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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0ing | Triangle $ABC$ is inscribed in circle $\omega$. The tangent lines to $\omega$ at $B$ and $C$ meet at $T$. Point $S$ lies on ray $BC$ such that $AS \perp AT$. Points $B_1$ and $C_1$ lie on ray $ST$ (with $C_1$ in between $B_1$ and $S$) such that $B_1T = BT = C_1T$. Prove that triangles $ABC$ and $AB_1C_1$ are similar to... | [
"**Solution 1.** (Based on work by Oleg Golberg) We start with an important geometric observation.\n\n**Lemma.** Triangle $ABC$ inscribed in circle $\\omega$. Lines $BT$ and $CT$ are tangent to $\\omega$. Let $M$ be the midpoint of side $BC$. Then $\\angle BAT = \\angle CAM$. (Line $AT$ is a symmedian of triangle.)... | United States | Team Selection Test | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Circle... | null | proof only | null | |
0kzn | Jerry likes to play with numbers. One day, he wrote all the integers from $1$ to $2024$ on the whiteboard. Then he repeatedly chose four numbers on the whiteboard, erased them, and replaced them by either their sum or their product. (For example, Jerry's first step might have been to erase $1$, $2$, $3$, and $5$, and t... | [
"Each time this operation was performed, the number of even integers on the whiteboard was reduced by at most $3$. There were $\\frac{2024}{2} = 1012$ even integers on the whiteboard initially. Because $\\frac{1011}{3} = 337$ and $\\frac{1014}{3} = 338$, Jerry needed at least $338$ operations to eliminate all of th... | United States | AMC 10 B | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | A | |
0ek6 | Problem:
Kolikšna je absolutna vrednost razlike rešitev enačbe $\left(1-\left(1+x^{-2}\right)^{-1}\right)^{-1}=3,25$?
(A) -3
(B) -1
(C) 1
(D) 3
(E) 0 | [
"Solution:\n\nEnačbo prevedemo v obliko $x^{-2}=\\frac{4}{9}$. Rešitvi sta $\\frac{3}{2}=1 \\frac{1}{2}$ in $-\\frac{3}{2}=-1 \\frac{1}{2}$. Števili se razlikujeta za 3. Pravilen je odgovor $D$."
] | Slovenia | 22. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | D | |
0g2s | Problem:
Sei $n$ eine natürliche Zahl. In einer Reihe stehen $n+1$ Schüsseln, die von links nach rechts mit den Zahlen $0, 1, \ldots, n$ nummeriert sind. Am Anfang liegen $n$ Steine in der Schüssel $0$ und kein Stein in den anderen Schüsseln. Sisyphus will diese $n$ Steine in die Schüssel $n$ bewegen. Dafür bewegt Sis... | [
"Solution:\n\nWir ordnen jedem Stein eine eindeutige Zahl $k \\in \\{1, 2, \\ldots, n\\}$ zu und nennen diese Zahl seine Priorität. Wenn Sisyphus einen Zug macht, wählt er ein Feld aus und bewegt von diesem Feld oBdA den Stein mit der grösstmöglichen Priorität (dies ist erlaubt, weil die Steine ununterscheidbar sin... | Switzerland | SMO 2019 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
05jo | Problem:
Soit $ABCD$ un quadrilatère convexe. Pour tout point $M$ à l'intérieur ou sur le bord de $ABCD$, on pose
$$
f(M) = MA + MB + MC + MD.
$$
Prouver que, pour tout $M$, on a
$$
f(M) \leqslant \max(f(A), f(B), f(C), f(D))
$$ | [
"Solution:\n\nLemme. Si le point $K$ est sur les bords ou à l'intérieur du triangle $XYZ$, on a\n$$\nXY + XZ \\geqslant KY + KZ.\n$$\nPreuve du lemme. Si $K = Y$, le résultat est immédiat d'après l'inégalité triangulaire. Sinon, on note $P$ l'intersection de $(YK)$ avec le côté $[XZ]$. D'après l'inégalité triangula... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellan... | null | proof only | null | |
0fk4 | Problem:
Sean $a, b, c, d$ números enteros positivos que satisfacen $a b = c d$. Demostrar que $a + b + c + d$ no es un número primo. | [
"Solution:\nUsando la hipótesis $a b = c d$ se escribe\n$$\na(a + b + c + d) = (a + c)(a + d)\n$$\nde donde se obtiene que si $a + b + c + d$ fuese primo debería dividir a $a + c$ o $a + d$ que son menores que él."
] | Spain | Spanish Mathematical Olympiad - Local Stage | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
08ew | Problem:
Una successione $x_{1}, x_{2}, \ldots, x_{n}, \ldots$ è costituita da un blocco iniziale di $p$ interi positivi distinti, che poi si ripetono periodicamente. Questo vuol dire che $\left\{x_{1}, x_{2}, \ldots, x_{p}\right\}$ sono $p$ interi positivi distinti, e $x_{n+p}=x_{n}$ per ogni intero positivo $n$.
I t... | [
"Solution:\n\nDividiamo la dimostrazione in 3 parti: prima esibiamo una strategia in quattro passi per il punto (a), poi esibiamo una strategia in cinque passi per il punto (b) con $k \\leq 52$, ed infine mostriamo l'ottimalità del numero dei passi di tali strategie.\n\nStrategia per il punto (a)\n\nUna possibile s... | Italy | XXXVII Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Modular Arithmetic > Chinese re... | null | proof and answer | a: 4; b: k ≤ 52 | |
05qv | Problem:
Soit $n$ un entier naturel impair, et soit $a_{1}, \ldots, a_{n}$ des entiers naturels non nuls. On note $A$ le produit des entiers $a_{i}$, et $d$ leur plus grand diviseur commun.
Montrer que
$$
\operatorname{PGCD}\left(a_{1}^{n}+A, a_{2}^{n}+A, \ldots, a_{n}^{n}+A\right) \leqslant 2 d^{n}
$$ | [
"Solution:\n\nPour tout $i$, on pose $b_{i}=a_{i} / d$, de sorte que $\\operatorname{PGCD}\\left(b_{1}, \\ldots, b_{n}\\right)=1$, et on pose également $B=b_{1} \\times \\ldots \\times b_{n}$ et $\\Delta=\\operatorname{PGCD}\\left(b_{1}^{n}+B, b_{2}^{n}+B, \\ldots, b_{n}^{n}+B\\right)$. Alors $\\operatorname{PGCD}\... | France | Préparation Olympique Française de Mathématiques | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0b1m | Problem:
If $a^{3}+b^{3}+c^{3}=3 a b c=6$ and $a^{2}+b^{2}+c^{2}=8$, find the value of
$$
\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}
$$ | [
"Solution:\nSince $a, b, c$ are distinct and $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)=0$, then $a+b+c=0$. From $a+b+c=0$ and $a^{2}+b^{2}+c^{2}=8$, we have $a b+b c+c a=-4$ and $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=(a b+b c+c a)^{2}-2 a b c(a+b+c)=(a b+b c+c a)^{2}=16$. Thus,\n$... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | -8 | |
0ksl | Problem:
Show that the sum $AP^{4} + BP^{4} + CP^{4}$ does not depend on $P$, where $P$ is a point on the circumcircle of equilateral triangle $\triangle ABC$. | [
"Solution:\nWLOG assume $P$ is between $A$ and $B$, and let $s$ be the side length of $\\triangle ABC$. By Ptolemy's Theorem,\n$$\n(PA + PB) \\cdot s = PA \\cdot BC + PB \\cdot AC = PC \\cdot AB = PC \\cdot s \\Longrightarrow PA + PB = PC.\n$$\nBy the law of cosines on $\\triangle PAB$,\n$$\nPA^{2} + PB^{2} - 2 PA ... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0eud | For a positive integer $m \ge 2$, define
$$
A_m := \{ m+1, 3m+2, 5m+3, 7m+4, 9m+5, \dots \}.
$$
(1) Prove that, for any given $m \ge 2$, there exists a positive integer $a$, $1 \le a < m$, such that either $2^a \in A_m$ or $2^a + 1 \in A_m$.
(2) Assume that, for some $m \ge 2$, there exist positive integers $a$ and $b... | [
"(1) An arbitrary element of $A_m$ can be written in the form $m + 1 + k(2m + 1)$, where $k = 0, 1, 2, \\dots$. Putting $\\epsilon = 0$ or $1$,\n$$\nm+1+k(2m+1) = 2^a + \\epsilon \\iff 2^a \\equiv m+1-\\epsilon \\pmod{2m+1}\n$$\n$$\n\\iff 2^{a+1} \\equiv 1-2\\epsilon \\pmod{2m+1}\n$$\n$$\n\\iff 2^{a+1} \\equiv \\be... | South Korea | Korean Mathematical Olympiad Final Round | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof and answer | b0 = (a0 - 1)/2 | |
0397 | Find all positive integers $n$ such that if $a, b, c \ge 0$ and $a+b+c=3$, then $abc(a^n + b^n + c^n) \le 3$. | [
"For $a=2$, $b=c=\\frac{1}{2}$ and $n \\ge 3$ the inequality is not satisfied. On the other hand, for $n=1$ it is equivalent to the AM-GM inequality. It remains to consider the case $n=2$.\nWe shall prove that the inequality is true for $n=2$.\n\n*First solution.* Set $x = bc$. Then\n$$\nabc(a^2 + b^2 + c^2) = ax(a... | Bulgaria | Winter Mathematical Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | n = 1 or n = 2 | |
02pg | Problem:
Num tabuleiro $123 \times 123$, cada casa é pintada de roxo ou azul de acordo com as seguintes condições:
- Cada casa pintada de roxo que não está na borda do tabuleiro tem exatamente 5 casas azuis dentre suas 8 vizinhas.
- Cada casa pintada de azul que não está na borda do tabuleiro tem exatamente 4 casas ro... | [
"Solution:\n\n(a) Observando um tabuleiro $3 \\times 3$, podemos claramente ver que seu centro não está na borda do tabuleiro. A casa do centro pode:\n- Estar pintada de roxo. Nesse caso, temos dentre suas 8 vizinhas, 5 azuis e 3 roxas. No total, há 4 casas roxas e 5 casas azuis nesse tabuleiro.\n- Estar pintada de... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | (a) 4 purple and 5 blue; (b) 6724 purple | |
0blq | Determine the integers $n \ge 2$ such that $a^2 - a + 2 = 0$ in $\mathbb{Z}_n$ for a unique $a$ in $\mathbb{Z}_n$. | [
"We show that $7$ is the sole integer satisfying the required conditions. If $a \\in \\mathbb{Z}_n$ and $a^2 - a + 2 = 0$ in $\\mathbb{Z}_n$, then $(1-a)^2 - (1-a) + 2 = a^2 - a + 2 = 0$ in $\\mathbb{Z}_n$, so uniqueness forces $a = 1-a$, i.e., $2a = 1$. In particular, $2$ is invertible in $\\mathbb{Z}_n$ and $a = ... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 7 | |
064u | If the number $4\nu + 3$, where $\nu$ is an integer, is a multiple of $11$, find:
(i) The form of the integer $\nu$,
(ii) The remainder of the division of $\nu^4$ with $11$. | [
"(i) Let $4\\nu + 3 = 11\\lambda$, $\\lambda \\in \\mathbb{Z}$. Then $\\nu = \\frac{11\\lambda - 3}{4} = 2\\lambda + \\frac{3(\\lambda - 1)}{4}$, whence $4$ must divide $3(\\lambda - 1)$. Since $(4,3) = 1$, we get that $4|(\\lambda - 1)$.\nHence $\\lambda - 1 = 4\\kappa$, $\\kappa \\in \\mathbb{Z}$, and therefore\n... | Greece | 24th Hellenic Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | (i) ν = 11k + 2 for integer k. (ii) The remainder is 5. | |
0930 | Problem:
Determine all pairs of polynomials $(P, Q)$ with real coefficients satisfying
$$
P(x+Q(y))=Q(x+P(y))
$$
for all real numbers $x$ and $y$. | [
"Solution:\nIf either $P$ or $Q$ is constant then clearly $P \\equiv Q$. Suppose neither of $P, Q$ is constant.\nWrite $P(x)=a x^{n}+b x^{n-1}+R(x)$ and $Q(x)=c x^{m}+d x^{m-1}+S(x)$ with $n, m \\geqslant 1, a \\neq 0 \\neq c$, $\\operatorname{deg} R<n-1$, $\\operatorname{deg} S<m-1$.\nThen the degree of $P(x+Q(y))... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | All pairs (P, Q) of real-coefficient polynomials such that either (i) P ≡ Q (any polynomial), or (ii) P(x) = x + a and Q(x) = x + b for some real constants a, b. | |
0ie4 | Problem:
Let $f(x) = x^{3} + a x + b$, with $a \neq b$, and suppose the tangent lines to the graph of $f$ at $x = a$ and $x = b$ are parallel. Find $f(1)$. | [
"Solution:\nSince $f'(x) = 3x^{2} + a$, we must have $3a^{2} + a = 3b^{2} + a$. Then $a^{2} = b^{2}$, and since $a \\neq b$, $a = -b$. Thus $f(1) = 1 + a + b = 1$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | proof and answer | 1 | |
0kf4 | Problem:
Positive real numbers $x$ and $y$ satisfy
$$
||\cdots|||x|-y|-x| \cdots-y|-x|=||\cdots|||y|-x|-y| \cdots-x|-y|
$$
where there are 2019 absolute value signs $|\cdot|$ on each side. Determine, with proof, all possible values of $\frac{x}{y}$. | [
"Solution:\n\nClearly $x = y$ works.\n\nElse, WLOG $x < y$, define $d = y - x$, and define $f(z) := ||z - y| - x|$ so our expression reduces to\n$$\nf^{1009}(x) = \\left|f^{1009}(0) - y\\right|\n$$\nNow note that for $z \\in [0, y]$, $f(z)$ can be written as\n$$\nf(z) = \\begin{cases} d - z, & 0 \\leq z \\leq d \\\... | United States | HMMT February 2020 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 1/3, 1, 3 | |
0d5z | Hamza and Majid play a game on a horizontal $3 \times 2015$ white board. They alternate turns, with Hamza going first. A legal move for Hamza consists of painting three unit squares forming a horizontal $1 \times 3$ rectangle. A legal move for Majid consists of painting three unit squares forming a vertical $3 \times 1... | [
"Hamza has a winning strategy.\nWe divide the rectangle into $671$ squares of size $3 \\times 3$ and a small rectangle of size $3 \\times 2$.\nHamza will play as follows: He will paint at each time a horizontal $1 \\times 3$ rectangle in a white $3 \\times 3$ square (all unit squares in this $3 \\times 3$ square ar... | Saudi Arabia | SAMC 2015 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English, Arabic | proof and answer | Hamza. Strategy: Always paint a horizontal three-square segment entirely within an untouched three-by-three block until none remain; this claims each block and prevents any vertical move there, guaranteeing at least 1008 moves for Hamza and ensuring he makes the last move. | |
0c4k | The midpoints of the sides $AB$, $BC$, $CD$, $DA$ of the convex quadrilateral $ABCD$ are $E$, $F$, $G$, respectively $H$.
a) Prove that there exists an unique point $O$ inside the quadrilateral so that the areas of the quadrilaterals $AEOH$, $BFOE$, $CGOF$, $DHOG$ are equal.
b) Locate $O$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | O is the unique intersection of the line through the midpoint of diagonal AC parallel to BD and the line through the midpoint of diagonal BD parallel to AC. | |
07l2 | Prove that if $n$ is a positive integer either $3n$ or $7n$ contains an odd digit. | [
"Suppose a number $n$ exists with the property that $3n$ and $7n$ have only even digits and let $N$ be the smallest such. If $N$ is divisible by $10$, $\\frac{N}{10}$ is a smaller such number. Hence $N$ ends in one of $2, 4, 6, 8$. But $3N + 7N = 10N$ and $10N$ has the tens digit odd as a result of the carry from t... | Ireland | Irska | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0i9p | Let $a, b, c$ be real numbers in the interval $(0, \frac{\pi}{2})$. Prove that
$$
\frac{\sin a \sin(a-b) \sin(a-c)}{\sin(b+c)} + \frac{\sin b \sin(b-c) \sin(b-a)}{\sin(c+a)} + \frac{\sin c \sin(c-a) \sin(c-b)}{\sin(a+b)} \ge 0.
$$ | [
"By the **Product-to-sum formulas** and the **Double-angle formulas**, we have\n$$\n\\begin{aligned}\n\\sin(\\alpha - \\beta) \\sin(\\alpha + \\beta) &= \\frac{1}{2}[\\cos 2\\beta - \\cos 2\\alpha] \\\\\n&= \\sin^2 \\alpha - \\sin^2 \\beta.\n\\end{aligned}\n$$\nHence, we obtain\n$$\n\\begin{aligned}\n& \\sin a \\si... | United States | USA IMO 2003 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0bq4 | Problem:
Determinați valorile întregi ale lui $x$ și $y$ astfel încât
$$
x - 3y + 4 = 0 \quad \text{și} \quad \sqrt{x^{2} + 7y^{2} + 8x + 8y + 4} \in \mathbb{Q}
$$ | [] | Romania | OLIMPIADA DE MATEMATICĂ - ETAPA LOCALĂ | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (x, y) = (-10, -2) | |
04a8 | In how many ways can the number $\frac{2011}{2010}$ be represented as a product of two fractions of the form $\frac{n+1}{n}$, where $n$ is a positive integer? (Order of the factors is not important.) | [
"Let $p$ and $q$ be positive integers such that $\\frac{2011}{2010} = \\frac{p+1}{p} \\cdot \\frac{q+1}{q}$.\nThen $2011pq = 2010(pq + p + q + 1)$ i.e. $pq = 2010(p + q + 1)$.\n\nFrom the last equation we find\n$$\np = \\frac{2010(q+1)}{q-2010} = \\frac{2010(q-2010)+2010 \\cdot 2011}{q-2010} = 2010 + \\frac{2010 \\... | Croatia | CroatianCompetitions2011 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 16 | |
05x1 | Problem:
Soit $ABCDE$ un pentagone cyclique convexe tel que $AB = BD$. Soit $P$ le point d'intersection des droites $(EB)$ et $(AC)$. Soit $Q$ le point d'intersection des droites $(BC)$ et $(DE)$. Montrer que $(PQ)$ et $(AD)$ sont parallèles. | [
"Solution:\n\n\n\nSur la figure, il semble que $EPCQ$ est un quadrilatère cyclique. On le montre :\n$$\n\\begin{aligned}\n\\widehat{QCP} & = 180^{\\circ} - \\widehat{PCB} \\\\\n& = 180^{\\circ} - \\widehat{ACB} \\\\\n& = 180^{\\circ} - \\widehat{ADB} \\text{ par angle inscrit } \\\\\n& = 18... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02dy | $ABCD$ is any convex quadrilateral. Squares center $E, F, G, H$ are constructed on the outside of the edges $AB, BC, CD$ and $DA$ respectively. Show that $EG$ and $FH$ are equal and perpendicular. | [
"Consider $A, B, C, D$ to be the points $a, b, c, d$ in the complex plane. Then $E, F, G, H$ are the points $\\frac{a+b}{2} + i\\frac{b-a}{2}$, $\\frac{b+c}{2} + i\\frac{c-b}{2}$, $\\frac{c+d}{2} + i\\frac{d-c}{2}$, $\\frac{d+a}{2} + i\\frac{a-d}{2}$. Hence vector $\\vec{GE}$ is represented by the complex number $h... | Brazil | VI OBM | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof only | null | |
0dla | Determine all pairs $(m, n)$ of non-negative integers such that
$$
2n! = m!(m! + 2).
$$ | [
"The answer is $(3, 4)$.\n\nFor $m = 0, 1$ or $2$, we obtain the impossible $2n! = 3$ and $2n! = 8$. However, $m = 3$ works with $n = 4$.\n\nNow let $m \\ge 4$. Clearly, $2n! > 2m!$, which implies $n > m$, so we can write $n = m + a$ for some positive integer $a$. The equation transforms into\n$$\n2(m + 1)(m + 2)(m... | Saudi Arabia | Saudi Booklet | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (3, 4) | |
0bnh | Determine the continuous increasing functions $f: [0, \infty) \to \mathbb{R}$ satisfying
$$
\int_{0}^{x+y} f(t) dt = \int_{0}^{x} f(t) dt + \int_{0}^{y} f(t) dt,
$$
for all non-negative real numbers $x$ and $y$. | [
"Clearly, every constant function satisfies the required conditions. Conversely, write the condition in the statement in the equivalent form\n$$\n\\int_{x}^{x+y} f(t) \\, dt \\le \\int_{0}^{y} f(t) \\, dt\n$$\nto infer that $\\int_{0}^{y} f(t+x) \\, dt \\le \\int_{0}^{y} f(t) \\, dt$ for all non-negative $x$ and $y... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All constant functions f(x) = c for x ≥ 0 | |
0dqf | Given $a_1 \ge 1$ and $a_{k+1} \ge a_k + 1$ for all $k = 1, 2, \dots, n$, show that
$$
a_1^3 + a_2^3 + \dots + a_n^3 \ge (a_1 + a_2 + \dots + a_n)^2.
$$ | [
"We will prove it by induction. First, it is clear that $a_1^3 \\ge a_1^2$ since $a_1 \\ge 1$.\n\nNext, suppose it is true for $n$ terms. Then\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n+1} a_k^3 &\\ge a_{n+1}^3 + \\sum_{k=1}^n a_k^3 \\ge a_{n+1}^3 + \\left(\\sum_{k=1}^n a_k\\right)^2 \\\\\n&= \\left(\\sum_{k=1}^{n+1} a_... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0jqw | Problem:
Let $ABCD$ be a convex quadrilateral whose diagonals $AC$ and $BD$ meet at $P$. Let the area of triangle $APB$ be $24$ and let the area of triangle $CPD$ be $25$. What is the minimum possible area of quadrilateral $ABCD$? | [
"Solution:\n\nNote that $\\angle APB = 180^{\\circ} - \\angle BPC = \\angle CPD = 180^{\\circ} - \\angle DPA$ so $4[ BPC ][ DPA ] = (PB \\cdot PC \\cdot \\sin BPC)(PD \\cdot PA \\cdot \\sin DPA) = (PA \\cdot PB \\cdot \\sin APB)(PC \\cdot PD \\cdot \\sin CPD) = 4[ APB ][ CPD ] = 2400 \\Longrightarrow [ BPC ][ DPA ]... | United States | HMMT November 2015 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / P... | null | proof and answer | 49 + 20*sqrt(6) | |
0j6t | Problem:
Let $n$ be a positive integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be a set of positive integers such that $a_{1}=2$ and $a_{m}=\varphi\left(a_{m+1}\right)$ for all $1 \leq m \leq n-1$, where, for all positive integers $k$, $\varphi(k)$ denotes the number of positive integers less than or equal to $k$ that ... | [
"Solution:\n\nWe first note that $\\varphi(s)<s$ for all positive integers $s \\geq 2$, so $a_{m}>2$ for all $m>1$.\n\nFor integers $s>2$, let $A_{s}$ be the set of all positive integers $x \\leq s$ such that $\\gcd(s, x)=1$. Since $\\gcd(s, x)=\\gcd(s, s-x)$ for all $x$, if $a$ is a positive integer in $A_{s}$, so... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0k97 | Problem:
Find all ordered pairs $(a, b)$ of positive integers such that $2a + 1$ divides $3b - 1$ and $2b + 1$ divides $3a - 1$. | [
"Solution:\n\nThis is equivalent to the existence of nonnegative integers $c$ and $d$ such that $3b - 1 = c(2a + 1)$ and $3a - 1 = d(2b + 1)$. Then\n$$\nc d = \\frac{(3b - 1)(3a - 1)}{(2a + 1)(2b + 1)} = \\frac{3a - 1}{2a + 1} \\cdot \\frac{3b - 1}{2b + 1} < \\frac{3}{2} \\cdot \\frac{3}{2} = 2.25.\n$$\nNeither $c$... | United States | HMMT November 2019 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2, 2), (12, 17), (17, 12) | |
05ai | Let $n$ be a positive integer divisible by $4$. Prove that
$$\sin^2\left(1 \cdot \frac{360^\circ}{n}\right) + \sin^2\left(2 \cdot \frac{360^\circ}{n}\right) + \sin^2\left(3 \cdot \frac{360^\circ}{n}\right) + \dots + \sin^2\left(n \cdot \frac{360^\circ}{n}\right) = \frac{n}{2}.$$ | [
"Let $n = 4k$. For each $i = 1, 2, \\dots, n$, denote $\\alpha_i = i \\cdot \\frac{360^\\circ}{n}$ and $x_i = \\sin^2 \\alpha_i$. We pair the terms as $(x_1, x_{k+1}), (x_2, x_{k+2}), \\dots, (x_k, x_{2k})$ and $(x_{2k+1}, x_{3k+1}), (x_{2k+2}, x_{3k+2}), \\dots, (x_{3k}, x_{4k})$. Since\n$$\n\\alpha_{i+k} = (i + k... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0gdx | 設整數 $N > 2^{5000}$。試證: 若 $1 \le a_1 < \cdots < a_k < 100$ 為相異正整數, 則
$$
\prod_{i=1}^{k} (N^{a_i} + a_i)
$$
有至少 $k$ 個相異質因數。
Let $N > 2^{5000}$ be a positive integer. Prove that if $1 \le a_1 < \dots < a_k < 100$ are distinct positive integers then the number
$$
\prod_{i=1}^{k} (N^{a_i} + a_i)
$$
has at least $k$ distinc... | [
"First, we prove the following lemma.\n\n**Lemma.** For any positive integers $X, A$ and $B$ with $A \\neq B$ we have\n$$\n\\text{gcd}(X^A + A, X^B + B) < A^B + B^A.\n$$\n\n*Proof of lemma.* Note that $X^B + B$ divides $X^{AB} - (-B)^A$, and that\n$$\n\\text{gcd}(X^A + A, (X^A)^B - (-B)^A) = \\text{gcd}(X^A + A, (-... | Taiwan | 2020 Taiwan IMO 1J | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0jp5 | Problem:
Let $ABCD$ be a quadrilateral with $\angle BAD = \angle ABC = 90^{\circ}$, and suppose $AB = BC = 1$, $AD = 2$. The circumcircle of $ABC$ meets $\overline{AD}$ and $\overline{BD}$ at points $E$ and $F$, respectively. If lines $AF$ and $CD$ meet at $K$, compute $EK$. | [
"Solution:\n\nAnswer: $\\frac{\\sqrt{2}}{2}$\n\nAssign coordinates such that $B$ is the origin, $A$ is $(0,1)$, and $C$ is $(1,0)$. Clearly, $E$ is the point $(1,1)$. Since the circumcenter of $ABC$ is $\\left(\\frac{1}{2}, \\frac{1}{2}\\right)$, the equation of the circumcircle of $ABC$ is $\\left(x-\\frac{1}{2}\\... | United States | HMMT February 2015 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof and answer | √2/2 | |
0dmg | Problem:
Доказати да за позитивне реалне бројеве $a, b$ и $c$, такве да је $a+b+c=1$, важи неједнакост
$$
\frac{1}{b c+a+\frac{1}{a}}+\frac{1}{c a+b+\frac{1}{b}}+\frac{1}{a b+c+\frac{1}{c}} \leqslant \frac{27}{31}
$$
(Марко Радовановић са сарадницима) | [
"Solution:\n\nТражена неједнакост је очигледно еквивалентна неједнакости\n$$\n\\frac{a}{p+a^{2}}+\\frac{b}{p+b^{2}}+\\frac{c}{p+c^{2}} \\leq \\frac{27}{31}\n$$\nгде је $a+b+c=1$ и $p=a b c+1$. Посматраћемо функцију\n$$\nf(x)=\\frac{3(a+b+c)}{3 x+a^{2}+b^{2}+c^{2}}-\\frac{a}{x+a^{2}}-\\frac{b}{x+b^{2}}-\\frac{c}{x+c... | Serbia | СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
03zv | Find a triple $(l, m, n)$ ($1 < l < m < n$) of positive integers such that $\sum_{k=1}^{l} k$, $\sum_{k=l+1}^{m} k$, $\sum_{k=m+1}^{n} k$ form a geometric sequence in order. (posed by Tao Pingsheng) | [
"$\\sum_{k=1}^{l} k = S_{l}$, $\\sum_{k=l+1}^{m} k = S_{m} - S_{l}$, $\\sum_{k=m+1}^{n} k = S_{n} - S_{m}$,\nform a geometric sequence in order. Then\n$$\nS_{l}(S_{n} - S_{m}) = (S_{m} - S_{l})^2, \\qquad \\textcircled{1}\n$$\nthat is $S_{l}(S_{n} + S_{m} - S_{l}) = S_{m}^2$. Thus, $S_{l} \\mid S_{m}^2$, that is\n$... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | (3, 11, 36) | |
07c0 | Suppose that $a_1, a_2, a_3, \dots$ and $b_1, b_2, b_3, \dots$ are two sequences of real numbers. These sequences are said to be Co-Algebraic if a non-zero two-variable polynomial $P(x, y)$ with real coefficients exists such that for each natural number $n$, $P(a_n, b_n) = 0$.
a) Prove that sequences $n$ and $2^n$ (fo... | [
"a) Assume to the contrary that there exists a non-zero polynomial $P(x, y) \\in \\mathbb{R}[x, y]$ such that for every $n \\in \\mathbb{N}$, $P(n, 2^n) = 0$. Let $d$ be the degree of $P$ with respect to its second variable, $y$. $P(x, y)$ can be written as,\n$$\nP(x, y) = p_d(x)y^d + \\cdots + p_1(x)y + p_0(x),\n$... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Intermediate Algebra > Exponential functions",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | a) Not co-algebraic. b) No. c) Such an index exists; for example one can take n = m + 4·5^k constructed as in the proof. | |
0cg1 | Points $D$ and $E$ are considered on the ($BC$) side of triangle $ABC$, with $D$ between $B$ and $E$.
About a point $R$ of the segment ($AE$) we will say that it is *remarkable* if the lines $PQ$ and $BC$ are parallel, where $\{P\} = DR \cap AC$ and $\{Q\} = CR \cap AB$.
About a point $R'$ of the segment ($AD$) we will... | [
"a) Applying Menelaus' theorem in the triangle $ABE$ with transversal $Q - R - C$, we get that $\\frac{AQ}{QB} \\cdot \\frac{BC}{CE} \\cdot \\frac{ER}{RA} = 1$, therefore $\\frac{AQ}{QB} = \\frac{CE}{BC} \\cdot \\frac{RA}{ER}$. Similarly, applying Menelaus' theorem in the triangle $AEC$ with transversal $P - R - D$... | Romania | 74th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem"
] | English | proof and answer | BD = CE = φ · DE, where φ = (1 + √5) / 2 | |
0kbi | Problem:
Let $P(x) = x^{3} + x^{2} - r^{2} x - 2020$ be a polynomial with roots $r, s, t$. What is $P(1)$? | [
"Solution:\n\nPlugging in $x = r$ gives $r^{2} = 2020$. This means $P(1) = 2 - r^{2} - 2020 = -4038$.",
"Solution:\n\nVieta's formulas give the following equations:\n$$\n\\begin{aligned}\nr + s + t & = -1 \\\\\nr s + s t + t r & = -r^{2} \\\\\nr s t & = 2020 .\n\\end{aligned}\n$$\nThe second equation is $(r + t)(... | United States | HMMT February 2020 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | -4038 | |
0dvx | Problem:
Poišči vsa petmestna števila $\overline{abcde}$, ki so deljiva z $9$ in za katera velja $\overline{ace} - \overline{bda} = 760$. | [
"Solution:\n\nEnačbo $\\overline{ace} - \\overline{bda} = 760$ preoblikujemo v $100a + 10c + e - 100b - 10d - a = 760$, od koder sledi, da je $e = a$. Zdaj lahko enačbo delimo z $10$ in dobimo $10(a - b) + (c - d) = 76$. Ločimo 2 možnosti, in sicer $c - d = 6$ ali $c - d = -4$.\n\nV prvem primeru je $c = d + 6$ in ... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 81828, 91269 | |
09do | $a$, $b$, $c$ нь $a + b + c = 3$ байх сөрөг биш бодит тоонууд байх.
$$
\left(a^2 b + \frac{2}{3}\right) \left(b^2 c + \frac{2}{3}\right) + \left(b^2 c + \frac{2}{3}\right) \left(c^2 a + \frac{2}{3}\right) \\
+ \left(c^2 a + \frac{2}{3}\right) \left(a^2 b + \frac{2}{3}\right) \le \frac{75}{4}
$$
tönygtéel bish biélén... | [
"Лемм. $a + b + c = 3$ бол $ab^2 + bc^2 + ca^2 \\le 4 - abc$ байна.\n\n$$\n\\left(a^2 b + \\frac{2}{3}\\right) \\left(b^2 c + \\frac{2}{3}\\right) + \\left(b^2 c + \\frac{2}{3}\\right) \\left(c^2 a + \\frac{2}{3}\\right) \\\\\n+ \\left(c^2 a + \\frac{2}{3}\\right) \\left(a^2 b + \\frac{2}{3}\\right) \\leq \\frac{75... | Mongolia | ММО-48 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | Mongolian | proof only | null | |
0hz2 | Problem:
If $f(x)$ is a monic quartic polynomial such that $f(-1) = -1$, $f(2) = -4$, $f(-3) = -9$, and $f(4) = -16$, find $f(1)$. | [
"Solution:\n\nThe given data tells us that the roots of $f(x) + x^{2}$ are $-1$, $2$, $-3$, and $4$. Combining with the fact that $f$ is monic and quartic, we get\n$$\nf(x) + x^{2} = (x + 1)(x - 2)(x + 3)(x - 4).\n$$\nHence\n$$\nf(1) = (1 + 1)(1 - 2)(1 + 3)(1 - 4) - 1 = (2)(-1)(4)(-3) - 1 = 2 \\times (-1) \\times 4... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 23 | |
0aus | Problem:
Inside square $A B C D$, a point $E$ is chosen so that triangle $D E C$ is equilateral. Find the measure of $\angle A E B$.
 | [
"Solution:\n\nSince $\\triangle D E C$ is an equilateral triangle, then $|D E| = |C E|$ and each angle has a measure of $60^{\\circ}$. This implies that $\\angle A D E$ has measure of $30^{\\circ}$. Since $|A D| = |D E|$, then $\\triangle A D E$ is an isosceles triangle. Thus, $\\angle D A E = \\angle D E A = 75^{\... | Philippines | 18th PMO National Stage Oral Phase | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 150° | |
0hc1 | Find the smallest positive integer of the form $\overline{30x070y03}$, which is divisible by $37$, where $x$, $y$ are digits. | [
"Let us rewrite this number as follows:\n$$\n\\overline{30x070y03} = 300070003 + 10^6x + 10^2y = 37 \\cdot (8110000 + 27027x + 3y) + (3 + x - 11y).\n$$\nThus, $3 + x - 11y$ must be divisible by $37$ for the smallest $x$. Since $x$, $y$ are digits, then this expression can only take values $0$, $-37$, $-74$. For eac... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | 300070703 | |
0l47 | In how many ways can 6 juniors and 6 seniors form 3 disjoint teams of 4 people so that each team has 2 juniors and 2 seniors?
(A) 720 (B) 1350 (C) 2700 (D) 3280 (E) 8100 | [
"Select the first junior and call this person $A$. There are 5 other juniors and $\\binom{6}{2} = 15$ pairs of seniors who could team up with $A$. Select the next junior not yet on a team, say $B$. There are 3 other juniors and $\\binom{4}{2} = 6$ pairs of seniors who could team up with $B$. The third team consists... | United States | AMC 10 A | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | B | |
001e | Sea $M = \{1,2,...,49\}$ el conjunto de los primeros 49 enteros positivos. Determine el máximo entero $k$ tal que el conjunto $M$ tiene un subconjunto de $k$ elementos en el que no hay 6 números consecutivos. Para ese valor máximo de $k$, halle la cantidad de subconjuntos de $M$, de $k$ elementos, que tienen la propied... | [] | Argentina | XVIII Olimpiada Iberoamericana de Matemática | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | español | proof and answer | k = 41; number of such subsets = 495 | |
07ko | Find $a_3, a_4, \dots, a_{2008}$, such that $a_i = \pm 1$ for $i = 3, \dots, 2008$ and
$$
\sum_{i=3}^{2008} a_i 2^i = 2008,
$$
and show that the numbers $a_3, a_4, \dots, a_{2008}$ are uniquely determined by these conditions. | [
"Existence: Dividing both sides by $8$, we require\n$$\n251 = \\sum_{i=0}^{2005} a_{i+3} 2^i.\n$$\nNow $251 = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7$. Also $-(1+2+\\dots+2^{m-1}) + 2^m = 1$, for $m = 1, 2, \\dots$ So\n$$\n251 = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7(-1 - 2 - \\dots - 2^{1997} + 2^{199... | Ireland | Irish Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | a3 = a5 = a6 = a7 = a8 = a9 = a2008 = +1 and a4 = a10 = a11 = a12 = ... = a2007 = -1; this assignment exists and is unique. | |
0h8g | Given three pairwise distinct positive integers $a$, $b$, $c$, whose product is $320$. Determine the smallest possible prime sum of these numbers. | [
"Clearly, their sum is greater than $2$, so the prime sum has to be odd. Since all three numbers can't be odd simultaneously, since their product is $320$, then two numbers are even and one is odd. There are exactly two odd divisors of $320 = 2^6 \\cdot 5$: $1$ and $5$. Consider these cases.\n\n$c = 1$, the followi... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 37 | |
0hs4 | Problem:
Let $0<\alpha<1$. Prove that there exists a real number $x$, $0<x<1$, such that $\alpha^{n}<\{n x\}$ for every positive integer $n$. (Here $\{n x\}$ is the fractional part of $n x$.) | [
"Solution:\nIt suffices to find some real (noninteger) $x$ with this last property; we then replace $x$ by $\\{x\\}$ to satisfy the $0<x<1$ condition (note that the other condition is still satisfied, since $\\{n x\\}=\\{n\\{x\\}\\}$). We claim that $x=\\sqrt{m^{2}-1}$ for a sufficiently large integer $m$ meets our... | United States | Berkeley Math Circle | [
"Number Theory > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
04uw | Given a scalene acute triangle $ABC$, let $M$ be the midpoint of its side $BC$ and $N$ the midpoint of the arc $BAC$ of its circumcircle. Let $\omega$ be the circle with diameter $BC$ and $D, E$ its intersections with the bisector of angle $BAC$. Points $D', E'$ lie on $\omega$ such that $DED'E'$ is a rectangle. Prove ... | [] | Czech Republic | Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gp4 | Let $a$, $b$, $c$ be the lengths of the sides of a triangle; $r_a$, $r_b$, $r_c$ be the corresponding exradii, respectively, and $r$ be the inradius. Prove that
$$
\frac{a+b+c}{2\sqrt{a^2+b^2+c^2}} \leq \frac{\sqrt{r_a^2+r_b^2+r_c^2}}{r_a+r_b+r_c-3r}
$$ | [
"Note that\n$$\n\\frac{a+b+c}{2}r = \\frac{b+c-a}{2}r_a = \\frac{a+b+c}{2}r_a - ar_a = \\frac{a+b+c}{2}r_b - br_b = \\frac{a+b+c}{2}r_c - cr_c.\n$$\nTherefore, we have $\\frac{3r(a+b+c)}{2} = \\frac{a+b+c}{2}(r_a + r_b + r_c) - ar_a - br_b - cr_c$ and hence\n$$\nar_a + br_b + cr_c = \\frac{a+b+c}{2}(r_a + r_b + r_c... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0ax4 | Problem:
A point $P$ is chosen randomly inside the triangle with sides $13$, $20$, and $21$. Find the probability that the circle centered at $P$ with radius $1$ will intersect at least one of the sides of the triangle. | [
"Solution:\nLet $ABC$ be a triangle with sides $BC = 13$, $CA = 20$, and $AB = 21$ and let $S$ be the set of points $P$ such that the circle $\\omega$ with radius $1$ centered at $P$ intersects at least one of the sides of $ABC$. For a fixed side of $ABC$ (say $\\ell$), $\\omega$ intersects $\\ell$ if and only if $... | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 75/196 | |
09c4 | **ББ-А1.** (Д.Ганзориг) $p$ анхны тоо. $p^2 = 2^n \cdot 3^m + 1$; $m, n \in \mathbb{N}$ хэлбэртэй бол $p \le 17$ гэж батал. | [
"$\\forall p > 3$ хувьд $p^2 - 1$-нь $2$, $3$, $24$-т хуваагдана гэдгийг хялбар харуулж болно. $p > 17$ үед $p^2 - 1$-нь $3$-аас их анхны хуваагчтай гэж харуульяа. Эсрэгээс нь тийм биш гэвэл $p-1$, $p+1$ нь $2 \\cdot 3^k$, $2^k$-тай тэнцүү байна. (харгалзан байх албагүй). $p+1 = 2^k$ байг. $n$-тэгш бол $2^n - 1 = (... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Mongolian | proof only | null | |
0gps | In an equilateral triangle $ABC$, let $D$ be a point on the side $[BC]$ other than the vertices. Let $I$ be the excenter of the triangle $ABD$ opposite to the side $[AB]$ and $J$ be the excenter of the triangle $ACD$ opposite to the side $[AC]$. Let the circumcircles of the triangles $AIB$ and $AJC$ intersect at the po... | [
"First note that $\\angle IBA = \\angle JCA = 60^\\circ$. Then $\\angle IEA = \\angle JEA = 60^\\circ$ and $\\angle IEJ = 120^\\circ$. Therefore A is on the angle bisector of the angle IEJ. Next observe that $\\angle IAD = 90^\\circ + \\angle BAD/2$ and $\\angle JAD = 90^\\circ + \\angle CAD/2$. Hence $\\angle IAJ ... | Turkey | Team Selection Test for JBMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08nc | Problem:
Let $AB$ and $CD$ be chords in a circle of center $O$ with $A, B, C, D$ distinct, and let the lines $AB$ and $CD$ meet at a right angle at point $E$. Let also $M$ and $N$ be the midpoints of $AC$ and $BD$ respectively. If $MN \perp OE$, prove that $AD \parallel BC$. | [
"Solution:\n\n$E$ can be inside, or outside the circle (Figure 3) but the proof below holds in both cases; notice that $E$ cannot be on the circle as $A, B, C, D$ are distinct. Let lines $AC$ and $NE$ meet at point $P$. Then $EN = DN = BN$ (median in a right triangle), so $\\angle PEC = \\angle NED = \\angle NDE = ... | JBMO | JBMO Shortlist | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
07mb | Let $n$ and $k$ be positive integers and suppose that $k$ is odd. Let $r$ be the greatest integer for which $2^r$ divides $n$. Let $h$ be the greatest integer for which $2^h$ divides $1 + k + k^2 + \dots + k^{n-1}$. Prove
(i) if $k \equiv 1 \pmod 4$, then $h = r$;
(ii) if $k \equiv 3 \pmod 4$ and $n$ is odd, then $h = ... | [
"The key observation for the proof is that for an integer $y$ with $y - 1$ divisible by 4,\n$$\n\\frac{y^2 - 1}{y - 1} = y + 1 \\quad \\text{is divisible by 2 but not by 4.}\n$$\n\nIf we write $n = 2^r m$, then $m$ is odd. First note that that\n$$\n1 + k + k^2 + \\dots + k^{n-1} = \\frac{k^n - 1}{k-1} = \\frac{(k^m... | Ireland | Irish Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
08y6 | Five points $A$, $B$, $C$, $D$, $E$, $P$ lie on a plane. Points $A$, $B$, $C$, $D$ lie on a straight line in this order. Furthermore, it is known that $AB = BC = CD = 6$, $PB = 5$ and $PC = 7$ are satisfied. Here we denote the length of the line segment $XY$ also by $XY$. Let $Q$ be the point of intersection, different... | [
"$\\frac{55\\sqrt{7}}{14}$\n\nLet $M$ be the point of intersection of line segments $BC$ and $PQ$. By using the well-known theorem on the power of a point with respect to a circle for the point $M$ and the two given circles, we get $AM \\cdot CM = PM \\cdot QM = DM \\cdot BM$. Combining this with the fact $AM + CM ... | Japan | Japan 2015 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 55√7/14 | |
00ul | Let $ABCD$ be a circumscribed quadrilateral (i.e. convex with sides that are all tangent to a single circle) and let $X$ be the intersection point of its diagonals $AC$ and $BD$. Let $I_1, I_2, I_3, I_4$ be the incenters of $\triangle DXC$, $\triangle BXC$, $\triangle AXB$, and $\triangle DXA$, respectively. The circum... | [
"We will prove the following auxiliary result.\n*Lemma.* Let $ABC$ be an arbitrary triangle, and $D$ be an arbitrary point on the segment $AB$. Denote by $O_1$ and $O_2$ the incenters of $\\triangle ADC$ and $\\triangle BDC$, respectively. If the circumcircle $\\omega$ of $\\triangle CO_1O_2$ intersects the sides $... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscel... | null | proof only | null | |
071d | Problem:
Let $A$, $B$ be two sets of $N$ consecutive integers. If $N=2003$, can we form $N$ pairs $(a,b)$ with $a \in A$, $b \in B$ such that the sums of the pairs are $N$ consecutive integers? What about $N=2004$? | [
"Solution:\nWithout loss of generality, let $A = B = \\{1, 2, \\ldots, N\\}$—if we have a solution for $A = \\{a+1, a+2, \\ldots, a+N\\}$ and $B = \\{b+1, b+2, \\ldots, b+N\\}$, then subtracting $a$ from every element of $A$ and $b$ from every element of $B$ gives a solution for $A = B = \\{1, 2, \\ldots, N\\}$.\n\... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | Yes for N equals 2003; No for N equals 2004. | |
0lbc | Let $a$, $b$ be two odd natural numbers. Suppose that $a$ is a divisor of $b^2 + 2$ and $b$ is a divisor of $a^2 + 2$. Prove that $a$ and $b$ belong to the sequence $(v_n)$ defined by
$$
v_1 = v_2 = 1 \quad \text{and} \quad v_n = 4v_{n-1} - v_{n-2} \quad \forall n \ge 3.
$$ | [
"From the given assumptions, we have\n$$\n(a, b) = 1, \\text{ and} \\tag{1a}\n$$\n$$\na^2 + b^2 + 2 \\equiv 0 \\pmod{ab}. \\tag{2a}\n$$\nBy (2a), there exists an integer $k$ such that $a^2 + b^2 + 2 = kab$. We will prove that $k = 4$.\nIf $a = b$, by (1a), we have $a = b = 1$. This implies that $k = 4$.\nConsider t... | Vietnam | Vietnam Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | Vietnamese | proof only | null | |
0490 | Determine all $a \in \mathbb{R}$ such that for every $x \in \mathbb{R}$ the following inequality holds:
$$
\frac{x}{x^2 + 2x + 3} > \frac{x+a}{1+x+x^2}.
$$ | [] | Croatia | CroatianCompetitions2011 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a ≤ -1 | |
0cz9 | Let $f = a X^{2} + b X + c \in \mathbb{Z}[X]$ be a polynomial such that for every positive integer $n$, $f(n)$ is a perfect square. Prove that $f = g^{2}$ for some polynomial $g \in \mathbb{Z}[X]$. | [
"The sequence $x_{n} = \\sqrt{f(n+1)} - \\sqrt{f(n)}$, $n \\geq 1$, contains only integers. We have\n$$\n\\lim_{n \\rightarrow \\infty} x_{n} = \\lim_{n \\rightarrow \\infty} \\frac{f(n+1) - f(n)}{\\sqrt{f(n+1)} + \\sqrt{f(n)}} = \\frac{2a}{2\\sqrt{a}} = \\sqrt{a}\n$$\nIt follows that $\\sqrt{a}$ is an integer, hen... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof only | null | |
05lk | Problem:
Dans un cirque, il y a plusieurs clowns. Chacun utilise au moins 5 couleurs, parmi 12 possibles, pour se peindre. Une même couleur est utilisée par au plus 20 clowns. Deux clowns n'ont jamais exactement les mêmes couleurs. Combien y a-t-il de clowns au maximum? | [
"Solution:\n\nSoit $n$ le nombre de clowns et $N$ le nombre de paires $(L, c)$ où $L$ est un clown et $c$ une couleur qu'il porte. D'une part, chaque clown ayant au moins 5 couleurs, $N \\geq 5 \\times n$. D'autre part, chacune des 12 couleurs peut apparaître dans au plus 20 paires. Donc $N \\leq 20 \\times 12$. On... | France | Envoi de combinatoire | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 48 | |
0cmk | Konstantin had two sets of 17 coins: the first set consists of 17 genuine coins, while the second set consists of 12 genuine and 5 false coins. (All the coins look equally; the genuine coins have equal weights, and the false coins have equal weights different from the weight of genuine ones; but it is not known if the ... | [
"Назовем набор из 17 настоящих монет хорошим, а набор из 12 настоящих и 5 фальшивых монет — плохим. Опишем возможные действия Кости.\n\nРазобьем 17 монет из оставшегося у него набора на четыре группы $A, B, C, D$, содержащие 2, 3, 5, и 7 монет соответственно. При первом взвешивании на одну чашу весов положим группы... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English; Russian | proof and answer | Yes | |
0cy4 | Let $a$ and $b$ be real numbers such that $a+b \neq 0$. Solve the equation
$$
\frac{1}{(x+a)^2-b^2}+\frac{1}{(x+b)^2-a^2}=\frac{1}{x^2-(a+b)^2}+\frac{1}{x^2-(a-b)^2}
$$ | [] | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | x = (a^2 + b^2) / (a + b) (with no solution if ab = 0 due to domain restrictions) | |
024k | Problem:
Cinco piratas encontraram um cofre do tesouro cheio de moedas de ouro e as dividiram entre si. Sabe-se que:
- O que o primeiro pirata recebeu é equivalente à metade do que receberam os outros quatro em conjunto.
- O que o segundo pirata recebeu é equivalente à terça parte do que receberam os outros quatro em ... | [
"Solution:\n\nSejam $a, b, c, d, e$ as quantidades de moedas recebidas pelos cinco piratas. O número total de moedas é $S = a + b + c + d + e$.\n\nO primeiro pirata recebeu metade do que receberam os outros quatro em conjunto, isto é, $a = (b + c + d + e)/2 = (S - a)/2$. Então $a = S/3$.\n\nO segundo pirata recebeu... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 1800 | |
0atn | Problem:
The base $AB$ of a triangular piece of paper $ABC$ is $16~\mathrm{cm}$ long. The paper is folded down over the base, with the crease $DE$ parallel to the base of the paper, as shown. The area of the triangle that projects below the base (shaded region) is $16\%$ that of the area of $\triangle ABC$. What is th... | [] | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 11.2 | |
0906 | Show that there are no positive integer solutions $(a, b, c, d, n)$ satisfying the equation $a^2 + b^2 + c^2 + d^2 - 4\sqrt{abcd} = 7 \cdot 2^{2n-1}$. | [
"Assume the existence of a positive integer solution $(a, b, c, d, n)$ such that $a^2 + b^2 + c^2 + d^2 - 4\\sqrt{abcd} = 7 \\cdot 2^{2n-1}$, and take the one with the minimum value of $n$. Since $\\sqrt{abcd}$ is rational, $abcd$ must be a square number. First, we establish the following two lemmas. Note that $a^2... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Greatest c... | English | proof only | null | |
0044 | Consideramos en el plano una cantidad finita de rectas tales que no hay dos paralelas ni tres concurrentes. Estas rectas dividen al plano en regiones finitas e infinitas. En cada región finita se escribe un número $1$ o un número $-1$. La operación permitida consiste en elegir un triángulo cualquiera formado por tres d... | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | Español | proof only | Yes, it is always possible to obtain all ones. | |
02ex | Given positive real numbers $x_1, x_2, \dots, x_n$ find the polygon $A_0A_1 \dots A_n$ with $A_0A_1 = x_1$, $A_1A_2 = x_2$, $\dots$, $A_{n-1}A_n = x_n$ which has the greatest area. | [
"The answer is the convex polygon inscribed in a semicircle of diameter $A_0A_n$, that is, $\\angle A_0A_iA_n = 90^\\circ$ for all $i = 1, 2, \\dots, n-1$.\n\nFirst of all, the polygon must be convex because if it isn't the case one can obtain a polygon of greater area by reflecting the sides of the polygon inside ... | Brazil | XIV OBM | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | The unique maximizer is the convex polygon whose vertices lie on a semicircle with endpoints at the first and last vertices, so that each interior vertex subtends a right angle with these endpoints (that is, it is inscribed in a semicircle with diameter A0An). | |
0eyx | Problem:
A regular $n$-gon is inscribed in a circle radius $R$. The distance from the center of the circle to the center of a side is $h_{n}$. Prove that $(n + 1) h_{n + 1} - n h_{n} > R$. | [] | Soviet Union | 3rd ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
04e2 | The segment $\overline{AB}$ is the longer side of the rectangle $ABCD$. Let $E$ be the intersection of the perpendicular from $B$ to $\overline{AC}$ and the line $AD$, and let $F$ be the intersection of the circle with centre $A$ which passes through $B$ and $\overline{CD}$. Prove that the lines $AF$ and $EF$ are perpe... | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
0atv | Problem:
Solve the inequality $\log \left(5^{\frac{1}{x}}+5^{3}\right)<\log 6+\log 5^{1+\frac{1}{2 x}}$. | [] | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | (1/4, 1/2) | |
0c1x | Consider the rectangle $ABCD$ and the arbitrary points $E \in (CD)$ and $F \in (AD)$. The perpendicular line from $E$ on the line $FB$ intersects $BC$ in $P$. The perpendicular line from $F$ on the line $EB$ intersects $AB$ in $Q$. Show that the points $P, D$ and $Q$ are collinear. | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0aj5 | Let $ABC$ be an acute triangle. The lines $l_1$ and $l_2$ are perpendicular to $AB$ at the points $A$ and $B$ respectively. The perpendicular lines from the midpoint $M$ of $AB$ to the lines $AC$ and $BC$ intersect $l_1$ and $l_2$ at the points $E$ and $F$, respectively. If $D$ is the intersection point of the lines $E... | [
"Let the circles with diameter $EM$ and $FM$ intersect for second time at $D'$ and let them intersect the sides $CA$, $CB$ at points $G,K$ respectively. Since\n$$\n\\angle ED'M = \\angle FD'M = 90^{\\circ},\n$$\nwe have that $E,D',F$ are collinear.\nSince $EM$ is a diameter and $AG$ is a chord perpendicular to it, ... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09sr | Problem:
Zij $\triangle ABC$ een gelijkbenige driehoek met $|AB| = |AC|$. Laat $D$, $E$ en $F$ punten zijn op de respectievelijke lijnstukken $BC$, $CA$ en $AB$ zodat $|BF| = |BE|$ en zodat $ED$ de binnenbissectrice van $\angle BEC$ is.
Bewijs dat $|BD| = |EF|$ dan en slechts dan als $|AF| = |EC|$. | [
"Solution:\n\nUit het gegeven en de bissectricestelling volgt dat\n$$\n\\frac{|BF|}{|BD|} = \\frac{|BE|}{|BD|} = \\frac{|CE|}{|CD|}\n$$\nVanwege de gelijkbenigheid van $\\triangle ABC$ geldt verder $\\angle FBD = \\angle ECD$, wat samen geeft dat $\\triangle BFD \\sim \\triangle CED$. Hieruit volgt dat $\\angle BFD... | Netherlands | IMO-selectietoets II | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06xr | Let $ABC$ be an acute-angled triangle with $AB < AC$. Denote its circumcircle by $\Omega$ and denote the midpoint of $\operatorname{arc} CAB$ by $S$. Let the perpendicular from $A$ to $BC$ meet $BS$ and $\Omega$ at $D$ and $E \neq A$ respectively. Let the line through $D$ parallel to $BC$ meet line $BE$ at $L$ and deno... | [
"Let $S'$ be the midpoint of $\\operatorname{arc} BC$ of $\\Omega$, diametrically opposite to $S$ so $SS'$ is a diameter in $\\Omega$ and $AS'$ is the angle bisector of $\\angle BAC$. Let the tangent of $\\omega$ at $P$ meet $\\Omega$ again at $Q \\neq P$, then we have $\\angle SQS' = 90^\\circ$.\nWe will show that... | IMO | International Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry ... | null | proof only | null | |
08q8 | Problem:
Let $A B C$ be a non-isosceles triangle with incenter $I$. Let $D$ be a point on the segment $B C$ such that the circumcircle of $B I D$ intersects the segment $A B$ at $E \neq B$, and the circumcircle of $C I D$ intersects the segment $A C$ at $F \neq C$. The circumcircle of $D E F$ intersects $A B$ and $A C$... | [
"Solution:\nSince $B D I E$ is cyclic, and $B I$ is the bisector of $\\angle D B E$, then $I D = I E$. Similarly, $I D = I F$, so $I$ is the circumcenter of the triangle $D E F$.\n\nWe also have\n$$\n\\angle I E A = \\angle I D B = \\angle I F C\n$$\nwhich implies that $A E I F$ is cyclic. We can assume that $A, E,... | JBMO | Junior Balkan 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"
] | null | proof only | null | |
06zj | Problem:
Find a number $N$ with five digits, all different and none zero, which equals the sum of all distinct three digit numbers whose digits are all different and are all digits of $N$. | [
"Solution:\nThere are $4 \\times 3 = 12$ numbers with a given digit of $n$ in the units place. Similarly, there are $12$ with it in the tens place and $12$ with it in the hundreds place. So the sum of the 3 digit numbers is $12 \\times 111(a+b+c+d+e)$, where $n=abcde$. So $8668a=332b+1232c+1322d+1331e$. We can easi... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Other"
] | null | proof and answer | 35964 | |
09af | Given are sequences $\{x_n\}$ and $\{y_n\}$ by the following recurrent relations: $x_1 = 1$, $y_1 = 39$ and
$$
x_{n+1} = 23x_n + y_n + 2, \quad y_{n+1} = 551x_n + 24y_n + 64.
$$
Prove that $x_n$ is a perfect square for all positive integer $n$. | [
"$x_{n+1} = 23x_n + y_n + 2$, $y_{n+1} = 551x_n + 24y_n + 64$. Multiplying the first equation by $24$ then subtracting these two, we get $y_{n+2} = 24x_{n+1} - x_n + 18$. And we have $x_{n+2} = 23x_{n+1} + y_{n+1} + 2$. Adding last two equations, we have\n$$\nx_{n+2} = 47x_{n+1} - x_n + 1 \\quad (*)\n$$\nThus, $x_{... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0huv | Problem:
Two dice are loaded so that the numbers $1$ through $6$ come up with various (possibly different) probabilities on each die. Is it possible that, when both dice are rolled, each of the possible totals $2$ through $12$ has an equal probability of occurring? | [
"Solution:\n\nThe answer is no. Suppose that each of the totals from $2$ to $12$ has an equal probability, which must be $1/11$ since the sum of all probabilities is $1$. Let $a$ and $b$ be the probabilities of a $1$ and a $6$, respectively, on the first die, and let $c$ and $d$ be the corresponding probabilities o... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | No | |
028y | Problem:
Existe um número de 8 algarismos da forma
$$
9999 * * * *
$$
que é um quadrado perfeito? | [
"Solution:\nSeja $x$ um número de oito algarismos da forma\n$$\nx = 9999 * * * *\n$$\nComo o menor desses números é $99990000$ e o maior é $99999999$, temos que:\n$$\n99990000 \\leq x \\leq 99999999\n$$\nObservemos que $10^{8} = 100000000 = 99999999 + 1$. Então $99990000 \\leq x < 10^{8}$. Como $10^{8} = (10^{4})^{... | Brazil | Nível 3 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | No | |
0f6y | Problem:
A $99 \times 100$ chess board is colored in the usual way with alternate squares black and white. What fraction of the main diagonal is black? What if the board is $99 \times 101$? | [] | Soviet Union | 20th ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | For 99×100: 1/2; For 99×101: 5000/9999 | |
06y2 | Let $\mathbb{Z}_{>0}$ be the set of all positive integers. Determine all subsets $\mathcal{S}$ of $\left\{2^{0}, 2^{1}, 2^{2}, \ldots\right\}$ for which there exists a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that
$$
\mathcal{S}=\left\{f(a+b)-f(a)-f(b) \mid a, b \in \mathbb{Z}_{>0}\right\} .
$$ | [
"$$\nf(a+b)=2^{e(a, b)}+f(a)+f(b) .\n$$\n\nSolution 1. Clearly $\\mathcal{S}$ must be nonempty. We start with constructions when $1 \\leqslant|\\mathcal{S}| \\leqslant 2$.\n- If $\\mathcal{S}=\\left\\{2^{k}\\right\\}$, then take $f(x)=c x-2^{k}$ for any integer $c>2^{k}$.\n- If $\\mathcal{S}=\\left\\{2^{k}, 2^{\\el... | IMO | IMO2024 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | All and only the nonempty subsets of {2^0, 2^1, 2^2, …} with at most two elements. Equivalently, S = {2^k} or S = {2^k, 2^ℓ} for some integers k, ℓ ≥ 0 with k ≠ ℓ. | |
0cjq | Turbo the snail is in the lower left cell of an $n \times n$ array, $n \ge 2$, and aims to reach the upper right cell by moving one cell rightwards or one cell upwards. Some cells contain monsters, visible to Turbo, and they must be avoided. Assume there is a unique way for Turbo to achieve his goal. In terms of $n$, d... | [
"Let $(i, j)$ denote the cell on the $i$-th row and the $j$-th column, where $(1, 1)$ is the lower left cell. The required minimum is $n-1$ and is achieved by placing the monsters in the cells $(k, n-k+1), k=2, \\dots, n$, to force Turbo move rightwards from $(1, 1)$ to $(1, n)$ and thence upwards to $(n, n)$.\n\nT... | Romania | 75th NMO Selection Tests | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n-1 | |
0e4b | Find all real $x$ and $y$, such that $x + y^2 = xy + 1$ and $xy = 4 + y$. | [
"The first equation implies $x + y^2 = xy + 1$, or $x - xy = 1 - y^2$, so $x(1 - y) = 1 - y^2$. This can be rewritten as $(1 - y)(x - 1 - y) = 0$.\n\nIf $y = 1$, then this equation is satisfied. The second equation then implies $x \\cdot 1 = 4 + 1$, so $x = 5$.\n\nIf $y \\ne 1$, then $x - 1 - y = 0$, so $x = 1 + y$... | Slovenia | National Math Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (5, 1), (3, 2), (-1, -2) | |
0ka4 | Problem:
Let $ABCD$ be an isosceles trapezoid with $AD = BC = 255$ and $AB = 128$. Let $M$ be the midpoint of $CD$ and let $N$ be the foot of the perpendicular from $A$ to $CD$. If $\angle MBC = 90^{\circ}$, compute $\tan \angle NBM$. | [
"Solution:\n\nConstruct $P$, the reflection of $A$ over $CD$. Note that $P$, $M$, and $B$ are collinear. As $\\angle PNC = \\angle PBC = 90^{\\circ}$, $PNBC$ is cyclic. Thus, $\\angle NBM = \\angle NCP$, so our desired tangent is $\\tan \\angle ACN = \\frac{AN}{CN}$. Note that $NM = \\frac{1}{2} AB = 64$. Since $\\... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous... | null | proof and answer | 120/353 | |
0cgo | Let $(\mathbb{K}, +, \cdot)$ be a field having the property that $x^2y = yx^2$, for all $x, y \in \mathbb{K}$. Prove that $(\mathbb{K}, +, \cdot)$ is commutative.
Sorin Rădulescu and Mihai Piticari | [
"We have $2 \\cdot 1 = (a+1)^2 - a^2 - 1 \\in Z(\\mathbb{K})$, for $a \\in \\mathbb{K}$. If the characteristic $char(\\mathbb{K})$ of $\\mathbb{K}$ is not $2$, then $2 \\cdot 1$ is an invertible element in $Z(\\mathbb{K})$, such that\n$$\na = (2 \\cdot 1)^{-1} \\cdot (2 \\cdot a) \\in Z(\\mathbb{K}), \\text{ for an... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Field Theory"
] | English | proof only | null | |
0hxe | Problem:
Let $ABC$ be a scalene triangle inscribed in circle $\Gamma$. The internal bisector of $\angle A$ meets $\overline{BC}$ and circle $\Gamma$ at points $D$ and $E$. The circle with diameter $\overline{DE}$ meets $\Gamma$ at a second point $F$. Prove that $\frac{AB}{AC}=\frac{FB}{FC}$. | [
"Solution:\nLet $K$ be the midpoint of $\\operatorname{arc} \\widehat{BC}$ containing $A$. Let $M$ be the midpoint of $\\overline{BC}$.\n\nSince $\\angle DME=\\angle DFE=90^\\circ$, the points $D, M, E, F$ are concyclic. Moreover, since $\\angle EFK=\\angle EFD=90^\\circ$, the points $F, D,... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0h3f | Let $ABCD$ be a parallelogram. The incircle of triangle $ABC$ touches its sides $AB$, $BC$ and $AC$ at points $M$, $N$ and $K$ respectively. The lines $MK$ and $CD$ intersect at $P$, the lines $NK$ and $AD$ intersect at $Q$. Prove that the points $B$, $E$ and $F$ are collinear, where $E$ and $F$ are the midpoints of $M... | [
"Оскільки $AM = AK$, $CN = CK$ як дотичні, то трикутники $MAK$ та $NCK$ рівнобедрені. Звідси $\\angle AMK = \\angle AKM$, і $\\angle CNK = \\angle CKN$.\n\nДалі, $\\angle AMK = \\angle CPK$ і $\\angle AQK = \\angle CNK$ як різносторонні при паралельних прямих. Крім того, $\\angle AKM = \\angle CKP$ і $\\angle AKQ =... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0hax | Find all real values of $k$, such that all solutions of the equation $k(2-k)x^2 - (k+4)x + 6 = 0$ are positive integers. | [
"First, we go through cases when this equation is not quadratic.\nFor $k=0$, equation becomes $-4x+6=0$, and has a non-integer solution.\nFor $k=2$, equation becomes $-6x+6=0$, and has a single solution $x=1$, which satisfies the statement.\n\nNow, suppose $k(2-k) \\neq 0$. We find the discriminant of this quadrati... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | k = 1/2, 1, 2 | |
09jx | Find all integers $a, b$ such that $a \le b$ and $(a+b)^2 = a^3 + b^3$. | [
"Answer: $(a, b) = (0, 1)$, $(1, 2)$, $(2, 2)$ and $(-n, n)$ for any integer $n \\ge 0$.\nIf $a+b=0$ then $(a,b) = (-n,n)$ for an integer $n \\ge 0$, so now assume that $a+b \\ne 0$. Then $a+b = a^2-ab+b^2$ and hence $(a-b)^2 + (a-1)^2 + (b-1)^2 = 2$. By setting $x = a-1, y = b-1$ we get $x \\le y$ and $x^2+y^2+(x-... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (a, b) = (0, 1), (1, 2), (2, 2), and (-n, n) for any integer n >= 0 | |
06ts | Find all positive integers $n$ for which we can fill in the entries of an $n \times n$ table with the following properties:
- each entry can be one of $I$, $M$ and $O$;
- in each row and each column, the letters $I$, $M$ and $O$ occur the same number of times; and
- in any diagonal whose number of entries is a multiple... | [
"We first show that such a table exists when $n$ is a multiple of $9$. Consider the following $9 \\times 9$ table.\n$$\n\\left(\\begin{array}{ccccccccc}\nI & I & I & M & M & M & O & O & O \\\\\nM & M & M & O & O & O & I & I & I \\\\\nO & O & O & I & I & I & M & M & M \\\\\nI & I & I & M & M & M & O & O & O \\\\\nM... | IMO | IMO 2016 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All positive integers divisible by 9 | |
0h2a | Find all functions $f: \mathbb{R} \to \mathbb{R}$, such that:
1) For all real $x, y$ the following equality holds
$$
f(2x) = f(x+y)f(y-x) + f(x-y)f(-x-y).
$$
2) $f(x) \ge 0$ for all $x$. | [
"Take $x = y$, we get:\n$$\nf(2x) = f(0)f(2x) + f(0)f(-2x). \\quad (1)\n$$\nTake in (1) $x = 0$ and we get $f(0) = 2f^2(0)$. So $f(0) = 0$ or $f(0) = \\frac{1}{2}$.\nIf $f(0) = 0$, (1) implies that $f(2x) = 0$, hence $f = 0$.\n\nIf $f(0) = \\frac{1}{2}$, (1) implies that $f(2x) = \\frac{1}{2}f(2x) + \\frac{1}{2}f(-... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0 for all x, or f(x) = 1/2 for all x | |
0kqf | A right square pyramid with volume $54$ has a base with side length $6$. The five vertices of the pyramid all lie on a sphere with radius $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | [] | United States | AIME II | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Other 3D problems"
] | null | final answer only | 21 | |
0j50 | Problem:
Let $a$, $b$, and $c$ be the side lengths of a triangle, and assume that $a \leq b$ and $a \leq c$. Let $x=\frac{b+c-a}{2}$. If $r$ and $R$ denote the inradius and circumradius, respectively, find the minimum value of $\frac{a x}{r R}$. | [
"Solution:\nIt is well-known that both $\\frac{a b c}{4 R}$ and $\\frac{r(a+b+c)}{2}$ are equal to the area of triangle $A B C$. Thus $\\frac{a b c}{4 R}=\\frac{r(a+b+c)}{2}$, and\n$$\nR r=\\frac{a b c}{2(a+b+c)}\n$$\nSince $a \\leq b$ and $a \\leq c$, we have $\\frac{a^{2}}{b c} \\leq 1$. We thus obtain that\n$$\n... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 3 | |
0jc8 | Problem:
Let $\otimes$ be a binary operation that takes two positive real numbers and returns a positive real number. Suppose further that $\otimes$ is continuous, commutative $(a \otimes b = b \otimes a)$, distributive across multiplication $(a \otimes (b c) = (a \otimes b)(a \otimes c))$, and that $2 \otimes 2 = 4$.... | [
"Solution:\n\nAnswer: $y = \\sqrt{2}$\n\nWe note that $\\left(a \\otimes b^{k}\\right) = (a \\otimes b)^{k}$ for all positive integers $k$. Then for all rational numbers $\\frac{p}{q}$ we have $a \\otimes b^{\\frac{p}{q}} = \\left(a \\otimes b^{\\frac{1}{q}}\\right)^{p} = (a \\otimes b)^{\\frac{p}{q}}$. So by conti... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | y = sqrt(2) | |
0f1q | Problem:
Call a triangle big if each side is longer than $1$. Show that we can draw $100$ big triangles inside an equilateral triangle with side length $5$ so that all the triangles are disjoint. Show that you can draw $100$ big triangles with every vertex inside or on an equilateral triangle with side $3$, so that th... | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | null | proof only | null | |
0acd | The teacher ordered 20 students from one class in a row and gave 800 candies to them. Every student had to calculate the ratio $\frac{x}{x+2k-1}$, where $x$ is the number of candies that he got and $k$ is the number of his position in the row, counting from left to right. After calculating they all got the same result.... | [
"Let the first student received $x_1$ candies, the second $x_2$ candies, ..., and the twentieth received $x_{20}$ candies. So $x_1 + x_2 + \\dots + x_{20} = 800$. Let $\\frac{x_k}{x_k + 2k-1} = M$ from where $x_k = (2k-1)\\frac{M}{1-M}$. For the sum $x_1 + x_2 + \\dots + x_{20}$ we obtain\n$$\nx_1 + x_2 + \\dots + ... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 46 |
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