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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0avy | Problem:
In parallelogram $ABCD$, $AB = 1$, $BC = 4$, and $\angle ABC = 60^\circ$. Suppose that $AC$ is extended from $A$ to a point $E$ beyond $C$ so that triangle $ADE$ has the same area as the parallelogram. Find the length of $DE$. | [] | Philippines | 19th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 2√3 | |
057p | Find the largest remainder that can be left over when dividing the number $2019$ by a three-digit natural number. | [
"If $673 < m < 1000$ then dividing $2019$ by $m$ gives quotient $2$ and remainder $2019 - 2m$. Obviously the remainder increases when $m$ decreases. Thus in the case $m = 674$ we obtain the largest remainder $671$.\n\nDividing $2019$ by $673$ gives remainder $0$. Dividing $2019$ by $672$ or any smaller number gives... | Estonia | Open Contests | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 671 | |
06ou | In triangle $A B C$, let $J$ be the centre of the excircle tangent to side $B C$ at $A_{1}$ and to the extensions of sides $A C$ and $A B$ at $B_{1}$ and $C_{1}$, respectively. Suppose that the lines $A_{1} B_{1}$ and $A B$ are perpendicular and intersect at $D$. Let $E$ be the foot of the perpendicular from $C_{1}$ to... | [
"Let $K$ be the intersection point of lines $J C$ and $A_{1} B_{1}$. Obviously $J C \\perp A_{1} B_{1}$ and since $A_{1} B_{1} \\perp A B$, the lines $J K$ and $C_{1} D$ are parallel and equal. From the right triangle $B_{1} C J$ we obtain $J C_{1}^{2}=J B_{1}^{2}=J C \\cdot J K=J C \\cdot C_{1} D$ from which we in... | IMO | IMO 2006 Shortlisted Problems | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"G... | English | proof and answer | ∠BEA1 = 90°, ∠AEB1 = 90° | |
07nz | Find the smallest positive integer $N$ for which the equation
$$(x^2 - 1)(y^2 - 1)(z^2 - 1) = N$$
has at least two different integer solutions $(x, y, z)$ satisfying $1 < x \le y \le z$. | [
"Let us first determine the smallest possible $N$ for which we find two solutions of the form $(2, y, z)$. If we let $M = N/3$, this amounts to finding the smallest possible positive integer $M$ for which the equation $(y^2 - 1)(z^2 - 1) = M$ has at least two solutions with $2 \\le y \\le z$. This is solved in prob... | Ireland | Ireland | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1080 | |
0jl6 | Problem:
$$
\sum_{k=0}^{100}\left\lfloor\frac{2^{100}}{2^{50}+2^{k}}\right\rfloor
$$
(Here, if $x$ is a real number, then $\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$.) | [
"Solution:\n\nLet $a_{k}=\\frac{2^{100}}{2^{50}+2^{k}}$. Notice that, for $k=0,1, \\ldots, 49$,\n\n$$\na_{k}+a_{100-k}=\\frac{2^{100}}{2^{50}+2^{k}}+\\frac{2^{100}}{2^{50}+2^{100-k}}=\\frac{2^{100}}{2^{50}+2^{k}}+\\frac{2^{50+k}}{2^{k}+2^{50}}=2^{50}\n$$\n\nIt is clear that for $k=0,1, \\ldots, 49$, $a_{k}, a_{100-... | United States | HMMT 2014 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 101*2^49-50 | |
0glo | Let $p$ be a prime. Show that $\sqrt[3]{p} + \sqrt[3]{p^5}$ is irrational. | [
"Let $r = \\sqrt[3]{p} + \\sqrt[3]{p^5}$. Then,\n$$\nr^3 = (\\sqrt[3]{p} + \\sqrt[3]{p^5})^3 = p + p^5 + 3p^2(\\sqrt[3]{p} + \\sqrt[3]{p^5}) = p + p^5 + 3p^2 r.\n$$\nHence $r$ is a root of the polynomial $x^3 - 3p^2x - p^5 - p$. Assume to the contrary that $r$ is rational. By the rational root theorem we have $r$ i... | Thailand | The 14th Thailand Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
03e6 | Let $x_0, x_1, \dots$ be a sequence of real numbers such that $x_0 = 1$ and $x_{n+1} = \sin(x_n) + \frac{\pi}{2} - 1$ for all $n \ge 0$. Show that the sequence converges and find its limit.
(Kristyan Vasilev) | [
"We will first prove that the sequence is strictly increasing. Note that for each $n$ we have $x_n \\le \\frac{\\pi}{2}$, since $\\sin x \\le 1$ for each $x \\in \\mathbb{R}$. Besides we have that the function $f(x) = \\sin x - x$ is decreasing for $x \\in \\mathbb{R}$, because $f'(x) = \\cos x - 1 \\le 0$ for each... | Bulgaria | Autumn tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | π/2 | |
04z0 | Let $D$ be the midpoint of side $BC$ of triangle $ABC$. Prove that the intersection point of medians of triangle $ABD$ and that of triangle $ACD$ are equidistant from line $AD$. (Grade 11.) | [
"Triangles $ABD$ and $ACD$ have equal area since $|BD| = |CD|$ and the altitudes drawn from $A$ coincide (Fig. 11). As these triangles have a common side $AD$, also the altitudes drawn from vertices $B$ and $C$, respectively, must be equal. Thus $B$ and $C$ are equidistant from line $AD$. Since the point\n\n![](att... | Estonia | Estonija 2010 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
070q | Problem:
Find all positive integers $n < 1000$ such that the cube of the sum of the digits of $n$ equals $n^{2}$. | [
"Solution:\n\n$n < 1000$, so the sum of the digits is at most $27$, so $n^{2}$ is a cube not exceeding $27^{3}$. So we are looking for $m^{3}$ which is also a square. That implies $m$ is a square. So the only possibilities are $m = 1, 4, 9, 16, 25$. Giving $n = 1, 8, 27, 64, 125$. The corresponding cubes of the dig... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Other"
] | null | proof and answer | 1, 27 | |
0ame | Problem:
For what values of $k$ will the system below have no solution?
$$
\begin{aligned}
(k-3)x + 2y &= k^2 - 1 \\
x + \left(\frac{k-4}{3}\right)y &= 0
\end{aligned}
$$ | [] | Philippines | Area Stage | [
"Algebra > Linear Algebra > Determinants"
] | null | proof and answer | k = 6 | |
0j30 | Problem:
Let $a_{1}$, $a_{2}$, and $a_{3}$ be nonzero complex numbers with non-negative real and imaginary parts. Find the minimum possible value of
$$
\frac{\left|a_{1}+a_{2}+a_{3}\right|}{\sqrt[3]{\left|a_{1} a_{2} a_{3}\right|}}
$$ | [
"Solution:\nAnswer: $\\sqrt{3} \\sqrt[3]{2}$\n\nWrite $a_{1}$ in its polar form $r e^{i \\theta}$ where $0 \\leq \\theta \\leq \\frac{\\pi}{2}$. Suppose $a_{2}, a_{3}$ and $r$ are fixed so that the denominator is constant. Write $a_{2}+a_{3}$ as $s e^{i \\phi}$. Since $a_{2}$ and $a_{3}$ have non-negative real and ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | sqrt(3)*2^(1/3) | |
0cp9 | Two runners started simultaneously from one point on the street. Each of them ran along the street to the stadium, and then three laps around the stadium. Each runner ran at a constant speed along the whole distance, and the first runner had overtaken the second runner twice. Prove that the speed of the first runner is... | [
"Первый мог обогнать второго только на кольцевой дорожке стадиона. Так как он вбежал на стадион первым, на своём первом круге он обогнать второго не мог. Стало быть, обгоны случились, когда первый бежал по стадиону свои второй и третий круги. Пока первый бежал эти два круга, он обогнал второго по крайней мере на кр... | Russia | Russian Mathematical Olympiad | [
"Math Word Problems"
] | English, Russian | proof only | null | |
03qn | Find all polynomials $P(x)$ with real coefficients, which satisfy the equation
$$
P(a-b) + P(b-c) + P(c-a) = 2P(a+b+c)
$$
for all real numbers $a, b, c$ such that $ab + bc + ca = 0$. | [
"Let $P(x)$ satisfy the given equation.\n$$\n\\text{If } a = b = c = 0, \\text{ then } P(0) = 0.\n$$\n$$\n\\text{If } b = c = 0, \\text{ then } P(-a) = P(a) \\text{ for all real } a.\n$$\nHence $P(x)$ is even. Without loss of generality, we may assume that\n$$\nP(x) = a_n x^{2n} + \\cdots + a_1 x^2,\\ a_n \\neq 0.\... | China | International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | P(x) = Ax^4 + Bx^2 for real A, B | |
064z | Given are 50 points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least 130 scalene triangles with vertices of that color. | [
"Since $50 = 4 \\cdot 12 + 2$, according to the pigeonhole principle we will have at least 13 points colored in the same color. Using these 13 points we construct $\\binom{13}{3} = 286$ different triangles, since there are no three collinear points.\n\nWe will prove that there are at most $12 \\cdot 13 = 156$ isosc... | Greece | 11th Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
03ku | Problem:
Determine the number of real solutions $a$ to the equation
$$
\left[\frac{1}{2} a\right]+\left[\frac{1}{3} a\right]+\left[\frac{1}{5} a\right]=a
$$
Here, if $x$ is a real number, then $[x]$ denotes the greatest integer that is less than or equal to $x$. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 30 | |
083o | Problem:
Trovare tutte le coppie $(p, q)$ di numeri primi (positivi) tali che l'equazione
$$
x^{2}-(6p-4q)x+3pq=0
$$
abbia due radici intere. | [] | Italy | UNIONE MATEMATICA ITALIANA Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (p, q) ∈ {(5, 3), (7, 5), (23, 5)} | |
03mi | The sequence $a_1, a_2, \dots, a_n$ consists of the numbers $1, 2, \dots, n$ in some order. For which positive integers $n$ is it possible that the $n+1$ numbers $0, a_1, a_1+a_2, a_1+a_2+a_3, \dots, a_1+a_2+\dots+a_n$ all have different remainders when divided by $n+1$? | [
"**Solution:** It is possible if and only if $n$ is odd.\nIf $n$ is even, then $a_1 + a_2 + \\dots + a_n = 1 + 2 + \\dots + n = \\frac{n}{2} \\cdot (n + 1)$, which is congruent to $0$ mod $n + 1$. Therefore, the task is impossible.\n\nNow suppose $n$ is odd. We will show that we can construct $a_1, a_2, \\dots, a_n... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n is odd | |
0krk | Problem:
Compute the sum of all 2-digit prime numbers $p$ such that there exists a prime number $q$ for which $100q + p$ is a perfect square.
Proposed by: Sheldon Kieren Tan | [
"Solution:\n\nAll squares must end with $0, 1, 4, 5, 6$, or $9$, meaning that $p$ must end with $1$ and $9$. Moreover, since all odd squares are $1 \\bmod 4$, we know that $p$ must be $1 \\bmod 4$. This rules all primes except for $41, 61, 29, 89$. Since $17^2 = 289$, $19^2 = 361$, $23^2 = 529$, $89, 61$, and $29$ ... | United States | HMMT February 2022 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | 179 | |
07en | Consider triangle $ABC$ with orthocenter $H$. Let points $M$ and $N$ be the midpoints of segments $BC$ and $AH$. Point $D$ lies on line $MH$ so that $AD \parallel BC$ and point $K$ lies on line $AH$ so that $DNMK$ is cyclic. Points $E$ and $F$ lie on lines $AC$ and $AB$ such that $\angle EHM = \angle C$ and $\angle FHM... | [
"First we prove that $E$, $M$ and $F$ are collinear. Let $H'$ be the reflection of $H$ with respect to $M$. It is known that $AH'$ is diameter of circumcircle of triangle $ABC$, and so $\\angle H'CA = 90^\\circ$.\n\n\n\nLet $E'$ be the intersection point of perpendicular line to $HH'$ throu... | Iran | Iranian 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 > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous... | English | proof only | null | |
000y | Sea $ABC$ un triángulo inscrito en una circunferencia $\Gamma$. Sea $\Gamma_a$ una circunferencia tangente internalmente a $\Gamma$ y a los lados $AB$ y $AC$. Sea $A'$ el punto de tangencia de $\Gamma$ y $\Gamma_a$. Defina $B'$ y $C'$ de modo análogo. Pruebe que $AA'$, $BB'$ y $CC'$ son concurrentes. | [] | Argentina | XII Olimpíada Matemática Rioplatense | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | español | proof only | null | |
08mc | Problem:
Show that there are infinitely many positive integers $c$, such that both of the following equations have solutions in positive integers:
$$
\left(x^{2}-c\right)\left(y^{2}-c\right)=z^{2}-c
$$
and
$$
\left(x^{2}+c\right)\left(y^{2}-c\right)=z^{2}-c
$$ | [
"Solution:\nThe first equation always has solutions, namely the triples $\\{x, x+1, x(x+1)-c\\}$ for all $x \\in \\mathbb{N}$. Indeed,\n$$\n\\left(x^{2}-c\\right)\\left((x+1)^{2}-c\\right)=x^{2}(x+1)^{2}-2c\\left(x^{2}+(x+1)^{2}\\right)+c^{2}=(x(x+1)-c)^{2}-c.\n$$\n\nFor the second equation, we try $z=|xy-c|$. We n... | JBMO | 2009 Shortlist JBMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0cd4 | The isosceles triangle $ABC$ has $AB = AC$ and $\angle CAB = 30^\circ$. Let $F \in [AB]$, $E \in [AC]$, $P \in [AB]$, and $M \in [BC]$ be points such that $\angle FCA = 15^\circ$, $\angle FEA = 45^\circ$, $CP \perp CF$ and $[PM]$ is the bisector of $\angle EPC$. Prove that $E$ is the midpoint of the segment $AC$ and $M... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0gt2 | Let $ABCD$ be a parallelogram. Suppose that a point $P$ is chosen on the arc of the circumcircle of $ABC$ not containing $A$; a point $Q$ is chosen on the extension of the segment $AC$ on the side $C$ such that $\angle PBC = \angle CDQ$. Show that the circumcircle of $APQ$ is tangent to the line $AB$. | [
"The equalities $\\angle APB = \\angle ACB = \\angle QAD$ and $\\angle ABP = \\angle QDA$ imply the similarity $APB \\sim QAD$. Hence $AP/AQ = PB/AD = BP/BC$, then the equality $\\angle PBC = \\angle PAQ$ implies the similarity $BPC \\sim APQ$. Therefore $\\angle APQ = \\angle BPC = \\angle BAQ$, thus the circle $(... | Turkey | 30th Junior Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07n0 | Prove that the product of six consecutive positive integers cannot be a perfect cube. | [
"Because $6! = 720$ is not a perfect cube, we only need to consider products $t = n(n+1)(n+2)(n+3)(n+4)(n+5)$ with $n \\ge 2$. Let $a = n(n+5)$, $b = (n+1)(n+4) = a+4$ and $c = (n+2)(n+3) = a+6$, then\n$$\nt = a(a + 4)(a + 6) = a^3 + 10a^2 + 24a.\n$$\nBecause $a \\ge 14$, we have $(a+3)^3 = a^3 + 9a^2 + 27a + 27 = ... | Ireland | Ireland | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0evc | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that for all $x, y \in \mathbb{R}$
$$
f(x^{2015} + f(y)^{2015}) = f(x)^{2015} + y^{2015}.
$$ | [
"Put $x = 0$ in the given equation\n$$\nf(x^{2015} + f(y)^{2015}) = f(x)^{2015} + y^{2015}. \\qquad (1)\n$$\nThen, we have\n$$\nf(f(y))^{2015} = f(0)^{2015} + y^{2015}. \\qquad (2)\n$$\nIt implies that $f$ is a bijective function. By putting $f(x)$ to $x$ in (1), we have\n$$\nf(f(x))^{2015} + f(y)^{2015} = f(f(x))^... | South Korea | Korean Mathematical Olympiad Final Round | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x and f(x) = -x | |
026k | Problem:
Multiplicação com letras - Na operação abaixo, as letras $a, b$ e $c$ são algarismos distintos e diferentes de 1.
$$
\begin{array}{r}
a b b \\\times \quad c \\\hline b c b 1
\end{array}
$$
Determine os valores de $a, b$ e $c$. | [
"Solution:\n\nSe o produto de $b$ por $c$ termina em $1$, então $b \\times c$ pode ser $21$ ou $81$, segue que $b \\times c = 3 \\times 7$ ou $9 \\times 9$. A única possibilidade de escrever o produto de dois números distintos menores que $10$ é $21 = 3 \\times 7$. Assim temos dois possíveis casos:\n\n1o caso: $b =... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a=5, b=3, c=7 | |
0kqv | Problem:
Michel starts with the string $H M M T$. An operation consists of either replacing an occurrence of $H$ with $H M$, replacing an occurrence of $M M$ with $M O M$, or replacing an occurrence of $T$ with $M T$. For example, the two strings that can be reached after one operation are $H M M M T$ and $H M O M T$.... | [
"Solution:\n\nEach final string is of the form $H M x M T$, where $x$ is a string of length 10 consisting of $M$ s and $O$ s. Further, no two $O$ s can be adjacent. It is not hard to prove that this is a necessary and sufficient condition for being a final string.\n\nLet $f(n)$ be the number of strings of length $n... | United States | HMMT February 2022 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 144 | |
00bm | Consider the product $P_n = 1! \cdot 2! \cdot 3! \cdot \dots \cdot n!$, where $n! = 1 \cdot 2 \cdot 3 \cdot \dots \cdot n$, for every positive integer $n$.
a) Find all possible values of positive integers $m$ such that $\frac{P_{2020}}{m!}$ is a perfect square.
b) Prove that there exist infinitely many values of $n$ ... | [
"a) First, note that\n$$\n\\begin{aligned}\nP_{2020} &= 1 \\cdot (1 \\cdot 2) \\cdot (1 \\cdot 2 \\cdot 3) \\cdots (1 \\cdot 2 \\cdot 3 \\cdots 2020) = 1^{2020} \\cdot 2^{2019} \\cdot 3^{2018} \\cdots 2019^2 \\cdot 2020 \\\\\n&= (1^{1010} \\cdot 2^{1009} \\cdot 3^{1009} \\cdots 2018 \\cdot 2019)^2 \\cdot (2 \\cdot ... | Argentina | 29° Olimpiada Matemática del Cono Sur | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a) m = 1010.
b) Infinitely many values exist; for example, for n = 8(k^2 + k), both m = 4(k^2 + k) and m = 4(k^2 + k) + 1 yield a perfect square. | |
01mm | Prove that if positive numbers $a$, $b$, $x$, $y$ satisfy the inequalities $ab \ge xa + yb$ and $a \ge b$, then they satisfy the inequality $x + y \le a$. | [] | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
05dr | Problem:
Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the condition
$$
f\left(y^{2}+2 x f(y)+f(x)^{2}\right)=(y+f(x))(x+f(y))
$$
for all real numbers $x$ and $y$. | [
"Solution:\nIt can be easily checked that the functions $f(x)=x$, $f(x)=-x$ and $f(x)=\\frac{1}{2}-x$ satisfy the given condition. We will show that these are the only functions doing so. Let $y=-f(x)$ in the original equation to obtain\n$$\nf\\left(2 f(x)^{2}+2 x f(-f(x))\\right)=0\n$$\nfor all $x$. In particular,... | European Girls' Mathematical Olympiad (EGMO) | European Girls' Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x; f(x) = -x; f(x) = 1/2 - x | |
0koy | For positive integers $a, b$, and $c$ with $a < b < c$, consider collections of postage stamps in denominations $a, b$, and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to 1000 cents, let $f(a, b, c)$... | [] | United States | AIME II | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | null | |
0fip | Problem:
Halla todos los pares de números naturales $x, y$ ($x<y$) tales que la suma de todos los números naturales comprendidos estrictamente entre ambos es igual a $1999$. | [
"Solution:\n\nTenemos que sumar del número $x+1$ hasta el número $y-1$ y obtener el número $1999$. Esta suma: $(x+1)+(x+2)+\\ldots+(y-1)=1999$ corresponde a la de una progresión aritmética de diferencia $1$ y con $y-x-1$ términos, por tanto es:\n$$\n\\left(\\frac{(x+1)+(y-1)}{2}\\right)(y-x-1)=1999\n$$\nDe donde se... | Spain | XXXV Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (x, y) = (1998, 2000) and (998, 1001) | |
0lby | Solve the following system of equations on $\mathbb{R}$
$$
\begin{cases} \sqrt{(\sin x)^2 + \frac{1}{(\sin x)^2}} + \sqrt{(\cos y)^2 + \frac{1}{(\cos y)^2}} = \sqrt{\frac{20y}{x+y}}, \\ \sqrt{(\sin y)^2 + \frac{1}{(\sin y)^2}} + \sqrt{(\cos x)^2 + \frac{1}{(\cos x)^2}} = \sqrt{\frac{20x}{x+y}}. \end{cases}
$$ | [
"Note that $x \\neq \\frac{k\\pi}{2}$, $y \\neq \\frac{m\\pi}{2}$ ($k, m \\in \\mathbb{Z}$) and $xy > 0$. From the given condition, one can get\n$$\nA = 20 \\sqrt{\\frac{xy}{(x+y)^2}},\n$$\nwhere $A$ equals to the product of two left hand sides of the given system. Using the Cauchy-Schwarz and AM-GM inequality\n$$\... | Vietnam | VMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | x = y = pi/4 + k*pi/2, for any integer k | |
0hvq | Problem:
The numbers $1, 2, \ldots, 2014$ are arranged evenly around a circle in arbitrary order. We are permitted to swap two adjacent numbers, as long as they do not sum to $2015$. Prove that it is impossible to perform finitely many swaps so that each number ends up diametrically opposite from its starting point. | [
"Solution:\n\nAssume for contradiction it's possible. Each time a number moves one slot counterclockwise, assign a score of $+1$; each time a number moves one slot clockwise, assign a score of $-1$. Thus every move produces a total score of $1 + (-1) = 0$.\n\nLet $S(a)$ denote the total score accumulated by moves o... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof only | null | |
0i5x | Problem:
The unknown real numbers $x$, $y$, $z$ satisfy the equations
$$
\frac{x+y}{1+z}=\frac{1-z+z^{2}}{x^{2}-x y+y^{2}} ; \quad \frac{x-y}{3-z}=\frac{9+3 z+z^{2}}{x^{2}+x y+y^{2}}
$$
Find $x$. | [
"Solution:\n\n$\\sqrt[3]{14}$\n\nCross-multiplying in both equations, we get, respectively,\n$$\nx^{3}+y^{3}=1+z^{3}, \\quad x^{3}-y^{3}=27-z^{3}.\n$$\nNow adding gives $2 x^{3}=28$, or $x=\\sqrt[3]{14}$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 14^(1/3) | |
0d83 | There are $64$ towns in a country, and some pairs of towns are connected by roads but we do not know these pairs. We may choose any pair of towns and find out whether they are connected by a road. Our aim is to determine whether it is possible to travel between any two towns using roads. Prove that there is no algorith... | [
"In general, we can replace $64$ and $2016$ by $n$ and $\\frac{n(n-1)}{2}$.\nThe given problem can be converted into the graph theory by considering each town as a vertex in the graph $G$ and each road connects two towns as the edge. So there are exactly $\\frac{n(n-1)}{2}$ undirected edges in this graph.\nWe need ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Algorithms"
] | English | proof only | null | |
0h9l | Several representatives of competing companies that produce the game "Overwatch" arrived at the conference. Consider all representatives of different companies to be competitors. It is known that each participant of the conference has exactly $2018$ competitors among all other participants. What is the largest possible... | [
"Notice that from the problem statement it follows that the same number of participants arrived from each company. Indeed, if there were two different numbers of participants from some two companies, they would have a different number of competitors. Let $m$ be the number of companies participating in the conferenc... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 4036 | |
01si | Is it possible to partition a set $M$ of positive integers into some disjoint subsets so that in each subset one of the numbers is equal to the sum of all other numbers of this subset if
$$
\text{a) } M = \{1, 2, \dots, 20\} \text{?}
$$ | [
"a) Yes, it is possible; the desired partition exists. For example, see the following table:\n\n| 20 | 19 | 18 | 17 | 16 | 15 |\n|----|----|----|----|----|----|\n| 14 | 12 | 10 | 13 | 11 | 9 |\n| 6 | 7 | 8 | 4 | 5 | 1, 2, 3 |\n\nIn each subset, one of the numbers is equal to the sum of the others."
] | Belarus | FINAL ROUND | [
"Discrete Mathematics > Other"
] | English | proof only | null | |
0cz5 | Find all positive integers $x$ and $y$ such that
$$
\binom{x}{y}=1432
$$ | [
"We have $1432=2^{3} \\cdot 179$. The equation is equivalent to\n$$\n\\frac{x!}{y!(x-y)!}=2^{3} \\cdot 179\n$$\nor $y!(x-y)! \\cdot 2^{3} \\cdot 179 = x!$. It follows $179 \\mid x!$, hence $x \\geq 179$.\n\nIt is clear that $(x, y) = (1432, 1), (1432, 1431)$ are solutions. We shall prove that there are no other sol... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | [(1432, 1), (1432, 1431)] | |
090e | Let $S = \{0, 1, 2, \dots, 8\}$. How many functions $f : S \to S$ satisfy the following condition: For any elements $x, y, z \in S$, if $x+y-z$ is divisible by $9$, then $f(x)f(y) - f(f(z))$ is also divisible by $9$? | [
"$858$\n\nIn this solution, $x \\equiv y$ always means that $x-y$ is divisible by $9$. Furthermore, for any integer $x$, we denote by $[x]$ the unique element $x' \\in S$ satisfying $x \\equiv x'$. Note that for any integers $s$ and $t$, we have $[[s-t]+t] = [s]$ and $[s[t]] = [st]$.\n\nUnder this notation, the con... | Japan | The 35th Japanese Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | 858 | |
05rn | Problem:
Soit $a, b, c$ des nombres réels positifs ou nuls tels que $a+b+c=1$.
Démontrer que
$$
\frac{5+2 b+c^{2}}{1+a}+\frac{5+2 c+a^{2}}{1+b}+\frac{5+2 a+b^{2}}{1+c} \geqslant 13
$$ | [
"Solution:\nSoit $S$ la somme\n$$\n\\frac{5+2 b+c^{2}}{1+a}+\\frac{5+2 c+a^{2}}{1+b}+\\frac{5+2 a+b^{2}}{1+c}\n$$\nNotons également $x_{i}$ le $i^{\\text{ème}}$ plus petit élément de l'ensemble $\\{a, b, c\\}$. L'inégalité du réordonnement indique que\n$$\n\\begin{aligned}\n& \\frac{b}{1+a}+\\frac{c}{1+b}+\\frac{a}... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0gmz | Find the largest real number $t$ such that, in any school with $2006$ students and $14$ teachers where every student is acquainted with at least one teacher, a student and a teacher can be found such that they are acquainted with each other, and the ratio of the number of students who are acquainted with the teacher to... | [
"$t = 2006/14$.\nIf every student is acquainted with every teacher, then all relevant ratios are $2006/14$. This means $t \\le 2006/14$. Now we will show that $t \\ge 2006/14$.\nFor $1 \\le i \\le 14$, let $a_i$ denote the number of students who are acquainted with the $i$\\text{th} teacher, and for $1 \\le j \\le ... | Turkey | 14th Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2006/14 | |
09qn | Problem:
De cirkels $\Gamma_{1}$ en $\Gamma_{2}$ snijden elkaar in $D$ en $P$. De gemeenschappelijke raaklijn van de twee cirkels het dichtste bij punt $D$ raakt $\Gamma_{1}$ in $A$ en $\Gamma_{2}$ in $B$. De lijn $A D$ snijdt $\Gamma_{2}$ voor de tweede keer in $C$. Zij $M$ het midden van lijnstuk $B C$.
Bewijs dat ... | [
"Solution:\n\nOplossing I. Zij $S$ het snijpunt van $P D$ en $A B$. Dan ligt $S$ op de machtlijn van de twee cirkels en geldt dus $|S A|=|S B|$. Dus $P S$ is een zwaartelijn in driehoek $P A B$.\nVanwege de raaklijnomtrekshoekstelling op $\\Gamma_{1}$ met koorde $A P$ geldt $\\angle B A P=180^{\\circ}-\\angle A D P... | Netherlands | Dutch TST toets 8 juni | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02ug | Problem:
a) Todo número real ao quadrado é maior ou igual a $0$, sendo $0$ apenas se o número elevado ao quadrado for o próprio $0$. Consequentemente, para quaisquer números reais $a$ e $b$ temos $(a-b)^2 \geq 0$. Prove que
$$
\frac{a^2+b^2}{2} \geq a b
$$
com igualdade ocorrendo somente quando $a=b$.
b) A figura a se... | [
"Solution:\na) Desenvolvendo o produto notável, temos\n$$\n(a-b)^2 \\geq 0 \\Leftrightarrow a^2-2 a b+b^2 \\geq 0 \\Leftrightarrow a^2+b^2 \\geq 2 a b \\Leftrightarrow \\frac{a^2+b^2}{2} \\geq a b\n$$\nVeja que a igualdade da última expressão acontece apenas quando há igualdade na primeira, ou seja, quando $a=b$.\n... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 3/2 | |
091t | Problem:
Let $ABC$ be an acute triangle. Denote by $B_{0}$ and $C_{0}$ the feet of the altitudes from vertices $B$ and $C$, respectively. Let $X$ be a point inside the triangle $ABC$ such that the line $BX$ is tangent to the circumcircle of the triangle $AXC_{0}$ and the line $CX$ is tangent to the circumcircle of the... | [
"Solution:\n\n\nLet $A_{0}$ be the foot of the altitude from $A$. The quadrilateral $ACA_{0}C_{0}$ is cyclic because $\\varangle AA_{0}C=\\varangle AC_{0}C=90^{\\circ}$. By the power of the point $B$ with respect to that circle we have $|BA||BC_{0}|=|BA_{0}||BC|$.\nThe power of the point $B... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0arx | Problem:
Let $x$ and $y$ be the integral and fractional parts of $\sqrt{37-20 \sqrt{3}}$. Find the value of $x+y+\frac{4}{y}$. | [
"Solution:\n$\\sqrt{37-2(10) \\sqrt{3}} = \\sqrt{(5-2 \\sqrt{3})^{2}} = 5-2 \\sqrt{3} = 1 + (4-2 \\sqrt{3}) = 1 + 2(2-\\sqrt{3})$\n\n$\\Rightarrow x=1,\\ y=4-2 \\sqrt{3}$. Thus, $x+y+\\frac{4}{y} = 1+4-2 \\sqrt{3}+\\frac{4}{4-2 \\sqrt{3}} = 5-2 \\sqrt{3} + (4+2 \\sqrt{3}) = 9$."
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 9 | |
02eo | Each face of a tetrahedron is a triangle with sides $a$, $b$, $c$ and the tetrahedron has circumradius $1$. Find $a^2 + b^2 + c^2$. | [
"Opposite edges of the tetrahedron have the same length. Let $AB$, $CD$ have length $a$. Take parallel planes through $AB$ and $CD$. Project $A$, $B$ onto the other plane, and $C$, $D$ onto the other plane. Then we get $8$ points at the vertices of a rectangular block. If the sides have lengths $x$, $y$, $z$, then ... | Brazil | XII OBM | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof and answer | 8 | |
031e | Problem:
Find all positive integers $n$ for which there exists $n$ points in the plane such that any of them lies on exactly $\frac{1}{3}$ of the lines determined by these $n$ points. | [
"Solution:\nWe shall prove that $n=6$. If we take 6 points in general position (no three are collinear), then the lines are 15 and any point lies on 5 lines, i.e. $n=6$ is a solution of the problem.\n\nDenote by $k$ the number of the lines defined by the given $n$ points. Assume that there is a line $l$ containing ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | n = 6 | |
0dce | Let be given a positive integer $n > 1$. Find all polynomials $P(x)$ non constant, with real coefficients such that
$$
P(x) P\left(x^{2}\right) \ldots P\left(x^{n}\right) = P\left(x^{\frac{n(n+1)}{2}}\right)
$$
for all $x \in \mathbb{R}$. | [
"Denote $m$ as degree of $P(x)$ then by comparing the leading coefficients $a$ of two sides, we get $a^{n} = a$. We consider two cases based on the parity of $n$.\n\n1. If $n$ is even then $a = 1$. Put $P(x) = x^{m} + Q(x)$ with $Q \\in \\mathbb{R}[x]$ and if $Q(x) \\equiv 0$, we obtain $P(x) = x^{m}$ as a solution... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | If n is even: P(x) = x^m for some positive integer m. If n is odd: P(x) = ± x^m for some positive integer m. | |
04sh | A triangle $ABC$ is given in which $BC$ is the shortest side. Denote $M$ its midpoint. On the sides $AB$ and $AC$ take the points $X$ and $Y$, respectively, in such a way that $|BX| = |BC| = |CY|$. Denote $Z$ the intersection point of lines $CX$ and $BY$. Prove that the line $ZM$ passes through the centre of the excirc... | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinate... | English | proof only | null | |
000r | a) Se tienen dos sucesiones, cada una de $2003$ enteros consecutivos, y un tablero de $2$ filas y $2003$ columnas.
| | | | | ... | |
|---|---|---|---|-----|---|
| | | | | ... | |
¿Decida si siempre es posible distribuir los números de la primera sucesión en la primera fila y los de la segunda suce... | [] | Argentina | XVIII Olimpíada Iberoamericana de Matemática | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | español | proof and answer | a) Yes. Arrange both rows in increasing order and cyclically shift the second row by 1001 positions; the column sums are consecutive. b) No. For an even number of columns, the required average of a block of consecutive integers is a half-integer, but the column-sum average is an integer, so it is impossible. | |
0ktl | Problem:
An $E$-shape is a geometric figure in the two-dimensional plane consisting of three rays pointing in the same direction, along with a line segment such that
- the endpoints of the rays all lie on the segment,
- the segment is perpendicular to all three rays,
- both endpoints of the segment are endpoints of ra... | [
"Solution:\n\nDefine a $C$-shape to be an $E$-shape without the middle ray. Then, an $E$-shape consists of a ray and a $C$-shape. Two $C$-shapes can intersect at most $6$ times, a $C$-shape and a ray can intersect at most $2$ times, and two rays can intersect at most $1$ time. Thus, the number of intersections of t... | United States | HMMT February | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 11 | |
0a1f | A maths teacher has $10$ cards with the numbers $1$ to $10$ on them, one number per card. She places these cards in some order in a line next to each other on the table. The students come to the table, one at a time. The student whose turn it is goes once through the line of cards from left to right and removes every c... | [
"Consider a sequence of cards for which exactly nine students get a turn. Now replace all cards $k$ by $11-k$. We will prove that this gives a sequence for which two students get a turn. We can repeat this process and that gives back the original sequence. Next, consider a sequence for which exactly two students ge... | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0ixj | Let $a_1, a_2, \dots, a_n$ be distinct positive integers and let $M$ be a set of $n-1$ positive integers not containing $s = a_1 + a_2 + \dots + a_n$. A grasshopper is to jump along the real axis, starting at the point $0$ and making $n$ jumps to the right with lengths $a_1, a_2, \dots, a_n$ in some order. Prove that t... | [
"(By Anton Mellit, observer from the Ukraine delegation, and Ilya Bogdanov, observer from the Russia delegation).\n\nWe will represent a route of the grasshopper by a (ordered) sequence of indices $(i_1, i_2, \\dots, i_n)$ if it makes consecutive jumps $(a_{i_1}, a_{i_2}, \\dots, a_{i_n})$.\n\nTo simplify our argum... | United States | IMO 2009 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
05tv | Problem:
Morgane et Bosphore jouent au jeu suivant. Morgane a écrit les entiers de 1 à 8 sur les sommets d'un octogone régulier : chaque entier est écrit sur un des huit sommets de l'octogone. Bosphore choisit ensuite un sommet et calcule la somme des nombres écrits sur ce sommet et sur ses deux voisins. Il note $s$ c... | [
"Solution:\n\nTout d'abord, si Morgane répartit les entiers de 1 à 8 comme suit, on constate en effet que Bosphore devra lui donner au moins 12 bonbons; on a indiqué, à côté de chaque sommet, la somme $s$ que calculerait Bosphore s'il choisissait ce sommet.\n\n\n\nRéciproquement, quel que s... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 12 | |
01kd | Basil considers all quadratic polynomials $y = ax^2 + bx + c$ with negative discriminants, where $a$, $b$, $c$ are positive integers not greater than $50$. For each polynomial Basil writes its minimal value.
Find the largest and the smallest numbers among the written numbers. | [
"The distance between the parabola $y = ax^2 + bx + c$ and the axis of abscissae is equal to\n$$\nd = \\frac{4ac - b^2}{4a}\n$$\nTherefore, the problem is equivalent to the following problem: find the greatest and the smallest value of the expression above if positive integers $a$, $b$, $c$ satisfy the inequalities... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | largest = 50 - 1/200, smallest = 3/196 | |
0hdk | Let $a$ and $b$ be two different natural numbers such that $a^2 + b^2 + 1$ is divisible by $2ab + 1$. Prove that $2ab + 1$ is a perfect square (is a square of an integer). | [
"Assume the contrary. Let $(a, b)$ be a pair of natural numbers for which this does not hold, such that $a < b$ and this pair has a minimal sum of $a+b$.\n\n**Lemma 1.** $a^2 + b^2 + 1$ is divisible by $2ab + 1$ if and only if $(2a^2 + 1)^2$ is divisible by $2ab + 1$.\n\n*Proof.* Since $(a^2, 2ab+1)=1$, then $a^2+... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Grea... | null | proof only | null | |
0em8 | Can you enumerate the edges of a tetrahedron with numbers from $1$ to $6$ so that all four triangular faces have the same sum, where the sum of a face is calculated by adding the numbers on its edges? | [
"Label the vertices as shown.\n\n\n\nLet $f(XY)$ denote the number on edge $XY$. Notice that\n$$\n\\begin{align*}\nf(BA) + f(AD) + f(DB) &= f(AD) + f(DC) + f(AC) \\\\\n&= f(BC) + f(CD) + f(BD) \\\\\n&= f(AB) + f(BC) + f(AC).\n\\end{align*}\n$$\naccording to the condition. This means that $f... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | No, it cannot be done. | |
0ace | A rhomb with length of one diagonal equal to $7.2\, ext{cm}$ has area of $34.56\, ext{cm}^2$. Calculate the perimeter of the rhomb and the radius of its inscribed circle. | [
"In the problem it is given that $d_1 = 7.2\\,\\text{cm}$ and $P = 34.56\\,\\text{cm}^2$. From the formula for area $P = \\frac{d_1 \\cdot d_2}{2}$ we have $d_2 = \\frac{2 \\cdot P}{d_1} = \\frac{2 \\cdot 34.56}{7.2} = 9.6\\,\\text{cm}$. For the side length of the rhomb we have $a = \\sqrt{\\left(\\frac{d_1}{2}\\ri... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | null | proof and answer | Perimeter = 24 cm; inradius = 2.88 cm | |
06h9 | Let $f$ be a function defined on nonzero real numbers such that for any nonzero $x$, $$
\frac{2}{x}f(-x) + f\left(\frac{1}{x}\right) = x.
$$ Find the value of $f(8)$. | [
"We have $f(8) = \\frac{205}{8}$.\nIt is given that\n$$\n\\frac{2}{x}f(-x) + f\\left(\\frac{1}{x}\\right) = x \\quad (1)\n$$\nfor any $x \\neq 0$. Replacing $x$ by $-\\frac{1}{x}$, we obtain\n$$\n-2x f\\left(\\frac{1}{x}\\right) + f(-x) = -\\frac{1}{x}. \\quad (2)\n$$\nUsing (1) and (2), we can solve for $f(-x)$. I... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | final answer only | 205/8 | |
01i4 | Let $I$ be the incenter of a triangle $ABC$. Let $F$ and $G$ be the feet of the perpendiculars drawn from $A$ to the lines $BI$ and $CI$, respectively. Rays $AF$ and $AG$ intersect the circumcircles of the triangles $CFI$ and $BGI$ second times at points $K$ and $L$, respectively. Prove that line $AI$ bisects the segme... | [
"Since $\\angle IFK = 90^\\circ$, then $IK$ is the diameter of the circumcircle of $CFI$, hence also $\\angle ICK = 90^\\circ$. Similarly is $IL$ the diameter of the circumcircle of $BGI$ and $\\angle IBL = 90^\\circ$. Therefore are the lines $CK$ and $GL$ parallel, also $BL$ and $FK$ are parallel."
] | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0aqg | Problem:
Find all pairs $(a, b)$ of integers such that $\sqrt{2010+2 \sqrt{2009}}$ is a solution of the quadratic equation $x^{2}+a x+b=0$. | [] | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a = -2, b = -2008 | |
0d06 | For any positive integer $k$ define
$$
H_k = 1 + \frac{1}{2} + \dots + \frac{1}{k}.
$$
Prove the relation
$$
1 + \frac{1}{n+1} \sum_{k=1}^{n} H_k = H_{n+1}.
$$ | [
"We have\n$$\n\\begin{align*}\n\\sum_{k=1}^{n} H_k &= 1 + \\left(1 + \\frac{1}{2}\\right) + \\left(1 + \\frac{1}{2} + \\frac{1}{3}\\right) + \\dots + \\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) \\\\\n&= n + \\frac{n-1}{2} + \\frac{n-2}{3} + \\dots + \\frac{1}{n} \\\\\n&= (n+1) - 1 + \\frac{n+1-2}{2} + \... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0gj8 | Find all functions $f : \mathbb{N} \to \mathbb{N}$ such that, for any set $A$ containing exactly 2024 distinct positive integers, the set
$$
S_A := \{f^{(k)}(x) \mid 1 \le k \le 2024,\ x \in A\}
$$
is also a set containing exactly 2024 distinct positive integers.
*Remark*: $f^{(k)}$ denotes the $k$-th iteration of $f$... | [
"顯然 $f(x) = x$ 為一解。此外，對於任何由 2024 個相異正整數所成集合 $A = \\{a_1, a_2, \\dots, a_{2024}\\}$，任何滿足\n$$\n\\begin{cases} f(a_i) = a_{i-1}, & \\text{其中 } a_0 = a_{2024}, \\\\ f(x) \\in A, & \\text{對於所有 } x \\notin A, \\end{cases}\n$$\n的函數 $f$ 皆滿足題意。讓我們證明以上便是全部的解。\n\n用所有正整數做點，並將 $x$ 連向 $f(x)$ 構成有向圖 $G$ (允許自環)。注意若 $x$ 可通往 2025 個點，... | Taiwan | IMO 2J, Independent Study 1 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Chinese; English | proof and answer | Exactly the following functions: (1) the identity function f(n) = n for all positive integers n; or (2) there exists a fixed set Y of size 2024 and a cyclic ordering y1, y2, ..., y2024 of its elements such that f(yi) = yi+1 for i = 1, 2, ..., 2023 and f(y2024) = y1 (i.e., f permutes Y as a single 2024-cycle), and for e... | |
0gxa | There are discounts in the supermarket. An active customer will be given a discount of $4\%$, if he buys goods for the sum of over $300$; and if the same customer buys goods for the sum of over $600$, he will be given a discount of $10\%$. What is the biggest amount of money that he can spend for goods if he has
a) $5... | [
"a) If a customer buys goods for the sum of $x \\ge 600$, he'll pay $10\\%$ less the cost of these goods. Consequently, he can buy goods for the sum $x$ if $0.90x \\le 594 \\Rightarrow x \\le 660$ at the most.\n\nb) Analogously to the item a) we obtain that a customer will be given a discount of $4\\%$, to buy good... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | English | final answer only | a) 660; b) 556.25 | |
079n | Find all increasing functions $f: \mathbb{R}^+ \cup \{0\} \to \mathbb{R}^+ \cup \{0\}$, such that for every $x, y \in \mathbb{R}^+ \cup \{0\}$ we have
$$
f\left(\frac{x + f(x)}{2} + y\right) = 2x - f(x) + f(f(y))
$$
(note that $f$ is not necessarily strictly increasing.) | [
"Suppose that $f(0) = a$. Let $x = y = 0$, we have\n$$\nf\\left(\\frac{a}{2}\\right) = f(a) - a\n$$\nLet $x = \\frac{a}{2}, y = 0$, we have\n$$\nf\\left(\\frac{f(a) - \\frac{a}{2}}{2}\\right) = 2a\n$$\nLet $x = a, y = 0$, we have\n$$\nf\\left(\\frac{f(a) + a}{2}\\right) = 2a\n$$\n\nSince $f$ is increasing, so for e... | Iran | 27th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x for all x ≥ 0 | |
04zv | Every face of a unit cube has one of numbers $-1$, $0$, $1$ written on it in such a way that every two faces with a common edge contain different numbers. Is it possible to form
a) a cube of size $2 \times 2 \times 2$;
b) a cube of size $3 \times 3 \times 3$
so that in the grids that come up on the faces, every two ... | [
"First note that the placement of the numbers on the faces of the unit cube is unique. Indeed, let a number $x$ be written on some face; then the neighboring faces contain alternately the other numbers $y$ and $z$, while the opposite face again contains $x$. This means that each of the numbers $-1$, $0$, $1$ occurs... | Estonia | Selected Problems from the Final Round of National Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) yes; b) yes | |
0jt2 | Problem:
Starting with an empty string, we create a string by repeatedly appending one of the letters $H$, $M$, $T$ with probabilities $\frac{1}{4}$, $\frac{1}{2}$, $\frac{1}{4}$, respectively, until the letter $M$ appears twice consecutively. What is the expected value of the length of the resulting string? | [
"Solution:\n\nLet $E$ be the expected value of the resulting string. Starting from the empty string,\n- We have a $\\frac{1}{2}$ chance of not selecting the letter $M$; from here the length of the resulting string is $1+E$.\n- We have a $\\frac{1}{4}$ chance of selecting the letter $M$ followed by a letter other th... | United States | HMMT February 2016 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 6 | |
0bnr | Find all complex numbers $z$ such that
$$
|z| + |z - 5i| = |z - 2i| + |z - 3i|.
$$ | [
"We notice that $|z - 2i| = \\left|\\frac{2}{5}(z - 5i) + \\frac{3}{5}z\\right| \\le \\frac{2}{5}|z - 5i| + \\frac{3}{5}|z|$.\n\nIn the same way, $|z - 3i| = \\left|\\frac{3}{5}(z - 5i) + \\frac{2}{5}z\\right| \\le \\frac{3}{5}|z - 5i| + \\frac{2}{5}|z|$, whence $|z| + |z - 5i| \\ge |z - 2i| + |z - 3i|$.\n\nEqualit... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof and answer | z = a i with a ∈ (-∞, 0] ∪ [5, +∞) | |
0dfy | Two players, $A$ and $B$ play the following game. On a $1 \times n$ board, where fields are labeled in order from $1$ to $n$, a coin is placed at position $k$. Players take turns moving the coin, with player $A$ starting first. Each player can move a coin one or two fields in either direction, with the restriction that... | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Player A has a winning strategy if and only if n is even; Player B has a winning strategy if and only if n is odd (independent of k). | |
01rq | Solve the equation in nonnegative integers $a$, $b$, $c$:
$$
3^a + 2^b + 2015 = 3c!
$$ | [
"(Solution by A. Sheremet.) Since $3c! > 2015$, we have $c \\ge 6$.\n\n1) For $a = 0$ from the given equation\n$$\n3^a + 2^b + 2015 = 3c! \\quad (1)\n$$\nwe have $2^b = 3(c! - 672)$, which is impossible.\n\n2) Let $a = 1$. If $c = 6$, then $2^b = 142$ -- there are no solutions. If $c \\ge 7$, then (1) implies $2^b ... | Belarus | SELECTION and TRAINING SESSION | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof and answer | (4, 6, 6) | |
0767 | Problem:
Written on a blackboard is the polynomial $x^{2} + x + 2014$. Calvin and Hobbes take turns alternatively (starting with Calvin) in the following game. During his turn, Calvin should either increase or decrease the coefficient of $x$ by $1$. And during his turn, Hobbes should either increase or decrease the co... | [
"Solution:\n\nFor $i \\geq 0$, let $f_{i}(x)$ denote the polynomial on the blackboard after Hobbes' $i$-th turn. We let Calvin decrease the coefficient of $x$ by $1$. Therefore $f_{i+1}(2) = f_{i}(2) - 1$ or $f_{i+1}(2) = f_{i}(2) - 3$ (depending on whether Hobbes increases or decreases the constant term). So for s... | India | Indian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
06bl | Let $\theta$ be a fixed angle. Let two circles touch each other internally at point $A$. Let $AB$ be a chord of the outer circle and such that $AB$ is not a diameter. Let $M$ be a variable point on the major arc $AB$ of the outer circle. $MA$ meets the inner circle at $N$. $P$ is a point on the segment $MB$ such that $... | [
"Let $AB$ meet the inner circle at $C$. Let $D$ be the point on the tangent at $B$ to the outer circle such that $\\angle BDC = \\theta$, and $D$ lies on opposite side of $AB$ as $M$. We claim that the locus of $P$ is $\\overarc{ABD}$.\n\n\n\nFirstly, since $A$ is the homothetic centre of t... | Hong Kong | 1997-2023 IMO HK TST | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | The locus is the arc from the tangency point to the point on the tangent at the chord endpoint subtending the given angle with the chord endpoint and the intersection point on the inner circle; equivalently, the arc ABD on the circumcircle of triangle ABD from A to D. | |
05iz | Problem:
Existe-t-il des réels $a$, $b$, $c$, $d > 0$ et $e$, $f$, $g$, $h < 0$ vérifiant simultanément
$$
a e + b c > 0,\quad e f + c g > 0,\quad f d + g h > 0 \text{ et } d a + h b > 0 ?
$$ | [
"Solution:\nNon, il n'en existe pas. Par l'absurde : supposons qu'il existe de tels réels. On commence par réécrire les inégalités, mais avec uniquement des termes positifs. On a donc\n$$\nb c > a(-e) \\text{ et } (-e)(-f) > c(-g) \\text{ et } (-g)(-h) > (-f)d \\text{ et } d a > (-h) b\n$$\nSi l'on multiplie toutes... | France | Olympiades Françaises de Mathématiques, Envoi Numéro 3 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | No, such real numbers do not exist. | |
0e4d | Problem:
Poišči vse polinome $P$ s celimi koeficienti, za katere velja: za vsako celo število $a$ in vsako praštevilo $p$, ki deli $P(a)$, velja, da $p$ deli $a$. | [
"Solution:\n\nNaj bo $P$ polinom, ki ustreza pogojem naloge in $p$ poljubno praštevilo. Če je $q$ praštevilo, ki deli $P(p)$, potem deli tudi $p$, torej je $q=p$. Zato za vsako praštevilo $p$ velja $P(p)= \\pm p^{m_{p}}$ za neko nenegativno celo število $m_{p}$, ki je lahko odvisno od $p$.\n\nPolinoma $P(x)= \\pm 1... | Slovenia | 55. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | P(x) = ± x^n for any nonnegative integer n | |
09ju | Kate wants to write a set of distinct positive integers on the board, with each integer not exceeding $2023$, such that the sum of any two numbers written on the board is not divisible by their difference. For example, $4 \nmid 10$, $13 \nmid 27$, and $17 \nmid 23$, so Kate can write the numbers $3$, $7$, and $20$.
(1... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 675 | |
06hv | Let $ABC$ be an acute triangle and $H$ be the perpendicular foot at the side $BC$ from $A$. Let $D$ and $E$ be points on the segments $AB$ and $AC$, $F$ and $G$ be the perpendicular feet at $BC$ from $D$ and $E$ respectively. Let $P$ be the perpendicular foot at the segment $DH$ from $E$. Suppose that $DG, EF, AH$ are ... | [
"Suppose $DG, EF, AH$ are concurrent at $X$. Let $Q$ be the point on the line $CP$ such that $QH \\parallel PG$. We only work on the configuration as shown since the other cases are similar.\n\nFirstly, by the parallel lines $AH \\parallel EG$ and $QH \\parallel PG$, we have $\\triangle CAH \\sim \\triangle CEG$ an... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0ih4 | Problem:
Let $T$ be a triangle with side lengths $26$, $51$, and $73$. Let $S$ be the set of points inside $T$ which do not lie within a distance of $5$ of any side of $T$. Find the area of $S$. | [
"Solution:\nNote that the sides of $S$ are parallel to the sides of $T$, so $S$ is a triangle similar to $T$.\n\nThe semiperimeter of $T$ is $s = \\frac{1}{2}(26 + 51 + 73) = 75$.\n\nBy Heron's formula, the area of $T$ is $\\sqrt{75 \\cdot 49 \\cdot 24 \\cdot 2} = 420$.\n\nIf $r$ is the inradius of $T$, then the ar... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | 135/28 | |
0k45 | Problem:
Jerry has ten distinguishable coins, each of which currently has heads facing up. He chooses one coin and flips it over, so it now has tails facing up. Then he picks another coin (possibly the same one as before) and flips it over. How many configurations of heads and tails are possible after these two flips? | [
"Solution:\n\nWe have two cases:\n\nCase 1: Jerry picks the same coin twice. Then, the first time he flips the coin, it becomes tails, and then the second time, it becomes heads again, giving us the original state of all heads.\n\nCase 2: Jerry picks two different coins. In this case, there are two coins with tails... | United States | HMMT November 2018 | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 46 | |
01m2 | Positive real $a$, $b$, $c$ satisfy the condition
$$
\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 1 + \frac{1}{6} \left( \frac{a}{c} + \frac{b}{a} + \frac{c}{b} \right).
$$
Prove that
$$
\frac{a^3bc}{b+c} + \frac{b^3ca}{c+a} + \frac{c^3ab}{a+b} \ge \frac{1}{6}(ab + bc + ca)^2.
$$
(I. Voronovich) | [
"By the Cauchy-Bunyakovskii inequality, we have\n$$\n(a^2b + b^2c + c^2a)(a+b+c) \\ge (ab + bc + ca)^2. \\quad (*)\n$$\n\nNow consider the given equality. Multiplying both its sides by $(a+b+c)abc$, we get\n$$\nA = (a+b+c)\\left(\\frac{a^2bc}{b+c} + \\frac{b^2ca}{c+a} + \\frac{c^2ab}{a+b}\\right) = abc(a+b+c) + \\f... | Belarus | Selection and Training Session | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0iao | Problem:
Two differentiable real functions $f(x)$ and $g(x)$ satisfy
$$
\frac{f^{\prime}(x)}{g^{\prime}(x)}=e^{f(x)-g(x)}
$$
for all $x$, and $f(0)=g(2003)=1$. Find the largest constant $c$ such that $f(2003)>c$ for all such functions $f, g$. | [
"Solution: $1-\\ln 2$\n\nRearranging the given equation gives $f^{\\prime}(x) e^{-f(x)}=g^{\\prime}(x) e^{-g(x)}$ for all $x$, so\n$$\n\\frac{d}{d x}\\left(e^{-f(x)}-e^{-g(x)}\\right)=-f^{\\prime}(x) e^{-f(x)}+g^{\\prime}(x) e^{-g(x)}=0.\n$$\nThus, $e^{-f(x)}-e^{-g(x)}$ is a constant, and it must be less than $e^{-... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Equations > ODEs"
] | null | proof and answer | 1 - ln 2 | |
0ch4 | Let $\triangle ABC$ be a triangle with $\angle BAC = 30^\circ$ and $\angle ABC = 100^\circ$. Let $m$ be the perpendicular bisector of $AC$, $E$ be the intersection of $m$ and $AB$, and $D$ be the point on $m$, inside the triangle $ABC$, such that $\angle CAD = 10^\circ$. Let $M$ be the intersection of the lines $AD$ an... | [
"a) As $m$ is the perpendicular bisector, we have $DA = DC$ and $EA = EC$. Thus $\\triangle DEA \\equiv \\triangle DEC$.\n\n\n\nFrom $DA = DC$, we get $\\angle DCA = \\angle DAC = 10^\\circ$. We have $\\angle DCE = \\angle DAE = 20^\\circ$, so $\\angle BCE = 20^\\circ$ and $CE$ is the bisec... | Romania | 74th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
060v | Problem:
Déterminer le plus grand entier $n$ pair ayant la propriété suivante : quelle que soit la façon de paver une grille $n \times n$ avec des dominos, il existe une ligne coupant le tableau en deux parties non vides et n'intersectant aucun domino. | [
"Solution:\n\nRéponse: $n=6$\n\nMentionnons pour commencer que la réponse peut se deviner en essayant l'exercice pour des petites valeurs de $n$.\nLe problème demande de trouver le plus grand entier $n$ vérifiant une certaine propriété, il comporte donc nécessairement deux parties. Nous allons montrer d'abord que $... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 6 | |
0k3k | Problem:
Find the sum of the digits of $11 \cdot 101 \cdot 111 \cdot 110011$. | [
"Solution:\nThere is no regrouping, so the answer is $2 \\cdot 2 \\cdot 3 \\cdot 4=48$. The actual product is 13566666531 ."
] | United States | HMMT February 2018 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 48 | |
085z | Problem:
Sia $A B C D E F G H I L M N$ un dodecagono regolare. Sia $P$ il punto di intersezione delle diagonali $A F$ e $D H$. Sia $S$ la circonferenza passante per $A$ e $H$, congruente a quella circoscritta al dodecagono e distinta da essa. Dimostrare che:
(a) $P$ appartiene a $S$;
(b) il centro di $S$ appartiene al... | [
"Solution:\n\nSia $T$ la circonferenza circoscritta al dodecagono, e sia $O$ il suo centro; sia $O_{S}$ il centro della circonferenza $S$ e sia $Q$ il punto medio di $A H$. I punti $O$ e $O_{S}$ sono sull'asse di $A H$ e, avendo la stessa distanza da $A$ ed $H$, sono simmetrici rispetto ad $A H$, e anche simmetrici... | Italy | Cesenatico | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Circles"
] | null | proof only | null | |
08jl | Problem:
Pentru orice număr natural $n$ notăm cu $C(n)$ suma cifrelor sale în baza 10. Arătaţi că oricare ar fi numărul natural $k$ există un număr natural $m$ astfel încât ecuaţia $x+C(x)=m$ are cel puţin $k$ soluţii. | [] | JBMO | Olimpiada Naţională de Matematică | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0e3y | Find the greatest common divisor of all even numbers of the form $n^4 - 4n^2$ where $n$ is a positive integer. | [] | Slovenia | Selection Examinations for the IMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 192 | |
0aw5 | Problem:
A small class of nine boys are to change their seating arrangement by drawing their new seat numbers from a box. After the seat change, what is the probability that there is only one pair of boys who have switched seats with each other and only three boys who have unchanged seats? | [] | Philippines | 19th Philippine Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 1/48 | |
0dco | The altitudes $A A_{1}$ and $B B_{1}$ of an acute-angled triangle $A B C$ meet at point $O$. Let $A_{1} A_{2}$ and $B_{1} B_{2}$ be the altitudes of triangles $O B A_{1}$ and $O A B_{1}$ respectively. Prove that $A_{2} B_{2}$ is parallel to $A B$. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0aym | Problem:
Let $S = \{1, 2, 3, \ldots, 12\}$. Find the number of nonempty subsets $T$ of $S$ such that if $x \in T$ and $3x \in S$, then it follows that $3x \in T$. | [] | Philippines | 21st PMO Area Stage | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1151 | |
0kjb | For each integer $n \ge 2$, let $S_n$ be the sum of all products $jk$, where $j$ and $k$ are integers and $1 \le j < k \le n$. What is the sum of the 10 least values of $n$ such that $S_n$ is divisible by 3?
(A) 196 (B) 197 (C) 198 (D) 199 (E) 200 | [] | United States | 2021 AMC 10 B Fall | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | B | |
0knb | Problem:
Let scalene triangle $A B C$ have circumcenter $O$ and incenter $I$. Its incircle $\omega$ is tangent to sides $B C, C A$, and $A B$ at $D, E$, and $F$, respectively. Let $P$ be the foot of the altitude from $D$ to $E F$, and let line $D P$ intersect $\omega$ again at $Q \neq D$. The line $O I$ intersects the... | [
"Solution:\n\nLet $H$ be the orthocenter of $\\triangle D E F$. We first claim that $O, I, H$ are collinear. We present two proofs.\n\nProof 1. Invert about $\\omega$. Circle $(A B C)$ inverts to a circle with center on $O I$, but $A, B, C$ invert to the midpoints of $E F, F D, D E$, respectively, so the nine-point... | United States | HMMT Spring 2021 Team Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations >... | null | proof only | null | |
08ut | Let $ABCDEFG$ be a regular heptagon. Suppose $P$, $Q$, $R$, $S$ are points on the line segments $AB$, $BC$, $CD$, $EF$, respectively, for which $BP = CQ = DR = FS = \frac{1}{3}$ are satisfied. Let $T$ be the point of intersection of the line segments $PR$ and $QS$. Find the value of $\angle PTS$.
Here by $XY$ we mean ... | [
"$\\boxed{\\frac{720^{\\circ}}{7}}$\n\nLet $U$, $V$, $W$ be points on the sides $DE$, $FG$, $GA$, respectively, dividing respective side in $2 : 1$ ratio. Then, it is easy to see that $PQRUSVW$ becomes also a regular heptagon. Hence these $7$ points lie on the circumference of a same circle, and divides the circumf... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 720/7 degrees | |
06a0 | Let $AB\Gamma$ be an acute angled triangle with $AB < A\Gamma$. Let $\Delta$ be the midpoint of the side $B\Gamma$ and $BE$, $\Gamma Z$ altitudes of the triangle $AB\Gamma$. The line $ZE$ meets the line $B\Gamma$ at point $\Theta$.
a. Find the angles of the triangle $Z\Delta E$ with respect to the angle $\hat{A}$ of $... | [
"a.\nThe triangle $BZ\\Gamma$ is right angled at $Z$ and $Z\\Delta$ is median.\nHence $Z\\Delta = \\frac{B\\Gamma}{2} = B\\Delta$. From the isosceles triangle $B\\Delta Z$ it follows that:\n$$\n\\hat{\\Delta}_1 = B\\hat{\\Delta}Z = 180^\\circ - 2\\hat{B}. \\qquad (1)\n$$\nSimilarly we get $\\Delta E = \\frac{B\\Gam... | Greece | 37th Hellenic Mathematical Olympiad 2020 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | Angles of triangle ZΔE: at Z equals A, at Δ equals 180 − 2A, at E equals A. Angle BΘZ equals B − Γ. | |
0ioh | Problem:
A circle inscribed in a square,
Has two chords as shown in a pair.
It has radius $2$,
And $P$ bisects $TU$.
The chords' intersection is where?
Answer the question by giving the distance of the point of intersection from the center of the circle. | [
"Solution:\nAnswer: $2\\sqrt{2} - 2$.\nLet $OB$ intersect the circle at $X$ and $Y$, and the chord $PM$ at $Q$, such that $O$ lies between $X$ and $Q$. Then $MNXQ$ is a parallelogram. For, $OB \\parallel NM$ by homothety at $C$ and $PM \\parallel NX$ because $MNXP$ is an isosceles trapezoid. It follows that $QX = M... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | null | final answer only | 2√2 - 2 | |
0jyu | Problem:
Let $w = w_{1} w_{2} \ldots w_{n}$ be a word. Define a substring of $w$ to be a word of the form $w_{i} w_{i+1} \ldots w_{j-1} w_{j}$, for some pair of positive integers $1 \leq i \leq j \leq n$. Show that $w$ has at most $n$ distinct palindromic substrings.
For example, aaaaa has 5 distinct palindromic subs... | [
"Solution:\n\nFor each palindrome substring appearing in $w$, consider only the leftmost position in which it appears. I claim that now, no two substrings share the same right endpoint. If some two do, then you can reflect the smaller one about the center of the larger one to move the smaller one left."
] | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
09v0 | A four digit number $\overline{aabb}$, that is, the number whose digits are $a, a, b$, and $b$, is the square of an integer.
Of which integer is $\overline{aabb}$ the square? | [
"$88$"
] | Netherlands | Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | final answer only | 88 | |
04jx | In an acute triangle $ABC$ we have $|AB| > |BC|$, and the points $A_1$ and $C_1$ are the feet of altitudes from the vertices $A$ and $C$, respectively. Let $D$ be the second intersection of the circumcircles of triangles $ABC$ and $A_1BC_1$ (different from $B$). Let $Z$ be the intersection of the tangents of the circum... | [
"\nSince $\\angle CAZ = \\angle ZCA = \\beta$ (angle between the tangent and the chord), it follows that $\\angle XZY = \\angle AZC = 180^\\circ - 2\\beta$.\n\nQuadrilateral $CAC_1A_1$ is cyclic, so we have $\\angle C_1A_1B = \\angle YA_1C = \\alpha$ and $\\angle BC_1A_1 = \\angle AC_1X = \... | Croatia | Croatian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00ub | Let $a$, $b$, $c$, $d$ be non-negative real numbers such that
$$
\frac{1}{a+3} + \frac{1}{b+3} + \frac{1}{c+3} + \frac{1}{d+3} = 1.
$$
Prove that there is a permutation $(x_1, x_2, x_3, x_4)$ of the sequence $(a, b, c, d)$ such that
$$
x_1x_2 + x_2x_3 + x_3x_4 + x_4x_1 \ge 4.
$$ | [
"Assume that $a \\ge b \\ge c \\ge d$. We will show that the sequence $(x_1, x_2, x_3, x_4) = (a, b, d, c)$ satisfies the requirement $x_1x_2 + x_2x_3 + x_3x_4 + x_4x_1 \\ge 4$, i.e. $(a+d)(b+c) \\ge 4$. Denoting\n$$\nx = \\frac{a+d}{2}, \\quad y = \\frac{b+c}{2},\n$$\nwe need to show that $xy \\ge 1$ using $a \\ge... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
06fg | The incircle of a scalene and acute $\triangle ABC$ touches $BC$, $CA$ and $AB$ at $D$, $E$ and $F$ respectively. $H$ is a point on the segment $EF$ such that $DH \perp EF$. Suppose $AH \perp BC$, prove that $H$ is the orthocentre of $\triangle ABC$. | [
"Let $P$ be the midpoint of $DF$. Then $\\angle FPB = 90^\\circ$. Since $\\angle BFP = \\angle DEH$, we have $\\triangle BPF \\sim \\triangle DHE$. This implies $\\frac{BF}{PF} = \\frac{DE}{HE}$, which yields\n$$\nBF \\times HE = DE \\times PF = \\frac{1}{2}DE \\times DF.\n$$\nBy symmetry,\n$$\nCE \\times HF = \\fr... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null |
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