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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0btq | Let $a$, $b$, $c$ be positive numbers with $abc \ge 1$. Prove that
$$
\frac{1}{a^3 + 2b^3 + 6} + \frac{1}{b^3 + 2c^3 + 6} + \frac{1}{c^3 + 2a^3 + 6} \le \frac{1}{3}.
$$ | [
"Notice that $a^3 + b^3 + 1 \\ge 3ab$ and $b^3 + 1 + 1 \\ge 3b$ to obtain that $\\frac{1}{a^3+2b^3+6} \\le \\frac{1}{3ab+3b+3}$, hence it is sufficient to show that $\\frac{1}{ab+b+1} + \\frac{1}{bc+c+1} + \\frac{1}{ca+a+1} \\le 1$.\n\nTo this end, observe that $\\frac{1}{ab+b+1} + \\frac{1}{bc+c+1} + \\frac{1}{ca+... | Romania | 67th NMO Selection Tests for JBMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
07ca | Ali is given a piece of paper shaped like an equilateral triangle. He cuts the paper into two pieces with one cut. Then, he puts the pieces arbitrarily on a table and cuts them both with one cut to get four pieces of paper. Lastly, he puts all four pieces on the table and cuts them all with one cut to get eight pieces.... | [
"a) Note that every polygon produced during this process is convex. That said, the idea here is to find an upper bound for the sum of angles of them. After cutting a polygon $P$ into two polygons $P_1$ and $P_2$, three possibilities arise: $(\\sigma(Q)$ denotes the sum of angles of polygon $Q)$\n* The cutting line ... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Yes | |
0fcn | Problem:
Determina los dos valores de $x$ más próximos (por defecto y por exceso) a $2003^{\circ}$ que cumplen la siguiente ecuación trigonométrica:
$$
\frac{1}{\operatorname{sen}^{2} x}-\frac{1}{\cos ^{2} x}-\frac{1}{\operatorname{tg}^{2} x}-\frac{1}{\operatorname{cotg}^{2} x}-\frac{1}{\sec ^{2} x}-\frac{1}{\operatorn... | [
"Solution:\nLa expresión se puede escribir así\n$$\n\\begin{gathered}\n\\operatorname{cosec}^{2} x-\\sec ^{2} x-\\cot ^{2} x-\\operatorname{tg}^{2} x-\\cos ^{2} x-\\operatorname{sen}^{2} x=-3 \\\\\n\\left(1+\\cot ^{2} x\\right)-\\left(1+\\operatorname{tg}^{2} x\\right)-\\cot ^{2} x-\\operatorname{tg}^{2} x-1=-3\n\\... | Spain | Fase Local | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 1935° and 2025° | |
04p8 | 299 zeros and one one are written circular. The following moves are allowed:
* You can select all numbers simultaneously and subtract both of its neighbouring numbers from each number.
* You can select two numbers such that there are exactly two numbers between them, and either increase both selected numbers by 1 or de... | [
"Let us analyse how the moves affect certain sums. Denote by $a_k$ the numbers written at a certain moment, so that $a_1$ is the number written in the place where the one was written before any move was made, and $a_2, \\dots, a_{300}$ are the numbers written clockwise from $a_1$. After applying the first type of m... | Croatia | Croatian Mathematical Society Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) No; b) No | |
05r9 | Problem:
Trouver toutes les fonctions $f: \mathbb{R} \rightarrow \mathbb{R}$ telles qu'il existe un réel $A$ vérifiant $A>f(x)$ pour tout $x \in \mathbb{R}$, et telles que pour tous réels $x, y$, on ait :
$$
f(x f(y)) + y f(x) = x f(y) + f(x y)
$$ | [
"Solution:\n\nLa fonction nulle est clairement solution. Supposons désormais que $f$ n'est pas nulle sur tout $\\mathbb{R}$.\n\nEn posant $x=0$, on a $y f(0) = f(0)$ pour tout réel $y$ donc $f(0) = 0$.\n\nEn posant $y=1$, on a $f(x f(1)) = x f(1)$ pour tout réel $x$, donc si $f(1) \\neq 0$, on contredit l'énoncé en... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 5 : Pot-POURRI | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | Two functions: (1) f(x) = 0 for all x; (2) f(x) = 2x for x < 0 and f(x) = 0 for x ≥ 0. | |
0c0m | Prove that a positive integer $A$ is a perfect square if and only if, for all positive integers $n$, at least one of the numbers
$$
(A+1)^2 - A, (A+2)^2 - A, (A+3)^2 - A, \dots, (A+n)^2 - A
$$
is a multiple of $n$. | [
"If $A$ is a perfect square, i.e. there exists $B \\in \\mathbb{N}$ such that $A = B^2$, then $(A+k)^2 - A = (B^2+k)^2 - B^2 = (B^2+B+k)(B^2-B+k)$ for all $k = 1, n$, and exactly one of the (consecutive) numbers $B^2+B+1, B^2+B+2, \\dots, B^2+B+n$ is a multiple of $n$.\n\nConversely, if $A$ is not a perfect square,... | Romania | 69th NMO Selection Tests for JBMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
09tz | Every day, Maurits bikes to school. He can choose between two different routes. Route $B$ is $1.5$ km longer than route $A$. However, because he encounters fewer traffic lights, his average speed along route $B$ is $2$ km/h higher than along route $A$. This makes that travelling along the two routes takes exactly the s... | [
"45 minutes"
] | Netherlands | First Round, January 2019 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 45 minutes | |
06ll | Let $a$, $b$ and $c$ be positive real numbers satisfying $abc = 1$. Prove that
$$
\frac{1}{a^3 + 2b^2 + 2b + 4} + \frac{1}{b^3 + 2c^2 + 2c + 4} + \frac{1}{c^3 + 2a^2 + 2a + 4} \le \frac{1}{3}.
$$ | [
"By the AM-GM inequality, we have\n$$a^3 + b^2 + b \\ge 3\\sqrt[3]{a^3b^3} = 3ab,$$\n$$b^2 + b + 1 \\ge 3\\sqrt[3]{b^3} = 3b.$$ \nThis gives $a^3 + 2b^2 + 2b + 4 \\ge 3ab + 3b + 3 = 3(ab + b + 1)$. Since $abc = 1$, we can let $a = \\frac{x}{y}$, $b = \\frac{y}{z}$, $c = \\frac{z}{x}$ for some positive real numbers ... | Hong Kong | CHKMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
02c4 | Problem:
Os doze pontos - Doze pontos estão marcados numa folha de papel quadriculada, conforme mostra a figura. Qual o número máximo de quadrados que podem ser formados unindo quatro desses pontos?
 | [
"Solution:\n\nOs doze pontos - No total, temos 11 possíveis quadrados como mostrado a seguir.\n\n"
] | Brazil | Lista 4 | [
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | 11 | |
0jm8 | Problem:
Let $\mathcal{P}$ be a parabola with focus $F$ and directrix $\ell$. A line through $F$ intersects $\mathcal{P}$ at two points $A$ and $B$. Let $D$ and $C$ be the feet of the altitudes from $A$ and $B$ onto $\ell$, respectively. Given that $A B=20$ and $C D=14$, compute the area of $A B C D$. | [
"Solution:\nObserve that $A D + B C = A F + F B = 20$, and that $A B C D$ is a trapezoid with height $B C = 14$. Hence the answer is $\\frac{1}{2}(A D + B C)(14) = 140$."
] | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 140 | |
0e12 | In the triangle $ABC$ let $D$ be the foot of the altitude to the side $AB$. Given the points $E$ and $F$ on the sides $AD$ and $BC$, such that $\angle BAF = \angle ACE$, let the segments $AF$ and $CE$ intersect at $G$ and let the segments $AF$ and $CD$ intersect at $T$. Find the angles of the triangle $ABC$, given that... | [
"Denote $\\angle BAF = \\angle ACE = \\varphi$. Since $CGF$ is an equilateral triangle, we have $\\angle CGA = 120^\\circ$. Thus, $\\angle GAC = 180^\\circ - \\angle CGA - \\angle ACG = 60^\\circ - \\varphi$ and $\\angle BAC = \\angle BAF + \\angle GAC = \\varphi + 60^\\circ - \\varphi = 60^\\circ$.\n\nSince the tr... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | ∠A = 60°, ∠B = 45°, ∠C = 75° | |
0bn7 | Show that the edges of a connected finite simple graph can be oriented so that the number of edges leaving each vertex is even if and only if the total number of edges is even. | [
"Given any orientation, the total number of edges equals the sum of all out-degrees. If the latter are all even, then so is the former.\n\nTo establish the converse, induct on the number of edges to show that any connected simple finite graph with an even number of edges splits into edge-disjoint paths of length $2... | Romania | 2015 Danube Mathematical Competition | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0669 | Let $ABCD$ be a convex quadrilateral. Let $E$ and $F$ be points on the sides $AB$ and $CD$, respectively, such that $AB : AE = CD : DF = n$. If $S$ is the area of the quadrilateral $AEFD$, show that
$$
S \le \frac{AB \cdot CD + n(n-1)DA^2 + nDA \cdot BC}{2n^2}.
$$ | [] | Greece | 15th Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof only | null | |
0ipk | Problem:
Solve the following system of equations for $w$.
$$
\begin{aligned}
& 2w + x + y + z = 1 \\
& w + 2x + y + z = 2 \\
& w + x + 2y + z = 2 \\
& w + x + y + 2z = 1
\end{aligned}
$$ | [
"Solution:\nAdd all the equations together to find that $5x + 5y + 5z + 5w = 6$, or $x + y + z + w = \\frac{6}{5}$. We can now subtract this equation from the first equation to see that $w = \\frac{-1}{5}$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | -1/5 | |
0jnn | Problem:
Let $P(x) = x^{3} + a x^{2} + b x + 2015$ be a polynomial all of whose roots are integers. Given that $P(x) \geq 0$ for all $x \geq 0$, find the sum of all possible values of $P(-1)$. | [
"Solution:\nSince all the roots of $P(x)$ are integers, we can factor it as $P(x) = (x - r)(x - s)(x - t)$ for integers $r, s, t$. By Vieta's formula, the product of the roots is $r s t = -2015$, so we need three integers to multiply to $-2015$.\n\n$P(x)$ cannot have two distinct positive roots $u, v$ since otherwi... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 9496 | |
08lc | Problem:
Every cell of table $4 \times 4$ is colored into white. It is permitted to place the cross (pictured below) on the table such that its center lies on the table (the whole figure does not need to lie on the table) and change colors of every cell which is covered into opposite (white and black). Find all $n$ suc... | [
"Solution:\nThe cross covers at most five cells so we need at least 4 steps to change the color of every cell. If we place the cross 4 times such that its center lies in the cells marked below, we see that we can turn the whole square black in $n=4$ moves.\n\n\n\nFurthermore, applying the s... | JBMO | 2008 Shortlist JBMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | all even numbers at least 4 | |
0atr | Problem:
Let $P(x) = 1 + 8x + 4x^{2} + 8x^{3} + 4x^{4} + \cdots$ for values of $x$ for which this sum has finite value. Find $P(1/7)$. | [
"Solution:\nLet us write $P(x)$ as an infinite series:\n\n$$\nP(x) = 1 + 8x + 4x^2 + 8x^3 + 4x^4 + 8x^5 + 4x^6 + \\cdots\n$$\n\nNotice the coefficients alternate between $8$ and $4$ starting from $8x$.\n\nLet us group the terms:\n\n$P(x) = 1 + (8x + 4x^2) + (8x^3 + 4x^4) + (8x^5 + 4x^6) + \\cdots$\n\nLet us factor ... | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 9/4 | |
09u5 | Starting with a positive integer, a *fragment* of that number is any positive number obtained by removing one or more digits from the beginning and/or end of that number. For example: the numbers $2$, $1$, $9$, $20$, $19$, and $201$ are the fragments of $2019$.
What is the smallest positive integer $n$ such that the fo... | [
"$1836$"
] | Netherlands | First Round, January 2019 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | 1836 | |
04ez | Players $A$, $B$ and $C$ roll the die in consecutive order. What is the probability that $C$ rolls a greater number than the other two players? | [] | Croatia | Mathematica competitions in Croatia | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 55/216 | |
063c | Problem:
Es sei $\mathbb{Z}^{+}$ die Menge der positiven ganzen Zahlen.
Man bestimme alle Funktionen $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$ mit der Eigenschaft, dass für alle positiven ganzen Zahlen $m$ und $n$ gilt: $m^{2}+f(n) \mid m f(m)+n$. | [
"Solution:\n\nFür eine beliebige positive ganze Zahl $n$ wählen wir $m$ so, dass $f(n) \\mid m$ gilt. Dann folgt $f(n) \\mid m^{2}+f(n)$ und $f(n) \\mid m f(m)+n$. Weil $m$ durch $f(n)$ teilbar ist, muss auch $n$ durch $f(n)$ teilbar sein. Es gilt also $f(n) \\mid n$, das bedeutet $f(n) \\leq n$ für alle $n \\in \\... | Germany | Germany TST | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | f(n) = n for all positive integers n | |
04at | Each side and each diagonal of a convex $n$-gon is colored in one of $k$ colors. It is known that there doesn't exist a closed monochrome broken line whose vertices are also the vertices of the given $n$-gon. Determine the largest possible value of $n$. (Russia 1990) | [
"It is known that a graph with $n$ vertices that contains no cycles has at most $n-1$ edges. That fact is easily proven by induction by the number of vertices $n$. Hence, at most $n-1$ edges (i.e. sides and diagonals) can be colored with the same color.\n\nSince the number of colors is $k$ and the total number of e... | Croatia | CroatianCompetitions2011 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 2k | |
0ac0 | From a point inside the equilateral triangle $ABC$ are drawn perpendicular lines to the sides $AB$, $BC$ and $CA$. The lengths of these lines are $m$, $n$, $p$ and the length of the side of the triangle $ABC$ is equal to $a$. Find the ratio of the areas of the triangle $ABC$ and the triangle formed with the points of t... | [
"Let $P$ be arbitrary point inside the triangle $ABC$. The points $K$, $L$ and $M$ are points of intersection of the perpendicular lines passing through the point $P$ to the sides $AB$, $BC$ and $CA$ respectively. Since\n$$\n\\angle PKB = \\angle PLB = \\angle PLC = \\angle PMC = \\angle PMA = \\angle AKP = 90^\\ci... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | a^2 / (mp + mn + np) | |
0fca | Problem:
Halla todas las ternas $(x, y, z)$ de números reales que son solución de la ecuación
$$
\sqrt{3^{x}\left(5^{y}+7^{z}\right)}+\sqrt{5^{y}\left(7^{z}+3^{x}\right)}+\sqrt{7^{z}\left(3^{x}+5^{y}\right)}=\sqrt{2}\left(3^{x}+5^{y}+7^{z}\right)
$$ | [
"Solution:\n\nPoniendo $a=3^{x}$, $b=5^{y}$ y $c=7^{z}$ la ecuación anterior se convierte en:\n$$\n\\sqrt{a(b+c)}+\\sqrt{b(c+a)}+\\sqrt{c(a+b)}=\\sqrt{2}(a+b+c)\n$$\nAplicando la desigualdad entre las medias aritmética y geométrica resulta:\n$$\n\\begin{aligned}\n& \\sqrt{a(b+c)} \\leq \\sqrt{2}\\left(\\frac{a}{2}+... | Spain | null | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | All solutions are x = t / ln 3, y = t / ln 5, z = t / ln 7 for any real t. | |
0dik | Let $n > 3$ be an integer. Suppose that $n$ children are arranged in a circle, and $n$ coins are distributed between them (some children may have no coins). At every step, a child with at least 2 coins may give 1 coin to each of their neighbours on the right and left. Determine all initial distributions of coins from w... | [
"The answer is: all distributions where $\\sum_{i=1}^n i c_i = \\frac{n(n+1)}{2} \\pmod n$, where $c_i$ denotes the number of coins the $i$-th child starts with.\n\nEncode the sequence $c_i$ as polynomial $p(x) = \\sum_i a_i x_i$. The cyclic nature of the problem makes it natural to work modulo $x^n - 1$. Child $i$... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | All initial distributions with sum of i times c_i congruent to n(n+1)/2 modulo n (with the total number of coins equal to n). | |
0463 | Given two fixed points $A$ and $B$ on the unit circle $\omega$, satisfying $\sqrt{2} < AB < 2$. Let $P$ be a moving point on $\omega$ such that $\triangle ABP$ is an acute-angled triangle and $AP > AB > BP$.
For this moving point $P$, let $H$ be the orthocenter of $\triangle ABP$. Take a point $S$ on the arc $\widehat... | [
"We prove that the midpoint $M$ of $AB$ satisfies the given condition.\n\n\n\nLet $P_1$ be the antipodal point of $P$ on the circle $\\omega$, and let $H_1$ be the intersection of the extension of $AH$ with $\\omega$. We will show that $QP_1 = QH_1$.\n\nDenote $O$ as the center of $\\omega$... | China | China-TST-2023B | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | English | proof and answer | the midpoint of the chord joining the two fixed points | |
0l2r | In the following expression, Melanie changed some of the plus signs to minus signs:
$$
1 + 3 + 5 + 7 + \dots + 97 + 99.
$$
When the new expression was evaluated, it was negative. What is the least number of plus signs that Melanie could have changed to minus signs?
(A) 14 (B) 15 (C) 16 (D) 17 (E) 18 | [
"**Answer (B):** To minimize the number of minus signs needed to make the expression negative, minus signs should be chosen for all of the largest numbers. Hence the first $k$ numbers of the expression will stay positive and the last $50-k$ will be made negative for the greatest value of $k$ that gives a negative v... | United States | 2024 AMC 12 B | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | B | |
01sq | Determine all positive integers $M$ for which the sequence $a_0, a_1, a_2, \ldots$, defined by $a_0 = 2M + \frac{1}{2}$ and $a_{k+1} = a_k[a_k]$ for $k = 0, 1, 2, \ldots$, contains at least one integer term. | [] | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | All positive integers M | |
02pn | Problem:
Sejam $a$ e $b$ números reais tais que existam números reais distintos $m$, $n$ e $p$, satisfazendo as igualdades abaixo:
$$
\left\{\begin{array}{l}
m^{3}+a m+b=0 \\
n^{3}+a n+b=0 \\
p^{3}+a p+b=0
\end{array}\right.
$$
Mostre que $m+n+p=0$. | [
"Solution:\n\nSubtraindo a segunda equação da primeira, obtemos\n$$\n\\begin{gathered}\nm^{3}-n^{3}+a m-a n=0 \\Longleftrightarrow \\\\\n(m-n)\\left(m^{2}+m n+n^{2}\\right)+a(m-n)=0 \\Longleftrightarrow \\\\\n(m-n)\\left(m^{2}+m n+n^{2}+a\\right)=0\n\\end{gathered}\n$$\ne como $m-n \\neq 0$, temos que $m^{2}+m n+n^... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0dgw | Let $n > 3$, $x_1, x_2, \dots, x_n > 0$ and $x_1x_2 \dots x_n = 1$. Prove that
$$
\frac{1}{1+x_1+x_1x_2} + \frac{1}{1+x_2+x_2x_3} + \dots + \frac{1}{1+x_n+x_nx_1} > 1.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
03fn | The reals $x, y$ satisfy $x(x - 6) \le y(4 - y) + 7$. Find the minimal and maximal values of the expression $x + 2y$. | [
"Let $a = x + 2y$. Then $x = a - 2y$ and\n$$\n\\begin{aligned}\n(a - 2y)(a - 2y - 6) &\\le y(4 - y) + 7 \\\\\na^2 - 2ay - 6a - 2ay + 4y^2 + 12y &\\le 4y - y^2 + 7 \\\\\n5y^2 - 2(2a - 4)y + (a^2 - 6a - 7) &\\le 0 \\\\\nD &= (2a - 4)^2 - 5(a^2 - 6a - 7) \\ge 0 \\\\\n4a^2 - 16a + 16 - 5a^2 + 30a + 35 &\\ge 0 \\\\\na^2... | Bulgaria | Bulgarian Spring Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | min = -3, max = 17 | |
07bo | Let $x$, $y$ and $z$ be nonnegative real numbers. Knowing that
$$
2(xy + yz + zx) = x^2 + y^2 + z^2,
$$
prove
$$
\frac{x + y + z}{3} \geq \sqrt[3]{2xyz}.
$$ | [
"There is no loss of generality in assuming that $x \\ge y \\ge z$. We have\n$$\n\\begin{aligned}\nx^2 + y^2 + z^2 &= 2(xy + yz + zx) \\\\\n\\Rightarrow \\quad x^2 + x(-2y - 2z) + y^2 + z^2 - 2yz &= 0 \\\\\n\\Rightarrow \\quad x = (y + z) \\pm \\sqrt{(y + z)^2 - y^2 - z^2 + 2yz} = (y + z) \\pm 2\\sqrt{yz} = (\\sqrt... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
00tb | Let $K$ and $N > K$ be fixed positive integers. Let $n$ be a positive integer and let $a_1, a_2, \dots, a_n$ be distinct integers. Suppose that whenever $m_1, m_2, \dots, m_n$ are integers, not all equal to $0$, such that $|m_i| \le K$ for each $i$, then the sum
$$
\sum_{i=1}^{n} m_i a_i
$$
is not divisible by $N$. Wha... | [
"The answer is $n = \\lceil \\log_{K+1} N \\rceil$.\n\nNote first that for $n \\le \\lceil \\log_{K+1} N \\rceil$, taking $a_i = (K+1)^{i-1}$ works. Indeed let $r$ be maximal such that $m_r \\ne 0$. Then on the one hand we have\n$$\n\\left| \\sum_{i=1}^{n} m_i a_i \\right| \\le \\sum_{i=1}^{n} K (K+1)^{i-1} = (K+1)... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | ceil(log_{K+1} N) | |
01b1 | Positive numbers $a$, $b$, $c$ satisfy $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 3$. Prove the inequality
$$
\frac{1}{\sqrt{a^3 + b}} + \frac{1}{\sqrt{b^3 + c}} + \frac{1}{\sqrt{c^3 + a}} \le \frac{3}{\sqrt{2}}
$$ | [
"Apply a lot of AM-GM inequalities:\n$$\n\\begin{align*}\n\\frac{1}{\\sqrt{a^3+b}} + \\frac{1}{\\sqrt{b^3+c}} + \\frac{1}{\\sqrt{c^3+a}} &\\le \\frac{1}{\\sqrt{2a\\sqrt{ab}}} + \\frac{1}{\\sqrt{2b\\sqrt{bc}}} + \\frac{1}{\\sqrt{2c\\sqrt{ca}}} \\\\\n&= \\frac{1}{\\sqrt{2}} \\left( \\frac{\\sqrt{a\\sqrt{ab}}}{a\\sqrt... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0gxn | In triangle $ABC$ points $M$ and $N$ are the midpoints of the sides $BC$ and $AC$ respectively. Inside $\triangle ABC$ a point $P$ is taken such that $\angle BAP = \angle PBC = \angle PCA$. It is known that $\angle PNA = \angle AMB$. Prove that $ABC$ is an isosceles triangle. | [
"Let us draw the line $l \\parallel BC$ through the point $A$ and denote $W = BP \\cap l$. Then\n\n$\\angle BPC = 180^\\circ - (\\angle PBC + \\angle PCB) =$\n$180^\\circ - (\\angle PCA + \\angle PCB) = 180^\\circ - \\angle BCA \\Rightarrow$\n$\\angle CPW = \\angle CAW$. And so points $A, P, C, W$ are cyclic. Thus ... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof only | null | |
0hde | For arbitrary integer $a, b, c$, we define a function $f(x) = a x^2 + b x + c$. Prove that the expression
$$
f(2020) + f(2019) + \dots + f(1011) - f(1010) - f(1009) - \dots - f(1)
$$
is divisible by $2020$. | [
"Let us denote the expressions as follows:\n$$\n\\begin{aligned}\nS &= f(2020) + f(2019) + \\dots + f(1011) - f(1010) - f(1009) - \\dots - f(1) \\\\\n&= a(2020^2 + 2019^2 + \\dots + 1011^2 - 1010^2 - 1009^2 - \\dots - 1^2) \\\\\n&\\quad + b(2020 + 2019 + \\dots + 1011 - 1010 - 1009 - \\dots - 1) \\\\\n&\\quad + c(1... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
03ol | If real numbers $x$ and $y$ satisfy $(x+5)^2 + (y-12)^2 = 14^2$, then the minimum value of $x^2 + y^2$ is ( ).
(A) 2
(B) 1
(C) $\sqrt{3}$
(D) $\sqrt{2}$ | [
"Let $x+5 = 14\\cos\\theta$ and $y-12 = 14\\sin\\theta$, for $\\theta \\in [0, 2\\pi)$.\nHence\n$$\n\\begin{align*}\nx^2 + y^2 &= (14\\cos\\theta - 5)^2 + (14\\sin\\theta + 12)^2 \\\\\n&= 14^2 + 5^2 + 12^2 - 140\\cos\\theta + 336\\sin\\theta \\\\\n&= 365 + 28(12\\sin\\theta - 5\\cos\\theta)\n\\end{align*}\n$$\n$$\n... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | MCQ | B | |
026o | Problem:
O percurso de um atleta - Um atleta resolveu fazer uma corrida de $15~\mathrm{km}$. Começou correndo $5~\mathrm{km}$ na direção Sul, depois virou para direção Leste, correndo mais $5~\mathrm{km}$ e, novamente, virou para a direção Norte, correndo os $5~\mathrm{km}$ restantes. Após esse percurso, constatou, pa... | [
"Solution:\n\nO Polo Norte da Terra é o ponto mais fácil de ser identificado como solução: Saindo o atleta do Polo Norte, correndo $5~\\mathrm{km}$ para o sul, depois $5~\\mathrm{km}$ para o leste e finalmente $5~\\mathrm{km}$ para o norte, ele volta novamente para o Polo Norte.\n\n\n\nVamo... | Brazil | Nível 3 | [
"Geometry > Non-Euclidean Geometry > Spherical Geometry"
] | null | proof and answer | The North Pole; and any point on the parallel that is five kilometers north of a parallel whose circumference is five kilometers near the South Pole, so that the five-kilometer eastward leg loops around back to the same point. | |
0d7l | Let $ABC$ be a non-isosceles triangle with circumcircle $(O)$ and incircle $(I)$. Denote $(O_{1})$ as the circle that is internally tangent to $(O)$ at $A_{1}$ and also tangent to segments $AB$, $AC$ at $A_{b}$, $A_{c}$ respectively. Define the circles $(O_{2})$, $(O_{3})$ and the points $B_{1}$, $C_{1}$, $B_{c}$, $B_{... | [
"1) We use inversion to solve this problem.\n\nSuppose that $AI$, $A_{1}I$ intersect $(O)$ at $A_{0}$, $A_{1}$. Because $AI$ is the bisector then $A_{0}$ is the midpoint of the minor $\\operatorname{arc} BC$. Based on the property of the Mixtilinear circle, we also have $A_{1}$ as the midpo... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0b4c | Problem:
Let $a, b, c$ be real numbers with $1 < a < b < c$ that satisfy the equations
$$
\begin{gathered}
\log_{a} b + \log_{b} c + \log_{c} a = 6.5 \\
\log_{b} a + \log_{c} b + \log_{a} c = 5
\end{gathered}
$$
Then $\max \left\{ \log_{a} b, \log_{b} c, \log_{c} a \right\}$ can be written in the form $\sqrt{x} + \sqrt... | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | 16 | |
08nx | Problem:
Let $AB$ be a diameter of a circle $\omega$ with center $O$ and $OC$ be a radius of $\omega$ which is perpendicular to $AB$. Let $M$ be a point on the line segment $OC$. Let $N$ be the second point of intersection of the line $AM$ with $\omega$, and let $P$ be the point of intersection of the lines tangent to ... | [
"Solution:\nSince the lines $PN$ and $BP$ are tangent to $\\omega$, $NP = PB$ and $OP$ is the bisector of $\\angle NOB$. Therefore the lines $OP$ and $NB$ are perpendicular. Since $\\angle ANB = 90^\\circ$, it follows that the lines $AN$ and $OP$ are parallel. As $MO$ and $PB$ are also parallel and $AO = OB$, the t... | JBMO | 17th Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01qt | Do there exist functions $f$ and $g$, $f: \mathbb{R} \to \mathbb{R}$, $g: \mathbb{R} \to \mathbb{R}$, such that $f(x + f(y)) = y^2 + g(x)$ for all real $x$ and $y$? | [
"Answer: there are no such functions.\n\nSuppose that there exist functions $f, g$ satisfying the equality\n$$\nf(x + f(y)) = y^2 + g(x). \\quad (*)\n$$\nFirst, suppose that $f(y) = f(z)$ for some $y, z \\in \\mathbb{R}$. Then $z^2 + g(x) = f(x + f(z)) = f(x + f(y)) = y^2 + g(x)$, whence\n$$\nz^2 = y^2. \\quad (1)\... | Belarus | Selection and Training Session | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | No, there are no such functions. | |
07ml | The maximum of two real numbers $a$ and $b$ is defined as follows:
$$
\max\{a, b\} = \begin{cases} a, & \text{if } a \ge b, \\ b, & \text{otherwise.} \end{cases}
$$
For any two positive real numbers $x_0 > 0$, $x_1 > 0$ a sequence of real numbers $x_n$ is defined recursively as given below. Find $x_{2010}$.
$$
x_{n+1} ... | [
"The recurrence is a version of what is sometimes called the *Lyness* max equation. All solutions are periodic with period 5 which can be established by computation. Hence $x_{2010} = x_0$.\nThe change of variable $x_n = 4y_n$ puts the recurrence in standard form\n$$\ny_{n+1} = \\frac{\\max\\{y_n, 1\\}}{y_{n-1}}\n$... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | x_0 | |
0haw | There is an equilateral trapezoid with bases $BC$ and $AD$ and known angles: $\angle BDC = 10^\circ$ and $\angle BDA = 70^\circ$. Prove that the following equality holds true: $AD^2 = BC(AD + AB)$. | [
"It is clear that $\\angle AMD = 20^\\circ$, $\\angle DBM = 150^\\circ$. Let's construct equilateral triangle $\\triangle KMD$, then points $M$, $B$, $D$ are on the circle centered at $K$ (fig. 8.76). After that $KM = KB$, $\\angle KMB = 80^\\circ$, from where $\\angle MBK = 80^\\circ$, but where $\\angle MBC = 80^... | Ukraine | Ukrainian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0fqm | Find all pairs of integers $(x, y)$ such that
$$
3^4 2^3 (x^2 + y^2) = x^3 y^3.
$$ | [
"First note that if $xy = 0$, then $x^2 + y^2 = 0$ and $x = y = 0$ is a solution. Also note that if $xy < 0$, then $x^2 + y^2 < 0$, impossible; thus $x, y$ are both positive or both negative. Changing simultaneously the sign of $x$ and $y$ does not change the equation, hence we may assume without loss of generality... | Spain | SPANISH MATHEMATICAL OLYMPIAD (FINAL ROUND) | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (-6, -6), (0, 0), (6, 6) | |
0bub | Problem:
a) Fie $H_{1}$ și $H_{2}$ subgrupuri ale grupului $(G, \cdot)$. Arătați că $H_{1} \cap H_{2}$ este subgrup al grupului $(G, \cdot)$.
b) Dacă $f: \mathbb{R} \rightarrow \mathbb{R}$ este o funcție bijectivă cu $f^{-1}(1)=2$, să se determine elementul neutru al legii de compoziție definite prin:
$$
x * y = f\le... | [] | Romania | Olimpiada Națională de Matematică | [
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | 1 | |
01ly | a) Does there exist a function $f : \mathbb{R} \to \mathbb{R}$, such that for any real $x$ the following equality holds $f(\sin x) + f(\cos x) = 2$?
b) The same question for $f(\sin x) + f(\cos x) = \sin 2x$. | [] | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | a) Yes: f(x) = 1 for all real x. b) No, such a function does not exist. | |
03a1 | Find all real numbers $r$ such that the inequality
$$
r(ab + bc + ca) + (3 - r) \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \ge 9
$$
holds true for arbitrary positive numbers $a, b$ and $c$. | [
"Taking $a = b = c$ we obtain\n$$\nra^2 + (3 - r) \\frac{1}{a} \\ge 3 \\iff (a - 1)(r(a^2 + a + 1) - 3) \\ge 0\n$$\nfor any $a > 0$. Then it easily follows that $r = 1$.\n\nConversely, let $r = 1$. Then we write the inequality as $\\frac{2 + abc}{3} \\ge \\frac{3}{\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}}$. The G... | Bulgaria | Fall Mathematical Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | r = 1 | |
0ihg | Problem:
Let $AC$ be a diameter of a circle $\omega$ of radius $1$, and let $D$ be the point on $AC$ such that $CD=1/5$. Let $B$ be the point on $\omega$ such that $DB$ is perpendicular to $AC$, and let $E$ be the midpoint of $DB$. The line tangent to $\omega$ at $B$ intersects line $CE$ at the point $X$. Compute $AX$. | [
"Solution:\nWe first show that $AX$ is perpendicular to $AC$. Let the tangent to $\\omega$ at $A$ intersect $CB$ at $Z$ and $CE$ at $X'$. Since $ZA$ is parallel to $BD$ and $BE=ED$, $ZX' = X'A$. Therefore, $X'$ is the midpoint of the hypotenuse of the right triangle $ABZ$, so it is also its circumcenter. Thus $X'A ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
... | null | proof and answer | 3 | |
00yd | Problem:
The Wonder Island Intelligence Service has 16 spies in Tartu. Each of them watches on some of his colleagues. It is known that if spy $A$ watches on spy $B$ then $B$ does not watch on $A$. Moreover, any 10 spies can be numbered in such a way that the first spy watches on the second, the second watches on the ... | [
"Solution:\n\nWe call two spies $A$ and $B$ neutral to each other if neither $A$ watches on $B$ nor $B$ watches on $A$.\n\nDenote the spies $A_{1}, A_{2}, \\ldots, A_{16}$. Let $a_{i}, b_{i}$ and $c_{i}$ denote the number of spies that watch on $A_{i}$, the number of that are watched by $A_{i}$ and the number of sp... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
03ds | In an acute triangle $ABC$, point $M$ is the midpoint of $AB$ and $AH$ is the altitude. Let $CP$ be the perpendicular to the line $MH$. If $AB = 21$, $BH = 7$ and $BP = CP$ find the length of $AC$. | [
"Answer: $AC = \\sqrt{473}$.\n\nLet $PE \\perp BC$ and denote $\\angle ABC = \\beta$. Since $HM$ is a median in the right triangle we have $\\angle MHA = 90^\\circ - \\beta$, $\\angle PHC = \\beta$ and $\\angle PCH = 90^\\circ - \\beta$. Therefore $\\triangle ABH \\sim \\triangle PHE \\sim \\triangle CPH$ and thus\... | Bulgaria | Bulgaria 2022 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | √473 | |
0730 | Problem:
In a triangle $ABC$ right-angled at $C$, the median through $B$ bisects the angle between $BA$ and the bisector of $\angle B$. Prove that
$$
\frac{5}{2}<\frac{AB}{BC}<3
$$ | [
"Solution:\n\nSince $E$ is the mid-point of $AC$, we have $AE = EC = b/2$. Since $BD$ bisects $\\angle ABC$, we also know that $CD = ab/(a+c)$. Since $BE$ bisects $\\angle ABD$, we also have\n$$\n\\frac{BD^{2}}{BA^{2}} = \\frac{DE^{2}}{EA^{2}}\n$$\nHowever,\n$$\n\\begin{aligned}\nBD^{2} & = BC^{2} + CD^{2} = a^{2} ... | India | INMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic /... | null | proof only | null | |
0g6k | 考慮一多項式 $P(x) = (x + d_1)(x + d_2) \cdots (x + d_9)$，其中 $d_1, d_2, \cdots, d_9$ 是 9 個相異正整數。試證存在一個正整數 $N$ 使得對所有的整數 $x \ge N$，$P(x)$ 能被一個大於 20 的質數整除。 | [
"首先觀察到，對每個足標 $i \\in \\{1, 2, \\cdots, 9\\}$，\n$$\nD_i = \\prod_{1 \\le j \\le 9,\\ j \\ne i} |d_i - d_j|\n$$\n為正數。令 $N = \\max\\{D_1 - d_1, D_2 - d_2, \\cdots, D_9 - d_9\\}$。以下我們將證明 $N$ 滿足題設。\n\n假設存在一整數 $x \\ge N$ 使得 $P(x)$ 之所有質因數皆小於 20。$\\forall\\ 1 \\le i \\le 9$，考慮 $(x+d_i)/D_i$ 的最簡分式，記為 $A_i/B_i$。注意到 $A_i|P(x)... | Taiwan | 二〇一二數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
028n | Problem:
Um caminho retangular - Janete passeia por um caminho de forma retangular $ABCD$ com largura $AB = 1992~\mathrm{m}$. Ela gasta 24 minutos para percorrer a largura $AB$. Depois, com a mesma velocidade, ela percorre o comprimento $BC$ e a diagonal $CA$ em 2 horas e 46 minutos. Qual é o comprimento $BC$? | [] | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 6745 | |
0k8e | Problem:
A cactus is a finite simple connected graph where no two cycles share an edge. Show that in a nonempty cactus, there must exist a vertex which is part of at most one cycle. | [
"Solution:\nLet $C$ be the original cactus. For every cycle in $C$, arbitrarily remove one of its edges, yielding a new graph $T$. Observe that since the cycles are edge-disjoint, we removed exactly one edge from every cycle, meaning the graph stays connected. However, there are no longer any cycles, so $T$ is a tr... | United States | HMIC | [
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
02o6 | The sidelengths and area of a triangle are all integer numbers. Find the minimum value of its area. | [
"The $3$–$4$–$5$ triangle has area $\\frac{3 \\cdot 4}{2} = 6$. We will prove that no other triangle with integer sidelengths and area has smaller area.\nLet $a$, $b$, $c$ be the sidelengths. Then its area is $S = \\sqrt{s(s-a)(s-b)(s-c)}$, where $s = \\frac{a+b+c}{2}$. Since the area is also an integer, $a+b+c$ is... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 6 | |
01av | Let $x$ and $y$ be non-negative integer solutions of the equation
$$
2x^2 - 17xy + y^2 + x = 0.
$$
Prove that $x$ is a perfect square. | [
"If $x$ is divisible by $p$, then it is easy to see that $y$ is also divisible by $p$. Substitute $x = p^a x_1$, $y = p^b y_1$ in the equation. Then we obtain\n$$\np^{2b}y_1^2 = p^a(p^b17x_1y_1 - p^a2x_1^2 - x_1),\n$$\nhence $a = 2b$ is even. (For $p = 2$ we have analogous observations.) Therefore $x$ is a perfect ... | Baltic Way | Baltic Way 2013 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0fq3 | Problem:
Se dispone de una fila de 2018 casillas, numeradas consecutivamente de 0 a 2017. Inicialmente, hay una ficha colocada en la casilla 0. Dos jugadores $A$ y $B$ juegan alternativamente, empezando $A$, de la siguiente manera: En su turno, cada jugador puede, o bien hacer avanzar la ficha 53 casillas, o bien hace... | [
"Solution:\n\nVamos a probar que el jugador $A$ tiene estrategia ganadora. Comienza de la única forma posible: llevando la ficha hasta la casilla 53. A partir de ahí, durante 38 turnos dobles $B A$, el jugador $A$ hará lo contrario de $B$: si $B$ avanza 53, $A$ retrocede 2, y viceversa. De este modo, la ficha queda... | Spain | LIII Olimpiada matemática Española | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Player A has a winning strategy: start by moving forward to the first reachable forward step, then on each subsequent turn do the opposite of B’s previous move so that each pair of turns produces a fixed net advance, steering to a position where only backward moves are forced, and finally complete with a forward move t... | |
0801 | Problem:
Su un lago c'è un villaggio di capanne poste su palafitte nei nodi di un reticolo rettangolare $m \times n$ (vedi esempio in figura). Dalla piattaforma di ogni capanna partono esattamente $p$ ponti, che la collegano ad una o più delle capanne contigue (rispetto al reticolo, quindi non in diagonale). Per quali... | [
"Solution:\n\nSe $m=1$ o $n=1$ una capanna posta in un'estremità ha una sola capanna contigua, che quindi non può essere collegata a nessun'altra capanna (i $p$ ponti che partono dalla prima capanna devono necessariamente raggiungere la capanna contigua). Quindi si hanno solo le possibilità $(m, n)=(1,2)$ o $(m, n)... | Italy | XV Gara Nazionale di Matematica | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | Possible configurations are:
- If min(m, n) = 1: only when max(m, n) = 2, with any positive p.
- If m ≥ 2 and n ≥ 2: exactly when m·n is even and p ≥ 2. | |
0jb5 | Problem:
Find the sum of all real solutions for $x$ to the equation $$\left(x^{2}+2x+3
ight)^{\left(x^{2}+2x+3
ight)^{\left(x^{2}+2x+3
ight)}} = 2012.$$ | [
"Solution:\nLet $y = x^{2} + 2x + 3$. Note that there is a unique real number $y$ such that $y^{y^{y}} = 2012$ because $y^{y^{y}}$ is increasing in $y$.\n\nThe sum of the real distinct solutions of the equation $x^{2} + 2x + 3 = y$ is $-2$ by Vieta's Formula as long as $2^{2} + 4(y - 3) > 0$, which is equivalent to... | United States | HMMT November 2012 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | -2 | |
0dlf | Let $ABC$ be a triangle with $AC > AB > BC$ and angle bisector $AD$ ($D \in BC$). Denote by $\omega_1$ and $\omega_2$ the circumcircles of triangles $ABD$ and $ACD$, respectively. The line $AC$ intersects $\omega_1$ again at $F$ and the line $AB$ intersects $\omega_2$ again at $E$. The line $DE$ intersects $\omega_1$ a... | [] | Saudi Arabia | Saudi Booklet | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Miquel po... | null | proof only | null | |
0jbu | In cyclic quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $P$. Let $E$ and $F$ be the respective feet of the perpendiculars from $P$ to lines $AB$ and $CD$. Segments $BF$ and $CE$ meet at $Q$. Prove that lines $PQ$ and $EF$ are perpendicular to each other.
![](attached_image_2.png... | [
"Let $G$, $X$, and $Y$ be the respective feet of the perpendiculars from $P$ to $EF$, $EC$, and $FB$. Note that $EPYB$ and $FPXC$ are cyclic quadrilaterals, so\n$$\n\\begin{aligned} \\angle EYF &= 90^\\circ + \\angle EYP = 90^\\circ + \\angle EBP = 90^\\circ + \\angle ABP \\\\\n&= 90^\\circ + \\angle DCP = 90^\\cir... | United States | Team Selection Test | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
0an1 | Problem:
Solve for all real numbers $x$ satisfying
$$
x + \sqrt{x-1} + \sqrt{x+1} + \sqrt{x^{2}-1} = 4
$$ | [] | Philippines | Area Stage | [
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 5/4 | |
0c4d | Problem:
Fie $a < b$ numere reale şi $f : (a, b) \rightarrow \mathbb{R}$ o funcţie astfel încât funcţiile $g : (a, b) \rightarrow \mathbb{R}$, $g(x) = (x - a) f(x)$ şi $h : (a, b) \rightarrow \mathbb{R}$, $h(x) = (x - b) f(x)$ să fie crescătoare. Arătaţi că funcţia $f$ este continuă pe $(a, b)$. | [
"Solution:\n\nFie $a < c < b$. Pentru $x \\in (c, b)$ avem $g(x) \\geq g(c)$ şi, cum $x - a > 0$, $f(x) \\geq \\frac{c - a}{x - a} f(c)$. Apoi, din $h(x) \\geq h(c)$ şi $x - b < 0$ rezultă $f(x) \\leq \\frac{c - b}{x - b} f(c)$. Deoarece $\\lim_{x \\rightarrow c} \\frac{c - a}{x - a} = \\lim_{x \\rightarrow c} \\fr... | Romania | Olimpiada Nationala de Matematica | [
"Precalculus > Functions",
"Precalculus > Limits"
] | null | proof only | null | |
0joa | Problem:
Victor has a drawer with 6 socks of 3 different types: 2 complex socks, 2 synthetic socks, and 2 trigonometric socks. He repeatedly draws 2 socks at a time from the drawer at random, and stops if the socks are of the same type. However, Victor is "synthetic-complex type-blind", so he also stops if he sees a s... | [
"Solution:\n\nLet the socks be $C_1$, $C_2$, $S_1$, $S_2$, $T_1$, $T_2$, where $C$, $S$ and $T$ stand for complex, synthetic and trigonometric respectively. The possible stopping points consist of three pairs of socks of the same type plus four different complex-synthetic $(C$-$S)$ pairs, for a total of $7$. So the... | United States | HMMT February 2015 | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 3/7 | |
0hjq | Problem:
Is there a positive integer $k$ such that
$$
(\cdots((4 \underbrace{!}_{k}) !) ! \cdots) !>(\cdots((3 \underbrace{!}_{k+1}) !) ! \cdots) ! ?
$$ | [
"Solution:\n\nThe answer is no. Since $3 != 6$, we have\n$$\n(\\cdots((3 \\underbrace{!}_{k+1}) !) ! \\cdots) !=(\\cdots((6 \\underbrace{!}_{k}) !) ! \\cdots) !>(\\cdots((4 \\underbrace{!}_{k}) !) ! \\cdots) !\n$$\nwhere the last step follows by using the obvious lemma that if $x > y$ then $x ! > y !$ (for positive... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No | |
05n2 | Problem:
On note $S$ l'ensemble des entiers de $1$ à $2016$. Combien y a-t-il de manières de partitionner $S$ en deux sous-ensembles $A$ et $B$ de telle manière que ni $A$ ni $B$ ne contient deux entiers dont la somme est une puissance de $2$ ? | [
"Solution:\n\nLa réponse est $2^{11} = 2048$. Plus précisément, on va montrer qu'il est possible de répartir comme on veut les nombres $1, 2, 4, 8, \\ldots, 1024 = 2^{10}$ comme on veut entre $A$ et $B$ et qu'une fois ces nombres placés il existe une unique manière de répartir les autres. On peut alors conclure car... | France | OCympiades Françaises de Mathématiques - Envoi Numéro 4 - Combinatoire | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2048 | |
0d1w | Tarik wants to choose some distinct numbers from the set $S = \{2, \ldots, 111\}$ in such a way that each of the chosen numbers cannot be written as the product of two other distinct chosen numbers. What is the maximum number of numbers Tarik can choose? | [
"First, we see that it is possible for Tarik to choose the 101 numbers $11, 12, \\ldots, 111$, since the product $11 \\times 12 > 111$.\n\nAssume that Tarik has chosen $k$ numbers and let $d$ be the smallest among these numbers. If $d \\geq 11$, then clearly, $k \\leq 101$.\n\nIf $2 \\leq d \\leq 6$, from each of t... | Saudi Arabia | Selection tests for the Gulf Mathematical Olympiad 2013 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 101 | |
06f7 | a. Given five points on a plane such that no three of the points are collinear, show that among the triangles which are drawn using any three of these five points as vertices, at least three of the triangles formed are not acute-angled triangles. (An acute-angled triangle is one in which all the three interior angles a... | [
"(a) We first show that there must be a non-acute triangle among any 4 points. Consider the convex hull of $A$, $B$, $C$, $D$.\n\n* If the convex hull is a quadrilateral $ABCD$, then since\n$$ \\angle ABC + \\angle BCD + \\angle CDA + \\angle DAB = 360^\\circ, $$\none of these angles is at least $90^\\circ$. This g... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
01p5 | Given real positive numbers $x_1, \dots, x_n$ ($n \ge 3$) such that $x_1 \cdot \dots \cdot x_n = 1$, prove that
$$
\frac{x_1^8}{(x_1^4 + x_2^4)x_2} + \frac{x_2^8}{(x_2^4 + x_3^4)x_3} + \dots + \frac{x_n^8}{(x_n^4 + x_1^4)x_1} \ge \frac{n}{2}.
$$
(I. Voronovich) | [
"We use the following\n\n**Lemma.** For any positive $a$ and $b$ the following inequality is valid\n$$\n(a^3 + b^3)^2 \\geq 2ab(a^4 + b^4). \\qquad (*)\n$$\nIndeed,\n$$\n(*) \\Leftrightarrow a^6 - 2a^5b + 2a^3b^3 - 2ab^5 + b^6 \\geq 0 \\Leftrightarrow (a-b)^2(a^4 - a^2b^2 + b^4) \\geq 0\n$$\nwhich is true.\n\nNow, ... | Belarus | BelarusMO 2013_s | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0caf | Problem:
În triunghiul isoscel fix $ABC$, punctul $M$ este mijlocul bazei $BC$. Punctul $P$ este variabil în interiorul triunghiului, astfel încât $\angle CBP = \angle PCA$. Arătaţi că suma măsurilor unghiurilor $\angle BPM$ şi $\angle APC$ este constantă. | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ Barajul 3 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | null | proof only | null | |
0b4a | Problem:
In square $ABCD$, $P$ lies on the ray $AD$ past $D$ and lines $PC$ and $AB$ meet at $Q$. Point $X$ is the foot of the perpendicular from $B$ to $DQ$, and the circumcircle of triangle $APX$ meets line $AB$ again at $Y$. Suppose that $DP = \frac{16}{3}$ and $BQ = 27$. The length of $BY$ can be written in the fo... | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 65 | |
09ua | An artist has an extraordinary working rhythm. He works for $3$ hours very intensively on his art, and then he sleeps for $8$ hours before starting to work again. Suppose that he starts working at midnight in the night from $31$ July to $1$ August.
Which day of August is the first day after $1$ August on which the arti... | [
"$7$ August"
] | Netherlands | Junior Mathematical Olympiad | [
"Math Word Problems"
] | English | proof and answer | 7 August | |
0fum | Problem:
Finde den grösstmöglichen Wert des Ausdrucks
$$
\frac{x y z}{(1+x)(x+y)(y+z)(z+16)}
$$
wobei $x, y, z$ positive reelle Zahlen sind. | [
"Solution:\n\nSei $A$ der Nenner des Bruchs. Es gilt nach AM-GM\n$$\n\\begin{aligned}\nA & =\\left(1+\\frac{x}{2}+\\frac{x}{2}\\right)\\left(x+\\frac{y}{2}+\\frac{y}{2}\\right)\\left(y+\\frac{z}{2}+\\frac{z}{2}\\right)(z+8+8) \\\\\n& \\geq 81 \\sqrt[3]{x^{2} / 4} \\cdot \\sqrt[3]{x y^{2} / 4} \\cdot \\sqrt[3]{y z^{... | Switzerland | IMO Selektion | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 1/81 | |
0gig | 設三角形 $ABC$ 的內心為 $I$，內切圓為 $\omega$。令 $E, F$ 為 $\omega$ 與 $CA, AB$ 的切點，$X, Y$ 為三角形 $BIC$ 的外接圓與 $\omega$ 的交點。在 $BC$ 上取一點 $T$ 使得 $\angle AIT$ 為直角。令 $G$ 為 $EF$ 與 $BC$ 的交點，$Z$ 為 $XY$ 與 $AT$ 的交點。證明 $AZ$, $ZG$, $AI$ 圈成一個等腰三角形。
Let $I$ be the incenter of triangle $ABC$, and let $\omega$ be its incircle. Let $E$ and $F$ be the ... | [
"**解. 解法一：**令 $M$ 為 $AT$ 與外接圓 $\\odot(ABC)$ 的交點。那麼 $TM \\cdot TA = TB \\cdot TC = TI^2$，所以 $\\angle AMI = \\angle AIT = 90^\\circ$，即 $M \\in \\odot(AEF) \\cap \\odot(ABC)$。由密克定理 (或 $\\angle MFG = \\angle MAC = \\angle MBG$)，$M$ 位於 $\\odot(BFG)$ 上，因此，若令 $W$ 為 $XY$ 與 $BC$ 的交點，則\n$$\n\\angle WGM = \\angle AFM = \\angl... | Taiwan | IMO 3J, Mock Exam 1 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Mique... | Chinese; English | proof only | null | |
02pc | Problem:
Cada uma das placas das bicicletas de Quixajuba contém três letras. A primeira letra é escolhida dentre os elementos do conjunto $\mathcal{A}=\{\mathrm{G}, \mathrm{H}, \mathrm{L}, \mathrm{P}, \mathrm{R}\}$, a segunda letra é escolhida dentre os elementos do conjunto $\mathcal{B}=\{\mathrm{M}, \mathrm{I}, \math... | [
"Solution:\nInicialmente, é possível fazer o emplacamento de $5 \\times 3 \\times 4 = 60$ bicicletas. Vamos analisar as duas situações possíveis:\n\n- Aumentamos duas letras num dos conjuntos. Com isso, podemos ter\n\n| $\\mathcal{A} \\times \\mathcal{B} \\times \\mathcal{C}$ | Número de Placas |\n| :---: | :---: |... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 40 | |
01wp | The numbers $1, 2, \dots, 49, 50$ are written on the blackboard. Ann performs the following operations: she chooses three arbitrary numbers $a, b, c$ from the board, replaces them by their sum $a + b + c$ and writes the number $(a + b)(b + c)(c + a)$ to her notebook. Ann performs such operations until only two numbers ... | [
"Answer: $\\frac{A}{B} = 4$.\n\n(Solution by P. Verigo.) Let us solve the problem in more general case. Replace $50$ by an arbitrary positive integer $n > 2$ of the form $n = 4k + 2$ and let the numbers initially written on the blackboard be $1, 2, \\dots, n-1, n$.\n\nFor any $\\ell$ numbers $a_1, a_2, \\dots, a_\\... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 4 | |
0inj | Problem:
Five marbles of various sizes are placed in a conical funnel. Each marble is in contact with the adjacent marble(s). Also, each marble is in contact all around the funnel wall. The smallest marble has a radius of $8$, and the largest marble has a radius of $18$. What is the radius of the middle marble? | [
"Solution:\n\nAnswer: $12$. One can either go through all of the algebra, find the slope of the funnel wall and go from there to figure out the radius of the middle marble. Or one can notice that the answer will just be the geometric mean of $18$ and $8$ which is $12$.\n\n"
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | final answer only | 12 | |
0dg7 | Prove there exist infinitely many positive integers divisible by $2021$ and each of them containing the same number of digits $0, 1, \ldots, 9$. | [
"Let $k = 2021$. We want to construct infinitely many positive integers divisible by $k$ such that in their decimal representation, each digit $0, 1, \\ldots, 9$ appears the same number of times.\n\nLet $n$ be a positive integer. Consider the number $N$ whose decimal representation consists of $n$ copies of each di... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0c3z | Problem:
Triunghiul $ABC$ este înscris în cercul $\mathcal{C}(O, 1)$. Fie $G_1, G_2, G_3$ centrele de greutate ale triunghiurilor $OBC, OAC$ şi respectiv $OAB$. Demonstraţi că triunghiul $ABC$ este echilateral dacă şi numai dacă $AG_1 + BG_2 + CG_3 = 4$. | [
"Solution:\n\nDacă triunghiul $ABC$ este echilateral, avem $AG_1 = BG_2 = CG_3 = \\frac{4}{3}$, de unde obţinem $AG_1 + BG_2 + CG_3 = 4$.\n\nReciproc, considerăm planul complex $ABC$ cu originea în $O$. Notăm cu $p$ afixul unui punct $P$ din planul complex considerat. Avem $g_1 = \\frac{b + c}{3}$, $g_2 = \\frac{c ... | Romania | Olimpiada Naţională de Matematică | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0jc6 | Let $\triangle ABC$ be a triangle. Its excircles touch sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Prove that the perimeter of triangle $\triangle ABC$ is at most twice that of triangle $DEF$.
 | [
"We consider the configuration shown in the diagram below. (Our proof uses directed lengths and can be easily modified for different configurations.)\n\nLet $a, b, c$ denote the side lengths of $BC, CA, AB$, and let $A, B, C$ denote $\\angle A, \\angle B, \\angle C$, respectively. Suppose t... | United States | Team Selection Test Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate M... | null | proof only | null | |
0d8y | Let $BC$ be a chord of a circle $(O)$ such that $BC$ is not a diameter. Let $AE$ be a diameter perpendicular to $BC$ such that $A$ belongs to the larger $\operatorname{arc} BC$ of $(O)$. Let $D$ be a point on the larger $\operatorname{arc} BC$ of $(O)$ which is different from $A$. Suppose that $AD$ intersects $BC$ at $... | [
"1) First, note that $\\angle ADE = 90^\\circ$ then $DO$, $DF$ are two medians of the triangles $ADE$, $SDT$.\nThen $\\angle ODF = \\angle ODT + \\angle FDT = \\angle OET + \\angle FTD = 90^\\circ$. Hence, $\\angle OIF = 180^\\circ - \\angle ODF = 90^\\circ$ which implies that $I$ is the midpoint of $BC$.\nSince $O... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | MN^2/4 | |
0hi0 | In a far, far galaxy there are 225 inhabited planets. Between some pairs of inhabited planets there is a two-way space connection, and from each planet you can get to any other planet (possibly with several transfers). The influence of a planet is defined as the number of other planets with which this planet has a dire... | [
"Let's reformulate this problem in terms of graphs. Planets are vertices of the graph, direct flights are edges. It is known that in any graph, any two vertices that are not connected by an edge have different degrees. The question is how many edges can be in such a graph, the number of vertices being irrelevant. F... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | 1593 | |
031i | Problem:
Let $M$, $N$ and $P$ be points on the sides $AB$, $BC$ and $CA$ of $\triangle ABC$, respectively. The lines through $M$, $N$ and $P$, parallel to $BC$, $AC$ and $AB$, respectively, meet at a point $T$. Prove that:
a) if $\frac{AM}{MB} = \frac{BN}{NC} = \frac{CP}{PA}$, then $T$ is the centroid of $\triangle AB... | [
"Solution:\nSet $PT \\cap BC = P_1$, $NT \\cap AB = N_1$ and $MT \\cap AC = M_1$. The triangles $N_1MT$, $PTM_1$ and $TP_1N$ are similar to $\\triangle ABC$. Set $k_1 = \\frac{N_1M}{AB}$, $k_2 = \\frac{PT}{AB}$ and $k_3 = \\frac{TP_1}{AB}$. Then\n$$\nk_1 + k_2 + k_3 = 1\n$$\nsince\n$$\nk_1 + k_2 + k_3 = \\frac{N_1M... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
085g | Problem:
Un triangolo equilatero ha lo stesso perimetro di un rettangolo di dimensioni $b$ ed $h$ (con $b > h$). L'area del triangolo è $\sqrt{3}$ volte l'area del rettangolo. Quanto vale $\frac{b}{h}$?
(A) $\sqrt{3}$
(B) 2
(C) $\frac{3+\sqrt{3}}{2}$
(D) $\frac{3+\sqrt{5}}{2}$
(E) $\frac{7+3 \sqrt{5}}{2}$. | [
"Solution:\n\nLa risposta è $\\mathbf{(E)}$. Detto $a$ il lato del triangolo equilatero, si hanno le relazioni $3a = 2(b + h)$, $\\frac{a^{2} \\sqrt{3}}{4} = \\sqrt{3} b h$, da cui\n$$\n\\left\\{\n\\begin{array}{l}\nb + h = \\frac{3a}{2} \\\\\nbh = \\frac{a^{2}}{4}\n\\end{array}\n\\right.\n$$\nL'equazione risolvent... | Italy | Progetto Olimpiadi di Matematica 2007 GARA di SECONDO LIVELLO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Geometry > Plane Geometry > Triangles"
] | null | MCQ | E | |
0h3b | A square $11 \times 11$ is divided into parts of sizes $4 \times 4$, $1 \times 3$, $3 \times 1$ (not necessarily all these sizes must be present). Prove that there is a row of the initial square intersecting an odd number of these parts. | [
"Оскільки рядок початкового квадрата містить непарну кількість клітинок, він буде перетинати непарну кількість частин $1 \\times 3$ та $3 \\times 1$. Якщо твердження задачі неправильне, то кожен рядок має перетинати непарну кількість частин $4 \\times 4$, тобто одну. Відтак, один квадрат $4 \\times 4$ буде лежати в... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
00rs | Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Prove that
$$
\frac{1}{a^5+b^5+c^2} + \frac{1}{b^5+c^5+a^2} + \frac{1}{c^5+a^5+b^2} \le 1
$$ | [
"First we remark that\n$$\na^5 + b^5 \\ge ab(a^3 + b^3).\n$$\nIndeed\n$$\n\\begin{aligned}\na^5 + b^5 \\ge ab(a^3 + b^3) &\\Leftrightarrow a^5 - a^4b - ab^4 + b^5 \\ge 0 \\\\\n&\\Leftrightarrow (a-b)(a^4 - b^4) \\ge 0 \\\\\n&\\Leftrightarrow (a-b)^2(a^2 + b^2)(a+b) \\ge 0.\n\\end{aligned}\n$$\nWe rewrite the inequa... | Balkan Mathematical Olympiad | BMO 2017 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
0bqk | For a positive integer $n$ denote $d(n)$ the number of its positive divisors and $s(n)$ their sum. It is known that $n + d(n) = s(n) + 1$, $m + d(m) = s(m) + 1$ and $nm + d(nm) + 2016 = s(nm)$. Find $n$ and $m$.
Mihai Bunget | [] | Romania | 67th NMO Shortlisted Problems | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | (n, m) = (2, 2017) or (2017, 2) | |
068e | Let $AB\Gamma\Delta$ be a square of side $\alpha$. On the side $A\Gamma\Delta$ we get points $E$ and $Z$ such that $\Delta E = \frac{\alpha}{3}$ and $AZ = \frac{\alpha}{4}$.
If the lines $BZ$ and $\Gamma E$ intersect at point $H$, express the area of the triangle $B\Gamma H$ as a function of $\alpha$. | [
"\nFigure 1\nWe draw the altitude $H\\Lambda$ of the triangle $B\\Gamma H$. Let it intersect $A\\Gamma$ at $K$. We put $EK = x$, $KZ = y$ and $KH = z$. Then $H\\Lambda = \\alpha + z$ and\n$$\nE_{B\\Gamma H} = \\frac{1}{2} \\alpha (\\alpha + z) \\qquad (1)\n$$\nThe triangles $\\Gamma\\Delta ... | Greece | 34th Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 6α^2/7 | |
0h50 | The road between $A$ and $B$ is $15$ km long, firstly the road goes up, then it is flat, and lastly it goes down. It is known that every part is no less than $1$ km. The path made by a pedestrian takes exactly $3$ hours. What are the minimum and the maximum amount of time that is taken by the path in opposite direction... | [
"Mark the up, flat and down parts on the way from $A$ to $B$ as $x$, $y$, $z$ respectively. Then:\n$$\nx + y + z = 15, \\frac{x}{4} + \\frac{y}{5} + \\frac{z}{6} = 3,\\ 1 \\le x, y, z \\le 13.\n$$\nFrom the first equation: $y = 15 - x - z$, substitute it into the second equation:\n$$\n\\frac{x}{4} + \\frac{15-z-x}{... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour) | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Minimum time = 73/24 hours, Maximum time = 97/30 hours | |
04gf | The rhombus $AKLM$ is inscribed in the triangle $ABC$, so that point $K$ is on $\overline{AB}$, point $L$ is on $\overline{BC}$ and point $M$ is on $\overline{CA}$. If the rhombus has side of length $2\sqrt{2}$, the area of triangle $LMC$ is $3$, and the area of triangle $KLB$ is $4$, prove that $\angle BAC = 60^\circ$... | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0la6 | Let $m$ and $n$ be positive integers. Prove that $(2m + 3)^n + 1$ is a multiple of $6m$ if and only if $3^n + 1$ is a multiple of $4m$. | [] | Vietnam | Vijetnam 2008 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
060a | Problem:
Soient $a$ et $b$ deux réels. Supposons que $2a + a^{2} = 2b + b^{2}$. Montrer que si $a$ est un entier (pas forcément positif), alors $b$ est aussi un entier. | [
"Solution:\n\nOn réécrit l'égalité\n$$\n\\begin{gathered}\n2a - 2b = b^{2} - a^{2} \\\\\n2(a - b) = -(a + b)(a - b)\n\\end{gathered}\n$$\nAlors soit $b = a$, et alors $b$ est un entier, soit\n$$\n2 = -a - b\n$$\nd'où\n$$\nb = -2 - a\n$$\net $b$ est encore un entier."
] | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0d22 | Find all values of $n$ for which there exists a convex cyclic non-regular polygon with $n$ vertices such that the measures of all its internal angles are equal. | [
"Let $P_{1} P_{2} \\cdots P_{n}$ be a convex cyclic non-regular polygon with the measures of all its internal angles equal and let $O$ be its circumcenter. Because all the angles are equal, all the arcs $\\widehat{P_{i} P_{i+2}}$, for $i=1, \\ldots, n$, have the same length. Hence, $\\angle P_{i} O P_{i+2} = \\frac... | Saudi Arabia | Selection tests for the Gulf Mathematical Olympiad 2013 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles"
] | English | proof and answer | All even integers at least 4 | |
089i | Problem:
Quante sono le coppie di interi positivi $(m, n)$ tali che la frazione $\frac{m}{n}$ sia ridotta ai minimi termini e strettamente minore di 1, e che il prodotto $mn$ sia uguale a $1 \cdot 2 \cdot 3 \cdot \ldots \cdot 24 \cdot 25$ (ovvero al prodotto dei primi 25 interi positivi)?
(A) $2^{7}$
(B) $2^{8}-1$
(C... | [
"Solution:\n\nLa risposta è $\\mathbf{( C )}$. Se $m$ è multiplo di un certo primo $p$, allora deve essere divisibile per la massima potenza di $p$ che divida $25!$ (dove per $25!$ intendiamo il prodotto degli interi da 1 a 25) affinché $n=\\frac{25!}{m}$ non abbia fattori $p$ (altrimenti la frazione $\\frac{m}{n}$... | Italy | Olimpiadi della Matematica - Gara di Febbraio | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | C | |
0193 | Consider a right angled triangle $ABC$ with sides of length $3$, $4$, and $5$. Determine the greatest possible radius of a circle that is tangent to two among the lines $BC$, $CA$, and $AB$ and that in addition passes through at least one of the points $A$, $B$, and $C$. | [
"Consider a general triangle $ABC$. Suppose we have a circle that touches the lines $AB$ and $AC$. Since it cannot also pass through the point $A$, we may suppose it passes through the point $C$. The centre of the circle will then lie either on the internal, or the external, bisector of the angle at $A$.\n\nAssume ... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 15 | |
07go | Find the maximum number of 5-element subsets of the set $\{1, 2, \dots, 20\}$ such that the intersection of any pair of these subsets has exactly one element. | [
"The answer is $16$. For the example consider the following family of $5$-element subsets.\n$$\n\\begin{aligned}\n\\{1, 2, 3, 4, 5\\} & & \\{2, 6, 10, 14, 18\\} \\\\\n\\{2, 7, 11, 15, 19\\} & & \\{2, 8, 12, 16, 20\\} \\\\\n\\{1, 6, 7, 8, 9\\} & & \\{3, 8, 13, 15, 18\\} \\\\\n\\{3, 9, 12, 14, 19\\} & & \\{3, 6, 11, ... | Iran | 38th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 16 | |
029s | Problem:
Numa divisão, aumentando o dividendo de $1989$ e o divisor de $13$, o quociente e o resto não se alteram. Qual é o quociente? | [
"Solution:\n\n$153$."
] | Brazil | Desafios | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 153 | |
080s | Problem:
Si vuole misurare la lunghezza di un circuito automobilistico usando un'auto che ha il contachilometri inizialmente azzerato e che misura solo i chilometri e non le centinaia di metri. Qual è il minimo $n$ tale che, guardando solamente quanto segna il contachilometri alla fine dell'n-esimo giro, il pilota pos... | [] | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | B | |
0fhd | Problem:
Supongamos que la ecuación $x^{3}+p x^{2}+q x+r=0$ con $r \neq 0$, admite tres raíces reales y positivas. Determinar la relación que debe ligar los números reales $p, q$ y $r$ a fin de que las tres raíces puedan ser las longitudes de los lados de un triángulo. | [
"Solution:\n\nIdentificando coeficientes entre $x^{3}+p x^{2}+q x+r$ y $(x-x_{1})(x-x_{2})(x-x_{3})$ se obtienen las llamadas relaciones de Cardano-Vieta entre las raíces y los coeficientes del polinomio (los primeros miembros son las funciones simétricas elementales de las raíces)\n$$\n\\begin{aligned}\n-x_{1} x_{... | Spain | OME 27 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | p < 0, q > 0, r < 0, and p^3 − 4 p q + 8 r > 0 | |
01f2 | Determine all positive integers $n$ with the following property: For any integer $k$ the polynomial $x^n + k$ is either irreducible or has an integer root. | [
"The answer is $n = 1$ or $n$ prime. It is clear that $n = 1$ has the desired property.\n\nAssume that $n > 1$. To see that $n$ must be a prime, assume that $n$ is composite, and let $d$ be a non-trivial divisor of $n$.\nWe consider $k = -2^d$. Since $2$ is a prime number, and $d < n$, $2^d$ is not a perfect $n$th-... | Baltic Way | Baltic Way 2019 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers",
... | English | proof and answer | n = 1 or n is prime |
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