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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
084v | Problem:
I membri di una tribù hanno dieci dita alle mani e nove ai piedi e quindi contano indifferentemente in base 10 o 19. Nella loro cultura matematica, un numero intero positivo è detto "sacro" se in entrambe le basi si scrive con le stesse due cifre (comprese tra 1 e 9). Quanti sono i numeri sacri? | [
"Solution:\n\nLa risposta è 4. Sia $n$ un numero sacro e sia $A B$ la scrittura del numero in base 10. Allora in base 19 per la sacralità esso si scrive come $A B$ o come $B A$. Il primo caso non genera soluzioni poichè $A$ e $B$ sono diversi da 0, e quindi $A B$ in base 19 è maggiore di $A B$ in base 10 per ogni $... | Italy | Progetto Olimpiadi di Matematica 2006 GARA di SECONDO LIVELLO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 4 | |
0ktc | Problem:
For any positive integer $n$, let $\tau(n)$ denote the number of positive divisors of $n$. If $n$ is a positive integer such that $\frac{\tau\left(n^{2}\right)}{\tau(n)}=3$, compute $\frac{\tau\left(n^{7}\right)}{\tau(n)}$. | [
"Solution:\n\nLet the prime factorization of $n$ be $n=p_{1}^{e_{1}} p_{2}^{e_{2}} \\cdots p_{k}^{e_{k}}$. Then, the problem condition is equivalent to\n$$\n\\prod_{i=1}^{k} \\frac{2 e_{i}+1}{e_{i}+1}=3\n$$\nNote that since $\\frac{2 x+1}{x+1} \\geq 1.5$ for $x \\geq 1$, and $1.5^{3}>3$, we have $k \\leq 2$. Also, ... | United States | HMMT February | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 29 | |
02an | Problem:
Na figura, $ABCD$ é um quadrado cuja área é $\frac{7}{32}$ da área do triângulo $XYZ$. Qual é a razão entre $XA$ e $XY$?
 | [
"Solution:\n\nSejam $l$ o comprimento do lado do quadrado, $h$ a altura do triângulo $\\triangle XAB$, $H$ a altura do triângulo $\\triangle XYZ$ e $b$ o comprimento do lado $YZ$.\nA área do quadrado é $l^2$ e a área do triângulo $\\triangle XYZ$ é $\\frac{bH}{2}$. Como os triângulos $XYZ$ e $ABC$ são semelhantes, ... | Brazil | Lista 6 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 7/8 or 1/8 | |
0cyj | Prove that
$$
\frac{\sin^{3} a}{\sin b} + \frac{\cos^{3} a}{\cos b} \geq \frac{1}{\cos(a-b)},
$$
for all $a$ and $b$ in the interval $\left(0, \frac{\pi}{2}\right)$. | [
"Since $a, b \\in \\left(0, \\frac{\\pi}{2}\\right)$ we have $\\sin a, \\sin b, \\cos a, \\cos b > 0$.\nThen\n$$\n\\begin{aligned}\n\\frac{\\sin^{3} a}{\\sin b} + \\frac{\\cos^{3} a}{\\cos b} &= \\frac{\\sin^{4} a}{\\sin a \\sin b} + \\frac{\\cos^{4} a}{\\cos a \\cos b} \\\\\n&= \\frac{(\\sin^{2} a)^2}{\\sin a \\si... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
01tk | An infinite sequence $(a_n)$, $n \in \mathbb{N}$, of positive numbers is called *lacunar* if there exists a number $q > 1$ such that $a_{n+1}/a_n \ge q$ for all $n \in \mathbb{N}$. Also, the sequence is called *rare* if there exists a positive integer $k$ such that the interval $(x, 2x)$ contains at most $k$ terms of t... | [
"a) Let the sequence $(a_n)$, $n \\in \\mathbb{N}$, be lacunar. Then there exists a number $q > 1$ such that\n$$\na_{n+1} \\ge q a_n \\quad \\forall n \\in \\mathbb{N}. \\qquad (1)\n$$\nIn particular, any lacunar sequence is increasing. From (1) it follows that any half-interval $(x, qx]$ contains at most one term ... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series"
] | English | proof and answer | a) Yes. b) No. | |
0e97 | When the third grade pupil Benjamin calculated the sum $1+2+3+\ldots+2012$, he forgot to add some terms, and he got an incorrect sum that was divisible by $2011$. When Anika calculated the sum $A = 1+2+3+\ldots+2013$, she forgot to add the same terms as Benjamin, and she got an incorrect sum $N$ that was divisible by $... | [
"Let us denote the sum of the terms omitted by Benjamin by $x$. Since $1 + 2 + 3 + \\ldots + 2012 = \\frac{2012 \\cdot 2013}{2} = 1006 \\cdot 2013$, Benjamin's result was equal to $1006 \\cdot 2013 - x$. So, there exists a non-negative integer $m$, such that $1006 \\cdot 2013 - x = 2011m$.\n\nSince $A = 1 + 2 + 3 +... | Slovenia | National Math Olympiad 2013 - First Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2/3 | |
0j4o | Problem:
Mike and Harry play a game on an $8 \times 8$ board. For some positive integer $k$, Mike chooses $k$ squares and writes an $M$ in each of them. Harry then chooses $k+1$ squares and writes an $H$ in each of them. After Harry is done, Mike wins if there is a sequence of letters forming "$H M M$" or "$M M H$", w... | [
"Solution:\n\nAnswer: 16\n\nSuppose Mike writes $k$ $M$'s. Let $a$ be the number of squares which, if Harry writes an $H$ in, will yield either $H M M$ or $M M H$ horizontally, and let $b$ be the number of squares which, if Harry writes an $H$ in, will yield either $H M M$ or $M M H$ vertically. We will show that $... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 16 | |
04vs | Suppose we have a 9-digit number $N$ in which all the digits are distinct and non-zero. Then we consider all the sums of adjacent triples of digits of $N$ and order them in a non-decreasing sequence. For the following sequences, determine whether there exists an $N$ for which we get them as a result:
a) $11, 15, 16, 18... | [
"a) Yes, the number $N = 137658942$ works, since the sums of consecutive triples are (left to right) 11, 16, 18, 19, 22, 21, 15.\n\nb) We shall prove that no such number exists by showing that for any $N$, the sum of the 7 numbers is at most 122, while the numbers given to us have sum 123. Denote the $i$-th digit o... | Czech Republic | First Round of the 73rd Czech and Slovak Mathematical Olympiad (take-home part) | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a) Yes; for example N = 137658942. b) No. | |
09jc | Consider a triangle whose sides have length $3$, $4$ and $5$. Prove that the angle against the side with length $3$ is between $36^\circ$ and $38^\circ$. | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0331 | Problem:
Find all real numbers $a$ such that the graphs of the functions $x^{2}-2 a x$ and $-x^{2}-1$ have two common tangent lines and the perimeter of the quadrilateral with vertices at the tangent points is equal to $6$. | [
"Solution:\nThe equation of a common tangent line to the graphs of $f(x)$ and $g(x)$ at points $(x_{1}, f(x_{1}))$ and $(x_{2}, g(x_{2}))$ has the form\n$$\ny = f(x_{1}) + f'(x_{1})(x - x_{1}) = g(x_{2}) + g'(x_{2})(x - x_{2})\n$$\nHence $f'(x_{1}) = g'(x_{2})$ and $f(x_{1}) - f'(x_{1}) x_{1} = g(x_{2}) - g'(x_{2})... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a = ±√2/2 | |
04cl | Let $p_1$ and $q_1$ be integers such that the equation $x^2 + p_1x + q_1 = 0$ has two integer solutions. For all $n \in \mathbb{N}$ we define the numbers $p_{n+1}$ and $q_{n+1}$ by the formulas
$$
p_{n+1} = p_n + 1, \quad q_{n+1} = q_n + \frac{1}{2}p_n.
$$
Prove that there exists an infinite number of positive integers... | [
"Let $D_n$ be a discriminant of quadratic equation $x^2 + p_n x + q_n = 0$, for each $n \\in \\mathbb{N}$, i.e.\n$$D_n = p_n^2 - 4q_n.$$ \nBy assumption we conclude that $D_1$ is a square of an integer.\nFurther, we have:\n$$\nD_{n+1} = p_{n+1}^2 - 4q_{n+1} = (p_n + 1)^2 - 4 \\left( q_n + \\frac{1}{2}p_n \\right) =... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0fuj | Problem:
An einem Fussballturnier nehmen $n>10$ Mannschaften teil. Dabei spielt jede Mannschaft genau einmal gegen jede andere. Ein Sieg gibt zwei Punkte, ein Unentschieden einen Punkt, und eine Niederlage keinen Punkt. Nach dem Turnier stellt sich heraus, dass jede Mannschaft genau die Hälfte ihrer Punkte in den Spie... | [
"Solution:\n\nWir nennen die 10 schlechtesten Mannschaften die Verlierer, die $n-10$ besten Mannschaften die Gewinner. Wir verwenden wiederholt folgende Tatsache: Spielen $k$ Manschaften gegeneinander, dann ist die Gesamtzahl gewonnener Punkte genau $k(k-1)$. Wir zählen die Gesamtzahl gewonnener Punkte auf zwei Art... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 25 | |
0a9j | Problem:
In a triangle $A B C$ assume $A B = A C$, and let $D$ and $E$ be points on the extension of segment $B A$ beyond $A$ and on the segment $B C$, respectively, such that the lines $C D$ and $A E$ are parallel. Prove that $C D \geq \frac{4 h}{B C} C E$, where $h$ is the height from $A$ in triangle $A B C$. When d... | [
"Solution:\n\nBecause $A E \\parallel D C$, the triangles $A B E$ and $D B C$ are similar. So\n$$\nC D = \\frac{B C}{B E} \\cdot A E\n$$\n\n$$\nC D = \\frac{A E \\cdot B C}{B E \\cdot C E} \\cdot C E\n$$\n\n\n\nLet $A F$ be an altitude of $A B C$. Then $A E \\geq A F = h$, and equality hold... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 25 | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | Equality holds if and only if the point on the base is its midpoint (equivalently, the foot of the altitude from the apex). | |
04z3 | The diagonals of trapezoid $ABCD$ with bases $AB$ and $CD$ meet at $P$. Prove the inequality $S_{PAB} + S_{PCD} > S_{PBC} + S_{PDA}$, where $S_{XYZ}$ denotes the area of triangle $XYZ$. | [
"Let $a = |AB|$ and $b = |CD|$ and let $h_a$ and $h_b$ be the altitudes of triangles $PAB$ and $PCD$ drawn from $P$ (see Fig. 7). Denote $S_1 = S_{PAB} + S_{PCD}$ and $S_2 = S_{PBC} + S_{PDA}$. Then $S_1 = \\frac{1}{2}(a h_a + b h_b)$ and $S_1 + S_2 = \\frac{1}{2}(a + b)(h_a + h_b)$, whence $S_2 = \\frac{1}{2}(a h_... | Estonia | Estonija 2010 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
09l7 | Calculate the total number of acute-angled triangles that can be formed using the vertices of a regular $30$-gon. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 910 | |
03mz | Find all polynomials $P(x)$ with integer coefficients such that $P(P(n)+n)$ is a prime number for infinitely many integers $n$. | [
"Note that if $P(n) = 0$ then $P(P(n) + n) = P(n) = 0$ which is not prime. Let $P(x)$ be a degree $k$ polynomial of the form $P(x) = a_k x^k + a_{k-1} x^{k-1} + \\dots + a_0$ and note that if $P(n) \\ne 0$ then\n$$\nP(P(n) + n) - P(n) = a_k[(P(n) + n)^k - n^k] + a_{k-1}[(P(n) + n)^{k-1} - n^{k-1}] + \\dots + a_1 P(... | Canada | Kanada | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All polynomials are either P(x) = p where p is a prime number, or P(x) = -2x + b where b is an odd integer. | |
02gu | Let $ABCD$ be a convex quadrilateral. Prove that the incircles of the triangles $ABC$, $BCD$, $CDA$ and $DAB$ have a point in common if, and only if, $ABCD$ is a rhombus. | [
"If $ABCD$ is a rhombus the incircle touch the respective triangle in the midpoint of the diagonals, which belongs to all four circles.\n\nNow suppose the circles have a common point $P$. Since the incircle are contained in its triangle and the intersection of the triangles $ABC$, $BCD$, $CDA$, $DAB$ is the interse... | Brazil | XXVI OBM | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > M... | English | proof only | null | |
0icc | Problem:
$P$ is a point inside triangle $ABC$, and lines $AP$, $BP$, $CP$ intersect the opposite sides $BC$, $CA$, $AB$ in points $D$, $E$, $F$, respectively. It is given that $\angle APB = 90^{\circ}$, and that $AC = BC$ and $AB = BD$. We also know that $BF = 1$, and that $BC = 999$. Find $AF$.
![](attached_image_1.... | [
"Solution:\n\n\nLet $AC = BC = s$, $AB = BD = t$. Since $BP$ is the altitude in isosceles triangle $ABD$, it bisects angle $B$. So, the Angle Bisector Theorem in triangle $ABC$ gives $AE / EC = AB / BC = t / s$. Meanwhile, $CD / DB = (s - t) / t$. Now Ceva's theorem gives us\n$$\n\\begin{ga... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 499/500 | |
0bxe | Let $m$ be a positive integer, let $p$ be a prime, let $a_1 = 8p^m$, and let $a_n = (n+1)^{a_{n-1}/n}$, $n = 2, 3, \dots$. Determine the primes $p$ for which the products $a_n\left(1 - \frac{1}{a_1}\right)\left(1 - \frac{1}{a_2}\right)\dots\left(1 - \frac{1}{a_n}\right)$, $n = 1, 2, 3, \dots$, are all integral. | [
"The required primes are $2$ and $5$. Begin by noticing that if $a_1$ is any even integer, then $a_{n-1}$ is an integral power of $n$ for all $n \\ge 3$. Thus, if $n \\ge 3$, then $(a_n - 1)/a_{n-1} = ((n+1)^{n_k} - 1)/n^{k+1}$ for some non-negative integer $k$. An easy induction on $k$ shows that $(n+1)^{n_k} - 1$... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Algebra > Algebraic ... | English | proof and answer | p = 2 or p = 5 | |
0aqc | Problem:
The perimeter of a right triangle is $90~\mathrm{cm}$. The squares of the lengths of its sides sum up to $3362~\mathrm{cm}^2$. What is the area of the triangle? | [
"Solution:\n\nLet the sides of the right triangle be $a$, $b$, and $c$, where $c$ is the hypotenuse.\n\nGiven:\n$$\na + b + c = 90\n$$\n$$\na^2 + b^2 + c^2 = 3362\n$$\nSince the triangle is right, $a^2 + b^2 = c^2$.\n\nSo,\n$$\na^2 + b^2 + c^2 = c^2 + c^2 = 2c^2\n$$\nTherefore,\n$$\n2c^2 = 3362 \\implies c^2 = 1681... | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Geometry > Plane Geometry > Triangles",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 180 | |
0cok | A quadrilateral $ABCD$ is inscribed into a circle $\omega$. Its diagonals intersect at point $K$. Let $M_1, M_2, M_3, M_4$ be the midpoints of arcs $AB$, $BC$, $CD$, $DA$ (not containing the other vertices of $ABCD$), respectively. Let $I_1, I_2, I_3, I_4$ be the incenters of triangles $ABK$, $BCK$, $CDK$, $DAK$, respe... | [
"Заметим, что точка $I_1$ лежит на биссектрисах $AM_2$ и $BM_4$ углов $BAC$ и $ABD$, поэтому $I_1 = AM_2 \\cap BM_4$ (см. рис. 22). Аналогично $I_2 = BM_3 \\cap CM_1$, $I_3 = CM_4 \\cap DM_2$, $I_4 = DM_1 \\cap AM_3$.\n\n\nРис. 22\n\nРис. 23\n\nПоскольку $\\overrigh... | Russia | Final round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous ... | English; Russian | proof only | null | |
0fvn | Problem:
Ein Kreis $k_{1}$ liegt innerhalb eines zweiten Kreises $k_{2}$ und berührt diesen im Punkt $A$. Eine Gerade durch $A$ schneide $k_{1}$ nochmals in $B$ und $k_{2}$ in $C$. Die Tangente an $k_{1}$ durch $B$ schneide $k_{2}$ in den Punkten $D$ und $E$. Die Tangenten an $k_{1}$ durch $C$ berühren $k_{1}$ in den ... | [
"Solution:\n\nWir zeigen, dass die vier Punkte alle den gleichen Abstand zu $C$ haben.\n\nLemma 2. Sei $P$ ein Punkt auf einem Kreis $k$. Schneidet eine zu der Tangenten von $k$ durch $P$ parallele Gerade den Kreis $k$ in den Punkten $R$ und $S$, so gilt $P R = P S$.\n\nBeweis des Lemmas. Wir betrachten die Gerade ... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07nb | Suppose $x, y, z$ are positive numbers such that
$$
1 = 2xyz + xy + yz + zx.
$$
Prove that there are positive numbers $a, b, c$ such that
$$
x = \frac{a}{b+c}, \quad y = \frac{b}{c+a}, \quad z = \frac{c}{a+b}.
$$ | [
"Since\n$$\n\\det \\begin{pmatrix} 1 & -x & -x \\\\ -y & 1 & -y \\\\ -z & -z & 1 \\end{pmatrix} = 1 - 2xyz - xy - yz - zx,\n$$\nthe determinant of the system of equations\n$$\na - xb - xc = 0, \\quad ya - b + yc = 0, \\quad za + zb - c = 0, \\quad (2)\n$$\nin the variables $a, b, c$, is equal to zero by hypothesis.... | Ireland | Ireland | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
0d2q | Let $T$ be a real number satisfying the property: For any nonnegative real numbers $a, b, c, d, e$ with their sum equal to $1$, it is possible to arrange them around a circle such that the products of any two neighboring numbers are no greater than $T$. Determine the minimum value of $T$. | [
"Assume, without loss of generality, that\n$$\n0 \\leq a \\leq b \\leq c \\leq d \\leq e\n$$\nBecause\n$$\ne a \\leq \\max \\{e b, c d\\} \\leq e c \\leq e d\n$$\nto get the smallest possible maximum, the best arrangement around the circle is\n\nIn this case, the maximum is given by\n$$\n\\... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 1/9 | |
093v | Problem:
Let $n \geqslant 3$ be an integer. Zagi the squirrel sits at a vertex of a regular $n$-gon. Zagi plans to make a journey of $n-1$ jumps such that in the $i$-th jump, it jumps by $i$ edges clockwise, for $i \in\{1, \ldots, n-1\}$. Prove that if after $\left\lceil\frac{n}{2}\right\rceil$ jumps Zagi has visited ... | [
"Solution:\n\nNumber the vertices $0,1, \\ldots, n-1$ clockwise starting at the vertex Zagi is on. After his $i$-th jump Zagi will be at a vertex numbered $1+2+\\cdots+i=\\frac{i(i+1)}{2} \\pmod n$. We need to prove that if for all $k \\in\\left\\{0,1,2, \\ldots,\\left\\lceil\\frac{n}{2}\\right\\rceil\\right\\}$ th... | Middle European Mathematical Olympiad (MEMO) | 15th Middle European Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0giv | 令 $m$ 與 $n$ 為大於 1 的正整數。在 $m \times n$ 方格紙上的每一格都有一枚背面向上的硬幣。每一步，我們依次進行以下動作：
(1) 選擇一個 $2 \times 2$ 的區域；
(2) 將該區域左上角與右下角的硬幣翻面；
(3) 將該區域左下角與右上角的硬幣擇一翻面。
試求所有 $(m, n)$，使得我們能透過有限步將硬幣全部翻成正面。
Let $m$ and $n$ be positive integers greater than 1. In each square of a $m \times n$ grid lies a coin with its tail-side up. A *move* c... | [
"答案為滿足 $3 \\mid mn$ 的所有 $(m, n)$。\n\n構造：不失一般性假設 $3 \\mid m$。當 $2 \\mid n$ 時，我們可用左圖方式將所有硬幣翻面：\n\n\n當 $2 \\nmid n$ 時，先用上述的方法將左邊的 $m \\times (n-1)$ 全部翻成正面，剩下最右邊一排是背面。接著，令 $L(i, j)$ 為將 $(i, j)$，$(i+1, j)$ 和 $(i, j+1)$ 位置翻面的 $L$ 型操作，而 $R(i, j)$ 為將 $(i, j)$，$(i, j-1)$ 和 $(i-1, j)$ 位置翻面的操作。則注意到 $R... | Taiwan | IMO 1J, Mock Exam 2 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Chinese; English | proof and answer | All pairs (m, n) such that 3 divides mn | |
0cmy | A sequence $a_1, a_2, \dots$ satisfies the following conditions: $a_1 \in (1, 2)$, and $a_{k+1} = a_k + \frac{k}{a_k}$ for every positive integer $k$. Prove that there is at most one (unordered) pair of distinct indices $(i, j)$ such that $a_i + a_j$ is an integer. (A. Golovanov) | [
"Let $b_k = a_k - k$. Then\n$$\nb_{k+1} = b_k - 1 + \\frac{k}{k + b_k} = b_k - \\frac{b_k}{k + b_k} = b_k \\left(1 - \\frac{1}{k + b_k}\\right).\n$$\nFrom here, by obvious induction on $k$, we get that $b_k > 0$ (since $b_1 > 0$). Moreover, $b_{k+1} = b_k - \\frac{b_k}{k + b_k} < b_k$. Hence, in particular, $b_k \\... | Russia | Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English; Russian | proof only | null | |
0g2c | Problem:
Déterminer tous les nombres naturels $n \geq 2$ tels que pour tous les nombres entiers $0 \leq i, j \leq n$ :
$$
i+j \equiv \binom{n}{i}+\binom{n}{j} \quad(\bmod 2)
$$ | [
"Solution:\nEn posant $j=i+1$ (pour $i<n$ ), on obtient\n$$\n\\binom{n+1}{i+1}=\\binom{n}{i}+\\binom{n}{i+1} \\equiv i+i+1 \\equiv 1 \\quad(\\bmod 2)\n$$\nAinsi, puisque $\\binom{n+1}{0}=1$, cela signifie que pour tout $0 \\leq i \\leq n+1$ le nombre $\\binom{n+1}{i}$ doit être impair. Pour tout $0 \\leq i \\leq n+... | Switzerland | SMO-Selektion | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | n = 2^k - 2 for k ≥ 2 | |
098e | Problem:
Fie $P(X)$ un polinom cu coeficienți reali, astfel încât
$x \cdot P(x+2022)-(x+2023) \cdot P(x)=2022 \cdot x-2023$, pentru orice $x \in \mathbb{R}$. Aflaţi restul împărțirii polinomului $P(X)$ la polinomul $Q(X)=X^{2}+X$. | [
"Solution:\n\nPresupunem că polinomul $P(X)$ verifică relația din enunț. Deoarece această egalitate este adevărată pentru orice număr real $x$, atunci pentru $x=0$, obținem $-2023 \\cdot P(0) = -2023 \\Leftrightarrow P(0) = 1$.\n\nPentru $x = -2023$ obținem $-2023 \\cdot P(-1) = -2023^{2} \\Leftrightarrow P(-1) = 2... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | -2022 X + 1 | |
06lr | Let $f(x)$ be a polynomial with rational coefficients, and let $\alpha$ be a real number. If
$$
\alpha^3 - 2019\alpha = (f(\alpha))^3 - 2019f(\alpha) = 2021,
$$
prove that $(f^n(\alpha))^3 - 2019f^n(\alpha) = 2021$ for any positive integer $n$.
(Here, we define $f^n(x) = \underbrace{f(f(\dots f(x)\dots))}_{n \text{ tim... | [
"We claim that $g(x) = x^3 - 2019x - 2021$ is irreducible over $\\mathbb{Q}$. Since $\\deg g = 3$, if $g$ were reducible, then it must have a linear factor. In other words, $g$ would have a rational root $c$. Since $g$ is monic, $c$ must be an integer by the rational root theorem. However, $c^3 - 2019c$ must be eve... | Hong Kong | Year 2021 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Abstract Algebra > Field Theory",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
02tm | Problem:
Seja $n$ um número inteiro positivo. Se, para cada divisor primo $p$ de $n$, o número $p^{2}$ não divide $n$, dizemos então que $n$ é livre de quadrados. Mostre que todo número livre de quadrados tem uma quantidade de divisores que é igual a uma potência de 2. | [
"Solution:\n\nSuponha que $n$ é um número livre de quadrados e considere sua fatoração em primos:\n$$\nn = p_{1}^{\\alpha_{1}} \\cdot p_{2}^{\\alpha_{2}} \\cdot \\ldots \\cdot p_{k}^{\\alpha_{k}}\n$$\nComo $n$ é livre de quadrados, os expoentes $\\alpha_{i}$ são todos iguais a $1$. Portanto,\n$$\nn = p_{1} \\cdot p... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0iqp | Problem:
Suppose that $A B C$ is an isosceles triangle with $A B = A C$. Let $P$ be the point on side $A C$ so that $A P = 2 C P$. Given that $B P = 1$, determine the maximum possible area of $A B C$. | [
"Solution:\nAnswer: $\\sqrt{\\frac{9}{10}}$\n\nLet $Q$ be the point on $A B$ so that $A Q = 2 B Q$, and let $X$ be the intersection of $B P$ and $C Q$. The key observation is that, as we will show, $B X$ and $C X$ are fixed lengths, and the ratio of areas $[A B C] / [B C X]$ is constant. So, to maximize $[A B C]$, ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 9/10 | |
0bly | Given a triangle $ABC$ with $m(\angle A) = 90^\circ$, $AC < AB$, consider on the rays $BA$ and $AC$ points $E$ and $D$ respectively, such that $A \in (BE)$, $C \in (AD)$, $AE = AC$ and $AD = AB$. Denote by $M$ and $N$ the midpoints of $[BC]$ and $[DE]$ respectively, and let $\{R\} = EC \cap BD$. Show that $MN = RA$. | [
"The hypothesis implies that triangles $\\triangle ACE$ and $\\triangle ABD$ are right angled and isosceles, so the triangle $\\triangle RBE$ is also right angled and isosceles, that is $m(\\widehat{RBE}) = m(\\widehat{REB}) = 45^\\circ$. By the equality of triangles $\\triangle ABC$ and $\\triangle ADE$ (CC) we ge... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0fn1 | ¿Existen infinitos enteros positivos que no pueden representarse de la forma
$$
a^3 + b^5 + c^7 + d^9 + e^{11},
$$
donde $a$, $b$, $c$, $d$, $e$ son enteros positivos? Razónese la respuesta. | [] | Spain | Olimpiada Matemática Española | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | Yes, there are infinitely many positive integers that cannot be represented in that form. | |
09a4 | Let $A$ be one of the intersection points of circles $\omega_1(O_1)$ and $\omega_2(O_2)$. $\ell$ is a line that touches $\omega_1$ and $\omega_2$ at $B$ and $C$ respectively. Let $O_3$ be the center of the circumscribed circle of $ABC$. And we choose $D$ as $A$ is midpoint of $O_3D$. If $M$ is midpoint of $O_1O_2$, the... | [
"Let us denote by $Q$, $R$ and $S$ the midpoints of $BC$, $AB$ and $AC$ respectively. Then $O_3R \\perp AB$, $O_1R \\perp AB$, $O_3S \\perp AC$, $O_2S \\perp AC$, $O_3Q \\perp BC$ and $MQ \\perp BC$. Assuming $\\angle ABC = \\beta$ and $\\angle BCA = \\gamma$, we get $\\angle BO_3O_1 = \\gamma$ and $\\angle CO_3O_2... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Method... | null | proof only | null | |
0fte | Problem:
$n$ sei eine positive ganze Zahl mit mindestens vier verschiedenen positiven Teilern. Die vier kleinsten unter diesen Teilern seien $d_{1}, d_{2}, d_{3}, d_{4}$. Finde alle solchen Zahlen $n$, für die gilt
$$
d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}=n
$$ | [
"Solution:\n\nEs ist $d_{1}=1$. Nehme an, $n$ sei ungerade, dann wären auch $d_{1}, d_{2}, d_{3}, d_{4}$ ungerade und daher $n=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}$ gerade, Widerspruch. Also ist $n$ gerade und $d_{2}=2$. Nehme an, $n$ sei durch 4 teilbar, dann ist eine der Zahlen $d_{3}, d_{4}$ gleich 4 und die ... | Switzerland | IMO Selektion | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 130 | |
02tn | Problem:
Cinco crianças sentam-se ao redor de uma mesa circular. Cada criança escolhe um número inteiro positivo e o relata para as outras. Em seguida, cada criança faz a seguinte conta: soma os números das duas crianças à sua esquerda, subtrai a soma dos números das outras duas crianças à sua direita e multiplica ess... | [
"Solution:\n\nVamos supor que os números em sentido horário são $a, b, c, d$ e $e$. Os valores obtidos como resultados finais são:\n$$\n\\begin{aligned}\na((e+d)-(b+c)) & = a e + a d - a b - a c \\\\\nb((a+e)-(c+d)) & = b a + b e - b c - b d \\\\\nc((b+a)-(d+e)) & = c b + c a - c d - c e \\\\\nd((c+b)-(e+a)) & = d ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | 0 | |
00hl | Show that $r=2$ is the largest real number $r$ which satisfies the following condition:
If a sequence $a_{1}, a_{2}, \ldots$ of positive integers fulfills the inequalities
$$
a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+r a_{n+1}}
$$
for every positive integer $n$, then there exists a positive integer $M$ such that $a_{n+2}... | [
"First, let us assume that $r>2$, and take a positive integer $a \\geq 1 /(r-2)$.\nThen, if we let $a_{n}=a+\\lfloor n / 2\\rfloor$ for $n=1,2, \\ldots$, the sequence $a_{n}$ satisfies the inequalities\n$$\n\\sqrt{a_{n}^{2}+r a_{n+1}} \\geq \\sqrt{a_{n}^{2}+r a_{n}} \\geq \\sqrt{a_{n}^{2}+\\left(2+\\frac{1}{a}\\rig... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2 | |
0e6g | Problem:
Janja se je odločila za ekološko vzrejo kokoši in gosk. Ograditi namerava parcelo v obliki pravokotnika s ploščino $1632~\mathrm{m}^2$. Goske in kokoši bo imela ločene, tako da bo parcelo po širini pregradila na pol (glej sliko). Za ograditev bo skupno potrebovala $198~\mathrm{m}$ ograde. Kolikšna je širina i... | [
"Solution:\n\nDolžino parcele označimo z $x$ in širino z $y$. Ploščina parcele (ploščina pravokotnika) je $x \\cdot y = 1632$.\n\nOgrado sestavljajo vse 4 stranice in prečna ograda: $2x + 3y = 198$.\n\nDobili smo sistem enačb, najlažje ga rešimo z zamenjalnim načinom. Iz prve enačbe izrazimo npr. $y = \\frac{1632}{... | Slovenia | Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | Length 51 and width 32, or length 48 and width 34 | |
0hba | For positive numbers $x$, $y$, $z$, satisfying equality $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3$, prove that the following inequality is true:
$$(x-1)(y-1)(z-1) \le \frac{1}{4}(xyz-1).$$
(Vadim Mitrofanov) | [
"From the description of the problem we have that $xy + yz + zx = 3xyz$. The given inequality is equivalent to\n$$\n(x-1)(y-1)(z-1) = (xyz-1) + (x+y+z-xy-yz-xz) = (xyz-1) + (x+y+z-3xyz) = x+y+z-1-2xyz \\le \\frac{1}{4}(xyz-1) \\Leftrightarrow \\\\\n x+y+z \\le \\frac{9}{4}xyz + \\frac{3}{4}.\n$$\n\nLet us get rid o... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0ey6 | Problem:
Prove that there exists a number divisible by $5^{1000}$ with no zero digit. | [
"Solution:\nWe first find a multiple of $5^{1000}$ which has no zeros in the last $1000$ digits. Suppose that we have a multiple $n \\cdot 5^{1000}$ whose last zero is in place $r$ (treating the last place as place $0$, the next to last as place $1$ and so on). Then $n(10^{r} + 1)$ has the same digits in places $0$... | Soviet Union | 1st ASU | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
05ap | Find all positive integers $n$ for which it is possible to write all positive integers from $1$ to $n$ in some order on a circle, each exactly once, such that every number is coprime with the number that is two positions away from it. | [
"Let $n$ be odd. Write the number $1$ at some point on the circle and, after every $\\frac{2}{n}$ full turns, write the next number by size (Fig. 45 shows the case for $n = 11$). In this way, all the numbers from $1$ to $n$ can be written in different points. Choosing any number on the circle and the number two pos... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | All positive integers n with n ≡ 0, 1, or 3 mod 4 (i.e., n not ≡ 2 mod 4). | |
0cgy | Consider an odd prime $p$ and a positive integer $N < 50p$. Let $a_1, a_2, \dots, a_N$ be a list of positive integers less than $p$ such that any specific value occurs at most $\frac{51}{100}N$ times and $a_1 + a_2 + \dots + a_N$ is not divisible by $p$. Prove that there exists a permutation $b_1, b_2, \dots, b_N$ of t... | [
"**Lemma.** Let $n$ be a positive integer and let $c_1, c_2, \\dots, c_n$ be a list of positive integers less than $p$ such that each specific value occurs at most $\\frac{1}{2}(n+1)$ times. Fix a residue $r \\neq c_1 + c_2 + \\dots + c_n \\pmod{p}$. Then there exists a permutation $d_1, d_2, \\dots, d_n$ of the $c... | Romania | Romanian Master of Mathematics | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
00al | Let $n \ge 3$ an integer. Find the number of arrangements $a_1, a_2, ..., a_n$ of $1, 2, ..., n$ around a circle, in clockwise direction, such that $|a_1 - a_2| + |a_2 - a_3| + \dots + |a_{n-1} - a_n| + |a_n - a_1| = 2n - 2$. | [
"First we clarify the meaning of the given equality. Let $a_1, a_2, ..., a_n$ be an arbitrary circular arrangement of $1, 2, ..., n$, $n \\ge 3$, in clockwise direction. The extremal numbers $1$ and $n$ separate the remaining numbers into two groups. For convenience denote them by $b_1, ..., b_k$ and $c_1, ..., c_l... | Argentina | Argentine National Olympiad 2016 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 2^(n-2) | |
0k1s | Problem:
Let $A$ be the number of unordered pairs of ordered pairs of integers between $1$ and $6$ inclusive, and let $B$ be the number of ordered pairs of unordered pairs of integers between $1$ and $6$ inclusive. (Repetitions are allowed in both ordered and unordered pairs.) Find $A - B$. | [
"Solution:\nThere are $6 \\cdot 6$ ordered pairs of integers between $1$ and $6$ inclusive and $21$ unordered pairs of integers ($\\binom{6}{2} = 15$ different pairs and $6$ doubles). Then, $A = \\binom{36}{2} + 36 = 666$ and $B = 21 \\cdot 21 = 441$. Therefore $A - B = 225$.\n\nFor general $n$, there are $n^{2}$ o... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 225 | |
0c18 | Let $\mathcal{F}$ be the set of continuous functions $f : \mathbb{R} \to \mathbb{R}$ which satisfy the condition
$$
e^{f(x)} + f(x) \geq x + 1,$$
for any real number $x$. Find the minimum value attained by the integral
$$
I(f) = \int_{0}^{e} f(x) \, dx,
$$
for $f$ lying in $\mathcal{F}$. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Functions"
] | null | proof and answer | 3/2 | |
0jgu | Problem:
Let $P(x)$ be a polynomial such that for all integers $x \geq 1$,
$$
P(x)=\sum_{n=1}^{x} n^{2012}
$$
(a) Find $P(-2)$.
(b) Find $P(1 / 2)$. | [
"Solution:\n(a) Consider the relation\n$$\nP(x)=P(x-1)+x^{2012} .\n$$\nIf $x$ is an integer greater than $1$, then (1) clearly holds. Therefore (1) holds for all real $x$ (two polynomials cannot have infinitely many common values unless they are the same). We can therefore use the recurrence $P(x-1)=P(x)-x^{2012}$ ... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | final answer only | P(-2) = -1; P(1/2) = 1/(2^2012) | |
0jq2 | Problem:
We are given $n$ identical cubes, each of size $1 \times 1 \times 1$. We arrange all of these $n$ cubes to produce one or more congruent rectangular solids, and let $B(n)$ be the number of ways to do this. For example, if $n=12$, then one arrangement is twelve $1 \times 1 \times 1$ cubes, another is one $3 \t... | [
"Solution:\n\nThe exact value of $B\\left(2015^{100}\\right)$ is $921,882,251,894,177$. Thus $m=14$.\n\nLet us estimate $B(n)$. First we note that the actual primes do not matter, just the exponents. Since $2015=5 \\cdot 13 \\cdot 31$, we need to find $B\\left(p^{100} q^{100} r^{100}\\right)$, where $p, q, r$ are d... | United States | BAMO | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | m=14 | |
0c9d | Problem:
a) Scrieți numărul $2021$ ca sumă de puteri distincte cu baza $(-2)$.
b) Arătați că numărul $2021$ nu se poate scrie ca sumă de puteri distincte cu baza $(-3)$. | [
"Solution:\n\na)\n$2021 = (-2)^{12} + (-2)^{11} + (-2)^{5} + (-2)^{2} + (-2)^{0}$\n\nb)\n$(-3)^{k} = M_{3}$, pentru orice $k \\geq 1$.\nO sumă de puteri distincte cu baza $(-3)$ ar putea avea ca termen pe $(-3)^{0}$ sau nu. Prin urmare, o sumă de puteri distincte cu baza $(-3)$ poate avea următoarele forme:\n$M_{3}... | Romania | Olimpiada Națională GAZETA MATEMATICĂ Etapa a III-a | [
"Number Theory > Modular Arithmetic",
"Number Theory > Other"
] | null | proof and answer | a) 2021 = (-2)^12 + (-2)^11 + (-2)^5 + (-2)^2 + (-2)^0. b) Impossible to write 2021 as a sum of distinct powers with base -3. | |
0l78 | From an unlimited supply of 1-cent coins, 10-cent coins, and 25-cent coins, Silas wants to find a collection of coins that has a total value of $N$ cents, where $N$ is a positive integer. He uses the so-called greedy algorithm, successively choosing the coin of greatest value that does not cause the value of his collec... | [] | United States | AIME II | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 610 | |
02gj | $ABCD$ is a cyclic quadrilateral and $M$ a point on the side $CD$ such that $ADM$ and $ABCM$ have the same area and the same perimeter. Show that two sides of $ABCD$ have the same length. | [
"\n\nSince the perimeter of $ADM$ and $ABCM$ are equal and share the side $AM$, $a + b + m = n + c \\iff n - m = a + b - c$. Since they also have the same area, the area of $ADM$ is half the area of $ABCD$. If $\\angle ABC = \\theta$, then $\\angle CDA = 180^\\circ - \\theta$ and\n$$\n\\beg... | Brazil | XXIV OBM | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
04j6 | Let $ABCD$ be a tetrahedron such that $\angle BAC = \angle CAD = \angle DAB = 90^\circ$, $|AD| = 2\sqrt{2}$ and $|AB| = |AC| = 3$ holds. Determine the radius of the inscribed sphere of the tetrahedron.
(Mea Bombardelli) | [
"MATHEMATICAL COMPETITIONS IN CROATIA IN 2015\nIf $a_1$ and $z_1$ are connected by bus, let us put $A' = \\{z_1, g, a_1, \\dots, a_n\\}$ and $Z' = \\{z_2, \\dots, z_m\\}$. Then $(A', Z')$ is a good pair and the number of elements of $A' \\cup Z'$ is greater than the number of elements of $A \\cup Z$, which contradi... | Croatia | First round – City competition | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area"
] | English | proof and answer | (3(√2 - 1))/2 | |
058d | There is a rectangular grid with 3 rows and $n$ columns on a blackboard.
a. Find the number of ways to write exactly one of numbers $1, 2, \ldots, 3n$ into each square in such a way that all the following conditions are met:
(1) Different squares contain different numbers.
(2) For each $i = 1, 2, \ldots, 3n - 1$, the ... | [
"(a) Note that all three squares in the leftmost and rightmost columns must be traversed consecutively because along with both corner squares of either of the columns one has to traverse the middle square of the same column and this cannot be done twice. In no other column can one traverse all three squares consecu... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | a) 0 if n is odd; if n is even, 3n · 2^(n/2).
b) 2^n. | |
0izw | Problem:
A triangle with side lengths $5,7,8$ is inscribed in a circle $C$. The diameters of $C$ parallel to the sides of lengths $5$ and $8$ divide $C$ into four sectors. What is the area of either of the two smaller ones? | [
"Solution:\n\nLet $\\triangle PQR$ have sides $p=7$, $q=5$, $r=8$. Of the four sectors determined by the diameters of $C$ that are parallel to $PQ$ and $PR$, two have angles equal to $P$ and the other two have angles equal to $\\pi - P$. We first find $P$ using the law of cosines:\n\n$$\n49 = 25 + 64 - 2(5)(8) \\co... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 49π/18 | |
036t | Problem:
Consider the set $A = \{1, 2, 3, 4, \ldots, 2^{n}\}$, $n \geq 2$. Find the number of the subsets $B$ of $A$, such that if the sum of two elements of $A$ is a power of $2$ then exactly one of them belongs to $B$. | [
"Solution:\n\n1. Let $B$ be a subset of $A$ having the given property. Since $1 + 3 = 2^{2}$, we have that exactly one of the numbers $1$ or $3$ belongs to $B$.\nIf $1 \\in B$ then $3 \\notin B$. We prove by induction that for any integer $t$, $0 \\leq t < 2^{n-2}$, the integers of the form $4t+1$ belong to $B$ and... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2^{n+1} | |
05q7 | Problem:
Pour tout entier naturel $n$, on note $S(n)$ la somme des chiffres de l'écriture décimale de $n$. On dit qu'un entier naturel $n$ est joli si $S(n)=S\left(n^{2}\right)$. Déterminer toutes les valeurs possibles de $S(n)$ pour les entiers jolis $n$. | [
"Solution:\n\nModulo $9$, $S(n)$ vérifie l'équation $x^{2}=x$. On vérifie en testant $x=0,1, \\ldots, 8$ que les seules solutions (modulo $9$) sont $0$ et $1$, donc $S(n)$ est nécessairement congru à $0$ ou $1$ modulo $9$.\n\nRéciproquement, on vérifie que si $n=10^{k}-1=99 \\cdots 9$ ($k$ fois) alors $n^{2}=99 \\c... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All nonnegative integers congruent to 0 or 1 modulo 9 | |
0ksa | Problem:
Let $ABCD$ be a rectangle inscribed in circle $\Gamma$, and let $P$ be a point on minor arc $AB$ of $\Gamma$. Suppose that $PA \cdot PB = 2$, $PC \cdot PD = 18$, and $PB \cdot PC = 9$. The area of rectangle $ABCD$ can be expressed as $\frac{a \sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integ... | [
"Solution:\nWe have\n$$\nPD \\cdot PA = \\frac{(PA \\cdot PB)(PD \\cdot PC)}{PB \\cdot PC} = \\frac{2 \\cdot 18}{9} = 4\n$$\nLet $\\alpha = \\angle DPC = 180^\\circ - \\angle APB$ and $\\beta = \\angle APD = \\angle BPC$. Note that $\\alpha + \\beta = 90^\\circ$. We have, letting $x = AB = CD$ and $y = AD = BC$,\n$... | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 21055 | |
0e1x | Problem:
Za cela števila $a, b, c$ in $d$ velja $a > b > c > d$ in
$$
(1-a)(1-b)(1-c)(1-d) = 10
$$
Katere vrednosti lahko zavzame izraz $a + b - c - d$? | [
"Solution:\n\nKer so števila $a, b, c$ in $d$ različna, so različna tudi števila $1-a, 1-b, 1-c$ in $1-d$. Ker je $10$ produkt le dveh praštevil, lahko $10$ zapišemo kot produkt štirih celih števil le, če sta dve izmed teh števil $1$ in $-1$. Preostali števili sta tedaj $-2$ in $5$ ali pa $2$ in $-5$.\n\nZaradi $a ... | Slovenia | 54. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 9 | |
0htl | Problem:
It's a week before Thanksgiving, and a family is trying to find their turkey. There are 5 boxes in a row, and the turkey is hiding in one of the 5 boxes.
Every day, the family is allowed to check one box to try to find the turkey, and every night, the turkey moves to a box right next to the box it was in. Fo... | [
"Solution:\n\nOne strategy is as follows:\n\nFirst assume the turkey starts in an even numbered box (box 2 or 4). On day 1, check box 2. If the turkey is not there, he must have been in box 4 to begin with, so tomorrow he will be in box 3 or box 5. On day 2, check box 3. If the turkey is not there, then he must cur... | United States | Berkeley Math Circle: Monthly Contest 8 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
02kf | Problem:
Na figura abaixo temos dois quadrados. O maior tem lado $a+b$ e o menor lado $a$.
Qual é a área da região em cinza?
A) $b$
B) $a+b$
C) $a^{2}+2 a b$
D) $b^{2}$
E) $2 a b+b^{2}$ | [
"Solution:\n\nSolução 1 - Usaremos que $(a+b)^{2}=a^{2}+2 a b+b^{2}$. Lembre que a área de um quadrado de lado $l$ é $l^{2}$. Note que a área da região cinza é a diferença entre as áreas do maior e do menor quadrado. O lado do maior é $a+b$, portanto sua área é $(a+b)^{2}$. Já o lado do menor é $a$, logo sua área é... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | E | |
0ifq | Problem:
A cuboctahedron is a polyhedron whose faces are squares and equilateral triangles such that two squares and two triangles alternate around each vertex, as shown.
What is the volume of a cuboctahedron of side length $1$? | [
"Solution: $5 \\sqrt{2} / 3$\nWe can construct a cube such that the vertices of the cuboctahedron are the midpoints of the edges of the cube.\n\nLet $s$ be the side length of this cube. Now, the cuboctahedron is obtained from the cube by cutting a tetrahedron from each corner. Each such tet... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes"
] | null | final answer only | 5 sqrt(2) / 3 | |
0dxs | Problem:
Dan je paralelogram $ABCD$. Naj bo $E$ razpolovišče daljice $CD$, $F$ razpolovišče $DA$ ter $G$ razpolovišče $AB$. Trikotniku $DFE$ očrtana krožnica se dotika daljice $AB$ v točki $G$. Dokaži, da je $|AB| = \sqrt{2} |AD|$. | [
"Solution:\n\n1. način\nOznačimo $\\angle BAD = \\alpha$. Potem je $\\angle EGB = \\alpha$ ter $\\angle ADE = \\pi - \\alpha$. Ker je štirikotnik $DEFG$ tetiven, je $\\angle FGE = \\alpha$. Zato je $\\angle FGA = \\pi - \\angle FGE - \\angle EGB = \\pi - 2\\alpha$ in $\\angle GFA = \\pi - \\angle GAF - \\angle FGA ... | Slovenia | 51. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | |AB| = sqrt(2) * |AD| | |
0kg2 | Let $A$, $B$, $C$, $D$ be four points such that no three are collinear and $D$ is not the orthocenter of triangle $ABC$. Let $P$, $Q$, $R$ be the orthocenters of $\triangle BCD$, $\triangle CAD$, $\triangle ABD$, respectively. Suppose that lines $AP$, $BQ$, $CR$ are pairwise distinct and are concurrent. Show that the f... | [
"\n\n*Proof.* Note that $\\overline{AQ} \\parallel \\overline{BP}$, as both are perpendicular to $\\overline{CD}$. Since lines $AP$ and $BQ$ are distinct, lines $AQ$ and $BP$ are distinct.\nBy symmetric reasoning, we get that $AQCPBR$ is a hexagon with opposite sides parallel and concurrent... | United States | USA TSTST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Concurrency and... | English | proof only | null | |
04qg | Find all positive integers $n$ such that
$$
\frac{n^{3n-2} - 3n + 1}{3n - 2}
$$
is an integer. | [
"Since $3n-2 \\mid n^{3n-2}-3n+1$, it follows that $3n-2 \\mid n^{3n-2}-1$, hence $n^{3n-2} \\equiv 1 \\pmod{3n-2}$.\n\nNote that $n$ must be odd (otherwise the even $3n-2$ would divide the odd $n^{3n-2}-3n+1$, which is impossible).\n\nLet $p > 2$ be the smallest prime factor of the odd $3n-2$. Obviously $p \\nmid ... | Croatia | Croatian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 1 | |
0eol | When $a$ is increased by $20\%$ and $b$ is decreased by $20\%$ the resulting values are equal. The ratio of $a$ to $b$ is
(A) $\frac{1}{2}$ (B) $1$ (C) $\frac{2}{3}$ (D) $\frac{3}{4}$ (E) $\frac{4}{3}$ | [
"$120\\%$ of $a$ is $\\frac{6}{5}a$ while $80\\%$ of $b$ is $\\frac{4}{5}b$.\nWe thus have $\\frac{6}{5}a = \\frac{4}{5}b$ from which it follows that $\\frac{a}{b} = \\frac{4}{5} \\times \\frac{5}{6} = \\frac{2}{3}$"
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | C | |
05lz | Problem:
Soit $ABC$ un triangle tel que $\widehat{A}=40^{\circ}$ et $\widehat{B}=60^{\circ}$. Soient $D$ et $E$ des points de $[AC]$ et $[AB]$ tels que $\widehat{CBD}=40^{\circ}$ et $\widehat{ECB}=70^{\circ}$. On note $F$ l'intersection de $(BD)$ et $(CE)$. Montrer que $(AF) \perp (BC)$. | [
"Solution:\n\n\n\nOn a $\\widehat{BFC}=180^{\\circ}-\\widehat{CBF}-\\widehat{FCB}=180^{\\circ}-40^{\\circ}-70^{\\circ}=70^{\\circ}=\\widehat{FCB}$ donc $BCF$ est isocèle en $B$. Il vient $BC=BF$.\n\nSoit $M$ le point d'intersection entre la bissectrice de $\\widehat{CBF}$ et $(AC)$. On a $\... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0l6p | Problem:
Let $\triangle ABC$ be an acute triangle with orthocenter $H$. Points $E$ and $F$ are on segments $\overline{AC}$ and $\overline{AB}$, respectively, such that $\angle EHF = 90^\circ$. Let $X$ be the foot of the altitude from $H$ to $\overline{EF}$. Prove that $\angle BXC = 90^\circ$. | [
"Solution:\n\n\n\nWe use $\\angle$ to denote directed angles. Let $Y$ and $Z$ be the feet of the altitudes from $B$ and $C$ to $AC$ and $AB$, respectively. Then $\\angle HZF = \\angle HXF = 90^\\circ$, so $HZFX$ is cyclic. Similarly, $HYEX$ is cyclic. Therefore,\n\n$$\n\\angle BYX = \\angle... | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Confi... | null | proof only | null | |
0i7k | Problem:
You have $2003$ switches, numbered from $1$ to $2003$, arranged in a circle. Initially, each switch is either ON or OFF, and all configurations of switches are equally likely. You perform the following operation: for each switch $S$, if the two switches next to $S$ were initially in the same position, then you... | [
"Solution:\n$1 / 2^{2002}$\n\nThere are $2^{2003}$ equally likely starting configurations. All switches end up ON if and only if switches $1,3,5,7, \\ldots, 2003,2,4, \\ldots, 2002$—i.e., all $2003$ of them—were initially in the same position. This initial position can be ON or OFF, so this situation occurs with pr... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 1/2^{2002} | |
04wu | Find out the number of sequences $(a_n)_{n=1}^{\infty}$ of integer numbers, which satisfy
$$
a_n \neq -1 \quad \text{and} \quad a_{n+2} = \frac{a_n + 2006}{a_{n+1} + 1}
$$
for every positive integer $n$. | [
"Every sequence satisfying given conditions is determined by first two terms. Thus we are looking for integer pairs $(a_1, a_2)$, for which all the other terms are integers. Writing out the formula for several small values of $n$ and multiplying it we obtain\n$$\n\\begin{align*}\na_3(a_2 + 1) &= a_1 + 2006, \\\\\na... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko 2006 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 14 | |
0703 | Problem:
Show that given any integer $0 < n \leq 2^{1000000}$ we can find a set $S$ of at most $1100000$ positive integers such that $S$ includes $1$ and $n$ and every element of $S$ except $1$ is a sum of two (possibly equal) smaller elements of $S$. | [] | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0943 | Problem:
Initially, two positive integers $a$ and $b$ with $a \neq b$ are written on a blackboard. At each step, Andrea picks two numbers $x$ and $y$ on the blackboard with $x \neq y$ and writes the number
$$
\operatorname{gcd}(x, y)+\operatorname{lcm}(x, y)
$$
on the blackboard as well. Let $n$ be a positive integer.... | [
"Solution:\n\nWrite $a = d x$, $b = d y$ with $d = \\operatorname{gcd}(a, b)$. Then\n$$\n\\operatorname{gcd}(a, b) + \\operatorname{lcm}(a, b) = d + d x y = d(1 + x y) = d \\cdot (\\operatorname{gcd}(x, y) + \\operatorname{lcm}(x, y))\n$$\nand $1 + x y$ is coprime with both $x$ and $y$. This means that without loss... | Middle European Mathematical Olympiad (MEMO) | MEMO Individual Competition | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0c7n | Given a positive integer $k$, a loop of length $k$ in a graph is a list $v_1, e_1, v_2, e_2, \dots, v_k, e_k$, where the $v_i$ are (not necessarily distinct) vertices, the $e_i$ are (not necessarily distinct) edges, and each $e_i$ joins $v_i$ and $v_{i+1}$ (indices are reduced modulo $k$); the loop traces an edge $e$ i... | [
"Let $G$ be a connected graph with vertex set $V$ and edge set $E$; $G$ may have loops and/or multiple edges. The idea is to expand at most $|V|-1$ suitable edges to pairs of edges to make $G$ into a graph $\\hat{G}$ on $V$ each vertex of which has an even degree. A maximal loop in $\\hat{G}$ tracing each edge at m... | Romania | IMAR Mathematical Competition | [
"Discrete Mathematics > Graph Theory"
] | English | proof only | null | |
0b5p | Given any $n$ positive integers, and a sequence of $2^n$ integers (with terms among them), prove there exists a subsequence made of consecutive terms, such that the product of its terms is a perfect square.
Also show that we cannot replace $2^n$ with any lower value (therefore $2^n$ is the threshold value for this prop... | [
"Since the integers could well be distinct primes, this is equivalent to proving the stricter problem of, given a finite alphabet $A$ of $n$ letters $a_1, a_2, \\dots, a_n$, and a word $w$ of length $2^n$ on this alphabet, to show it contains a nonempty contiguous subword $x$ ($w = \\overline{uxv}$, where $u, v$ co... | Romania | Local Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Linear Algebra > Vectors"
] | English | proof only | null | |
0aap | In the right triangle $\lg \frac{a-b}{2} = \frac{1}{2}(\lg a + \lg b - \lg 2)$, where $a > b$, $a, b$ are the sides of the triangle. Find the angles of the triangle. | [
"The given equation is equivalent to $\\lg \\frac{a-b}{2} = \\lg \\sqrt{\\frac{ab}{2}}$. Therefore\n$$\n\\frac{a-b}{2} = \\sqrt{\\frac{ab}{2}} \\Leftrightarrow \\frac{a^2-2ab+b^2}{4} = \\frac{ab}{2} \\Leftrightarrow a^2-4ab+b^2=0.\n$$\nSince $a > b > 0$, by dividing with $b^2$, we have $(\\frac{a}{b})^2 - 4\\frac{a... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 15°, 75°, 90° | |
01ee | Let $\omega_1$ and $\omega_2$ be two circles with centers $O_1$ and $O_2$, respectively, with $O_2$ lying on $\omega_1$. Let $A$ be a common point of $\omega_1$ and $\omega_2$. A line through $A$ intersects $\omega_1$ in $B \neq A$ and $\omega_2$ in $C \neq A$ such that $A$ lies between $B$ and $C$. The ray $O_2O_1$ in... | [
"Let $F$ be the second intersection of $\\omega_1$ and $\\omega_2$. Notice that\n$$\n\\angle BFO_2 = 180^\\circ - \\angle BAO_2 = \\angle CAO_2 = \\angle O_2CA\n$$\nand since $AO_2 = EO_2$ that $\\angle FBO_2 = \\angle O_2BA$. It follows that $F$ is the reflection of $C$ over $BO_2$, thus it suffices to prove that ... | Baltic Way | Baltic Way shortlist | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English | proof only | null | |
0bqs | Prove that $2n^2 + 27n + 91$ is a perfect square for infinitely many $n \in \mathbb{N}$. | [
"Let $2n^2 + 27n + 91 = k^2$ for some $k \\in \\mathbb{N}$. Rearranging, we get:\n\n$$\n2n^2 + 27n + 91 - k^2 = 0\n$$\n\nThis is a quadratic in $n$:\n\n$$\n2n^2 + 27n + (91 - k^2) = 0\n$$\n\nThe discriminant must be a perfect square for $n$ to be integer:\n\n$$\n\\Delta = 27^2 - 4 \\cdot 2 \\cdot (91 - k^2) = 729 -... | Romania | 67th NMO Shortlisted Problems | [
"Number Theory > Diophantine Equations > Pell's equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
00fx | Let $A B C$ be a triangle. Let $M$ and $N$ be the points in which the median and angle bisector, respectively, at $A$ meet the side $B C$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $N A$ meets $M A$ and $B A$, respectively, and $O$ be the point in which the perpendicular at $P$ to $B A$ meets $... | [
"Let $A N$ meet the circumcircle of $A B C$ at point $K$, the midpoint of arc $B C$ that does not contain $A$.\n\n\n\nThe orthogonal projection of $K$ onto side $B C$ is $M$. Let $R$ and $S$ be the orthogonal projections of $K$ onto lines $A B$ and $A C$, respectively. Points $R, M$, and $S... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0h1x | Three circles touch externally, $A_1A_2$, $B_1B_2$, $C_1C_2$ are their diameters that have the same directions. Prove that $A_1B_2$, $B_1C_2$, $C_1A_2$ are concurrent. | [
"Let circles with diameters $B_1B_2$, $C_1C_2$ touch at point $A$. By analogy we define $B$, $C$ (fig. 22).\n\nSince touching point is a center of homothety that transforms one circle into another, then $A = B_1C_2 \\cap B_2C_1$ and so on. Let $D = A_2C_1 \\cap A_1B_2$. Points $A$, $B$, $C$, $D$ are cyclic, because... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0et3 | Determine all pairs $(m, n)$ of non-negative integers that satisfy the equation
$$
20^m - 10m^2 + 1 = 19^n.
$$ | [
"Let the pair $(m, n)$ satisfy the equation. $m = 0$ implies that $2 = 19^n$, an impossibility. So we must have $m > 0$. Taking both sides modulo $10$ gives $1 \\equiv (-1)^n \\pmod{10}$, implying that $n$ must be even. Taking both sides modulo $20$ gives $-10m^2 + 1 \\equiv (-1)^n \\pmod{20}$, so that $2 \\mid m^2... | South Africa | The South African Mathematical Olympiad, Third Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (2, 2) | |
0d7g | A lock has 16 keys arranged in a $4 \times 4$ array, each key oriented either horizontally or vertically. In order to open it, all the keys must be vertically oriented. When a key is switched to another position, all the other keys in the same row and column automatically switch their positions too. Show that no matter... | [
"The problem is solved if there is a way to change the orientation of any specified single key, without changing any of the others. This is equivalent to finding a way to switch the chosen key an odd number of times, while switching all other keys an even number of times.\nThis can be done by switching all keys on ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms"
] | English | proof only | null | |
09r3 | Problem:
Bepaal alle paren $(x, y)$ van positieve gehele getallen die voldoen aan
$$
x + y + 1 \mid 2 x y \quad \text{en} \quad x + y - 1 \mid x^{2} + y^{2} - 1
$$ | [
"Solution:\nEr geldt\n$$\n\\left(x^{2} + y^{2} - 1\\right) - (x + y + 1)(x + y - 1) = \\left(x^{2} + y^{2} - 1\\right) - \\left(x^{2} + y^{2} + 2 x y - 1\\right) = -2 x y.\n$$\nOmdat $x + y - 1$ een deler is van $x^{2} + y^{2} - 1$ en natuurlijk ook van $(x + y + 1)(x + y - 1)$, zien we dat $x + y - 1$ een deler is... | Netherlands | Dutch TST | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All pairs with the numbers differing by one: (x, x+1) for x ≥ 1 and (x, x−1) for x ≥ 2. | |
08yi | Some of the vertices of a regular $120$-gon are labeled by $*$. What is the maximal possible number of vertices labeled with $*$ if the following condition must be satisfied?
There exists no isosceles triangle with the angle at the top vertex of $18^\circ$ formed by $*$-labeled vertices of the $120$-gon. | [
"$78$\n\nLabel the vertices of the regular $120$-gon as $A_1, A_2, \\dots, A_{120}$ clockwise. Form six subsets of the set of all the vertices $A_i$, $i = 1, 2, \\dots, 120$ by putting the vertex $A_i$ into the $r$-th group for $r = 1, 2, \\dots, 6$ if the remainder obtained is $r-1$ when $i$ is divided by $6$. Not... | Japan | 2019 Japan Mathematical Olympiad First Stage | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 78 | |
0kyy | How many ordered pairs of integers $(m, n)$ satisfy $\sqrt{n^2 - 49} = m$?
(A) 1 (B) 2 (C) 3 (D) 4 (E) infinitely many | [
"Notice that $m \\ge 0$, and if $(m, n)$ is a solution, then so is $(m, -n)$. Assume $n \\ge 0$. Squaring both sides of the given equation gives $n^2 - 49 = m^2$, so $n^2 - m^2 = (n - m)(n + m) = 49$. Because $n - m$ and $n + m$ are positive integers, either $n - m = n + m = 7$, or $n - m = 1$ and $n + m = 49$. The... | United States | AMC 10 A | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
0fk2 | Problem:
Sea $ABC$ un triángulo y $D$, $E$ y $F$ puntos situados en los segmentos $AC$, $BA$ y $CB$ respectivamente, de forma que los segmentos $AF$, $BD$, $CE$ concurren en un punto $P$ interior al triángulo. Sabemos que $BP = 6$, $PD = 6$, $PC = 9$, $PE = 3$ y $AF = 20$. Hallar el área del triángulo $ABC$. | [
"Solution:\n\n\n\nUtilizamos varias veces que las áreas de dos triángulos de la misma base son proporcionales a las alturas y que las de dos triángulos de la misma altura son proporcionales a las bases.\n\nLas letras $x, y, z, \\ldots$ de la figura de la izquierda indican las áreas de los t... | Spain | Spanish Mathematical Olympiad - Local Stage | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 108 | |
00rt | Find all the functions $f: \mathbb{N} \to \mathbb{N}$ such that:
$$
n + f(m) \mid f(n) + n f(m) \quad (1)
$$
for any $m, n \in \mathbb{N}$ | [
"We will consider 2 cases, whether the range of the functions is infinite or finite or in other words the function takes infinite or finite values.\n\n**Case 1.** The function has an infinite range. Let's fix a random natural number $n$ and let $m$ be any natural number. Then using (1) we have\n$$\nn + f(m) \\mid f... | Balkan Mathematical Olympiad | BMO 2017 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | f(n) = n^2 for all natural n, or f(n) = 1 for all natural n | |
0l6i | Problem:
Sophie is at $(0,0)$ on a coordinate grid and would like to get to $(3,3)$. If Sophie is at $(x,y)$, in a single step she can move to one of $(x + 1, y)$, $(x, y + 1)$, $(x - 1, y + 1)$, or $(x + 1, y - 1)$. She cannot revisit any points along her path, and neither her $x$-coordinate nor her $y$-coordinate ca... | [
"Solution:\n\nLet a lateral move refer to one which is either up or right. Then the lateral moves are the only ones which increase Kelvin's sum of coordinates by $1$, while all other moves do not change the sum, so Kelvin must make $6$ of them, one to increase this sum from $i$ to $i + 1$ for each $i \\in [0, 5]$.\... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2304 | |
0ahg | Let $a$, $b$ and $c$ be positive real numbers for which the equality $a+b+c+2=abc$ holds. Prove that the inequality $\frac{a}{b+1} + \frac{b}{c+1} + \frac{c}{a+1} \ge 2$ holds. When does equality hold? | [
"At first we notice that the equality\n$$\n\\begin{aligned} (a+1)(b+1) + (a+1)(c+1) + (b+1)(c+1) &= a+b+c+(a+b+c+2)+ab+ac+bc+1 \\\\ &= a+b+c+abc+ab+ac+bc+1 = (a+1)(b+1)(c+1) \\end{aligned}\n$$\nholds. Now from the inequality between the arithmetic and geometric mean we get:\n$$\n\\begin{aligned} \\frac{a}{b+1} + \\... | North Macedonia | XVI Junior Macedonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds if and only if a = b = c = 2. | |
01lq | Given natural number $a > 1$ and different odd prime numbers $p_1, \ldots, p_n$, with
$$
a^{p_1} \equiv 1 \pmod{p_2},\ a^{p_2} \equiv 1 \pmod{p_3},\ \ldots,\ a^{p_n} \equiv 1 \pmod{p_1}.
$$
Prove that
a) $(a-1)$ is divisible by $p_i$ for some $i = 1, \ldots, n$.
b) Can $(a-1)$ be divisible by $p_i$ for exactly one $i... | [
"a) (Solution of A.Ivanin, O.Volod'ko, A.Zhuk.) (Further we put $p_{n+1} = p_1$.)\nLet $p_k$ be the greatest number among all $p_i$ ($i = 1, 2, \\ldots, n$). The condition $a^{p_i} \\equiv 1 \\pmod{p_{i+1}}$ implies that $a$ and $p_i$ are relatively prime for all $i$. Then $a^{p_{k+1}-1} \\equiv 1 \\pmod{p_{k+1}}$ ... | Belarus | Selection and Training Session | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Th... | English | proof and answer | a) There exists an index i such that p_i divides a−1. b) Yes; it is possible for a−1 to be divisible by exactly one of the primes. | |
081z | Problem:
È noto che i Marziani maschi dicono sempre la verità, mentre le Marziane mentono sempre; al contrario i Venusiani maschi mentono e le Venusiane dicono sempre il vero. Atterra un'astronave piena di Marziani e Venusiani; all'ufficio immigrazione due degli occupanti, Ark e Bark, fanno le seguenti dichiarazioni:
... | [
"Solution:\n\nLa risposta è $\\mathbf{( E )}$. Se Ark dice il vero, allora Bark è un venusiano maschio, quindi mente, ma allora Ark deve essere un venusiano maschio, assurdo. Dunque Ark mente, quindi Bark è una femmina marziana, che mente, e dunque Ark è effettivamente un venusiano maschio (che mente). Quindi tutto... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Logic"
] | null | MCQ | E | |
08bb | Problem:
Dato il triangolo $ABC$ rettangolo in $A$ costruiamo sull'ipotenusa il quadrato $BCDE$ (con $D, E$ dalla parte opposta di $A$ rispetto a $BC$). Sapendo che le aree dei triangoli $ABE$ e $ACD$ valgono rispettivamente $6~\mathrm{m}^2$ e $27~\mathrm{m}^2$, quanto vale l'area del triangolo $ABC$?
(A) $3 \sqrt{2}... | [
"Solution:\n\nMostriamo che il prodotto delle due aree fornite è uguale al quadrato dell'area del triangolo $ABC$ cercata. Siano $P$ il piede dell'altezza uscente dal vertice $A$ del triangolo $ACD$, $Q$ il piede dell'altezza uscente dal vertice $A$ del triangolo $ABE$, $H$ il piede dell'altezza uscente dal vertice... | Italy | Progetto Olimpiadi della Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | D | |
053l | Denote the number of all positive divisors of a positive integer $n$ by $\delta(n)$ and the sum of all positive divisors of a positive integer $n$ by $\sigma(n)$. Prove that $\sigma(n) > \frac{\delta(n)^2}{2}$. | [
"Let $a_1, a_2, \\dots, a_{\\delta(n)}$ be the positive divisors of $n$ in increasing order. We obtain\n$$\n\\sigma(n) = a_1 + a_2 + \\dots + a_{\\delta(n)} \\ge 1 + 2 + \\dots + \\delta(n) = \\frac{\\delta(n) \\cdot (\\delta(n)+1)}{2} > \\frac{\\delta(n)^2}{2}.\n$$",
"For every positive divisor $d$ of a positive... | Estonia | Estonian Math Competitions | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0fb9 | Problem:
Demuestra que no existen enteros $a, b, c, d$ tales que el polinomio $P(x)=a x^{3}+b x^{2}+c x+d$ ($a \neq 0$) cumpla que $P(4)=1$ y $P(7)=2$. | [
"Solution:\n\nSupongamos que tal polinomio existe.\nPor el teorema del resto $P(x) = (x-4) Q(x) + 1$, siendo $Q(x)$ un polinomio de grado dos con coeficientes enteros.\nEntonces $P(7) = 2 = (7-4) Q(7) + 1 \\Rightarrow Q(7) = \\frac{1}{3}$ que no es entero, en contra de la hipótesis."
] | Spain | null | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0gm7 | A positive real number $a$ and an angle with vertex $A$ are given. Show that all the circles which pass through $A$ and intersect the sides of the given angle at points $B$, $C$ with $|AB| + |AC| = a$ have a second common point in addition to $A$. | [] | Turkey | VIIth NATIONAL MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
06ah | The positive real numbers $x, y, z$ are such that $x + y + z = 4$ and
$x, y, z \in [0,2]$. Find the minimal value of the algebraic expression:
$$
A = \sqrt{2+x} + \sqrt{2+y} + \sqrt{2+z} + \sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x}.
$$ | [
"We will prove first that $\\sqrt{2+x} + \\sqrt{y+z} = \\sqrt{2+x} + \\sqrt{4-x} \\ge 2 + \\sqrt{2}$.\nIndeed, this is equivalent to\n$$\n2 + x + 4 - x + \\sqrt{(2 + x)(4 - x)} \\ge 4 + 2 + 2\\sqrt{2} \\Leftrightarrow \\\\ x(2 - x) \\ge 0,\n$$\nwhich is true, since $x \\in [0,2]$.\nSimilarly we have $\\sqrt{2+y} + ... | Greece | Selection examinations | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 6 + 3*sqrt(2) | |
09kd | Let $X$ be a finite set of positive integers such that no two subsets have the same sum. Prove that the sum of the inverses of the elements of $X$ satisfy
$$
\sum_{x \in X} \frac{1}{x} < \frac{5}{2}.
$$
For example, for $X = \{1, 2, 5\}$ the sums $1, 2, 5, 1+2, 1+5, 2+5, 1+2+5$ are all distinct and $1/1 + 1/2 + 1/5 = 1... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0ex6 | Problem:
Find all possible integer solutions for $\sqrt{x + \sqrt{x + \ldots + \sqrt{x}}} = y$, where there are 1998 square roots. | [
"Solution:\nLet $s_{1} = \\sqrt{x}$, $s_{2} = \\sqrt{x + s_{1}}$, $s_{3} = \\sqrt{x + s_{2}}$ and so on. So the equation given is $y = s_{1998}$.\n\nWe show first that all $s_{n}$ must be integral for $1 \\leq n \\leq 1998$. $y$ is integral, so $s_{1998}$ is integral. Now suppose $s_{n}$ is integral. Then $s_{n-1} ... | Soviet Union | 4th ASU | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x = 0, y = 0 | |
06qf | Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all real $x, y$ the identity
$$
f(x f(x+y))=f(y f(x))+x^{2} .
$$ | [
"It is no hard to see that the two functions given by $f(x)=x$ and $f(x)=-x$ for all real $x$ respectively solve the functional equation. In the sequel, we prove that there are no further solutions.\nLet $f$ be a function satisfying the given equation. It is clear that $f$ cannot be a constant. Let us first show th... | IMO | IMO Problem Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x and f(x) = -x | |
08cz | Problem:
Nel villaggio di Asip tutti gli abitanti appartengono ad uno dei due gruppi seguenti: i Cavalieri, che dicono sempre la verità, e i Furfanti, che mentono sempre. Oggi, per il censimento annuale, è arrivato dalla capitale il Gran Notaio e tutti gli abitanti si mettono in fila indiana davanti a lui. Ciascuno di... | [
"Solution:\n\nLa risposta è (E). Come prima cosa osserviamo che il primo della fila, davanti al quale ci sono $0$ persone, è necessariamente un Cavaliere, perché dice per forza la verità. Dunque ci sono $999$ abitanti ad Asip e almeno uno di loro è un Cavaliere. Concentriamoci sull'affermazione riguardante la parit... | Italy | GARA di FEBBRAIO | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | MCQ | E | |
0azq | Problem:
In $\triangle ABC$, the length of $AB$ is $12$ and its incircle $O$ has radius $4$. Let $D$ be the point of tangency of circle $O$ with $AB$. If $AD : AB = 1 : 3$, find the area of $\triangle ABC$. | [
"Solution:\n\nLet $AB = 12$, $r = 4$ (incircle radius), and $AD : AB = 1 : 3$ so $AD = 4$, $DB = 8$.\n\nLet $AC = b$, $BC = c$.\n\nLet $s$ be the semiperimeter of $\\triangle ABC$.\n\nRecall that the incircle touches $AB$ at $D$, and $AD = s - a$, $DB = s - b$, where $a = BC$, $b = AC$, $c = AB$.\n\nBut here $AB = ... | Philippines | 21st PMO Area Stage | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | final answer only | 96 | |
026n | Problem:
Menor inteiro - Sejam $p$ e $q$ inteiros positivos tais que $\frac{5}{8}<\frac{p}{q}<\frac{7}{8}$. Qual é o menor valor de $p$ para que $p+q=2005$ ? | [
"Solution:\n\n$$\n\\frac{5}{8}<\\frac{p}{2005-p}<\\frac{7}{8}\n$$\ndo qual segue que\n$$\n5(2005-p)<8p \\quad \\text{e} \\quad 8p<7(2005-p)\n$$\nLogo,\n$$\n\\frac{5 \\times 2005}{13}<p<\\frac{7 \\times 2005}{15} \\Rightarrow 771,15<p<935,66\n$$\nLogo 772 é o menor valor de $p$ que satisfaz as condições do problema.... | Brazil | Nível 3 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 772 |
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