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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
02xv | Problem:
Às margens de um lago circular, existem pedras numeradas de 1 a 10, no sentido horário. O sapo Frog parte da pedra 1 e salta no sentido horário apenas nestas 10 pedras.
a) Se Frog salta de 2 em 2 pedras, ou seja, ele vai da pedra 1 para a 3, da 3 para a 5 e assim por diante, após 100 saltos em que pedra estar... | [
"Solution:\n\na) Depois de 5 saltos, Frog volta para a pedra 1 e inicia a mesma sequência. Como 100 é múltiplo de 5, no $100^{\\circ}$ salto ele vai para a pedra 1.\n\nb) No $1^{\\circ}$ salto ele se desloca 1 pedra; no $2^{\\circ}$, 2 pedras; no $3^{\\circ}$, 3 pedras e assim até o último salto quando se desloca 1... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | a) 1; b) 1 | |
044l | Let the side length of the base and height of regular pyramid $P-ABCD$ be equal. Point $G$ is the centroid of face $\triangle PBC$. Then the sine of the angle between line $AG$ and base $ABCD$ is ______. | [
"$O$ and $H$, respectively. Then $O$ is the centre of the base square, $H$ lies on $OM$, and $\\frac{GH}{PO} = \\frac{HM}{OM} = \\frac{GM}{PM} = \\frac{1}{3}$.\n\nFor the sake of convenience, let $AB = PO = 6$. Thus, $GH = \\frac{PO}{3} = 2$, $OH = \\frac{2}{3}OM = 2$. And since $AO = 3\\sqrt{2}$, $\\angle AOH = 13... | China | China Mathematical Competition | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(38)/19 | |
0dvh | Problem:
Za ulomek $\frac{m}{n}$, kjer sta $m$ in $n$ naravni števili, velja $\frac{1}{3}<\frac{m}{n}<1$. Če števcu prištejemo naravno število, imenovalec pa s tem številom pomnožimo, se vrednost ulomka ne spremeni. Poišči vse take ulomke $\frac{m}{n}$. | [
"Solution:\n\nIz $\\frac{m}{n}=\\frac{m+k}{n \\cdot k}$ izrazimo $m=\\frac{k}{k-1}$. Ker je $m$ naravno število, mora biti $k=2$, tako da je tudi $m=2$. Zaradi $\\frac{1}{3}<\\frac{2}{n}<1$ mora biti $2<n<6$. Vse možne rešitve so $\\frac{2}{3}, \\frac{2}{4}$ in $\\frac{2}{5}$."
] | Slovenia | 47. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2/3, 2/4, 2/5 | |
07ru | A designer plans to decorate the opening ceremony for this year's IMO contest in Romania with rows of coloured clock faces. Each clock face consists of twelve coloured rectangles glued to a round white base, as illustrated below.
The following design rules apply:
1. Each rectangle is blue, ye... | [
"Consider first the set $S_m$ of all possible clock faces using $m$ colours, where we consider two to be equal only if the colours match *without any rotation*: there are $m^{12}$ elements in $S_m$. Let $f$ be the map that rotates an element of $S_m$ by $1/12$ of a complete circle, i.e. 30 degrees. ($f$ depends on ... | Ireland | Irish | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 43315 | |
07ic | Find all injective functions $f : \mathbb{Z}^{\ge 0} \rightarrow \mathbb{Z}^{\ge 0}$ such that for all $n \in \mathbb{N}$ and every $(n+1)$-tuple of real numbers like $(a_0, \dots, a_n)$ where not all of them are equal to zero, the polynomial $\sum_{i=0}^{n} a_i x^i$ has a real root if and only if the polynomial $\sum_... | [
"As $P(x) = 1$ has no real root, $x^{f(0)}$ has also no real root. So $f(0) = 0$.\nNow considering $x^i + 1$, it has a real root if and only if $i$ is odd. So $i$ and $f(i)$ have the same parity.\nFor even $n$ consider the following polynomial in which $a, \\epsilon$ are positive numbers:\n$$\nP(x) = x^n + n a^{n-1... | Iran | 40th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | All functions of the form f(n) = c n for n ≥ 0, where c is an odd positive integer. | |
013l | Problem:
Let $p, q, r$ be positive real numbers and $n \in \mathbb{N}$. Show that if $p q r=1$, then
$$
\frac{1}{p^{n}+q^{n}+1}+\frac{1}{q^{n}+r^{n}+1}+\frac{1}{r^{n}+p^{n}+1} \leq 1
$$ | [
"Solution:\n\nThe key idea is to deal with the case $n=3$. Put $a=p^{n / 3}$, $b=q^{n / 3}$, and $c=r^{n / 3}$, so $a b c=(p q r)^{n / 3}=1$ and\n$$\n\\frac{1}{p^{n}+q^{n}+1}+\\frac{1}{q^{n}+r^{n}+1}+\\frac{1}{r^{n}+p^{n}+1}=\\frac{1}{a^{3}+b^{3}+1}+\\frac{1}{b^{3}+c^{3}+1}+\\frac{1}{c^{3}+a^{3}+1} .\n$$\nNow\n$$\n... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
07ao | Suppose that $x$, $y$ and $z$ are positive real numbers and $x^2 + y^2 + z^2 = x^2 y^2 + y^2 z^2 + z^2 x^2$. Prove that
$$
(x - y)^2 (y - z)^2 (z - x)^2 \le (x^2 - y^2)^2 + (y^2 - z^2)^2 + (z^2 - x^2)^2.
$$ | [
"Because of the problem's assumption, it is enough to prove that\n$$\n(\\prod (x-y))^2 (\\sum x^2) \\le \\sum (x^2 - y^2)^2 (\\sum (xy)^2).\n$$\nBy Cauchy-Schwarz inequality we have\n$$\n(\\sum xy(x^2 - y^2))^2 \\le \\sum (x^2 - y^2)^2 (\\sum (xy)^2). \\quad (1)\n$$\nOn the other hand, an easy calculation shows tha... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
020f | Problem:
Does there exist a prime number whose decimal representation is of the form $3811 \cdots 11$ (that is, consisting of the digits $3$ and $8$ in that order followed by one or more digits $1$)? | [
"Solution:\nWrite\n$$\na(n) = 38 \\underbrace{11 \\cdots 11}_{n \\text{ digits } 1}.\n$$\nThere are three cases to consider, depending on the remainder of $n$ upon division by three.\n\n- If $n = 3k + 1 \\equiv 1 \\pmod{3}$, then the sum of the digits of $a(n)$ is equal to $3(k+4)$, i.e. divisible by $3$, and hence... | Benelux Mathematical Olympiad | BxMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | No | |
018l | Given a rectangular grid, split into $m \times n$ squares, a colouring of the squares in two colours (black and white) is called *valid* if it satisfies the following conditions:
* all squares touching the border of the grid should be coloured black.
* No four squares forming a $2 \times 2$-square should be coloured in... | [
"There exist a valid colouring iff $n$ or $m$ is odd.\n\n**Proof.** If, without loss of generality, the number of rows is odd, colour every second row black, as well as the boundary, and all other squares white. It is easy to check that this coloring is valid.\n\nIf both $n$ and $m$ are even, there is no valid colo... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | A valid coloring exists if and only if at least one of the grid dimensions is odd. | |
039b | Prove that there are no distinct positive integers $x$ and $y$ such that
$$
x^{2007} + y! = y^{2007} + x!.
$$ | [
"Assume, for the sake of contradiction, that there exist distinct positive integers $x$ and $y$ such that\n$$\nx^{2007} + y! = y^{2007} + x!.\n$$\nWithout loss of generality, suppose $x > y$.\n\nThen,\n$$\nx^{2007} - y^{2007} = x! - y!.\n$$\nLet us estimate the size of both sides for large $x$ and $y$.\n\nNote that... | Bulgaria | First selection test for IMO 2007, Vietnam | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
01q8 | All vertices of triangles $ABC$ and $A_1B_1C_1$ lie on the hyperbola $y = 1/x$. It is known that $AB \parallel A_1B_1$ and $BC \parallel B_1C_1$. Prove that $AC_1 \parallel A_1C$. | [
"Let the coordinates of the given points be $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$, $A_1(a_1; 1/a_1)$, $B_1(b_1; 1/b_1)$, $C_1(c_1; 1/c_1)$. It is easy to calculate the slope of the line $AB$: $k = -1/(ab)$. Similarly, the slopes of $A_1B_1$, $BC$, $B_1C_1$ are $-1/(a_1b_1)$, $-1/(bc)$, $-1/(b_1c_1)$, respectively. ... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
08k5 | Problem:
Let $a, b, c, d, e$ be real numbers such that $a+b+c+d+e=0$. Let, also $A=ab+bc+cd+de+ea$ and $B=ac+ce+eb+bd+da$.
Show that
$$
2005 A+B \leq 0 \text{ or } \quad A+2005 B \leq 0
$$ | [
"Solution:\nWe have\n$$\n0=(a+b+c+d+e)^2=a^2+b^2+c^2+d^2+e^2+2A+2B\n$$\nThis implies that\n$$\nA+B \\leq 0 \\text{ or } 2006(A+B)=(2005 A+B)+(A+2005 B) \\leq 0\n$$\nThis implies the conclusion.\n\nWe have\n$$\n\\begin{aligned}\n2A+2B &= a(b+c+d+e)+b(c+d+e+a)+c(d+e+a+b) \\\\\n&\\quad +d(e+a+b+c)+e(a+b+c+d) \\\\\n&= ... | JBMO | OJBM | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
05vv | Problem:
Un ensemble $E$ d'entiers strictement positifs est dit intéressant si pour tout $n \geqslant 1$ et pour tous $x_{1}, \ldots, x_{n}$ des éléments de $E$ deux à deux distincts, leur moyenne arithmétique $\frac{1}{n}\left(x_{1}+\ldots+x_{n}\right)$ et leur moyenne géométrique $\left(x_{1} \cdot \ldots \cdot x_{n... | [
"Solution:\n\n1) Pour commencer, on peut remarquer que si $n \\geqslant 1$, alors dès que $x_{1}, \\ldots, x_{n} \\geqslant 1$ sont des entiers tous multiples de $n$, chacun des nombres $\\frac{x_{k}}{n}$ est un entier, de sorte que leur somme $\\frac{1}{n}\\left(x_{1}+\\ldots+x_{n}\\right)$ est un entier. De même,... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Yes: for example, the set of 2022 numbers {(2022!)^(k·2022!) for k from 1 to 2022)} works. No: there is no infinite interesting set. | |
0jbo | Problem:
Given any positive integer, we can write the integer in base 12 and add together the digits of its base 12 representation. We perform this operation on the number $7^{6^{5^{4^{3^{2^{1}}}}}}$ repeatedly until a single base 12 digit remains. Find this digit. | [
"Solution:\n\nFor a positive integer $n$, let $s(n)$ be the sum of digits when $n$ is expressed in base 12. We claim that $s(n) \\equiv n \\pmod{11}$ for all positive integers $n$. Indeed, if $n = d_{k} 12^{k} + d_{k-1} 12^{k-1} + \\cdots + d_{0}$ with each $d_{i}$ an integer between 0 and 11, inclusive, because $1... | United States | HMMT November 2012 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 4 | |
0d7n | Given two circles $O_{1}$ and $O_{2}$ intersect at $A$ and $B$. Let $d_{1}$ and $d_{2}$ be two lines through $A$ and be symmetric with respect to $AB$. The line $d_{1}$ cuts $O_{1}$ and $O_{2}$ at $G$, $E$ ($E \neq A$), respectively; the line $d_{2}$ cuts $O_{1}$ and $O_{2}$ at $F$, $H$ ($F \neq A$), respectively; such... | [
"\n\nWe have\n$$\n\\begin{aligned}\n\\angle AKO_{1} & = 90^{\\circ} - \\frac{1}{2} \\angle AO_{1}K = 90^{\\circ} - \\angle ABK = 90^{\\circ} - \\angle ABL \\\\\n& = 90^{\\circ} - \\frac{1}{2} \\angle AO_{2}L = \\angle ALO_{2}\n\\end{aligned}\n$$\nTherefore $ALNK$ is a cyclic quadrilateral.\... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
089h | Problem:
I numeri $a, b$ sono interi positivi. Qual è il minimo valore di $a+b$ affinché $21 a b^{2}$ e $15 a b$ siano entrambi cubi perfetti?
(A) 160
(B) 260
(C) 360
(D) 460
(E) 560 . | [
"Solution:\n\nLa risposta è (E). I più piccoli cubi perfetti della forma $21 a b^{2}$ e $15 a b$ sono tali per cui $a$ e $b$ possono contenere solo i fattori 3, 5 e 7 (quelli contenuti in 15 o in 21). Poniamo $a=3^{l} \\cdot 5^{m} \\cdot 7^{n}$ e $b=3^{p} \\cdot 5^{q} \\cdot 7^{r}$ (con esponenti interi non negativ... | Italy | Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | E | |
0b88 | Let $ABC$ be a scalene triangle, let $I$ be its incenter, and let $A_1$, $B_1$ and $C_1$ be the points of contact of the excircles with the sides $BC$, $CA$ and $AB$, respectively. Prove that the circumcircles of the triangles $AIA_1$, $BIB_1$ and $CIC_1$ have a common point different from $I$. | [
"The problem amounts to showing collinearity of the antipodes $A_2$, $B_2$ and $C_2$ of $I$ in the circles $AIA_1$, $BIB_1$ and $CIC_1$, respectively. In the sequel, we use the following standard notations: $I_a$, $I_b$ and $I_c$ are the centers of the excircles tangent to the sides $BC$, $CA$ and $AB$, respectivel... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Circles... | English | proof only | null | |
09pm | Problem:
Zij $m$ een positief geheel getal. Bewijs dat voor alle positieve reële getallen $a$ en $b$ geldt:
$$
\left(1+\frac{a}{b}\right)^{m}+\left(1+\frac{b}{a}\right)^{m} \geq 2^{m+1}
$$ | [] | Netherlands | TOETS TRAININGSKAMP | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
00k3 | Let $n \ge 3$ be an integer. For a convex $n$-gon $A_1A_2\dots A_n$ we consider a line $g$ through $A_1$ that does not contain any other point of the $n$-gon. Let $h$ be the orthogonal to $g$ through $A_1$. We orthogonally project the $n$-gon onto $h$. For $j = 1, \dots, n$ let $B_j$ denote the image of $A_j$. The line... | [
"Each arrangement of $B_1, \\dots, B_n$ begins with $B_1$ and ends with $B_k$ for some $k$ with $2 \\le k \\le n$. From $B_1$ through $B_k$ the projections are arranged from \"top to bottom\" and from $B_k$ through $B_n$ and back to $B_1$ from \"bottom to top\". The $k-2$ projections $B_2, \\dots, B_{k-1}$ assume s... | Austria | AustriaMO2013 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 2^{n-2} | |
07j8 | In the triangle $ABC$, the angle $\angle A$ is obtuse. Points $E$ and $F$ are the feet of altitudes from $B$ and $C$, respectively. Tangents to the circumcircle of $ABC$ at $B$ and $C$ intersect the line $EF$ at points $K$ and $L$, respectively. Let $\angle CLB = 135^\circ$. Point $R$ lies on the segment $BK$ such that... | [
"Since $\\angle BLC = 135^\\circ = \\angle BLS$ we have $\\angle CLS = 90^\\circ$. We shall provide and prove several lemmas.\n\n**Lemma 1.** Let $P$ be the intersection of the lines $LS, BC$ then $\\angle PRB = 90^\\circ$\n*Proof.* Since the line $LC$ is tangent to the circumcircle $ABC$\n$$\n\\angle LCE = \\angle... | Iran | 41th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis ... | null | proof only | null | |
0hxi | Problem:
Let $ABC$ be an acute scalene triangle with centroid $G$. The rays $BG$ and $CG$ meet the circumcircle of $ABC$ again at points $P$ and $Q$. Let $D$ denote the foot of the altitude from $A$ to $BC$. Suppose ray $GD$ meets the circumcircle of $ABC$ again at $E$. Show that the circumcircle of triangle $ADE$ lie... | [
"Solution:\n\nLet $M$ and $N$ be the midpoints of $CA$ and $AB$. By Pascal's theorem on $AABPQC$, we find that the tangent to the circumcircle at $A$, the line $MN$, and the line $PQ$ are concurrent at a single point $X$.\n\n\n\nOur claim is that $X$ is the desired circumcenter.\n\nFirst no... | United States | Berkeley Math Circle: Monthly Contest 8 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"... | null | proof only | null | |
0kyg | Problem:
Suppose $r$, $s$, and $t$ are nonzero reals such that the polynomial $x^{2} + r x + s$ has $s$ and $t$ as roots, and the polynomial $x^{2} + t x + r$ has $5$ as a root. Compute $s$. | [
"Solution:\n\nThe first equation implies $s t = s$, so $t = 1$. Then $x^{2} + x + r$ has $5$ as a root, so $r + 30 = 0$, implying $r = -30$. Finally, $x^{2} - 30 x + s$ has $1$ as a root, so $s = 29$."
] | United States | HMMT February 2024 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | 29 | |
03mm | Let $O$ denote the circumcentre of an acute-angled triangle $ABC$. Let point $P$ on side $AB$ be such that $\angle BOP = \angle ABC$, and let point $Q$ on side $AC$ be such that $\angle COQ = \angle ACB$. Prove that the reflection of $BC$ in the line $PQ$ is tangent to the circumcircle of triangle $APQ$. | [
"Let the circumcircle of triangle $OBP$ intersect side $BC$ at the points $R$ and $B$ and let $\\angle A$, $\\angle B$ and $\\angle C$ denote the angles at vertices $A$, $B$ and $C$, respectively. Now note that since $\\angle BOP = \\angle B$ and $\\angle COQ = \\angle C$, it follows that\n$$\n\\angle POQ = 360^\\{... | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0fun | Problem:
Zeige, dass es in jedem konvexen 9-Eck zwei verschiedene Diagonalen gibt, sodass die beiden Geraden, auf denen diese Diagonalen liegen, entweder parallel sind, oder sich in einem Winkel von weniger als $7^{\circ}$ schneiden. | [
"Solution:\n\nEin konvexes 9-Eck besitzt $\\binom{9}{2}-9=27$ Diagonalen. Wir verschieben die Diagonalen parallel, sodass alle durch einen festen Punkt $P$ gehen. Nun wählen wir eine beliebige Diagonale aus und nennen sie $d_{1}$. Dreht man $d_{1}$ im Gegenuhrzeigersinn um $P$, dann werden die anderen Diagonalen ei... | Switzerland | Vorrundenprüfung | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Rotation",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0g1t | Problem:
Alle Felder eines $8 \times 8$ Quadrats sind anfangs weiss gefärbt. In einem Zug darf man alle Felder eines horizontalen oder vertikalen $1 \times 3$ Rechtecks umfärben (alle weissen Felder werden schwarz und alle schwarzen Felder weiss). Ist es möglich, dass nach einer endlichen Anzahl Zügen alle Felder schw... | [
"Solution:\n\nWir verwenden die Standardfärbung für 3 Farben.\n\nOffensichtlich hat man 22 weisse und 21 gelbe Felder, sowie 21 blaue und 21 rote Felder. Wenn man nun einen $3 \\times 1$ Block umfärbt, färbt man immer einen gelben, einen blauen und einen weissen Block um. Um alle weissen Qu... | Switzerland | SMO - Finalrunde | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
07mt | Prove that
$$
\frac{2}{3} + \frac{4}{5} + \frac{6}{7} + \dots + \frac{2010}{2011}
$$
is not an integer. | [
"Let $S$ be the sum. Then\n$$\n1005 - S = \\frac{1}{3} + \\frac{1}{5} + \\frac{1}{7} + \\frac{1}{9} + \\dots + \\frac{1}{2011} = T.\n$$\nThen $S$ is an integer if and only if $T$ is an integer. Let $M = 3 \\cdot 5 \\cdot 7 \\cdot 9 \\dots 2009$. If $T$ is an integer then $MT$ is an integer.\n$$\nMT = \\frac{M}{3} +... | Ireland | Ireland | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
03bw | Find all prime numbers $p$ and $q$ such that
$p^2|q^3+1$ and $q^2|p^6-1$. | [
"If $p=3$, then $q^2|3^6-1 = 728 = 2^3 \\cdot 7 \\cdot 11$ and therefore $q=2$ which gives a solution. Let $p \\neq 3$. Since $(q+1, q^2-q+1) = 1$ or $3$, we have $p^2|q+1$ or $p^2|q^2-q+1$, which implies that $p < q$. If $p+1=q$ then $p=2$ and $q=3$ which is another solution. In the sequel we assume that $q \\ge p... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | (p, q) = (2, 3) or (3, 2) | |
0by2 | Determine all positive integers $n$ satisfying the following condition: for every monic polynomial $P$ of degree at most $n$ with integer coefficients, there exists a positive integer $k \le n$, and $k+1$ distinct integers $x_1, x_2, \dots, x_{k+1}$ such that $P(x_1) + P(x_2) + \dots + P(x_k) = P(x_{k+1})$.
*Note.* A ... | [
"To rule out all other values of $n$, it is sufficient to exhibit a monic polynomial $P$ of degree at most $n$ with integer coefficients, whose restriction to the integers is injective, and $P(x) \\equiv 1 \\pmod{n}$ for all integers $x$. This is easily seen by reading the relation in the statement modulo $n$, to d... | Romania | THE Tenth ROMANIAN MASTER OF MATHEMATICS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 2 | |
04wa | Given two different real numbers $a$, $b$ such that the expressions $a^3 + b$ and $a + b^3$ have the same value, prove that $-1 \le ab < \frac{1}{3}$. | [] | Czech Republic | District Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
040g | Let $S_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n}$, where $n$ is a positive integer. Prove that for any real numbers $a, b$ with $0 \le a < b \le 1$, there are infinite many terms in the sequence $\{S_n - [S_n]\}$ that are within $(a, b)$. (Here $[x]$ denotes the largest integer not greater than real number $x$.) | [
"For any $n \\in \\mathbb{N}^*$, we have\n$$\n\\begin{align*}\nS_{2^n} &= 1 + \\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{2^n} = 1 + \\frac{1}{2} + \\left( \\frac{1}{2^1} + \\frac{1}{2^2} \\right) + \\\\\n& \\quad \\left( \\frac{1}{2^{n-1}} + \\frac{1}{2^n} \\right) \\\\\n&> 1 + \\frac{1}{2} + \\left( \\frac{... | China | China Mathematical Competition (Complementary Test) | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0cep | A triangle is tiled with a finite number of triangles whose sides all have an odd length. Prove that the perimeter of the triangle is an integer of the same parity as the number of triangles in the tiling.
Marius Cavachi | [
"Every inner edge of a triangle is subdivided into one or more 'short' segments by (the boundaries of) some other triangles on the opposite side. Each short segment is shared by exactly two triangles. Notice further that every short segment lies along a\n\nunique segment of maximal length which is a concatenation o... | Romania | Seventeenth Stars of Mathematics Competition | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
06ky | Determine all sequences $p_1, p_2, p_3, \dots$ of prime numbers for which there exists an integer $k$ such that the recurrence relation
$$
p_{n+2} = p_{n+1} + p_n + k
$$
holds for all positive integers $n$. | [
"The sequence can be any constant sequence $p, p, p, \\dots$ where $p$ is a prime.\n\nThe recurrence relation can be rewritten as\n$$\np_{n+2} + k = (p_{n+1} + k) + (p_n + k).\n$$\nSince the characteristic equation $\\lambda^2 - \\lambda - 1 = 0$ has roots $\\frac{1 \\pm \\sqrt{5}}{2}$, we have\n$$\np_n = A \\left(... | Hong Kong | HKG TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | All constant sequences p, p, p, … where p is prime, with k = −p. | |
00c6 | Each cell in an $8 \times 8$ board is painted white or black, in such a way that every $2 \times 3$ or $3 \times 2$ rectangle contains at least two black cells having a common edge. What is the minimum number of black cells that there can be in the board? | [
"We claim that at least two of the cells $A, B, C, D$ in a block as in the picture are black.\n\nOtherwise, there are at most one black cell among them; then, there are at least 3 white cells. Without loss of generality, assume that $A, B, C$ are white. Then, the following $2 \\times 3$ rec... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 24 | |
0083 | Find the minimum and the maximum of the sum $S = \frac{a}{b} + \frac{c}{d}$ where $a, b, c, d \in \mathbb{N}$ satisfy $a + c = 20202$, $b + d = 20200$. | [
"For clarity we write $p$ and $p+2$ for $20200$ and $20202$ whenever possible. The conditions are $a + c = p + 2$, $b + d = p$. By symmetry assume $b \\le d$, then $1 \\le b \\le \\frac{p}{2}$. In each sum $S = \\frac{a}{b} + \\frac{c}{d}$ replace $a$ and $c$ by their extremal values $a = 1, c = p + 1$ and $a = p +... | Argentina | Mathematical Olympiad Rioplatense | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Minimum = 1/141 + 20201/20059; Maximum = 20201 + 1/20199 | |
0ayf | Problem:
A standard deck of 52 cards has the usual 4 suits and 13 denominations. What is the probability that two cards selected at random, and without replacement, from this deck will have the same denomination or have the same suit? | [
"Solution:\nLet $A$ be the event that the 2 chosen cards will have the same denomination; and let $B$ be the event that the 2 chosen cards will have the same suit. Note that $A \\cap B = \\emptyset$. So that $\\mathbb{P}(A \\cup B) = \\mathbb{P}(A) + \\mathbb{P}(B)$.\n\nSince there are 4 suits to choose from, then ... | Philippines | 20th Philippine Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 5/17 | |
0g2u | Problem:
Eine Gruppe von Kindern sitzt im Kreis. Am Anfang hat jedes Kind eine gerade Anzahl Bonbons. In jedem Schritt muss jedes Kind die Hälfte seiner Bonbons dem Kind zu seiner Rechten abgeben. Sollte ein Kind nach einem Schritt eine ungerade Anzahl Bonbons haben, bekommt es vom Kindergärtner ein zusätzliches Bonbo... | [
"Solution:\n\nDer Vollständigkeit halber bemerken wir kurz, dass nach jedem Zug alle Kinder stets eine gerade Anzahl an Bonbons haben. Wir bezeichnen mit $2 m_{i}$ die Anzahl Bonbons, die das Kind mit den wenigsten Bonbons vor dem $i$-ten Schritt hat und analog dazu $2 M_{i}$ die Anzahl Bonbons, die das Kind mit de... | Switzerland | SMO 2019 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0219 | Problem:
a.
Let $a_{0}, a_{1}, \ldots, a_{2024}$ be real numbers such that $\left|a_{i+1}-a_{i}\right| \leqslant 1$ for $i=0,1, \ldots, 2023$.
Find the minimum possible value of
$$
a_{0} a_{1}+a_{1} a_{2}+\cdots+a_{2023} a_{2024}
$$
b.
Does there exist a real number $C$ such that
$$
a_{0} a_{1}-a_{1} a_{2}+a_{2} a_{3... | [
"Solution:\n\na.\nThe minimum value is $-506$. Note that from $\\left|a_{i}-a_{i-1}\\right| \\leq 1$ it follows that\n$$\na_{i} a_{i-1}=\\frac{\\left(a_{i}+a_{i-1}\\right)^{2}-\\left(a_{i}-a_{i-1}\\right)^{2}}{4} \\geq-\\frac{\\left(a_{i}-a_{i-1}\\right)^{2}}{4} \\geq-\\frac{1}{4}\n$$\nAdding this for $i=1,2, \\ldo... | Benelux Mathematical Olympiad | 16th Benelux Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a: -506; b: No, such a constant does not exist. | |
0h00 | Natural numbers $a, b$ are chosen in such a way, that the number $m = a + b + 2\sqrt{ab+1}$ is natural. Prove that $m$ is composite. | [
"Let us suppose that we can find such numbers $a, b$, that $p = a + b + 2\\sqrt{ab+1}$ is prime. Then $ab+1$ is complete square, $a$ and $b$ are distinct. By Cauchy-Schwartz inequality $a + b \\ge 2\\sqrt{ab}$. Besides this, we have $2\\sqrt{ab} + 1 > 2\\sqrt{ab+1} \\Rightarrow a + b \\ge 2\\sqrt{ab} + 1$. $p$ is a... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010) | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0cxd | Let $ABC$ be a triangle with $\widehat{B} \geq 2 \widehat{C}$. Denote by $D$ the foot of the altitude from $A$ and by $M$ the midpoint of $BC$. Prove that $DM \geq \frac{AB}{2}$. | [
"\nDenote by $a, b, c$ the length sides of triangle $ABC$. In triangle $ADM$ we have\n$$\n\\begin{gathered}\nDM^{2} = AM^{2} - AD^{2} = \\frac{2\\left(b^{2} + c^{2}\\right) - a^{2}}{4} - \\frac{4K^{2}}{a^{2}} \\\\\n= \\frac{2\\left(b^{2} + c^{2}\\right) - a^{2}}{4} - \\frac{16K^{2}}{4a^{2}}... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
0hb4 | Andrew and Olesya in turn cut some squares from the rectangle $4000 \times 2019$, following the lines in such a way that after every turn the remaining figure stays connected. The one who cannot make a move loses. Who is going to win if both children play the best they can, and Olesya is first?
(Bogdan Rublyov)
![](a... | [
"In her first turn Olesya cuts the square $2018 \\times 2018$ as shown in fig. 34.\n\nAfter this move Olesya can follow the symmetric strategy, with the only not trivial moves regarding the black unit squares.\n\nHowever, if Andrew is able to cut (w.l.o.g.) the left black square then it implies that all squares to ... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Olesya | |
068c | Let $(x_n)$, $n \in \mathbb{N}^*$ be a sequence which is recursively defined by
$x_{n+1} = 3x_n^3 + x_n,$
where $x_1 = \frac{a}{b}$, and $a, b$ are positive integers such that $3$ doesn't divide $b$. If for some positive integer $m$ we have that $x_m$ is a perfect square of a rational, prove that $x_1$ is a perfect squ... | [
"We will prove that if $x_{n+1}$ is a perfect square of a rational, then $x_n$ is also a perfect square of a rational, and the desired result is obtained by a simple induction.\n\nNote first that since $3$ doesn't divide $b$, it will not divide any of the denominators of the sequence terms.\n\nFrom the recursive re... | Greece | Hellenic Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0elw | A sequence $(a_n)$ is defined by
$$
a_1 = 1, \quad a_n = 3a_{n-1} + 2^{n-1}, \quad \text{for } n \ge 2.
$$
Find a formula for the general term $a_n$ in terms of $n$. | [
"Evaluating the first few terms, one finds $a_1 = 1$, $a_2 = 5$, $a_3 = 19$, $a_4 = 65$, $a_5 = 211$ and $a_6 = 665$. The values always increase by a factor of $3$, plus a little bit. A guess is that the terms are similar to $3^n$, and computing the difference one fits it to be $2^n$. We'll show by induction that $... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | a_n = 3^n - 2^n | |
0bq7 | Problem:
Fie $a$, $b$, $c$ numere complexe astfel încât $|a-1| = |b-2| = |c+3|$ şi $a + b + c = 0$. Arătaţi că $|a-b+1| = |a-c-4| = |b-c-5|$. | [] | Romania | Olimpiada Națională de Matematică | [
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof only | null | |
05hg | Problem:
Montrer qu'il existe une infinité de nombres entiers strictement positifs $a$ tels que $a^{2}$ divise $2^{a}+3^{a}$. | [
"Solution:\n\nRemarquons que $a=1$ convient puis construisons par récurrence une suite $(u_{n})$ strictement croissante d'entiers impairs vérifiant la propriété.\n\nTout d'abord, on vérifie que $a=1$ et $a=5$ conviennent. Posons ainsi $u_{0}=1$ et $u_{1}=5$.\n\nConsidérons un entier $n \\geqslant 1$ et supposons la... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES, ENVOI No. 3 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof only | null | |
0e85 | Problem:
Izberimo neki osni presek enakostraničnega stožca s polmerom $2\ \mathrm{dm}$ in nanj postavimo pravokotni koordinatni sistem tako, da je koordinatno izhodišče v središču osnovne ploskve stožca, vrh pa leži na pozitivnem delu ordinatne osi (enoti na abscisni in ordinatni osi sta dolgi $1\ \mathrm{dm}$). Izrač... | [] | Slovenia | Državno tekmovanje | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | Apex coordinates: (0, 2√3).
Side lines (intercept form): x/2 + y/(2√3) = 1 and x/(-2) + y/(2√3) = 1.
Surface area (total): 12π dm^2.
Volume: (8√3/3)π dm^3. | |
0194 | Let $\{x_n\}$ be a sequence of integers such that $x_0 = a$, $x_1 = 3$ and
$$
x_n = 2x_{n-1} - 4x_{n-2} + 3 \text{ for all } n > 1.
$$
Determine the largest integer $k$ for which there exists a prime $p$ such that $p^k$ divides $x_{2011} - 1$. | [
"Let $y_n = x_n - 1$. Hence\n$$\ny_n = x_n - 1 = 2(y_{n-1}+1)-4(y_{n-2}+1)+3-1 = 2y_{n-1}-4y_{n-2} = 2(2y_{n-2}-4y_{n-3})-4y_{n-2} = -8y_{n-3}\n$$\nfor all $n > 2$. Hence\n$$\nx_{2011}-1=y_{2011}=-8y_{2008}=\\cdots=(-8)^{670}y_1=2^{2011}.\n$$\nHence $k = 2011$."
] | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 2011 | |
05d7 | Problem:
Let $n \geq 1$ be an integer and let $t_{1}<t_{2}<\ldots<t_{n}$ be positive integers. In a group of $t_{n}+1$ people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following conditions to hold at the same time:
i) The number of games played by ... | [
"Solution:\n\nLet $\\mathcal{T}=\\{t_{1}, \\ldots, t_{n}\\}$. The proof proceeds by induction on $n=|\\mathcal{T}|$. If $n=1$ and $\\mathcal{T}=\\{t\\}$, choose a group of $t+1$ people and let every pair of two persons play against each other. Then every person has played $t$ games and the conditions of the problem... | European Girls' Mathematical Olympiad (EGMO) | null | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0ad8 | In the isosceles triangle $ABC$, $M$ is the middle point of the base $AB$. Let $N$ be a point from the leg $BC$, such that $MN \perp BC$ and $S$ be the middle point of the segment $MN$. Prove that $AN$ is perpendicular with $CS$. | [
"From the conditions in the problem we have $\\overline{AM} = \\overline{MB}$, $CM \\perp AB$, $MN \\perp BC$, $\\overline{MS} = \\overline{SN}$. Let $P$ be a point on $BC$ such that $MP \\parallel AN$. From $\\triangle ANB$ we have $\\overline{AM} = \\overline{MB}$ and $MP \\parallel AN$, which implies that $MP$ i... | North Macedonia | Macedonian Mathematical Competitions | [
"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"
] | null | proof only | null | |
0abz | Let $ABC$ be an isosceles triangle with $\overline{AB} = \overline{AC}$. Let $D$ be the midpoint of $BC$, $M$ the midpoint of $AD$ and $N$ the projection of $D$ to $BM$. Prove that $\angle ANC = 90^\circ$. | [
"Let $S$ be the point so that $ABCD$ is a parallelogram. Then $ADCS$ is a rectangle and $R$ is the intersection point of the diagonals $AC$ and $DS$. The point $N$ lies on the diagonal $BS$ of the parallelogram $ABDS$ from where we obtain that $SND$ is a right triangle. The point $R$ is a circumcenter for the trian... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
04j2 | Determine the largest positive integer $n$ such that
$$
n + 5 \mid n^4 + 1395.
$$ | [
"$$\n4(2^{m-1} + 1)(2^m + 1)(2^{2m-1} + 2^m + 1) = (p - 1)(p + 1).\n$$\nSince $p$ is odd, the right-hand side is the product of two consecutive even numbers, so it is divisible by 8. The left-hand side is not divisible by 8, unless $m = 1$.\nIt follows that the only solution is $(p, m, n) = (11, 1, 3)$.\n**3.3.** L... | Croatia | First round – City competition | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | 2015 | |
06z9 | Problem:
The function $f$ on the positive integers satisfies $f(1)=1$, $f(2n+1)=f(2n)+1$ and $f(2n)=3 f(n)$. Find the set of all $m$ such that $m=f(n)$ for some $n$. | [
"Solution:\n\nWe show that to obtain $f(n)$, one writes $n$ in base 2 and then reads it in base 3. For example, $12 = 1100_2$, so $f(12) = 1100_3 = 36$. Let $g(n)$ be defined in this way. Then certainly $g(1) = 1$. Now $2n+1$ has the same binary expansion as $2n$ except for a final 1, so $g(2n+1) = g(2n) + 1$. Simi... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | All positive integers whose base-three representation contains only the digits 0 and 1 (i.e., no digit 2). | |
0c3k | Problem:
Determinați numerele prime $p$ pentru care numărul $a = 7^{p} - p - 16$ este pătrat perfect. | [
"Soluție:\n\n$p = 2$ nu verifică.\n\n$p = 3$ este soluție: $a = 7^{3} - 3 - 16 = 324 = 18^{2}$.\n\nArătăm că nu avem alte soluții. Fie $p \\geq 5$ un număr prim.\n\nDacă $p \\equiv 1 \\pmod{4}$, atunci $a \\equiv 2 \\pmod{4}$, deci $a$ nu este pătrat perfect.\n\nSe constată ușor că $p = 7$ nu este soluție (calculân... | Romania | Al treilea test de selecţie pentru OBMJ | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 3 | |
0dho | From a point $A$ lying outside the circle $(O)$, draw two tangent lines $AB$, $AC$ of $(O)$ with $B$, $C$ are tangent points. A line passes through $A$, lies inside the angle $OAC$, cuts $(O)$ at $R$, $S$ ($R$ is between $A$ and $S$). The segments $BR$, $BS$ cut the ray $AO$ respectively at $D$, $E$. Denote $H$ as orth... | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chas... | English | proof and answer | 1:1 | |
0ev5 | Let $a$, $b$ be relatively prime positive integers and let $a_n$ and $b_n$ be integer sequences satisfying $(a + b\sqrt{2})^{2n} = a_n + b_n\sqrt{2}$. Find all primes $p$ such that there is a positive integer $n$ less than or equal to $p$ satisfying $b_n \equiv 0 \pmod{p}$. | [
"Let $p$ be a prime. First suppose that $p$ is an odd prime dividing $a^2 - 2b^2$. Since $a$ and $b$ are relatively prime, $b_1 = 2ab$ is not divisible by $p$. Suppose that there is a positive integer $n$ such that $b_n$ is divisible by $p$. Let $r$ be the smallest positive integer such that $b_r$ is divisible by $... | South Korea | The 26th Korean Mathematical Olympiad Final Round | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
... | null | proof and answer | All primes p with p = 2 or p ∤ (a^2 - 2b^2). | |
0b43 | Problem:
Suppose $a, b, c$ are the roots of the polynomial $x^{3} + 2x^{2} + 2$. Let $f$ be the unique monic polynomial whose roots are $a^{2}, b^{2}, c^{2}$. Find $f(1)$.
(a) -17
(b) -16
(c) -15
(d) -14 | [] | Philippines | 24th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | MCQ | c | |
0b7d | Let $a$, $b$, $c$ be positive real numbers. Prove that
$$
\frac{a^2 b (b-c)}{a+b} + \frac{b^2 c (c-a)}{b+c} + \frac{c^2 a (a-b)}{c+a} \geq 0.
$$ | [
"By clearing denominators (brute force), the inequality becomes\n$$\na^3 b^3 + b^3 c^3 + c^3 a^3 \\geq a^2 b c^3 + b^2 c a^3 + c^2 a b^3. \\quad (1)\n$$\nIn order to justify (1), use the AM-GM inequality. Thus\n$$\na^3 b^3 + b^3 c^3 + c^3 a^3 \\geq 3ab^3c^2.\n$$\nBy summation with the two other analogous inequaliti... | Romania | Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0gyu | Find all prime $p$ and natural $m$, that satisfy the equation:
$$
2p^2 + p + 9 = m^2.
$$ | [
"Let us rewrite our equation in the following way: $p(2p+1) = (m-3)(m+3)$. Since $p$ is prime, we have that $(m-3) \\nmid p$ or $(m+3) \\nmid p$.\n\n$$1) \\quad (m-3) \\nmid p \\quad \\Rightarrow \\quad m-3 = kp \\quad \\Rightarrow \\quad (m+3) > kp \\quad \\text{and}$$\n$$3p^2 > p(2p+1) = (m-3)(m+3) > k^2p^2, \\qu... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | p=5, m=8 | |
01un | A rectangle $R$ with odd integer side lengths is divided into small rectangles with integer side lengths.
Prove that there is at least one rectangle among the small rectangles whose distances from the four sides of $R$ are either all odd or all even. | [
"**1.** See IMO-2017 Shortlist, Problem C1."
] | Belarus | Selection and Training Session | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0h2z | Real numbers $x, y \in (0, \pi)$ satisfy the equality
$$
\cos 2x \cos y - \cos 2y \cos x = \cos y - \cos x.
$$
Show that $x = y$. | [
"Запишемо дану рівність у вигляді $\\cos^2 x \\cos y - \\cos^2 y \\cos x = \\cos y - \\cos x$, $(\\cos x \\cos y + 1)(\\cos x - \\cos y) = 0$. Оскільки для $x \\in (0; \\pi)$ і $y \\in (0; \\pi)$ $\\cos x \\cos y > -1$, то $\\cos x = \\cos y$, і тому, враховуючи спадання функції $f(t) = \\cos t$ на проміжку $(0; \\... | Ukraine | Ukrainian Mathematical Olympiad | [
"Precalculus > Trigonometric functions"
] | English | proof only | null | |
0jfh | Problem:
Triangle $ABC$ is inscribed in a circle $\omega$ such that $\angle A = 60^\circ$ and $\angle B = 75^\circ$. Let the bisector of angle $A$ meet $BC$ and $\omega$ at $E$ and $D$, respectively. Let the reflections of $A$ across $D$ and $C$ be $D'$ and $C'$, respectively. If the tangent to $\omega$ at $A$ meets l... | [
"Solution:\n\nWe will show that $CE^2 = (CF)(CC')$. By a simple computation using the given angles, one may find that this is equivalent to $CF = AC - AB$, or $AF = 2AC - AB$.\n\nWe compute $AF$ by trigonometry. Assume for simplicity that $AC = \\frac{1}{2}$, so $AD' = 2AD = 2AC = 1$ because $\\triangle ACD$ is iso... | United States | HMIC | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kv3 | Problem:
Points $X$, $Y$, and $Z$ lie on a circle with center $O$ such that $XY = 12$. Points $A$ and $B$ lie on segment $XY$ such that $OA = AZ = ZB = BO = 5$. Compute $AB$. | [
"Solution:\n\nLet the midpoint of $XY$ be $M$. Because $OAZB$ is a rhombus, $OZ \\perp AB$, so $M$ is the midpoint of $AB$ as well. Since $OM = \\frac{1}{2} OX$, $\\triangle OMX$ is a $30$-$60$-$90$ triangle, and since $XM = 6$, $OM = 2\\sqrt{3}$. Since $OA = 5$, the Pythagorean theorem gives $AM = \\sqrt{13}$, so ... | United States | HMMT February | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2*sqrt(13) | |
09z9 | A whiteboard contains a calculation $1?2?3?4?5?6$, where each question mark is either a $+$ or a $\times$. The correct outcome of the calculation is written on the back of the board. Jaap copies the calculation but accidentally turns one of the plus signs into a times sign. The outcome is now $58$ more than the number ... | [] | Netherlands | Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 68 | |
0dk2 | Find all odd integer $n$ such that the number of integers $k$ with $0 < k < \frac{n}{4}$ and $\gcd(n, k) = 1$ is odd. | [
"We claim that the only integers that work are prime powers $p^k$ in which $p \\equiv 5$ or $7$ modulo $8$. Let define $\\omega, \\Omega$ as function from the set of odd integers to $\\{0; 1\\}$ by\n$$\n\\omega(n) = \\begin{cases} 0 & \\text{if } n \\equiv 1,3 \\pmod 8 \\\\ 1 & \\text{if } n \\equiv 5,7 \\pmod 8 \\... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | English | proof and answer | All odd integers that are prime powers with the base prime congruent to five or seven modulo eight, i.e., n = p^a with p ≡ 5 or 7 mod 8 and a ≥ 1. | |
0agy | Find all natural numbers $n$ for which each natural number having $n-1$ digits '1' and one digit '7' in its decimal representation is prime. | [
"A number $B$ having $n-1$ digits '1' and one digit '7' in decimal representation is of the form $B = A_n + 6 \\cdot 10^k$ where $A_n$ is a number having $n$ digits '1', and $0 \\le k < n$. Notice that if $3|n$ then the sum of the digits of $B$ is $3n+6$. Notice that\n$$\nA_1 = 1,\\ A_2 = 4,\\ A_3 = 6,\\ A_4 = 5,\\... | North Macedonia | XVIII-th Macedonian mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | n = 1 and n = 2 | |
09vd | Problem:
Gegeven is een kwadratisch polynoom $P(x)$ met twee verschillende reële nulpunten. Voor alle reële getallen $a$ en $b$ met $|a|,|b| \geq 2017$ geldt dat $P\left(a^{2}+b^{2}\right) \geq P(2 a b)$. Bewijs dat minstens één van de nulpunten van $P$ negatief is. | [
"Solution:\n\nSchrijf $P(x)=c(x-d)(x-e)$, waarbij $d$ en $e$ de nulpunten zijn, dus $d \\neq e$. Verder geldt $c \\neq 0$, anders is $P$ niet kwadratisch. Voor $|a|,|b| \\geq 2017$ volgt nu uit $P\\left(a^{2}+b^{2}\\right) \\geq P(2 a b)$ dat\n$$\nc\\left(a^{2}+b^{2}-d\\right)\\left(a^{2}+b^{2}-e\\right) \\geq c(2 ... | Netherlands | IMO-selectietoets | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
07yv | Problem:
Tre amici possiedono ciascuno tre gettoni.
Dopo ogni partita il vincitore riceve un gettone da ognuno degli altri due amici.
Qual è la probabilità che il gioco non si debba interrompere entro cinque partite poiché uno dei giocatori rimane senza gettoni?
(A) $\frac{2}{9}$
(B) $\frac{1}{6}$
(C) $\frac{1}{3}$
(D... | [] | Italy | Progetto Olimpiadi di Matematica | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | MCQ | A | |
0j4i | Problem:
Evaluate
$$
\int_{1}^{\infty}\left(\frac{\ln x}{x}\right)^{2011} dx.
$$ | [
"Solution:\n\nAnswer: $\\frac{2011!}{2010^{2012}}$\n\nBy the chain rule, $\\frac{d}{dx}(\\ln x)^{n}=\\frac{n (\\ln x)^{n-1}}{x}$.\n\nWe calculate the definite integral using integration by parts:\n$$\n\\int_{x=1}^{\\infty} \\frac{(\\ln x)^{n}}{x^{2011}} dx=\\left[\\frac{(\\ln x)^{n}}{-2010 x^{2010}}\\right]_{x=1}^{... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Derivatives"
] | null | proof and answer | 2011! / 2010^{2012} | |
08wz | There are 2013 cards numbered $0$, $1$, $2$, $\ldots$, $2012$. Initially, all the cards are placed with the face with a written number down. Then, we perform for each $i = 1, 2, \dots, 2013$ the following operation $i$ starting with $i = 1$ and with increasing order ending up with $i = 2013$:
Operation $i$: Flip each ... | [
"Let $n = 2013$ throughout the subsequent discussion on this problem. For any real number $r$ denote by $\\lfloor r \\rfloor$ the smallest integer greater than or equal to $r$. In order to obtain the desired solution, we prove the following two lemmas.\n\n**Lemma 1.** For $1 \\le i \\le n$ and $0 \\le x \\le n-1$, ... | Japan | Japan 2013 Initial Round | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 793 | |
05ba | Find the least number of buttons that can be placed on the squares of a $5 \times 5$ grid so that no two buttons are on the same square or on squares with a common side (buttons may be on squares with a common vertex) and no buttons can be added to the grid under the same conditions. | [
"We say that a button *covers* a square if the button lies on either the square itself or one of its neighbors. Consider the $2 \\times 2$ corner areas and the central cross consisting of 5 squares (colored with green and red respectively in Fig. 17).\n\nEvery button covering a corner has to be located in the corre... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 7 | |
0j0u | Problem:
In the game of set, each card has four attributes, each of which takes on one of three values. A set deck consists of one card for each of the $81$ possible four-tuples of attributes. Given a collection of $3$ cards, call an attribute good for that collection if the three cards either all take on the same val... | [
"Solution:\n\nIn counting the number of sets of $3$ cards, we first want to choose which of our two attributes will be good and which of our two attributes will not be good. There are $\\binom{4}{2} = 6$ such choices.\n\nNow consider the two attributes which are not good, attribute $X$ and attribute $Y$. Since thes... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 25272 | |
0i4s | Problem:
Determine the number of four-digit integers $n$ such that $n$ and $2 n$ are both palindromes. | [
"Solution:\n\nLet $n = \\underline{a} \\underline{b} \\underline{b} \\underline{a}$. If $a, b \\leq 4$ then there are no carries in the multiplication $n \\times 2$, and $2 n = (2 a)(2 b)(2 b)(2 a)$ is a palindrome. We shall show conversely that if $n$ and $2 n$ are palindromes, then necessarily $a, b \\leq 4$. Hen... | United States | HMMT 2002 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 20 | |
0cgl | Consider a triangle $ABC$ with $\angle BAC = 120^\circ$ and the isosceles triangles $PAB$ and $NAC$ such that $\angle APB = \angle ANC = \angle BAC$, with line $AB$ separating points $P$ and $C$, and line $AC$ separating points $N$ and $B$. Prove that, if $G$ is the centroid of triangle $ABC$, then $GP = GN = \frac{AB+... | [
"\nOn the other hand, from $\\angle APB = 120^\\circ$ and $\\angle APF = 30^\\circ$, it follows that $\\angle FPB = 90^\\circ$, hence the triangle $PBF$ is a $30^\\circ - 60^\\circ - 90^\\circ$ triangle, from which $BF = 2FP$, (2). Using (1) and (2) we conclude $BF = 2FA$, (3).\nDenote $BB'... | Romania | 74th 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 > Distance chasing"
] | English | proof only | null | |
0bhw | Prove that the product of every three odd consecutive positive integers can be written as the sum of three consecutive integers. | [
"Let the three odd consecutive numbers be $2p + 1$, $2p + 3$ and $2p + 5$, where $p$ is a positive integer. Then one of these numbers is divisible by $3$:\n\nIf $p = 3k$, with integer $k$, then $2p + 3 = 2(3k) + 3 = 6k + 3 = 3(2k + 1)$;\n\nIf $p = 3k + 1$, with integer $k$, then $2p + 1 = 2(3k + 1) + 1 = 6k + 3 = 3... | Romania | 65th Romanian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
02r3 | Problem:
Um algarismo é afilhado de um número natural se ele é o algarismo das unidades de algum divisor desse número. Por exemplo, os divisores de $56$ são $1$, $2$, $4$, $7$, $8$, $14$, $28$ e $56$, logo os afilhados de $56$ são $1$, $2$, $4$, $6$, $7$ e $8$.
a) Quais são os afilhados de $57$?
b) Ache um número qu... | [
"Solution:\n\na) Os divisores de $57$ são $1$, $3$, $19$ e $57$, donde seus afilhados são $1$, $3$, $9$ e $7$.\n\nb) O exemplo mais simples é $49$, cujos afilhados são $1$, $7$ e $9$.\n\nc) Se um número tem um divisor terminado em $0$ então este número é múltiplo de $10$. Logo ele é múltiplo de $2$ e de $5$, e port... | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | a) The affiliates of 57 are 1, 3, 7, 9. b) One example is 49, whose affiliates are 1, 7, 9. c) Any number that has 0 among its affiliates also has 2 and 5 as affiliates. d) Any number that has 0 and 9 among its affiliates also has 8 as an affiliate. | |
08de | Problem:
Siano $p, q$ numeri primi. Dimostrare che, se $p+q^{2}$ è un quadrato perfetto, allora il numero $p^{2}+q^{n}$ non è un quadrato perfetto per nessun intero positivo $n$. | [
"Solution:\n\nScriviamo $p+q^{2}=a^{2}$ con $a$ intero positivo. Allora $p=a^{2}-q^{2}=(a-q)(a+q)$, e siccome $p$ è un numero primo i fattori $a-q$ e $a+q$ devono essere uguali a $\\pm 1$ o a $\\pm p$. Siccome $a+q$ è un numero positivo, anche $a-q$ deve esserlo, ed inoltre $a+q>a-q$, quindi l'unica possibilità è $... | Italy | XXXV Olimpiade Italiana di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
02wu | Problem:
Em uma classe com 35 estudantes, pesquisou-se sobre os gostos relativos a matemática e literatura e constatou-se que:
- 7 homens gostam de matemática;
- 6 homens gostam de literatura;
- 5 homens e 8 mulheres disseram não gostar de ambos;
- há 16 homens na classe;
- 5 estudantes gostam de ambos; $\mathrm{e}$
-... | [
"Solution:\n\nSejam $H$ o conjunto dos homens e $U$ o conjunto total de pessoas, portanto $U-H$ é o conjunto das mulheres. Além deles, considere os conjuntos Mat e $L$ das pessoas que gostam de matemática e literatura, respectivamente. Se $x$ representa a quantidade de homens que gostam de matemática e literatura e... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | a) 2; b) 2 | |
007i | Evaluate the sum
$$
\left\lfloor \frac{1}{13} \right\rfloor + \left\lfloor \frac{3}{13} \right\rfloor + \left\lfloor \frac{3^2}{13} \right\rfloor + \dots + \left\lfloor \frac{3^{101}}{13} \right\rfloor.
$$
Here $[\dots]$ denotes the integer part of a number. | [
"Ignore the integer parts of three consecutive summands with numerators $3^{3k}$, $3^{3k+1}$, $3^{3k+2}$. The sum of three such fractions is an integer; moreover it equals $3^{3k}$:\n$$\n\\frac{3^{3k}}{13} + \\frac{3^{3k+1}}{13} + \\frac{3^{3k+2}}{13} = \\frac{3^k(1+3+3^2)}{13} = 3^{3k} \\quad \\text{for } 0 \\le k... | Argentina | Mathematical Olympiad Rioplatense | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | (27^34 - 1)/26 - 34 | |
05oi | Problem:
Existe-t-il un sous-ensemble infini $A$ de $\mathbb{N}$ qui vérifie la propriété suivante : toute somme finie d'éléments distincts de $A$ n'est jamais une puissance d'un entier (c'est-à-dire un entier de la forme $a^{b}$ avec $a$ et $b$ entiers supérieurs ou égaux à 2) ? | [
"Solution:\n\nOn cherche à construire un tel ensemble $A=\\{a_{0}, a_{1}, \\ldots\\}$, avec $(a_{0}, a_{1}, \\ldots)$ une suite croissante d'entiers. On choisit comme premier élément $a_{0}=0$.\n\nSoit $n \\in \\mathbb{N}$. On suppose qu'on a déjà trouvé $a_{0}<\\cdots<a_{n}$ tels que l'ensemble $B$ des sommes d'él... | France | Envoi 1: Arithmétique | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
02wi | Problem:
Um ciclo de três conferências teve sucesso constante, isto é, em cada sessão havia o mesmo número de participantes. No entanto, a metade dos que compareceram à primeira não voltou mais; um terço dos que compareceram à segunda conferência assistiu apenas a ela, e um quarto dos que compareceram à terceira não a... | [
"Solution:\n\na) Chamemos de $P$ o número de presentes em cada conferência, $x, y, t$ e $z$ serão os números inteiros dos que foram às três conferências; para às primeira e segunda, apenas; primeira e terceira, apenas; e segunda e terceira, apenas, respectivamente.\n\n\n\nPodemos então escr... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | a) 156 people attended each conference. b) 37 people attended all three conferences. | |
0ail | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y \in \mathbb{R}$ the following holds:
$$
f(x^2) + f(2y^2) = (f(x+y) + f(y))(f(x-y) + f(y)).
$$ | [
"Let $P(x,y)$ be the assertion $f(x^2) + f(2y^2) = (f(x+y) + f(y))(f(x-y) + f(y))$.\n\n$P(0,x)$ gives us\n$$\nf(0) + f(2x^2) = 2f(x)(f(x) + f(-x)) \\quad (1)\n$$\nand $P(0, -x)$ gives us\n$$\nf(0) + f(2x^2) = 2f(-x)(f(x) + f(-x)). \\quad (2)\n$$\nBy combining (1) and (2) we get\n$$\nf(x)^2 = f(-x)^2. \\quad (3)\n$$... | North Macedonia | European Mathematical Cup | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0 for all x; f(x) = 1/2 for all x; f(x) = x^2 for all x | |
0ivk | Problem:
The roots of $z^{6}+z^{4}+z^{2}+1=0$ are the vertices of a convex polygon in the complex plane. Find the sum of the squares of the side lengths of the polygon.
Answer: $12-4 \sqrt{2}$ | [
"Solution:\n\nFactoring the polynomial as $(z^{4}+1)(z^{2}+1)=0$, we find that the 6 roots are $e^{ \\pm i \\pi / 4}$, $e^{ \\pm i \\pi / 2}$, $e^{ \\pm i 3 \\pi / 4}$. The calculation then follows from the Law of Cosines or the distance formula."
] | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | final answer only | 12-4*sqrt(2) | |
09g5 | Find all real polynomials of degree $n$ satisfying
$$
P(P(x) + x) = P(P(x)) + P(x)^n + 1.
$$ | [
"*Answer: $P(x) = a x - 1$, $a \\neq 0$.*\n\nFirst note that $P$ cannot be a constant. Now let $n = 1$ and $P(x) = a x + b$. Then\n$$\n\\begin{cases}\nP(P(x) + x) = a(a x + b + x) + b = (a^2 + a)x + a b + b, \\\\\nP(P(x)) + P(x) + 1 = a(a x + b) + b + a x + b + 1 = (a^2 + a)x + a b + 2b + 1.\n\\end{cases}\n$$\nHenc... | Mongolia | 2015 Mongolian IMO Team Selection Tests | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Exactly the linear polynomials of the form P(x) = a x − 1 with a nonzero real constant (so solutions occur only when the degree is one; there are no solutions for higher degree). | |
05o9 | Problem:
On place quatre points dans le plan, trois jamais alignés, et on les relie tous deux à deux. On colorie chacun des six segments obtenus soit en bleu soit en rouge. Montrer qu'il existe deux triangles différents coloriés de la même façon. Par exemple, dans l'exemple suivant (où l'on a remplacé rouge par épais ... | [
"Solution:\n\nIl y a quatre façons différentes de colorier un triangle (avec zéro, un, deux ou trois segments rouges) et il y a quatre triangles dessinés. Donc, soit deux triangles sont coloriés de la même façon, soit tous les triangles sont différents. Montrons que le deuxième cas n'est pas possible. En effet, deu... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0g8j | 是否能找到十個集合 $A_1, A_2, \dots, A_{10}$, 同時滿足下列條件:
(i) 每個集合有三個元素, 形如 $\{a, b, c\}$, 其中 $a \in \{1, 2, 3\}$, $b \in \{4, 5, 6\}$, $c \in \{7, 8, 9\}$。
(ii) 任兩集合都不相等。
(iii) 將這十個集合依次圍成一圈 ($A_1, A_2, \dots, A_{10}$), 則任意相鄰的兩集合沒有共同元素, 但是任意不相鄰的兩集合都有共同元素(註. $A_{10}$ 與 $A_1$ 相鄰。)
Can we find ten sets $A_1, A_2, \dots, A_{10}$ suc... | [
"可以。考慮\n$(1,4,9)$, $(2,5,7)$, $(3,4,8)$, $(1,5,9)$, $(2,4,8)$, $(3,5,9)$, $(2,4,7)$, $(1,5,8)$, $(3,4,7)$, $(2,5,8)$。"
] | Taiwan | 二〇一四數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Other"
] | null | final answer only | (1,4,9), (2,5,7), (3,4,8), (1,5,9), (2,4,8), (3,5,9), (2,4,7), (1,5,8), (3,4,7), (2,5,8) | |
07mq | A Dutch hillwalking club with $4n$ members arranges a series of walks over a number of weekends, according to the following rules.
(a) Two walks take place each weekend – one takes place on Saturday, and the other on Sunday.
(b) Exactly $2n$ members of the club participate in each walk.
(c) On any weekend, no club memb... | [
"Suppose that each pair of club members participates together in $r$ walks. The total number $m$ of walks organised by the club is then given by\n$$\nm\\binom{2n}{2} = r\\binom{4n}{2},\n$$\nsince the right-hand side counts the total number of walks by considering all pairs of club members and using rule (d), but co... | Ireland | Ireland | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0hla | Problem:
Prove that if two medians in a triangle are equal in length, then the triangle is isosceles. | [
"Solution:\n\nLet equal medians $AD$ and $BE$ in triangle $ABC$ meet at $F$. It is well known that\n$$\nAF : FD = BF : FE = 2 : 1.\n$$\nTriangle $ABC$ is isosceles if we can show $AC = BC$ or equivalently, that $AE = BD$. But triangles $AFE$ and $BFD$ are congruent with vertical angles plus sides that are $1/3$ and... | United States | null | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
06d6 | A triangle $ABC$ is given. A circle $\Gamma$ passes through vertex $A$ and is tangent to side $BC$ at point $P$. The circle $\Gamma$ intersects sides $AB$ and $AC$ at points $M$ and $N$, respectively. Prove that (minor) arcs $\widehat{MP}$ and $\widehat{NP}$ are equal if and only if $\Gamma$ is tangent to the circumcir... | [
"If $AB = AC$, the result is obvious due to symmetry (both statements are equivalent to $P$ being the midpoint of $BC$). WLOG assume $AB < AC$. Let $D$ be the intersection of the line $BC$ and the tangent at $A$ to the circumcircle $(ABC)$.\n\nIf $\\overline{MP}$ and $\\overline{NP}$ are equal, then $MN$ is paralle... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0j3h | Problem:
Allison has a coin which comes up heads $\frac{2}{3}$ of the time. She flips it 5 times. What is the probability that she sees more heads than tails? | [
"Solution:\n\nThe probability of flipping more heads than tails is the probability of flipping 3 heads, 4 heads, or 5 heads. Since 5 flips will give $n$ heads with probability $\\binom{5}{n}\\left(\\frac{2}{3}\\right)^{n}\\left(\\frac{1}{3}\\right)^{5-n}$, our answer is\n\n$\\binom{5}{3}\\left(\\frac{2}{3}\\right)^... | United States | Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Combinations",
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 64/81 | |
0cfq | Solve in $[1, \infty) \times \mathbb{R}$ the following system of equations
$$
\begin{cases}
x + y = 2^x \\
x^2 + y^2 = 2^{[y]}.
\end{cases}
$$ | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | (1, 1) | |
06om | A cake has the form of an $n \times n$ square composed of $n^2$ unit squares. Strawberries lie on some of the unit squares so that each row or column contains exactly one strawberry; call this arrangement $\mathcal{A}$.
Let $\mathcal{B}$ be another such arrangement. Suppose that every grid rectangle with one vertex at ... | [
"We use capital letters to denote unit squares; $O$ is the top left corner square. For any two squares $X$ and $Y$ let $[X Y]$ be the smallest grid rectangle containing these two squares. Strawberries lie on some squares in arrangement $\\mathcal{A}$. Put a plum on each square of the target configuration $\\mathcal... | IMO | IMO 2006 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0k9f | Problem:
Let $x_{1}, x_{2}, \ldots, x_{n}$ and $y_{1}, y_{2}, \ldots, y_{n}$ be nonnegative real numbers such that $x_{i}+y_{i}=1$ for each $i=1,2, \ldots, n$. Prove that
$$
\left(1-x_{1} x_{2} \ldots x_{n}\right)^{m}+\left(1-y_{1}^{m}\right)\left(1-y_{2}^{m}\right) \ldots\left(1-y_{n}^{m}\right) \geq 1,
$$
where $m$ ... | [
"Solution:\n\nSuppose we have $n$ coins that we flip $m$ times each, where $x_{i}$ is the probability the $i$th coin comes up heads and $y_{i}=1-x_{i}$ is the probability the $i$th coin comes up tails. Then the first term is the probability that every time we toss the $n$ coins, at least one comes up tails, and the... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof only | null | |
06f9 | The sequence $\{a_n\}$ is defined by $a_1 = 0$ and $(n+1)^3 a_{n+1} = 2n^2(2n+1)a_n + 2(3n+1)$ for all integers $n \ge 1$. Show that infinitely many members of the sequence are positive integers. | [
"Let $b_n = n^2a_n + 2$ for all $n \\ge 1$. The recurrence relation becomes\n$$\n(n + 1)(b_{n+1} - 2) = 2(2n + 1)(b_n - 2) + 2(3n + 1).\n$$\nThis is the same as\n$$\nb_{n+1} = \\frac{2(2n + 1)}{n + 1}b_n.\n$$\nIt follows that\n$$\nb_n = \\frac{2(2n-1)}{n} b_{n-1} = \\frac{2^2(2n-1)(2n-3)}{n(n-1)} b_{n-2} = \\dots =... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0gl1 | Let $M$ and $N$ be positive integers. Mr. Pisut starts walking from the point $(0, N)$ to the point $(M, 0)$ in such a way that:
* each of his steps is of 1 unit length in the direction parallel to either the X-axis or the Y-axis;
* for each point $(x, y)$ on his path, $x \ge 0$ and $y \ge 0$.
For each step, he measure... | [
"Suppose that Mr. Pisut walks $k$ steps in total and the $i$-th step is from the point $(x_{i-1}, y_{i-1})$ to the point $(x_i, y_i)$. Notice that if the $i$-th step is parallel to the X-axis, then $y_i = y_{i-1}$ and he records $y_{i-1}(x_i - x_{i-1})$. Likewise, if the $i$-th step is parallel to the Y-axis, then ... | Thailand | Tajland 2014 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0l13 | Problem:
Let $\ell$ and $m$ be two non-coplanar lines in space, and let $P_{1}$ be a point on $\ell$. Let $P_{2}$ be the point on $m$ closest to $P_{1}$, $P_{3}$ be the point on $\ell$ closest to $P_{2}$, $P_{4}$ be the point on $m$ closest to $P_{3}$, and $P_{5}$ be the point on $\ell$ closest to $P_{4}$. Given that $... | [
"Solution:\nThe figure below shows the situation of the problem when projected appropriately, which will be explained later.\n\nLet $a$ be the answer. By taking the $z$-axis to be the cross product of these two lines, we can let the lines be on the planes $z=0$ and $z=h$, respectively. Then... | United States | HMMT February 2024 Guts Round | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | sqrt(39)/4 | |
058e | Find the least positive integer $n$ such that $\sqrt[5]{5n}$, $\sqrt[6]{6n}$ and $\sqrt[7]{7n}$ are integers. | [
"Let $n = 2^\\alpha \\cdot 3^\\beta \\cdot 5^\\gamma \\cdot 7^\\delta \\cdot s$, where $s$ is not divisible by $2$, $3$, $5$ or $7$; then $5n = 2^\\alpha \\cdot 3^\\beta \\cdot 5^{\\gamma+1} \\cdot 7^\\delta \\cdot s$, $6n = 2^{\\alpha+1} \\cdot 3^{\\beta+1} \\cdot 5^\\gamma \\cdot 7^\\delta \\cdot s$ and $7n = 2^\... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English | proof and answer | 2^35 * 3^35 * 5^84 * 7^90 | |
06bd | Define a $k$-clique to be a set of $k$ people such that every pair of them know each other (knowing is mutual). At a certain party, there are two or more $3$-cliques, but no $5$-clique. Every pair of $3$-cliques has at least one person in common. Prove that there exists at least one, and not more than two persons at th... | [
"We consider two cases.\n\n**Case 1.** There exist two $3$-cliques sharing two common people.\nSuppose the two $3$-cliques are $\\{A, B, C\\}$ and $\\{A, B, D\\}$. If all $3$-cliques contain $A$ or $B$, we are done as we can remove $A$ and $B$. If there exists a $3$-clique without $A$ and $B$, it must be $\\{C, D, ... | Hong Kong | 1997-2023 IMO HK TST | [
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
050p | Find all pairs $(x, y)$ of positive integers such that
$$
\frac{1}{x^2} + \frac{249}{xy} + \frac{1}{y^2} = \frac{1}{2012}.
$$ | [
"Let $\\gcd(x, y) = d$ and $x = ad$, $y = bd$. Then the equation can be written as $\\frac{a^2+249ab+b^2}{a^2b^2d^2} = \\frac{1}{2012}$ or\n$$\na^2 b^2 d^2 = 2012(a^2 + 249ab + b^2).\n$$\nAs $a$ and $b$ are relatively prime, $a^2$ and $b^2$ are both relatively prime to $a^2 + 249ab + b^2$ and therefore both they mu... | Estonia | Estonian Math Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (503, 1006) and (1006, 503) | |
01ad | $D$ is a point inside triangle $ABC$. The circle $S_1$ inscribed in the triangle $ABD$ touches the circle $S_2$ inscribed in the triangle $CBD$. Prove that the intersection point of outer common tangent lines of circles $S_1$ and $S_2$ lies on the line $AC$. | [
"Let the rays $CD$ and $AD$ intersect sides $AB$ and $BC$ in the points $X$ and $Y$ correspondingly. Denote by $K$, $L$, $M$, $P$, $Q$ the tangent points of circles $S_1$ and $S_2$ and segments $BD$, $AD$, $CD$, $AB$, $BC$ (see the picture).\n\nThen\n$$\nAD + BC = AL + LD + BQ + CQ = AP + DM + BP + CM = AB + CD.\n$... | Baltic Way | Baltic Way 2013 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | null | proof only | null | |
0eju | Problem:
Naj bo $ABC$ ostrokotni trikotnik. Krožnica s središčem v $A$, ki se dotika stranice $BC$, seka stranico $AB$ v točki $B_{1}$ in stranico $CA$ v točki $C_{2}$. Krožnica s središčem v $B$, ki se dotika stranice $CA$, seka stranico $BC$ v točki $C_{1}$ in stranico $AB$ v točki $A_{2}$. Krožnica s središčem v $C... | [
"Solution:\n\nDokažimo, da sta premici $AB$ in $C_{1}C_{2}$ vzporedni. Naj bodo $D, E$ in $F$ zaporedoma nožišča višin iz $A, B$ in $C$ trikotnika $ABC$. Kot med tangento in tetivo je enak obodnemu kotu nad tetivo, ta pa je enak polovici središčenega kota nad tetivo. Ker je stranica $AC$ tangenta na krožnico $s$ sr... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
01iy | In the land of Flensburg there is a single, infinitely long, street with houses numbered $2, 3, \ldots$ The police in Flensburg is trying to catch a thief which every day moves from the house where he is currently hiding to one of its neighbouring houses.
To taunt the local law enforcement the thief reveals every day t... | [
"We will prove that the police are always able to catch the thief in finite time.\nLet $h_i$ denote the house the thief stays at the $i$-th night and $p_i$ denote the greatest prime divisor of $h_i$.\nThe police know that he stays at different neighbouring houses every night, so $|h_{i+1} - h_i| = 1$ for all non-ne... | Baltic Way | Baltic Way 2023 Shortlist | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null |
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