one of $m$, $n$ is $0$, the other has to be $0$ too, and $(m,n) = (0,0)$ is one solution.\n\nIf $mn \\neq 0$, let $d = \\gcd(m,n)$ and we write $m = da$, $n = db$, $a, b \\in \\mathbb{Z}$ with $(a,b) = 1$. Then, the given equation is transformed into\n$$\nd^3 a^5 - d^3 b^5 = 16ab \\quad (1)\n$$\nSo, by the abov... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | (m, n) = (0, 0) and (m, n) = (-2, 2) | |
0gay | 三角形 $ABC$ 中, $\angle A = 60^\circ$. 設點 $O$, $H$ 分別為 $\triangle ABC$ 的外心及垂心。在 $BH$ 線段上取一點 $M$, 並在直線 $CH$ 上取一點 $N$, 使得 $H$ 位於 $C$, $N$ 之間, 且 $BM = CN$. 試求
$$
\frac{MH + NH}{OH}
$$ | [
"$$\n\\frac{MH + NH}{OH} = \\sqrt{3}.\n$$\n在 $BH$ 線段上取 $K$ 點使得 $BK = CH$. 連 $OK$, $OB$, $OC$ 等線段。\n\n因為 $O$ 是 $\\triangle ABC$ 的外心, 所以 $\\angle BOC = 2\\angle A = 120^\\circ$. 又因為 $H$ 是 $\\triangle ABC$ 的垂心, 所以 $\\angle BHC = 180^\\circ - \\angle A = 120^\\circ$. 因此 $\\angle BOC = 120^\\cir... | Taiwan | 二〇一七數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | sqrt(3) | |
0ggg | 令 $a_1, a_2, \dots, a_n$ 為滿足 $a_1 + a_2 + \dots + a_n = 1$ 的正實數 ($n \ge 2$)。證明：
$$
\sum_{k=2}^{n} \frac{a_k}{1 - a_k} (a_1 + a_2 + \dots + a_{k-1})^2 < \frac{1}{3}
$$ | [
"$$\ns_k = a_1 + a_2 + \\dots + a_k \\quad \\text{and} \\quad b_k = \\frac{a_k s_{k-1}^2}{1 - a_k},\n$$\nwith the convention that $s_0 = 0$. Note that $b_k$ is exactly a summand in the sum we need to estimate. We shall prove the inequality\n$$\nb_k < \\frac{s_k^3 - s_{k-1}^3}{3}. \\qquad (1)\n$$\nIndeed, it suffice... | Taiwan | 2022 數學奧林匹亞競賽第三階段選訓營, 獨立研究(一) | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | Chinese; English | proof only | null | |
0cvu | Let $BE$ and $CF$ be altitudes in a tetrahedron $ABCD$. A plane $\alpha$ through the midpoint of $AD$ is perpendicular to $AD$. Assume that the points $A$, $C$, $D$, and $E$ lie on a circle, and the points $A$, $B$, $D$, and $F$ also lie on a circle. Show that the points $E$ and $F$ are equidistant from $\alpha$. (A. K... | [
"The line $CF$ is perpendicular to the plane $ABD$, so $CF \\perp AD$. Similarly, $BE \\perp AD$. Therefore, the lines $CF$ and $BE$ are parallel to the plane $\\alpha$ or lie in it. The points $B$, $C$, $E$, and $F$ lie on the sphere $\\omega$ circumscribed about the tetrahedron $ABCD$. Also, since $\\angle BEC = ... | Russia | Regional round | [
"Geometry > Solid Geometry > Other 3D problems"
] | English; Russian | proof only | null | |
0gbz | 一張月曆是一個長方形的方格紙。我們稱一張月曆符合法規, 若且唯若它滿足以下三點:
(1) 月曆的每一格都被塗成白色或紅色, 且恰有 $10$ 個紅色格子。
(2) 若月曆的一橫排共有 $N$ 格, 則當我們從最左上角的格子, 依次填入 $1, 2, \dots$, 由左而右, 然後由上而下, 我們將找不到連續 $N$ 個數字, 它們所在的格子都是白色的。
(3) 若月曆的一直列共有 $M$ 格, 則當我們從最左下角的格子, 依次填入 $1, 2, \dots$, 由下而上, 然後由左而右, 我們將找不到連續 $M$ 個數字, 它們所在的格子都是白色的(也就是說, 如果我們將整張月曆順時鐘旋轉 $90$ 度, 它仍然滿足條件 (2))... | [
"答案：$10! = 3628800$ 種。\n注意到在一個 $10 \\times 10$ 的方格表中塗紅 $10$ 格，使得每行每列都恰有一紅格的塗法共有 $10!$ 種。以下建立塗 $10 \\times 10$ 方格表與符合法規的月曆之間的一一對應。\n\n\n\n– 方格表 $\\rightarrow$ 符合法規的月曆：\n如上圖進行以下操作：\n\n1. 先觀察相鄰兩行，如果左行紅格比右行高，則將兩行間的格線加粗。\n2. 再觀察相鄰兩列，如果上列紅格比下列右，則將兩列間的格線加粗。\n3. 把加粗的格線全部擦掉，便得到一個符合法規的月曆：\n\n(證明)\n* ... | Taiwan | 二〇一八數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 10! = 3628800 | |
0fgc | Problem:
Hallar los valores de $n \in \mathbb{N}$ tales que $5^{n}+3$ es una potencia de 2 de exponente natural. | [
"Solution:\n\nPrimera solución\nConsideremos la ecuación\n$$\n5^{\\alpha}+3=2^{\\beta}\n$$\nCalculemos los posibles valores de $\\alpha$ dando valores bajos de $\\beta$. Los valores $\\beta=0,1$ no dan solución. Para $\\beta=2$ nos sale $\\alpha=0$. Los valores $\\beta=4,5,6,8,9$ tampoco dan solución, y para $\\bet... | Spain | OME 22 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | n = 0, 1, 3 | |
0d67 | Given two non-constant polynomials $P(x), Q(x)$ with real coefficients. For a real number $a$, we define
$$
P_{a} = \{ z \in \mathbb{C} : P(z) = a \} ; \quad Q_{a} = \{ z \in \mathbb{C} : Q(z) = a \} .
$$
Denote by $K$ the set of real numbers $a$ such that $P_{a} = Q_{a}$. Suppose that the set $K$ contains at least two... | [
"First we see that if the polynomial $P(x)-a$ has a root $\\alpha$ with multiplicity $k$, then $P'(x)$ also has the root $\\alpha$ with multiplicity $k-1$.\n\nAssume that $a, b$ are two distinct elements of $K$ and $r_{1}, r_{2}, \\ldots, r_{i}$ are the roots of $P(x)-a$ with multiplicity $k_{1}, k_{2}, \\ldots, k_... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
052m | 20 students participated on a field trip. They all wanted to climb on top of a lighthouse, but only one person was allowed to the lighthouse at once. The order of climbing was determined by a lottery such that in the beginning every student is randomly assigned a number of 1 through 20 (such that no number is repeated)... | [
"Let the number of students be $n$. The last student to climb the lighthouse has got all the numbers 1 through $n$ with the lottery. As number $n$ is only available in the first round, that student had to get $n$ in the first round. As number $n-1$ is only available in 1st and 2nd round and in the 1st round that st... | Estonia | Open Contests | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 6 | |
0ai9 | Consider an acute triangle $ABC$ with area $S$. Let $CD \perp AB$ ($D \in AB$), $DM \perp AC$ ($M \in AC$) and $EN \perp BC$ ($N \in BC$). Denote by $H_1$ and $H_2$ the orthocentres of the triangles $MNC$ and $MND$ respectively. Find the area of the quadrilateral $AH_1BH_2$ in terms of $S$.
 | [
"Let $O$, $P$, $K$, $R$ and $T$ be the mid-points of the segments $CD$, $MN$, $CN$, $CH_1$ and $MH_1$, respectively. From $\\triangle MNC$ we have that $\\overline{PK} = \\frac{1}{2}\\overline{MC}$ and $PK \\parallel MC$.\n\nAnalogously, from $\\triangle MH_1C$ we have that $\\overline{TR} = \\frac{1}{2}\\overline{... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof and answer | S | |
03m7 | Let $ABCD$ be a cyclic quadrilateral whose opposite sides are not parallel, $X$ the intersection of $AB$ and $CD$, and $Y$ the intersection of $AD$ and $BC$. Let the angle bisector of $\angle AXD$ intersect $AD$, $BC$ at $E$, $F$ respectively and let the angle bisector of $\angle AYB$ intersect $AB$, $CD$ at $G$, $H$ r... | [
"Since $ABCD$ is cyclic, $\\triangle XAC \\sim \\triangle XDB$ and $\\triangle YAC \\sim \\triangle YBD$. Therefore,\n$$\n\\frac{XA}{XD} = \\frac{XC}{XB} = \\frac{AC}{DB} = \\frac{YA}{YB} = \\frac{YC}{YD}.\n$$\nLet $s$ be this ratio. Therefore, by the angle bisector theorem,\n$$\n\\frac{AE}{ED} = \\frac{XA}{XD} = \... | Canada | Kanada 2011 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
050j | Integers $a$, $b$, $c$ are such that $a + b + c$ is divisible by $6$, and $a^2 + b^2 + c^2$ is divisible by $36$. Does it imply that $a^3 + b^3 + c^3$ is divisible by
a) $8$;
b) $27$? | [
"a) As the sum of $a$, $b$, and $c$ is divisible by $6$, and is therefore even, there must be either $0$ or $2$ odd numbers among the three. If we had $2$ odd numbers, the sum of the squares $a^2 + b^2 + c^2$ would give a remainder of $0 + 1 + 1 = 2$ when dividing by $4$. But this is not possible, since the sum is ... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | a) Yes; b) No | |
0gox | Let $D$ be a point different from the vertices on the side $BC$ of a triangle $ABC$. Let $I, I_1$ and $I_2$ be the incenters of the triangles $ABC$, $ABD$ and $ADC$, respectively. Let $E$ be the second intersection point of the circumcircles of the triangles $AI_1I$ and $ADI_2$, and $F$ be the second intersection point... | [
"Let $F'$ be the point of intersection of the bisectors of the angles $\\angle ABD$ and $\\angle ADC$. Since $\\angle F'BD = \\angle ABD/2$ and $\\angle F'DC = (\\angle ABD + \\angle BAD)/2$, we have $\\angle I_1F'D = \\angle BF'D = \\angle BAD/2 = \\angle I_1AD$. Therefore $A, I_1, D, F'$ are concyclic. We also ha... | Turkey | Team Selection Test for IMO 2011 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geom... | English | proof only | null | |
0e05 | The base $AB$ of the trapezoid $ABCD$ is longer than the base $CD$, and $\angle ADC$ is a right angle. The diagonals $AC$ and $BD$ are perpendicular. Let $E$ be the foot of the altitude from $D$ to the line $BC$. Prove that $\frac{|AE|}{|BE|} = \frac{|AC| \cdot |CD|}{|AC|^2 - |CD|^2}$. | [
"\n\n1st solution. Since $ACD$ is a right triangle, we have $|AC|^2 - |CD|^2 = |AD|^2$. The diagonals of the trapezoid intersect at a right angle, so $\\angle DCA = \\frac{\\pi}{2} - \\angle BDC = \\angle ADB$. This, together with $\\angle ADC = \\frac{\\pi}{2} = \\angle BAD$, implies that ... | Slovenia | Selection Examinations for the IMO | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0fki | Problem:
Determinar el mayor número de planos en el espacio tridimensional para los que existen seis puntos con las siguientes condiciones:
i) Cada plano contiene al menos cuatro de los puntos.
ii) Cuatro puntos cualesquiera no pertenecen a una misma recta. | [
"Solution:\n\nSean $r$ y $s$ dos rectas que se cruzan en el espacio. Sean $A, B$ y $C$ tres puntos distintos de $r$ y sean $P, Q$ y $R$ tres puntos distintos en $s$. Cada uno de los puntos de $r$ define con $s$ un plano, y análogamente cada punto de $s$ con $r$. Estos 6 planos cumplen las condiciones del problema, ... | Spain | XLV Olimpiada Matemática Española | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 6 | |
00bb | Prove that every positive integer can be expressed as a sum of powers of $3$, $4$ and $7$ in such a way that the representation does not contain two powers with the same base and the same exponent.
For example, $2 = 7^0 + 7^0$ and $22 = 3^2 + 3^2 + 4^1$ are not valid sums, but $2 = 3^0 + 7^0$ and $22 = 3^2 + 3^0 + 4^1 ... | [
"Consider the powers of $3$, $4$ and $7$ in increasing order $x_1^{\\alpha_1} \\le x_2^{\\alpha_2} \\le x_3^{\\alpha_3} \\le \\dots$\nWe will prove, by induction on $n$, that it is possible to represent all the integers from $1$ to $x_1^{\\alpha_1} + x_2^{\\alpha_2} + \\dots + x_n^{\\alpha_n}$ in the desired way us... | Argentina | 29° Olimpiada Matemática del Cono Sur | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0kuu | Problem:
Prove that there do not exist pairwise distinct complex numbers $a$, $b$, $c$, and $d$ such that
$$
a^{3}-b c d=b^{3}-c d a=c^{3}-d a b=d^{3}-a b c .
$$ | [
"Solution:\n\nFirst suppose none of them are $0$. Let the common value of the four expressions be $k$, and let $a b c d = P$. Then for $x \\in \\{a, b, c, d\\}$,\n$$\nx^{3} - \\frac{P}{x} = k \\Longrightarrow x^{4} - k x - P = 0\n$$\nHowever, Vieta's tells us $a b c d = -P$, meaning $P = -P$, so $P = 0$, a contradi... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers",
"Discrete Mathematics > Combinatorics > Pigeonhol... | null | proof only | null | |
0js5 | Problem:
We have 10 points on a line $A_{1}, A_{2}, \ldots, A_{10}$ in that order. Initially there are $n$ chips on point $A_{1}$. Now we are allowed to perform two types of moves. Take two chips on $A_{i}$, remove them and place one chip on $A_{i+1}$, or take two chips on $A_{i+1}$, remove them, and place a chip on $... | [
"Solution:\n\nAnswer: 46\n\nWe claim that $n=46$ is the minimum possible value of $n$. As having extra chips cannot hurt, it is always better to perform the second operation than the first operation, except on point $A_{1}$. Assign the value of a chip on point $A_{i}$ to be $i$. Then the total value of the chips in... | United States | HMMT November 2016 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 46 | |
01hj | Find all triples $(a, b, c)$ of real numbers such that
$$ \cos(ax) + \cos(bx) = 2 \cos(cx) $$
holds for all $x \in \mathbb{R}$. | [
"The triples we are looking for have the forms $(t, t, t)$, $(-t, t, t)$, $(t, -t, t)$, $(t, t, -t)$ where $t \\in \\mathbb{R}$.\nIf $c = 0$ then $\\cos(ax) + \\cos(bx) = 2$ for all $x \\in \\mathbb{R}$. Since $\\cos(ax) \\le 1$ and $\\cos(bx) \\le 1$, we must have $\\cos(ax) = 1$ and $\\cos(bx) = 1$ for any $x$. T... | Baltic Way | Baltic Way 2021 Shortlist | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All real triples with |a| = |b| = |c|; equivalently, all triples of the forms (t, t, t), (−t, t, t), (t, −t, t), (t, t, −t) for real t. | |
041q | Let the sequence $\{a_n\}$ be defined by $a_1 = 1$, $a_2 = 2$, $a_{n+1} = \frac{a_n^2 + (-1)^n}{a_{n-1}}$ $(n = 2, 3, \dots)$.
Prove that the sum of squares of any two adjacent terms of the sequence is also in the sequence. | [
"By $a_{n+1} = \\frac{a_n^2 + (-1)^n}{a_{n-1}}$, we have $a_{n+1}a_{n-1} = a_n^2 + (-1)^n$ $(n = 2, 3, \\dots)$, so\n$$\n\\begin{align*}\n\\frac{a_n - a_{n-2}}{a_{n-1}} &= \\frac{a_n a_{n-2} - a_{n-2}^2}{a_{n-1} a_{n-2}} = \\frac{a_{n-1}^2 + (-1)^{n-1} - a_{n-2}^2}{a_{n-1} a_{n-2}} \\\\\n&= \\frac{a_{n-1}^2 - a_{n-... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
061h | Problem:
Es sei $P$ die Menge aller geordneter Paare $(p, q)$ von nichtnegativen ganzen Zahlen. Man bestimme alle Funktionen $f: P \rightarrow \mathrm{IR}$ mit der Eigenschaft
$$
f(p, q)=\left\{\begin{array}{c}
0 \quad \text{ wenn } p q=0 \\
1+\frac{1}{2} f(p+1, q-1)+\frac{1}{2} f(p-1, q+1) \text{ sonst }
\end{array} ... | [
"Solution:\n\nDie einzige solche Funktion ist $f(p, q)=p \\cdot q$.\nDiese Funktion erfüllt offensichtlich für $p=0$ oder $q=0$ die erste Bedingung. Für $p q \\neq 0$ gilt\n$$\n1+\\frac{1}{2} f(p+1, q-1)+\\frac{1}{2} f(p-1, q+1)=1+\\frac{1}{2}(p+1)(q-1)+\\frac{1}{2}(p-1)(q+1)=p q,\n$$\nso dass auch die zweite Bedin... | Germany | Auswahlwettbewerb zur IMO 2002 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | f(p, q) = p*q | |
04pv | Prove that the number
$$
\overbrace{222\ldots2}^{n \text{ digits}} - 3^n + 1
$$
is divisible by $7$ for any positive integer $n$. (Matko Ljulj) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0abo | If one of the sides of a square is increased two times and the other one is increased by $22\text{ mm}$, then the new rectangle has a perimeter that is $2000\text{ mm}$ greater than the perimeter of the square. What is the side length of the square? | [
"If we denote the side of the square by $a$, then the side lengths of the rectangle are $2a$ and $a+22$. The sum of the two side lengths of the rectangle with length $2a$ is equal to the perimeter of the square. Hence the sum of the other two side lengths, with length $a+22$, is $2000\\text{ mm}$. Now we obtain $2a... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 978 mm | |
00n6 | Let $x$ and $y$ be real numbers satisfying $(x + 1)(y + 2) = 8$. Show that
$$
(xy - 10)^2 \geq 64.
$$
Furthermore, determine all pairs $(x, y)$ of real numbers for which equality holds. | [
"The inequality $(2x - y)^2 \\geq 0$ (with equality if and only if $y = 2x$) is equivalent to\n$$\n(2x + y)^2 \\geq 8xy.\n$$\nThe constraint $(x + 1)(y + 2) = 8$ gives $2x + y = 6 - xy$. Substituting this into the inequality above yields\n$$\n(6 - xy)^2 \\geq 8xy,\n$$\nwhich is equivalent to\n$$\n(xy - 10)^2 \\geq ... | Austria | Austria2019 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (1, 2) and (-3, -6) | |
0dia | There are $n > 2022$ cities in the country. Some pairs of cities are connected with straight two-ways airlines. Call the set of the cities unlucky, if it is impossible to color the airlines between them in two colors without monochromatic triangle. The set containing all the cities is unlucky. Is there always an unluck... | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Yes |
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