id stringlengths 4 4 | problem_markdown stringlengths 36 3.59k | solutions_markdown listlengths 0 10 | images images listlengths 0 15 | country stringclasses 58
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values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0ep7 | Human hair grows at a rate of about $1$ centimetre per month. This is equivalent to about how many millimetres every ten years?
(A) $12$
(B) $120$
(C) $1\,200$
(D) $12\,000$
(E) $120\,000$ | [
"**C** $1$ cm per month $= 10$ mm per month $= 120$ mm per year $= 1\\,200$ mm in ten years"
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | MCQ | C | |
0ghs | A positive integer is said to be **so-last-year** if it has three distinct positive divisors whose sum is equal to $2022$. Determine the smallest so-last-year number.
稱一個正整數為**去年老梗**如果這個數字有三個相異正因數的和為 $2022$。試求最小的去年老梗數。 | [
"Observe $1344$ is a solution as $6 + 672 + 1344 = 2022$. Claim that $1344$ is the smallest one. Towards contradiction, assume $N < 1344$ is also so-last-year, then there exist $a < b < c$, such that\n$$\n2022 = N \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) < 1344 \\left( \\frac{1}{a} + \\frac{1}{b}... | Taiwan | 2023 數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Chinese (Traditional) | proof and answer | 1344 | |
08it | Problem:
The real numbers $a_{1}, a_{2}, \ldots, a_{2003}$ satisfy simultaneously the relations: $a_{i} \geq 0$ for all $i = 1, 2, \ldots, 2003$; $a_{1} + a_{2} + \ldots + a_{2003} = 2$; $a_{1} a_{2} + a_{2} a_{3} + \ldots + a_{2003} a_{1} = 1$. Find the smallest value of the sum $a_{1}^{2} + a_{2}^{2} + \ldots + a_{2... | [] | JBMO | The third selection test for IMO 2003 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 3/2 | |
06wq | Prove that there are only finitely many quadruples $(a, b, c, n)$ of positive integers such that
$$
n! = a^{n-1} + b^{n-1} + c^{n-1}
$$ | [
"For fixed $n$ there are clearly finitely many solutions; we will show that there is no solution with $n > 100$. So, assume $n > 100$. By the AM-GM inequality,\n$$\n\\begin{aligned}\nn! & = 2 n(n-1)(n-2)(n-3) \\cdot (3 \\cdot 4 \\cdots (n-4)) \\\\\n& \\leqslant 2(n-1)^{4}\\left(\\frac{3+\\cdots+(n-4)}{n-6}\\right)^... | IMO | IMO 2021 Shortlisted Problems | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
00o0 | Let $ABC$ be an acute triangle, with $AC \neq BC$. Let $M$ be the midpoint of segment $AB$. Let $H$ be the orthocenter of triangle $ABC$, $D$ the footpoint of the altitude through $A$ on $BC$ and $E$ the footpoint of the altitude through $B$ on $AC$.
Prove that lines $AB$, $DE$ and the orthogonal to $MH$ through $C$ in... | [
"\n\nLet $\\angle ACB = \\gamma$ and $F$ be the foot of $C$ on $MH$. We will first demonstrate that $F$ lies on the circumcircle $k$ of triangle $ABC$.\nLet $H_1$ denote the symmetric point to $H$ with respect to $M$. The quadrilateral $AH_1BH$ is a parallelogram, and since we have $\\angle... | Austria | AUT_ABooklet_2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneo... | English | proof only | null | |
06f1 | Let $A$, $B$ and $C$ be real numbers such that
(i) $\sin A \cos B + |\cos A \sin B| = \sin A |\cos A| + |\sin B \cos B|$,
(ii) $\tan C$ and $\cot C$ are defined.
Find the minimum value of $(\tan C - \sin A)^2 + (\cot C - \cos B)^2$. | [
"The minimum value is $3 - 2\\sqrt{2}$.\nCondition (i) can be rewritten as\n$$\n(\\sin A - |\\sin B|)(\\cos B - |\\cos A|) = 0.\n$$\nIf $\\sin A = |\\sin B|$, then we have $\\sin^2 A = \\sin^2 B = 1 - \\cos^2 B$.\nIf $\\cos B = |\\cos A|$, then we have $\\cos^2 B = \\cos^2 A = 1 - \\sin^2 A$.\nTherefore, in any cas... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 3 - 2√2 | |
0dn0 | Problem:
Нека је $P$ тачка на дијагонали $BD$ паралелограма $ABCD$ таква да је $\varangle PCB = \varangle ACD$. Кружница описана око троугла $ABD$ сече праву $AC$ у тачкама $E$ и $A$. Доказати да је
$$
\varangle AED = \varangle PEB. \quad \text{(Марко Ђикић)}
$$ | [
"Solution:\n\nДоказ изводимо у случају када је $\\angle BAC \\leq 90^\\circ$. Други случај је аналоган.\nНека се праве $DE$ и $BC$ секу у $L$. Четвороугао $CDPL$ је тетиван јер је $\\angle PDL = \\angle PCL$, одакле имамо $\\angle PLE = \\angle PCD = \\angle BCA = \\angle DAC = \\angle DBE = \\angle PBE$, па је и ч... | Serbia | Serbian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
037r | Problem:
Ivan and Peter play the following game. Ivan chooses a secret number from the set $A=\{1,2, \ldots, 90\}$. Then Peter chooses a subset $B$ of $A$ and Ivan tells Peter whether his number is in the set $B$ or not. If the answer is "yes" then Peter pays Ivan 2 leva, and if the answer is "no" then he pays Ivan 1 ... | [
"Solution:\n\nWe shall solve the problem for $A=\\{1,2, \\ldots, t\\}$. Let $F_{0}=F_{1}=1$, $F_{n+1}=F_{n}+F_{n-1}$ for $n \\geq 1$ be the Fibonacci sequence. We shall prove by induction that if $F_{n-1}<t \\leq F_{n}, n \\geq 2$, then the desired sum equals $n$.\n\nSince for $t=2$ and $t=3$ Peter needs 2 or 3 lev... | Bulgaria | Team selection test for 23. BMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 11 | |
0fgb | Problem:
Un segmento $d$ divide al segmento $s$ si existe un natural $n$ tal que
$$
n d = d + d + \stackrel{n}{n}^{*} + d = s
$$
a) Demostrar que si el segmento $d$ divide a los segmentos $s$ y $s'$ con $s < s'$, entonces divide al segmento diferencia $s' - s$.
b) Demostrar que ningún segmento divide al lado $s$ y a l... | [
"Solution:\n\na) Si tenemos $s = n d$ y $s' = n' d$, con $n, n'$ números naturales, entonces evidentemente, $s' - s = (n' - n) d$.\n\nb) Llamamos $d$ a la diagonal $AC$ del pentágono y $s$ al lado. El paralelismo entre diagonales y lados opuestos produce cinco rombos interiores al pentágono, formados, cada uno, por... | Spain | OME 22 | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0aeo | Човек и пол, за два и пол дена јаде три и пол леба. Колку леба ќе изедат 100 луѓе за 45 дена? | [
"Човек и пол, за два и пол дена јаде три и пол леба. Значи тројца луѓе за 5 дена јадат $2 \\cdot 2 \\cdot 3,5 = 14$ леба. Еден човек за 1 ден јаде $\\frac{1}{3} \\cdot \\frac{1}{5} \\cdot 14 = \\frac{14}{15}$ леба. Еден човек за 45 дена јаде $45 \\cdot \\frac{14}{15} = 42$ леба, а 100 луѓе за 45 дена јадат $42 \\cd... | North Macedonia | Републички натпревар по математика за основно образование | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | Macedonian, English | final answer only | 4200 | |
0e7v | Problem:
Naj bo $D$ razpolovišče stranice $AB$, $E$ presečišče stranice $BC$ in simetrale kota $\angle BAC$, $F$ pa pravokotna projekcija točke $E$ na stranico $AB$ trikotnika $ABC$. Denimo, da je $\angle CDA = \angle ACB$ in $|CE| = |BF|$. Določi velikost kotov trikotnika $ABC$. | [
"Solution:\n\nKer je $\\angle CDA = \\angle ACB$, sta si trikotnika $ADC$ in $ACB$ podobna, zato je\n$$\n\\frac{|AC|}{|AB|} = \\frac{|AD|}{|AC|} = \\frac{|AB|}{2|AC|}\n$$\nod koder sledi $|AB| = |AC| \\sqrt{2}$. Ker je $AE$ simetrala kota $\\angle BAC$, je\n$$\n\\frac{|BE|}{|CE|} = \\frac{|AB|}{|AC|} = \\sqrt{2}\n$... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | ∠ACB = π/2, ∠BAC = π/4, ∠CBA = π/4 | |
0iwc | Problem:
Let $f$ be a polynomial with integer coefficients such that the greatest common divisor of all its coefficients is $1$. For any $n \in \mathbb{N}$, $f(n)$ is a multiple of $85$. Find the smallest possible degree of $f$. | [
"Solution:\n\nNotice that, if $p$ is a prime and $g$ is a polynomial with integer coefficients such that $g(n) \\equiv 0 \\pmod{p}$ for some $n$, then $g(n + m p)$ is divisible by $p$ as well for any integer multiple $m p$ of $p$. Therefore, it suffices to find the smallest possible degree of a polynomial $f$ for w... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 17 | |
0e0i | Problem:
Poišči vsa realna števila $x, y$ in $z$, ki rešijo sistem enačb
$$
\begin{aligned}
x + y + 2z &= 0 \\
x y - z^{2} &= 0 \\
y^{2} + 5z + 6 &= 0
\end{aligned}
$$ | [
"Solution:\n\nIz prve enačbe izrazimo $x = -2z - y$ in vstavimo v drugo. Dobimo $(-2z - y)y - z^{2} = 0$ oziroma $-(z + y)^{2} = 0$. Od tod sledi $z = -y$. Vstavimo v tretjo enačbo. Dobimo $y^{2} - 5y + 6 = 0$ oziroma $(y - 2)(y - 3) = 0$. Torej je $y = 2$ ali $y = 3$.\n\nSistem enačb rešijo $x = 2$, $y = 2$, $z = ... | Slovenia | Slovenian Secondary School Mathematical Competition | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (2, 2, -2) and (3, 3, -3) | |
0d68 | Given that the polynomial $P(x) = x^{5} - x^{2} + 1$ has $5$ roots $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$. Find the value of the product
$$
Q(r_{1}) Q(r_{2}) Q(r_{3}) Q(r_{4}) Q(r_{5}),
$$
where $Q(x) = x^{2} + 1$. | [
"Since $r_{1}, \\ldots, r_{5}$ are the roots of $P(x) = x^{5} - x^{2} + 1$, by factorization theorem we have\n$$\nP(x) = \\prod_{i=1}^{5} (x - r_{i}) .\n$$\nIt follows that\n$$\n\\prod_{j=1}^{5} Q(r_{j}) = \\prod_{j=1}^{5} (r_{j}^{2} + 1) = \\prod_{j=1}^{5} (r_{j} + i) \\prod_{j=1}^{5} (r_{j} - i) = P(i) P(-i),\n$$... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | 5 | |
05q3 | Problem:
Soit $n$ un entier positif. Montrer que dans un ensemble $A$ de $2^{n}$ nombres strictement positifs, on peut choisir un sous-ensemble $B$ de taille $n+1$ tel que la somme de deux nombres différents dans $B$ ne soit jamais dans $A$. | [
"Solution:\n\nSoient $n$ et $A$ comme dans l'énoncé. Nous allons montrer, par récurrence sur $m \\leqslant n+1$, que l'algorithme glouton consistant à toujours prendre le plus grand nombre qui ne pose pas de problème convient.\n\nPosons $A_{0}=A$, et $B_{0}=B$. Supposons que, pour $m<n+1$, on a construit $A_{m}$ et... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0iqu | Problem:
John M. is sitting at $(0, 0)$, looking across the aisle at his friends sitting at $(i, j)$ for each $1 \leq i \leq 10$ and $0 \leq j \leq 5$. Unfortunately, John can only see a friend if the line connecting them doesn't pass through any other friend. How many friends can John see? | [
"Solution:\n\n36\n\nThe simplest method is to draw a picture and count which friends he can see. John can see the friend on point $(i, j)$ if and only if $i$ and $j$ are relatively prime."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 36 | |
07gn | Let $a, b > 1$ be positive integers. Prove that there are infinitely many positive integers $n$ such that the following equation has no solution $(k, t)$ in positive integers.
$$
\varphi(a^n - 1) = b^k - b^t.
$$ | [
"We first prove the following two lemmas.\n\n**Lemma.** Let $p$ be a prime number and $b > 1$ be an integer such that $\\gcd(p, b) = 1$. Then, for all positive integers $n$;\n$$\n\\nu_p(b^n - 1) \\leq \\nu_p(n) + C,\n$$\nfor some constant positive integer $C$.\n*Proof.* Let $d$ be the order of $b$ modulo $p$ then i... | Iran | 38th Iranian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, ineq... | null | proof only | null | |
0axp | Problem:
Let $f(x) = \sqrt{4 \sin^4 x - \sin^2 x \cos^2 x + 4 \cos^4 x}$ for any $x \in \mathbb{R}$. Let $M$ and $m$ be the maximum and minimum values of $f$, respectively. Find the product of $M$ and $m$. | [
"Solution:\nLet us simplify the expression inside the square root:\n\nLet $s = \\sin^2 x$, $c = \\cos^2 x$. Then $s + c = 1$.\n\n$4 \\sin^4 x + 4 \\cos^4 x = 4(s^2 + c^2)$\n\nBut $s^2 + c^2 = (s + c)^2 - 2sc = 1 - 2sc$\n\nSo $4(s^2 + c^2) = 4(1 - 2sc) = 4 - 8sc$\n\nAlso, $-\\sin^2 x \\cos^2 x = -sc$\n\nSo the expre... | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | null | proof and answer | sqrt(7) | |
0fny | En el cuadrilátero convexo $ABCD$, se tiene $\angle ABC = \angle CDA = 90^\circ$. La perpendicular a $BD$ desde $A$ corta a $BD$ en el punto $H$. Los puntos $S$ y $T$ están en los lados $AB$ y $AD$, respectivamente, y son tales que $H$ está dentro del triángulo $SCT$ y
$$
\angle CHS - \angle CSB = 90^\circ, \quad \angl... | [
"Claramente, $B$, $D$ están en la circunferencia de diámetro $AC$. Luego la altura $AH$ del triángulo $ABD$ mide $AH = \\frac{AB \\cdot AD}{AC}$.\n\nConsideremos la circunferencia circunscrita a $CSH$, con centro en $O_C$. Por ángulo central, tenemos que\n$$\n\\angle COCS = 2(180^\\circ - \\angle CHS) = 180^\\circ ... | Spain | LV Olimpiada Internacional de Matemáticas | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quad... | Spanish | proof only | null | |
0j2v | Problem:
When flipped, a coin has a probability $p$ of landing heads. When flipped twice, it is twice as likely to land on the same side both times as it is to land on each side once. What is the larger possible value of $p$? | [
"Solution:\n\nThe probability that the coin will land on the same side twice is $p^{2} + (1-p)^{2} = 2p^{2} - 2p + 1$.\n\nThe probability that the coin will land on each side once is $p(1-p) + (1-p)p = 2p(1-p) = 2p - 2p^{2}$.\n\nWe are told that it is twice as likely to land on the same side both times, so\n$$\n2p^... | United States | Harvard-MIT November Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | (3 + sqrt(3)) / 6 | |
0431 | Suppose $a, b > 0$. The equation $\sqrt{|x|} + \sqrt{|x+a|} = b$ for $x$ has exactly three different real solutions, namely $x_1, x_2, x_3$, and $x_1 < x_2 < x_3 = b$. Then the value of $a+b$ is ______. | [
"Let $t = x + \\frac{a}{2}$. Then the equation $\\sqrt{|t - \\frac{a}{2}|} + \\sqrt{|t + \\frac{a}{2}|} = b$ for $t$ has exactly three different real solutions $t_i = x_i + \\frac{a}{2}$ ($i = 1, 2, 3$).\n\nSince $f(t) = \\sqrt{|t - \\frac{a}{2}|} + \\sqrt{|t + \\frac{a}{2}|}$ is an even function, the three real so... | China | China Mathematical Competition | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 144 | |
0f43 | Problem:
A convex polygon is drawn inside the unit circle. Someone makes a copy by starting with one vertex and then drawing each side successively. He copies the angle between each side and the previous side accurately, but makes an error in the length of each side of up to a factor $1 \pm \mathbf{p}$. As a result the... | [] | Soviet Union | 15th ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0d7z | On a checkered square $10 \times 10$ the cells of the upper left $5 \times 5$ square are black and all the other cells are white. What is the maximal $n$ such that the original square can be dissected (along the borders of the cells) into $n$ polygons such that in each of them the number of black cells is three times l... | [
"The answer is $9$. We can see that there are only $9$ cells on the border of the black square that connect to the white area. Each of them belongs to at most $1$ polygon, so there are at most $9$ polygons.\n\nAn example as follows (each of cells belongs to one part that has the ratio of black:white is $1:3$ )"
] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 9 | |
01aq | $$
a^b b^c c^a = p.
$$
Find all triples $(a, b, c)$ of integers (not necessarily positive) such that the equation holds, where $p$ is a prime number. | [
"**Answer:** if $p > 2$ then $(p, 1, 1)$ and $(-p, 1, -1)$ together with cyclic permutations; if $p = 2$ then $(2, 1, 1)$, $(2, 1, -1)$, $(2, 2, -1)$ and $(-2, 2, -1)$ together with cyclic permutations.\n\nSolution:\n\nSuppose $a, b, c$ satisfy the equation. As $p$ is positive, this implies that $|a|^b |b|^c |c|^a ... | Baltic Way | Baltic Way 2013 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis... | null | proof and answer | If p > 2: all cyclic permutations of (p, 1, 1) and (−p, 1, −1). If p = 2: all cyclic permutations of (2, 1, 1), (2, 1, −1), (2, 2, −1), and (−2, 2, −1). | |
0h0u | Find maximal natural number, with all distinct digits such that the difference between any two consecutive digits is at least $2$. | [
"It is clear that our number has to have $10$ digits.\nWe start with $975$ (it is clear that $8$ cannot be neither second nor third digit).\n$975864$ - are first six digits. We can't have $3$ after, so we have $2$, and finally the answer is: $9758642031$."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Algorithms"
] | English | proof and answer | 9758642031 | |
05sq | Problem:
Une grille de taille $n \times n$ contient $n^2$ cases. Chaque case contient un entier naturel compris entre $1$ et $n$, de telle sorte que chaque entier de l'ensemble $\{1, \ldots, n\}$ apparaît exactement $n$ fois dans la grille. Montrer qu'il existe une colonne ou une ligne de la grille contenant au moins ... | [
"Solution:\n\nEssayons d'abord de regarder ce qui se passe dans ce qui semble être le pire cas : si chaque ligne/colonne ne contient pas beaucoup de numéros, chaque numéro va y être beaucoup présent. En particulier, pour tout $i$, le nombre de $i$ semble être écrit dans peu de lignes et de colonnes. Mais à priori s... | France | Envoi 5: Pot Pourri | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0a2a | How many zeros does the number
$$
2^{35} \times 3^{52} \times 5^{23}
$$
end with? In this problem, $a^{b^c}$ is the number you get by first calculating
the power
$$
b^c = \underbrace{b \times b \times \dots \times b}_{c \text{ times } a \ b}
$$
and then raising *a* to this power.
A) 6 B) 8 C) 25 D) 243 E) 256 | [] | Netherlands | Junior Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | B | |
09k1 | Let $p$, $q$, $r$ be prime numbers such that $p < q < r$ and at least two of $p+n$, $q+n$, $r+n$ are relatively prime for any positive integer $n$. Find all possible values of the pair $(p, q)$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | (2, 3) | |
05z1 | Problem:
On dit qu'un entier $k > 1$ est superbe s'il existe $m, n, a$ trois entiers strictement positifs tels que
$$
5^{m} + 63 n + 49 = a^{k}
$$
Déterminer le plus petit entier superbe. | [
"Solution:\n\nSupposons que $k = 2$ est superbe : il existe trois entiers strictement positifs $m, n, a$ tels que $5^{m} + 63 n + 49 = a^{2}$. En regardant modulo $3$, $5^{m} + 1 \\equiv 2^{m} + 1 \\equiv a^{2} \\pmod{3}$. Or les puissances de $2$ modulo $3$ valent alternativement $2$ si $m$ est impair, puis $1$ si... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | 5 | |
08pq | Problem:
Let $H$ be the orthocentre of an acute triangle $A B C$ with $B C > A C$, inscribed in a circle $\Gamma$. The circle with centre $C$ and radius $C B$ intersects $\Gamma$ at the point $D$, which is on the arc $A B$ not containing $C$. The circle with centre $C$ and radius $C A$ intersects the segment $C D$ at t... | [
"Solution:\nWe use standard notation for the angles of triangle $A B C$. Let $P$ be the midpoint of $C H$ and $O$ the centre of $\\Gamma$. As\n$$\n\\alpha = \\angle B A C = \\angle B D C = \\angle D K L\n$$\nthe quadrilateral $A C K L$ is cyclic. From the relation $C B = C D$ we get $\\angle B C D = 180^\\circ - 2 ... | JBMO | Junior Balkan Mathematical Olympiad | [
"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 | |
019i | For a positive integer $n$ assume that $n$ numbers have been chosen from the table
$$
\begin{array}{cccc}
0 & 1 & \cdots & n-1 \\
n & n+1 & \cdots & 2n-1 \\
\vdots & \vdots & \ddots & \vdots \\
(n-1)n & (n-1)n+1 & \cdots & n^2-1
\end{array}
$$
with no two of them from the same row or the same column. What is the maxima... | [
"The product is\n$$\n\\prod_{i=1}^{n} (a_i + b_{p(i)}),\n$$\nfor some permutation $p$ of $1, \\dots, n$, where $a_i = (i-1)n$ and $b_i = i-1$. Assume that $i + p(i) \\neq n+1$ for some $i$ and let the least such $i$ be chosen. Since $j + p(j) = n+1$ for $j < i$ then $k = p(i) < n+1-i$ and $l = p^{-1}(n+1-i) > i$. R... | Baltic Way | Baltic Way 2013 | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | (n-1)^n n! | |
07ku | Let $p(x)$ be a polynomial with rational coefficients. Prove that there exists a positive integer $n$ such that the polynomial $q(x)$ defined by
$$
q(x) = p(x + n) - p(x)
$$
has integer coefficients. | [
"Each term in $p(x)$ is of the form $a_i x^i$, where $a_i$ is rational. Expanding the expression $a_i(x+n)^i - a_i x^i$, we see that $n$ is a factor in all terms. Thus it suffices to pick $n$ to equal the least common multiple of the denominators of the coefficients $a_i$."
] | Ireland | Irska | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof only | null | |
01hw | There are $2021$ points on a plane, no three of which are collinear. For every $5$ points there exists at least $4$ among them which are concyclic. Is it necessarily true that at least $2020$ of the points are concyclic? | [
"Answer: Yes.\n\nLet us first prove a lemma that if $4$ points $A$, $B$, $C$, $D$ all lie on circle $\\Gamma$ and some two points $X$, $Y$ do not lie on $\\Gamma$, then these $6$ points are pairs of intersections of three circles, circle $\\Gamma$ and two other circles. Indeed, according to the problem statement th... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | Yes | |
00xl | Problem:
Let $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1}, b_{2}, \ldots, b_{n}$ be two finite sequences consisting of $2n$ different real numbers. Rearranging each of the sequences in the increasing order we obtain $a_{1}^{\prime}, a_{2}^{\prime}, \ldots, a_{n}^{\prime}$ and $b_{1}^{\prime}, b_{2}^{\prime}, \ldots, b_{n}^... | [
"Solution:\nLet $m$ be such index that $\\left| a_{m}^{\\prime} - b_{m}^{\\prime} \\right| = \\max_{1 \\leq i \\leq n} \\left| a_{i}^{\\prime} - b_{i}^{\\prime} \\right| = c$. Without loss of generality we may assume $a_{m}^{\\prime} > b_{m}^{\\prime}$. Consider the numbers $a_{m}^{\\prime}, a_{m+1}^{\\prime}, \\ld... | Baltic Way | Baltic Way 1993 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof only | null | |
0kg7 | What is the value of $1234 + 2341 + 3412 + 4123$?
(A) 10,000 (B) 10,010 (C) 10,110 (D) 11,000 (E) 11,110 | [
"**Answer (E):** When the four numbers are added, the digits $1$, $2$, $3$, and $4$ appear exactly once in each column of the addition problem, as shown in the figure.\n$$\n\\begin{array}{@{}r@{}l}\n & 1234 \\\\\n & 2341 \\\\\n & 3412 \\\\\n + & 4123 \\\\\n \\hline\n\\end{array}\n$$\nThe digits sum to $10$, produci... | United States | Fall 2021 AMC 10 B | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | E | |
0c7j | Integers from $1$ to $49$ are placed arbitrarily on a quadratic table formed by $7 \times 7$ squares. Show that one can find a $2 \times 2$ square on the table, formed by $4$ neighboring cells, such that the sum of numbers inside is at least $81$. | [] | Romania | 70th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
071r | Problem:
If $a, b, c$ are positive real numbers such that $a b c = 1$, prove that
$$
a^{b+c} b^{c+a} c^{a+b} \leq 1
$$ | [
"Solution:\nNote that the inequality is symmetric in $a, b, c$ so that we may assume that $a \\geq b \\geq c$. Since $a b c = 1$, it follows that $a \\geq 1$ and $c \\leq 1$. Using $b = 1 / (a c)$, we get\n$$\na^{b+c} b^{c+a} c^{a+b} = \\frac{a^{b+c} c^{a+b}}{a^{c+a} c^{c+a}} = \\frac{c^{b-c}}{a^{a-b}} \\leq 1\n$$\... | India | INMO | [
"Algebra > Intermediate Algebra > Other"
] | null | proof only | null | |
00zl | Problem:
A sequence of integers $a_{1}, a_{2}, \ldots$, is such that $a_{1}=1$, $a_{2}=2$ and for $n \geq 1$
$$
a_{n+2}= \begin{cases}5 a_{n+1}-3 a_{n} & \text{ if } a_{n} \cdot a_{n+1} \text{ is even, } \\ a_{n+1}-a_{n} & \text{ if } a_{n} \cdot a_{n+1} \text{ is odd. }\end{cases}
$$
Prove that $a_{n} \neq 0$ for all ... | [
"Solution:\nConsidering the sequence modulo $6$ we obtain $1, 2, 1, 5, 4, 5, 1, 2, \\ldots$ The conclusion follows."
] | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0bmq | In a math contest take part 50 students and 3 problems are submitted to the contestants. It is known that each student solved at least a problem and the total number of correct solutions is 100. Prove that at most 25 students solved all the three problems. | [
"Denote $a$, $b$, $c$ the number of the students which solved exactly one, two, respectively three problems. Then $a + b + c = 50$ and $a + 2b + 3c = 100$. It follows $b + 2c = 50$, whence $2c \\le 50$, yielding $c \\le 25$."
] | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0gtl | Find all prime numbers $p$ for which the number
$$
3^p + 4^p + 5^p + 9^p - 98
$$
has at most 6 positive divisors. | [
"Answer: $p = 2, 3, 5$.\nLet $f(p) = 3^p + 4^p + 5^p + 9^p - 98$. The primes $2, 3, 5$ satisfy the following conditions: $f(2) = 3 \\cdot 11$, $f(3) = 7 \\cdot 11^2$, $f(5) = 7 \\cdot 9049$.\n\nLet $p > 5$. Since $7$ divides $3^p + 4^p$, $5^p + 9^p$ and $98$ we get $7 \\mid f(p)$. Now since\n$$\np \\equiv 1 \\pmod{... | Turkey | Team Selection Test | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 2, 3, 5 | |
0b2c | Problem:
Suppose $A=\{1,2, \ldots, 20\}$. Call $B$ a visionary set of $A$ if $B \subseteq A$, $B$ contains at least one even integer, and $|B| \in B$, where $|B|$ is the cardinality of set $B$. How many visionary sets does $A$ have? | [
"Solution:\n\nIgnoring the constraint that all visionary sets must contain at least one even integer, we count a total of\n$$\n\\sum_{i=0}^{19}\\binom{19}{i}=2^{19}\n$$\nsets. From this total, we subtract the number of sets that only contain odd numbers. Evidently, $|B|$ has to be odd. Since there are only $10$ odd... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 524032 | |
0j7m | Problem:
Determine the remainder when
$$
2^{\frac{1 \cdot 2}{2}} + 2^{\frac{2 \cdot 3}{2}} + \cdots + 2^{\frac{2011 \cdot 2012}{2}}
$$
is divided by $7$. | [
"Solution:\nWe have that $2^{3} \\equiv 1 \\pmod{7}$. Hence, it suffices to consider the exponents modulo $3$. We note that the exponents are the triangular numbers and upon division by $3$ give the pattern of remainders $1, 0, 0, 1, 0, 0, \\ldots$, so what we want is\n$$\n\\begin{aligned}\n2^{\\frac{1 \\cdot 2}{2}... | United States | Harvard-MIT November Tournament | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | 1 | |
0368 | Problem:
Find all values of the real parameter $p$ such that the equation $|x^{2}-p x-2 p+1|=p-1$ has four real roots $x_{1}, x_{2}, x_{3}$ and $x_{4}$ such that
$$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=20
$$ | [
"Solution:\nAnswer: $p=2$. The condition $p>1$ is necessary (but not sufficient!) for existence of four roots. We consider two cases:\n\nCase 1. If $x^{2}-p x-2 p+1=p-1 \\Longleftrightarrow x^{2}-p x-3 p+2=0$ then by the Vieta theorem we obtain $x_{1}^{2}+x_{2}^{2}=p^{2}-2(2-3 p)=p^{2}+6 p-4$.\n\nCase 2. If $x^{2}-... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 2 | |
0d4m | Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be an injective function such that $f(1)=2$, $f(2)=4$ and
$$
f(f(m)+f(n))=f(f(m))+f(n)
$$
for all $m, n \in \mathbb{N}$. Prove that $f(n)=n+2$ for all $n \geq 2$. | [
"Taking $n=1$, we obtain $f(f(m)+2)=f(f(m))+2$, for all $m \\in \\mathbb{N}$. Taking $m=1$, we obtain $f(2+f(n))=4+f(n)$, for all $n \\in \\mathbb{N}$. Therefore\n$$\nf(f(n))=f(n)+2, \\quad \\text{for all } n \\in \\mathbb{N} .\n$$\nBecause $f(2)=f(f(1))=f(1)+2=4$, we prove by induction that\n$$\nf(n)=n+2\n$$\nfor ... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English, Arabic | proof and answer | f(n) = n + 2 for all n ≥ 2 | |
0e0f | Problem:
Dokaži enakost
$$
1005^{\ln 121}=11^{\ln (1+3+5+\ldots+2009)}
$$ | [
"Solution:\n\nKer je\n$$\n\\begin{aligned}\n& 1+3+5+\\ldots+2009= \\\\\n& \\quad=(1+2009)+(3+2007)+\\ldots+(1003+1007)+1005 \\\\\n& \\quad=502 \\cdot 2010+1005=1005^{2}\n\\end{aligned}\n$$\nje dovolj pokazati $1005^{\\ln (121)}=11^{\\ln \\left(1005^{2}\\right)}$. Če dobljeno enakost logaritmiramo, dobimo\n$$\n\\ln ... | Slovenia | Slovenian Secondary School Mathematical Competition | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
09d1 | Самбарт $\overline{abcdefxyz}$ гэсэн 9 оронтой тоо бичигдсэн байв. Бат энэ тоог $abc, def, xyz$ гэж гурав, гурван орцоор нь салган хооронд нь дурын байдлаар эвлүүлэн 9 оронтой тоонуудыг үүсгэж байв. /Жишээ нь: $\overline{abcdefabc}$ тоонуудыг үүсгэж болно/. Харин Цэдэг эдгээр тоонуудыг харж байснаа "Энэ бүх тоонуудыг 2... | [
"$$\n\\begin{array}{l}\n\\blacktriangle \\ abcdefxyz \\text{ тоог } \\overline{abc} \\cdot 10^6 + \\overline{def} \\cdot 10^3 + xyz = \\overline{abc} \\cdot 999999 + \\overline{def} \\cdot 999 + (\\overline{abc} + \\overline{def} + \\overline{xyz}) \\text{ болно.} \\\\\n\\qquad 999999 \\equiv 0 \\pmod{27},\\ 999 \\... | Mongolia | ММО-48 | [
"Number Theory > Modular Arithmetic"
] | Mongolian | proof and answer | Yes | |
07ll | Let $ABCD$ be a quadrilateral inscribed in a circle and let $M$ be a point on the circle. Consider the projections of the point $M$ on two opposite sides of the quadrilateral, and on its diagonals. Show that there exists a circle passing through these four points if and only if the quadrilateral is a trapezoid. | [
"Let $F, G, H, I$ and $J$ be the projections of $M$ on the diagonal $AC$, the side $BC$, the side $CD$, the side $AD$ and the diagonal $BD$ respectively. The points $G, H, J$ are collinear, because they are on Simons's line for $\\triangle BCD$. The points $F, H, I$ are on Simons's line for $\\triangle ACD$.\n\nBec... | Ireland | Irska | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03m3 | Three speed skaters have a friendly “race” on a skating oval. They all start from the same point and skate in the same direction, but with different speeds that they maintain throughout the race. The slowest skater does $1$ lap a minute, the fastest one does $3.14$ laps a minute, and the middle one does $L$ laps a minu... | [
"Assume that the length of the oval is one unit. Let $x(t)$ be the difference of distances that the slowest and the fastest skaters have skated by time $t$. Similarly, let $y(t)$ be the difference between the middle skater and the slowest skater. The path $(x(t), y(t))$ is a straight ray $R$ in $\\mathbb{R}^2$, sta... | Canada | Kanada 2010 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 16 | |
0fb4 | Problem:
Hallad el número mínimo de apuestas de quiniela que debemos rellenar para asegurar que obtenemos, al menos, 5 aciertos en una de ellas. (Una apuesta de quiniela consiste en un pronóstico de resultado para 14 partidos, en cada partido hay 3 posibles resultados). | [
"Solution:\n\nHay que rellenar 3 apuestas:\nEn 14 partidos, hay un resultado (1, $\\mathrm{X}$ o 2) que se repite al menos 5 veces (en caso contrario, el número de partidos sería menor o igual que $4 \\cdot 3 = 12$, pero $14 > 12$). Hacemos las tres apuestas que siguen: todo 1, todo X, todo 2. En una de ellas tenem... | Spain | null | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 3 | |
0c69 | Given an integer $n \ge 2$, colour red exactly $n$ cells of an infinite sheet of grid paper. A rectangular grid array is called *special* if it contains at least two red opposite corner cells; single red cells and 1-row or 1-column grid arrays whose end-cells are both red are special. Given a configuration of exactly $... | [
"The required minimum is $1 + \\lceil (n+1)/5 \\rceil$ and is achieved by the configuration described in the second block of the proof.\n\nCounting multiplicities, the cells $a$ and $c$ are both covered by three of these special rectangular grid arrays, the cells $b$ and $d$ are both covered by two, and all other r... | Romania | SELECTION TESTS FOR THE 2019 BMO AND IMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 1 + ceil((n+1)/5) | |
03ca | In a triangle $\triangle ABC$ points $L$, $P$ and $Q$ lie on the segments $AB$, $AC$ and $BC$, respectively, and are such that $PCQL$ is a parallelogram. The circle with center the midpoint $M$ of the segment $AB$ and radius $CM$ and the circle of diameter $CL$ intersect for the second time at point $T$. Prove that the... | [
"Since $AC \\parallel LQ$ and $BC \\parallel LP$, we have $S_{ALQ} = S_{CLQ} = S_{PLC} = S_{PLB}$. Let the point $K$ be such that $AKBC$ is a parallelogram. By analogy we have $S_{AKQ} = S_{AKC} = S_{CKB} = S_{PKB}$.\n\nThe locus of the points $X$ such that $\\triangle AXQ$ and $\\triangle PXB$ are oriented in one ... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0bcq | Problem:
Fie $ABC$ un triunghi ascuţitunghic înscris în cercul $\Gamma$ şi având ortocentrul $H$. Fie $K$ un punct pe cercul $\Gamma$, de cealaltă parte a lui $BC$ decât $A$. Fie $L$ simetricul lui $K$ faţă de dreapta $AB$, şi fie $M$ simetricul lui $K$ faţă de dreapta $BC$. Fie $E$ al doilea punct de intersecţie al c... | [] | Romania | Olimpiada europeana de matematica a fetelor | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09gn | The point $M$ is chosen inside of the convex quadrilateral $ABCD$ such that $\angle BMC = \angle AMD$. The triangles $ABE$ and $CDF$ are erected outwardly on the sides $AB$, $CD$ of the quadrilateral $ABCD$ such that
$$
\angle BAE = \angle DAM, \quad \angle ABE = \angle CBM, \quad \angle CDF = \angle ADM, \quad \angle ... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coord... | English | proof only | null | |
06cb | Recall that $n$ is perfect if the sum of the divisors of $n$ is $2n$. Suppose now $n$ is an odd perfect number. Show that $n$ has at least 3 distinct prime factors. | [
"If $n = p^k$ for some odd prime $p$ and positive integer $k$, then we have\n$$\n\\frac{\\sigma(n)}{n} = \\frac{1 + p + \\cdots + p^k}{p^k} = 1 + \\frac{1}{p} + \\cdots + \\frac{1}{p^k} < \\frac{1}{1 - \\frac{1}{p}} = \\frac{p}{p-1} < 2.\n$$\nIf $n = p^k q^\\ell$ for some odd primes $p, q$ and positive integers $k,... | Hong Kong | HKG TST | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0969 | Problem:
Determinați toate funcțiile continue $f: \mathbb{R}_{+}^{*} \rightarrow \mathbb{R}$, care verifică relația
$$
f\left(\frac{x}{y}\right)=\frac{f(x)}{y}-\frac{f(y)}{x}, \quad \forall x, y \in \mathbb{R}_{+}^{*}
$$ | [
"Solution:\n1) $f(1)=f\\left(\\frac{x}{x}\\right)=\\frac{f(x)}{x}-\\frac{f(x)}{x}=0$.\n\n2) $f\\left(\\frac{1}{y}\\right)=\\frac{f(1)}{y}-\\frac{f(y)}{1}=\\frac{0}{y}-f(y)=-f(y)$.\n\n3) $f(x \\cdot y)=f\\left(\\frac{x}{1 / y}\\right)=\\frac{f(x)}{1 / y}-\\frac{f(1 / y)}{x}=y \\cdot f(x)+\\frac{f(y)}{x}$.\n\n4)\n$$\... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = C (x - 1/x) for any real constant C | |
0lak | Solve the following system of equations:
$$
\begin{cases} x^4 - y^4 = 240 \\ x^3 - 2y^3 = 3(x^2 - 4y^2) - 4(x - 8y). \end{cases}
$$ | [
"Reformulate the given system in the following form:\n$$\n\\begin{cases} x^4 + 16 = y^4 + 256 \\\\ x^3 - 3x^2 + 4x = 2y^3 - 12y^2 + 32y \\end{cases} \\quad (1) \\quad (2)\n$$\nMultiplying both sides of equation (2) with $-8$, then adding it to (1), side by side, we obtain equation:\n$$(x-2)^4 = (y-4)^4 \\quad (3).$... | Vietnam | Vietnamese Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | (x, y) = (-4, -2) or (4, 2) | |
0hfk | You are given a triangle $ABC$, $\omega$ is its circumscribed circle, $I$ is its incenter. Let $K$ be any point on the arc $AC$ of $\omega$, not containing point $B$. Point $P$ is symmetrical to the point $I$ with respect to the point $K$. Point $T$ on the arc $AC$ of the circle $\omega$, which contains point $B$, is s... | [] | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry >... | English | proof only | null | |
0i36 | Let $a, b, c$ be positive real numbers such that
$$
a + b + c \ge abc.
$$
Prove that at least two of the inequalities
$$
\frac{2}{a} + \frac{3}{b} + \frac{6}{c} \ge 6, \quad \frac{2}{c} + \frac{3}{a} + \frac{6}{b} \ge 6,
$$
are true. | [
"**First Solution.** Assume, for the sake of contradiction, that at least two of the numbers\n$$\n\\frac{2}{a} + \\frac{3}{b} + \\frac{6}{c}, \\quad \\frac{2}{b} + \\frac{3}{c} + \\frac{6}{a}, \\quad \\frac{2}{c} + \\frac{3}{a} + \\frac{6}{b}\n$$\nare less than $6$. Without loss of generality, we may further assume... | United States | USA IMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0e6q | At most how many interior angles of an $n$-gon can be greater than $180^\circ$?
(A) $n-1$ (B) $n-2$ (C) $n-3$ (D) $n-4$ (E) $n-5$ | [
"The answer is $n - 3$. It can be seen from the picture that only $n - 3$ interior angles of an $n$-gon can be greater than $180^\\circ$. The sum of all the interior angles of an $n$-gon is equal to $(n - 2) \\cdot 180^\\circ$. If at least $n - 2$ interior angles were greater than $180^\\circ$, the sum of all the i... | Slovenia | National Math Olympiad 2012 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | C | |
0gaq | 定義 $f_k(n)$ 為正整數 $n$ 的所有正因數之 $k$ 次方總和, 也就是
$$
f_k(n) := \sum_{m|n, m>0} m^k.
$$
試找出所有正整數對 $(a, b)$, 使得 $f_a(n) \mid f_b(n)$ 對於所有正整數 $n$ 均成立。 | [
"$a = b$ 為所有可能。\n\n首先帶入 $n = 2$, 必須有 $1 + 2^a \\mid 1 + 2^b$。設 $b = aq + r$, 其中 $0 \\le r < a$, 可知\n$$\n1 + 2^b = 1 + 2^{aq + r} \\equiv 1 + (-1)^q \\times 2^r \\pmod{1 + 2^a}.\n$$\n由於 $|1 + (-1)^q \\times 2^r| < 1 + 2^a$, 必須有 $1 + (-1)^q \\times 2^r = 0$, 故 $q$ 為奇數且 $r = 0$。換言之, $b$ 必須為 $a$ 的奇數倍。\n\n以下證明 $q = 1$。不... | Taiwan | 二〇一七數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a=b | |
0cm3 | Problem:
Given a positive integer $n$, determine the largest real number $\mu$ satisfying the following condition: for every $4n$-point configuration $C$ in an open unit square $U$, there exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area ... | [
"Solution:\nThe required maximum is $\\frac{1}{2n+2}$. To show that the condition in the statement is not met if $\\mu > \\frac{1}{2n+2}$, let $U = (0,1) \\times (0,1)$, choose a small enough positive $\\epsilon$, and consider the configuration $C$ consisting of the $n$ four-element clusters of points $\\left(\\fra... | Romanian Master of Mathematics (RMM) | The 7th Romanian Master of Mathematics Competition | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 1/(2n+2) | |
00nw | Let $\alpha$ be a nonzero real number.
Determine all functions $f: \mathbb{R} \to \mathbb{R}$ with
$$
f(f(x + y)) = f(x + y) + f(x)f(y) + \alpha xy
$$
for all $x, y \in \mathbb{R}$. | [
"*Answer.* For $\\alpha = -1$, the identity is the only solution. For other values of $\\alpha$, there is no solution.\n\nThe functional equation immediately implies that $f$ cannot be a constant function, as $\\alpha xy$ would then have to be constant. In the following, we let $(F)$ denote the given functional equ... | Austria | AUT_ABooklet_2023 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | For alpha equal to minus one, the only solution is f(x) = x for all real x; for all other alpha, there is no solution. | |
0ksd | Problem:
A regular octagon is inscribed in a circle of radius $2$. Alice and Bob play a game in which they take turns claiming vertices of the octagon, with Alice going first. A player wins as soon as they have selected three points that form a right angle. If all points are selected without either player winning, the... | [
"Solution:\n\nA player ends up with a right angle iff they own two diametrically opposed vertices. Under optimal play, the game ends in a draw: on each of Bob's turns he is forced to choose the diametrically opposed vertex of Alice's most recent choice, making it impossible for either player to win. At the end, the... | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 2√2 and 4 + 2√2 | |
0671 | Let $(x_n)$, $n \ge 1$, be a sequence of real numbers with $x_1 = 1$, such that $2x_{n+1} = 3x_n + \sqrt{5x_n^2 - 4}$, for $n=1,2,3,...$
a. Prove that all terms of the sequence are natural numbers.
β. Examine if there exist a term of the sequence divisible by $2011$. | [
"a.\nFrom the given recurrence relation we get\n$$\n(2x_{n+1} - 3x_n)^2 = 5x_n^2 - 4 \\Rightarrow 4x_{n+1}^2 - 12x_{n+1}x_n + 4x_n^2 = -4 \\Rightarrow x_{n+1}^2 - 3x_{n+1}x_n + x_n^2 = -1 \\quad (1)\n$$\nwhich can be written also in the form\n$$\nx_{n+2}^2 - 3x_{n+2}x_{n+1} + x_{n+1}^2 = -1 \\quad (2).\n$$\nWe cons... | Greece | SELECTION EXAMINATION | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Numb... | English | proof and answer | All terms are natural numbers, and no term is divisible by 2011. | |
0clo | Determine all positive integers $a, b, c$ such that the numbers $ab + c$, $bc + a$, and $ca + b$ are powers of 2. | [
"We will prove that there are only two families of solutions:\n$$\n(1, 1, 2^x - 1), \\quad (1, 2^x - 1, 2^x + 1),\n$$\nwhere $x$ is a positive integer.\n\n*Step I:* $a, b, c$ are odd and pairwise coprime in pairs.\nIt is easy to see that\n$$\na \\equiv b \\equiv c \\pmod{2}.\n$$\nIf $a, b, c$ are all even, denote b... | Romania | 75th NMO Selection Tests | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | (1, 1, 2^x − 1) and (1, 2^x − 1, 2^x + 1) for any positive integer x | |
03tz | Find the smallest positive integer $n$ such that any sequence of positive integers $a_1, a_2, \dots, a_n$ satisfying $\sum_{i=1}^{n} a_i = 2007$ must have several consecutive terms whose sum is 30. | [
"Firstly, we could construct a sequence of positive integers with 1017 terms $a_1, a_2, \\dots, a_{1017}$, such that we cannot find consecutive terms whose sum is 30. Hence, we could set $a_1 = a_2 = \\cdots = a_{29} = 1$, $a_{30} = 31$ and $a_{30+m+i} = a_i$, $i \\in \\{1, 2, \\dots, 30\\}$, $m \\in \\mathbb{N}$, ... | China | China Southeastern Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 1018 | |
08u8 | Let $ABC$ be a triangle for which $\angle BAC = 60^\circ$. Let $P$ ($Q$) be the point of intersection of the bisector of $\angle ABC$ ($\angle ACB$) and the side $AC$ ($AB$), respectively. Denote by $r_1$ ($r_2$) the radius of the in-circle of the triangle $ABC$ ($APQ$), respectively. Determine the radius of the circum... | [
"Let us denote by $I$, the intersection of $BP$ and $CQ$, which is the in-center of the triangle $ABC$. Let $I'$ be the in-center of the triangle $APQ$. Since both $AI$ and $AI'$ are the bisector of $\\angle BAC$, the points $A, I, I'$ lie on the same line. If we denote by $D$ the point of tangency of the in-circle... | Japan | Japan Junior 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"
] | null | proof and answer | 2(r1 - r2) | |
0jdu | Problem:
Find the smallest positive integer $n$ such that
$$
\frac{5^{n+1} + 2^{n+1}}{5^{n} + 2^{n}} > 4.99.
$$ | [
"Solution:\nWriting $5^{n+1} = 5 \\cdot 5^{n}$ and $2^{n+1} = 2 \\cdot 2^{n}$ and cross-multiplying yields $0.01 \\cdot 5^{n} > 2.99 \\cdot 2^{n}$, and re-arranging yields $(2.5)^{n} > 299$. A straightforward calculation shows that the smallest $n$ for which this is true is $n = 7$."
] | United States | HMMT 2013 | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 7 | |
0jto | Problem:
Kelvin the frog jumps along the number line starting at $0$. Every time he jumps, he jumps either one unit left or one unit right. For example, one sequence of jumps might be $0 \rightarrow -1 \rightarrow 0 \rightarrow 1 \rightarrow 2 \rightarrow 3 \rightarrow 2$.
How many ways are there for Kelvin to make ex... | [
"Solution:\n\nFirst, note that every time Kelvin jumps, he must jump from an even number to an odd one or vice-versa. Thus after ten jumps, he must land on an even number. So, if that number is to be prime, it must be $2$.\n\nThis means Kelvin must make $6$ jumps right and $4$ jumps left. That means the answer is $... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | final answer only | 210 | |
0im3 | A square grid on the Euclidean plane consists of all points $(m, n)$, where $m$ and $n$ are integers. Is it possible to cover all grid points by an infinite family of discs with non-overlapping interiors if each disc in the family has radius at least $5$? | [
"It is not possible. The proof is by contradiction. Suppose that such a covering family $\\mathcal{F}$ exists. Let $D(P, \\rho)$ denote the disc with center $P$ and radius $\\rho$. Start with an arbitrary disc $D(O, r)$ that does not overlap any member of $\\mathcal{F}$. Then $D(O, r)$ covers no grid point. Take th... | United States | USAMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | No | |
0hob | Problem:
Let $A_{1}, B_{1}, C_{1}$ be the points on the sides $BC, CA, AB$ (respectively) of the triangle $ABC$. Prove that the three circles circumscribed about the triangles $\triangle AB_{1}C_{1}$, $\triangle BC_{1}A_{1}$, and $\triangle CA_{1}B_{1}$ intersect at one point. | [
"Solution:\nDenote by $\\alpha$, $\\beta$, and $\\gamma$ the angles of $\\triangle ABC$. Assume that the circumcircles of $\\triangle AB_{1}C_{1}$ and $BC_{1}A_{1}$ intersect at the point $M$. Assume that $M$ is in the interior of $\\triangle ABC$ (the other cases are similar).\n\nBy the properties of the inscribed... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
050g | Let $a, b, c$ be positive real numbers such that $2a^2 + b^2 = 9c^2$. Prove that
$$
\frac{2c}{a} + \frac{c}{b} \ge \sqrt{3}.
$$ | [
"Using the AM-GM inequality for three terms twice, one gets\n$$\n\\begin{aligned}\n\\frac{2c}{a} + \\frac{c}{b} &= \\frac{(2b+a)c}{ab} = \\frac{(2b+a)\\sqrt{2a^2+b^2}}{3ab} = \\frac{(b+b+a)\\sqrt{a^2+a^2+b^2}}{3ab} \\\\\n&\\ge \\frac{3\\sqrt[3]{b^2a}\\sqrt{3\\sqrt[3]{a^4b^2}}}{3ab} = \\frac{3\\sqrt{3}\\sqrt[3]{b^2a... | Estonia | IMO Team Selection Contest | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0d5q | We color each unit square of a $8 \times 8$ table into green or blue such that there are $a$ green unit squares in each $3 \times 3$ square and $b$ green unit squares in each $2 \times 4$ rectangle. Find all possible values of ($a, b$). | [
"By tiling our $8 \\times 8$ table by eight $2 \\times 4$ rectangles like in Tiling (1) we find that the total number of green unit squares in the table is $8b$.\n\nBy tiling our $8 \\times 8$ table by four $3 \\times 3$ squares, three $2 \\times 4$ rectangles and one $2 \\times 2$ square, like in Tiling (2), we fi... | Saudi Arabia | SAMC 2015 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English, Arabic | proof and answer | a = 0, b = 0 or a = 9, b = 8 | |
0gxw | Prove for any positive integer $n$:
$$
\sum_{k=0}^{n-1} \frac{\left(\binom{n-1}{k}\right)^2}{k+1} = \frac{\binom{2n-1}{n}}{2n}
$$ | [
"Since $\\frac{n \\binom{n-1}{k}}{k+1} = \\frac{n(n-1)!}{(k+1)k!(n-k-1)!} = \\frac{n!}{k!(n-k-1)!} = \\binom{n}{k+1}$, the left-hand-side sum takes the form\n$$\n\\sum_{k=0}^{n-1} \\binom{n-1}{k} \\binom{n}{k+1} = \\sum_{k=0}^{n-1} \\binom{n-1}{k} \\binom{n}{n-k-1} = \\binom{2n-1}{n}.\n$$\nThe latter equality follo... | Ukraine | The Problems of Ukrainian Authors | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0cub | Pasha chose 2017 (not necessarily distinct) positive integers $a_1, a_2, \dots, a_{2017}$, and then he plays a solitaire game. Initially, he has 2017 empty large boxes and an unbounded supply of small stones. By a move, Pasha adds $a_1$ stones into some box by his choice, $a_2$ stones into any other box by his choice, ... | [
"Yes.\n\nNotice that $2017 = 43 \\cdot 46 + 39$. One example of Pasha's numbers consists of 39 twos, 46 numbers equal to 44, and ones as the remaining numbers.\n\nTo achieve the goal in 43 moves, Pasha chooses 39 boxes in which he always puts 2 stones—after 43 moves, each of these will contain $43 \\cdot 2 = 86$ st... | Russia | XLIII Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English; Russian | proof and answer | Yes | |
07y5 | Problem:
Sia $ABC$ un triangolo rettangolo, con ipotenusa $AC$, e sia $H$ il piede dell'altezza condotta da $B$ ad $AC$. Sapendo che le lunghezze $AB$, $BC$ e $BH$ costituiscono i lati di un nuovo triangolo rettangolo, determinare i possibili valori di $\frac{AH}{CH}$. | [
"Solution:\n\nPoiché i triangoli $ABH$ e $BCH$ sono rettangoli, si ha $BH < AB$ e $BH < HC$. Dunque il lato maggiore fra $AB$ e $BC$ deve essere l'ipotenusa del nuovo triangolo rettangolo.\n\nSupponiamo dapprima $BC > AB$. Allora abbiamo $AB^2 + BH^2 = BC^2$ e, analizzando il triangolo rettangolo $BCH$, $CH^2 + BH^... | Italy | null | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (sqrt(5) - 1)/2 and (sqrt(5) + 1)/2 | |
05um | Problem:
Soit $n$ un entier naturel. Démontrer que l'écriture de l'entier $n\left(2^{n}-1\right)$ en base 2 compte exactement $n$ occurrences du chiffre 1. | [
"Solution:\n\nNotons $s_{n}$ l'entier $n\\left(2^{n}-1\\right)$. Puisque $s_{0}=0$ et $s_{1}=1$, l'énoncé est bien vérifié lorsque $n \\leqslant 1$. On suppose donc désormais que $n \\geqslant 2$.\n\nOn pose alors $t=2^{n}-n$ et $s=n-1$, de sorte que $n\\left(2^{n}-1\\right)=2^{n} s+t$ et que $s+t=2^{n}-1$. En vert... | France | Préparation Olympique Française de Mathématiques - Test du 14 et du 21 Février 2021 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof only | null | |
0995 | Let $t$, $k$, $m$ be positive integers and $t > \sqrt{k} m$. Prove that
$$ \binom{2m}{0} + \binom{2m}{1} + \dots + \binom{2m}{m-t-1} < \frac{2^{2m}}{2k} $$
(proposed by B. Amarsanaa, folklore) | [
"**Lemma.** For all integers $0 \\leq t, s \\leq m$ such that $t + s \\leq m$, $\\frac{\\binom{2m}{m-s}}{\\binom{2m}{m-t-s}} > \\frac{t^2}{m}$.\n**Proof of lemma.**\n$$\nA = \\frac{\\binom{2m}{m-s}}{\\binom{2m}{m-t-s}} = \\frac{(m-t-s)!(m+t+s)!}{(m-s)!(m+s)!} =\n$$\n$$\n= \\left( \\frac{(m+t+1)(m+t+2)\\dots(m+t+s)}... | Mongolia | International Mathematical Olympiad 51 Selection Test | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Mongolian | proof only | null | |
0ji7 | Problem:
Let $N$ be the largest positive integer that can be expressed as a 2013-digit base $-4$ number. What is the remainder when $N$ is divided by $210$? | [
"Solution:\nThe largest is $$\\sum_{i=0}^{1006} 3 \\cdot 4^{2i} = 3 \\frac{16^{1007}-1}{16-1} = \\frac{16^{1007}-1}{5}.$$\n\nThis is $1 \\pmod{2}$, $0 \\pmod{3}$, $3 \\cdot 1007 \\equiv 21 \\equiv 1 \\pmod{5}$, and $3\\left(2^{1007}-1\\right) \\equiv 3\\left(2^{8}-1\\right) \\equiv 3\\left(2^{2}-1\\right) \\equiv 2... | United States | HMMT November 2013 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 51 | |
0ffw | Problem:
Sea $\mathcal{P}$ el conjunto de los puntos del plano y $f: \mathcal{P} \rightarrow \mathcal{P}$ una aplicación que cumple las tres condiciones siguientes:
a) $f$ es biyectiva.
b) Para cada recta $r$ del plano, $f(r)$ es una recta.
c) Para cada recta $r$, la recta $f(r)$ es paralela o coincidente con $r$.
¿Qu... | [
"Solution:\n\nVamos a denotar por $(P Q)$ la recta determinada por dos puntos distintos $P$ y $Q$, y abreviaremos $f(P)$ por $P^*$. De acuerdo con la condición (a), ($f$ es inyectiva), $P \\neq Q \\Rightarrow P^* \\neq Q^*$, y de acuerdo con la condición (b) se tiene, cuando $P \\neq Q$ que $f((P Q)) = (P^* Q^*)$. ... | Spain | OME 21 | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
05t8 | Problem:
Soit $ABC$ un triangle et soit $\omega$ son cercle circonscrit. Soit $\ell_{B}$ et $\ell_{C}$ deux droites parallèles l'une à l'autre, passant respectivement par les points $B$ et $C$. On note $D$ le point d'intersection, autre que $B$, entre $\omega$ et la droite $\ell_{B}$. De même, on note $E$ le point d'i... | [
"Solution:\n\nCommençons par tracer une figure, en prenant soin de faire apparaître le triangle $OO_{1}O_{2}$, puisque $P$ en est le centre du cercle circonscrit. On en profite pour noter $\\omega_{1}, \\omega_{2}$ et $\\omega^{\\prime}$ les cercles circonscrits à $ADG$, à $AEF$ et à $OO_{1}O_{2}$.\n\n![](attached_... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incente... | null | proof only | null | |
0fac | Problem:
The real numbers $x_1, x_2, \ldots, x_{1991}$ satisfy
$$
|x_1 - x_2| + |x_2 - x_3| + \ldots + |x_{1990} - x_{1991}| = 1991.
$$
What is the maximum possible value of
$$
|s_1 - s_2| + |s_2 - s_3| + \ldots + |s_{1990} - s_{1991}|,
$$
where $s_n = (x_1 + x_2 + \ldots + x_n)/n$? | [] | Soviet Union | 25th ASU | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1990 | |
0ky5 | Problem:
Point $Y$ lies on line segment $X Z$ such that $X Y=5$ and $Y Z=3$. Point $G$ lies on line $X Z$ such that there exists a triangle $A B C$ with centroid $G$ such that $X$ lies on line $B C$, $Y$ lies on line $A C$, and $Z$ lies on line $A B$. Compute the largest possible value of $X G$. | [
"Solution:\n\nThe key claim is that we must have $\\frac{1}{X G}+\\frac{1}{Y G}+\\frac{1}{Z G}=0$ (in directed lengths).\n\nWe present three proofs of this fact.\n\nProof 1: By a suitable affine transformation, we can assume without loss of generality that $A B C$ is equilateral. Now perform an inversion about $G$ ... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof and answer | 20/3 | |
0f1t | Problem:
You are given a regular $n$-gon. Each vertex is marked $+1$ or $-1$. A move consists of changing the sign of all the vertices which form a regular $k$-gon for some $1 < k \leq n$. [A regular $2$-gon means two vertices which have the center of the $n$-gon as their midpoint.] For example, if we label the vertic... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | 2^{80} | |
0jrm | Problem:
There are 7 boxes arranged in a row and numbered 1 through 7. You have a stack of 2015 cards, which you place one by one in the boxes. The first card is placed in box #1, the second in box #2, and so forth up to the seventh card which is placed in box #7. You then start working back in the other direction, pla... | [
"Solution:\nThe answer is box #3. Card #1 is placed into box #1, and this gets visited again by card #13. Hence, box #1 is visited every 12 times. Since $2004 = 167 \\cdot 12$, we see that card #2005 will be placed into box #1. Now, counting \"by hand,\" we see that card #2015 will go into box #3."
] | United States | BAMO | [
"Discrete Mathematics > Other",
"Number Theory > Other"
] | null | final answer only | 3 | |
0hrn | Problem:
In the triangle $A B C$ the angle $B$ is not a right angle, and $A B : B C = k$. Let $M$ be the midpoint of $A C$. The lines symmetric to $B M$ with respect to $A B$ and $B C$ intersect $A C$ at $D$ and $E$. Find $B D : B E$. | [
"Solution:\n\nAs $B C$ is the angle bisector in the triangle $M B E$, we have $\\frac{C E}{B E} = \\frac{C M}{B M}$ (by a well-known property of the angle bisector). Similarly, $\\frac{A D}{B D} = \\frac{A M}{B M}$. Draw a line $B M'$ symmetric to $B M$ with respect to the angle bisector of $A B C$ (point $M'$ is o... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | k^2 | |
0k9i | Problem:
Find all ordered triples of non-negative integers $(a, b, c)$ such that $a^{2}+2b+c$, $b^{2}+2c+a$, and $c^{2}+2a+b$ are all perfect squares. | [
"Solution:\nWe have the trivial solutions $(a, b, c) = (0, 0, 0)$ and $(a, b, c) = (1, 1, 1)$, as well as the solution $(a, b, c) = (127, 106, 43)$ and its cyclic permutations.\n\nThe case $a = b = c = 0$ works. Without loss of generality, $a = \\max \\{a, b, c\\}$. If $b$ and $c$ are both zero, it's obvious that w... | United States | Berkeley Math Circle | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (0,0,0), (1,1,1), (127,106,43), (106,43,127), (43,127,106) | |
0l8q | Consider the function $g(x) = \frac{2x}{1+x^2}$. Find all function $f(x)$ defined and continuous on the interval $(-1; 1)$ which satisfy the condition:
$$
(1 - x^2)f(g(x)) = (1 + x^2)^2 f(x)
$$
for every $x \in (-1; 1)$. | [
"The function $f(x)$ satisfies the demanded conditions when and only when the function $\\varphi(x) = (1-x^2)f(x)$ is defined and continuous on $(-1;1)$ and satisfies the condition\n$$\n\\varphi(g(x)) = \\varphi(x) \\quad \\forall x \\in (-1; 1). \\quad (1)\n$$\nConsider the function $h(x)$ defined on $(0; +\\infty... | Vietnam | VIETNAMESE MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = a / (1 - x^2) for any real constant a | |
03uq | The minimum value of $f(x) = \frac{5-4x+x^2}{2-x}$ for $x \in (-\infty, 2)$ is ( ). | [
"Let $x < 2 \\Rightarrow 2-x > 0$. Then\n$$\n\\begin{aligned}\nf(x) &= \\frac{1 + (4 - 4x + x^2)}{2-x} = \\frac{1}{2-x} + (2-x) \\\\\n&\\geq 2 \\times \\sqrt{\\frac{1}{2-x} \\times (2-x)} = 2.\n\\end{aligned}\n$$\nThe equality holds if and only if $\\frac{1}{2-x} = 2-x$, and it is so when $x = 1 \\in (-\\infty, 2)$... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | MCQ | C | |
0ea1 | Let $D$ be a point on the side $AB$ of the triangle $ABC$. The circumcircles of the triangles $BCD$ and $ACD$ meet the lines $AC$ and $BC$ again at points $E$ and $F$, respectively. The bisector of the segment $EF$ intersects the line $AB$ at $M$ and meets the altitude to $AB$ from $D$ at $N$. Let $T$ be the intersecti... | [
"First, we will show that the points $E$, $D$, $M$, $F$ and $N$ are concyclic.\n\nLet the circumcircle of the triangle $EDF$ intersect the line $AB$ at $M_1$. The quadrilaterals $BCED$ and $AFDC$ are cyclic, so $\\angle M_1EF = \\angle M_1DF = \\angle ACB = \\angle EDA = \\angle EFM_1$. The triangle $EM_1F$ is isos... | Slovenia | National Math Olympiad in Slovenia | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
02xl | Problem:
De quantas maneiras podemos colocar 8 algarismos iguais a 1 e 8 algarismos iguais a $0$ em um tabuleiro $4 \times 4$ de modo que as somas dos números escritos em cada linha e coluna sejam as mesmas?
| 1 | 0 | 1 | 0 |
| :--- | :--- | :--- | :--- |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | | [
"Solution:\nComo a soma dos números de todas as casas do tabuleiro é 8, a soma dos números em cada linha e coluna é $8 / 4 = 2$. Ou seja, em cada linha e coluna temos exatamente dois algarismos iguais a 1 e dois algarismos iguais a 0. Podemos escolher a posição do primeiro 1 da primeira linha de 4 maneiras. Em segu... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 90 | |
0d3w | Let $x$, $y$ be positive real numbers. Find the minimum of
$$
x^{2} + x y + \frac{y^{2}}{2} + \frac{2^{6}}{x + y} + \frac{3^{4}}{x^{3}}.
$$ | [
"$$\nx^{2} + x y + \\frac{y^{2}}{2} + \\frac{2^{6}}{x + y} + \\frac{3^{4}}{x^{3}} = \\frac{x^{2}}{2} + \\frac{(x + y)^{2}}{2} + \\frac{2^{6}}{x + y} + \\frac{3^{4}}{x^{3}}.\n$$\nBy applying AM-GM inequality we have\n$$\n\\frac{x^{2}}{2} + \\frac{3^{4}}{x^{3}} = \\frac{x^{2}}{6} + \\frac{x^{2}}{6} + \\frac{x^{2}}{6}... | Saudi Arabia | SAMC | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English, Arabic | proof and answer | 63/2 (achieved at x=3, y=1) | |
0hdr | Let $a$, $b$, $c$ be the sides of a triangle. How many numbers (maximum) out of $\frac{a+b}{a+b-c}$, $\frac{b+c}{b+c-a}$ and $\frac{c+a}{c+a-b}$ can be greater than $2$? | [
"First, we show an example where two numbers are, indeed, greater than $2$:\n$$\na = b = 5,\\ c = 1,\\ \\text{ then } \\frac{a+b}{a+b-c} = \\frac{10}{9} < 2,\\ \\text{ and } \\frac{b+c}{b+c-a} = \\frac{c+a}{c+a-b} = \\frac{6}{1} > 2.\n$$\n\nSuppose all of these fractions are greater than $2$. Then, clearly, this tr... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof and answer | 2 | |
0ldt | On the Cartesian plane, let $C$ be the graph of the function $y = \sqrt[3]{x^2}$. A line $d$ varies on the plane such that $d$ always cuts $C$ at three distinct points with $x$-coordinates $x_1, x_2$ and $x_3$.
a. Prove that the following value is a constant:
$$
\sqrt[3]{\frac{x_1 x_2}{x_3^2}} + \sqrt[3]{\frac{x_2 x_3... | [
"a. It is easy to see that $d$ cannot be a line parallel to the $y$-axis (otherwise $d$ only intersects $C$ at at most one point), so $d$ has the form $y = a x + b$ where $a, b \\in \\mathbb{R}$ and $a \\neq 0$. The $x$-coordinates of intersections of $d$ and $C$ are roots of\n$$\n\\sqrt[3]{x^2} = a x + b.\n$$\nLet... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathema... | English | proof and answer | a: 3; b: the sum is less than -15/4 | |
0j85 | Problem:
Five people of heights $65, 66, 67, 68$, and $69$ inches stand facing forwards in a line. How many orders are there for them to line up, if no person can stand immediately before or after someone who is exactly $1$ inch taller or exactly $1$ inch shorter than himself? | [
"Solution:\n\nAnswer: $14$\n\nLet the people be $A, B, C, D, E$ so that their heights are in that order, with $A$ the tallest and $E$ the shortest. We will do casework based on the position of $C$.\n\n- Case 1: $C$ is in the middle. Then, $B$ must be on one of the two ends, for two choices. This leaves only one cho... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 14 | |
0824 | Problem:
Una capra (indicata in figura con $\Pi^{\alpha}$) è legata a un punto $A$ con una corda lunga 6 metri come mostrato nella figura (vista dall'alto). Il segmento $AB$ rappresenta una staccionata lunga 4 metri oltre la quale la capra non può saltare. Il quadrato rappresenta un edificio di lato 2 metri all'intern... | [
"Solution:\n\nLa risposta è $\\mathbf{(A)}$. Infatti semplici osservazioni permettono di vedere che l'area che può essere brucata dalla capra è quella indicata nella seguente figura:\n\n\n\nL'area $S_{1}$ è un quarto di quella di un cerchio di raggio 6 e cioè: $\\frac{1}{4}\\left(6^{2}\\rig... | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | MCQ | A | |
0g3c | Problem:
Let $(m, n)$ be a pair of positive integers. Julia has carefully planted $m$ rows of $n$ dandelions in an $m \times n$ array in her back garden. Now, Jana and Viviane decide to play a game with a lawnmower they just found. Taking alternating turns and starting with Jana, they can mow down all the dandelions i... | [
"Solution:\n\nFirst we make two general observations about the problem. Firstly, since the game always finishes in a finite number of moves there is always a winner and consequently a player who has a winning strategy at the start. Secondly, when a player mows down a row or a column, it does not matter which row or... | Switzerland | Final round | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Jana has a winning strategy for all pairs except when both dimensions exceed one and their sum is even; equivalently, she wins if the sum of the dimensions is odd or if the grid is a single cell. | |
0l5y | The 9 members of a baseball team went to an ice-cream parlor after their game. Each player had a single-scoop cone of chocolate, vanilla, or strawberry ice cream. At least one player chose each flavor, and the number of players who chose chocolate was greater than the number of players who chose vanilla, which was grea... | [
"The only triples of numbers of choices of chocolate, vanilla, and strawberry, in that order, that meet the given conditions are $(6, 2, 1)$, $(5, 3, 1)$, and $(4, 3, 2)$. Using the formula for the number of permutations with repetitions gives\n$$\n\\begin{aligned}\nN &= \\frac{9!}{6!2!1!} + \\frac{9!}{5!3!1!} + \\... | United States | AIME I | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 16 | |
0ex7 | Problem:
$ABCD$ is a convex quadrilateral. $A'$ is the foot of the perpendicular from $A$ to the diagonal $BD$, $B'$ is the foot of the perpendicular from $B$ to the diagonal $AC$, and so on. Prove that $A'B'C'D'$ is similar to $ABCD$. | [
"Solution:\n\nLet the diagonals meet at $O$. Then $CC'O$ is similar to $AA'O$ (because $CC'O = AA'O = 90^{\\circ}$, and $\\angle COC'$, $\\angle AOA'$ are opposite angles), so $A'O / C'O = AO / CO$. Similarly, $B'O / D'O = BO / DO$. $AA'O$ is also similar to $BB'O$, so $A'O / B'O = AO / BO$. Thus $OA':OB':OC':OD' =... | Soviet Union | 4th ASU | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
086d | Problem:
Un numero naturale $n$ è detto gradevole se gode delle seguenti proprietà:
- la sua espressione decimale è costituita da 4 cifre;
- la prima e la terza cifra di $n$ sono uguali;
- la seconda e la quarta cifra di $n$ sono uguali;
- il prodotto delle cifre di $n$ divide $n^{2}$.
Si determinino tutti i numeri gr... | [] | Italy | XXV OLIMPIADE ITALIANA DI MATEMATICA | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1111, 1212, 1515, 2424, 3636 |
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