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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0ejg | Problem:
Dokaži, da ne obstajata naravni števili $a$ in $b$, za kateri velja $\sqrt{a}+\sqrt{b}=\sqrt{2021}$. | [
"Solution:\n\nDenimo, da taki naravni števili obstajata. Tedaj iz dane enakosti očitno sledi $a, b<2021$. Enakost preoblikujemo v $\\sqrt{a}=\\sqrt{2021}-\\sqrt{b}$ in jo kvadriramo, da dobimo $a=2021-2 \\sqrt{2021 b}+b$. Ker sta $a$ in $b$ naravni števili, je $2 \\sqrt{2021 b}$ celo število in zato je $\\sqrt{2021... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0gab | 你受託幫神盾局辦餐會，但局內有若干對員工是仇人。對一群至少包含 3 個人且人數為奇數的員工們而言，只要可以讓他們圍著一個圓桌入坐，使得任何相鄰的兩個員工都是仇人的話，就稱這群員工為**復仇者聯盟**。
你發現：如果想要將所有員工分坐若干桌，使得同桌的任兩人都不是仇人的話，至少需要 11 張桌子。試證：神盾局內至少可找到 $2^{10} - 11$ 個復仇者聯盟。
You are responsible for arranging a banquet for an agency. In the agency, some pairs of agents are enemies. A group of agents are called... | [
"將每個員工視為一個頂點，並將每對仇人之間連線，從而將問題轉化到圖 $(V, E)$ 上。考慮用若干顏色去塗這些點的塗色法；一個塗法被稱為”合法的”，若且唯若任兩邊的兩端點的顏色不同。則原題等價於：若最少色的合法塗法為 $k$ 色，則圖 $(V, E)$ 上至少存在 $2^{k-1} - k$ 個環。\n\n對於任何 $k$ 色塗法, 以 $\\hat{V} = (V_1, V_2, \\dots, V_k)$ 來表示之, 其中 $V_i$ 為被塗成第 $i$ 色的點所成集合。讓我們用以下方式將塗法排序：對於兩種塗法 $\\hat{V} = (V_1, \\dots, V_k)$ 和 $\\hat{U} = (U_1, \\... | Taiwan | 二〇一六數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
0avo | Problem:
A function $g: \mathbb{N} \rightarrow \mathbb{N}$ satisfies the following:
a. If $m$ is a proper divisor of $n$, then $g(m)<g(n)$.
b. If $m$ and $n$ are relatively prime and greater than $1$, then
$$
g(m n)=g(m) g(n)+(n+1) g(m)+(m+1) g(n)+m+n
$$
Find the least possible value of $g(2016)$. | [
"Solution:\nConsider $h(x):=g(x)+x+1$. We have that, for $m, n$ coprime and greater than $1$,\n$$\n\\begin{aligned}\nh(m) h(n) & =(g(m)+m+1)(g(n)+n+1) \\\\\n& =g(m) g(n)+(n+1) g(m)+(m+1) g(n)+m n+m+n+1 \\\\\n& =g(m n)+m n+1 \\\\\n& =h(m n)\n\\end{aligned}\n$$\nRepeating this, we find that more generally, if $m_{1},... | Philippines | 19th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 3053 | |
0g4i | Problem:
Let $x$, $y$ and $a_{0}$, $a_{1}$, $a_{2}$, $\ldots$ be integers satisfying $a_{0}=a_{1}=0$ and
$$
a_{n+2}=x \cdot a_{n+1}+y \cdot a_{n}+1
$$
for all integers $n \geq 0$. Let $p$ be any prime number. Show that $\operatorname{gcd}\left(a_{p}, a_{p+1}\right)$ is either equal to $1$ or greater than $\sqrt{p}$. | [
"Solution:\nAssume that $\\operatorname{gcd}\\left(a_{p}, a_{p+1}\\right) \\neq 1$ and let $q$ be a prime dividing $\\operatorname{gcd}\\left(a_{p}, a_{p+1}\\right)$. Considering the sequence $a_{0}, a_{1}, a_{2}, \\ldots$ modulo $q$, any two consecutive terms still uniquely determine the subsequent term. The fact ... | Switzerland | Swiss Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0il1 | Problem:
Compute
$$
\int_{0}^{\pi / 3} x \tan^{2}(x) \, dx
$$ | [
"Solution:\n$$\n\\begin{aligned}\n\\int_{0}^{\\pi / 3} x \\tan^{2}(x) \\, dx & = \\int_{0}^{\\pi / 3} x\\left(-1+\\frac{1}{\\cos^{2}(x)}\\right) dx \\\\\n& = -\\left.\\frac{x^{2}}{2}\\right|_{0}^{\\pi / 3} + \\int_{0}^{\\pi / 3} \\frac{x \\, dx}{\\cos^{2}(x)} \\\\\n& = -\\left.\\frac{x^{2}}{2}\\right|_{0}^{\\pi / 3... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Trigonometric functions"
] | null | proof and answer | -π^2/18 + π√3/3 - ln(2) | |
0ib2 | Problem:
Right triangle $X Y Z$ has right angle at $Y$ and $X Y = 228$, $Y Z = 2004$. Angle $Y$ is trisected, and the angle trisectors intersect $X Z$ at $P$ and $Q$ so that $X, P, Q, Z$ lie on $X Z$ in that order. Find the value of $(P Y + Y Z)(Q Y + X Y)$. | [
"Solution:\n\nThe triangle's area is $(228 \\cdot 2004) / 2 = 228456$. All the angles at $Y$ are $30$ degrees, so by the sine area formula, the areas of the three small triangles in the diagram are $Q Y \\cdot Y Z / 4$, $P Y \\cdot Q Y / 4$, and $X Y \\cdot P Y / 4$, which sum to the area of the triangle. So expand... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 1370736 | |
07rh | Points $D$, $E$ and $F$ lie respectively on sides $BC$, $CA$ and $AB$ of triangle $ABC$ such that $BDEF$ is a parallelogram. Prove that the area of $BDEF$ is maximal when $D$, $E$ and $F$ are the mid-points of the sides. | [
"\n\nWe use the usual notation: $a = |BC|$, $b = |CA|$, $c = |AB|$ and $\\beta = \\angle CBA$.\n\nBecause $D$ is between $B$ and $C$, there exists a real number $\\lambda$ satisfying $0 < \\lambda < 1$ and $|BD| = \\lambda a$. Then $|DC| = (1 - \\lambda)a$ and because $AB \\parallel DE$, th... | Ireland | Irish | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0adp | Во овоштарникот на дедо Ване растат овошни дрвца и тоа: $240$ дрвца кајсии, $260$ дрвца праски и $130$ дрвца јаболка. Учениците од одделението на малиот Ване, внукот на дедо Ване, на денот на дрвото дошле во овоштарникот и засадиле уште $75$ дрвца кајсии, $126$ дрвца праски и $142$ дрвца јаболка.
a) Колку ученици има ... | [
"a) На денот на дрвото се засадени вкупно $75 + 126 + 142 = 343$ овошни дрвца. Бидејќи секој ученик засадил по $7$ дрвца, во одделението има $343 : 7 = 49$ ученици.\n\nб) Има: $240 + 75 = 315$ дрвца кајсии, $260 + 126 = 386$ дрвца праски и $130 + 142 = 272$ дрвца јаболка.\n\nв) Во овоштарникот има вкупно $315 + 386... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Macedonian, English | final answer only | Number of students: 49; Trees by type: apricots 315, peaches 386, apples 272; Total trees: 973 | |
0dt0 | In a scalene triangle $ABC$ with centroid $G$ and circumcircle $\omega$ centred at $O$, the extension of $AG$ meets $\omega$ at $M$; lines $AB$ and $CM$ intersect at $P$; and lines $AC$ and $BM$ intersect at $Q$. Suppose the circumcentre $S$ of the triangle $APQ$ lies on $\omega$ and $A$, $O$, $S$ are collinear. Prove ... | [
"Let $AD$, $BE$ and $CF$ be the medians of the triangle $ABC$. Since $A$, $O$, $S$ are collinear, $AS$ is a diameter of $\\omega$ with $AO = OS$. The triangles $AOB$ and $ASP$ are similar isosceles triangles. Thus $OB \\parallel SP$. Since $O$ is the midpoint of $AS$, we have $B$ is the midpoint of $AP$. Similarly,... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0fy3 | Problem:
Sei $P$ ein reelles Polynom, sodass für alle reellen $x$ die Gleichung $P(x)=P(1-x)$ gilt. Beweise, dass ein reelles Polynom $Q$ existiert mit
$$
P(x)=Q(x(1-x))
$$ | [
"Solution:\n\nDas Polynom $S(x)=P\\left(x+\\frac{1}{2}\\right)$ erfüllt nach Voraussetzung die Gleichung $S(x)=S(-x)$ für alle reellen $x$. Also ist $S$ gerade und damit von der Form $S(x)=R\\left(x^{2}\\right)$ für ein reelles Polynom $R$. Zur Begründung sei $S(x)=\\sum_{k=0}^{n} a_{k} x^{k}$, dann gilt nämlich\n$... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0iik | Let $\mathcal{P}$ be a regular 2006-gon. A diagonal of $\mathcal{P}$ is called *good segment* if its endpoints divide the boundary of $\mathcal{P}$ into two parts, each composed of an odd number of sides of $\mathcal{P}$. The sides of $\mathcal{P}$ are also called *good segment*.
Suppose $\mathcal{P}$ has been dissecte... | [
"**First Solution:** We start with the following lemma.\n**Lemma** Let $P_iP_j$ is a diagonal used in $\\mathcal{T}$, and $\\widehat{P_iP_j}$ is non-major and contains $n$ segments of $\\mathcal{P}$, then there are at most $\\lfloor \\frac{n}{2} \\rfloor$ good triangles with vertices on $\\widehat{P_iP_j}$. More pr... | United States | IMO | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 1003 | |
0174 | Find all functions from the set of the real numbers to the set of the real numbers which satisfy the functional equation
$$
f(f(x + y)) = f(x^2 - y^2) + 4xyf(x + y)
$$
for all real numbers $x$ and $y$. | [
"Let $u = x + y$ and $v = x - y$. If the function $f$ satisfies the equation, then\n$$\nf^2(u) = f(uv) + (u^2 - v^2)f(u).\n$$\nSetting $u = 1$ gives $f^2(1) = f(v) + (1 - v^2)f(1)$. Setting also $v = 1$ gives $f^2(1) = f(1)$ so we have\n$$\nf(v) = cv^2 \\quad \\text{where } c = f(1).\n$$\nFrom $f^2(1) = f(1)$ we no... | Baltic Way | BALTIC WAY | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 0, f(x) = x^2, or f(x) = -x^2 | |
0ibc | Problem:
A regular decagon $A_{0} A_{1} A_{2} \cdots A_{9}$ is given in the plane. Compute $\angle A_{0} A_{3} A_{7}$ in degrees. | [
"Solution:\n$54^{\\circ}$\n\nPut the decagon in a circle. Each side subtends an arc of $360^{\\circ} / 10 = 36^{\\circ}$. The inscribed angle $\\angle A_{0} A_{3} A_{7}$ contains 3 segments, namely $A_{7} A_{8}$, $A_{8} A_{9}$, $A_{9} A_{0}$, so the angle is $108^{\\circ} / 2 = 54^{\\circ}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 54° | |
064j | Problem:
Der größte gemeinsame Teiler zweier positiver ganzer Zahlen $m$ und $n$ sei mit $\operatorname{ggT}(m, n)$ bezeichnet.
Es sei eine unendliche Menge $S$ positiver ganzer Zahlen gegeben, sodass es vier paarweise verschiedene Zahlen $v, w, x, y \in S$ gibt, für die $\operatorname{ggT}(v, w) \neq \operatorname{gg... | [
"Solution:\n\nIm folgenden nennen wir eine dreielementige Teilmenge $\\{s, t, u\\} \\subset S$ ein ausgewogenes Dreieck, falls die Menge $\\{\\mathrm{ggT}(s, t), \\operatorname{ggT}(s, u), \\operatorname{ggT}(t, u)\\}$ genau zwei verschiedene Elemente hat. Es ist zu zeigen, dass es ein ausgewogenes Dreieck gibt.\n\... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade 2022 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0fj2 | Problem:
Determinar la función $f: \mathbb{N} \rightarrow \mathbb{N}$ (siendo $\mathbb{N}=\{1,2,3, \ldots\}$ el conjunto de los números naturales) que cumple, para cualesquiera $s, n \in \mathbb{N}$, las siguientes condiciones:
$f(1)=f\left(2^{s}\right)=1$ y si $n<2^{s}$, entonces $f\left(2^{s}+n\right)=f(n)+1$.
Cal... | [
"Solution:\n\nPara cada número natural $n$ definimos $f(n)$ como la suma de las cifras de la expresión de $n$ escrito en base 2. Está claro que esta función $f$ cumple las condiciones a) y b). Además, es la única función que las cumple, porque el valor de $f(n)$ viene determinado por las condiciones a) y b).\n\nPro... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | max_{n≤2001} f(n) = 10; the least n with f(n) = 2001 is n = 2^{2001} − 1 | |
06ej | Find the integral solutions of the equation $7(x + y) = 3(x^2 - xy + y^2)$. | [
"The solutions are $(x, y) = (0, 0), (4, 5), (5, 4)$.\nLet $a = x + y$ and $b = x - y$. The equation can be rewritten as\n$$\n28(x + y) = 3((x + y)^2 + 3(x - y)^2),\n$$\nwhich means $28a = 3(a^2 + 3b^2)$. Clearly, $3 \\mid a$. Let $a = 3c$. The equation becomes\n$$\n28c = 9c^2 + 3b^2.\n$$\nWe further find that $3 \... | Hong Kong | IMO HK TST | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | [(0, 0), (4, 5), (5, 4)] | |
0b7q | Let $p > 5$ be a prime. Find all positive integers $x$ such that $5p + x$ divides $5p^n + x^n$, for all $n \in \mathbb{N}^*$. | [
"$1^\\circ \\ 5p + x \\mid 5p^n + x^n \\text{ for all } n \\ge 1;$\n$2^\\circ \\ 5p + x \\mid 30p^2.$\nTo prove $1^\\circ \\Rightarrow 2^\\circ$, set $n = 2$ to obtain $5p+x \\mid 5p^2+x^2 \\Rightarrow 5p+x \\mid (x+5p)(x-5p) + 30p^2$, hence $5p+x \\mid 30p^2$, as needed. For the second implication, observe that $5... | Romania | NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | All positive integers x such that 5p + x divides 30p^2, equivalently x = d − 5p where d runs over the divisors of 30p^2 greater than 5p. Explicitly: x ∈ {p, 5p, 10p, 25p, p(p−5), p(2p−5), p(3p−5), 5p(p−1), p(6p−5), 5p(2p−1), 5p(3p−1), 5p(6p−1)}. | |
0keo | Problem:
Two sides of a regular $n$-gon are extended to meet at a $28^{\circ}$ angle. What is the smallest possible value for $n$? | [
"Solution:\n\nWe note that if we inscribe the $n$-gon in a circle, then according to the inscribed angle theorem, the angle between two sides is $\\frac{1}{2}$ times some $x-y$, where $x$ and $y$ are integer multiples of the arc measure of one side of the $n$-gon. Thus, the angle is equal to $\\frac{1}{2}$ times an... | United States | HMMT February 2020 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 45 | |
0glw | Prove or disprove that there exist $2017$ consecutive positive integers that cannot be written as $a^2 + b^2$ where $a$ and $b$ are integers. | [
"We will prove that such a sequence exists. First we prove the following lemma.\n\n*Lemma.* Let $q$ be a prime such that $q \\equiv 3 \\pmod 4$. If $n \\equiv q \\pmod{q^2}$ then $n$ cannot be written as the sum of two squares.\n\n*Proof.* Let $q$ be a prime such that $q \\equiv 3 \\pmod 4$, and $n$ be an integer s... | Thailand | The 14th Thailand Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0j2e | Problem:
Let the functions $f(\alpha, x)$ and $g(\alpha)$ be defined as
$$
f(\alpha, x)=\frac{\left(\frac{x}{2}\right)^{\alpha}}{x-1} \quad g(\alpha)=\left.\frac{d^{4} f}{d x^{4}}\right|_{x=2}
$$
Then $g(\alpha)$ is a polynomial in $\alpha$. Find the leading coefficient of $g(\alpha)$. | [
"Solution:\nWrite the first equation as $(x-1) f=\\left(\\frac{x}{2}\\right)^{\\alpha}$. For now, treat $\\alpha$ as a constant. From this equation, repeatedly applying derivative with respect to $x$ gives\n$$\n\\begin{aligned}\n(x-1) f^{\\prime}+f & =\\left(\\frac{\\alpha}{2}\\right)\\left(\\frac{x}{2}\\right)^{\\... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1/16 | |
01sl | Let $a$, $b$, $c$, $d$, $x$, $y$ denote the lengths of the sides $AB$, $BC$, $CD$, $DA$ and the diagonals $AC$, $BD$ of a cyclic quadrilateral $ABCD$, respectively.
$$
\left(\frac{1}{a} + \frac{1}{c}\right)^2 + \left(\frac{1}{b} + \frac{1}{d}\right)^2 \ge 8 \left(\frac{1}{x^2} + \frac{1}{y^2}\right).
$$ | [
"(Solution by Y. Dubovik, U. Kazlouski.) By the Cosine Law for the triangles $DAB$ and $DCB$,\n$$\ny^2 = a^2 + d^2 - 2ad \\cos A, \\quad (1)\n$$\n$$\ny^2 = b^2 + c^2 + 2bc \\cos A. \\quad (2)\n$$\nMultiplying (1) and (2) by $bc$ and $ad$, respectively, and summing the obtained equalities, we get\n$$\n(bc + ad)y^2 =... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0d5n | Let $ABC$ be a triangle, $H_{a}, H_{b}$ and $H_{c}$ the feet of its altitudes from $A, B$ and $C$, respectively, $T_{a}, T_{b}, T_{c}$ its intouch points on the sides $BC, CA$ and $AB$, respectively. The circumcircles of triangles $A H_{b} H_{c}$ and $A T_{b} T_{c}$ intersect again at $A'$. The circumcircles of triangl... | [
"\n\nLet $H$ and $I$ be the orthocenter and the incenter of triangle $ABC$, respectively. Because $\\angle A H_{b} H = \\angle H H_{c} A = 90^{\\circ}$, $AH$ is a diameter of the circumcircle of $A H_{c} H H_{b}$, and therefore $\\angle A A' H = 90^{\\circ}$. Because $\\angle A T_{b} I = \\... | Saudi Arabia | SAMC 2015 | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Arabic | proof only | null | |
04y0 | A strange calculator has only two buttons with positive integers, each consisting of two digits. It displays the number $1$ at the beginning. Whenever a button with number $N$ is pressed, the calculator replaces the displayed number $X$ with the number $X \cdot N$ or $X + N$. Multiplication and addition alternate, mult... | [
"Let $a, b$ be the numbers written on the buttons. Consider the sequence $(x_n)_{n=0}^{\\infty}$ such that $x_{n+1}$ is formed by the last four digits of $a(x_n + b)$ for each $n \\ge 0$, that is,\n$$\nx_{n+1} \\equiv a(x_n + b) \\pmod{10\\,000} \\quad \\text{and} \\quad 0 \\le x_{n+1} < 10\\,000.\n$$\nSince there ... | Czech-Polish-Slovak Mathematical Match | Czech-Polish-Slovak Mathematics Competition | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | Yes for both. For example: (a) choose 31 and 34 to obtain the target and then repeat the two-step cycle; (b) choose 47 and 62 to obtain the target and then repeat. Moreover, choosing 11 and 12 allows generating all four-digit endings infinitely often. | |
0e1t | Given 20 points in space so that no three of them are collinear, prove that the number of planes determined by these points is not equal to $1111$. | [
"Assume, to the contrary, that the number of planes is equal to $1111$. Now, $20$ points in space can define at most $\\binom{20}{3} = \\frac{20 \\cdot 19 \\cdot 18}{3 \\cdot 2 \\cdot 1} = 1140$ planes, so $1140 - 1111 = 29$ triplets of points lie in the planes already determined by another triplet. If one of the p... | Slovenia | Selection Examinations for the IMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
02b2 | Problem:
Cada um dos números $x_{1}, x_{2}, \ldots, x_{2004}$ pode ser igual a $\sqrt{2}-1$ ou a $\sqrt{2}+1$. Quantos valores inteiros distintos a soma
$$
\sum_{k=1}^{1002} x_{2k-1} x_{2k} = x_{1} x_{2} + x_{3} x_{4} + x_{5} x_{6} + \cdots + x_{2003} x_{2004}
$$
pode assumir? | [
"Solution:\nTemos que os possíveis produtos $x_{2k-1} x_{2k}$ onde $k \\in \\{1,2, \\ldots, 1002\\}$ são $(\\sqrt{2}-1)(\\sqrt{2}-1) = 3-2\\sqrt{2}$, $(\\sqrt{2}+1)(\\sqrt{2}+1) = 3+2\\sqrt{2}$ e $(\\sqrt{2}-1)(\\sqrt{2}+1) = 1$.\n\nSuponha que $a$ produtos são iguais a $3-2\\sqrt{2}$, $b$ produtos são iguais a $3+... | Brazil | Nível 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 502 | |
022s | Problem:
10. (N1) Contas do papagaio - Rosa tem um papagaio que faz contas de um modo estranho. Cada vez que Rosa diz dois números ele faz a mesma conta, veja:
- Se Rosa diz "4 e 2" o papagaio responde "9"
- Se Rosa diz "5 e 3" o papagaio responde "12"
- Se Rosa diz "3 e 5" o papagaio responde "14"
- Se Rosa diz "9 e ... | [
"Solution:\n\n1"
] | Brazil | null | [
"Math Word Problems"
] | null | final answer only | 1 | |
0jsi | Problem:
Compute $\tan \left(\frac{\pi}{7}\right) \tan \left(\frac{2 \pi}{7}\right) \tan \left(\frac{3 \pi}{7}\right)$. | [
"Solution:\n\nConsider the polynomial $P(z)=z^{7}-1$. Let $z=e^{i x}=\\cos x+i \\sin x$. Then\n$$\n\\begin{aligned}\nz^{7}-1= & \\left(\\cos ^{7} x-\\binom{7}{2} \\cos ^{5} x \\sin ^{2} x+\\binom{7}{4} \\cos ^{3} x \\sin ^{4} x-\\binom{7}{6} \\cos x \\sin ^{6} x-1\\right) \\\\\n& +i\\left(-\\sin ^{7} x+\\binom{7}{2... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | sqrt(7) | |
0csy | В республике математиков выбрали число $\alpha > 2$ и выпустили монеты достоинствами в 1 рубль, а также в $\alpha^k$ рублей при каждом натуральном $k$. При этом $\alpha$ было выбрано так, что достоинства всех монет, кроме самой мелкой, иррациональны. Могло ли оказаться, что любую сумму в натуральное число рублей можно ... | [
"Могло.\n\nПокажем, что математики могли выбрать число $\\alpha = \\frac{-1 + \\sqrt{29}}{2}$; это число является корнем уравнения $\\alpha^2 + \\alpha = 7$. Ясно, что $\\alpha > 2$. Нетрудно видеть, что при натуральных $m$ мы имеем $(2\\alpha)^m = a_m + b_m\\sqrt{29}$, где $a_m$ и $b_m$ — целые числа, причём $a_m ... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Algebra > Intermediate Algebra > Quadratic functions",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Yes; for example, choose alpha = (-1 + sqrt(29)) / 2, which makes every power irrational and ensures any integer sum can be formed with at most six coins of each denomination. | |
0jed | Problem:
A cafe has 3 tables and 5 individual counter seats. People enter in groups of size between 1 and 4, inclusive, and groups never share a table. A group of more than 1 will always try to sit at a table, but will sit in counter seats if no tables are available. Conversely, a group of 1 will always try to sit at t... | [
"Solution:\nAnswer: 16\n\nWe first show that $M+N \\geq 16$. Consider the point right before the last table is occupied. We have two cases:\n\nFirst, suppose there exists at least one open counter seat. Then, every table must contribute at least 3 to the value of $M+N$, because no groups of 1 will have taken a tabl... | United States | HMMT 2013 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 16 | |
0i3o | Problem:
How many integers between $1$ and $2000$ inclusive share no common factors with $2001$? | [
"Solution:\nTwo integers are said to be relatively prime if they share no common factors, that is if there is no integer greater than $1$ that divides evenly into both of them. Note that $1$ is relatively prime to all integers. Let $\\varphi(n)$ be the number of integers less than $n$ that are relatively prime to $... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 1232 | |
00eg | Let $\mathbb{R}^+$ be the set of positive real numbers. Find all non-negative real numbers $\alpha$ for which there exists a function $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that
$$
f(x^{\alpha} + y) = (f(x + y))^{\alpha} + f(y)
$$
for any $x, y \in \mathbb{R}^+$. | [
"The answer is $\\alpha = 0$. In this case, the function $f(x) = x$ satisfies the statement. From now on we assume $\\alpha > 0$. Note that if such a function $f$ exists, then it is strictly increasing: indeed, taking $z > y$, there is $x \\in \\mathbb{R}^+$ such that $z = x^{\\alpha} + y$, from where we obtain:\n$... | Argentina | Rioplatense Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | α = 0 | |
029f | Problem:
O número $1089$ tem uma propriedade interessante. Quando fazemos a multiplicação deste número por $9$, como é mostrado a seguir,
| 1 | 0 | 8 | 9 |
| :--- | :--- | :--- | :--- |
| | | $\times$ | 9 |
| 9 | 8 | 0 | 1 |
obtemos o número $9801$ que é o número $1089$ com os seus algarismos escritos da esquerd... | [
"Solution:\n\na) Estamos procurando um número de cinco algarismos $A B C D E$ tal que\n$$\nA B C D E \\times 9 = E D C B A\n$$\nSomando $A B C D E$ dos dois lados dessa igualdade, obtemos que $A B C D E \\times 10 = E D C B A + A B C D E$. Mas o número $A B C D E \\times 10$ pode ser representado ainda pelos algari... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | a) 10989; b) 1099989 | |
0c16 | Let $a$, $b$, $c$ be three non-negative real numbers such that $a + b + c = 3$. Prove the inequality:
$$
\frac{a}{1 + b} + \frac{b}{1 + c} + \frac{c}{1 + a} \ge \frac{1}{1 + a} + \frac{1}{1 + b} + \frac{1}{1 + c}.
$$ | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
037x | Problem:
Let $p > 2$ be a prime number. Find the number of the subsets $B$ of the set $\{1, 2, \ldots, p-1\}$ such that $p$ divides the sum of the elements of $B$. | [
"Solution:\nConsider the set $A' = A \\cup \\{0\\}$ instead of $A$. Let $B = \\{a_1, a_2, \\ldots, a_k\\}$ be a nonempty subset of $A'$. Set\n$$\ni + B = \\{i + a_1 \\pmod{p}, i + a_2 \\pmod{p}, \\ldots, i + a_k \\pmod{p}\\}$$\nNote that the sums of the elements of the sets $i + B$, $i = 0, 1, \\ldots, p-1$, are al... | Bulgaria | Team selection test for 47. IMO | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | (2^{p-1} - 1)/p | |
0hu1 | Problem:
Does there exist an integer such that its cube is equal to $3 n^{2} + 3 n + 7$, where $n$ is an integer? | [
"Solution:\n\nSuppose that there exist integers $n$ and $m$ such that $m^{3} = 3 n^{2} + 3 n + 7$. Then from $m^{3} \\equiv 1 \\pmod{3}$ it follows that $m = 3k + 1$ for some $k \\in \\mathbb{Z}$. Substituting into the initial equation we obtain $3k\\left(3k^{2} + 3k + 1\\right) = n^{2} + n + 2$. It is easy to chec... | United States | Berkeley Math Circle | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
09jq | A family has five children with distinct positive integer ages such that the sum of the ages of any two children is different from the sum of the ages of any other two children. What is the minimum possible age of the eldest child? | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 8 | |
00hp | Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$. | [
"By inspection, we see that the pairs $(a, b)$ with $a = b$ are solutions, and so too are the pairs $(a, 1)$. We will see that these are the only solutions.\n\n- Case 1. Consider the case $b < a$. Since $b-1$ is a multiple of $a-1$, it follows that $b = 1$. This yields the second set of solutions described above.\n... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All pairs are (n, n) and (n, 1) for any positive integer n. | |
0cv3 | An equilateral triangle $ABC$ is inscribed into a circle $\Omega$ and circumscribed about a circle $w$. Points $P$ and $Q$ are chosen on the sides $AC$ and $AB$, respectively, so that the segment $PQ$ is tangent to $w$. A circle $\Omega_b$ centered at $P$ passes through $B$, and a circle $\Omega_c$ centered at $Q$ pass... | [
"The second meeting point of $\\Omega$ and $\\Omega_b$ is symmetric to $B$ about $OP$; a similar statement holds for that of $\\Omega$ and $\\Omega_c$. To prove that these points coincide, one needs just to verify that $\\angle POQ = 60^\\circ$.\n\n\nПервое решение. Пусть $O$ — центр треугольника $ABC$, и пусть $w$... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English; Russian | proof only | null | |
0kl7 | Problem:
Let $k$ and $n$ be positive integers and let
$$
S=\left\{\left(a_{1}, \ldots, a_{k}\right) \in \mathbb{Z}^{k} \mid 0 \leq a_{k} \leq \cdots \leq a_{1} \leq n, a_{1}+\cdots+a_{k}=k\right\}
$$
Determine, with proof, the value of
$$
\sum_{\left(a_{1}, \ldots, a_{k}\right) \in S}\binom{n}{a_{1}}\binom{a_{1}}{a_{2}... | [
"Solution:\nLet\n$$\nT=\\left\\{\\left(b_{1}, \\ldots, b_{n}\\right) \\mid 0 \\leq b_{1}, \\ldots, b_{n} \\leq k, b_{1}+\\cdots+b_{n}=k\\right\\}\n$$\nThe sum in question counts $|T|$, by letting $a_{i}$ be the number of $b_{j}$ that are at least $i$. By stars and bars, $|T|=\\binom{k+n-1}{k}$.\n\nOne way to think ... | United States | HMMT Spring 2021 Team Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinator... | null | proof and answer | (k+n-1 choose k) | |
0fsv | Problem:
Mit Hilfe der drei Buchstaben I, M, O werden Wörter der Länge $n$ gebildet. Wieviele solche Wörter der Länge $n$ gibt es, in denen keine benachbarten M's vorkommen? | [
"Solution:\n\nWir nennen ein Wort zulässig, wenn es keine zwei benachbarten M's enthält. Sei $a_{n}$ die Anzahl zulässiger Wörter der Länge $n$ und $b_{n}$ die Anzahl zulässiger Wörter der Länge $n$, die nicht mit einem $M$ beginnen. Nehme ein zulässiges Wort der Länge $n$ und entferne den ersten Buchstaben, das ve... | Switzerland | IMO - Selektion | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | (1/6)*[(3+2*sqrt(3))*(1+sqrt(3))^n + (3-2*sqrt(3))*(1-sqrt(3))^n] | |
02s1 | Problem:
Considere o diagrama ilustrado abaixo:
Augusto gosta de contar caminhos partindo de algum ponto, chegando no ponto $A$ e nunca passando por um mesmo vértice duas vezes. Para isso, ele representa um caminho pela sequência dos pontos que o caminho visita. Por exemplo, o caminho pontilha... | [
"Solution:\na) Os caminhos inusitados saindo de $D$ e chegando em $A$ são $D C B A$, $D B A$ e $D C A$.\n\nb) Um caminho inusitado saindo de $E$, tem que visitar logo em seguida um dos pontos $D$ ou $C$. No caso em que ele visita o ponto $D$, a sua continuação até o ponto $A$ pode ser qualquer um dos caminhos inusi... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | a) 3; c) 89 | |
0kdg | Problem:
How many six-digit multiples of $27$ have only $3$, $6$, or $9$ as their digits? | [
"Solution:\nDivide by $3$. We now want to count the number of six-digit multiples of $9$ that only have $1$, $2$, or $3$ as their digits. Due to the divisibility rule for $9$, we only need to consider when the digit sum is a multiple of $9$. Note that $3 \\cdot 6 = 18$ is the maximum digit sum.\n\nIf the sum is $18... | United States | HMMO 2020 | [
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 51 | |
0h4g | 1. Compare $A$ to $0$, where:
$$
a)\ A = 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \dots + 2012 + 2013 - 2014 - 2015 + 2016;
$$
$$
b)\ A = \frac{1}{1} - \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + \frac{1}{9} - \frac{1}{10} - \frac{1}{11} + \dots + \frac{1}{2012... | [
"a.\nIf we split all numbers into $504$ groups of $4$, from left to right, each group will have numbers of the type:\n$$\n(4k+1)-(4k+2)-(4k+3)+(4k+4)=0.\n$$\nAs we can see, the sum of numbers in each group is $0$, therefore, $A = 0$.\n\nb.\nLike in the previous question, split all numbers into groups of $4$: $\\lef... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | a) A = 0; b) A > 0 | |
0g52 | Problem:
Determine all triples $(a, b, p)$ of positive integers where $p$ is prime and the equation
$$
(a+b)^{p} = p^{a} + p^{b}
$$
is satisfied. | [
"Solution:\n$(a, b, p) = (1, 1, 2)$ is the only solution. Let's split the problem into two cases.\n\n- Case 1: $a = b$\n\nThe equation simplifies into $2^{p} a^{p} = 2 p^{a}$, and since $4 \\mid 2^{p}$, $2 \\mid p^{a}$ which implies that $p = 2$. Plugging this new piece of information in the initial equation yields... | Switzerland | Second round 2023 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (a, b, p) = (1, 1, 2) | |
0ewa | Problem:
Given integers $a_0$, $a_1$, ..., $a_{100}$, satisfying $a_1 > a_0$, $a_1 > 0$, and $a_{r + 2} = 3 a_{r + 1} - 2 a_r$ for $r = 0$, $1$, ..., $98$. Prove $a_{100} > 2^{99}$. | [
"Solution:\n\nAn easy induction gives $a_r = (2^{r} - 1)a_1 - (2^{r} - 2)a_0$ for $r = 2$, $3$, ..., $100$. Hence, in particular, $a_{100} = (2^{100} - 2)(a_1 - a_0) + a_1$. But $a_1$ and $(a_1 - a_0)$ are both at least $1$. Hence result."
] | Soviet Union | 2nd ASU | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0l2j | Problem:
Compute
$$
\sum_{i=1}^{4} \sum_{t=1}^{4} \sum_{e=1}^{4}\left\lfloor\frac{i t e}{5}\right\rfloor
$$ | [
"Solution:\nNote that $5$ never divides $ite$ because $5$ is prime. Thus, we can pair up the terms: since $ite \\equiv -(5-i) t e \\pmod{5}$,\n$$\n\\left\\lfloor\\frac{i t e}{5}\\right\\rfloor+\\left\\lfloor\\frac{(5-i) t e}{5}\\right\\rfloor=\\frac{i t e}{5}+\\frac{(5-i) t e}{5}-1\n$$\nThus\n$$\n\\begin{aligned}\n... | United States | HMMT November 2024 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | 168 | |
0bap | The numbers $x$, $y$, $z$, $t$, $a$ and $b$ are positive integers, so that $xt - yz = 1$ and $\frac{x}{y} > \frac{a}{b} > \frac{z}{t}$. Prove that $ab \ge (x+z)(y+t)$. | [
"From $\\frac{x}{y} > \\frac{a}{b}$ follows that $xb > ya$, hence $xb - ya \\ge 1$. In the same way, $at - bz \\ge 1$. Multiplying the first inequality by $t$, the second one by $y$ and adding the two relations yields $bxt - byz \\ge t + y$, that is $b \\ge t + y$.\n\nIn the same way, $a \\ge x + z$. These two ineq... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0jxy | Problem:
Compute the number of functions $f:\{1,2, \ldots, 9\} \rightarrow \{1,2, \ldots, 9\}$ which satisfy $f(f(f(f(f(x))))) = x$ for each $x \in \{1,2, \ldots, 9\}$. | [
"Solution:\n\nAll cycle lengths in the permutation must divide $5$, which is a prime number. Either $f(x) = x$ for all $x$, or there exists exactly one permutation cycle of length $5$. In the latter case, there are $\\binom{9}{5}$ ways to choose which numbers are in the cycle and $4!$ ways to create the cycle. The ... | United States | HMMT November | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | 3025 | |
0jbr | Problem:
How many ways are there to arrange three indistinguishable rooks on a $6 \times 6$ board such that no two rooks are attacking each other? (Two rooks are attacking each other if and only if they are in the same row or the same column.) | [
"Solution:\n\nThere are $6 \\times 6 = 36$ possible places to place the first rook. Since it cannot be in the same row or column as the first, the second rook has $5 \\times 5 = 25$ possible places, and similarly, the third rook has $4 \\times 4 = 16$ possible places. However, the rooks are indistinguishable, so th... | United States | HMMT November 2012 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 2400 | |
0al2 | Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a function satisfying the following property: If $A$, $B$, $C \in \mathbb{R}^2$ are the vertices of an equilateral triangle with sides of length $1$, then $f(A) + f(B) + f(C) = 0$. Prove that $f(x) = 0$ for all $x \in \mathbb{R}^2$. | [] | North Macedonia | Team Selection Test for IMO | [
"Algebra > Algebraic Expressions > Functional Equations",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof only | null | |
00f9 | Given are two tangent circles and a point $P$ on their common tangent perpendicular to the lines joining their centres. Construct with ruler and compass all the circles that are tangent to these two circles and pass through the point $P$. | [
"Throughout this problem, we will assume that the given circles are externally tangent, since the problem does not have a solution otherwise.\nLet $\\Gamma_{1}$ and $\\Gamma_{2}$ be the given circles and $T$ be their tangency point. Suppose $\\omega$ is a circle that is tangent to $\\Gamma_{1}$ and $\\Gamma_{2}$ an... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1991 | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0e96 | Find all real numbers $x$ that satisfy the equation
$$
\log_{\sin x} \left( \frac{1}{2} \sin 2x \right) = 2.
$$ | [
"The base of the logarithm has to be positive and different from $1$, so $\\sin x > 0$ and $\\sin x \\ne 1$. Also, $\\frac{1}{2} \\sin 2x > 0$ since the logarithm is only defined for positive numbers. When all this holds we may use the definition of the logarithm to transform the equation into the equivalent equati... | Slovenia | National Math Olympiad 2013 - First Round | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | x = pi/4 + 2k*pi, where k is any integer | |
0inv | Problem:
The positive integer $n$ is such that the numbers $2^{n}$ and $5^{n}$ start with the same digit when written in decimal notation; determine this common leading digit. | [
"Solution:\n\nAnswer: $3$. Note $1 = 1^{2} < 2^{2} < 3^{2} < 10 < 4^{2} < \\cdots < 9^{2} < 10^{2} = 100$. Divide $2^{n}$ and $5^{n}$ by $10$ repeatedly until each is reduced to a decimal number less than $10$ but at least $1$; call the resulting numbers $x$ and $y$. Since $(5^{n})(2^{n}) = 10^{n}$, either $x y = 1... | United States | $10^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3 | |
0i6n | Problem:
If we pick (uniformly) a random square of area $1$ with sides parallel to the $x$- and $y$-axes that lies entirely within the $5$-by-$5$ square bounded by the lines $x=0$, $x=5$, $y=0$, $y=5$ (the corners of the square need not have integer coordinates), what is the probability that the point $(x, y) = (4.5, ... | [
"Solution:\n\nThe upper-left corner of the unit square is picked uniformly from the square $0 \\leq x \\leq 4$; $1 \\leq y \\leq 5$, and for it to contain the desired point it must lie in the square $3.5 \\leq x \\leq 4$; $1 \\leq y \\leq 1.5$. The answer is the ratio of the squares' areas, $\\frac{1}{4} / 16 = \\f... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 1/64 | |
06xp | Let $ABC$ be a triangle with $AC > BC$. Let $\omega$ be the circumcircle of triangle $ABC$ and let $r$ be the radius of $\omega$. Point $P$ lies on segment $AC$ such that $BC = CP$ and point $S$ is the foot of the perpendicular from $P$ to line $AB$. Let ray $BP$ intersect $\omega$ again at $D$ and let $Q$ lie on line ... | [
"Solution 1 (Similar Triangles).\n\nFirst observe that\n$$\n\\angle DPA = \\angle BPC \\stackrel{CP=CB}{=} \\angle CBP = \\angle CBD = \\angle CAD = \\angle PAD\n$$\nso $DP = DA$. Thus there is a symmetry in the problem statement swapping $(A, D) \\leftrightarrow (B, C)$.\nLet $O$ be the ce... | IMO | International Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
03s4 | Given real numbers $a$, $b$, $c$, satisfying $a + b + c = 1$, prove that
$$10(a^3 + b^3 + c^3) - 9(a^5 + b^5 + c^5) \ge 1.$$ (posed by Li Shenghong) | [
"Since $\\sum a^3 = 1 - 3\\Pi(a+b)$, $\\sum a^5 = 1 - 5\\Pi(a+b)[\\sum a^2 + \\sum ab]$, therefore, the original inequality holds\n$$\n\\begin{align*}\n&\\Leftrightarrow 10[1 - 3\\Pi(a+b)] - 9[1 - 5\\Pi(a+b)(\\sum a^2 + \\sum ab)] \\ge 1 \\\\\n&\\Leftrightarrow 45\\Pi(a+b)(\\sum a^2 + \\sum ab) \\ge 30\\Pi(a+b) \\\... | China | China Western Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0atd | Problem:
Suppose $\alpha, \beta \in (0, \pi / 2)$. If $\tan \beta = \frac{\cot \alpha - 1}{\cot \alpha + 1}$, find $\alpha + \beta$. | [] | Philippines | Philippine Mathematical Olympiad | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | π/4 | |
04ct | For which $n \in \mathbb{N}$ exist an angle $\alpha$ and a convex $n$-gon with angles $\alpha, 2\alpha, \dots, n\alpha$? | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | n = 3 or 4 | |
0d73 | Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that
$$
f(2 m+f(m)+f(m) f(n))=n f(m)+m
$$
for any integers $m, n$. | [
"Let $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ satisfy the given functional equation. Putting $n=0$ in this equation, we get\n$$\nf(2 m+f(m)+f(m) f(0))=m \\quad \\forall m \\in \\mathbb{Z} ;\n$$\ntherefore, $f$ is surjective, so there exists $u \\in \\mathbb{Z}$ such that $f(u)=-1$.\nWith $m=u$, the given equation ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | f(n) = n - 2 for all integers n | |
03vq | Find the maximum positive number $M$ such that for every $n \in \mathbb{N}^*$, there are positive numbers $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_n$ satisfying
$$
(a) \sum_{k=1}^{n} b_k = 1,\ 2b_k \ge b_{k-1} + b_{k+1},\ k = 2, 3, \dots, n-1,
$$
$$
(b) a_k^2 \le 1 + \sum_{i=1}^{k} a_i b_i,\ k = 1, 2, \dots, n,
$... | [
"Firstly, we prove that\n$$\n\\max_{1 \\le k \\le n} a_k < 2,\\ \\text{ and } \\max_{1 \\le k \\le n} b_k < \\frac{2}{n-1}.\n$$\nLet $L = \\max_{1 \\le k \\le n} a_k$. From (b) and $\\sum_{k=1}^{n} b_k = 1$, we get $L^2 \\le 1 + L$, so $L < 2$.\nLet $b_m = \\max_{1 \\le k \\le n} b_k$. Then by using $2b_k \\ge b_{k... | China | China National Team Selection Test | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 3/2 | |
096v | Problem:
Fie $a$ şi $b$ două drepte paralele. Cercul $\Omega$ este tangent la dreapta $a$ în punctul $A$ şi intersectează dreapta $b$ în punctele distincte $B$ şi $C$. Punctul $T$ este situat pe dreapta $a$. Dreptele $B T$ şi $C T$ intersectează din nou cercul $\Omega$ în punctele $M$ şi, respectiv, $N$. Să se arate, ... | [
"Solution:\n\n1) Fie $P$ punctul de intersecţie al dreptelor $a$ şi $M N$ (figura alăturată) şi fie $m(\\angle T B C)=\\alpha$. Cum $a \\| b$, rezultă $m(\\angle A T B)=\\alpha$.\n\n2) Patrulaterul $M B C N$ este înscris în cerc, deci $\\alpha + m(\\angle M N C)=180^{\\circ}$. Dar şi $m(\\angle M N C)+m(\\angle P N... | Moldova | A 63-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
070s | Problem:
Given points $P_1, P_2, \ldots, P_n$ on a line we construct a circle on diameter $P_i P_j$ for each pair $i, j$ and we color the circle with one of $k$ colors. For each $k$, find all $n$ for which we can always find two circles of the same color with a common external tangent. | [
"Solution:\n\nThere are $n-1$ circles with diameter $P_i P_{i+1}$. Obviously, each pair has a common tangent. If $n-1 > k$, then two of them must have the same color.\n\nIf $n-1 \\leq k$, then color all circles with diameter $P_i P_j$ and $i < j$ with color $i$. Then if two circles have the same color, then both ha... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All n such that n ≥ k + 2 | |
09ws | Problem:
Ward en Gabriëlle spelen een spel op een groot vel papier. In het begin staan er 999 enen op het papier geschreven. Ward en Gabriëlle zijn om en om aan de beurt, waarbij Ward begint. Een speler die aan de beurt is, mag twee getallen $a$ en $b$ van het papier uitkiezen waarvoor geldt $\operatorname{ggd}(a, b)=... | [
"Solution:\n\nGabriëlle kan winnen met de volgende strategie: ze kiest steeds de grootste twee getallen op het papier als $a$ en $b$. We bewijzen met inductie naar $k$ dat ze dit altijd mag doen en dat na haar $k$-de zet het getal $2 k+1$ en verder $998-2 k$ enen op het papier staan.\n\nIn de eerste beurt van Ward ... | Netherlands | IMO-selectietoets II | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | Gabriëlle | |
0b76 | Let $a$, $b$, $c$ be integers larger than $1$. Prove that
$$
a(a-1) + b(b-1) + c(c-1) \leq (a + b + c - 4)(a + b + c - 5) + 4.
$$ | [
"The inequality is equivalent to\n$$\na^2 + b^2 + c^2 - a - b - c \\leq (a + b + c)^2 - 9(a + b + c) + 24,\n$$\nor\n$$\n0 \\leq ab + bc + ca - 4(a + b + c) + 12.\n$$\nSince $ab - 2a - 2b + 4 = (a - 2)(b - 2)$, summing together with the similar relations implies $(ab - 2a - 2b + 4) + (bc - 2b - 2c + 4) + (ca - 2c - ... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0jtp | Problem:
The three points $A$, $B$, $C$ form a triangle. $AB = 4$, $BC = 5$, $AC = 6$. Let the angle bisector of $\angle A$ intersect side $BC$ at $D$. Let the foot of the perpendicular from $B$ to the angle bisector of $\angle A$ be $E$. Let the line through $E$ parallel to $AC$ meet $BC$ at $F$. Compute $DF$. | [
"Solution:\n\nAnswer: $\\frac{1}{2}$\n\nSince $AD$ bisects $\\angle A$, by the angle bisector theorem $\\frac{AB}{BD} = \\frac{AC}{CD}$, so $BD = 2$ and $CD = 3$.\n\nExtend $BE$ to hit $AC$ at $X$. Since $AE$ is the perpendicular bisector of $BX$, $AX = 4$.\n\nSince $B$, $E$, $X$ are collinear, applying Menelaus' T... | United States | HMMT November 2016 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 1/2 | |
055t | Is there a positive integer $n$ for which it is possible to write a number $-1$, $0$ or $1$ into each cell of an $n \times n$ table in such a way that every integer from $-n$ to $n$ occurs at least once among the row sums, column sums and the two sums of the numbers on one long diagonal? If yes then find the least such... | [
"Suppose that an $n \\times n$ table is filled with numbers $-1$, $0$, and $1$ in such a way that the conditions are met. The sums $n$ and $-n$ can be obtained only from a row, column or diagonal with all $1$s and all $-1$s, respectively, whence $n$ and $-n$ cannot arise as sums of different kind (one as a row sum ... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 6 | |
06ob | How many solutions are there to the equation $\log_3 x = 3 \sin(3\pi x)$? | [
"Answer: 81\nA sketch of the graphs of $y = \\log_3 x$ and $y = 3 \\sin (3\\pi x)$ is as follows:\n\n\n\nNote that the right hand side of the equation (i.e. $3\\sin(3\\pi x)$) must be between $-3$ and $3$. Hence any solution satisfies $-3 \\le \\log_3 x \\le 3$, or $\\frac{1}{27} \\le x \\l... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Precalculus > Functions",
"Precalculus > Trigonometric functions"
] | null | proof and answer | 81 | |
0kmm | Problem:
Let $P$ be a point on segment $BC$ of triangle $\triangle ABC$. Let $O_1$ and $O_2$ be the respective circumcenters of $\triangle ABP$ and $\triangle ACP$. Given $BC = O_1O_2$, show that some angle of $\triangle ABC$ is more than $75^\circ$. | [
"Solution:\n\nWLOG, assume $O_1$ is at least as close to $BC$ as is $O_2$. Let $M, N$, and $Q$ be the respective midpoints of $\\overline{BP}$, $\\overline{CP}$, and $\\overline{AP}$, and let $T$ be the foot of the altitude from $O_1$ to $\\overline{O_2N}$.\n\nAs $\\overline{O_1M}$ and $\\overline{O_2N}$ are the pe... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ffi | Problem:
Hallar todas las funciones $f$ definida en el conjunto de los números reales, que toman valores reales positivos y que satisfacen las condiciones
1) $\quad f(x f(y))=y f(x)$ para todo $x, y$ positivos,
2) $\quad f(x) \rightarrow 0$ si $x \rightarrow \infty$. | [] | Spain | International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | All such functions are given by f(x) = 1/x for x > 0; on x ≤ 0, f can be defined arbitrarily with positive values. | |
0glf | Each point on the plane is colored either red, green or blue. Prove that there exists an isosceles triangle where all vertices are the same color. | [
"Consider a circle $\\omega$ with center $O$. Without loss of generality, let $O$ be red. If there exist two red points on $\\omega$ not belonging to the same diameter, then these points together with $O$ form a red isosceles triangle.\n\n\n\nOn the other hand, if $\\omega$ contains at most... | Thailand | The 13th Thailand Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
02sd | Problem:
O pequeno Abel ganhou de presente um tabuleiro $2 \times n$ e $n$ fichas de tamanho $2 \times 1$. Por exemplo, a figura a seguir mostra o caso em que $n=10$, isto é, quando Abel tem um tabuleiro $2 \times 10$ e 10 fichas de tamanho $2 \times 1$.
Ele brinca de preencher o tabuleiro us... | [
"Solution:\n\na) Podemos contar facilmente os primeiros casos e observar que $a_{1}=1, a_{2}=2$ e $a_{3}=3$, como mostra a figura abaixo.\n\n\nb) Ao começar a preencher o seu tabuleiro $2 \\times 10$, Abel pode proceder de duas maneiras distintas:\n- Abel pode começar colocando uma ficha ve... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | a1=1, a2=2, a3=3; a10=89 | |
0737 | Let $a_1, a_2, \dots, a_n$ be an arbitrary arrangement of numbers $1, 2, \dots, n$ on a circle. Find
$$
\min \sum_{j=1}^{n} |a_j - a_{j+1}| \quad \text{and} \quad \max \sum_{j=1}^{n} |a_j - a_{j+1}|,
$$
where $a_{n+1} = a_1$ and the extrema are taken over all possible arrangements of $1, 2, \dots, n$. | [
"**Minimum:** Consider $1$ and $n$ on the circle. They divide the circle into two arcs. The sum of the numbers on either of the arc is at least $n-1$. Suppose for example the numbers $1 = b_1, b_2, \\dots, b_k = n$ appear on one of the arcs between $1$ and $n$, in that order. Then the sum of absolute differences of... | India | Indija TS 2007 | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Minimum: 2(n − 1). Maximum: n^2/2 if n is even; (n^2 − 1)/2 if n is odd. | |
0e4x | A $4 \times 4$ board is divided into 16 squares. Onto this board we place several tiles like the one in the picture
(the tiles can be rotated),
each tile covering two squares. At least how many tiles do we need to place onto the board, so that every uncovered square will be adjacent to at leas... | [
"We can place four tiles as shown in the first picture. Each uncovered square has at least one covered neighbour. Now, let us show that this cannot be the case if we use less than four tiles. Put a tile onto the board and mark all the neighbouring squares. Each row contains at most three squares that are either cov... | Slovenia | National Math Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4 | |
0d6y | Let $c$ be a given real number. Find all polynomials $P$ with real coefficients such that
$$
(x+1) P(x-1)-(x-1) P(x)=c \text{ for all } x \in \mathbb{R} .
$$ | [
"In terms of $G(x) := P(x) - \\frac{c}{2}$, the given condition can be rewritten as\n$$\n(x+1) G(x-1) = (x-1) G(x) \\text{ for all } x \\in \\mathbb{R} .\n$$\nIt follows immediately (with $x = \\pm 1$) that $G(0) = 0$ and $G(-1) = 0$. Therefore, $G(x) = x(x+1) Q(x)$ for some $Q \\in \\mathbb{R}[x]$. Then\n$$\nQ(x-1... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | P(x) = k x(x+1) + c/2 for any real constant k | |
09zw | Problem:
In een scherphoekige driehoek $A B C$ geldt $|A B|>|C A|>|B C|$. De punten $D, E$ en $F$ zijn de voetpunten van de hoogtelijnen vanuit respectievelijk $A, B$ en $C$. De lijn door $F$ evenwijdig aan $D E$ snijdt $B C$ in $M$. De bissectrice van $\angle M F E$ snijdt $D E$ in $N$. Bewijs dat $F$ het middelpunt ... | [
"Solution:\n\n\n\nVanwege de lengte-eis in de opgave ligt de configuratie vast: $M$ ligt op de halfrechte $C B$ voorbij $B$, en $N$ ligt op de halfrechte $E D$ voorbij $D$. Zie de figuur. Noem $\\alpha=\\angle B A C$ en $\\beta=\\angle A B C$. Noem verder $H$ het hoogtepunt van de driehoek ... | Netherlands | IMO-selectietoets II | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
09mo | Let $P(x)$ and $Q(x)$ be polynomials with non-negative real coefficients, and let $P'(x)$ denote the derivative of $P(x)$. Suppose that we have $P(0) = Q(0) = 0$ and $Q(1) \le 1 \le P'(0)$.
(1) Prove that $0 \le Q(x) \le x \le P(x)$ for all $0 \le x \le 1$.
(2) Prove that $P(Q(x)) \le Q(P(x))$ for all $0 \le x \le 1$.
... | [
"Since $P(0) = Q(0) = 0$ and the coefficients of $P(x)$ and $Q(x)$ are non-negative, we see that the functions $P(x)/x$ and $Q(x)/x$ are increasing for $x > 0$.\nLet $0 \\le x \\le 1$.\n(1) For $x = 0$, we have $Q(0) = 0 = P(0)$. For $0 < x \\le 1$, we have\n$$0 \\le Q(x)/x \\le Q(1)/1 \\le 1 \\le P'(0) \\le P(x)/x... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof only | null | |
0hil | Inside the triangle $ABC$ there exists a point $O$ such that $\angle BOC = 90^\circ - \angle BAC$. The rays $BO$ and $CO$ intersect the sides $AC$ and $AB$ at the points $K$ and $L$ respectively. The points $K_1$ and $L_1$ are selected on the segments $LC$ and $BK$ respectively, so that $BK_1 = K_1K$ and $CL_1 = L_1L$.... | [
"Let $K_2$ and $L_2$ be the midpoints of the segments $BK$ and $CL$ respectively (fig. 18). Then $MK_2 \\parallel AC$ and $ML_2 \\parallel AB$. Hence, $\\angle K_2ML_2 = \\angle BAC$. From the isosceles triangles $BK_1K$ and $CL_1L$ we have that $\\angle L_1L_2K_1 = \\angle K_1K_2L_1 = 90^\\circ$ and then $L_1, L_2... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
086j | Problem:
Francesco e Andrea decidono di consultare l'oracolo matematico per sapere se hanno delle coppie $(x, y)$ di numeri (reali) fortunati. Per determinare la coppia (o le coppie) di numeri fortunati, l'oracolo chiede sia a Francesco che a Andrea il giorno $(g)$ e mese $(m)$ di nascita, dopodiché per ciascuno di lo... | [
"Solution:\n\nLa risposta è (D). Esaminiamo la situazione di Francesco: perché il sistema ammetta infinite soluzioni è necessario che la seconda equazione sia equivalente alla prima, ovvero che differiscano al più per una costante moltiplicativa. Il termine noto della seconda equazione è $362 = 181 \\times 2$ quind... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | D | |
052y | A boardgame board consists of 10 squares in a row that are numbered 1 to 10. On some square there is a button. In one move it is allowed to move the button to a square whose number is either smaller by 2 or 2 times bigger. Does there exist an initial location for the button that allows the player to visit all squares o... | [
"No move allows the button to be placed to the square number 9. Therefore the button should start from there to have any hope. If on some later move the button is placed on an even-numbered square, then it will also stay on an even-numbered square on every move that follows. Therefore all the odd-numbered squares m... | Estonia | Open Contests | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
0kc9 | Problem:
The Fibonacci numbers are defined by $F_{0}=0$, $F_{1}=1$, and $F_{n}=F_{n-1}+F_{n-2}$ for $n \geq 2$. There exist unique positive integers $n_{1}$, $n_{2}$, $n_{3}$, $n_{4}$, $n_{5}$, $n_{6}$ such that
$$
\sum_{i_{1}=0}^{100} \sum_{i_{2}=0}^{100} \sum_{i_{3}=0}^{100} \sum_{i_{4}=0}^{100} \sum_{i_{5}=0}^{100} ... | [
"Solution:\nWe make use of the identity\n$$\n\\sum_{i=0}^{\\ell} F_{i}=F_{\\ell+2}-1\n$$\n(easily proven by induction) which implies\n$$\n\\sum_{i=k}^{\\ell} F_{i}=F_{\\ell+2}-F_{k+1}\n$$\nApplying this several times yields\n$$\n\\begin{aligned}\n& \\sum_{i_{1}=0}^{100} \\sum_{i_{2}=0}^{100} \\sum_{i_{3}=0}^{100} \... | United States | HMMT February 2020 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 1545 | |
05sp | Problem:
Soit $N > 1$. Alice et Bob jouent au jeu suivant. $2N$ cartes numérotées de $1$ à $2N$ sont mélangées puis disposées dans une ligne, de manière à ce que les faces numérotées soient visibles. Chacun à leur tour, Alice et Bob choisissent une carte, soit celle tout à droite soit celle tout à gauche de la ligne, ... | [
"Solution:\n\nOn commence par tester le jeu pour des petites valeurs, par exemple pour un jeu composé de $2 \\times 3$ cartes. Après plusieurs essais, on s'aperçoit par exemple que la stratégie qui consiste à prendre toujours la plus grande carte parmi la carte située la plus à gauche et la carte située la plus à d... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0hez | Say a pair of natural numbers $(a, b)$ is *interesting* if there exists natural number $n$, such that minimal prime divisor of $a + n$ equals maximal prime divisor of $b + n$. Find all pairs of natural numbers that are interesting. | [
"Let us now assume that $|a-b| \\neq 1$. We will show that the pair $(a, b)$ is interesting. Let's consider the smallest prime number $p$ dividing $|a-b|$. Denote by $p_1 < p_2 < \\ldots < p_s$ all prime numbers less than $p$. It is clear that they may not exist. From the choice of $p$ it follows that none of $p_1,... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All pairs (a, b) with |a − b| ≠ 1 | |
0h2b | Given natural number $k > 1$. Find all integer pairs $(x, y)$, that satisfy the following equation:
$$
y^k = x^2 + x.
$$ | [
"We have $y^k = x(x+1)$. If $x = 0$ and $x = -1$ then $y = 0$.\n\nSuppose that $x \\neq 0, -1$.\n\n$x$ and $x+1$ are coprime, therefore $x = a^k$ and $x+1 = b^k$ for some integer nonzero $a, b$, that are of the same sign. Then\n$$\n(x+1) - x = 1 = b^k - a^k = (b - a)(b^{k-1} + b^{k-2}a + \\dots + a^{k-1})\n$$\nWhic... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All solutions are (x, y) = (0, 0) and (−1, 0) for any integer exponent greater than one. | |
09qo | Problem:
Laat $\Gamma_{1}$ en $\Gamma_{2}$ twee snijdende cirkels met middelpunten respectievelijk $O_{1}$ en $O_{2}$ zijn, zodat $\Gamma_{2}$ het lijnstuk $O_{1} O_{2}$ snijdt in een punt $A$. De snijpunten van $\Gamma_{1}$ en $\Gamma_{2}$ zijn $C$ en $D$. De lijn $A D$ snijdt $\Gamma_{1}$ een tweede keer in $S$. De ... | [
"Solution:\n\nOplossing I. We gaan $\\angle O_{1} E A$ berekenen. Omdat driehoek $O_{1} E D$ gelijkbenig is met top $O_{1}$ en omdat $A E F D$ een koordenvierhoek is, geldt\n$$\n\\angle O_{1} E A=\\angle O_{1} E D-\\angle A E D=90^{\\circ}-\\frac{1}{2} \\angle E O_{1} D-\\angle A F D .\n$$\nVerder is\n$$\n\\begin{a... | Netherlands | Dutch TST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0lbl | Let $a, b, c \ge 2$ be real numbers, satisfying the condition $a + b + c = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 8$. Prove that we have the inequality
$$
3(ab + bc + ca) \le 81 \le (a + b + c)^2.
$$ | [
"Putting $x = a + b + c$ then we have $a + b + c = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + 8 \\ge \\frac{9}{a + b + c} + 8 \\Rightarrow x \\ge \\frac{9}{x} + 8$.\nDirect manipulation leads to $(x - 9)(x + 1) \\ge 0$ or $x \\ge 9$.\nMoreover, $a + b + c = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + 8 < \\frac{... | Vietnam | Vietnamese Mathematical Competitions | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0cdo | On the set $A = [0, \infty)$, of all nonnegative real numbers, we consider three functions $f, g, h : A \to A$ and the binary operation $*: A \times A \to A$, defined by
$$
x * y = f(x) + g(y) + h(x) \cdot |x - y|, \quad \text{for any } x, y \ge 0.
$$
If $(A, *)$ is a commutative monoid:
a) show that the function $h$ i... | [
"a) Let $e$ be the unit element of the monoid $(A, *)$. Then\n$$\nf(0) + g(e) + h(0) \\cdot e = 0 * e = 0 \\quad \\text{and} \\quad f(e) + g(0) + h(e) \\cdot e = e * 0 = 0,\n$$\nso that $f(e) = g(e) = f(0) = g(0) = h(e) \\cdot e = h(0) \\cdot e = 0$, whence $e = e * e = f(e) + g(e) = 0$.\nThen $0 * x = x$ and $x * ... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Abstract Algebra > Other"
] | null | proof and answer | There are exactly two possibilities. Case 1: h(x) = 0 for all x ≥ 0, f(x) = x, g(x) = x, and x*y = x + y. Case 2: h(x) = 1/2 for all x ≥ 0, f(x) = x/2, g(x) = x/2, and x*y = max(x, y). | |
0hzi | Problem:
Let $f(x) = x + \frac{1}{2x + \frac{1}{2x + \frac{1}{2x + \cdots}}}$ for $x > 0$. Find $f(99) f'(99)$. | [
"Solution:\nAssume that the continued fraction converges (it does) so that $f(x)$ is well defined. Notice that\n$$\nf(x) - x = \\frac{1}{x + f(x)}\n$$\nso\n$$\nf(x)^2 - x^2 = 1,\n$$\nor\n$$\nf(x) = \\sqrt{1 + x^2}\n$$\n(we need the positive square root since $x > 0$).\n\nThus\n$$\nf'(x) = \\frac{x}{\\sqrt{1 + x^2}}... | United States | Harvard-MIT Math Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Functions"
] | null | final answer only | 99 | |
0bsl | Let $A \in \mathcal{M}_n(\mathbb{C})$ be an invertible matrix, with lines $L_1, L_2, \dots, L_n$. Consider the matrices $B \in \mathcal{M}_n(\mathbb{C})$ with lines $O, L_2, \dots, L_n$ and $C \in \mathcal{M}_n(\mathbb{C})$ with lines $L_2, \dots, L_n, O$, where $O$ denotes a line all of whose entries are zero. Let $D ... | [
"a) Consider the matrix $P \\in \\mathcal{M}_n(\\mathbb{C})$,\n$$\nP = \\begin{pmatrix}\n0 & 0 & 0 & \\cdots & 0 \\\\\n0 & 1 & 0 & \\cdots & 0 \\\\\n0 & 0 & 1 & \\cdots & 0 \\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n0 & 0 & 0 & \\cdots & 1\n\\end{pmatrix}.\n$$\nIt is not difficult to see that $D = ... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices"
] | English | proof only | null | |
0dab | Prove that in every triangle, there are two sides $x, y$ such that
$$
\frac{\sqrt{5}-1}{2} \leq \frac{x}{y} \leq \frac{\sqrt{5}+1}{2} .
$$ | [
"Let $a, b, c$ be the side lengths of the given triangle and we may assume that $a \\geq b \\geq c$. Denote $m=\\frac{a}{b}$ and $n=\\frac{b}{c}$. Note that\n$$\nm, n \\geq 1 > \\frac{\\sqrt{5}-1}{2} .\n$$\nWe will show that at least one of $m, n$ is less than or equal to $\\frac{\\sqrt{5}+1}{2}$. Assume that both ... | Saudi Arabia | Team selection tests for JBMO 2018 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
0gqv | Let $a$, $b$, $c$ be nonnegative real numbers satisfying $a^2 + b^2 + c^2 = 1$. Prove that
$$
\sqrt{a+b} + \sqrt{b+c} + \sqrt{c+a} \ge 5abc + 2
$$ | [
"First of all let us show that $\\sqrt{a+b} + \\sqrt{b+c} + \\sqrt{c+a} \\ge \\sqrt{7(a+b+c)-3}$ (1). Let $a+b+c = x$, then $ab+bc+ca = \\frac{x^2-1}{2}$ and since by Cauchy-Schwarz inequality $1 = a^2+b^2+c^2 \\le (a+b+c)^2 \\le 3(a^2+b^2+c^2) = 3$ we get that $1 \\le x \\le \\sqrt{3}$. It can be readily shown tha... | Turkey | Team Selection Test | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0l2h | Problem:
Equilateral triangles $ABF$ and $BCG$ are constructed outside regular pentagon $ABCDE$. Compute $\angle FEG$. | [
"Solution:\n\nWe have $\\angle FEG = \\angle AEG - \\angle AEF$.\n\nSince $EG$ bisects $\\angle AED$, we get $\\angle AEG = 54^{\\circ}$.\n\nNow, $\\angle EAF = 108^{\\circ} + 60^{\\circ} = 168^{\\circ}$.\n\nSince triangle $EAF$ is isosceles, this means $\\angle AEF = 6^{\\circ}$, so the answer is $54^{\\circ} - 6^... | United States | HMMT February 2024 Guts Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 48° | |
0aju | Solve the equation $x + y^2 + (\text{gcd}(x, y))^2 = x y \cdot \text{gcd}(x, y)$ in the set of natural numbers. | [
"We introduce the substitution $z = \\gcd(x, y)$ and we get the equation $x + y^2 + z^2 = x y z$. There exist natural numbers $a$ and $b$ such that $x = a z$ and $y = b z$. Then the equation gets the form $a z + b^2 z^2 + z^2 = a b z^3$, i.e. $a + b^2 z + z = a b z^2$. Since the right-hand side is divisible by $z$ ... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (5, 2), (5, 3) | |
0cqp | Let $D$ be a point on the side $AC$ of a triangle $ABC$. Let $E$ and $F$ be the points symmetric to $D$ with respect to the angle bisectors of $\angle A$ and $\angle C$, respectively. Prove that the midpoint of the segment $EF$ lies on the line $A_0C_0$, where $A_0$ and $C_0$ are the points of tangency of the incircle ... | [
"Пусть $B_0$ — точка касания вписанной окружности со стороной $AC$. Можно считать, что точка $D$ лежит на отрезке $AB_0$. Точки $A_0$ и $C_0$ симметричны точке $B_0$ относительно биссектрис углов $C$ и $A$ соответственно. Следовательно, точка $E$ лежит на отрезке $AC_0$, а точка $F$ — на продолжении отрезка $CA_0$,... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | English; Russian | proof only | null | |
0a1i | A triangular number is the number of circles you can place in an equilateral triangle of a certain height. For example, in the figure on the right, you can see that the fourth triangular number is $10$. Jasmine writes down the first $1000$ triangular numbers: $1$, $3$, $6$, $10$, $15$, $\ldots$.
How many of these numbe... | [
"6."
] | Netherlands | Junior Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | final answer only | 200 | |
0ilf | Problem:
Evaluate the functions $\phi(n)$, $\sigma(n)$, and $\tau(n)$ for $n=12$, $n=2007$, and $n=2^{2007}$. | [
"Solution:\n\nFor $n=12=2^{2} \\cdot 3^{1}$,\n$$\n\\phi(12)=2(2-1)(3-1)=4, \\quad \\sigma(12)=(1+2+4)(1+3)=28, \\quad \\tau(12)=(2+1)(1+1)=6\n$$\n\nFor $n=2007=3^{2} \\cdot 223$,\n$$\n\\phi(2007)=3(3-1)(223-1)=1332, \\quad \\sigma(2007)=(1+3+9)(1+223)=2912, \\quad \\tau(2007)=(2+1)(1+1)=6\n$$\n\nFor $n=2^{2007}$,\n... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | For 12: φ=4, σ=28, τ=6. For 2007: φ=1332, σ=2912, τ=6. For 2^2007: φ=2^2006, σ=2^2008−1, τ=2008. | |
0kci | Problem:
How many positive integers at most $420$ leave different remainders when divided by each of $5$, $6$, and $7$? | [
"Solution:\n\nNote that $210 = 5 \\cdot 6 \\cdot 7$ and $5$, $6$, $7$ are pairwise relatively prime. So, by the Chinese Remainder Theorem, we can just consider the remainders $n$ leaves when divided by each of $5$, $6$, $7$. To construct an $n$ that leaves distinct remainders, first choose its remainder modulo $5$,... | United States | HMMT February 2020 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | final answer only | 250 | |
0fo9 | Hallar para qué valores del número real $a$ todas las raíces del polinomio, en la variable $x$,
$$
x^3 - 2x^2 - 25x + a
$$ | [
"Sean $\\alpha$, $\\beta$ y $\\gamma$ las raíces del polinomio. Aplicando las fórmulas de Cardano-Vieta se tiene\n$$\n\\alpha + \\beta + \\gamma = 2, \\quad \\alpha\\beta + \\alpha\\gamma + \\beta\\gamma = -25.\n$$\nAhora bien\n$$\n\\alpha^2 + \\beta^2 + \\gamma^2 = (\\alpha + \\beta + \\gamma)^2 - 2(\\alpha\\beta ... | Spain | L Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Spanish | proof and answer | a = 50 | |
07lp | Let
$$
A = \{ (x, y, z) \in \mathbb{R}^3 : xyz = 1,\ x + y + z = 3 \},
$$
under the function
$$
F(x, y, z) = xy + yz + zx.
$$
Determine the image of the set $A$ under $F$. | [
"Let $F$ stand for a value of the function $F$. Clearly, $x, y, z \\in A$ iff the cubic $t^3 - 3t^2 + Ft - 1$ has three real roots. Normalise this to the form ($s = t - 1$)\n$$\ns^3 - (3-F)s - (3-F) = 0.\n$$\nIf $a, b, c$ are the roots of this, then they are real iff\n$$\n0 \\le 4(3-F)^3 - 27(3-F)^2 = (F-3)^2(4(3-F... | Ireland | Irska | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (-∞, -15/4] ∪ {3} | |
01qg | Pairwise distinct prime numbers $p$, $q$, $r$ satisfy the equality
$$
rp^3 + p^2 + p = 2rq^2 + q^2 + q.
$$ | [
"Answer: 2014.\nBy condition, $rp^3 = -(p^2 + p) + 2rq^2 + (q^2 + q)$, and $p^2 + p$, $q^2 + q$, $2rq^2$ are even for all natural $p$ and $q$, then $rp^3$ is also even. Therefore, since $r$ and $p$ are prime, we see that either $p = 2$ or $r = 2$.\n\nIf $p = 2$, then the initial equality has the form $8r + 4 + 2 = ... | Belarus | Final Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Intermediate Algebra > Quadratic functions",
... | English | proof and answer | 2014 | |
0gp1 | Each student in the class has chosen one mathematics and one physics problem out of $20$ mathematics and $11$ physics problems such that different students choose different pairs of problems. Given that for each student, at least one of the problems chosen by him is chosen by at most one more student, determine the max... | [
"For $1 \\le i \\le 20$ and $1 \\le j \\le 11$ we define $a_{i,j}$ as follows; $a_{i,j} = 1$ if the $i$-th mathematics problem and $j$-th physics problem are chosen by some student, $a_{i,j} = 0$ otherwise. Now we can reformulate the problem: Find the maximal possible value of the expression\n$$\nA = \\sum_{i=1}^{2... | Turkey | 16th Junior Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 54 |
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