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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
02ci | Problem:
Mostre que existe um múltiplo de 2017 que termina em 2016. | [
"Solution:\nConsidere a seguinte lista contendo 2018 números:\n$$\n\\begin{aligned}\na_{1} & =2016 \\\\\na_{2} & =20162016 \\\\\na_{3} & =201620162016 \\\\\\vdots & \\vdots \\\\\na_{2018} & =\\underbrace{20162016 \\ldots 20162016}_{2018 \\text{ vezes}} .\n\\end{aligned}\n$$\nComo existem apenas 2017 restos possívei... | Brazil | NÍVEL 3 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
04ye | Let $ABCD$ be a quadrilateral, such that $AB = BC = CD$. There are points $X, Y$ on rays $CA, BD$, respectively, such that $BX = CY$. Let $P, Q, R, S$ be the midpoints of segments $BX, CY, XD, YA$, respectively. Prove that points $P, Q, R, S$ lie on a circle. | [
"Let $M$ be the midpoint of $XY$. Note that $PR$ is midline in triangles $XBD$ and $XBY$, hence $M$ lies on $PR$. Analogously $M$ lies on $QS$.\n\nLet $\\omega_1$ be a circle with center $B$ and radius $AB = BC$ and $\\omega_2$ be a circle with center $C$ and radius $BC = CD$.\nDistance of $X$ from center of $\\ome... | Czech-Polish-Slovak Mathematical Match | CAPS Match 2024 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | English | proof only | null | |
04a1 | The difference in reciprocal values of two consecutive positive integers is $0.0\dot{a} = 0.0aaa...$. Determine all possible values of the digit $a$. | [] | Croatia | CroatianCompetitions2011 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1, 3 | |
0ezg | Problem:
A room is an equilateral triangle side $100$ meters. It is subdivided into $100$ rooms, all equilateral triangles with side $10$ meters. Each interior wall between two rooms has a door. If you start inside one of the rooms and can only pass through each door once, show that you cannot visit more than $91$ room... | [] | Soviet Union | 4th ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | k^2 - k + 1 (which equals 91 when k = 10) | |
05wo | Problem:
Trouver toutes les triplets d'entiers positifs $(x, y, z)$ satisfaisant l'équation
$$
x! + 2^{y} = z!
$$ | [
"Solution:\nSoit $(x, y, z)$ un triplet solution.\nOn cherche dans un premier temps à réduire le nombre de valeurs que $x$ peut prendre. Pour ce faire notons que $x! < z!$ donc $x < z$. En particulier, $x!$ divise $z!$ et $x!$, donc $x!$ divise $z! - x! = 2^{y}$. Supposons par l'absurde que $x \\geqslant 3$, dans c... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | ((0, 0, 2), (1, 0, 2), (2, 2, 3)) | |
0eb5 | Does the equation $\cos(\sin x) = \sin(\cos x)$ have at least one real solution? | [
"Rewrite the equation as $\\cos(\\sin x) = \\cos\\left(\\frac{\\pi}{2} - \\cos x\\right)$. This implies that $\\frac{\\pi}{2} - \\cos x = \\pm \\sin x + 2k\\pi$ for some $k \\in \\mathbb{Z}$, so $\\cos x \\pm \\sin x = \\frac{\\pi}{2} - 2k\\pi$. Since $|\\cos x \\pm \\sin x| \\le |\\cos x| + |\\sin x| \\le 2$, we h... | Slovenia | National Math Olympiad in Slovenia | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | null | proof and answer | No | |
0gin | 芙莉蓮與寶箱怪用 $113$ 個寶箱玩一場遊戲。遊戲開始時，所有的寶箱都未上鎖且沒有寶石。每一年，芙莉蓮在其中一個未上鎖的寶箱裡增加一顆寶石，接著寶箱怪依據以下規則行動：
- 如果只剩一個寶箱未上鎖，則將所有寶箱解鎖；
- 如果多於一個寶箱未上鎖，則寶箱怪從未上鎖的寶箱中擇一上鎖。
試證：存在常數 $C > 0$，使得不論寶箱怪如何行動，芙莉蓮都總是能保證在任意年結束時，任兩個寶箱內的寶石數量相差不超過 $C$。（備註：寶箱怪不可以吃掉芙莉蓮。）
Frieren and a mimic use $113$ treasure chests to play a game. At the beginning, all chests ar... | [
"一般性地，對於 $n$ 個寶箱，可取 $C = n - 1$。芙莉蓮的策略很簡單：在每一年，從未上鎖的寶箱中，選一個寶石最少的寶箱增加寶石即可。（備註：這並非唯一可行的策略。）\n\n要證明此策略可行，令 $x_1^t \\le x_2^t \\le \\cdots \\le x_n^t$ 為在第 $t$ 年底時各寶箱內的寶石數量，並令 $x_1^0 = x_2^0 = \\cdots = x_n^0 = 0$。我們有以下觀察：\n\n(a) 對於任意 $t$，存在唯一的 $m = m(t)$ 使得 $x_m^{t+1} = x_m^t + 1$；\n\n(b) 對於任意 $j > m$，$x_m^t < x_m^{t+1... | Taiwan | IMO 1J, Mock Exam 1 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Chinese; English | proof only | null | |
0ckr | Non-negative real numbers $x$, $y$, $z$ satisfy the relations $xy + 4 \le 2(x+z)$, $yz + 4 \le 2(y+x)$, $zx + 4 \le 2(z+y)$. Prove that $x = y = z$. | [
"The hypothesis can be written in the form $x(y-2) \\le 2(z-2)$, $y(z-2) \\le 2(x-2)$, $z(x-2) \\le 2(y-2)$. If $x-2 < 0$, using the second relation we get $z-2 < 0$, then, according to the first, we obtain $y-2 < 0$, so $x$, $y$, $z \\in (0, 2)$ and $xyz < 8$. We also have $x(2-y) \\ge 2(2-z) > 0$, $y(2-z) \\ge 2(... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0ebo | Real numbers $x$ and $y$ satisfy $x^3 + x^2 + xy + x + y + 2 = 0$ and $y^3 - y^2 + 3y - x = 0$. Determine the value of the expression $x - y$. | [
"We subtract the equations to get\n$$\nx^3 - y^3 + x^2 + y^2 + xy + 2x - 2y + 2 = 0.\n$$\nWe rearrange the left side\n$$\n\\begin{align*}\nx^3 - y^3 + x^2 + y^2 + xy + 2x - 2y + 2 &= \\\\\n= (x - y)(x^2 + xy + y^2) + (x^2 + xy + y^2) + 2(x - y + 1) = \\\\\n= (x - y + 1)(x^2 + xy + y^2) + 2(x - y + 1) = \\\\\n= (x -... | Slovenia | National Math Olympiad 2015 – Final Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | -1 | |
01qf | Given a triangle $ABC$. Let $S$ be the circle passing through $C$ centered at $A$. Let $X$ be a variable point on $S$, and let $K$ be the midpoint of the segment $CX$.
Find the locus of the midpoints of $BK$, when $X$ moves along $S$. | [
"Let $R$ be the radius of the given circle, $M$ the midpoint of $AC$, $N$ the midpoint of $BM$, $T$ the midpoint of $BK$. Let $a, b, c, m, n, t, k$, $x$ be the (complex) coordinates of the points $A, B, C, M, N, T, X$, respectively. Then $m = (a+c)/2$, $n = (b+m)/2$, $k = (c+x)/2$, $t = (k+b)/2$. So we can easily f... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | A circle centered at N, the midpoint of BM (where M is the midpoint of AC), with radius equal to one quarter of AC (equivalently, R/4 where R is the radius of S). | |
0am2 | Problem:
From the $x y$-plane, select five distinct points that have integer coordinates. Find the probability that there is a pair of points among the five whose midpoint has integer coordinates. | [] | Philippines | AREA STAGE | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 1 | |
0ce4 | Find the pairs $(m, n)$ of positive integers with the property
$$(3m - 2)(5m - 1) = 6^n.$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (2, 2) | |
0bsh | Let $f: \mathbb{R} \to \mathbb{R}$ be a function with the following properties:
$$
\begin{aligned}
f(x+y) &\le f(x) + f(y), \\
f(tx + (1-t)y) &\le t f(x) + (1-t) f(y),
\end{aligned}
$$
for all $x, y \in \mathbb{R}$ and $t \in [0, 1]$.
a) Prove that for any $a \le b \le c \le d$, such that $d - c = b - a$, the followin... | [
"a) Let $t = \\frac{d-b}{a-d} = \\frac{c-a}{a-d} \\in [0, 1]$, then $b = t a + (1-t) d$ and $c = (1-t) a + t d$.\nUsing (P2) yields\n$$\nf(b) = f(t a + (1-t) d) \\le t f(a) + (1-t) f(d).\n$$\nSimilarly, $f(c) \\le (1-t) f(a) + t f(d)$. Adding up these two inequalities gives the result.\n\n\nb) We will prove the ine... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
056j | Medians *AD*, *BE*, and *CF* of triangle *ABC* intersect at point *M*. Is it possible that the circles with radii $MD$, $ME$, and $MF$
a) all have areas smaller than the area of triangle $ABC$;
b) all have areas greater than the area of triangle $ABC$;
c) all have areas equal to the area of triangle $ABC$? | [
"a) Let triangle $ABC$ be equilateral (Fig. 11). As the medians of this triangle all have equal lengths, the points $D$, $E$, and $F$ are located at equal distances from $M$, i.e., on a circle with centre $M$. It suffices to show that the area of this circle is less than the area of triangle $ABC$. Because the medi... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof and answer | a) yes; b) yes; c) no | |
0947 | Problem:
Let $\Omega$ be the circumcircle of a triangle $A B C$ with $\angle C A B = 90^\circ$. The medians through $B$ and $C$ meet $\Omega$ again at $D$ and $E$, respectively. The tangent to $\Omega$ at $D$ intersects the line $A C$ at $X$ and the tangent to $\Omega$ at $E$ intersects the line $A B$ at $Y$. Prove th... | [
"Solution:\n\nLet $G$ be the second intersection of the median through $A$ and $\\Omega$. We will show that $X Y$ is tangent to $\\Omega$ at $G$. Let $X'$ be the intersection of the tangents to $\\Omega$ at $D$ and $G$. We claim that $A$, $C$, and $X'$ are collinear.\n\n\n\nLet $O$ be the c... | Middle European Mathematical Olympiad (MEMO) | MEMO Team Competition | [
"Geometry > Plane Geometry > Circles > Tangents",
"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 | |
022b | Problem:
Na figura $ABCD$ é um quadrado cujo lado mede $1~\mathrm{cm}$, $E$ é o ponto médio da diagonal $BD$ e $F$ o ponto médio do segmento $BE$. Qual é a área do triângulo $\triangle BCF$?
 | [
"Solution:\n\nAs diagonais do quadrado $ABCD$ dividem o quadrado em 4 triângulos iguais, logo a área do triângulo $\\triangle BCE$ é\n\n\n\n$$\n1 \\div 4 = 0,25~\\mathrm{cm}^2\n$$\n\nComo o comprimento de $BF$ é a metade de $BE$ e a altura relativa aos lados $BF$ e $BE$ é $CE$, então a área... | Brazil | Lista 2 | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 0.125 cm^2 | |
0iak | Problem:
Let $ABC$ be an equilateral triangle of side length $2$. Let $\omega$ be its circumcircle, and let $\omega_{A}$, $\omega_{B}$, $\omega_{C}$ be circles congruent to $\omega$ centered at each of its vertices. Let $R$ be the set of all points in the plane contained in exactly two of these four circles. What is th... | [
"Solution:\n$\\omega_{A}$, $\\omega_{B}$, $\\omega_{C}$ intersect at the circumcenter; thus, every point within the circumcircle, and no point outside of it, is in two or more circles. The area inside exactly two circles is shaded in the figure. The two intersection points of $\\omega_{A}$ and $\\omega_{B}$, togeth... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 2 sqrt(3) | |
03j7 | Problem:
Prove that $2^{n-1}$ divides $n!$ if and only if $n=2^{k-1}$ for some positive integer $k$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0bkn | Let $ABC$ be an acute triangle and $D \in (BC)$, $E \in (AD)$ be mobile points. The circumcircle of triangle $CDE$ meets the median from $C$ of the triangle $ABC$ at $F$. Prove that the circumcenter of triangle $AEF$ lies on a fixed line. | [
"The quadrilateral $EFDC$ is cyclic, hence $\\angle FED \\equiv \\angle FCD$. Let $P$ be the reflection of point $C$ in the midpoint of the segment line $[AB]$; clearly $AP \\parallel BC$ and $\\angle FCD \\equiv \\angle FPA$.\n\nIt follows that $\\angle FPA \\equiv \\angle FED$, which means that the quadrilateral ... | Romania | 65th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0deq | Prove that for every polynomial $P(x)$ there exist polynomials $Q(x)$ and $R(x)$ such that $P(x) = Q(x^2) + R((x+1)^2)$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Linear Algebra > Linear transformations"
] | English | proof only | null | |
06p2 | Let $P$ be a polynomial of degree $n>1$ with integer coefficients and let $k$ be any positive integer. Consider the polynomial $Q(x)=P(P(\ldots P(P(x)) \ldots))$, with $k$ pairs of parentheses. Prove that $Q$ has no more than $n$ integer fixed points, i.e. integers satisfying the equation $Q(x)=x$.
(Romania) | [
"The claim is obvious if every integer fixed point of $Q$ is a fixed point of $P$ itself. For the sequel assume that this is not the case. Take any integer $x_{0}$ such that $Q\\left(x_{0}\\right)=x_{0}$, $P\\left(x_{0}\\right) \\neq x_{0}$ and define inductively $x_{i+1}=P\\left(x_{i}\\right)$ for $i=0,1,2, \\ldot... | IMO | IMO 2006 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
0knh | Determine all integers $s \ge 4$ for which there exist positive integers $a, b, c, d$ such that $s = a+b+c+d$ and $s$ divides $abc + abd + acd + bcd$. | [
"The answer is $s$ composite.\n\n**Composite construction** Write $s = (w+x)(y+z)$, where $w, x, y, z$ are positive integers. Let $a = wy$, $b = wz$, $c = xy$, $d = xz$. Then\n$$\nabc + abd + acd + bcd = wxyz(w+x)(y+z)\n$$\nso this works.\n\n**Prime proof** Choose suitable $a, b, c, d$. Then\n$$\n(a+b)(a+c)(a+d) = ... | United States | USA IMO TST | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Abstract Algebra > Field Theory"
] | English | proof and answer | All composite integers at least 4 | |
0gww | There are $n$ cities in a country. A monopoly company intends to establish air traffic between some of these cities. The government demands that the air traffic satisfies the following conditions. People should be able to get from each city to any other of these cities by one or several flights. Besides there must be a... | [
"a) Let's divide all the cities into two groups consisting of $1003$ and $1005$ cities respectively. Let's connect all the cities in each group in cycle. Thus, we obtain the degree $2$ of each vertex. Next lets select one city in each component and connect the two cities with a flight \"007\". Let's connect the res... | Ukraine | Ukrajina 2008 | [
"Discrete Mathematics > Graph Theory > Menger's theorem / max-flow, min-cut",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a) 1; b) 2 | |
0a2p | Inaya puts $88$ fours in a row and gets a very big number:
$$
\underbrace{4444\cdots4444}_{\text{88 fours}}
$$
She writes the fours as a plus sign with a slash in the upper left corner. When she removes that slash, she gets a plus sign. So she can make additions like
$$
44444444+44\cdots44444444
$$
The addition may not... | [] | Netherlands | Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | E | |
045q | As shown in the figure, points $K$ and $L$ are in the interior of a triangle $ABC$, point $D$ lies on the side $AB$. It is known that points $B$, $K$, $L$, $C$ are concyclic, and $\angle AKD = \angle BCK$, $\angle ALD = \angle BCL$. Prove that $AK = AL$.
 | [
"*Proof*. As shown in the following picture,\n\nlet the extensions of $AK$ and $AL$ intersect the circle passing through $B$, $K$, $L$, $C$ at points $X$ and $Y$, respectively. Connect $BX$ and $BY$. Considering the given conditions, we have $\\angle AKD = \\angle BCK = \\angle BXK$ and $\\... | China | 2022 CGMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
04xd | Determine all triples $(a, b, c)$ of positive real numbers satisfying the following system of equations
$$
\begin{aligned}
a\sqrt{b} - c &= a, \\
b\sqrt{c} - a &= b, \\
c\sqrt{a} - b &= c.
\end{aligned}
$$ | [
"Without loss of generality we can suppose $a = \\max\\{a, b, c\\}$. From the first equation of the given system we obtain\n$$\nc(\\sqrt{b} - 1) \\le a(\\sqrt{b} - 1) = c, \\quad \\text{i.e.} \\quad b \\le 4.\n$$\nSimilarly, from the second equation of the given system we have\n$$\nb(\\sqrt{c} - 1) = a \\ge b, \\qu... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | (4, 4, 4) | |
0f99 | Problem:
We are given $1998$ normal coins, $1$ heavy coin and $1$ light coin, which all look the same. We wish to determine whether the average weight of the two abnormal coins is less than, equal to, or greater than the weight of a normal coin. Show how to do this using a balance $4$ times or less. | [] | Soviet Union | 23rd ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
06qn | Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelograms.
Prove that $GR = GS$. | [
"Denote by $k$ the incircle and by $k_{a}$ the excircle opposite to $A$ of triangle $ABC$. Let $k$ and $k_{a}$ touch the side $BC$ at the points $X$ and $T$, respectively, let $k_{a}$ touch the lines $AB$ and $AC$ at the points $P$ and $Q$, respectively. We use several times the fact that opposing sides of a parall... | IMO | IMO Problem Shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle c... | English | proof only | null | |
0ext | Problem:
a. $B$ and $C$ are on the segment $AD$ with $AB = CD$. Prove that for any point $P$ in the plane: $PA + PD \geq PB + PC$.
b. Given four points $A$, $B$, $C$, $D$ on the plane such that for any point $P$ on the plane we have $PA + PD \geq PB + PC$. Prove that $B$ and $C$ are on the segment $AD$ with $AB = CD$... | [
"Solution:\n\na.\nSuppose the points lie in the order $A$, $B$, $C$, $D$. If $P$ lies on $AD$, then the result is trivial, and we have equality if $P$ lies outside the segment $AD$. So suppose $P$ does not lie on $AD$.\n\nLet $M$ be the midpoint of $AD$. Take $P'$ so that $P$, $M$, $P'$ are collinear and $PM = MP'$... | Soviet Union | 6th ASU | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
0hww | Problem:
Prove that there are infinitely many positive integers $n$ for which $n^{2}+1$ has no repeated prime factors (that is, $n^{2}+1$ is squarefree). | [
"Solution:\n\nBy Fermat's Christmas theorem, the only primes which may divide $n^{2}+1$ other than $2$ are those which are $1 \\pmod{4}$, and moreover $2^{2} \\nmid n^{2}+1$ for any $n$.\n\nConsider primes $p \\equiv 1 \\pmod{4}$. Observe that for any $x$, we have\n$$\n\\#\\{n \\leq x \\mid n^{2}+1 \\equiv 0 \\pmod... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof only | null | |
04us | We say that a positive integer $k$ is *fair* if the number of $2021$-digit palindromes that are a multiple of $k$ is the same as the number of $2022$-digit palindromes that are a multiple of $k$. Does the set $M = \{1, 2, \dots, 35\}$ contain more numbers that are fair or those that are not fair?
(A palindrome is an i... | [] | Czech Republic | Final Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | more fair numbers | |
0kup | Problem:
Over all real numbers $x$ and $y$ such that
$$
x^{3}=3 x+y \quad \text{ and } \quad y^{3}=3 y+x
$$
compute the sum of all possible values of $x^{2}+y^{2}$. | [
"Solution:\nFirst, we eliminate easy cases.\n- if $x=-y$, then $x^{3}=3 x-x=2 x$, so $x \\in\\{0, \\sqrt{2},-\\sqrt{2}\\}$. Therefore, we get $(\\sqrt{2},-\\sqrt{2}),(-\\sqrt{2}, \\sqrt{2})$, and $(0,0)$.\n- if $x=y \\neq 0$, then $x^{3}=3 x+x=4 x$, so $x \\in\\{2,-2\\}$. Therefore, we get $(2,2)$ and $(-2,-2)$.\n\... | United States | HMMT November 2023 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | 15 | |
041k | Suppose there are $m$ questions in an examination attended by $n$ students, where $m, n \ge 2$ are given natural numbers. The marking rule for each question is as follows: if there are exactly $x$ students failing to answer the question correctly, then they will each get 0 marks, and those who answer it correctly will ... | [
"For any $k = 1, 2, \\dots, m$, assuming there are $x_k$ students failing to answer the $k$th question correctly, then there are $n - x_k$ ones who answer it correctly and each gets $x_k$ marks from it accordingly. Suppose the sum of the $n$ students' total marks is $S$. Then we have\n$$\n\\sum_{i=1}^{n} p_i = S = ... | China | China Mathematical Competition (Complementary Test) | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | m(n-1) | |
0201 | Problem:
Abby and Brian play the following game: They first choose a positive integer $N$. Then they write numbers on a blackboard in turn. Abby starts by writing a $1$. Thereafter, when one of them has written the number $n$, the other writes down either $n+1$ or $2n$, provided that the number is not greater than $N$... | [
"Solution:\n\na. Abby has a winning strategy for odd $N$: Observe that, whenever any player writes down an odd number, the other player has to write down an even number. By adding $1$ to that number, the first player can write down another odd number. Since Abby starts the game by writing down an odd number, she ca... | Benelux Mathematical Olympiad | THIRD BENELUX MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | a) Abby; b) 31 | |
037a | Problem:
Find the least positive integer $a$ such that the system
$$
\left\lvert\, \begin{aligned}
& x+y+z=a \\
& x^{3}+y^{3}+z^{2}=a
\end{aligned}\right.
$$
has no integer solution. | [
"Solution:\nWhen $a=1,2,3$ the system has solutions $(1,0,0)$, $(1,1,0)$ and $(1,1,1)$, respectively. We shall prove that when $a=4$ the system has no integer solution.\n\nSuppose the contrary. Then we have\n$$\n4-z^{2}=x^{3}+y^{3}=(x+y)\\left(x^{2}-x y+y^{2}\\right)=(4-z)\\left(x^{2}-x y+y^{2}\\right)\n$$\ngiving ... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 4 | |
0jti | Problem:
Let $r_{1}, r_{2}, r_{3}, r_{4}$ be the four roots of the polynomial $x^{4}-4 x^{3}+8 x^{2}-7 x+3$. Find the value of
$$
\frac{r_{1}^{2}}{r_{2}^{2}+r_{3}^{2}+r_{4}^{2}}+\frac{r_{2}^{2}}{r_{1}^{2}+r_{3}^{2}+r_{4}^{2}}+\frac{r_{3}^{2}}{r_{1}^{2}+r_{2}^{2}+r_{4}^{2}}+\frac{r_{4}^{2}}{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}... | [
"Solution:\nAnswer: $-4$\nAdd 1 to each fraction to get\n$$\n\\frac{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2}}{r_{2}^{2}+r_{3}^{2}+r_{4}^{2}}+\\frac{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2}}{r_{1}^{2}+r_{3}^{2}+r_{4}^{2}}+\\frac{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r_{4}^{2}}{r_{1}^{2}+r_{2}^{2}+r_{4}^{2}}+\\frac{r_{1}^{2}+r... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | -4 | |
0kua | Problem:
The Fibonacci numbers are defined recursively by $F_{0}=0$, $F_{1}=1$, and $F_{i}=F_{i-1}+F_{i-2}$ for $i \geq 2$. Given 15 wooden blocks of weights $F_{2}, F_{3}, \ldots, F_{16}$, compute the number of ways to paint each block either red or blue such that the total weight of the red blocks equals the total we... | [
"Solution:\nPartition the blocks into sets\n$$\n\\left\\{F_{2}, F_{3}, F_{4}\\right\\},\\left\\{F_{5}, F_{6}, F_{7}\\right\\}, \\ldots,\\left\\{F_{14}, F_{15}, F_{16}\\right\\}\n$$\nWe can show by bounding that $F_{16}$ belongs on the opposite side as $F_{15}$ and $F_{14}$, and, in general, that $F_{3k+1}$ is on th... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 32 | |
0fn8 | Let $\Gamma$ be the circumcircle of a triangle $ABC$ and let $E$ and $F$ be the intersections of the bisectors of $\angle ABC$ and $\angle ACB$ with $\Gamma$. If $EF$ is tangent to the incircle $\gamma$ of $\triangle ABC$, then find the value of $\angle BAC$. | [
"Let us denote by $I$ the incenter of $\\triangle abc$. From the figure immediately follows $\\angle IBC = \\angle IFE$ and $\\angle ICB = \\angle IEF$. So, $\\triangle IBC \\sim \\triangle IFE$. Since both have the same height (the radii of the incircle) because $BC$ and $EF$ are both tangent to $\\gamma$, then $\... | Spain | International Mathematical Arhimede Contest | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Spanish | proof and answer | 60° | |
0dqt | Let $x, y, z > 0$ such that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} < \frac{1}{xyz}$. Show that
$$
\frac{2x}{\sqrt{1+x^2}} + \frac{2y}{\sqrt{1+y^2}} + \frac{2z}{\sqrt{1+z^2}} < 3.
$$ | [
"Let $r = 1/x$, $s = 1/y$, $t = 1/z$. There exists $\\alpha < 1$ such that $r+s+t = \\alpha^2 rst$ or $\\alpha(r+s+t) = \\alpha^3 rst$. Let $a = \\alpha r$, $b = \\alpha s$, $c = \\alpha t$. Write $a = \\tan A$, $b = \\tan B$, $c = \\tan C$, then $A+B+C = \\pi$. It is clear that\n$$\n\\begin{aligned}\n\\frac{1}{2} ... | Singapore | Singapore Mathematical Olympiad (SMO) 2011 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
02pj | Problem:
Numa corrida de São Paulo a Fortaleza participam quatro carros $A$, $B$, $C$, $D$ que largaram na seguinte ordem: primeiro $A$, segundo $B$, terceiro $C$ e por último $D$. Durante a corrida, $A$ e $B$ trocaram de posição (ultrapassaram um ao outro) 9 vezes e $B$ e $C$ trocaram de posição 8 vezes.
Para saber ... | [
"Solution:\n\nInicialmente, observe que se dois carros trocaram de posição um número par de vezes, eles terminaram na mesma ordem em que começaram e se trocaram de posição um número ímpar de vezes, eles terminaram na ordem inversa. Isto nos leva a concluir que $B$ terminou a corrida na frente de $A$ e de $C$.\n\nFa... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Logic"
] | null | proof only | null | |
0j1v | Problem:
Compute $\lim_{n \rightarrow \infty} \frac{\sum_{k=1}^{n}|\cos (k)|}{n}$. | [
"Solution:\n\n$\\frac{2}{\\pi}$\n\nThe main idea lies on the fact that positive integers are uniformly distributed modulo $\\pi$. (In other words, if each integer $n$ is written as $q \\pi + r$ where $q$ is an integer and $0 \\leq r < \\pi$, the value of $r$ will distribute uniformly in the interval $[0, \\pi]$.) U... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2/π | |
06fa | Determine all pairs $(x, y)$ of positive integers such that $\frac{x^2y + x + y}{xy^2 + y + 11}$ are integers. | [
"(IMO Shortlist 1998 N1/Problem 4) The solutions are $(x, y) = (11t^2, 11t), (7, 1), (121, 1), (10, 2)$ where $t \\in \\mathbb{Z}^+$.\nThe condition means $xy^2 + y + 11 \\mid x^2y + x + y$. Since\n$$\ny(x^2y + x + y) - x(xy^2 + y + 11) = y^2 - 11x,\n$$\nwe have\n$$\nxy^2 + y + 11 \\mid y^2 - 11x.\n$$\n* If $y^2 - ... | Hong Kong | IMO HK TST | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (x, y) = (11t^2, 11t) for any positive integer t; and (7, 1), (121, 1), (10, 2). | |
02j2 | Problem:
A menor raiz da equação $\frac{|x-1|}{x^{2}}=6$ é:
(A) $-\frac{1}{3}$
(B) $-\frac{1}{2}$
(C) $\frac{1}{3}$
(D) $\frac{1}{4}$
(E) $\frac{3}{2}$ | [
"Solution:\nCaso 1: $x \\geq 1$\n\nNesse caso, $x-1 \\geq 0$, donde $|x-1|=x-1$. A equação toma a forma $\\frac{x-1}{x^{2}}=6$ ou $6x^{2}-x+1=0$. Essa equação não tem raízes reais porque $\\Delta = (-1)^{2} - 4 \\times 6 \\times 1 = 1 - 24$ é negativo.\n\n\nCaso 2: $x < 1$\n\nNesse caso, $x-1 < 0$, donde $|x-1|=-(x... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | B | |
0l9u | Find all polynomials $P(x)$ with real coefficients satisfying the condition
$$
P(x^2) + x(3P(x) + P(-x)) = (P(x))^2 + 2x^2
$$
for all real numbers $x$. | [
"Let $P(x)$ be a polynomial satisfying the condition of the problem. We have:\n$$\nP(x^2) + x(3P(x) + P(-x)) = (P(x))^2 + 2x^2, \\quad x \\in \\mathbb{R}, \\quad (1)\n$$\n\nIt is clear that $\\deg P > 0$. Consider the cases:\ni) $\\deg P = 1$.\n$P(x)$ has the form $P(x) = ax + b$, $a \\neq 0$. Substituting it into ... | Vietnam | Vijetnam 2006 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | All polynomials are: P(x) = x; and for any nonnegative integer n, P(x) = x^{2n} + 2x or P(x) = x^{2n+1} + x. | |
0jum | Problem:
Let $ABC$ be a triangle with $AB = 3$, $AC = 8$, $BC = 7$ and let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively. Point $T$ is selected on side $BC$ so that $AT = TC$. The circumcircles of triangles $BAT$, $MAN$ intersect at $D$. Compute $DC$. | [
"Solution:\n\nWe note that $D$ is the circumcenter $O$ of $ABC$, since $2 \\angle C = \\angle ATB = \\angle AOB$. So we are merely looking for the circumradius of triangle $ABC$. By Heron's Formula, the area of the triangle is $\\sqrt{9 \\cdot 6 \\cdot 1 \\cdot 2} = 6\\sqrt{3}$, so using the formula $\\frac{abc}{4R... | United States | HMMT February 2016 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 7√3/3 | |
0gcc | 設 $\triangle ABC$ 為各邊長不相同的銳角三角形，點 $O$, $H$ 分別為其外心與垂心。
直線 $OA$ 分別與 $\triangle ABC$ 中過點 $B$ 的高及過點 $C$ 的高交於點 $P$, $Q$。
試證：三角形 $PQH$ 的外心，落在三角形 $ABC$ 的某一條中線上。
(註: 三角形中，頂點與對邊中點的連線，稱為中線。) | [
"Suppose, without loss of generality, that $AB < AC$. We have\n$$\n\\begin{aligned}\n\\angle PQH &= 90^\\circ - \\angle QAB = 90^\\circ - \\angle OAB \\\\\n&= \\frac{1}{2} \\angle AOB = \\angle ACB,\n\\end{aligned}\n$$\nand similarly $\\angle QPH = \\angle ABC$. Thus triangles $ABC$ and $HPQ$ are similar. Let $\\Om... | Taiwan | 二〇一八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 > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0gzn | The diagonals of a cyclic quadrilateral $ABCD$ are perpendicular. Points $K, L, M, Q$ are the orthocenters of the triangles $ABD$, $ACD$, $BCD$, $ABC$ respectively. Prove that quadrilaterals $KLMQ$ and $ABCD$ are equal. | [
"Let the diagonals of the quadrilateral $ABCD$ intersect at $O$. Altitudes of the triangles $BCD$ and $ACD$ lie on $AC$, so their orthocenters $K$ and $M$ too. Analogously, points $L$, $Q$ lie on $BC$ (Fig.06).\n\nNote that $BK \\parallel CL$, because $BK \\perp AD$ and $CL \\perp AD$. It follows that $\\angle BKC ... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry ... | English | proof only | null | |
0jt6 | Problem:
Tim starts with a number $n$, then repeatedly flips a fair coin. If it lands heads he subtracts $1$ from his number and if it lands tails he subtracts $2$. Let $E_{n}$ be the expected number of flips Tim does before his number is zero or negative. Find the pair $(a, b)$ such that
$$
\lim_{n \rightarrow \infty... | [
"Solution:\n\nWe have the recurrence $E_{n} = \\frac{1}{2}\\left(E_{n-1} + 1\\right) + \\frac{1}{2}\\left(E_{n-2} + 1\\right)$, or $E_{n} = 1 + \\frac{1}{2}\\left(E_{n-1} + E_{n-2}\\right)$, for $n \\geq 2$.\n\nLet $F_{n} = E_{n} - \\frac{2}{3} n$. By directly plugging this into the recurrence for $E_{n}$, we get t... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | (2/3, 2/9) | |
09fx | Find all pairs of positive integers $(a, b)$ such that $a^p - b^p - 1$ is divisible by $43$ for every prime $p \ge 5$. | [
"*Answer: $(a, b) \\in \\{(43n, 43m-1),\\ (43n-42, 43m),\\ (43n-6, 43m-7),\\ (43n+7-43, 43m+6-43) \\mid n, m \\in \\mathbb{Z}_{\\ge 1}\\}.$*\n\nIf $p = 43$ then it follows from Fermat's little theorem that $a \\equiv b+1 \\pmod{43}$. Hence if $a$ is divisible by $43$ then $b \\equiv -1 \\pmod{43}$ and if $b$ is div... | Mongolia | 2015 Mongolian IMO Team Selection Tests | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | {(43n, 43m-1), (43n-42, 43m), (43n-6, 43m-7), (43n+7-43, 43m+6-43) | n, m ∈ Z_{≥ 1}} | |
0guk | Let the line $l_0$ be tangent to the circumcircle of a scalene triangle $ABC$ at point $A$. A variable line $l$ parallel to $l_0$ intersects the sides $AB$ and $AC$ at points $X$ and $Y$, respectively. Let $T$ be the point of intersection of $BY$ and $CX$. Let $Z$ be the second intersection point of the line $AT$ and t... | [] | Turkey | Team Selection Test for EGMO 2023 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0jn3 | Problem:
Triangle $ABC$ is isosceles, and $\angle ABC = x^{\circ}$. If the sum of the possible measures of $\angle BAC$ is $240^{\circ}$, find $x$. | [
"Solution:\n\nThere are three possible triangles: either $\\angle ABC = \\angle BCA$, in which case $\\angle BAC = 180 - 2x$; $\\angle ABC = \\angle BAC$, in which case $\\angle BAC = x$; or $\\angle BAC = \\angle BCA$, in which case $\\angle BAC = \\frac{180 - x}{2}$. These sum to $\\frac{540 - 3x}{2}$, so we have... | United States | HMMT November 2015 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 20 | |
09ep | Find the number of words using $2014$ letters $a$ and $2014^2$ letters $b$ with the following properties:
i) Among the first $k$ letters of the word, number of $b$'s is not less than number of $a$'s for any $k$ satisfying the condition $i \le k \le 2014 \cdot 2015$;
ii) Among the last $k$ letters of the word, number of... | [
"In a rectangular grid, let's call right move the route from point $(x, y)$ to point $(x + 1, y)$ and up move the route to point from $(x, y)$ to point $(x, y + 1)$.\nFurthermore when we will count number of routes always will mean every route consists of only right and up moves.\nNow consider correspondence betwee... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | binom(2014*2015,2014) - 2*binom(2014*2015,2013) + binom(2014*2015,2012) | |
0dlu | Prove that there are infinitely many primes $p$ such that each of them divides an integer of the form $3^n - 2$, but does not divide any integer of the form $2^m - 3$, where $m$ and $n$ are positive integers. | [
"We show the stronger result that the set of prime $p \\equiv \\pm 5 \\pmod{12}$ such that $p \\mid 9^n - 2$ for some $n$ (since $9 = 3^2$, we can work with $9$ instead of $3$, by restricting to even exponents), but $p \\nmid 2^m - 3$ for any $m$, is infinite.\n\nNote that if $p \\mid 9^n - 2$ for some $n$, then $9... | Saudi Arabia | Saudi Booklet | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
03je | Problem:
Find all solutions of $a^{2} + b^{2} = n!$ for positive integers $a$, $b$, $n$ with $a \leq b$ and $n < 14$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Algebraic Number Theory > Quadratic forms",
"Number Theory > Algebraic Number Theory > Unique factorization",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All solutions with n < 14 are: (a, b, n) = (1, 1, 2) and (12, 24, 6). | |
0812 | Problem:
Sono date le tre quantità $X = a + 7b$, $Y = 2a + 5b$, $Z = 4a + 2b$, dove $a$ e $b$ sono numeri reali positivi. Allora:
(A) $X < Y < Z$
(B) $Z < Y < X$
(C) $Y < X < Z$
(D) $Y < Z < X$
(E) l'ordine di $X, Y, Z$ dipende dai valori di $a$ e $b$. | [
"Solution:\n\nLa risposta è (E). Ponendo $a = 3$ e $b = 1$ si ottiene $X = 10$, $Y = 11$, $Z = 14$ (dunque $X < Y < Z$), mentre ponendo $a = b = 1$ si ottiene $X = 8$, $Y = 7$, $Z = 6$ (dunque $X > Y > Z$). Questo permette di concludere che la relazione tra due qualunque dei tre numeri $X, Y, Z$ dipende dai partico... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | E | |
0af6 | Реши ја равенката $\left(\frac{x^3 + x}{3}\right)^3 + \frac{x^3 + x}{3} = 3x$. | [
"Ако воведеме смена $\\frac{x^3+x}{3} = y$, тогаш $y^3 + y = 3x$ и $x^3+x=3y$. Според тоа, ако $x_0$ е решение на равенката и $\\frac{x_0^3+x_0}{3} = y_0$, тогаш $(x_0,y_0)$ е решение на системот равенки $\\begin{cases} x^3 + x = 3y \\\\ y^3 + y = 3x \\end{cases}$. Точно е и обратното кое што не е тешко да се докаж... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | Macedonian, English | proof and answer | x = 0, ±sqrt(2) | |
0h5z | Find all pairs of real $(x; y)$, such that:
$$
\{x\} + \{y\} = [x+y].
$$ | [
"From the statement of the problem we conclude that $\\{x\\} + \\{y\\}$ is an integer, but $0 \\le \\{x\\} < 1$, so: $\\{x\\} + \\{y\\} = 0$ or $\\{x\\} + \\{y\\} = 1$.\n\n**Case 1:** $\\{x\\} + \\{y\\} = 0$. So $\\{x\\} = \\{y\\} = 0$, hence $x, y \\in \\mathbb{Z}$. But then $[x+y] = x+y=0$, so the solutions are $... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | All pairs are: (n, −n) for any integer n, and (x, 1 − x) where x is a non-integer real. | |
0imi | Problem:
Two real numbers $x$ and $y$ are such that $8 y^{4} + 4 x^{2} y^{2} + 4 x y^{2} + 2 x^{3} + 2 y^{2} + 2 x = x^{2} + 1$. Find all possible values of $x + 2 y^{2}$. | [
"Solution:\n\nAnswer: $\\frac{1}{2}$.\n\nWriting $a = x + 2 y^{2}$, the given quickly becomes $4 y^{2} a + 2 x^{2} a + a + x = x^{2} + 1$. We can rewrite $4 y^{2} a$ for further reduction to $a(2 a - 2 x) + 2 x^{2} a + a + x = x^{2} + 1$, or\n$$\n2 a^{2} + (2 x^{2} - 2 x + 1) a + (-x^{2} + x - 1) = 0\n$$\nThe quadr... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1/2 | |
0ecc | Problem:
Dana je pravokotna mreža velikosti $7 \times 9$ (glej sliko). V spodnjem levem vozlišču mreže je kolonija mravelj, v zgornjem desnem vozlišču pa je njihovo mravljišče. V vseh ostalih vozliččih mreže je po eno zrno riža. Vsaka mravlja iz kolonije se na poti do mravljišča sprehaja po povezavah mreže, vendar le ... | [
"Solution:\n\nKolonija mora vsebovati najmanj 8 mravelj.\n\nDokažimo najprej, da 8 mravelj lahko pobere vsa zrna riža. To lahko storijo na primer tako, da se $i$-ta od teh 8 mravelj sprehodi najprej navpično navzgor do $i$-te vrstice mreže, nato po $i$-ti vrstici v desno do skrajnega desnega roba mreže in nazadnje ... | Slovenia | 59. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 8 | |
0bge | Problem:
Fie $(A,+, \cdot)$ un inel care îndeplineşte simultan următoarele două condiţii:
(1) $A$ nu este corp,
(2) oricare ar fi $x$ un element neinversabil al lui $A$, există un număr întreg $m \geq 1$, care depinde de $x$, astfel încât
$$
x = x^{2} + x^{3} + \cdots + x^{2^{m}}
$$
Să se arate că:
a. $x + x = 0$, or... | [
"Solution:\n\na.\nEste suficient să demonstrăm că $1 + 1 = 0$. Fie $x$ un element neinversabil, fie $m \\in \\mathbb{N}^{*}$, astfel încât $x = x^{2} + x^{3} + \\cdots + x^{2^{m}}$, şi $y = x + x^{2} + \\cdots + x^{2^{m} - 1}$. În mod evident, $x y = x$, deci $x y^{k} = x$, oricare ar fi $k \\in \\mathbb{N}^{*}$. Î... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Algebra > Abstract Algebra > Ring Theory"
] | null | proof only | null | |
027h | Problem:
A seguinte figura mostra um quadrado $ABCD$, e dois pontos $P$ e $Q$ sobre os lados $\overline{BC}$ e $\overline{DA}$, respectivamente.
Dobramos agora o quadrado ao longo do segmento $\overline{PQ}$, levando o vértice $B$ até o ponto médio do segmento $\overline{CD}$.
![](attached_ima... | [
"Solution:\na) Chamemos $x$ o comprimento do segmento $\\overline{PC}$. Como o lado do quadrado mede 24, então $|PB|=24-x$ e também $|PB'|=24-x$.\n\nUsando o Teorema de Pitágoras no triângulo $PCB'$, obtemos a relação\n$$\nx^2 + 12^2 = (24-x)^2\n$$\nAssim, mostramos que $|PC|=9$. Observe ta... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | PC = 9, AQ = 3, PQ = 12 sqrt(5) | |
07g9 | Find the number of ways that we can write $1, 2, \dots, n$ on the vertices of a regular $n$-gon ($n \ge 4$) such that if the diagonals connecting $(a, b)$ and $(c, d)$ cut each other then
$$
a + b \not\equiv c + d \pmod{n}.
$$
(Two arrangements of numbers matching after a rotation are considered the same.) | [
"The answer is $\\phi(n)$.\n\n**Lemma.** if $a, b, c$ are three consecutive numbers, then $a + c \\equiv 2b \\pmod{n}$.\n\n*Proof.* If $a + c \\not\\equiv 2b \\pmod{n}$, then there should be $d \\in \\{a, b, c\\}$ such that $a + c \\equiv b + d \\pmod{n}$, but $ac$ intersect with $bd$ which is a contradiction.\n\n*... | Iran | 38th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | φ(n) | |
01ny | $k$ pawns stand in some $k$ cells of an $n \times n$ table, $n \ge 2$, (exactly one pawn stands in each cell of these $k$ cells) so that each $2 \times 2$ square contains exactly $2$ pawns.
Determine all possible values of $k$. | [
"If $n = 2k$ is even, the table can be partitioned into $k^2 = n^2/4$ of $2 \\times 2$ squares, and there are exactly two pawns in each of them. So the total number of the pawns in the table is equal to $2k^2 = n^2/2$.\n\nFig. 1\n\nFig. 2\n\nFig. 3\n![](attached_ima... | Belarus | Belorusija 2012 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | If n is even, the only possible value is k = n^2 / 2. If n is odd, all integers k with n(n − 1)/2 ≤ k ≤ n(n + 1)/2 are possible. | |
0d3i | Let $A$, $B$, $C$ be colinear points in this order, $\mathcal{C}$ an arbitrary circle passing through $B$ and $C$, and $l$ an arbitrary line different from $BC$, passing through $A$ and intersecting $\mathcal{C}$ at $M$ and $N$. The bisectors of the angles $\angle C M B$ and $\angle C N B$ intersect $BC$ at $P$ and $Q$... | [
"Let $D$ be the midpoint of the arc $\\overparen{BC}$ of the circle $\\mathcal{C}$ opposite to both $M$, $N$. Because $\\overparen{CD} = \\overparen{DB}$, the bisectors of angles $\\angle C M B$ and $\\angle C N B$, both, intersect the arc $\\widehat{CB}$ at $D$.\n\n\n\nTherefore, we have\n... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Arabic | proof only | null | |
0gor | In a triangle $ABC$, $|AB| = |AC|$, $D$ is the midpoint of $[BC]$ and $E$ is the foot of the perpendicular from $D$ to the line $AC$. Let $F$ be the second point where the line $BE$ intersects the circumcircle of the triangle $ABD$. If $G$ is the intersection point of the lines $DE$ and $AF$, then show that $|DG| = |GE... | [
"Let $\\angle ACB = \\alpha$. Since $AB = AC$, we have $\\angle ABD = \\angle ABC = \\alpha$. $DE \\perp AC$ and $AD \\perp BC$ imply that $\\angle EDC = 90^\\circ - \\alpha$ and $\\angle ADE = \\alpha$. Therefore, we get $\\angle ABD = \\angle ADE = \\alpha$ which implies that $DE$ is tangent to the circuncircle o... | Turkey | 16th Junior Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gbe | 設兩直線 $BC$, $EF$ 互相平行，$D$ 為在 $BC$ 線段上且與 $B$, $C$ 相異的一點。直線 $BF$, $CE$ 交於 $I$ 點。將 $\triangle CDE$, $\triangle BDF$ 的外接圓分別記為 $K$, $L$。圓 $K$, $L$ 分別與 $EF$ 切於 $E$, $F$ 點。令 $A$ 為圓 $K$, $L$ 異於 $D$ 的另一交點。設直線 $DF$ 與圓 $K$ 再交於 $Q$ 點，直線 $DE$ 與圓 $L$ 再交於 $R$ 點。令直線 $EQ$ 與 $FR$ 交於 $M$ 點。
證明: $I$, $A$, $M$ 三點共線。 | [
"1. 因 $BC \\parallel EF$，且圓 $K$, $L$ 分別切 $EF$ 於 $E$, $F$ 點，知點 $E$, $F$ 分別為 $CDE$、$BFD$ 的中點。得 $BF = DF$，$CE = DE$。因此，\n$$\n\\angle EFD = \\angle FDB = \\angle DBF = \\angle EFI.\n$$\n同理, $\\angle FED = \\angle FEI$。由此知 $\\triangle DEF \\sim \\triangle IEF$。\n\n2. $A$, $E$, $I$, $F$ 共圓：\n$$\n\\begin{align*}\n\\angle ... | Taiwan | 二〇一七數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous ... | null | proof only | null | |
0hpd | Problem:
Aerith has written down two letters on a piece of paper. Bob will pick a positive integer and she'll try to factor Bob's positive integer into two others, such that when spelled in English, one contains her first letter and the other her second.
For example, if Aerith had chosen letters "v" and "w" and Bob ch... | [
"Solution:\n\nWe can narrow down her possible choices quickly:\n- If Bob chose \"one,\" Aerith would have to have to say \"one times one\", so her two letters must both be in the word \"one\".\n- If Bob chose \"three,\" she would have to say \"one times three\", so one of her letters must be in \"three\" and thus m... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | o and e | |
08bw | Problem:
Un'urna contiene 8 palline, sulle quali sono scritti i numeri da 1 a 8. Federica pesca due palline di seguito, cancella il numero scritto sulla prima e lo sostituisce con il suo doppio, cancella il numero sulla seconda pallina e lo sostituisce con il quadruplo di esso. Reinserisce quindi le due palline nell'u... | [
"Solution:\n\nLa risposta è (A). Supponiamo che la pallina pescata alla fine da Federica riporti il numero 8. Questo può accadere per tre motivi diversi (che sono tra loro mutuamente esclusivi):\n- la pallina era numerata 8 fin dall'inizio, e non è stata estratta nelle prime due pescate. Questo evento ha probabilit... | Italy | Gara di Febbraio | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | A | |
0i3b | Problem:
What is the radius of the smallest sphere in which 4 spheres of radius $1$ will fit? | [
"Solution:\n\nThe centers of the smaller spheres lie on a tetrahedron. Let the points of the tetrahedron be $(1,1,1)$, $(-1,-1,1)$, $(-1,1,-1)$, and $(1,-1,-1)$. These points have distance $\\sqrt{3}$ from the center, and $\\sqrt{2}$ from each other, so the radius of the smallest sphere in which $4$ spheres of radi... | United States | Harvard-MIT Math Tournament | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | (2 + sqrt(6)) / 2 | |
0kxg | Problem:
Compute all ordered triples $(x, y, z)$ of real numbers satisfying the following system of equations:
$$
\begin{array}{r}
x y+z=40 \\
x z+y=51 \\
x+y+z=19 .
\end{array}
$$ | [
"Solution:\nBy adding the first two equations, we can get\n$$\nx y+z+x z+y=(x+1)(y+z)=91\n$$\nFrom the third equation we have\n$$\n(x+1)+(y+z)=19+1=20\n$$\nso $x+1$ and $y+z$ are the two roots of $t^{2}-20 t+91=0$ by Vieta's theorem. As the quadratic equation can be decomposed into\n$$\n(t-7)(t-13)=0\n$$\nwe know t... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Linear Algebra > Determinants",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (12, 3, 4) and (6, 5.4, 7.6) | |
018i | Prove that the number of lines which go through the origo, and precisely one other point with integer coordinates $(x, y)$, $0 \le x, y \le n$ is at least $2n$, when $n$ is sufficiently large. | [
"Notice first that the number of lines going through a point $(k, \\ell)$ with $k = m$ or $\\ell = m$ for some $m$ such that the line does not go through any point $(k', \\ell')$ with $0 \\le k', \\ell' < m$ is $2\\varphi(m)$.\n\nNow, the number of lines in the problem is\n$$\n2 \\sum_{n/2 < k \\le n} \\varphi(k).\... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
06c4 | The base of a pyramid is a convex polygon with $9$ sides. Each of the diagonals of the base and each of the edges on the lateral surface of the pyramid is coloured either black or white. Both colours are used. (Note that the sides of the base are not coloured.) Prove that there are three segments coloured the same colo... | [
"Consider the $9$ edges on the lateral surface. By the pigeonhole principle, $5$ of these edges $PA_1, PA_2, PA_3, PA_4, PA_5$ are of the same colour, say black. WLOG assume $A_1, A_2, A_3, A_4, A_5$ lie in this order at the base. Since the base has $9$ sides, WLOG assume $A_1A_2$ is a diagonal. Then $A_1A_4$ and $... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0haz | Determine if there exists a pair of proper fractions that cannot be simplified so that their difference equals their product and one of denominators is $2019$. If such pair exists, find at least two of such pairs of fractions. | [
"The solution is based on a simple fact: $\\frac{1}{n} - \\frac{1}{n+1} = \\frac{1}{n(n+1)} = \\frac{1}{n} \\cdot \\frac{1}{n+1}$."
] | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | (1/2019, 1/2020) and (1/2018, 1/2019) | |
0874 | Problem:
Nell'ultimo capodanno, andavano molto di moda degli occhiali con la forma del numero "2009" e le lenti al posto dei due zeri. Per fabbricare occhiali simili, è necessario che nel numero che rappresenta l'anno vi siano due o più zeri consecutivi (per esempio 3500 va bene, 2010 no). Quanti anni compresi tra l'a... | [
"Solution:\n\nSi noti che se invece avessimo contato il numero di modi in cui si può sostituire una cifra da 1 a 9 a $X$ e una cifra da 0 a 9 a $Y$ nei due pattern $X00Y$ e $XY00$, avremmo ottenuto un risultato errato perché avremmo contato due volte gli anni della forma $X000$.\n\nIl problema può essere risolto co... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | MCQ | A | |
0ij8 | Problem:
A moth starts at vertex $A$ of a certain cube and is trying to get to vertex $B$, which is opposite $A$, in five or fewer "steps," where a step consists in traveling along an edge from one vertex to another. The moth will stop as soon as it reaches $B$. How many ways can the moth achieve its objective? | [
"Solution:\n\nLet $X$, $Y$, $Z$ be the three directions in which the moth can initially go. We can symbolize the trajectory of the moth by a sequence of $X$'s, $Y$'s, and $Z$'s in the obvious way: whenever the moth takes a step in a direction parallel or opposite to $X$, we write down $X$, and so on.\n\nThe moth ca... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 48 | |
0fto | Problem:
Finde alle natürlichen Zahlen $a$, $b$ und $n$, sodass die folgende Gleichung gilt:
$$
a! + b! = 2^{n}
$$ | [
"Solution:\nDie Gleichung ist symmetrisch in $a$ und $b$, daher können wir oBdA $a \\leq b$ annehmen.\n\nIst $a \\geq 3$, dann ist die linke Seite der Gleichung durch $3$ teilbar, die rechte Seite aber nicht, Widerspruch.\n\nIst $a=1$, dann folgt $1 + b! = 2^{n}$. Für $b \\geq 2$ ist die linke Seite ungerade und gr... | Switzerland | Vorrundenprüfung | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | (1,1,1), (2,2,2), (2,3,3), (3,2,3) | |
055j | Let $\mathbb{R}^+$ be the set of positive real numbers. Determine all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ satisfying the equation
$$
xf(x^2)f(f(y)) + f(yf(x)) = f(xy)(f(f(x^2)) + f(f(y^2))).
$$
for all $x, y \in \mathbb{R}^+$. | [
"See IMO 2016 shortlist, problem A4. (That problem was proposed by Estonia.)"
] | Estonia | IMO Team Selection Contest I | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = 1/x for all positive x | |
0gz4 | In the expression
$$
12 - 11 - 10 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1
$$
there are somehow placed brackets and the value is calculated. Find the maximum value that can be reached. Justify your answer.
*Notice.* Left bracket can be placed only before a number and right - only after. For example, expressions $-4(-3-2)$ ... | [
"$$\n12 - (11 - 10 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1) = 46\n$$\n\nConsider an arbitrary arrangement of the brackets. Numbers $12$ and $11$ always have signs '+' and '-', respectively, that's why the sum reaches the maximum value when the other terms will have '+' sign."
] | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 56 | |
0ash | Problem:
Considered to be the most prolific mathematician of all time, he published, in totality, the most number of mathematical pages in history. Undertaken by the Swiss Society of Natural Sciences, the project of publishing his collected works is still going on and will require more than 75 volumes. Who is this gre... | [
"Solution:\n\nLeonhard Euler"
] | Philippines | Philippines Mathematical Olympiad | [
"Math Word Problems"
] | null | final answer only | Leonhard Euler | |
0c98 | A positive integer $A$ has 73 digits, all different from zero. Prove that we can erase 64 digits such that the new number is divisible by 37. | [
"After the elimination of the 64 digits the final number has $73 - 64 = 9$ digits.\nIf in the composition of $A$, every digit appears at most 8 times, then $A$ would have at most $9 \\cdot 8 = 72$ digits, which is false.\nTherefore, there is a digit different from 0, let that be $a$, that appears 9 times in $A$. Th... | Romania | RMC 2020 | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0cuv | Let $ABCD$ be a convex quadrilateral. Let $I_A, I_B, I_C, \text{ and } I_D$ be the incenters of the triangles $DAB$, $ABC$, $BCD$, and $CDA$, respectively. Given that $\angle B I_A A + \angle I_C I_A I_D = 180^\circ$, prove that $\angle B I_B A + \angle I_C I_B I_D = 180^\circ$.
Пусть $ABCD$ — выпуклый четырёхугольник... | [
"Let $P \\in AD$ be the center of positive homothety mapping $\\omega_A$ to $\\omega_D$, and $R \\in BD$ be the center of negative homothety mapping $\\omega_A$ to $\\omega_C$. We consider the case shown in Fig. 16. The given equality $\\angle B I_A A + \\angle I_C I_A I_D = 180^\\circ$ is equivalent to $\\angle P ... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > 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 > Homothety",
"Geometry > Pla... | English; Russian | proof only | null | |
01m3 | Points $L$ and $H$ are marked on the sides $AB$ of an acute-angled triangle $ABC$ so that $CL$ is a bisector and $CH$ is an altitude. Let $P$, $Q$ be the feet of the perpendiculars from $L$ to $AC$ and $BC$ respectively.
Prove that $AP \cdot BH = BQ \cdot AH$. (I. Gorodnin) | [
"We have\n$$\n\\begin{align*}\n\\frac{AP}{PH} &= \\frac{AL \\cos \\angle A}{AC \\cos \\angle A} = \\frac{AL}{AC} = [\\text{since } CL \\text{ is a bisectrix}] = \\\\\n&= \\frac{BL}{BC} = \\frac{BL \\cos \\angle B}{BC \\cos \\angle B} = \\frac{BQ}{BH}.\n\\end{align*}\n$$\nIt follows that $AP \\cdot BH = BQ \\cdot AH... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0axs | Problem:
How many ordered 5-tuples $(a, b, c, d, e)$ of integers satisfy $10 < a < b < c < d < e < 20$? | [
"Solution:\n\nWe simply need to choose which five of the integers $11, 12, \\ldots, 19$ will comprise the 5-tuple. Once the integers are chosen, there is only one way to assign them as $a, b, c, d, e$ to satisfy the given inequalities. Thus, there are $\\binom{9}{5} = 126$ possible 5-tuples."
] | Philippines | Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics"
] | null | final answer only | 126 | |
07te | It is Ann's birthday. She, her brother Jim, and 8 of her friends sit around a table in the shape of a regular decagon (a 10-sided polygon). Jim sits directly opposite Ann at the table. The 10 children play a game called *throw the parcel*, which works as follows. At the beginning, Ann has the parcel. The game then cons... | [
"Ann's claim is correct for all even $n$ greater than or equal to 18.\n\nDenote by $A_n, B_n, C_n, D_n, E_n$ and $F_n$ the number of ways for the parcel to end at a person at a distance (in either the clockwise or anticlockwise direction) of 0, 1, 2, 3, 4, and 5 from Ann, respectively, where Ann is at a distance of... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | All even n with n ≥ 18 | |
0dm3 | Find the smallest positive integer $k \ge 2$ such that for all $x, y, z \ne 1$ satisfying $x + y + z = 3$ and $xyz = 1$ then the $k$-th root of numbers $|x|, |y|$ and $|z|$ are the side lengths of some triangle. | [
"Suppose that $x = \\max\\{x, y, z\\}$. We see that if $x, y, z > 0$ then based on the AM-GM inequality, we have $x + y + z \\ge 3\\sqrt[3]{xyz} = 3$ and the equality must occur, which means that $x = y = z = 1$, is not satisfied. \n\nNext, consider $y, z < 0$ and put $u = -y > 0$, $v = -z > 0$. One can rewrite the... | Saudi Arabia | Saudi Booklet | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 4 | |
043t | Given vectors $\vec{a} = (1 + 2^m, 1 - 2^m)$, $\vec{b} = (4^m - 3, 4^m + 5)$, suppose $m$ is real. Then the minimum of the dot product of $\vec{a} \cdot \vec{b}$ is ______. | [
"Let $t = 2^m$, and then $\\vec{a} = (1 + t, 1 - t)$, $\\vec{b} = (t^2 - 3, t^2 + 5)$. Thus,\n$$\n\\begin{aligned}\n\\vec{a} \\cdot \\vec{b} &= (1 + t)(t^2 - 3) + (1 - t)(t^2 + 5) \\\\\n&= 2(t - 2)^2 - 6 \\geq -6.\n\\end{aligned}\n$$\nWhen $t = 2$, namely, $m = 1$, $\\vec{a} \\cdot \\vec{b}$ takes the minimum $-6$.... | China | China Mathematical Competition | [
"Algebra > Linear Algebra > Vectors",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | -6 | |
0btl | Let $O$ be the circumcircle of a triangle $ABC$. A circle $k$ is tangent to the lines $BC$, $CA$, $AB$ at points $D, E, F$, respectively, such that $A$ is on the other side of the line $BC$ with respect to the circle. Suppose the circle $k$ is equal to the circumcircle of the triangle. Prove that lines $OD$ and $EF$ ar... | [
"Let $I_a$ be the center of the circle $k$ and let $T$ be the midpoint of the arc $BC$ - not containing $A$ - of the circumcircle $ABC$. Notice that $OT$ is the perpendicular bisector of the line segment $BC$ to deduce that $OT \\perp BC$. As $I_a D \\perp BC$ and $OT = I_a D = R$, the quadrangle $ODI_aT$ is a para... | Romania | 67th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06e3 | Let $S = \{1, 2, \dots, 100\}$. Determine the number of functions $f : S \to S$ satisfying the following conditions.
(i) $f(1) = 1$;
(ii) $f$ is bijective (i.e. for every $y$ in $S$, the equation $f(x) = y$ has exactly one solution);
(iii) $f(n) = f(g(n))f(h(n))$ for every $n$ in $S$.
Here $g(n)$ and $h(n)$ denote the ... | [
"There are 348364800 such functions.\nWe first claim that condition (iii) can be replaced by the completely multiplicative condition. This means\n$$\nf(p_1^{a_1} p_2^{a_2} \\cdots p_s^{a_s}) = f(p_1)^{a_1} f(p_2)^{a_2} \\cdots f(p_s)^{a_s}\n$$\nfor any distinct primes $p_1, p_2, \\dots, p_s$ and positive integers $... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 348364800 | |
0k62 | Problem:
For dessert, Melinda eats a spherical scoop of ice cream with diameter $2$ inches. She prefers to eat her ice cream in cube-like shapes, however. She has a special machine which, given a sphere placed in space, cuts it through the planes $x=n$, $y=n$, and $z=n$ for every integer $n$ (not necessarily positive)... | [
"Solution:\n\nNote that if we consider the division of $\\mathbb{R}^3$ into unit cubes by the given planes, we only need to compute the sum of the probabilities that the ice cream scoop intersects each cube. There are three types of cubes that can be intersected:\n\n- The cube $0 \\leq x, y, z \\leq 1$ in which the... | United States | HMMT November 2019 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 7 + 13π/3 | |
03a9 | Let $a$, $b$ and $c$ be real numbers such that $a + b + c = 0$ and $a^4 + b^4 + c^4 = 50$. Find $ab + bc + ca$. | [
"Let $a + b + c = 0$. Then $c = -a - b$.\n\nSubstitute into $a^4 + b^4 + c^4$:\n$$\na^4 + b^4 + (-a - b)^4 = 50\n$$\nExpand $(-a - b)^4$:\n$$\n(-a - b)^4 = (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\n$$\nSo:\n$$\na^4 + b^4 + [a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4] = 50\n$$\n$$\n2a^4 + 2b^4 + 4a^3b + 4ab^3 + 6a^2... | Bulgaria | Fall Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | -5 | |
03od | The interval on which the function $f(x) = \log_{\frac{1}{2}}(x^2 - 2x - 3)$ is monotone increasing is ( ).
(A) $(-\infty, -1)$
(B) $(-\infty, 1)$
(C) $(1, +\infty)$
(D) $(3, +\infty)$ | [
"First, we will find the domain of $f(x)$. From $x^2 - 2x - 3 > 0$, we obtain $x < -1$ or $x > 3$. So the domain of definition for $f(x)$ is $(-\\infty, -1) \\cup (3, +\\infty)$.\n\nBut $u = x^2 - 2x - 3 = (x-1)^2 - 4$ is monotone decreasing on $(-\\infty, -1)$, and monotone increasing on $(3, +\\infty)$.\n\nSo $f(... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | MCQ | A | |
0ayy | Problem:
Let $\triangle ABC$ be a right triangle with legs $AB = 6$ and $BC = 8$. Let $R$ and $r$ be the circumradius and the inradius of $\triangle ABC$, respectively. Find the sum of $R$ and $r$. | [
"Solution:\nSince $\\triangle ABC$ is a right triangle, its circumradius is just half of its hypotenuse. By Pythagorean theorem, $AC = \\sqrt{6^{2} + 8^{2}} = 10$ and so $R = \\frac{AC}{2} = 10 / 2 = 5$.\n\nNow, we can compute the inradius by using the following identity:\n$$\n\\left(r \\times \\frac{AB + BC + AC}{... | Philippines | 20th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 7 | |
024y | Problem:
Papai Noel chegou à casa de Arnaldo e Bernaldo carregando dez brinquedos distintos e enumerados de 1 a 10 e disse a eles: "o brinquedo número 1 é para você, Arnaldo e o brinquedo número 2 é para você, Bernaldo. Mas esse ano, vocês podem escolher ficar com mais brinquedos contanto que deixem ao menos um para m... | [
"Solution:\n\nPara cada um dos 8 brinquedos, do número 3 ao número 10, devemos decidir se ele vai pertencer a Arnaldo, a Bernaldo ou deve ser deixado para Papai Noel. Se multiplicarmos então\n$$\n\\underbrace{3 \\times 3 \\times \\cdots \\times 3}_{8 \\text{ vezes }}\n$$\ncontaremos as formas de dividir os brinqued... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 6305 | |
03eq | The first, seventh, and seventeenth terms of an arithmetic progression are distinct and consecutive terms of a geometric progression. To find the difference of the arithmetic progression if its first term is a solution of the equation
$$
x^2 - 9x + x\sqrt{12-x} - 9\sqrt{12-x} = 0.
$$ | [
"Let $a_1$ and $d$ be the first term and the difference of the arithmetic progression, respectively. From the condition $a_1$, $a_1 + 6d$ and $a_1 + 16d$ are consecutive members of a geometric progression, i.e.\n$$\n(a_1 + 6d)^2 = a_1 \\cdot (a_1 + 16d) \\iff d \\cdot (a_1 - 9d) = 0.\n$$\nSince $d \\ne 0$, we get t... | Bulgaria | Bulgarian Winter Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | d = 1 or d = -4/9 | |
0jm9 | Problem:
Find the largest real number $c$ such that
$$
\sum_{i=1}^{101} x_{i}^{2} \geq c M^{2}
$$
whenever $x_{1}, \ldots, x_{101}$ are real numbers such that $x_{1}+\cdots+x_{101}=0$ and $M$ is the median of $x_{1}, \ldots, x_{101}$. | [
"Solution:\nAnswer: $\\frac{5151}{50}$ OR 103.02 OR $103 \\frac{1}{50}$\n\nSuppose without loss of generality that $x_{1} \\leq \\cdots \\leq x_{101}$ and $M = x_{51} \\geq 0$.\n\nNote that $f(t) = t^{2}$ is a convex function over the reals, so we may \"smooth\" to the case $x_{1} = \\cdots = x_{50} \\leq x_{51} = ... | United States | HMMT 2014 | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 5151/50 | |
0ijl | Problem:
Suppose that $n$ squares of an infinite square grid are colored grey, and the rest are colored white. At each step, a new grid of squares is obtained based on the previous one, as follows. For each location in the grid, examine that square, the square immediately above, and the square immediately to the right... | [
"Solution:\n\n(Sketch) Use strong induction. Consider the smallest rectangle $R$ that contains all the black squares. Suppose this rectangle contains $r$ squares. Assume any rectangle that contains $k < r$ squares will convert to all white squares after $k$ steps. Since $R$ is the smallest rectangle, its left-most ... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0fvw | Problem:
Sei $M \subset \{1,2,3, \ldots, 2007\}$ eine Menge mit folgender Eigenschaft: Unter je drei Zahlen aus $M$ kann man stets zwei auswählen, sodass die eine durch die andere teilbar ist. Wieviele Zahlen kann $M$ höchstens enthalten? | [
"Solution:\n\nZerlege die Menge $\\{1,2,3, \\ldots, 2007\\}$ wie folgt in 11 Teilmengen:\n$$\n\\{1\\} \\cup\\{2,3\\} \\cup\\{4,5,6,7\\} \\cup \\ldots \\cup\\{256, \\ldots, 511\\} \\cup\\{512, \\ldots, 1023\\} \\cup\\{1024, \\ldots, 2007\\}\n$$\nAus jeder dieser Teilmengen kann $M$ höchstens zwei Zahlen enthalten, d... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 21 | |
0hjm | Problem:
Ten different points are marked on a circle. Two players $A$ and $B$ play the following game. $A$ moves first and the players alternate their moves. In each of the moves a player connects two of the points with a straight line segment. A player whose segment crosses a segment previously drawn will lose the ga... | [
"Solution:\n\nNotice that a player can draw a line segment if and only if the $10$-gon is not partitioned into triangles. Since there is a total of $17$ segments for any partition of $10$-gon into triangles, the first player will win the game no matter how he plays."
] | United States | Berkeley Math Circle Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | null | proof and answer | Player A. Any maximal noncrossing drawing on ten points has exactly seventeen segments (a triangulation), so there are seventeen safe moves. Since A makes all odd-numbered moves, A makes the seventeenth safe move and leaves B with only crossing moves, so A wins. | |
0j46 | Problem:
Find the number of positive integers $x$ less than 100 for which
$$
3^{x}+5^{x}+7^{x}+11^{x}+13^{x}+17^{x}+19^{x}
$$
is prime. | [
"Solution:\nAnswer: 0\nWe claim that our integer is divisible by 3 for all positive integers $x$. Indeed, we have\n$$\n\\begin{aligned}\n3^{x}+5^{x}+7^{x}+11^{x}+13^{x}+17^{x}+19^{x} & \\equiv (0)^{x}+(-1)^{x}+(1)^{x}+(-1)^{x}+(1)^{x}+(-1)^{x}+(1)^{x} \\\\\n& \\equiv 3\\left[(1)^{x}+(-1)^{x}\\right] \\\\\n& \\equiv... | United States | Harvard-MIT November Tournament | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 0 | |
0fcv | Problem:
Consideremos los polinomios $P(x)=x^{3}+a x^{2}+b x+c, Q(x)=x^{3}+A x^{2}+B x+C$ ($x$ es la variable, $a, b, c, A, B, C$ son parámetros). Sabemos que las tres raíces de $P$ son positivas y que las raíces de $Q$ son los números inversos de las raíces de $P$. Probad que $a \cdot A>9, b \cdot B>9$. | [
"Solution:\nDenotamos por $x_{1}, x_{2}, x_{3}$ las raíces de $P$. Las raíces de $Q$ son $\\frac{1}{x_{1}}, \\frac{1}{x_{2}}, \\frac{1}{x_{3}}$. Utilizamos las fórmulas de Cardano para conseguir\n$$\n\\begin{gathered}\na=-\\left(x_{1}+x_{2}+x_{3}\\right) ; \\quad A=-\\left(\\frac{1}{x_{1}}+\\frac{1}{x_{2}}+\\frac{1... | Spain | null | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.