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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0bzp | A triangle $ABC$ has circumcenter $O$ and incenter $I$. The straight lines $AI$, $BI$, $CI$ meet again the circumcircle at $D$, $E$, respectively $F$. Denote $M$, $N$, $P$ the circumcenters of the triangles $IEF$, $IDF$, respectively $IED$. Prove that the straight lines $DM$, $EN$, $FP$ and $OI$ are concurrent.
Marin I... | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advan... | null | proof only | null | |
0l49 | Problem:
Compute the remainder when
$$
1002003004005006007008009
$$
is divided by $13$. | [
"Solution:\n\nNote that $13 \\mid 1001$. Thus we can repeatedly subtract any multiple of $1001$ from this number without changing the remainder. In particular, we can repeatedly subtract multiples of $1001$ from the left to the right, as follows.\n\n$$\n\\begin{aligned}\n1002003004005006007008009 & \\longrightarrow... | United States | HMMT November 2024 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 5 | |
05xi | Problem:
Soit $\alpha \neq 0$. Trouver toutes les fonctions $f: \mathbb{R} \rightarrow \mathbb{R}$ telles que pour tous réels $x, y$,
$$
f\left(x^{2}+y+f(y)\right)=f(x)^{2}+\alpha y
$$ | [
"Solution:\nPosons $x=0$. Alors l'équation devient $f(y+f(y))=f(0)^{2}+\\alpha y$ pour tout $y$. Comme $\\alpha$ est non nul, le côté droit décrit tous les réels lorsque $y$ varie, et ainsi $f$ est surjective.\n\nPosons $y=0$. Alors on peut écrire pour tout réel $x$,\n$$\nf(x)^{2}=f\\left(x^{2}+f(0)\\right)=f\\left... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | If alpha equals 2, the unique solution is f(x) = x for all real x; if alpha is not 2, there is no solution. | |
002k | Una pulga salta sobre puntos enteros de la recta numérica. En su primer movimiento salta desde el punto $0$ y cae en el punto $1$. Luego, si en un movimiento la pulga saltó desde el punto $a$ y cayó en el punto $b$, en el siguiente movimiento salta desde el punto $b$ y cae en uno de los puntos $b + (b - a) - 1$, $b + (... | [] | Argentina | XX Olimpiada Iberoamericana de Matemáticas | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | Español | proof only | null | |
0ciy | Find the natural numbers $m$, $n$, $p$ ($m, n \ge 1$, $p \ge 2$, $n$ odd), which simultaneously satisfy the equations
$$
p + p^2 + \dots + p^n = m^2 \quad \text{and} \quad \lfloor \sqrt{p} \rfloor + \lfloor \sqrt{p^2} \rfloor + \dots + \lfloor \sqrt{p^n} \rfloor = m^2 - 5m.
$$ | [] | Romania | 75th NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | (m, n, p) = (6, 1, 36) | |
07uj | The country of Cathay has four cities with airports and two Airlines Cathay Red and Cathay Blue. Each of the six routes is assigned to only one of the airlines. No airline is assigned all the routes. An assignment of routes is called good if there are two cities for which the airline that operates the route between the... | [
"The problem can be expressed in graph theory terms as colouring the edges of the graph $K_4$, the complete graph on 4 vertices. There is essentially one colouring which is not good, namely:\n\n\n\nThe dashed line represents one colour and the solid line the other. Interchanging the top ver... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0cua | Let $O$ be the circumcenter of an acute-angled isosceles triangle $ABC$ with $AB = AC$. The rays $BO$ and $CO$ meet the sides $AC$ and $AB$ at $B'$ and $C'$, respectively. Let $l$ be the line through $C'$ parallel to $AC$. Prove that $l$ is tangent to the circumcircle of the triangle $B'OC$. | [
"Пусть прямая $AO$ пересекает $l$ в точке $T$ (см. рис. 13).\n\nИз симметрии относительно $AO$ имеем $\\angle B'TO = \\angle C'TO$.\n\nПоскольку $l \\parallel AC$, получаем $\\angle C'TO = \\angle OAC = \\angle OCA$.\n\nИтак, $\\angle B'TO = \\angle B'CO$, то есть $T$ лежит на окружности $\\omega$, описанный около ... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
00hj | Determine all the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{2}+f(y)\right)=f(f(x))+f\left(y^{2}\right)+2 f(x y)
$$
for all real number $x$ and $y$. | [
"By substituting $x=y=0$ in the given equation of the problem, we obtain that $f(0)=0$. Also, by substituting $y=0$, we get $f\\left(x^{2}\\right)=f(f(x))$ for any $x$.\n\nFurthermore, by letting $y=1$ and simplifying, we get\n$$\n2 f(x)=f\\left(x^{2}+f(1)\\right)-f\\left(x^{2}\\right)-f(1)\n$$\nfrom which it follo... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x)=0 for all real x; f(x)=x^2 for all real x | |
0icw | Problem:
A tetrahedron has all its faces triangles with sides $13, 14, 15$. What is its volume? | [
"Solution:\nLet $ABC$ be a triangle with $AB = 13$, $BC = 14$, $CA = 15$. Let $AD$, $BE$ be altitudes. Then $BD = 5$, $CD = 9$. (If you don't already know this, it can be deduced from the Pythagorean Theorem: $CD^{2} - BD^{2} = (CD^{2} + AD^{2}) - (BD^{2} + AD^{2}) = AC^{2} - AB^{2} = 56$, while $CD + BD = BC = 14$... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 42√55 | |
088n | Problem:
Tre persone $A, B, C$ si trovano in prossimità di un incrocio stradale tra due strade perpendicolari. $A$ si trova esattamente sull'incrocio, mentre $B$ e $C$ si trovano su due strade distinte. Nel campo nei pressi dell'incrocio (all'interno dell'angolo retto $C \widehat{A} B$ ) c'è un cartellone pubblicitari... | [
"Solution:\n\nLa risposta è $(\\mathbf{D})$. Poiché $\\widehat{D A E}=\\widehat{D B E}=\\widehat{D C E}$ e $A, B, C$ si trovano dalla stessa parte della retta $D E$, i punti $A, B, C, D, E$ si trovano su di una circonferenza. Inoltre, dato che gli angoli alla circonferenza che insistono su $D E$ hanno un'ampiezza d... | Italy | Italian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | D | |
08b3 | Problem:
Sia $ABCDE$ un pentagono regolare di lato $1$ e sia $P$ l'intersezione tra le diagonali $AC$ e $BE$. Quanto misura il segmento $PC$?
(A) $1$
(B) $\frac{\sqrt{5}}{2}$
(C) $\sqrt{5}-1$
(D) $4(\sqrt{5}-2)$
(E) Nessuna delle precedenti. | [
"Solution:\n\nGli angoli interni di un $n$-agono regolare misurano $\\frac{n-2}{n} 180^{\\circ}$. Quindi gli angoli interni di un pentagono regolare misurano $\\frac{3}{5} 180^{\\circ}=108^{\\circ}$. I triangoli $EAB$ e $CBA$ sono isosceli, quindi $\\widehat{AEB}=\\widehat{EBA}=\\widehat{BCA}=\\widehat{CAB}=\\frac{... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | MCQ | A | |
095g | Problem:
Se consideră un cerc $C(O)$ şi o coardă $[MN]$ a acestui cerc. În unul dintre segmentele circulare determinate de coarda $[MN]$ se înscriu două cercuri $C_1(O_1)$ şi $C_2(O_2)$, $O_1 \neq O_2$, care sunt tangente la cercul $C(O)$ în punctele $A$ şi $B$, iar la coarda $[MN]$ - în punctele $C$ şi $D$, respectiv.... | [
"Solution:\nRealizăm desenul corespunzător datelor problemei. Fie $AC \\cap C = \\{P\\}$. Întrucât cercurile $C$ şi $C$ au o tangentă comună ce trece prin punctul $A$, punctele $A, O_1$ şi $O$ sunt coliniare. Triunghiurile $ACO_1$ şi $APO$ sunt isoscele şi au un unghi de la bază comun, atunci ele sunt asemenea. În ... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kxb | The Bank of Pittsburgh issues coins that have a heads side and a tails side. Vera has a row of $2023$ such coins alternately tails-up and heads-up, with the leftmost coin tails-up.
In a *move*, Vera may flip over one of the coins in the row, subject to the following rules:
* On the first move, Vera may flip over any ... | [
"The answer is $\\boxed{4044}$. In general, replacing $2023$ with $4n + 3$, the answer is $8n + 4$.\n\n**Bound**\n\nObserve that the first and last coins must be flipped, and so every coin is flipped at least once. Then, the $2n + 1$ even-indexed coins must be flipped at least twice, so they are flipped at least $4... | United States | USA TSTST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 4044 | |
0ajg | A trapezoid $ABCD$ ($AB \parallel CD$, $AB > CD$) is circumscribed. The incircle of the triangle $ABC$ touches the lines $AB$ and $AC$ at the points $M$ and $N$, respectively. Prove that the incenter of the trapezoid $ABCD$ lies on the line $MN$. | [
"**Version 1.** Let $I$ be the incenter of triangle $ABC$ and $R$ be the common point of the lines $BI$ and $MN$. Since\n$$ m(\\widehat{ANM}) = 90^\\circ - \\frac{1}{2} m(\\widehat{MAN}) \\quad \\text{and} \\quad m(\\widehat{BIC}) = 90^\\circ + \\frac{1}{2} m(\\widehat{MAN}) $$\nthe quadrilateral $IRNC$ is cyclic. ... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocente... | English | proof only | null | |
0hy0 | Problem:
Given that $r$ and $s$ are relatively prime positive integers such that $\frac{r}{s}=\frac{2(\sqrt{2}+\sqrt{10})}{5(\sqrt{3+\sqrt{5}})}$, find $r$ and $s$. | [
"Solution:\nSquaring both sides of the given equation yields\n$$\n\\frac{r^{2}}{s^{2}}=\\frac{4(12+4 \\sqrt{5})}{25(3+\\sqrt{5})}=\\frac{16(3+\\sqrt{5})}{25(3+\\sqrt{5})}=\\frac{16}{25}.\n$$\nBecause $r$ and $s$ are positive and relatively prime, then by inspection, $r=4$ and $s=5$."
] | United States | HMMT 1998 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | null | final answer only | r = 4, s = 5 | |
088s | Problem:
La differenza di due interi positivi $a$ e $b$ è un numero primo $p$ e il loro prodotto è un quadrato perfetto non superiore a $10000$. Qual è il massimo valore che può assumere $p$? | [
"Solution:\n\nLa risposta è $19$. Infatti, supponiamo senza perdita di generalità che sia $a > b$; allora $a = b + p$ e $ab = b(b + p)$.\n\nSe ora $p$ dividesse $b$ potremmo porre $b = p b'$, e avremmo che\n$$\na b = p b' (p b' + p) = p^2 b'(b' + 1)\n$$\nsarebbe un quadrato perfetto; ma poiché $b'$ e $b' + 1$ sono ... | Italy | Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 19 | |
0gli | Let $k$ be a positive integer. Show that if there exists a sequence $a_0, a_1, \dots$ of integers satisfying the condition
$$
a_n = \frac{a_{n-1} + n^k}{n} \text{ for all } n \ge 1,
$$
then $k-2$ is divisible by 3. | [
"Denote $f: \\mathbb{Z}_{\\ge 0}^2 \\to \\mathbb{Z}$ such that\n$$\nx^n = f(0, n) + f(1, n)[(x+1)-1] + f(2, n)[(x+1)(x+2) - (x+1)] + \\dots\n$$\nfor every non-negative integers $x, n$.\nThe above function always exists because the expression of $x^n$ can be uniquely represented in that particular way. Moreover, $f(... | Thailand | The first T3MO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0e6r | Jakob is reading a book with $630$ pages. On the first day he read one third of the book. The sum of the numbers labelling pages that Jakob read on the second day is $4400$. How many pages does Jakob still need to read to come to the end of the book? (The first page of the book is numbered $1$.)
(A) $210$
(B) $211$
(C... | [
"On the first day, Jakob read $210$ pages of the book. Suppose that on the second day he read $n$ pages. Then the sum of the numbers on these pages was equal to $211 + 212 + 213 + \\ldots + (210 + n) = 210n + \\frac{n(n+1)}{2}$, hence $210n + \\frac{n(n+1)}{2} = 4410$. From this we get $210n \\leq 4410$ or $n \\leq... | Slovenia | National Math Olympiad 2012 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | E | |
0028 | En el pizarrón hay escrito un número de 100 dígitos con los últimos tres dígitos de la derecha iguales a $999$. Debajo de este número, y usando exactamente los mismos dígitos, pero en otro orden, Luciano escribe un nuevo número de 100 dígitos: deja los tres últimos $999$, e intercambia a voluntad los primeros $97$ dígi... | [] | Argentina | XX OLIMPIADA MATEMÁTICA ARGENTINA | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | español | proof and answer | 45 | |
0foc | Hallar las soluciones enteras de la ecuación
$$
x^{4} + y^{4} = 3x^{3}y.
$$ | [
"Supongamos, en primer lugar, que $x = 0$. En este caso se tiene $y^4 = 0$, por lo que $y$ también tiene que ser 0. Así pues, una solución es $x = y = 0$.\n\nSi $x \\neq 0$ dividimos toda la ecuación por $x^4$, quedando\n$$\n1 + \\left(\\frac{y}{x}\\right)^{4} = 3\\frac{y}{x}.\n$$\nSea $t = y/x$, entonces, las solu... | Spain | L Olimpiada Matemática Española | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | Spanish | proof and answer | (0, 0) | |
0ko1 | Problem:
Alice and Bob are playing in an eight-player single-elimination rock-paper-scissors tournament. In the first round, all players are paired up randomly to play a match. Each round after that, the winners of the previous round are paired up randomly. After three rounds, the last remaining player is considered t... | [
"Solution:\n\nAlice's opponent is chosen randomly in the first round. If Alice's first opponent is Bob, then she will lose immediately to him. Otherwise, Bob will not face Alice in the first round. This means he faces someone who plays scissors, so Bob will lose in the first round. Also, this means Alice will never... | United States | HMMT November 2022 | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 6/7 | |
09pu | Problem:
Zij $n$ een geheel getal zo dat $\sqrt{1+12 n^{2}}$ een geheel getal is.
Bewijs dat $2+2 \sqrt{1+12 n^{2}}$ het kwadraat van een geheel getal is. | [
"Solution:\n\nStel $k = \\sqrt{1 + 12 n^{2}}$, waarbij $k$ een geheel getal is.\n\nDan geldt:\n$$\nk^{2} = 1 + 12 n^{2}\n$$\n$$\nk^{2} - 12 n^{2} = 1\n$$\nDit is een Pell-vergelijking van de vorm $x^{2} - 12 y^{2} = 1$ met $x = k$, $y = n$.\n\nWe moeten bewijzen dat $2 + 2k$ een kwadraat is van een geheel getal.\n\... | Netherlands | TOETS TRAININGSKAMP | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0hm4 | Problem:
We wish to distribute $12$ indistinguishable stones among $4$ distinguishable boxes $B_{1}$, $B_{2}$, $B_{3}$, $B_{4}$. (It is permitted some boxes are empty.)
a. Over all ways to distribute the stones, what fraction of them have the property that the number of stones in every box is even?
b. Over all ways ... | [
"Solution:\n\nWe will invoke the following famous theorem, colloquially called stars and bars: the number of ways to distribute $a$ indistinguishable objects among $b$ boxes is $\\binom{a+b-1}{b-1}$. One can see this by considering a string of $a$ objects (\"stars\") and $b-1$ dividers (\"bars\"), which give a dist... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | a. 12/65; b. 1/13 | |
06l0 | Let $S = \{1, 2, \dots, 100\}$. Consider a partition of $S$ into $S_1, S_2, \dots, S_n$ for some $n$, i.e. $S_i$ are nonempty, pairwise disjoint and $S = \bigcup_{i=1}^n S_i$. Let $a_i$ be the average of elements of the set $S_i$. Define the score of this partition by
$$
\frac{a_1 + a_2 + \dots + a_n}{n}.
$$
Among all ... | [
"The minimum possible score is $10$.\nSuppose we have a partition $S_1, S_2, \\dots, S_n$ with score $x$.\n\n**Claim 1.** If there are at least two subsets, say $S_1, S_2$, both with averages at least $x$, then merging these two subsets into one subset will not increase the score.\n*Proof.* Let $a'$ be the new aver... | Hong Kong | CHKMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 10 | |
0dti | Find the smallest positive integer $n$ for which there exist integers $x_1 < x_2 < \dots < x_n$ such that every integer from $1000$ to $2000$ can be written as a sum of some of the integers from $x_1, x_2, \dots, x_n$, without repetition. | [
"The answer is $n = 11$. The $11$ integers $2^0, 2^1, \\dots, 2^{10}$ certainly work. Now we show that $n = 10$ does not work.\n\nSuppose on the contrary that the $10$ integers $x_1 < x_2 < \\dots < x_{10}$ satisfy the requirement. They can form $2^{10} = 1024$ sums, not necessarily distinct. There are $1001$ integ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 11 | |
0er0 | In the figure, $ABCD$ is a rectangle.
The area of triangle $DEC$ is $9\ \text{cm}^2$ and $BE = \frac{2}{5} BC$.
What is the area of $ABCD$ in $\text{cm}^2$?
 | [
"Let $BE = 2x$; then with $BE = \\frac{2}{5} BC$ we must have $EC = 3x$.\nNow for $\\triangle DEC$, $\\frac{1}{2}(3x)(DC) = 9$, giving $DC = \\frac{6}{x}$, and since $BC = 5x$,\nthe area of rectangle $ABCD$ is $5x \\cdot \\frac{6}{x} = 30$."
] | South Africa | South African Mathematics Olympiad Second Round | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 30 | |
042q | Let the general term of sequence $\{a_n\}$ be
$$
a_n = \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right), \quad n = 1, 2, \dots
$$
Prove that there exist infinite positive integers $m$ such that $a_{m+4}a_m - 1$ are perfect squares. | [
"Denote $q_1 = \\frac{1+\\sqrt{5}}{2}$, $q_2 = \\frac{1-\\sqrt{5}}{2}$, and then $q_1 + q_2 = 1$, $q_1q_2 = -1$. Thus,\n$$\na_n = \\frac{1}{\\sqrt{5}}(q_1^n - q_2^n), \\quad n = 1, 2, \\dots\n$$\nHence $a_1 = 1, a_2 = 1$. Also note that $q_i + 1 = q_i^2$ ($i = 1, 2$), and there is\n$$\n\\begin{aligned}\na_{n+1} + a... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0ck3 | Let $x, y, z, t$ be pairwise distinct positive integers and $n = 2^x + 2^y + 2^z + 2^t$. The division of $n$ by $305$ gives the quotient $2^a$ and leaves the remainder $0$, where $a$ is a non-negative integer.
Find the remainder of the division of $a + x + y + z + t$ by $5$. | [
"Assume, without loss of generality, that $x < y < z < t$. Then $2^x + 2^y + 2^z + 2^t = 305 \\cdot 2^a$, whence\n$$\n2^x \\cdot (1 + 2^{y-x} + 2^{z-x} + 2^{t-x}) = 305 \\cdot 2^a \\tag{1}\n$$\nSince $x < y < z < t$, the numbers $y-x$, $z-x$ and $t-x$ are positive integers, hence $2^{y-x}$, $2^{z-x}$, $2^{t-x}$ are... | Romania | 75th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 2 | |
043c | Let $m, n$ be positive integers, $a_{ij}$ ($1 \le i \le m, 1 \le j \le n$) be nonnegative real numbers such that for any $i, j$, the inequalities
$$
a_{i,1} \ge a_{i,2} \ge \dots \ge a_{i,n}, \quad a_{1,j} \ge a_{2,j} \ge \dots \ge a_{m,j}
$$
hold. For $i = 1, 2, \dots, m$ and $j = 1, 2, \dots, n$, define
$$
X_{i,j} = ... | [
"The problem conditions imply that\n$$\nX_{i,j} \\ge (i + j - 1) \\cdot a_{i,j}, \\quad Y_{i,j} \\le (m + n - i - j + 1) \\cdot a_{i,j}.\n$$\nHence, we have\n$$\n\\begin{align*}\n\\prod_{i=1}^{m} \\prod_{j=1}^{n} X_{i,j} &\\ge \\prod_{i=1}^{m} \\prod_{j=1}^{n} (i+j-1) \\cdot a_{i,j} \\\\\n&= \\prod_{i=1}^{m} \\prod... | China | China National Team Selection Test | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0hpp | Problem:
Given a square table $n \times n$, two players $A$ and $B$ are playing the following game: At the beginning all cells of the table are empty, and the players alternate playing with coins. Player $A$ has the first move, and in each of the moves a player will put a coin on some of the cells that doesn't contain... | [
"Solution:\n\nIf $n$ is even, $B$ has the winning strategy - he just needs to put the coin in the centrally symmetric cell to the cell where $A$ has put the coin in his previous move. Following this strategy, $B$ will always have a free cell to put his coin, and thus he can never lose. Since the game has to end, $A... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | If the side length is even, the second player wins by always placing a coin on the square centrally symmetric to the opponent’s last move. If the side length is odd, the first player wins by first placing a coin on the center square and then always responding on the centrally symmetric square to the opponent’s last mov... | |
03by | Find all functions $f : \mathbb{Q}^+ \to \mathbb{R}^+$ such that
$$
f(xy) = f(x + y)(f(x) + f(y)), \text{ for any } x, y \in \mathbb{Q}^{+}.
$$ | [
"If $g(x) = 1/f(x)$ then $g : \\mathbb{Q}^+ \\to \\mathbb{R}^+$ and the given equation becomes\n$$\n(*) \\quad g(x+y)g(x)g(y) = g(xy)(g(x) + g(y)).\n$$\nLet $f(1) = c > 0$ i.e. $g(1) = \\frac{1}{c}$. From (*) we get $g(x+1) = c g(x) + 1$ and hence $g(2) = 2$, $g(3) = 2c+1$, $g(4) = 2c^2+c+1$, $g(5) = 2c^3+c^2+c+1$,... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 1/x for all positive rational x, and f(x) = 1/2 for all positive rational x | |
05nf | Problem:
Soit $a_{1}, a_{2}, a_{3}$ des entiers strictement positifs. Pour tout entier $n \geqslant 3$, on pose
$$
a_{n+1}=\operatorname{ppcm}\left(a_{n}, a_{n-1}\right)-\operatorname{ppcm}\left(a_{n-1}, a_{n-2}\right)
$$
étant entendu que l'on a $\operatorname{ppcm}(0, x)=0$ pour tout entier $x$.
Prouver qu'il exist... | [
"Solution:\n\nSi $x$ et $y$ sont des entiers, on note $x \\vee y=\\operatorname{ppcm}(x, y)$ et $x \\wedge y=\\operatorname{pgcd}(x, y)$, avec $\\operatorname{pgcd}(0, x)=x$ pour tout $x$. On rappelle qu'alors $x y=(x \\vee y)(x \\wedge y)$.\n\nPour tout entier $n \\geqslant 3$, $a_{n-1}$ divise $a_{n} \\vee a_{n-1... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
... | null | proof only | null | |
0f41 | Problem:
$ABC$ is a triangle. $A'$ lies on the side $BC$ with $BA' / BC = 1 / 4$. Similarly, $B'$ lies on the side $CA$ with $CB' / CA = 1 / 4$, and $C'$ lies on the side $AB$ with $AC' / AB = 1 / 4$. Show that the perimeter of $A'B'C'$ is between $1/2$ and $3/4$ of the perimeter of $ABC$. | [] | Soviet Union | 15th ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
0i5c | Problem:
Express, as concisely as possible, the value of the product
$$
\left(0^{3}-350\right)\left(1^{3}-349\right)\left(2^{3}-348\right)\left(3^{3}-347\right) \cdots\left(349^{3}-1\right)\left(350^{3}-0\right)
$$ | [
"Solution:\n0 . One of the factors is $7^{3}-343=0$, so the whole product is zero."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 0 | |
037z | Problem:
Let $D$ and $E$ be points on the sides $AB$ and $AC$ of $\triangle ABC$ such that $DE \parallel BC$. The circumcircle $k$ of $\triangle ADE$ meets the segments $BE$ and $CD$ at points $M$ and $N$. The lines $AM$ and $AN$ meet $BC$ at points $P$ and $Q$ such that $BC = 2PQ$ and $P$ lies between $B$ and $Q$. Pro... | [
"Solution:\nSince $\\angle PBM = \\angle MED = \\angle BAP$ we have $PB^{2} = PM \\cdot PA$. Analogously $QC^{2} = QN \\cdot QA$. Since $BC = 2PQ$ and $P$ lies between $B$ and $Q$, there exists a point $L$ on $PQ$ such that $PB = PL$ and $QC = QL$. Thus, $PL^{2} = PM \\cdot PA$, i.e. $M$ lies on the circle $k'$ thr... | Bulgaria | Team selection test for 47. IMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0gs2 | Let $P$ be a point inside the triangle $ABC$. The lines $AP$, $BP$ and $CP$ intersect the sides $BC$, $CA$ and $AB$ at points $D$, $E$ and $F$, respectively. Let $Q$ be a point on the ray $[BE$ such that $E \in [BQ]$ and $\angle EDQ = \angle FDB$. Suppose that $BE \perp AD$ and $|DQ| = 2|BD|$. Prove that $\angle FDE = ... | [
"Let $FD \\cap BP = T$ and $AT \\cap BC = R$. Using Menelaus and Ceva theorems, we get\n$$\n\\frac{BC}{DC} = \\frac{BF}{AF} \\cdot \\frac{AP}{PD} = \\frac{BR}{DR}.\n$$\n\nThis shows that the points $B$, $R$, $D$, $C$ are harmonic. The lines $AB$, $AR$, $AD$, $AC$ form a harmonic pencil and ... | Turkey | 26th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadri... | English | proof only | null | |
000k | Un policía intenta capturar a un ladrón en un tablero de $2001 \times 2001$. Ellos juegan alternadamente. Cada jugador, en su turno, debe moverse una casilla en uno de los tres siguientes sentidos:
$\downarrow$ (abajo); $\rightarrow$ (derecha); $\vee$ (diagonal superior izquierda).
Si el policía se encuentra en la casi... | [] | Argentina | XVII Olimpíada Iberoamericana de Matemática | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | español | proof only | null | |
09bf | $1 \times n$ хүснэгтийн нүд бүрийг $a_1, a_2, \dots, a_n$ өнгүүдийн аль нэгээр будахдаа $i_1, i_2, \dots, i_k$ дугаартай нүд-нүүд ялгаатай өнгөтэй бөгөөд зөвхөн $a_{i_1}, a_{i_2}, \dots, a_{i_k}$ өнгүүдээр будагдсан байх бүх ялгаатай будалтын тоог ол. | [
"$i_1, i_2, \\dots, i_k$ дугаартай нүднүүд хос хоорондоо ялгаатай өнгөтэй бөгөөд зөвхөн $a_{i_1}, \\dots, a_{i_k}$ өнгүүдээр будагдсан байх боломжийн тоо $k!$ бөгөөд үлдсэн $n-k$ нүдийг өгөгдсөн $n$ өнгөөр будах боломжийн тоо нь $n^{n-k}$ болно. Иймд бодлогын нөхцөлийг хангах бүх ялгаатай будалтын тоо нь $n^{n-k} \... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Mongolian | proof and answer | n^{n-k} * k! | |
0jot | Problem:
Let $a$, $b$ be integers chosen independently and uniformly at random from the set $\{0,1,2, \ldots, 80\}$. Compute the expected value of the remainder when the binomial coefficient $\binom{a}{b}=\frac{a!}{b!(a-b)!}$ is divided by $3$. (Here $\binom{0}{0}=1$ and $\binom{a}{b}=0$ whenever $a<b$.) | [
"Solution:\n\nAnswer: $\\frac{1816}{6561}$\n\nBy Lucas' Theorem we're looking at\n$$\n\\prod_{i=1}^{4} \\binom{a_{i}}{b_{i}}\n$$\nwhere the $a_{i}$ and $b_{i}$ are the digits of $a$ and $b$ in base $3$. If any $a_{i}<b_{i}$, then the product is zero modulo $3$.\n\nOtherwise, the potential residues are $\\binom{2}{0... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | final answer only | 1816/6561 | |
0c8u | Let $k$ be a positive integer. Determine the least integer $n \ge k + 1$ for which the game below can be played indefinitely:
Consider $n$ boxes, labelled $b_1, b_2, \dots, b_n$. For each index $i$, box $b_i$ contains initially exactly $i$ coins. At each step, the following three substeps are performed in order:
(1) ... | [
"The required minimum is $n = 2^k + k - 1$.\n\nIn this case the game can be played indefinitely by choosing the last $k+1$ boxes, $b_{2^k-1}, b_{2^k}, \\dots, b_{2^k+k-1}$, at each step: At step $r$, if box $b_{2^k+i-1}$ has exactly $m_i$ coins, then $\\lceil m_i/2 \\rceil$ coins are removed from that box, unless $... | Romania | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | n = 2^k + k - 1 | |
0du5 | Problem:
Naj bodo $k$, $m$ in $n$ poljubna naravna števila, ki niso deljiva s $5$. Dokaži, da je vsaj eno izmed števil $k^{2}-m^{2}$, $m^{2}-n^{2}$ in $n^{2}-k^{2}$ deljivo s $5$. | [
"Solution:\n\nKer števila $k$, $m$ in $n$ niso deljiva s $5$, so oblike $5t \\pm 1$ ali $5t \\pm 2$. Iz $(5t \\pm 1)^{2} = 5t' + 1$ in $(5t \\pm 2)^{2} = 5t' + 4$ pa vidimo, da imajo števila $k^{2}$, $m^{2}$ in $n^{2}$ le dva možna ostanka pri deljenju s $5$. Torej je vsaj ena izmed razlik $k^{2}-m^{2}$, $m^{2}-n^{... | Slovenia | 45. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0dq6 | Let $CD$ be a chord of a circle $\Gamma_1$ and $AB$ a diameter of $\Gamma_1$ perpendicular to $CD$ at $N$ with $AN > NB$. A circle $\Gamma_2$ centred at $C$ with radius $CN$ intersects $\Gamma_1$ at points $P$ and $Q$. The line $PQ$ intersects $CD$ at $M$ and $AC$ at $K$; and the extension of $NK$ meets $\Gamma_2$ at $... | [
"1. Extend $DC$ meeting $\\Gamma_2$ at $H$. Let the radius of $\\Gamma_2$ be $r$. Note that $DN = NC = CH = r$. Since $M$ is of equal power with respect to $\\Gamma_1$ and $\\Gamma_2$. Thus $MN \\cdot NH = MC \\cdot MD$. That is $MN(MC + r) = MC(MN + r)$ giving $MN = MC$. Thus $M$ is the midpoint of $NC$.\n\n | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05jh | Problem:
La suite de Fibonacci est construite ainsi : chaque terme est la somme des deux précédents. Ses premiers termes valent:
$$
1,1,2,3,5,8,13,21,34,55,89, \ldots
$$
Montrer que la suite de Fibonacci contient un multiple de 1000. | [
"Solution:\n\nNotons $F_{n}$ le $n$-ième terme de la suite de Fibonacci ($F_{1}=1$, $F_{2}=1$, $F_{3}=2$, etc.) et notons $u_{n}$ le reste de la division euclidienne de $F_{n}$ par $1000$ ($u_{n}$ est donc le nombre formé par les trois derniers chiffres de $F_{n}$). Comme il n'y a qu'un nombre fini de couples $(x, ... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
07op | Let $n \ge 2$ be an integer and $r = \sqrt[3]{2}$. Prove that there do not exist rational numbers $a_0, a_1, \dots, a_{n-1}$, not all of them equal to zero, such that
$$
a_0 + a_1 r + a_2 r^2 + \dots + a_{n-1} r^{n-1} = 0.
$$ | [
"Assume there exist rational numbers $a_0, a_1, \\dots, a_{n-1}$ such that\n$$\na_0 + a_1 r + a_2 r^2 + \\dots + a_{n-1} r^{n-1} = 0.\n$$\nConsider the polynomials\n$$\nf(x) = a_0 + a_1 x + \\dots + a_{n-1} x^{n-1} \\quad \\text{and} \\quad g(x) = x^n - 2.\n$$\nWe then have $f(r) = g(r) = 0$. By Eisenstein's criter... | Ireland | Irska 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Abstract Algebra > Field Theory"
] | null | proof only | null | |
0h5d | Prove that number $m^4 + 1$ has no divisors in interval $[m^2 - 2m, m^2 + 2m]$ for every natural $m > 2$. | [
"Suppose the opposite. Let\n$$\nm^2 + a \\in [m^2 - 2m, m^2 + 2m] \\text{ i } m^4 + 1 \\equiv m^2 + a.\n$$\nSince\n$$\n(m^4 + 1, m^2 + a) = (-a m^2 + 1, m^2 + a) = (a^2 + 1, m^2 + a), \\text{ then } a^2 + 1 \\equiv m^2 + a, \\\\\n a^2 + 1 = (m^2 + a) r.\n$$\nLet $a \\in [-2m+3, 0]$, then $\\frac{m^4+1}{m^2+a} \\le ... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour) | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0bus | Problem:
Demonstraţi că
$$
\log_{2016} \sqrt[2016]{1+\frac{a_{2}}{a_{1}}} + \log_{2016} \sqrt[2016]{1+\frac{a_{3}}{a_{2}}} + \cdots + \log_{2016} \sqrt[2016]{1+\frac{a_{1}}{a_{2016}}} \geq \frac{1}{5+\log_{2} 63}
$$
pentru orice $a_{1}, a_{2}, \ldots, a_{2016}$ numere reale strict pozitive. | [] | Romania | Olimpiada de Matematică Etapa Locală | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof only | null | |
0kyn | Quadrilateral $ABCD$ is a parallelogram, and $E$ is the midpoint of the side $\overline{AD}$. Let $F$ be the intersection of lines $EB$ and $AC$. What is the ratio of the area of quadrilateral $CDEF$ to the area of $\triangle CFB$?
(A) 5:4 (B) 4:3 (C) 3:2 (D) 5:3 (E) 2:1 | [
"Triangles $\\triangle AFE$ and $\\triangle CFB$ are similar by Angle-Angle. The ratio of corresponding sides is $1:2$, so the ratio of their areas is $1:4$. Furthermore, consider $\\triangle AFE$ and $\\triangle AFB$. Because $FE:FB = 1:2$ and the heights of the two triangles corresponding to bases $\\overline{FE}... | United States | AMC 10 B | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | A | |
0hhc | Let $K$ be the midpoint of the median $AM$ of the triangle $ABC$. On the side $AB$ there is a point $X$ such that $\angle KXM = \angle ACB$ and $AX > BX$, and on the side $AC$ there is a point $Y$ such that $\angle KYM = \angle ABC$ and $AY > CY$. Prove that the points $B, X, C, Y$ lie on the same circle.
(Mykhailo Sh... | [
"Let $N$ be the middle of $AB$, and $L$ be the middle of $AC$. Then the points $N$, $K$ and $L$ lie on the same line—the midline of $\\triangle ABC$ which is parallel to $BC$ (fig. 19).\n\nThen we have that $\\angle KXM = \\angle ACB = \\angle KNM$, so the quadrilateral $KNXM$ is inscribed, and similarly $KMYL$ is ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
0bem | Let $a$ and $b$ be real numbers different from $0$ and let $f: \mathbb{R} \to \mathbb{R}$ be a function defined by
$$
f(x) = \begin{cases} ax, & x \in \mathbb{Q} \\ bx, & x \in \mathbb{R} \setminus \mathbb{Q} \end{cases}
$$
Prove that $f$ is injective if and only if $f$ is surjective. | [
"The function $f$ is injective if and only if $\\frac{a}{b} \\in \\mathbb{Q}$. Indeed, suppose $f$ is injective and $\\frac{a}{b} \\in \\mathbb{R} \\setminus \\mathbb{Q}$. Then $f\\left(\\frac{a}{b}\\right) = a = f(1)$, a contradiction. Suppose now that $\\frac{a}{b} \\in \\mathbb{Q}$ and let $x_1, x_2$ such that $... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof only | null | |
0fqh | Problem:
Determinar si existe un conjunto finito $S$ formado por números primos positivos de manera que para cada entero $n \geq 2$, el número $2^{2}+3^{2}+\cdots+n^{2}$ sea múltiplo de algún elemento de $S$. | [
"Solution:\n\nEs bien conocido que\n$$\n1^{2}+2^{2}+\\cdots+n^{2}=\\frac{n(n+1)(2 n+1)}{6}=\\frac{2 n^{3}+3 n^{2}+n}{6}\n$$\ncon lo que restando 1 a ambos lados, resulta\n$$\n2^{2}+3^{2}+\\cdots+n^{2}=\\frac{2 n^{3}+3 n^{2}+n-6}{6}=\\frac{(n-1)\\left(2 n^{2}+5 n+6\\right)}{6}\n$$\nEsta expresión sugiere tomar $p=(n... | Spain | LV Olimpiada matemática Española | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0ffr | Problem:
Sea $ABCD$ un cuadrilátero convexo tal que la recta $CD$ es tangente al círculo de diámetro $AB$. Demostrar que la recta $AB$ es tangente al círculo de diámetro $CD$ si y sólo si las rectas $BC$ y $AD$ son paralelas. | [] | Spain | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08mp | Problem:
Determine the positive integers $a, b$ such that $a^{2} b^{2} + 208 = 4\{lcm[a, b] + gcd(a, b)\}^{2}$. | [
"Solution:\nLet $d = \\gcd(a, b)$ and $x, y \\in \\mathbb{Z}_{+}$ such that $a = d x$, $b = d y$. Obviously, $(x, y) = 1$. The equation is equivalent to $d^{4} x^{2} y^{2} + 208 = 4 d^{2} (x y + 1)^{2}$. Hence $d^{2} \\mid 208$ or $d^{2} \\mid 13 \\cdot 4^{2}$, so $d \\in \\{1, 2, 4\\}$. Take $t = x y$ with $t \\in... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | [(2, 12), (4, 6), (6, 4), (12, 2)] | |
0g1u | Problem:
Sei $k \geq 0$ eine ganze Zahl. Bestimme alle reellen Polynome $P$ von $\operatorname{Grad} k$ mit $k$ verschiedenen reellen Nullstellen, sodass für alle Nullstellen $a$ von $P$ gilt:
$$
P(a+1)=1
$$ | [
"Solution:\n\nAlle konstanten Polynome $P(x)=c \\neq 0$ sind Lösungen. Sei $P$ nun ein solches Polynom mit Grad mindestens 1. Das Polynom $Q(x)=P(x+1)-P(x)-1$ hat nach Voraussetzung mindestens $n$ verschiedene Nullstellen. Da $Q$ aber maximal $\\operatorname{Grad} n-1$ hat, folgt dass $Q(x)=0$ bzw. $P(x+1)=P(x)+1$ ... | Switzerland | SMO-Selektion | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All constant nonzero polynomials and all linear polynomials of the form x plus b with any real b. | |
08s4 | $n$ is a 4-digit integer with ten's place nonzero, and if we take the first 2 digits and the last 2 digits as two 2-digit integers, their product is a divisor of $n$. Determine all $n$ with this property. | [
"Let $A$ and $B$ be the first and the last 2 digits of $n$ respectively. We have to determine all $(A, B)$ such that $AB$ divides $100A + B$.\n\nSince $A$ divides $100A + B$, $A$ divides $B$. Let $k = \\frac{B}{A}$. Since $A$ and $B$ are 2-digit numbers, $10 \\leq A < \\frac{100}{k}$.\n\nThe condition is equivalent... | Japan | Japanese Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1734, 1352 | |
06u7 | A sequence of real numbers $a_{1}, a_{2}, \ldots$ satisfies the relation
$$
a_{n} = -\max_{i+j=n} (a_{i} + a_{j}) \quad \text{ for all } n > 2017
$$
Prove that this sequence is bounded, i.e., there is a constant $M$ such that $|a_{n}| \leqslant M$ for all positive integers $n$. | [
"Set $D = 2017$. Denote\n$$\nM_{n} = \\max_{k < n} a_{k} \\quad \\text{ and } \\quad m_{n} = -\\min_{k < n} a_{k} = \\max_{k < n} (-a_{k}) .\n$$\nClearly, the sequences $(m_{n})$ and $(M_{n})$ are nondecreasing. We need to prove that both are bounded.\n\nConsider an arbitrary $n > D$; our first aim is to bound $a_{... | IMO | International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0gqg | Find the total number of all permutations of $2014$-tuple $(1, 2, \dots, 2014)$ satisfying $i + a_i \leq j + a_j$ for all $1 \leq i < j \leq 2014$. | [
"Let $f(n)$ be the total number of all permutations of $n$-tuple $(1, 2, \\dots, n)$ satisfying $i + a_i \\le j + a_j$ for all $1 \\le i < j \\le n$. Readily $f(1) = 1$.\n\nIf for some permutation $a_1 = n$, then $a_2 = n-1, a_3 = n-2, \\dots, a_n = 1$ is the only possibility.\n\nIf for some permutation $a_{k+1} = ... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2^{2013} | |
0a1b | Problem:
In een driehoek $\triangle ABC$ met $\angle ABC < \angle BCA$ definiëren we $K$ als het middelpunt van de aangeschreven cirkel aan $AC$. De lijnen $AK$ en $BC$ snijden in een punt $D$. Laat $E$ het middelpunt zijn van de omgeschreven cirkel van $\triangle BKC$. Bewijs dat
$$
\frac{1}{|KA|} = \frac{1}{|KD|} + ... | [
"Solution:\n\nWe gebruiken de notatie $\\angle CAB = 2\\alpha$, $\\angle ABC = 2\\beta$ en $\\angle BCA = 2\\gamma$. Merk op dat $\\alpha + \\beta + \\gamma = \\frac{1}{2}(\\angle CAB + \\angle ABC + \\angle BCA) = \\frac{1}{2} \\cdot 180^{\\circ} = 90^{\\circ}$.\n\nWe bewijzen eerst dat $K$, $A$ en $E$ collineair ... | Netherlands | Selectietoets | [
"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 | |
0ap8 | Problem:
Container $A$ contained a mixture that is $40\%$ acid, while container $B$ contained a mixture that is $60\%$ acid. A chemist took some amount from each container, and mixed them. To produce 100 liters of mixture that is $17\%$ acid, she needed to pour 70 liters of pure water to the mixture she got from conta... | [
"Solution:\n\nLet $a$ be the amount (in liters) of mixture the chemist took from container $A$, and $b$ the amount she took from container $B$. Then $a + b + 70 = 100$. On the other hand, computing the amount of acid involved in the mixtures, we have $0.40 a + 0.60 b = 0.17 \\times 100$ or $4a + 6b = 170$. Solving ... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 5 | |
0jw8 | Problem:
Find the number of divisors of $2^{9} \cdot 3^{14}$. | [
"Solution:\n\nEach divisor is the product of some power of $2$ and some power of $3$. There are $10$ choices for the power of $2$ ($2^{0}, 2^{1}, \\ldots, 2^{9}$) and $15$ choices for the power of $3$ ($3^{0}, 3^{1}, \\ldots, 3^{14}$), and, since these choices are independent, there are a total of $10 \\cdot 15 = 1... | United States | Berkeley Math Circle | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | final answer only | 150 | |
0gep | 設 $ABC$ 為銳角三角形，$I$ 及 $I_A$ 分別為其內心及角 $A$ 內的旁心，且 $AB < AC$。設其內切圓交 $BC$ 於點 $D$。直線 $AD$ 分別與 $BI_A$ 及 $CI_A$ 交於點 $E$ 及 $F$。證明三角形 $AID$ 的外接圓與三角形 $IAEF$ 的外接圓相切。
Let $I$ and $I_A$ be the incenter and the $A$-excenter of an acute-angled triangle $ABC$, with $AB < AC$. Let the incircle meet $BC$ at $D$. The line $AD$ meets $BI_... | [
"Let $\\angle(p, q)$ denote the directed angle between lines $p$ and $q$.\nThe points $B, C, I$, and $I_A$ lie on the circle $\\Gamma$ with diameter $II_A$. Let $\\omega$ and $\\Omega$ denote the circles $(IAEF)$ and $(AID)$, respectively. Let $T$ be the second intersection point of $\\omega$ and $\\Gamma$. Then $T... | Taiwan | 2021 數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: c... | null | proof only | null | |
0jof | Problem:
What is the largest real number $\theta$ less than $\pi$ (i.e. $\theta<\pi$ ) such that
$$
\prod_{k=0}^{10} \cos \left(2^{k} \theta\right) \neq 0
$$
and
$$
\prod_{k=0}^{10}\left(1+\frac{1}{\cos \left(2^{k} \theta\right)}\right)=1 ?
$$ | [
"Solution:\nAnswer: $\\frac{2046 \\pi}{2047}$\n\nFor equality to hold, note that $\\theta$ cannot be an integer multiple of $\\pi$ (or else $\\sin =0$ and $\\cos = \\pm 1$).\n\nLet $z=e^{i \\theta / 2} \\neq \\pm 1$. Then in terms of complex numbers, we want\n$$\n\\prod_{k=0}^{10}\\left(1+\\frac{2}{z^{2^{k+1}}+z^{-... | United States | HMMT February | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 2046π/2047 | |
00xt | Problem:
In the triangle $ABC$ we have $|AB| = 15$, $|BC| = 12$ and $|AC| = 13$. Let the median $AM$ and bisector $BK$ intersect at point $O$, where $M \in BC$, $K \in AC$. Let $OL \perp AB$, where $L \in AB$. Prove that $\angle OLK = \angle OLM$. | [
"\nFigure 6"
] | Baltic Way | Baltic Way 1993 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
058h | There are $n$ lists of candidates taking part in elections. Let $h_i$ be the total number of votes given for the candidates of the $i$th list. There are $M$ seats in the representative assembly.
Anna proposes the following system for delivering mandates: For each list, one computes a reference number $v_i = \frac{h_i}... | [
"Let the total number of mandates given to the $i$th list be $m_i$ in the case of Bert's method and $m'_i$ in the case of Anna's method. Suppose that these methods result in different distribution of mandates. As the sum of numbers of mandates must be the same, we must have $m'_i > m_i$ for some $i = 1, 2, \\dots, ... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Other"
] | English | proof only | null | |
00th | Let $ABC$ be a triangle with $AB < AC$. Let $\omega$ be a circle passing through $B, C$ and assume that $A$ is inside $\omega$. Suppose $X, Y$ lie on $\omega$ such that $\angle BXA = \angle AYC$ and $X$ lies on the opposite side of $AB$ to $C$ while $Y$ lies on the opposite side of $AC$ to $B$.
Show that, as $X, Y$ var... | [
"Extend $XA$ and $YA$ to meet $\\omega$ again at $X'$ and $Y'$ respectively. We then have that:\n$$\n\\angle Y'YC = \\angle AYC = \\angle BXA = \\angle BXX'.\n$$\nso $BCX'Y'$ is an isosceles trapezium and hence $X'Y' \\parallel BC$.\n\nLet $\\ell$ be the line through $A$ parallel to $BC$ an... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0ftj | Problem:
In einer Gruppe von $n$ Leuten veranstaltet jedes Wochenende jemand eine Party, an der er alle seine Bekannten einander gegenseitig vorstellt. Nachdem jeder der $n$ Leute einmal eine Party gemacht hat, gibt es immer noch zwei Personen unter ihnen, die sich nicht kennen. Zeige, dass diese zwei sich auch in Zuk... | [
"Solution:\n\nBetrachte einen Graph, dessen Ecken die $n$ Leute sind, und verbinde zwei Leute mit einer Kante, wenn sie sich kennen. Immer wenn jemand eine Party macht, verändert sich dieser Graph. Nehme an $A$ macht eine Party, dann werden alle direkt mit $A$ verbundenen Ecken gegenseitig verbunden.\n\nWir zeigen ... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
018r | Let $E$ be an interior point in the convex quadrilateral $ABCD$. Let $F$, $G$, $H$, and $I$ be points opposite the quadrilateral with respect to the lines $AB$, $BC$, $CD$, and $DA$, respectively, such that $\triangle ABF \sim \triangle DCE$, $\triangle BCG \sim \triangle ADE$, $\triangle CDH \sim \triangle BAE$, and $... | [
"We consider oriented angles modulo $180^\\circ$. From the cyclic quadrilaterals $APES$, $BQEP$, $PQRS$, $CREQ$, $DSER$ and $\\triangle DCE \\sim \\triangle ABF$ we get\n$$\n\\begin{align*}\n\\angle AEB &= \\angle EAB + \\angle ABE = \\angle ESP + \\angle PQE \\\\\n&= \\angle ESR + \\angle RSP + \\angle PQR + \\ang... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0i3m | Problem:
Express $\frac{\sin 10+\sin 20+\sin 30+\sin 40+\sin 50+\sin 60+\sin 70+\sin 80}{\cos 5 \cos 10 \cos 20}$ without using trigonometric functions. | [
"Solution:\n\nWe will use the identities $\\cos a + \\cos b = 2 \\cos \\frac{a+b}{2} \\cos \\frac{a-b}{2}$ and $\\sin a + \\sin b = 2 \\sin \\frac{a+b}{2} \\cos \\frac{a-b}{2}$.\n\nThe numerator is\n$$(\\sin 10 + \\sin 80) + (\\sin 20 + \\sin 70) + (\\sin 30 + \\sin 60) + (\\sin 40 + \\sin 50)$$\n\n$= 2 \\sin 45 (\... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 4√2 | |
034w | Problem:
Find all values of the real parameter $a$ such that the equation
$$
\log_{a x}\left(3^{x}+4^{x}\right)=\log_{(a x)^{2}}\left(7^{2}\left(4^{x}-3^{x}\right)\right)+\log_{(a x)^{3}} 8^{x-1}
$$
has a solution. | [
"Solution:\nSince $a x>0$ and $4^{x}-3^{x}>0$ it follows that $a>0$ and $x>0$.\n\nFor $a>0$, $x>0$ and $a x \\neq 1$ the equation is equivalent to\n$$\n3^{x}+4^{x}=7 \\cdot 2^{x-1} \\sqrt{4^{x}-3^{x}} \\Longleftrightarrow 45\\left(\\frac{4}{3}\\right)^{2x}-57\\left(\\frac{4}{3}\\right)^{x}-4=0\n$$\nSetting $y=\\lef... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | a ∈ (0, ∞) \ {1} | |
0drd | A boy lives in a small island in which there are three roads at every junction. He starts from his home and walks along the roads. At each junction he would choose to turn to the road on his right or left alternately, i.e., his choices would be ..., left, right, left, .... Prove that he will eventually return to his ho... | [
"After walking for a long time, he must walk along a certain road 6 times and in the direction at least 3 times. When he reaches the junction along this direction, he must choose the same direction twice. Both times, tracing backwards, we conclude that both times the trails are the same. Since first one traces back... | Singapore | Singapur 2015 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Other"
] | null | proof only | null | |
0fim | Problem:
Las rectas $t$ y $t'$, tangentes a la parábola de ecuación $y = x^{2}$ en los puntos $A$ y $B$, se cortan en el punto $C$.
La mediana del triángulo $\triangle ABC$ correspondiente al vértice $C$ tiene longitud $m$.
Determinar el área del triángulo $\triangle ABC$ en función de $m$. | [
"Solution:\nSean $A\\left(a, a^{2}\\right)$; $B\\left(b, b^{2}\\right)$. Las ecuaciones de $t$ y $t'$ son:\n$$\nt: y=2 a x-a^{2}, \\quad t': y=2 b x-b^{2}\n$$\ny su intersección $C$ es: $C\\left(\\frac{a+b}{2}, a b\\right)$.\nLa mediana $CM$ está en la recta: $x=\\frac{a+b}{2}$, paralela al eje $OY$. Las coordenada... | Spain | Olimpíada Matemática Española | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | sqrt(m^3/2) | |
02ow | Problem:
Um número $n$ de dois algarismos é dividido pela soma de seus algarismos, obtendo resto $r$.
a. Encontre um número $n$ tal que $r=0$.
b. Mostre que $r$ não pode ser maior que 15.
c. Mostre que para qualquer $r$ menor ou igual a 12, existe um $n$ que deixa resto $r$ ao dividi-lo pela soma de seus algarismos. | [
"Solution:\n\na. Existem vários exemplos onde o resto da divisão é 0, sendo o menor deles $n=12$.\n\nb. Denotemos por $S$ a soma dos algarismos de $n$.\nObservemos que $S \\leqslant 18$ e a igualdade somente acontece se $n=99$, mas neste caso o resto da divisão é 9.\nSe $S=17$, temos dois possíveis valores de $n=89... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | a) One example is n = 12 (remainder 0).
b) The remainder cannot exceed 15; for instance, n = 79 gives remainder 15.
c) For each r from 1 to 12 there exists a two-digit number: with digit sum 10, the numbers 19, 28, 37, 46, 55, 64, 73, 82, 91 yield remainders 9, 8, 7, 6, 5, 4, 3, 2, 1 respectively; and 65, 76, 87 yield ... | |
0gsj | Suppose that for some positive integers $a$ and $b$, $\frac{a^2 + b^2}{(a-b)^2}$ is an integer number. Show that $\frac{a^3 + b^3}{(a-b)^3}$ is also an integer number. | [
"Since $\\frac{a^2 + b^2}{(a-b)^2}$ is an integer then\n$$\n\\frac{a^2 + b^2 - (a-b)^2}{(a-b)^2} = \\frac{2ab}{(a-b)^2}\n$$\nis also an integer. Without loss of generality we can assume that $\\gcd(a, b) = 1$. Then readily $\\gcd(a-b, a) = 1$ and $\\gcd(a-b, a) = 1$. Therefore, $(a-b)^2 \\mid 2$. Since $a-b$ can no... | Turkey | Team Selection Test | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0dvy | Problem:
Ali obstaja naravno število $n$, za katerega velja: če število $n$ pomnožimo z vsoto njegovih števk, je vsota števk dobljenega zmnožka enaka 3? | [
"Solution:\n\nOznačimo s $s(m)$ vsoto števk naravnega števila $m$. Naloga sprašuje, ali obstaja naravno število $n$, da je $s(n \\cdot s(n))=3$. Ker je naravno število deljivo s 3 natanko tedaj, ko je s 3 deljiva vsota njegovih števk, mora biti s 3 deljiv zmnožek $n \\cdot s(n)$. Torej mora biti s 3 deljivo vsaj en... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
01yf | Prove that all divisors of any positive integer can be arranged in the circle such that for any two consecutive divisors one would divide another.
(Mikhail Karpuk) | [
"Let's prove this statement by induction on the number of distinct prime divisors of a given number. In this case, this number will be denoted by $n$, the number of its divisors — by $d(n)$ and the divisors written in counterclockwise order — by $a_1, a_2, \\dots, a_{d(n)}$, where $a_1 = 1$.\n\n*Base:* let $n = p^k... | Belarus | Belarus2022 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
09sb | Problem:
Zij $n$ een positief geheel getal. We bekijken rijtjes getallen $a_{0}, a_{1}, \ldots, a_{k}$ en $b_{0}, b_{1}, \ldots, b_{k}$ die voldoen aan $a_{0}=b_{0}=1$ en $a_{k}=b_{k}=n$ en waarbij voor elke $i$ met $1 \leq i \leq k$ geldt dat $(a_{i}, b_{i})$ gelijk is aan ofwel $(1+a_{i-1}, b_{i-1})$ ofwel $(a_{i-1}... | [
"Solution:\n\nWe bewijzen met inductie naar $j$ dat $c_{1}+\\ldots+c_{j}=a_{j} b_{j}-1$. Voor $j=1$ staat hier $c_{1}=a_{1} b_{1}-1$ en dat is waar aangezien $(a_{1}, b_{1}) \\in\\{(1,2),(2,1)\\}$. Stel nu dat $c_{1}+\\ldots+c_{i-1}=a_{i-1} b_{i-1}-1$. Neem zonder verlies van algemeenheid aan dat $(a_{i}, b_{i})=(a... | Netherlands | IMO-selectietoets | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0hwp | Problem:
Find all primes $p$ such that there exist integers $a, b, c$, and $k$ satisfying the equations
$$
\begin{aligned}
& a^{2}+b^{2}+c^{2}=p \\
& a^{4}+b^{4}+c^{4}=k p .
\end{aligned}
$$ | [
"Solution:\nThe answers are $2$ and $3$. It is clear that for them we can use the values $a = b = k = 1$ and $c = 0$ or $1$ respectively. Assume $p \\geq 5$. Subtracting twice the second equation from the square of the first, we find that $p$ divides\n$$\n2 a^{2} b^{2} + 2 b^{2} c^{2} + 2 c^{2} a^{2} - a^{4} - b^{4... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequ... | null | proof and answer | 2 and 3 | |
05u5 | Problem:
Soit $ABC$ un triangle tel que $90^{\circ} > \widehat{ABC} > \widehat{BCA}$. Soit $D$ le point sur le segment $[BC]$ tel que $2 \widehat{DAC} = \widehat{ABC} - \widehat{BCA}$. On note $E$ le point d'intersection, autre que $A$, entre $(AB)$ et le cercle circonscrit à $ACD$, puis $P$ le point d'intersection en... | [
"Solution:\n\nComme d'habitude, on commence par une figure suffisamment belle et grande pour servir de support à la réflexion. À cette fin, on construit tout d'abord un triangle $A'BC'$ isométrique à $ABC$, tel que $C'$ se trouve sur la demi-droite $[BA)$, puis on trace la bissectrice de l'angle $\\widehat{CBA'}$, ... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09uz | A watchmaker installed the big and small hand of a clock in the wrong way. This makes the small hand go with the speed of the big hand and the big hand with the speed of the small hand. It is known that every day at $8{:}00$ the clock shows the right time.
How many times a day ($24$ hours) does the clock show the right... | [
"D) $22$"
] | Netherlands | Junior Mathematical Olympiad | [
"Math Word Problems"
] | English | MCQ | D | |
0fdt | Problem:
En un hexágono regular de lado unidad se sitúan 19 puntos. Demuestra que hay al menos un par de ellos separados por una distancia no mayor que $\sqrt{3} / 3$. | [
"Solution:\nDividamos el hexágono regular en 6 triángulos equiláteros iguales. Cada uno de ellos, si trazamos sus alturas, quedará dividido en 6 triángulos rectángulos. Uniendo estos triángulos rectángulos, dos a dos, por sus hipotenusas, habremos dividido el hexágono original en 18 regiones iguales.\n\n![](attache... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"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 | |
011s | Problem:
Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that $\sum_{i=1}^{n} a_{i}^{3}=3$ and $\sum_{i=1}^{n} a_{i}^{5}=5$. Prove that $\sum_{i=1}^{n} a_{i}>\frac{3}{2}$. | [
"Solution:\nBy Hölder's inequality,\n$$\n\\sum_{i=1}^{n} a^{3}=\\sum_{i=1}^{n}\\left(a_{i} \\cdot a_{i}^{2}\\right) \\leqslant\\left(\\sum_{i=1}^{n} a_{i}^{5 / 3}\\right)^{3 / 5} \\cdot\\left(\\sum_{i=1}^{n}\\left(a_{i}^{2}\\right)^{5 / 2}\\right)^{2 / 5} .\n$$\nWe will show that\n$$\n\\sum_{i=1}^{n} a_{i}^{5 / 3} ... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0frw | Sean $x_1 \le x_2 \le x_3 \le x_4$ cuatro números reales. Demuestra que existen $P(x)$ y $Q(x)$ polinomios de grado dos con coeficientes reales tales que $x_1, x_2, x_3$ y $x_4$ son las raíces de $P(Q(x))$ si y solamente si $x_1 + x_4 = x_2 + x_3$. | [
"Supongamos que las raíces de $P(x)$ son $r$ y $s$. Entonces las raíces de $P(Q(x))$ son las raíces de $Q(x) - r$ y de $Q(x) - s$. Por las relaciones de Cardano, la suma de las raíces de $Q(x) - r$ es la misma que la de $Q(x) - s$, con lo que las raíces de $P(Q(x))$ necesariamente satisfacen que la suma de dos de e... | Spain | LIX Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | Spanish | proof only | null | |
05wk | Problem:
Soient $a, b$ deux entiers. Montrer que $a b(a-b)$ est divisible par 2. | [
"Solution:\n\nOn s'intéresse à la parité d'un produit, il est donc pertinent de faire une disjonction de cas sur la parité des entiers en jeu, ici $a$ et $b$. On remarque alors la chose suivante :\n\n- Si $a$ ou $b$ est pair : alors $a b$ est pair (le produit d'un nombre pair par un entier quelconque étant pair) et... | France | Préparation Olympique Française de Mathématiques - Envoi 3: Arithmétique | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0bg9 | Problem:
Să se rezolve ecuaţia:
$$
2^{\sin^{4} x-\cos^{2} x}-2^{\cos^{4} x-\sin^{2} x}=\cos 2x
$$ | [
"Solution:\nEcuaţia se scrie echivalent\n$$\n2^{\\sin^{4} x+\\sin^{2} x}-2^{\\cos^{4} x+\\cos^{2} x}=2 \\cos 2x.\n$$\nSă observăm că\n$$\n\\begin{aligned}\n\\cos^{4} x+\\cos^{2} x-\\sin^{4} x-\\sin^{2} x &= (\\cos^{2} x+\\sin^{2} x)(\\cos^{2} x-\\sin^{2} x)+\\cos^{2} x-\\sin^{2} x \\\\\n&= 2 \\cos 2x\n\\end{aligned... | Romania | Olimpiada Naţională de Matematică | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | null | proof and answer | x = (2k+1)π/4, where k ∈ ℤ | |
0gyb | Prove that there exist infinitely many such prime numbers $p$ that amongst $\{0,1,2,\dots, p-1\}$ can be found at least 2008 numbers $x$ such that $x^a \equiv 1 \pmod{p}$. | [
"Let $p$ be some prime divisor of the number $2^{2^n} + 1$. Consider numbers $x$ of the form $2^k$. By the statement of the problem we need $x^x - 1 = 2^{2^s} - 1 \\equiv 0 \\pmod{p}$. If $s \\ge n+1$ then\n$$\n2^{2^{2^k}} - 1 = (2^{2^{n+1}})^{2^{2^k-(n+1)}} - 1 : 2^{2^{n+1}} - 1 = (2^{2^k} - 1)(2^{2^k} + 1),\n$$\n... | Ukraine | The Problems of Ukrainian Authors | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequaliti... | English | proof only | null | |
0aih | Let $ABCD$ be a cyclic quadrilateral with the intersection of internal angle bisectors of $\angle ABC$ and $\angle ADC$ lying on the diagonal $AC$. Let $M$ be the midpoint of $AC$. The line parallel to $BC$ that passes through $D$ intersects the line $BM$ in $E$ and the circumcircle of $ABCD$ at $F$ where $F \neq D$. P... | [
"We prove the problem in reverse as this is much more natural in this problem.\nWe note that if $BCEF$ is a parallelogram then the diagonales are bisecting each other so the point $G \\equiv BE \\cap CF$ should be the midpoint of $CE$.\nIf $G$ is the midpoint of $CE$ then $\\triangle GBC$ and $\\triangle GEF$ are c... | North Macedonia | European Mathematical Cup | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0f0y | Problem:
No triangle of area $1$ can be fitted inside a convex polygon. Show that the polygon can be fitted inside a triangle of area $4$. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0ixd | Problem:
Compute
$$
\lim_{h \rightarrow 0} \frac{\sin \left(\frac{\pi}{3}+4 h\right)-4 \sin \left(\frac{\pi}{3}+3 h\right)+6 \sin \left(\frac{\pi}{3}+2 h\right)-4 \sin \left(\frac{\pi}{3}+h\right)+\sin \left(\frac{\pi}{3}\right)}{h^{4}}
$$ | [
"Solution:\n\nThe derivative of a function is defined as $f'(x)=\\lim_{h \\rightarrow 0} \\frac{f(x+h)-f(x)}{h}$. Iterating this formula four times yields\n$$\nf^{(4)}(x)=\\lim_{h \\rightarrow 0} \\frac{f(x+4 h)-4 f(x+3 h)+6 f(x+2 h)-4 f(x+h)+f(x)}{h^{4}}.\n$$\nSubstituting $f=\\sin$ and $x=\\pi / 3$, the expressio... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Limits",
"Precalculus > Trigonometric functions"
] | null | proof and answer | sqrt(3)/2 | |
006g | Alrededor de una circunferencia están escritos $20$ números enteros.
Para cada uno de ellos, se calcula la suma de los $10$ números que le siguen en el sentido de las agujas del reloj. Terminado esto, cada uno de los $20$ números es sustituido por su correspondiente suma.
Demostrar que después de repetir varias veces e... | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | Spanish | proof only | null | |
0gx0 | (Malitskii Yrii) Find out what five-digit numbers are there more: (1) those in which all digits follow from left to right in the order of increasing, or (2) those in which each digit is not greater than $5$ and the order those digits follow from left to right is not descending (e.g. $12459$ satisfies the first conditio... | [
"In order to prove the answer let's associate each number in one set of numbers to a single number in the other set. It's enough to associate the number $abcde$ from the first set to the number $a(b-1)(c-2)(d-3)(e-4)$. This implies that both sets have equal power."
] | Ukraine | Ukrajina 2008 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | They are equal in number. | |
0ivv | Suppose that $s_1, s_2, s_3, \dots$ is a strictly increasing sequence of positive integers such that the subsequences $s_{s_1}, s_{s_2}, s_{s_3}, \dots$ and $s_{s_{1+1}}, s_{s_{2+1}}, s_{s_{3+1}}, \dots$ are both arithmetic progressions. Prove that the sequence $s_1, s_2, s_3, \dots$ is itself an arithmetic progression... | [
"Let $D$ be the common difference of the progression $s_{s_1}, s_{s_2}, s_{s_3}, \\dots$. Also, define $d_n = s_{n+1} - s_n$. For every $n$, $d_n \\ge 1$ because $s_1, s_2, s_3, \\dots$ is strictly increasing. Therefore, for all $i < j$, we have\n$$\ns_j - s_i = d_i + d_{i+1} + \\dots + d_{j-1} = \\sum_{k=i}^{j-1} ... | United States | IMO 2009 | [
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof only | null | |
0ke3 | Problem:
Ainsley and Buddy play a game where they repeatedly roll a standard fair six-sided die. Ainsley wins if two multiples of $3$ in a row are rolled before a non-multiple of $3$ followed by a multiple of $3$, and Buddy wins otherwise. If the probability that Ainsley wins is $\frac{a}{b}$ for relatively prime posi... | [
"Solution:\n\nWe let $X$ be the event of a multiple of $3$ being rolled and $Y$ be the event of a non-multiple of $3$ being rolled. In order for Ainsley to win, she needs event $X$ to happen consecutively; meanwhile, Buddy just needs $Y$ then $X$ to occur. Thus, if $Y$ occurs in the first two rolls, Buddy will be g... | United States | HMMO 2020 | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 109 | |
0hic | Suppose for a rhombus $ABCD$ there exists a point $T$ such that the following conditions are satisfied: $\angle ATC + \angle BTD = 180^\circ$ and the circumcircles of the triangles $ATC$ and $BTD$ are tangent to each other. Prove that the point $T$ is equidistant from the diagonals of the rhombus. | [
"Let $P$ be the point of intersection of the rhombus diagonals, $O_1$ and $O_2$ be the centers of the described circles $\\Delta ATC$ and $\\Delta BTD$ respectively (Fig. 1). Hence, the following equalities are true:\n$$\n\\angle A O_1 C = 2 \\angle A T C = 2 \\cdot (180^\\circ - \\angle B T D) = \\angle B O_2 D,\n... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > T... | English | proof only | null | |
0cpu | A pile of more than $n^2$ stones is put on the table. Pete and Basil make moves taking stones in turn; Pete moves first. By one move it is allowed to take $k$ stones from a pile if either $k = 1$, or $k$ is a prime number less than $n$, or $k$ is divisible by $n$. The one taking the last stone wins. Prove that Pete has... | [
"Предположим противное: Вася может всегда действовать так, чтобы помешать Пете; поскольку в любой ситуации, кроме конечной, можно сделать ход, это означает, что у Васи есть стратегия, поволяющая ему гарантированно взять последний камень. Пусть $d$ — изначальное количество камней, а $r$ — остаток от деления $d$ на $... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English, Russian | proof only | null | |
0399 | Let $A_1A_2A_3A_4A_5$ be a convex pentagon such that $S_{A_1A_2A_3} = S_{A_2A_3A_4} = S_{A_3A_4A_5} = S_{A_4A_5A_1} = S_{A_5A_1A_2}$. Prove that there exists a point $M$ such that $S_{A_1MA_2} = S_{A_2MA_3} = S_{A_3MA_4} = S_{A_4MA_5} = S_{A_5MA_1}$. | [] | Bulgaria | First selection test for IMO 2007, Vietnam | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0b23 | Problem:
Let $ABC$ be an acute triangle with circumcircle $\Gamma$ and $D$ the foot of the altitude from $A$. Suppose that $AD = BC$. Point $M$ is the midpoint of $DC$, and the bisector of $\angle ADC$ meets $AC$ at $N$. Point $P$ lies on $\Gamma$ such that lines $BP$ and $AC$ are parallel. Lines $DN$ and $AM$ meet at... | [] | Philippines | $22^{ ext {nd }}$ Philippine Mathematical Olympiad National Stage, Written Phase | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous ... | null | proof only | null | |
08sg | A polynomial $P(x)$ with integer coefficients satisfies $P(n^2) = 0$ for some nonzero integer $n$. Show that $P(a^2) \neq 1$ must hold for any nonzero rational number $a$. | [
"The assertion of the problem is obvious if $P(x)$ is a constant. Therefore, we assume in the sequel that $P(x)$ is a polynomial of degree $m$, where $m$ is a positive integer. As $P(x)$ is a polynomial with integer coefficients and satisfies $P(n^2) = 0$, there exists, by the factorization theorem, a polynomial $Q... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0ihx | Problem:
A positive integer $n$ is called "flippant" if $n$ does not end in $0$ (when written in decimal notation) and, moreover, $n$ and the number obtained by reversing the digits of $n$ are both divisible by $7$. How many flippant integers are there between $10$ and $1000$? | [
"Solution:\nWe use the notation \"$|$\" to mean \"divides.\"\n\nThere is only one flippant 2-digit number, namely $77$. Indeed, if $10a + b$ is flippant (where $a, b$ are integers $1$-$9$), then $7 \\mid 10a + b$ and $7 \\mid 10b + a$. Thus,\n$$\n7 \\mid 3(10a + b) - (10b + a) = 29a - 7b = a + 7(4a - b)\n$$\nso tha... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 17 | |
09ya | Problem:
Zij $ABC$ een rechthoekige driehoek met $\angle C = 90^\circ$ en zij $D$ het voetpunt van de hoogtelijn uit $C$. Zij $E$ het zwaartepunt van driehoek $ACD$ en zij $F$ het zwaartepunt van driehoek $BCD$. Het punt $P$ voldoet aan $\angle CEP = 90^\circ$ en $|CP| = |AP|$, terwijl het punt $Q$ voldoet aan $\angle... | [
"Solution:\n\nNoem $M, N, R$ en $S$ de middens van respectievelijk zijden $BC, CA, BD$ en $AD$. Zij $Z$ het zwaartepunt van $\\triangle ABC$. Vierhoek $QFMC$ is een koordenvierhoek wegens rechte hoeken $\\angle QFC = 90^\\circ = \\angle QMC$. Merk op dat $CQ$ dus de middellijn van de omgeschreven cirkel is. Analoog... | Netherlands | IMO-selectietoets III | [
"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 | |
08ol | Problem:
Vukasin, Dimitrije, Dusan, Stefan and Filip asked their professor to guess three consecutive positive integer numbers after they had told him these (true) sentences:
Vukasin: "Sum of the digits of one of them is a prime number. Sum of the digits of some of the other two is an even perfect number ($n$ is perf... | [
"Solution:\n\nLet the middle number be $n$, so the numbers are $n-1$, $n$ and $n+1$. Since $4$ does not divide any of them, $n \\equiv 2 \\pmod{4}$. Furthermore, neither $3$, $5$ nor $7$ divides $n$. Also $n+1+11 \\equiv 2 \\pmod{4}$ cannot be a square. Then $3$ must divide $n-1$ or $n+1$. If $n-1+11$ is a square, ... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Num... | null | proof and answer | 2013, 2014, 2015 | |
0894 | Problem:
Marco, Fabrizio e Giovanni, tre matematici, sfidano un gruppo di quattro fisici a un torneo di calcio balilla (costituito da un certo numero di partite) in cui alla fine vince il gruppo che ha segnato il maggior numero di goal totali.
In ogni partita i fisici segnano 2 goal in più di quanti ne avevano segnati... | [
"Solution:\n\nLa risposta è (D). I gol segnati dai fisici nelle prime $n$ partite sono pari alla somma dei primi $n$ numeri dispari, cioè sono $n^{2}$. Poiché i matematici hanno vinto, essi hanno segnato più di $330/2 = 165$ goal, quindi i fisici hanno segnato meno di $165$ goal. Il più grande quadrato perfetto min... | Italy | UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA GARA di SECONDO LIVELLO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | D |
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