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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0imp | Let $S$ be a set containing $n^2 + n - 1$ elements, for some positive integer $n$. Suppose that the $n$-element subsets of $S$ are partitioned into two classes. Prove that there are at least $n$ pairwise disjoint sets in the same class. | [
"In order to apply induction, we generalize the result to be proved so that it reads as follows:\n\n**Proposition.** If the $n$-element subsets of a set $S$ with $(n+1)m - 1$ elements are partitioned into two classes, then there are at least $m$ pairwise disjoint sets in the same class.\n\n*Proof:* Fix $n$ and proc... | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
051d | Let $a$ be a real number, $0 \le a \le 1$. Prove that for any nonnegative integer $n$ the inequality $(n+1)a \le n + a^{n+1}$ holds. | [
"The inequality is equivalent to the inequality $na - n \\le a^{n+1} - a$, or the inequality $n(a - 1) \\le a(a - 1)(a^{n-1} + a^{n-2} + \\dots + 1)$. If $a = 1$, then the inequality obviously holds. If $a < 1$ then $a - 1 < 0$, and dividing both sides of the inequality by $a - 1$ we get an equivalent inequality $n... | Estonia | Estonian Math Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0ld9 | Solve the system of equations
$$
\begin{cases} 6x - y + z^2 = 3 \\ x^2 - y^2 - 2z = -1 \\ 6x^2 - 3y^2 - z - 2z^2 = 0 \end{cases} \quad (x, y, z \in \mathbb{R})
$$ | [
"From the given system, it follows that\n$$\n(6x^2 - 3y^2 - z - 2z^2) - 3(x^2 - y^2 - 2z + 1) - (6x - y + z^2 - 3) = 0.\n$$\nThis equation can be rewritten as $(x-z)(x+z-2) = 0$. Hence, $x = z$ or $x+z = 2$.\n\n* If $x = z$. The given system is equivalent to\n$$\n\\begin{cases} x^2 + 6x - y = 3, \\\\ x^2 - 2x - y^2... | Vietnam | VMO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | ((-5 + sqrt(33))/2, (-7 + sqrt(33))/2, (-5 + sqrt(33))/2), ((-5 - sqrt(33))/2, (-7 - sqrt(33))/2, (-5 - sqrt(33))/2), ((-7 + sqrt(65))/2, (9 - sqrt(65))/2, (-7 + sqrt(65))/2), ((-7 - sqrt(65))/2, (9 + sqrt(65))/2, (-7 - sqrt(65))/2) | |
0dvd | Problem:
Naj za polinom $p(x)$ s celimi koeficienti velja $p(3)=2$. Ali je število $p(2003)$ lahko popolni kvadrat? | [
"Solution:\n\nKer ima polinom $p$ cele koeficiente, $x-y$ deli $p(x)-p(y)$. Potem je $p(2003)-p(3) = (2003-3)k$ in $p(2003) = 2000k + 2$. Torej da $p(2003)$ pri deljenju s $4$ ostanek $2$ in ne more biti popolni kvadrat."
] | Slovenia | 47. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0gn3 | Find the smallest positive integer $x_1$ such that $2006$ divides $x_{2006}$, if
$$
x_{n+1} = x_1^2 + x_2^2 + \dots + x_n^2 \text{ for each integer } n \ge 1.
$$ | [] | Turkey | Team Selection Examination for the International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 472 | |
05of | Problem:
On a $n$ lampes, chacune avec un interrupteur. Chaque lampe fonctionne normalement, appuyer sur son interrupteur l'éteint si elle est allumée, et l'allume si elle est éteinte. Toutes les minutes on touche aux interrupteurs.
À la première minute : on appuie sur tous les interrupteurs $(1,2,3, \ldots, n)$.
À ... | [
"Solution:\n\nSoit $i \\in \\{1, \\ldots, n\\}$ le numéro d'une des lampes. On appuie sur l'interrupteur $i$ à la $k$-ième minute si et seulement si $k$ est un diviseur de $i$, donc on a appuyé $d(i)$ fois sur l'interrupteur $i$, avec $d(i)$ le nombre de diviseurs de $i$.\n\nDécomposons $i$ en facteurs premiers : $... | France | Envoi 1: Arithmétique | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Exactly those lamps whose numbers are perfect squares | |
07eh | In acute triangle $ABC$, $H$ is the orthocenter. Let $\omega$ be the circumcircle of triangle $BHC$, and $O'$ be its center. $\Gamma$ is the circle that passes through the midpoints of the edges of triangle $ABC$. Point $X$ is an arbitrary point lying on arc $BHC$ of $\omega$. Line $AX$ intersects $\Gamma$ for the seco... | [
"Let $B'$ and $C'$ be the midpoints of $AB$ and $AC$ respectively. $\\Gamma$ is known to be the Euler's circle (nine-point circle) of triangle $ABC$, and passes through $H'$, the midpoint of $AH$. Thus $\\Gamma$ is the circumcircle of triangle $B'-H'-C'$. Now it's easy to see that $\\Gamma$ is obtained when $\\omeg... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ga9 | 令 $a, b, c$ 為非負實數, 滿足 $(a+b)(b+c)(c+a) \neq 0$. 試求
$$
(a + b + c)^{2016} \left( \frac{1}{a^{2016} + b^{2016}} + \frac{1}{b^{2016} + c^{2016}} + \frac{1}{c^{2016} + a^{2016}} \right).
$$
的最小值.
Let $a, b, c$ be non-negative real numbers such that $(a+b)(b+c)(c+a) \neq 0$. Find the minimum of
$$
(a + b + c)^{2016} \left(... | [
"When $xy + yz + zx = 1$ and $(x + y)(y + z)(z + x) \\neq 0$, we have\n$$\n\\frac{1}{x+y} + \\frac{1}{y+z} + \\frac{1}{z+x} \\ge \\frac{5}{2}. \\qquad (1)\n$$\nNow, WLOG we can assume $a^{2016}b^{2016}+b^{2016}c^{2016}+c^{2016}a^{2016} = 1$. Then, we have\n$$\n\\frac{1}{a^{2016} + b^{2016}} + \\frac{1}{b^{2016} + c... | Taiwan | 二〇一六數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 5 * 2^{2015} | |
0jtl | Problem:
Among citizens of Cambridge there exist $8$ different types of blood antigens. In a crowded lecture hall are $256$ students, each of whom has a blood type corresponding to a distinct subset of the antigens; the remaining of the antigens are foreign to them.
Quito the Mosquito flies around the lecture hall, p... | [
"Solution:\n\nLet $n=8$.\n\nFirst, consider any given student $S$ and an antigen $a$ foreign to him/her. Assuming $S$ has been bitten, we claim the probability $S$ will suffer due to $a$ is\n$$\n1 - \\frac{2^{2^{n-1}+1} - 1}{2^{2^{n-1}} (2^{n-1} + 1)}\n$$\nIndeed, let $N = 2^{n-1}$ denote the number of students wit... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | (65024 + 2^{-119})/129 hours | |
09nu | Let the sum of the elements of a set $X$ be denoted by $S(X)$. How many ways can we divide the numbers $2^1, 2^2, \dots, 2^{10}$ into sets $A$ and $B$ such that the equation
$$
x^2 - S(A)x + S(B) = 0
$$
has a positive integer solution? (The sets $A$ or $B$ may be empty.)
*To divide the numbers $c_1, c_2, \dots, c_n$ in... | [
"*Answer: 2.*\nThe numbers $2^1, 2^2, \\dots, 2^{10}$ can be used to produce only and all even numbers from $0$ to $2046$. Thus, let $S(A) = 2n$ for some integer $0 \\le n \\le 1023$. Then $S(B) = 2^1+2^2+\\dots+2^{10}-S(A) = 2046-2n$. Moreover $x^2-S(A)x+S(B) = x^2-2nx+(2046-2n) = 0$ implies\n$$\n23 \\cdot 89 = 20... | Mongolia | MMO2025 Round 4 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 2 | |
08bv | Problem:
Sia $ABC$ un triangolo acutangolo e sia $D$ il piede della bisettrice uscente da $A$. Sia $\omega$ la circonferenza per $A$ tangente a $BC$ in $D$, e siano $E, F$ le intersezioni di $\omega$ con $AB, AC$ rispettivamente. Le tangenti a $\omega$ in $E$ e $F$ si intersecano in $P$. Sapendo che $PE=3$ e che il ra... | [
"Solution:\n\nLa risposta è (B). Sia $O$ il centro di $\\omega$; per ipotesi si ha $\\widehat{EAD}=\\widehat{FAD}$, quindi $D$ è il punto medio dell'arco $EF$ e in particolare $OD$ passa per il punto medio del segmento $EF$. Ne segue che $OD$ è l'asse di $EF$, perché è ad esso perpendicolare e passa per il suo punt... | Italy | Gara di Febbraio | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | B | |
00ua | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,
$$
x f(x + f(y)) = (y - x) f(f(x)).
$$ | [
"Answer: For any real $c$, $f(x) = c - x$ for all $x \\in \\mathbb{R}$ and $f(x) = 0$ for all $x \\in \\mathbb{R}$.\n\nLet $P(x, y)$ denote the proposition that $x$ and $y$ satisfy the given equation. $P(0, 1)$ gives us $f(f(0)) = 0$.\n\nFrom $P(x, x)$ we get that $x f(x + f(x)) = 0$ for all $x \\in \\mathbb{R}$, w... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | All solutions are the zero function and the family f(x) = c − x for any real constant c. | |
09yd | Problem:
Vind alle functies $f: \mathbb{R} \rightarrow \mathbb{R}$ met
$$
f(x+y f(x+y))=y^{2}+f(x) f(y)
$$
voor alle $x, y \in \mathbb{R}$. | [
"Solution:\n\nOplossing I. Merk op dat de functie $f(x)=0$ voor alle $x$ niet voldoet. Er is dus een $a$ met $f(a) \\neq 0$. Vul $x=a$ en $y=0$ in, dat geeft $f(a)=f(a) f(0)$, dus $f(0)=1$. Vul nu $x=1$ en $y=-1$ in, dat geeft $f(1-f(0))=1+f(1) f(-1)$. Omdat $f(0)=1$, staat hier $1=1+f(1) f(-1)$, dus $f(1)=0$ of $f... | Netherlands | IMO-selectietoets III | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 1 - x for all real x; f(x) = x + 1 for all real x | |
0kjr | Triangle $ABC$ has side lengths $AB = 11$, $BC = 24$, and $CA = 20$. The bisector of $\angle BAC$ intersects $\overline{BC}$ in point $D$ and intersects the circumcircle of $\triangle ABC$ in point $E \neq A$. The circumcircle of $\triangle BED$ intersects the line $AB$ in points $B$ and $F \neq B$. What is $CF$?
(A) 2... | [] | United States | AMC 12 B | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | C | |
0gwr | Let $H$ be a point which altitudes of the acute-angled triangle $ABC$ intersect at. Points $A_1, B_1, C_1$ are midpoints of the sides $BC, CA, \text{and } AB$ respectively. Let $A_2$
and $C_2$ be such points for which $A_2A \perp AC$ and $A_2C_1 \perp AB$, $C_2C \perp AC$ and $C_2A_1 \perp BC$. Prove the following:
a... | [
"Let $H_2$ be a midpoint of the segment $BH$. Let $w_A$, $w_C$ be circles of radius $A_2A$ and $C_2C$ with centers at points $A_2$, $C_2$ respectively. Then they are tangent to line $AC$ at endpoints of the segment $AC$ and pass through point $B$, since points $A_2$, $C_2$ are on the respective perpendicular bisect... | Ukraine | Ukrajina 2008 | [
"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 > Transformations > Homothety",
... | English | proof only | null | |
03v2 | The real polynomial $\varphi(x) = ax^3 + bx^2 + cx + d$ has three positive roots, and $\varphi(0) < 0$. Prove that
$$
2b^3 + 9a^2d - 7abc \le 0. \quad \textcircled{1}
$$ | [
"**Proof** We denote by $x_1, x_2, x_3$ the three positive roots of the polynomial $\\varphi(x) = ax^3 + bx^2 + cx + d$. By Vieta's theorem, we have\n$$\n\\begin{align*}\nx_1 + x_2 + x_3 &= -\\frac{b}{a}, \\\\\nx_1x_2 + x_2x_3 + x_3x_1 &= \\frac{c}{a}, \\\\\nx_1x_2x_3 &= -\\frac{d}{a}.\n\\end{align*}\n$$\n---\nAs $... | China | China Girls' Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
05gf | Problem:
Soit $ABC$ un triangle et $H$ son orthocentre. Le cercle circonscrit de $ABH$, de centre $O$, recoupe la droite $(BC)$ en $D$, la droite $(DH)$ intersecte la droite $(AC)$ en $P$. Montrer que le centre du cercle circonscrit de $APD$ est sur le cercle circonscrit de $ODB$.
 | [
"Solution:\n\nUne construction soigneuse nous permet de conjecturer que le centre du cercle $APD$ est également sur la droite $(AB)$, c'est donc tout naturellement que l'on introduit $O_1$, le point de la droite $(AB)$ qui est à égale distance de $A$ et de $D$, $c'$ est alors un bon candidat pour être le centre du ... | France | Préparation Olympique Française de Mathématiques | [
"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 | |
0dy4 | Jaka chooses a three-digit number $x$, composed of three different non-zero digits. He then takes a piece of paper and writes down all other three-digit numbers he can form from those three digits. The sum of the numbers on the paper is $3434$. Find all possible $x$. | [
"Denote the digits of $x$ by $a$, $b$ and $c$, so that $x = \\overline{abc}$. Three-digit numbers we can form from $a$, $b$ and $c$ are $\\overline{abc}$, $\\overline{acb}$, $\\overline{bac}$, $\\overline{bca}$, $\\overline{cab}$, $\\overline{cba}$ and their sum is $100(2a + 2b + 2c) + 10(2a + 2b + 2c) + (2a + 2b +... | Slovenia | Slovenija 2008 | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 784 | |
0krq | Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles whose sides are three of these line segments and whose vertices are three distinct points from among the... | [] | United States | AIME II | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 72 | |
06k6 | In a committee there are $n$ members. Each pair of members are either friends or enemies. Each committee member has exactly three enemies. It is also known that for each committee member, an enemy of his friend is automatically his own enemy. Find all possible value(s) of $n$. | [
"Suppose $A$ and $B$ are enemies. By the given condition, every other member cannot be a friend of both $A$ and $B$. Since each of $A$ and $B$ has 2 more enemies other than themselves, we must have $n \\le 2 + 2 \\times 2 = 6$. Also, as each member has 3 enemies, we have $n \\ge 4$, and there are $\\frac{3n}{2}$ pa... | Hong Kong | HKG TST | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 4, 6 | |
0kx4 | Problem:
Convex quadrilateral $ABCD$ satisfies $\angle CAB = \angle ADB = 30^{\circ}$, $\angle ABD = 77^{\circ}$, $BC = CD$, and $\angle BCD = n^{\circ}$ for some positive integer $n$. Compute $n$. | [
"Solution:\n\nLet $O$ be the circumcenter of $\\triangle ABD$. From $\\angle ADB = 30^{\\circ}$, we get that $\\triangle AOB$ is equilateral. Moreover, since $\\angle BAC = 30^{\\circ}$, we have that $AC$ bisects $\\angle BAO$, and thus must be the perpendicular bisector of $BO$. Therefore, we have $CB = CD = CO$, ... | United States | HMMT February | [
"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"
] | null | proof and answer | 68 | |
03wy | Let $\{x_n\}$ be a sequence such that $x_1 \in \{5, 7\}$, and $x_{n+1} \in \{5^{x_n}, 7^{x_n}\}$, for $n = 1, 2, \dots$. Determine all the possible cases of the last two digits of $x_{2009}$. | [
"Let $n = 2009$. Then we have the following three cases:\n\n(1) If $x_n = 7^{5^{x_{n-2}}}$, then $x_n \\equiv 7 \\pmod{100}$.\nSince both $5$ and $7$ are odd, $x_k$ ($1 \\le k \\le n$) all are odd. $5^{x_{n-2}} \\equiv 1 \\pmod{4}$, i.e. $5^{x_{n-2}} = 4k+1$ for some positive integer $k$.\nIf $k, m \\ge 0$, it foll... | China | China Western Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 07, 25, 43 | |
0e8x | A herd of deer consists of harts and hinds. Hinds represent $55\%$ of the herd, and their weight is $45\%$ of the total weight of the herd. How many times is the average weight of a hart greater than the average weight of a hind?
(A) $\frac{81}{40}$
(B) $\frac{3}{2}$
(C) $\frac{121}{81}$
(D) $\frac{11}{9}$
(E) $\frac{6... | [
"Assume that the herd consists of $x$ animals with the combined weight of $y$. Then there are $\\frac{55}{100}x$ hinds with the combined weight of $\\frac{45}{100}y$ and $\\frac{45}{100}x$ harts with the combined weight of $\\frac{55}{100}y$. The average weight of a hind is then $\\frac{45}{100}y : \\frac{55}{100}x... | Slovenia | National Math Olympiad 2013 - First Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
083r | Problem:
Per motivi dietetici, Piero deve mangiare ad ogni pranzo una quantità fissa di carboidrati provenienti da pane e/o pasta. Per totalizzare tale quantità, Piero può mangiare 80 grammi di pasta e 40 grammi di pane, oppure 100 grammi di pasta e 30 grammi di pane. Se volesse mangiare solo pasta, quanti grammi ne d... | [] | Italy | Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO BIENNIO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
0esh | Which one of the following is not a rational number?
(A) $\sqrt{4}$
(B) $\pi^0$
(C) $\sqrt{6.25}$
(D) $\sqrt{14}$
(E) $\sqrt{0.49}$ | [
"D The others are, in order, $2$; $1$; $2.5$ and $0.7$"
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | MCQ | D | |
05wy | Problem:
Un rectangle est divisé en dominos $1 \times 2$ et $2 \times 1$. Dans chaque domino, une diagonale est tracée. Deux diagonales n'ont jamais d'extrémité commune. Montrez que exactement deux coins du rectangle sont des extrémités de ces diagonales. | [
"Solution:\n\nRemarquons que ce motif (ainsi que toutes ses rotations et symétries) est impossible :\n\n\nEn effet, le seul moyen de compléter le carré bas-gauche est d'ajouter ce domino :\n\n\nDe même, le seul moyen de compléter le carré bas-droite est d'ajouter ce... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
07gt | Point $D$ is chosen on the *Euler line* of triangle $ABC$, inside the triangle. Let $E, F$ be the intersection points of $BD, AC$ and $CD, AB$, respectively. Point $X$ lies on the line $AD$ such that $\angle EXF = 180^\circ - \angle A$, also $A$ and $X$ are on the same side of $EF$. If $P$ is the second intersection of... | [
"$$\n\\frac{BD}{CD} \\cdot \\frac{\\sin(\\angle A - \\angle ABD)}{\\sin(\\angle A - \\angle ACD)} = \\frac{\\cos \\angle B}{\\cos \\angle C} \\quad (\\heartsuit)\n$$\nLet $R$ and $Q$ be the intersections of the perpendicular bisectors of $AC$ and $AB$ with $AB$ and $AC$, respectively. It is well-known that $CR$ and... | Iran | 38th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > M... | null | proof only | null | |
0bch | Problem:
Fie matricele de ordin $2$ cu elemente reale $A$ şi $B$ astfel încât
$$
A B = A^{2} B^{2} - (A B)^{2} \quad \text{şi} \quad \operatorname{det}(B) = 2
$$
a) Arătaţi că matricea $A$ nu este inversabilă.
b) Calculaţi $\operatorname{det}(A+2 B) - \operatorname{det}(B+2 A)$. | [] | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof and answer | 6 | |
042k | Let $k$ be a fixed odd integer, $k > 3$. Prove: There exist infinitely many positive integers $n$, such that there are two positive integers $d_1, d_2$ satisfying $d_1, d_2$ each dividing $\frac{n^2+1}{2}$, and $d_1 + d_2 = n + k$. | [
"$$\n((k-2)^2 + 1)xy = (x + y - k)^2 + 1, \\qquad \\textcircled{1}\n$$\nwe prove ① has infinitely many positive odd solutions $(x, y)$.\nObviously $(1,1)$ is one positive odd solution, let $(x_1, y_1) = (1,1)$. Assume $(x_i, y_i)$ is one positive odd solution of ①, and $x_i \\le y_i$, let $x_{i+1} = y_i$, $y_{i+1} ... | China | China Team Selection Test | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof only | null | |
0hrf | Problem:
Let $a, b, c$ be positive real numbers satisfying $a b c=1$. Prove that
$$
a(a-1)+b(b-1)+c(c-1) \geq 0 .
$$ | [
"Solution:\nAt least two of $a, b, c$ are either not less than $1$ or not greater than $1$. Assume that $a$ and $b$ are on the same side of $1$. Next, transform the inequality as follows:\n$$\n\\begin{aligned}\na(a-1)+b(b-1)+c(c-1) &\\stackrel{?}{\\geq} 0 \\\\\na(a-1)+b(b-1)+c^{2}\\left(1-\\frac{1}{c}\\right) &\\st... | United States | Berkeley Math Circle | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0iio | Problem:
Compute the positive integer less than $1000$ which has exactly $29$ positive proper divisors. (Here we refer to positive integer divisors other than the number itself.) | [
"Solution:\n\nRecall that the number $N = p_{1}^{e_{1}} p_{2}^{e_{2}} \\cdots p_{k}^{e_{k}}$ (where the $p_{i}$ are distinct primes) has exactly $(e_{1}+1)(e_{2}+1) \\cdots (e_{k}+1)$ positive integer divisors including itself. We seek $N < 1000$ such that this expression is $30$. Since $30 = 2 \\cdot 3 \\cdot 5$, ... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 720 | |
01fz | Let $N$ be a positive integer. Determine the number of pairs, $(a, b)$, of positive integers $a$ and $b$ such that the number
$$
\frac{ab}{a+b}
$$
is a divisor of $N$. | [
"In what follows $d(n)$ denotes the number of positive divisors of a positive integer $n$. We show that the number of pairs is $d(N)^2$. Let $m$ be a fixed positive divisor of $N$. We claim that the number of pairs, $(a, b)$, such that $\\frac{ab}{a+b} = m$ is $d(m)^2$. This becomes clear if we rewrite the equation... | Baltic Way | Baltic Way 2019 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | d(N)^2 | |
03xq | Given an integer $k \ge 3$ and a sequence $\{a_n\}$ that satisfies $a_k = 2k$ and for each $n > k$,
$$
a_n = \begin{cases} a_{n-1} + 1, & \text{if } a_{n-1} \text{ and } n \text{ are coprime,} \\ 2n, & \text{otherwise.} \end{cases}
$$
Prove that $a_n - a_{n-1}$ is a prime for infinitely many $n$. (Posed by Zhu Huawei) | [
"Suppose that $a_l = 2l$, $l \\ge k$. Let $p$ be the least prime divisor of $k-1$. Then $(l-1, i) = \\begin{cases} 1, & 1 \\le i < p, \\\\ p, & i = p, \\end{cases}$ and thus\n$$\n(2l + i - 2, l + i - 1) = \\begin{cases} 1, & 1 \\le i < p, \\\\ p, & i = p. \\end{cases}\n$$\nFrom (1) we know that\n$$\na_{l+i-1} = \\b... | China | Chinese Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0asn | Problem:
How many ways can you choose four integers from the set $\{1,2,3, \ldots, 10\}$ so that no two of them are consecutive? | [
"Solution:\n\n35"
] | Philippines | Philippines Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 35 | |
0hfh | There is a central train station at point $O$, which is connected to other train stations $A_1, A_2, \dots, A_8$ with tracks. There is also a track between stations $A_i$ and $A_{i+1}$ for each $1 \le i \le 8$ (here $A_1 = A_9$). The length of each track $A_iA_{i+1}$ is equal to $1$, and the length of each track $OA_i$... | [
"**Answer:** yes.\n\nArrange the trains clockwise as following: $B_8, B_1, B_7, B_2, B_6, B_3, B_5$ and $B_4$. Now let's show how they should move to meet in time.\n\nLet $B_8$ move through the station of $B_1$ (at this time $B_1$ stays at place and is waiting for $B_8$), and they arrive to $O$ in time $t_8 = \\fra... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | yes | |
0b8x | Let $f : [0, 1] \to \mathbb{R}$ be a differentiable function with $f(0) = f(1)$, $\int_{0}^{1} f(x) dx = 0$ and $f'(x) \neq 1$, for every $x \in [0, 1]$.
a) Prove that the function $g : [0, 1] \to \mathbb{R}$ given by $g(x) = f(x) - x$ is decreasing.
b) Prove that for any integer $n \ge 1$ the following inequality ho... | [
"a) Since $f'$ has the intermediate value property, it follows that $f'(x) < 1$, for every $x \\in [0, 1]$, or $f'(x) > 1$, for every $x \\in [0, 1]$. But, in the second case $f$ would be strictly increasing, which contradicts $f(0) = f(1)$. So $f'(x) < 1$, for every $x \\in [0, 1]$, whence $g : [0, 1] \\to \\mathb... | Romania | Romanian Mathematical Olympiad | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications",
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | English | proof only | null | |
06jr | 2016 circles with radius $1$ are lying on the plane. Among these $2016$ circles, show that one can select a collection $C$ of $27$ circles satisfying the following: either every pair of two circles in $C$ intersects or every pair of two circles in $C$ does not intersect. | [
"Suppose there do not exist $27$ circles such that every pair of circles intersects.\n\nConsider a coordinate plane such that the line joining any pair of centres of the circles is not parallel to the coordinate axes. We label the circles as $\\Gamma_1, \\Gamma_2, \\dots, \\Gamma_{2016}$ such that the $x$-coordinat... | Hong Kong | Year 2016 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > P... | null | proof only | null | |
0gqu | For a given integer $n \ge 3$, let $S_1, S_2, \dots, S_m$ be distinct three-element subsets of the set $\{1, 2, \dots, n\}$ such that for each $1 \le i, j \le m; i \ne j$ the sets $S_i \cap S_j$ contain exactly one element. Determine the maximal possible value of $m$ for each $n$. | [
"For each $n$ let $f(n)$ be the maximal value of $m$. Readily $f(3) = f(4) = 1$ and $f(5) = 2$. If $n = 6$ each number belongs to at most 2 subsets. Thus, $f(6) \\le \\frac{6 \\cdot 2}{3} = 4$. Therefore $f(6) = 4$ since the example $\\{1, 2, 3\\}, \\{1, 4, 5\\}, \\{2, 4, 6\\}, \\{3, 5, 6\\}$ works.\n\nLet $n \\ge ... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | f(3)=1, f(4)=1, f(5)=2, f(6)=4, and for all n ≥ 7, f(n)=max{7, ⌊(n−1)/2⌋}. | |
05os | Problem:
Trouver tous les entiers strictement positifs $a, b$ et $c$ tels que $\operatorname{PGCD}(a, b, c)=1$,
$$
a\mid (b-c)^2, \quad b\mid (a-c)^2, \quad c \mid (a-b)^2
$$
et tels qu'il est possible de construire un triangle non aplati dont les côtés ont pour longueurs $a$, $b$ et $c$. | [
"Solution:\nTout d'abord, vérifions que $\\operatorname{PGCD}(a, b)=1$. Raisonnons par l'absurde en supposant que $p$ est un nombre premier qui divise à la fois $a$ et $b$. Alors $p$ divise $a$, qui lui même divise\n$$\n(b-c)^2 = b(b-2c) + c^2\n$$\ndonc $p$ divise $c^2$ et donc $p$ divise $c$. Ceci contredit le fai... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | a = b = c = 1 | |
0ajm | Let $ABCD$ be a parallelogram and let $E$, $F$, $G$ and $H$ be the midpoints of the sides $AB$, $BC$, $CD$ and $DA$, respectively. If $BH \cap AC = I$, $BD \cap EC = J$, $AC \cap DF = K$ and $AG \cap BD = L$, then prove that the quadrilateral $IJKL$ is a parallelogram. | [
"Let $AC \\cap BD = O$. Clearly, $AO$ and $BH$ are medians in the triangle $ABD$, hence $I$ is the centroid of $ABD$. Similarly $K$ is the centroid of $BCD$. If $\\overline{IO} = x$, then $\\overline{AI} = 2x$. Similarly, if $\\overline{KO} = y$, then $\\overline{CK} = 2y$. Therefore $3x = \\overline{AO} = \\overli... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
01a7 | In two endpoints of the main diagonal of the cube numbers $0$ and $2013$ are written, remaining $6$ vertices of the cube contain real numbers $x_1, \dots, x_6$. On each edge of the cube the difference between the numbers at its endpoints is written, $S$ is the sum of squares of the numbers written on the edges. For whi... | [
"$$\n\\{x_1, \\dots, x_6\\} = \\left\\{ \\frac{2 \\cdot 2013}{5}, \\frac{2 \\cdot 2013}{5}, \\frac{2 \\cdot 2013}{5}, \\frac{3 \\cdot 2013}{5}, \\frac{3 \\cdot 2013}{5}, \\frac{3 \\cdot 2013}{5} \\right\\}\n$$\nThe function\n$$\n(x-a)^2 + (x-b)^2 + (x-c)^2\n$$\nattains its minimum when $x = \\frac{a+b+c}{3}$. Let's... | Baltic Way | Baltic Way 2013 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | {2*2013/5, 2*2013/5, 2*2013/5, 3*2013/5, 3*2013/5, 3*2013/5} | |
02bz | Problem:
Uma loja estava vendendo um brinquedo por $R\$ 13{,}00$ a unidade. Para conseguir vender todo o seu estoque que não era superior a 100 unidades, resolveu abaixar o preço de um número inteiro de reais. Com isso, conseguiu vender todo o estoque por $R\$ 781{,}00$. Qual foi a redução do preço, por unidade? | [
"Solution:\n\nSe $x$ é o desconto em reais e $y$ é o número de peças, então\n$$\n(13-x) \\times y = 781 \\text{ e } y < 100\n$$\nAssim, $(13-x)$ e $y$ são divisores de $781$. Como $781 = 11 \\times 71$, a única solução é $y = 71$ e $13-x = 11$. Logo, a redução foi de $R\\$ 2,00$."
] | Brazil | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2 | |
0iaj | Problem:
A broken calculator has the $+$ and $\times$ keys switched. For how many ordered pairs $(a, b)$ of integers will it correctly calculate $a+b$ using the labelled $+$ key? | [
"Solution:\nWe need $a + b = a b$, or $a = \\frac{b}{b-1} = 1 - \\frac{1}{b-1}$, so $1/(b-1)$ is an integer. Thus $b$ must be $0$ or $2$, and $a$ is $0$ or $2$, respectively. So there are $2$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 2 | |
02fc | Call a super-integer an infinite sequence of decimal digits: $\ldots d_n \ldots d_3 d_2 d_1$. Given two super-integers $\ldots c_n \ldots c_3 c_2 c_1$ and $\ldots d_n \ldots d_3 d_2 d_1$, their product $\ldots p_n \ldots p_3 p_2 p_1$ is formed by taking $p_n \ldots p_3 p_2 p_1$ to be the last $n$ digits of the product ... | [
"The answer is yes. In fact, there exist two sequences $x = \\ldots a_n \\ldots$ and $y = \\ldots b_n \\ldots$ such that, for every $k \\ge 1$, $2^k$ divides $a_k \\ldots a_2 a_1$ and $5^k$ divides $b_k \\ldots b_2 b_1$; then $10^k$ divides $a_k \\ldots a_2 a_1 \\times b_k \\ldots b_2 b_1$ and $xy = 0$.\n\nConsider... | Brazil | XVI OBM | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Yes | |
0bu4 | Problem:
Fie $a$ şi $n$ două numere naturale nenule, astfel încât
$$
\{\sqrt{n+\sqrt{n}}\}=\{\sqrt{a}\}
$$
Arătaţi că $4 a+1$ este pătrat perfect. | [
"Solution:\n\nCondiţia din enunţ este echivalentă cu $\\sqrt{n+\\sqrt{n}}=\\sqrt{a}+k$, $k \\in \\mathbb{Z}$. Rezultă că $n+\\sqrt{n}=a+2k\\sqrt{a}+k^{2}$, deci $\\sqrt{n}=2k\\sqrt{a}+b$, unde $b=k^{2}-n+a$. Deducem că $n=4k^{2}a+b^{2}+4kb\\sqrt{a}$, de unde rezultă că $kb\\sqrt{a}$ este raţional.\n\nDacă $\\sqrt{a... | Romania | Olimpiada Naţională de Matematică, Etapa Judeţeană şi a Municipiului Bucureşti | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
02kh | Problem:
O café, o bolo e o gato - Dez minutos antes de colocar o bolo no forno, eu coloquei meu gato do lado de fora da casa. O bolo deve cozinhar por 35 minutos, então eu coloquei o despertador para tocar 35 minutos, após colocar o bolo no forno. Imediatamente fiz um café para mim, o que me tomou 6 minutos. Três min... | [
"Solution:\n\nVamos listar os eventos ocorridos e contar o tempo gasto em cada um. A primeira atividade foi colocar o gato fora de casa, logo nossa lista começa com essa atividade e o tempo é contado a partir dela.\n\n| Atividade | Tempo depois que o gato foi posto fora de casa |\n| :--- ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) 3:26; b) 45 minutes; c) 28 minutes | |
0d4b | Fatima and Asma are playing the following game. First, Fatima chooses $2013$ pairwise different numbers, called $a_{1}, a_{2}, \ldots, a_{2013}$. Then, Asma tries to know the value of each number $a_{1}, a_{2}, \ldots, a_{2013}$. At each time, Asma chooses $1 \leq i < j \leq 2013$ and asks Fatima "What is the set $\{a_... | [
"Let $n$ be the number of different questions Asma asks to know the values of each number $a_{1}, a_{2}, \\ldots, a_{2013}$. Let us make the following remarks:\n\na. Because Asma needs to know the value of each number $a_{i}$, each number $a_{i}$ must appear at least in one of the sets that Asma chooses to ask abou... | Saudi Arabia | SAMC | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English, Arabic | proof and answer | 1342 | |
059d | Priit's collection consists of $1111$ stamps which are all distributed into envelopes in such a way that every envelope contains more than one stamp, all envelopes contain the same number of stamps, and each envelope contains only stamps from one country. It is known that more than $40\%$ of stamps in this collection a... | [
"*Answer:* $41$ and $29$.\n\nAs $1111 = 11 \\cdot 101$ where the factors are prime, we have four cases:\n* $1$ envelope containing $1111$ stamps;\n* $1111$ envelopes, each containing $1$ stamp;\n* $11$ envelopes, each containing $101$ stamps;\n* $101$ envelopes, each containing $11$ stamps.\n\nThe first case is imp... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 41 and 29 | |
03pg | A circle with center $O$ and radius $R$ is drawn on a paper, and $A$ is a given point in the circle with $OA = a$. Fold the paper to make a point $A'$ on the circumference coincident with point $A$, then a crease line is left on the paper. Find out the set of all points on such crease lines, when $A'$ goes through ever... | [
"Establish an $xy$-coordinate system as in the diagram with $A(a, 0)$ given. Then the crease line $MN$ is the perpendicular bisector of segment $AA'$ when $A'$ ($R\\cos a$, $R\\sin a$) is made coincident with $A$ by folding the paper. Let $P(x, y)$ be any point on $MN$, then $|PA'| = |PA|$. That is\n\n | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | All points (x, y) satisfying ((2x - a)^2)/R^2 + (4y^2)/(R^2 - a^2) ≥ 1; equivalently, the set is the boundary and exterior of the ellipse ((2x - a)^2)/R^2 + (4y^2)/(R^2 - a^2) = 1. | |
041h | Suppose that $O$ and $I$ are the centres of the circumcircle and incircle of $\triangle ABC$ with radius $R$ and $r$, respectively. $P$ is the midpoint of arc $BAC$. Let $QP$ be the diameter of $O$. Let $PI$ intersect $BC$ at point $D$, and let the circumcircle of $\triangle AID$ intersect the extended line of $PA$ at ... | [
"$$\n\\begin{aligned}\nPF^2 &= AF \\cdot PF + PA \\cdot PF \\\\\n&= EF^2 + PI \\cdot PD.\n\\end{aligned} \n\\qquad \\textcircled{1}\n$$\nSince $PQ$ is the diameter of circle $O$ and point $I$ is on $AQ$, we see that $AI \\perp AP$. Consequently,\n$$\n\\angle IDF = \\angle IAP = 90^{\\circ}.\n$$\nThus, we have\n$$\n... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > D... | English | proof only | null | |
0dkx | Find all positive integers $n$ that cannot be written in the form
$$
n = \frac{a}{b} + \frac{a+1}{b+1},
$$
where $a$ and $b$ are positive integers. | [] | Saudi Arabia | Saudi Booklet | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All such n are exactly n = 1 and n = 2 + 2^k for k ≥ 0. | |
05d4 | Problem:
Let $ABCD$ be a convex quadrilateral with $\angle DAB = \angle BCD = 90^{\circ}$ and $\angle ABC > \angle CDA$. Let $Q$ and $R$ be points on the segments $BC$ and $CD$, respectively, such that the line $QR$ intersects lines $AB$ and $AD$ at points $P$ and $S$, respectively. It is given that $PQ = RS$. Let the... | [
"Solution:\n\nNote that $N$ is also the midpoint of $PS$. From right-angled triangles $PAS$ and $CQR$ we obtain $\\angle ANP = 2 \\angle ASP$, $\\angle CNQ = 2 \\angle CRQ$, hence\n$$\n\\angle ANC = \\angle ANP + \\angle CNQ = 2(\\angle ASP + \\angle CRQ) = 2(\\angle RSD + \\angle DRS) = 2 \\angle ADC.\n$$\nSimilar... | European Girls' Mathematical Olympiad (EGMO) | null | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0its | Given circles $\omega_1$ and $\omega_2$ intersecting at points $X$ and $Y$, let $\ell_1$ be a line through the center of $\omega_1$ intersecting $\omega_2$ at points $P$ and $Q$ and let $\ell_2$ be a line through the center of $\omega_2$ intersecting $\omega_1$ at points $R$ and $S$. Prove that if $P, Q, R$ and $S$ lie... | [
"Let $\\omega$ denote the circumcircle of $P, Q, R, S$ and let $O$ denote the center of $\\omega$. Line $XY$ is the radical axis of circles $\\omega_1$ and $\\omega_2$. It suffices to show that $O$ has equal power to the two circles; that is, to show that\n$$\nOO_1^2 - O_1S^2 = OO_2^2 - O_2Q^2 \\quad \\text{or} \\q... | United States | USAMO 2009 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0h5i | A committee has 4 subcommittees, each controlled by 3 leaders from the committee. For effective coordination, each two subcommittees must have exactly one leader in common. What is the least possible number of people in the committee? | [
"If we consider two subcommittees, they have exactly one leader in common, therefore, together they have exactly 5 members. Hence there are at least 5 people in the committee. Denote them by $A = \\{1, 2, 3, 4, 5\\}$. However it's impossible to choose leaders for another subcommittee out of them. Therefore, the com... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 6 | |
0h4h | Solve the equation for arbitrary distinct reals $a$, $b$, $c$:
$$
x^3 a - x a^3 + a^3 b - a b^3 + b^3 x - b x^3 = (x-a)(x-b)(x-c)(a-b).
$$ | [
"Obviously the equation can be solved by trivial transformations and reduction to a quadratic one. We suggest a different approach. Denote the left-hand and right-hand sides by $f(x)$ and $g(x)$ respectively. It's easy to verify that $f(a) = g(a) = 0$ and $f(b) = g(b) = 0$. But this implies that distinct numbers $a... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | x = a or x = b | |
0ikh | Problem:
Suppose $ABC$ is a triangle such that $AB = 13$, $BC = 15$, and $CA = 14$. Say $D$ is the midpoint of $\overline{BC}$, $E$ is the midpoint of $\overline{AD}$, $F$ is the midpoint of $\overline{BE}$, and $G$ is the midpoint of $\overline{DF}$. Compute the area of triangle $EFG$. | [
"Solution:\n\nBy Heron's formula, $[ABC] = \\sqrt{21(21-15)(21-14)(21-13)} = 84$. Now, unwinding the midpoint conditions yields $[EFG] = \\frac{[DEF]}{2} = \\frac{[BDE]}{4} = \\frac{[ABD]}{8} = \\frac{[ABC]}{16} = \\frac{84}{16} = \\frac{21}{4}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 21/4 | |
0cnv | Determine if there exists a coloring of all positive integers in $2009$ colors satisfying the following two conditions:
(i) there are infinitely many numbers of each color;
(ii) there is no triple of integers $(a, b, c)$ colored in three pairwise distinct colors such that $a = bc$.
$(N. Agakhanov)$ | [
"Можно.\n\nПриведем один из возможных вариантов такой раскраски. Пусть $p_1 < p_2 < \\dots < p_{2008}$ -- простые числа. Построим множества $A_1, A_2, \\dots, A_{2009}$ по следующему правилу: в $A_1$ включим все натуральные числа, делящиеся на $p_1$; в $A_2$ — все натуральные числа, делящиеся на $p_2$, но не делящи... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English; Russian | proof and answer | Yes, such a coloring exists. | |
0468 | Determine the largest real number $C$ such that
$$
\sum_{i=1}^{n} \sum_{j=1}^{n} (n - |i - j|) x_i x_j \ge C \sum_{i=1}^{n} x_i^2
$$
holds for every positive integer $n$ and any real numbers $x_1, x_2, \dots, x_n$. | [
"The needed constant $C = \\frac{1}{2}$. First show that $C = \\frac{1}{2}$ holds. Note that\n$$\n\\text{L.H.S. of (*)} = x_1^2 + (x_1 + x_2)^2 + \\dots + (x_1 + x_2 + \\dots + x_n)^2 \\\\\n\\qquad + (x_2 + \\dots + x_n)^2 + \\dots + (x_{n-1} + x_n)^2 + x_n^2. \\quad (1)\n$$\nUsing the inequality $a^2 + (a+b)^2 = a... | China | Chinese Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 1/2 | |
0e9d | Problem:
Izračunaj vrednosti koeficientov polinoma $p$ s predpisom $p(x) = a x^{3} + b x^{2} + c x + d$, če je $0$ njegova ničla in velja $p(1) = \frac{16}{3}$. Ničli odvoda polinoma $p$ sta $-1$ in $-3$. | [
"Solution:\n\nKer je $0$ ničla polinoma $p$, velja $p(0) = 0$.\n\n$$\np(0) = a \\cdot 0^3 + b \\cdot 0^2 + c \\cdot 0 + d = d = 0\n$$\nTorej je $d = 0$.\n\nPolinom je torej oblike:\n$$\np(x) = a x^3 + b x^2 + c x\n$$\n\nPodano je še $p(1) = \\frac{16}{3}$:\n$$\np(1) = a \\cdot 1^3 + b \\cdot 1^2 + c \\cdot 1 = a + ... | Slovenia | 14. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | a = 1/3, b = 2, c = 3, d = 0 | |
00d0 | En el pizarrón están escritos los 18 números enteros desde $1$ hasta $18$. Determinar la menor cantidad de números que hay que borrar para que entre los números restantes no haya dos tales que su suma sea un cuadrado perfecto. | [
"Borramos los siguientes $9$ números: $3$, $5$, $7$, $8$, $10$, $12$, $14$, $15$, $16$. Los números restantes son: $1$, $2$, $4$, $6$, $9$, $11$, $13$, $17$, $18$. Se verifica que la suma de cualesquiera dos de ellos no es un cuadrado perfecto. Por lo tanto, es posible lograr el objetivo borrando $9$ números.\n\nVe... | Argentina | Nacional OMA | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | Spanish | proof and answer | 9 | |
04kv | What is the maximum number of elements that a subset of $\{1, 2, 3, \dots, 2017\}$ can have so that for any two distinct elements $a$ and $b$ of that subset, the number $a + b$ is not divisible by $a - b$? (Hong Kong 1992) | [] | Croatia | Mathematical competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 673 | |
0cmv | A quadratic trinomial $f(x)$ is chosen so that the equation $(f(x))^5 - f(x) = 0$ has exactly 3 real roots. Find the $y$-coordinate of the vertex of the graph of $f(x)$. | [
"Так как $g(x) = (f(x))^5 - f(x) = f(x)(f(x) - 1)(f(x) + 1)((f(x))^2 + 1)$, то корнями нашего многочлена являются корни трехчленов $f(x)$, $f(x) - 1$ и $f(x) + 1$ (поскольку многочлен $(f(x))^2 + 1$ всюду положителен). Ясно, что любое число может быть корнем только одного из них.\n\nПусть $y_0$ — искомая ордината в... | Russia | Russian mathematical olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English; Russian | proof and answer | 0 | |
0jin | Problem:
Two circles $\omega$ and $\gamma$ have radii $3$ and $4$ respectively, and their centers are $10$ units apart. Let $x$ be the shortest possible distance between a point on $\omega$ and a point on $\gamma$, and let $y$ be the longest possible distance between a point on $\omega$ and a point on $\gamma$. Find t... | [
"Solution:\n\nLet $\\ell$ be the line connecting the centers of $\\omega$ and $\\gamma$. Let $A$ and $B$ be the intersections of $\\ell$ with $\\omega$, and let $C$ and $D$ be the intersections of $\\ell$ with $\\gamma$, so that $A, B, C$, and $D$ are collinear, in that order.\n\nThe shortest distance between a poi... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 51 | |
01tw | 6. Given a polynomial $p(x) = a_{2n}x^{2n} + a_{2n-1}x^{2n-1} + \dots + a_1x + a_0$ of even degree with positive coefficients $a_0, a_1, \dots, a_{2n}$.
a) Prove that there exists a permutation of these coefficients such that the polynomial obtained has no real roots.
b) Does the statement of a) remain true if some coe... | [
"**6. Answer : b) no, it does not.**\na) Let $b_0 \\le b_1 \\le \\dots \\le b_{2n}$ denote the coefficients $a_0, a_1, \\dots, a_{2n}$ of $p(x)$, which are arranged in non-decreasing order. Consider the permutation $c_0, c_1, \\dots, c_{2n}$ of the numbers $a_0, a_1, \\dots, a_{2n}$ such that $c_{2n} = b_{2n}, c_{2... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | no | |
03hf | Problem:
i) If $x = \left(1 + \frac{1}{n}\right)^n$ and $y = \left(1 + \frac{1}{n}\right)^{n+1}$, show that $y^x = x^y$.
ii) Show that, for all positive integers $n$,
$$
1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^n (n-1)^2 + (-1)^{n+1} n^2 = (-1)^{n+1} (1 + 2 + \cdots + n)
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0kpg | Problem:
Compute the number of positive integers that divide at least two of the integers in the set $\{1^{1}, 2^{2}, 3^{3}, 4^{4}, 5^{5}, 6^{6}, 7^{7}, 8^{8}, 9^{9}, 10^{10}\}$. | [
"Solution:\n\nFor a positive integer $n$, let $\\operatorname{rad} n$ be the product of the distinct prime factors of $n$. Observe that if $n \\mid m^{m}$, all prime factors of $n$ must divide $m$, so $\\operatorname{rad} n \\mid m$.\n\nTherefore, if $n$ is such an integer, $\\operatorname{rad} n$ must divide at le... | United States | HMMT February 2022 | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 22 | |
035y | Problem:
Let $f(x) = x^{2} + (2a - 1)x - a - 3$, where $a$ is a real parameter.
a) Prove that the equation $f(x) = 0$ has two distinct real roots $x_{1}$ and $x_{2}$.
b) Find all values of $a$ such that $x_{1}^{3} + x_{2}^{3} = -72$. | [
"Solution:\n\na) The discriminant of $f(x)$ is equal to $4a^{2} + 13 > 0$.\n\nb) We consecutively have\n$$\n\\begin{aligned}\n-72 & = x_{1}^{3} + x_{2}^{3} = (x_{1} + x_{2}) \\left[ (x_{1} + x_{2})^{2} - 3 x_{1} x_{2} \\right] \\\\\n& = (1 - 2a) \\left( 4a^{2} - a + 10 \\right ) = -8a^{3} + 6a^{2} - 21a + 10\n\\end... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | a = 2 | |
0jq7 | Problem:
Let $a, b, c$ be positive real numbers such that $a + b + c = 10$ and $ab + bc + ca = 25$. Let $m = \min \{ ab, bc, ca \}$. Find the largest possible value of $m$. | [
"Solution:\n\nAnswer: $\\sqrt{\\frac{25}{9}}$\n\nWithout loss of generality, we assume that $c \\geq b \\geq a$. We see that $3c \\geq a + b + c = 10$. Therefore, $c \\geq \\frac{10}{3}$.\n\nSince\n$$\n\\begin{aligned}\n0 &\\leq (a-b)^2 \\\\\n &= (a+b)^2 - 4ab \\\\\n &= (10-c)^2 - 4(25 - c(a+b)) \\\\\n &= (10-c)... | United States | HMMT February 2015 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 25/9 | |
0b10 | Problem:
How many nonempty subsets of $\{1,2, \ldots, 10\}$ have the property that the sum of its largest element and its smallest element is $11$? | [
"Solution:\n\nIf $a$ is the smallest element of such a set, then $11 - a$ is the largest element, and for the remaining elements we may choose any (or none) of the $10 - 2a$ elements $a + 1, a + 2, \\ldots, (11 - a) - 1$. Thus there are $2^{10 - 2a}$ such sets whose smallest element is $a$. We also require that $11... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Discrete Mathematics > Combinatorics"
] | null | final answer only | 341 | |
0j5v | Problem:
Rachel and Brian are playing a game in a grid with 1 row of 2011 squares. Initially, there is one white checker in each of the first two squares from the left, and one black checker in the third square from the left. At each stage, Rachel can choose to either run or fight. If Rachel runs, she moves the black ... | [
"Solution:\n\nBoth operations, run and fight, move the black checker exactly one square to the right, so the game will end after exactly 2008 moves regardless of Brian's choices. Furthermore, it is easy to see that if we view running and fighting as operations, they commute. So the order of the moves does not matte... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | final answer only | 2009 | |
0hs5 | Problem:
The tangents at $A$ and $B$ to the circumcircle of an acute triangle $ABC$ intersect at $T$. Point $D$ lies on line $BC$ such that $DA = DC$. Prove that $TD \parallel AC$. | [
"Solution:\n\nUsing the properties of tangent lines and chords, we have\n$$\n\\angle TAB = \\angle ACB = \\angle ABT = \\alpha\n$$\nMoreover, $\\triangle DAC$ is isosceles so $\\angle DAC = \\alpha$ as well. Now triangles $TAB$ and $DAC$ are similar, so\n$$\n\\frac{AT}{AB} = \\frac{AD}{AC} \\quad \\text{and} \\quad... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hzv | Problem:
$A, B, C, D$, and $E$ are relatively prime integers (i.e., have no single common factor) such that the polynomials $5A x^{4} + 4B x^{3} + 3C x^{2} + 2D x + E$ and $10A x^{3} + 6B x^{2} + 3C x + D$ together have 7 distinct integer roots. What are all possible values of $A$?
Your team has been given a sealed e... | [
"Solution:\n\nCall the negatives of the roots of the first polynomial $a, b, c, d$, and the negatives of the roots of the second polynomial $e, f, g$ (using the negatives avoids negative signs for the rest of the proof, thus preventing the possibility of dropping a sign). Then\n$$\n5A x^{4} + 4B x^{3} + 3C x^{2} + ... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | -3, -1, 1, 3 | |
056m | We call a positive integer $n$ whose all digits are distinct *bright*, if either $n$ is a one-digit number or there exists a divisor of $n$ which can be obtained by omitting one digit of $n$ and which is bright itself. Find the largest bright positive integer. (We assume that numbers do not start with zero.) | [
"First, we show by induction on the length of $n$ that if $10n$ is bright, then $n$ is bright as well. Assume that for one digit shorter numbers the statement holds. If after deleting $0$ we obtain a bright divisor, the statement holds trivially. If the bright divisor of $10n$ is obtained after deleting some other ... | Estonia | IMO Team Selection Contest | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | English | proof and answer | 146250 | |
0bb6 | A certain language uses an alphabet containing three letters. Some sequences of two or more letters are forbidden, and every two forbidden sequences have different lengths. Prove that there exists admissible words of every length. | [
"Let $a_n$ be the number of admissible words with $n$ letters; then $a_0 = 1$ (the empty word), $a_1 = 3$ and $a_2 = 8$.\n\nThen, if we add a letter at the end of a correct word with $n$ letters, we obtain either a correct word with $n+1$ letters, or a forbidden word of the form $XY$, with $Y$ a forbidden sequence ... | Romania | 2011 CLOCK-TOWER SCHOOL SENIORS COMPETITION | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0k21 | Problem:
Alice starts with the number $0$. She can apply $100$ operations on her number. In each operation, she can either add $1$ to her number, or square her number. After applying all operations, her score is the minimum distance from her number to any perfect square. What is the maximum score she can attain? | [
"Solution:\n\nAnswer: $94$\n\nNote that after applying the squaring operation, Alice's number will be a perfect square, so she can maximize her score by having a large number of adding operations at the end. However, her score needs to be large enough that the many additions do not bring her close to a larger squar... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 94 | |
0bk1 | Let $n \ge 2$ be a positive integer. Determine all possible values of the sum
$$
S = \lfloor x_2 - x_1 \rfloor + \lfloor x_3 - x_2 \rfloor + \dots + \lfloor x_n - x_{n-1} \rfloor,
$$
where $x_1, x_2, \dots, x_n$ are real numbers whose integer parts are $1, 2, \dots, n$, respectively. | [
"Let $a$ and $b$ be arbitrary reals. We have $\\lfloor b \\rfloor - \\lfloor a \\rfloor - 1 < b - a < \\lfloor b \\rfloor - \\lfloor a \\rfloor + 1$, hence $\\lfloor b \\rfloor - \\lfloor a \\rfloor - 1 \\le \\lfloor b - a \\rfloor \\le \\lfloor b \\rfloor - \\lfloor a \\rfloor$. Applying this to some consecutive t... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | {0, 1, 2, ..., n-1} | |
0jzb | Problem:
Prove that
$$
64 \frac{a b c d+1}{(a+b+c+d)^{2}} \leq a^{2}+b^{2}+c^{2}+d^{2}+\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}
$$
for $a, b, c, d>0$. | [
"Solution:\nBy Holder inequality, we have\n$$\n\\left(\\sum_{\\text{cyc}} a^{-2}\\right)\\left(\\sum_{\\text{cyc}} a\\right)^{2} \\geq (1+1+1+1)^{3} = 64\n$$\nAlso, we have\n$$\n\\left(\\sum_{\\text{cyc}} a^{2}\\right)\\left(\\sum_{\\text{cyc}} a\\right)^{2} \\geq \\left(4 \\sqrt[4]{(a b c d)^{2}}\\right) (4 \\sqrt... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0f1p | Problem:
Initially, there is one beetle on each square in the set $S$. Suddenly each beetle flies to a new square, subject to the following conditions:
(1) the new square may be the same as the old or different;
(2) more than one beetle may choose the same new square;
(3) if two beetles are initially in squares wit... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | Middle row and column only: No.
With the border squares also included: Yes.
All squares of the board: Yes. | |
05bm | The number $\sqrt{2}$ is on the computer screen. In one step, Juku can multiply the number currently on the screen by any positive rational number or add any natural number to it. Is there a positive integer $n$ such that every number that Juku can get after a finite number of steps can be obtained by at most $n$ steps... | [
"Any number that can be obtained after a finite number of steps is of the form $r\\sqrt{2} + s$, where $r$ is positive and $s$ is a nonnegative rational number. Indeed, the initial number $\\sqrt{2}$ is of the form $1\\sqrt{2} + 0$, and if we multiply a number of the form $r\\sqrt{2} + s$ by a positive rational num... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | 3 | |
053k | In the beginning, there are two positive integers on a blackboard. On each step, one chooses numbers $a$ and $b$ such that $a \le b$ from the numbers on the blackboard in all possible ways (equality means that one may take the same number twice), finds all corresponding sums $a+b+\gcd(a,b)$ and replaces all the numbers... | [
"If the numbers chosen from the blackboard are $x$ and $y$, then the number $x+y+\\gcd(x,y)$ will be on the blackboard on the next step. If $x$ is chosen together with itself, the number $x+x+\\gcd(x,x) = 3x$ will be on the blackboard on the next step. We show that there will be two equal numbers on the blackboard ... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
07ew | Let $S$ be an infinite subset of natural numbers. Define the set $S'$ as follows:
$$
S' = \{x^y + y^x \mid x, y \in S, x \ne y\}.
$$
Prove that there are infinitely many prime numbers $p$ such that $p$ divides at least one element in $S'$. | [
"Let $P$ be the set of all prime divisors of members of $S'$. Suppose, to the contrary, that $P$ is finite and $P = \\{p_1, p_2, \\dots, p_n\\}$. Define\n$$\nN = 4 \\prod_{i=1}^{n} p_i(p_i - 1).\n$$\n$S$ is infinite, hence there exists an infinite subset $S_1$ of $S$ such that every two members of $S_1$ are congrue... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Intermediate Alg... | English | proof only | null | |
0b5x | Let $D := \{1, 2, \dots, n\} \times \{1, 2, \dots, n\}$. Prove there exists a set $S \subset D$ with $|S| \ge \lfloor \frac{3}{5}n(n+1) \rfloor$, such that for any $(x_1, y_1), (x_2, y_2) \in S$ we have $(x_1 + x_2, y_1 + y_2) \notin S$. | [
"It is easy to find a weaker bound of $|S| = n\\lfloor\\frac{1}{2}n\\rfloor$ by taking $S = \\{(x, y) \\in D \\mid x > n/2\\}$. To find the bound asked, we need look at the diagonals of the tableau!\n\n\nDiagram for the selection $\\bullet$ of $S_u$ (exact for $n = 13$)."
] | Romania | 2010 Fourth STARS OF MATHEMATICS COMPETITION | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
05l2 | Problem:
Soit $n \geqslant 1$ un entier. Déterminer tous les entiers $p \geqslant 1$ pour lesquels il existe des entiers strictement positifs $x_{1}<x_{2}<\cdots<x_{n}$ tels que
$$
\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{n}{x_{n}}=p
$$ | [
"Solution:\nNous allons montrer que l'ensemble des solutions est l'ensemble $\\{1,2, \\ldots, n\\}$.\n\nSoient $p \\geqslant 1$ un entier et des entiers strictement positifs $x_{1}<x_{2}<\\cdots<x_{n}$ tels que\n$$\n\\frac{1}{x_{1}}+\\frac{2}{x_{2}}+\\cdots+\\frac{n}{x_{n}}=p\n$$\nOn remarque tout d'abord que $p \\... | France | Olympiades Françaises de Mathématiques - Test de Janvier | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All integers p with 1 ≤ p ≤ n | |
06yd | Let $ABCD$ be a cyclic quadrilateral such that $AC < BD < AD$ and $\angle DBA < 90^{\circ}$. Point $E$ lies on the line through $D$ parallel to $AB$ such that $E$ and $C$ lie on opposite sides of line $AD$, and $AC = DE$. Point $F$ lies on the line through $A$ parallel to $CD$ such that $F$ and $C$ lie on opposite side... | [
"Let $T$ be the midpoint of $\\operatorname{arc} \\overparen{BAC}$ and let lines $BA$ and $CD$ intersect $EF$ at $K$ and $L$, respectively. Note that $T$ lies on the perpendicular bisector of segment $BC$.\n\n\n\nSince $ABCD$ is cyclic, $\\frac{BD}{\\sin \\angle BAD} = \\frac{AC}{\\sin \\an... | IMO | IMO2024 Shortlisted Problems | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles >... | English | proof only | null | |
00wh | Problem:
Let $ABCD$ be a quadrangle, $|AD| = |BC|$, $\angle A + \angle B = 120^{\circ}$ and let $P$ be a point exterior to the quadrangle such that $P$ and $A$ lie at opposite sides of the line $DC$ and the triangle $DPC$ is equilateral. Prove that the triangle $APB$ is also equilateral. | [
"Solution:\n\nNote that $\\angle ADC + \\angle CDP + \\angle BCD + \\angle DCP = 360^{\\circ}$ (see Figure 1). Thus $\\angle ADP = 360^{\\circ} - \\angle BCD - \\angle DCP = \\angle BCP$. As we have $|DP| = |CP|$ and $|AD| = |BC|$, the triangles $ADP$ and $BCP$ are congruent and $|AP| = |BP|$. Moreover, $\\angle AP... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0886 | Problem:
(a) Dimostrare che per ogni intero positivo $n$ esistono due scalette di lunghezza $n$, senza elementi in comune, $a_{1}, a_{2}, \ldots, a_{n}$ e $b_{1}, b_{2}, \ldots, b_{n}$, tali che per ogni $i$ tra 1 ed $n$ il massimo comune divisore fra $a_{i}$ e $b_{i}$ è uguale a 1.
(b) Dimostrare che per ogni intero... | [
"Solution:\n\n(a) Fissiamo $n$ numeri consecutivi $a_{1}, a_{2}, \\ldots, a_{n}$. Sia ora $d$ un numero più grande di $n$ che non abbia fattori comuni con nessuno tra $a_{1}, a_{2}, \\ldots, a_{n}$ (ad esempio, un numero primo più grande di $a_{n}$), poniamo $b_{1} = a_{1} + d, b_{2} = a_{2} + d, \\ldots, b_{n} = a... | Italy | Cesenatico | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0ivc | Problem:
Let $T$ be a graph with one vertex $v_{n}$ for every integer $n$. An edge connects $v_{a}$ and $v_{b}$ if $|a-b|$ is a power of two. What is the chromatic number of $T$? Prove your answer. | [
"Solution:\n\nSince $v_{0}$, $v_{1}$, and $v_{2}$ are all connected to each other, three colors is necessary. Now, color $v_{n}$ red if $n \\equiv 0 \\pmod{3}$, blue if $n \\equiv 1 \\pmod{3}$, and green otherwise. Since $v_{a}$ and $v_{b}$ are the same color only if $3 \\mid (a-b)$, no two connected vertices are t... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3 | |
0aec | Даден е квадрат со страна $49$ cm. Раздели го дадениот квадрат на $2009$ помали квадрати од два типа коишто имаат целобројни страни (должините на страните на делбените квадратите можат да имаат една од две различни целобројни вредности). | [
"Ке разгледуваме квадрати со страни $1$ cm и $3$ cm. Бројот на квадратите со страна $3$ cm нека е $x$. Тогаш бројот на квадратите со страна $1$ cm ќе биде $2009-x$. Збирот на плоштините на помалите квадратчиња ќе биде иста со збирот на плоштините на големиот квадрат, т.е.\n$$\n\\begin{array}{l}\n(2009-x) \\cdot 1 +... | North Macedonia | Републички натпревар по математика за основно образование | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | Macedonian, English | proof and answer | 49 squares of side 3 cm and 1960 squares of side 1 cm | |
01q0 | Exactly one integer number from $1$ to $25$ is written in each cell of the $5 \times 5$ square table (see the figure). Per move it is allowed:
| 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|
| 6 | 7 | 8 | 9 | 10 |
| 11 | 12 | 13 | 14 | 15 |
| 16 | 17 | 18 | 19 | 20 |
| 21 | 22 | 23 | 24 | 25 |
1) to choose any two cells ... | [
"Answer: **a)** it is possible; **b)** it is impossible.\n\na) We show that using the allowed moves we can decrease any number in the table by $3$ so that all other numbers in the table keep their values. We will not consider the whole table but only the number $x$ which will be decreased by $3$ and three more numb... | Belarus | BelarusMO 2013_s | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0djo | Let $ABC$ be an isosceles acute triangle with $\ell$ is the bisector of angle $A$. A circle with center $O$ passes through $B, C$ and intersects the sides $AB, AC$ at $D, E$ respectively. Construct a parallelogram $DTEK$. Take $X, Y$ on $AB, AC$ respectively such that $AXKY$ is also a parallelogram. Construct $P, Q$ on... | [
"Since $BCED$ is cyclic, we have two similar triangles $ABC$ and $AED$. On the other hand\n$$\n\\angle TBC = \\angle TDE = \\angle KED \\text{ and } \\angle TCB = \\angle TED = \\angle KDE.\n$$\nThus two points $T, K$ are corresponding in the above pair of similar triangles, it can be deduced that\n$$\n\\angle TAB ... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triang... | English | proof only | null | |
007q | Let $n \ge 10^{2010}$ be an integer. Find the first digit after the decimal point of $\sqrt{n^2 + n + 200}$. | [
"The answer is $5$ for all $n \\ge 1000$. If $n \\ge 200$ we have $n^2+n+200 < n^2+2n+1 = (n+1)^2$, so $n < \\sqrt{n^2+n+200} < n+1$ and $\\lfloor \\sqrt{n^2+n+200} \\rfloor = n$. It also follows that $\\sqrt{n^2+n+200}$ is not an integer, moreover $\\sqrt{n^2+n+200}$ is irrational. Let $k$ be the first digit of $\... | Argentina | National Olympiad of Argentina | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 5 | |
07zn | Problem:
Un teatro ha $960$ posti, divisi nelle tre sezioni platea, palchi, galleria. In platea ci sono $370$ poltrone, mentre il numero di posti in galleria è inferiore di $290$ rispetto a quello dei palchi. Quanti sono i posti nei palchi?
(A) $150$
(B) $300$
(C) $315$
(D) $440$
(E) nessuna delle precedenti. | [] | Italy | Italian Mathematical Olympiad - Febbraio Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
0hoh | Problem:
Suppose that $ABC$ and $A'B'C'$ are two triangles such that $\angle A = \angle A'$, $AB = A'B'$, and $BC = B'C'$. Suppose also that $\angle C = 90^{\circ}$. Prove that triangles $ABC$ and $A'B'C'$ are congruent. | [
"Solution:\n\nConstruct a point $D$ on ray $AC$ such that $AD = A'C'$. Then $\\triangle ABD \\cong \\triangle A'B'C'$ by SAS. If $D = C$, then we are done, so assume that $D \\neq C$. We have $BD = B'C'$ and also $BC = B'C'$, so right triangle $BCD$ has hypotenuse $BD$ equal to leg $BC$, which is a contradiction."
... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0gve | A square board $6 \times 6$ is filled out with positive integer numbers (by one number in every cell). Each step, it is allowed to choose a square larger than $1 \times 1$ and increase by 1 all numbers in its cells. Is it always possible to reach, after several steps, a situation when all numbers on the board are divis... | [
"Будемо позначати через $S$ суму чисел, які містяться в клітинках зі знаком “+”, а через $Q$ — суму чисел, які містяться в клітинках зі знаком “-”. Нескладно перевірити, що на кожному кроці різниця $S-Q$ або не змінюється, або збільшується на 3, або ж — зменшується на 3. А тому, якщо спочатку $S-Q \\neq 0 \\pmod{3}... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | No | |
0e4w | Problem:
Dan je konveksen štirikotnik $ABCD$, točki $E$ in $F$ na stranici $AB$ ter točka $G$ na stranici $CD$, da so štirikotniki $ABCG$, $A F C D$ in $E F C G$ tetivni. Dokaži, da je $|AE|=|FB|$ natanko tedaj, ko sta stranici $AB$ in $CD$ vzporedni. | [
"Solution:\n\nOznačimo $\\angle DCA=\\gamma$. Zaradi tetivnosti štirikotnika $A F C D$ je\n$$\n\\angle DFA=\\angle DCA=\\gamma\n$$\nIz tetivnosti štirikotnika $ABCG$ sledi $\\angle GBA=\\angle GCA=\\gamma$. Od tod dobimo vzporednost premic $DF$ in $GB$.\n\nV tetivnem štirikotniku $A F C D$ velja $\\angle FAD=\\pi-\... | Slovenia | 55. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0blp | Let $f : [0, 1] \to [0, 1]$ be a function with the property that for every $y \in [0, 1]$ and every $\varepsilon > 0$ there exists $x \in [0, 1]$ so that $|f(x) - y| < \varepsilon$.
a) Prove that if $f$ is continuous on $[0, 1]$, then $f$ is surjective.
b) Give an example of a function $f$ with the given property which... | [
"a) Suppose a continuous function $f : [0, 1] \\to [0, 1]$ has the given property and take $y \\in [0, 1]$. From the hypothesis, there exists a sequence $(x_n)_{n \\ge 1}$, included in $[0, 1]$, so that $|f(x_n) - y| < 1/n$, $\\forall n \\ge 1$. The sequence $(x_n)_{n \\ge 1}$ is bounded, so it has a convergent sub... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Precalculus > Functions",
"Precalculus > Limits"
] | null | proof and answer | a) f is surjective. b) Define f on the unit interval by f equals the identity on rational points and equals zero on irrational points. | |
053b | Let $x_1, \dots, x_n$ be non-negative real numbers, not all of which are zeros.
(i) Prove that
$$
1 \le \frac{\left(x_1 + \frac{x_2}{2} + \frac{x_3}{3} + \dots + \frac{x_n}{n}\right) \cdot \left(x_1 + 2x_2 + 3x_3 + \dots + nx_n\right)}{\left(x_1 + x_2 + x_3 + \dots + x_n\right)^2} \le \frac{(n+1)^2}{4n}.
$$
(ii) Show t... | [
"Applying AM-GM gives\n$$\n\\begin{aligned} \\left(\\sum_{k=1}^{n} \\frac{x_k}{k}\\right) \\left(\\sum_{k=1}^{n} kx_k\\right) &= \\frac{1}{n} \\cdot \\left(\\sum_{k=1}^{n} \\frac{nx_k}{k}\\right) \\left(\\sum_{k=1}^{n} kx_k\\right) \\le \\\\ &\\le \\frac{1}{n} \\cdot \\frac{1}{4} \\left(\\sum_{k=1}^{n} \\frac{nx_k}... | Estonia | IMO Team Selection Contest | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
08os | Problem:
Positive integers are put into the following table
| 1 | 3 | 6 | 10 | 15 | 21 | 28 | 36 | | |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 | | |
| 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 | | |
| 7 | 12 | 18 | 25 | 33 | 42 | | | | ... | [
"Solution:\n\nWe shall observe straight lines as on the next picture. We can call these lines diagonals.\n\n| 1 | 3 | 6 | 10 | 15 | 21 | 28 | 36 | |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 | |\n| 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 | |\n... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | row 2, column 62 | |
09au | If $a_i \ge 0$, prove that
$$
\frac{n(n+1)}{2} \left( \prod_{i=1}^{n} A_i^i \right)^{\frac{2}{n(n+1)}} \ge \frac{n(n-1)}{2} \left( \prod_{i=1}^{n-1} A_i^i \right)^{\frac{2}{n(n-1)}} + nG_n,
$$
where $A_k = \sum_{i=1}^{k} \frac{2ia_i}{k(k+1)}$, $G_k = \left( \prod_{i=1}^{k} a_i^i \right)^{\frac{2}{k(k+1)}}$. | [
"The inequality is equivalent to\n$$\n\\frac{n(n+1)}{2} \\ge \\frac{n(n-1)}{2} \\left[ \\frac{\\prod_{i=1}^{n-1} A_i^{i(n+1)}}{\\prod_{i=1}^{n} A_i^{i(n-1)}} \\right]^{\\frac{2}{n(n-1)(n+1)}} + n \\left[ \\frac{\\prod_{i=1}^{n} a_i^i}{\\prod_{i=1}^{n} A_i^i} \\right]^{\\frac{2}{n(n+1)}}\n$$\nAccording to AM-GM ineq... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
03lb | Problem:
Find the last three digits of the number $2003^{2002^{2001}}$. | [
"Solution:\nWe must find the remainder when $2003^{2002^{2001}}$ is divided by $1000$, which will be the same as the remainder when $3^{2002^{2001}}$ is divided by $1000$, since $2003 \\equiv 3 \\pmod{1000}$. To do this we will first find a positive integer $n$ such that $3^{n} \\equiv 1 \\pmod{1000}$ and then try ... | Canada | 2003 CMO | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | 241 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.