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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
042u | Suppose that $a > 0$ and the minima of function $f(x) = x + \frac{100}{x}$ on intervals $(0, a]$ and $[a, +\infty)$ are $m_1, m_2$, respectively. If $m_1 m_2 = 2020$, then the value of $a$ is ______. | [
"Note that $f(x)$ is monotonically decreasing on $(0, 10]$ and monotonically increasing on $[10, +\\infty)$. When $a \\in (0, 10]$, $m_1 = f(a)$, $m_2 = f(10)$; when $a \\in [10, +\\infty)$, $m_1 = f(10)$, $m_2 = f(a)$. Therefore, there is always\n$$\nf(a)f(10) = m_1m_2 = 2020,\n$$\nnamely, $a + \\frac{100}{a} = \\... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 1 or 100 | |
01fc | A hacker is locked into an underground industrial complex. She is presented with a computer screen, on which appears a long message of length $72$, consisting of the symbols $E$, $X$, $I$, $T$, exactly $18$ letters of each kind in some seemingly random order. The message may be manipulated by inserting any one of the c... | [
"Let $18 = n$, so that the initial message has length $4n$, with exactly $n$ symbols of each kind.\nWe first establish an invariant. Assign\n$E = 3$, $X = -3$, $I = 2$, $T = -2$,\nand let $S$ denote the sum of the values of all symbols appearing in the message. Initially, $S = 0$, and the sum stays invariant under ... | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | English | proof only | null | |
0key | Problem:
In triangle $ABC$ with $AB=8$ and $AC=10$, the incenter $I$ is reflected across side $AB$ to point $X$ and across side $AC$ to point $Y$. Given that segment $XY$ bisects $AI$, compute $BC^{2}$. (The incenter $I$ is the center of the inscribed circle of triangle $ABC$.)
Proposed by: Carl Schildkraut | [
"Solution:\n\n\n\nLet $E, F$ be the tangency points of the incircle to sides $AC, AB$, respectively. Due to symmetry around line $AI$, $AXIY$ is a rhombus. Therefore\n$$\n\\angle XAI = 2 \\angle EAI = 2\\left(90^{\\circ} - \\angle EIA\\right) = 180^{\\circ} - 2 \\angle XAI,\n$$\nwhich impli... | United States | HMMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | null | proof and answer | 84 | |
00z1 | Problem:
Prove that for positive $a, b, c, d$
$$
\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a} \geq 4.
$$ | [
"Solution:\nThe inequality between the arithmetic and harmonic mean gives\n$$\n\\begin{aligned}\n& \\frac{a+c}{a+b}+\\frac{c+a}{c+d} \\geq \\frac{4}{\\frac{a+b}{a+c}+\\frac{c+d}{c+a}} = 4 \\cdot \\frac{a+c}{a+b+c+d} \\\\\n& \\frac{b+d}{b+c}+\\frac{d+b}{d+a} \\geq \\frac{4}{\\frac{b+c}{b+d}+\\frac{d+a}{d+b}} = 4 \\c... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
07o6 | $A$, $B$ and $C$ are points on the circumference of a circle with centre $O$, such that $\triangle ABC$ is not a right-angled triangle. The point $P$ lies on the circumcircle $\Gamma_1$ of the triangle $OAB$ such that $OP$ is a diameter of $\Gamma_1$. The point $Q$ lies on the circumcircle $\Gamma_2$ of the triangle $O... | [
"Case 1: $X$ and $P$ lie on the same side of the line $AO$.\n\n\n\n*Step 1:* $\\angle QAO = 90^\\circ$ (angle in a semicircle) and similarly $\\angle PAO = 90^\\circ$.\nTherefore $PAQ$ is a straight line and is a tangent to the circumcircle of $\\triangle ABC$ at the point $A$.\n\n*Step 2:*... | Ireland | Ireland | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
037n | Problem:
Find all pairs $(P, Q)$ of polynomials with real coefficients such that
$$
\frac{P(x)}{Q(x)}-\frac{P(x+1)}{Q(x+1)}=\frac{1}{x(x+2)}
$$
for infinitely many $x \in \mathbb{R}$. | [
"Solution:\nFirst solution. It suffices to consider the case when $P$ and $Q\\not\\equiv 0$ are relatively prime polynomials and the leading coefficient of $Q$ equals $1$. We have\n$$\nx(x+2)(P(x) Q(x+1)-Q(x) P(x+1))=Q(x) Q(x+1)\n$$\nfor infinitely many $x$, i.e. for every $x$. Thus the polynomials $Q(x)$ and $Q(x+... | Bulgaria | Team selection test for 47. IMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | All solutions are of the form Q(x) = x(x+1) R(x) and P(x) = (1/2 + x + c x(x+1)) R(x), where R is any nonzero polynomial and c is a real constant. | |
0f67 | Problem:
Two players play a game. Each takes it in turn to paint three unpainted edges of a cube. The first player uses red paint and the second blue paint. So each player has two moves. The first player wins if he can paint all edges of some face red. Can the first player always win? | [] | Soviet Union | 18th ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | No | |
04k9 | If $x$, $y$, $z$ and $w$ are real numbers such that
$$
\frac{x}{y+z+w} + \frac{y}{z+w+x} + \frac{z}{w+x+y} + \frac{w}{x+y+z} = 1,
$$
find
$$
\frac{x^2}{y+z+w} + \frac{y^2}{z+w+x} + \frac{z^2}{w+x+y} + \frac{w^2}{x+y+z}.
$$ | [
"If we multiply the condition by $x + y + z + w$, we get:\n$$\n\\frac{x^2 + x(y + z + w)}{y + z + w} + \\frac{y^2 + y(x + z + w)}{z + w + x} + \\frac{z^2 + z(x + y + w)}{w + x + y} + \\frac{w^2 + w(x + y + z)}{x + y + z} = x + y + z + w,\n$$\ni.e.\n$$\n\\frac{x^2}{y+z+w} + x + \\frac{y^2}{z+w+x} + y + \\frac{z^2}{w... | Croatia | Mathematical competitions in Croatia | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | 0 | |
0fzz | Problem:
Finde alle Tripel $(a, b, c)$ natürlicher Zahlen, sodass
$$
\frac{a+b}{c}, \frac{b+c}{a}, \frac{c+a}{b}
$$
ebenfalls natürliche Zahlen sind. | [
"Solution:\nWir unterscheiden drei Fälle, und zwar, dass die drei Zahlen gleich sind, dass zwei der drei Zahlen gleich sind und dass die drei Zahlen verschieden sind.\n\nFall 1: $a = b = c$\nDies ergibt die Lösung $(a, a, a)$.\n\nFall 2: Wir haben zwei gleiche und eine andere Zahl.\nNehme an, dass $a = b \\neq c$. ... | Switzerland | SMO - Vorrunde | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All triples that are permutations of (a, a, a), (a, a, 2a), and (a, 2a, 3a), where a is any natural number. | |
02dg | Two thieves stole a container of $8$ liters of wine. How can they divide it into two parts of $4$ liters each if all they have is a $3$ liter container and a $5$ liter container? Consider the general case of dividing $m+n$ liters into two equal amounts, given a container of $m$ liters and a container of $n$ liters (whe... | [
"Call the containers $L_8$, $L_5$, $L_3$. Fill $L_5$ from $L_8$, then fill $L_3$ from $L_5$, leaving $2$ in $L_5$. Empty $L_3$ into $L_8$. Empty $L_5$ into $L_3$ (so now $L_8$ has $6$, $L_5$ has $0$, $L_3$ has $2$). Fill $L_5$ from $L_8$. Fill $L_3$ from $L_5$. Empty $L_3$ into $L_8$. Now $L_5$ and $L_8$ each conta... | Brazil | III OBM | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0aqi | Problem:
Find the sum of all (numerical) coefficients in the expansion of $(x+y+z)^3$. | [
"Solution:\n\nTo find the sum of all numerical coefficients in the expansion of $(x+y+z)^3$, substitute $x=1$, $y=1$, $z=1$:\n\n$$(1+1+1)^3 = 3^3 = 27.$$\n\nTherefore, the sum of all coefficients is $27$."
] | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 27 | |
09t3 | Problem:
Zij $n$ een positief geheel getal. Gegeven zijn cirkelvormige schijven met stralen $1,2, \ldots, n$. Van elke grootte hebben we twee schijven: een doorzichtige en een ondoorzichtige. In elke schijf zit een gaatje, precies in het midden, waarmee we de schijven op een rechtopstaand staafje kunnen stapelen. We w... | [
"Solution:\n\nNoem een stapel geldig als hij aan de voorwaarden voldoet. Zij $a_{n}$ het aantal geldige stapels met $n$ schijven (met straal $1,2, \\ldots, n$ ). We bewijzen met inductie dat $a_{n}=(n+1)!$.\n\nVoor $n=1$ kunnen we twee stapels maken: met de doorzichtige schijf met straal $1$ en met de ondoorzichtig... | Netherlands | MO-selectietoets | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | (n+1)! | |
0cmm | 2009 nonnegative integers are arranged on a circle, each number does not exceed $100$. A positive integer $k$ is fixed. By one move, one can choose two neighboring positions on a circle and add $1$ to both numbers in these positions. It is allowed to make at most $k$ moves for each pair of neighboring positions. Find t... | [
"**Ответ.** $k = 100400$.\n\nОбозначим числа на окружности через $a_1, \\dots, a_{2009}$, и положим $a_{n+2009} = a_n = a_{n-2009}$. Пусть $N = 100400$.\n\n1. Положим $a_2 = a_4 = \\dots = a_{2008} = 100$ и $a_1 = a_3 = \\dots = a_{2009} = 0$. Пусть мы сумели сделать все числа равными при каком-то значении $k$. Рас... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English; Russian | proof and answer | 100400 | |
010n | Problem:
Can the points of a disc of radius $1$ (including its circumference) be partitioned into three subsets in such a way that no subset contains two points separated by distance $1$? | [
"Solution:\n\nAnswer: no.\nLet $O$ denote the centre of the disc, and $P_{1}, \\ldots, P_{6}$ the vertices of an inscribed regular hexagon in the natural order (see Figure 4).\nIf the required partitioning exists, then $\\{O\\}, \\{P_{1}, P_{3}, P_{5}\\}$ and $\\{P_{2}, P_{4}, P_{6}\\}$ are contained in different s... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | no | |
04wk | Let $n$ be a given positive integer. Solve the system of equations
$$
\begin{aligned}
x_1 + x_2^2 + x_3^3 + \dots + x_n^n &= n, \\
x_1 + 2x_2 + 3x_3 + \dots + nx_n &= \frac{n(n+1)}{2}
\end{aligned}
$$
in the set of nonnegative real numbers $x_1, x_2, \dots, x_n$. | [
"Suppose $x_1, x_2, \\dots, x_n$ satisfy the equations above. Then we have\n$$\n\\begin{aligned}\n0 &= x_1 + x_2^2 + x_3^3 + \\dots + x_n^n - n - (x_1 + 2x_2 + 3x_3 + \\dots + nx_n - \\frac{1}{2}n(n+1)) \\\\\n&= (x_2^2 - 2x_2 + 2 - 1) + (x_3^3 - 3x_3 + 3 - 1) + \\dots + (x_n^n - nx_n + n - 1).\n\\end{aligned}\n$$\n... | Czech-Polish-Slovak Mathematical Match | Czech-Polish-Slovak Match | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | x1 = x2 = ... = xn = 1 | |
0b1a | Problem:
In trapezoid $ABCD$, $AD$ is parallel to $BC$. If $AD = 52$, $BC = 65$, $AB = 20$, and $CD = 11$, find the area of the trapezoid. | [
"Solution:\n\nExtend $AB$ and $CD$ to intersect at $E$. Then $\\sqrt{\\frac{[EAD]}{[EBC]}} = \\frac{AD}{BC} = \\frac{4}{5} = \\frac{EA}{EB} = \\frac{ED}{EC}$. This tells us that $EB = 5 AB = 100$, and $EC = 5 CD = 55$. Triangle $EBC$ has semiperimeter $110$, and so by Heron's formula, the area of triangle $EBC$ is ... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 594 | |
03s1 | Find all non-negative integer solutions $(x, y, z, w)$ of the following equation
$$
2^x \cdot 3^y - 5^z \cdot 7^w = 1.
$$ | [
"Since $5^z \\cdot 7^w + 1$ is even, we have $x \\ge 1$.\n\nCase 1: $y = 0$. The equation to be solved becomes\n$$\n2^x - 5^z \\cdot 7^w = 1.\n$$\nIf $z \\neq 0$, then $2^x \\equiv 1 \\pmod{5}$. It follows that $4 \\mid x$. Thus $3 \\mid 2^x - 1$, which contradicts to $2^x - 5^z \\cdot 7^w = 1$.\nIf $z = 0$, then\n... | China | China Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | (1, 0, 0, 0), (3, 0, 0, 1), (1, 1, 1, 0), (2, 2, 1, 1) | |
02a7 | Problem:
Lados de um paralelepípedo - Se $x$ e $y$ são números inteiros positivos tais que $x y z=240$, $x y+z=46$ e $x+y z=64$, qual é o valor de $x+y+z$?
(a) 19
(b) 20
(c) 21
(d) 24
(e) 36 | [
"Solution:\n\nSolução 1: De $x y z=240$, segue que $x y=\\frac{240}{z}$. Substituindo em $x y+z=46$, obtemos $\\frac{240}{z}+z=46$, ou seja, $z^{2}-46z+240=0$. As raízes dessa equação são números cuja soma é 46 e cujo produto é 240, e é fácil verificar que essas raízes são 6 e 40. Logo, $z=6$ ou $z=40$. De maneira ... | Brazil | Nível 2 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | b | |
0e02 | Problem:
S tanko palico neznane dolžine želimo ugotoviti prav tako neznani širino in višino vrat. Če položimo palico vodoravno ob vratih, je ta za 2 laketa daljša od širine vrat. Če palico postavimo navpično, je za 1 laket daljša od višine vrat. Palica se natanko prilega odprtini vrat, če jo postavimo diagonalno med v... | [
"Solution:\n\nOznačimo dolžino palice z $d$, širino vrat z $x$ in višino vrat z $y$. Veljajo zveze $x = d - 2$, $y = d - 1$ in $x^{2} + y^{2} = d^{2}$.\n\nReševanje sistema treh enačb s tremi neznankami privede do enačbe $d^{2} - 6d + 5 = 0$ in rešitev $d_{1} = 1$ in $d_{2} = 5$. Rešitev $d = 1$ ne ustreza. Iz $d =... | Slovenia | Državno tekmovanje | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | width = 3, height = 4, rod length = 5 | |
0e04 | Let $n$ be a positive integer. $n \ge 3$. There are $n$ pairwise different numbers written on a blackboard. Show that we can choose two of those numbers so that no number from the blackboard multiplied by $3$ is equal to a multiple of their sum. | [
"Denote the numbers by $a_1, a_2, \\dots, a_n$. Without loss of generality we may assume that $a_1 > a_2 > \\dots > a_n$. Let us show we may also assume that not all of these numbers are divisible by $3$. If $b_1, \\dots, b_n$ are all divisible by $3$, then there exists a positive integer $k$ such that $3^k$ divide... | Slovenia | Selection Examinations for the IMO | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
03nr | Problem:
Assume that real numbers $a$ and $b$ satisfy
$$
a b + \sqrt{a b + 1} + \sqrt{a^{2} + b} \cdot \sqrt{b^{2} + a} = 0
$$
Find, with proof, the value of
$$
a \sqrt{b^{2} + a} + b \sqrt{a^{2} + b}
$$ | [
"Solution:\nLet us rewrite the given equation as follows:\n$$\na b + \\sqrt{a^{2} + b} \\sqrt{b^{2} + a} = -\\sqrt{a b + 1}.\n$$\nSquaring this gives us\n$$\n\\begin{aligned}\na^{2} b^{2} + 2 a b \\sqrt{a^{2} + b} \\sqrt{b^{2} + a} + (a^{2} + b)(b^{2} + a) & = a b + 1 \\\\\n(a^{2} b^{2} + a^{3}) + 2 a b \\sqrt{a^{2... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 1 | |
0jy3 | Problem:
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function satisfying $f(x) f(y) = f(x-y)$. Find all possible values of $f(2017)$. | [
"Solution:\nLet $P(x, y)$ be the given assertion. From $P(0,0)$ we get $f(0)^2 = f(0) \\Longrightarrow f(0) = 0, 1$.\nFrom $P(x, x)$ we get $f(x)^2 = f(0)$. Thus, if $f(0) = 0$, we have $f(x) = 0$ for all $x$, which satisfies the given constraints. Thus $f(2017) = 0$ is one possibility.\n\nNow suppose $f(0) = 1$. W... | United States | February 2017 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 0 or 1 | |
033p | Problem:
Let $A_{1}, A_{2}, \ldots, A_{n}$ be finite sets such that
$$
\left|A_{i} \cap A_{i+1}\right|>\frac{n-2}{n-1}\left|A_{i+1}\right|
$$
for any $i=1,2, \ldots, n$ ($A_{n+1} \equiv A_{1}$). Prove that their intersection is a nonempty set. | [
"Solution:\nWe may assume the set $A_{1}$ has maximal cardinality. Denote $A_{i} \\cap A_{i+1} = B_{i}$, $i=1,2, \\ldots, n$. Since $A_{n} \\supset B_{n-1} \\cup B_{n}$, then\n$$\n\\begin{aligned}\n\\left|A_{n}\\right| & \\geq \\left|B_{n-1} \\cup B_{n}\\right| = \\left|B_{n-1}\\right| + \\left|B_{n}\\right| - \\le... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
06pi | Given a convex $n$-gon $P$ in the plane. For every three vertices of $P$, consider the triangle determined by them. Call such a triangle good if all its sides are of unit length.
Prove that there are not more than $\frac{2}{3} n$ good triangles. | [
"Consider all good triangles containing a certain vertex $A$. The other two vertices of any such triangle lie on the circle $\\omega_{A}$ with unit radius and center $A$. Since $P$ is convex, all these vertices lie on an arc of angle less than $180^{\\circ}$. Let $L_{A} R_{A}$ be the shortest such arc, oriented clo... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0b3o | Problem:
Let $a$, $b$, $c$ be real numbers such that
$$
3 a b + 2 = 6 b, \quad 3 b c + 2 = 5 c, \quad 3 c a + 2 = 4 a
$$
Suppose the only possible values for the product $a b c$ are $r / s$ and $t / u$, where $r / s$ and $t / u$ are both fractions in lowest terms. Find $r+s+t+u$. | [
"Solution:\nThe three given equations can be written as\n\n$$\n3 a + \\frac{2}{b} = 12, \\quad 3 b + \\frac{2}{c} = 10, \\quad 3 c + \\frac{2}{a} = 8\n$$\n\nThe product of all the three equations gives us\n$$\n27 a b c + 6\\left(3 a + \\frac{2}{b}\\right) + 6\\left(3 b + \\frac{2}{c}\\right) + 6\\left(3 c + \\frac{... | Philippines | 24th Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 18 | |
01gu | Figure shows two non-intersecting circles $\alpha$ and $\beta$ in space. We say that circle $\alpha$ *devours* circle $\beta$ since one chord of $\beta$ (solid) is strictly contained in a chord of $\alpha$ (dashed).
The question is whether it is possible to place three circles $\alpha, \beta$ and $\gamma$ in space so a... | [
"Answer: No, it is impossible.\nConsider a point $X$ on a chord drawn in circle $\\alpha$, as in Figure ???. The power of $X$ with respect to $\\alpha$ is given by the familiar expression $p_{\\alpha}(X) = -xy$.\n\nLet us examine the case of two circles. The situation when $\\alpha$ devours $\\beta$ is represented ... | Baltic Way | Baltic Way 2020 | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof and answer | No, it is impossible. | |
05kp | Problem:
Soient $C$ et $C'$ deux cercles de centres $O$ et $O'$, extérieurs l'un à l'autre. Une tangente commune extérieure coupe les deux tangentes communes intérieures aux points $M$ et $N$.
Montrer que $(OM)$ est perpendiculaire à $(O'M)$ et que $(ON)$ est perpendiculaire à $(O'N)$. | [
"Solution:\n\nSoit $(T)$ la tangente commune extérieure de l'énoncé. Soit $(T')$ la tangente commune intérieure passant par $M$. Les droites $(T)$ et $(T')$ sont donc les deux tangentes à $C$ issues de $M$.\n\nComme ces deux tangentes sont symétriques par rapport à $(OM)$, la droite $(OM)$ est une bissectrice de $(... | France | Olympiades Françaises de Mathématiques - Test de Janvier | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05sn | Problem:
Trouver tous les couples d'entiers positifs $(x, y)$ tels que $2^{x}+5^{y}+2$ est un carré parfait. | [
"Solution:\n\nIci on est face à un problème d'équation diophantienne avec un carré et une puissance de $2$. On peut se rendre compte en testant les petits cas que $(x, y) = (0, 0)$ et $(1, 1)$ sont solutions. Comme on a un carré et une puissance de $2$, on est très tenté de regarder modulo $4$ ou $8$. Regardons déj... | France | Envoi 5: Pot Pourri | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | ((0, 0), (1, 1)) | |
07ws | Let $K, L, M$ denote three points on the sides $BC, AB$ and $AC$ of $\triangle ABC$, so that $ALKM$ is a parallelogram. Points $S$ and $T$ are chosen on lines $KL$ and $KM$ respectively, so that the quadrilaterals $ASBK$ and $AKCT$ are both cyclic. Prove that $SLMT$ is cyclic if and only if $K$ is the midpoint of $BC$. | [
"The problem can be solved by angle chasing, by algebraic equations resulting from similar triangles, or by a combination of the two methods.\n\nWe first show that $S, A, T$ are collinear. Because $ASBK$ is cyclic, we have $\\angle SAB = \\angle SKB$. From $KS \\parallel AC$ we get $\\angle SKB = \\angle C$, so $\\... | Ireland | IRL_ABooklet_2024 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03qh | Let $M(-1, 2)$ and $N(1, 4)$ be two points in a plane rectangular coordinate system $xOy$. $P$ is a moving point on the $x$-axis. When $\angle MPN$ takes its maximum value, the $x$-coordinate of point $P$ is ________. | [
"The center of a circle passing through points $M$ and $N$ is on the perpendicular bisector $y = 3 - x$ of $MN$. Denote the center by $S(a, 3-a)$, then the equation of the circle $S$ is\n$$\n(x-a)^2 + (y-3+a)^2 = 2(1+a^2).\n$$\nSince for a chord with a fixed length, the angle at the circumference subtended by the c... | China | China Mathematical Competition (Hainan) | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 1 | |
03ko | Problem:
Let $r_{1}, r_{2}, \ldots, r_{m}$ be a given set of $m$ positive rational numbers such that $\sum_{k=1}^{m} r_{k}=1$. Define the function $f$ by $f(n)=n-\sum_{k=1}^{m}\left[r_{k} n\right]$ for each positive integer $n$. Determine the minimum and maximum values of $f(n)$. Here $[x]$ denotes the greatest integer... | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | minimum = 0, maximum = m - 1 | |
01ij | Let $a_1, a_2, \dots, a_{2023}$ be positive real numbers with
$$
a_1 + a_2^2 + a_3^3 + \dots + a_{2023}^{2023} = 2023.
$$
Show that
$$
a_1^{2023} + a_2^{2022} + \dots + a_{2022}^2 + a_{2023} > 1 + \frac{1}{2023}.
$$ | [
"Let us prove that conversely, the condition\n$$\na_1^{2023} + a_2^{2022} + \\dots + a_{2023} \\le 1 + \\frac{1}{2023}\n$$\nimplies that\n$$\nS := a_1 + a_2^2 + \\dots + a_{2023}^{2023} < 2023.\n$$\nThis is trivial if all $a_i$ are less than $1$. So suppose that there is an $i$ with $a_i \\ge 1$, clearly it is uniq... | Baltic Way | Baltic Way 2023 Shortlist | [
"Algebra > Intermediate Algebra > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
05bx | The incentre of a triangle $ABC$ is $I$. Points $D$ and $E$ on the sides $AB$ and $AC$, respectively, satisfy $DI \perp BI$ and $EI \perp CI$. Prove that the line $DE$ is tangent to the incircle of the triangle $ABC$. | [
"Let $X$ and $Y$ be the reflections of points $D$ and $E$, respectively, across the point $I$ (Fig. 32). Then $\\angle XBI = \\angle IBD = \\angle IBA = \\angle CBI$, implying that $X$ lies on the line $BC$. As $\\angle DIB = 90^\\circ$, points $D$, $I$ and $X$ lie on a line, i.e., $X$ is the point of intersection ... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilater... | English | proof only | null | |
0i9o | Let $a$, $b$, $c$ be positive real numbers. Prove that
$$
\frac{(2a + b + c)^2}{2a^2 + (b+c)^2} + \frac{(2b + c + a)^2}{2b^2 + (c+a)^2} + \frac{(2c + a + b)^2}{2c^2 + (a+b)^2} \le 8.
$$ | [
"**First Solution.** (Based on work by Matthew Tang and Anders Kaseorg)\nBy multiplying $a$, $b$, and $c$ by a suitable factor, we reduce the problem to the case when $a + b + c = 3$. The desired inequality reads\n$$\n\\frac{(a+3)^2}{2a^2+(3-a)^2} + \\frac{(b+3)^2}{2b^2+(3-b)^2} + \\frac{(c+3)^2}{2c^2+(3-c)^2} \\le... | United States | USA IMO 2003 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
01xf | Positive integers $a$, $b$ and $c$ satisfy the equality
$$
\frac{a^2 - a - c}{b} + \frac{b^2 - b - c}{a} = a + b + 2.
$$
Prove that $a + b + c$ is a square of a positive integer. | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0ldx | Let $ABC$ be an acute, non-isosceles triangle with $H, O, O'$ as its orthocenter, circumcenter, nine-point center, and $D, E, F$ as the midpoints of the segments $BC, CA, AB$, respectively. $P$ is an arbitrary point inside triangle $DEF$. Let $DP, EP, FP$ intersect $(O')$ again at $D', E', F'$, respectively. $A'$ is th... | [
"(a) Let $I$ be the reflection of $O$ with respect to $P$. Since $O'$ is the midpoint of $OH$, it follows that $O'P \\parallel IH$. Moreover, we have $PO = PO'$, thus $IO = IH$.\nLet $S, G$ be midpoints of $AI$ and $AH$, respectively. We have\n$$\nSP = \\frac{1}{2}AO = \\frac{1}{2}R = O'D\n$$\nand $SP \\parallel AO... | Vietnam | VN IMO Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Pla... | English | proof only | null | |
09ld | What is the minimum perimeter of a scalene and acute-angled triangle whose sides are square numbers? | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 245 | |
042n | Given geometric sequence $\{a_n\}$, $a_9 = 13$, $a_{13} = 1$, then the value of $\log_{a_1} 13$ is ______. | [
"By the properties of geometric sequence, we have $\\frac{a_1}{a_9} = \\left(\\frac{a_9}{a_{13}}\\right)^2$, and thus $a_1 = \\frac{a_9^3}{a_{13}^2} = 13^3$.\n\nConsequently, $\\log_{a_1} 13 = \\frac{1}{3}$. $\\square$"
] | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 1/3 | |
0ckw | Determine the positive real numbers $a, b, c, d$ such that $a + b + c + d = 80$ and
$$
a + \frac{b}{1+a} + \frac{c}{1+a+b} + \frac{d}{1+a+b+c} = 8.
$$ | [
"Adding $4$ to both sides of the second equation, we write:\n$$\n1 + a + \\frac{1+a+b}{1+a} + \\frac{1+a+b+c}{1+a+b} + \\frac{1+a+b+c+d}{1+a+b+c} = 12.\n$$\nApplying the AM-GM inequality successively, we obtain:\n$$\n\\begin{aligned}\n1 + a + \\frac{1+a+b}{1+a} &\\ge 2\\sqrt{(1+a) \\cdot \\frac{1+a+b}{1+a}} = 2\\sq... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a = 2, b = 6, c = 18, d = 54 | |
0le4 | An integer sequence $(x_n)$ is defined as follows: $0 \le x_0 < x_1 \le 100$ and
$$
x_{n+2} = 7x_{n+1} - x_n + 280, \forall n \ge 0.
$$
a. Prove that if $x_0 = 2, x_1 = 3$ then for each positive integer $n$, the sum of divisors of the following number is divisible by 24
$$
x_n x_{n+1} + x_{n+1} x_{n+2} + x_{n+2} x_{n+... | [
"**Lemma 1.** If a positive integer $n$ satisfies $24|n+1$ then the sum of its positive divisors $\\sigma(n)$ is divisible by 24.\n\n*Proof.* Indeed, if $d$ is a divisor of $n$ then $\\frac{n}{d}$ is also a divisor of $n$. Because $n \\equiv 2 \\pmod{3}$ so it cannot be a perfect square, which means the sum of its ... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | (2, 3) | |
08g7 | Problem:
Siano $a, b, c$ tre numeri reali (positivi, negativi o nulli) tali che $a^{2}+b^{2}+c^{2}=6$.
a) Determinare il massimo valore possibile per l'espressione
$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2} .
$$
b) Determinare il massimo valore possibile per l'espressione
$$
(a-b)^{2} \cdot(b-c)^{2} \cdot(c-a)^{2} .
$$
In en... | [
"Solution:\n\nIl massimo valore possibile è $18$, e viene realizzato da tutte e sole le terne che, oltre alla condizione $a^{2}+b^{2}+c^{2}=6$, verificano anche $a+b+c=0$ (ad esempio la terna con $a=b=1$ e $c=-2$).\nPer dimostrarlo basta osservare che\n$$\n\\begin{aligned}\n(a-b)^{2}+(b-c)^{2}+(c-a)^{2} & =2\\left(... | Italy | Olimpiade Italiana di Matematica | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | a) Maximum value: 18, attained exactly by all triples with a+b+c=0 and a^2+b^2+c^2=6. b) Maximum value: 108, attained exactly by all permutations of (√3, 0, −√3). | |
070h | Problem:
$n > 1$ is an integer. $D_n$ is the set of lattice points $(x, y)$ with $|x|, |y| \leq n$. If the points of $D_n$ are colored with three colors (one for each point), show that there are always two points with the same color such that the line containing them does not contain any other points of $D_n$. Show th... | [
"Solution:\n\n\n\nConsider the 4 points shown in the diagram. In each case the segment joining them is the diagonal of an $m \\times 1$ parallelogram or rectangle, so it cannot contain any other lattice points. The next points along each line are obviously outside set $D_n$. That proves the... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
00az | A circle of length $999$ is divided into unit arcs by $999$ black points. Then $d$ arcs with lengths $1$, $2$, ..., $d$ are placed on the circle, with their end points black, so that none of these arcs contains another (otherwise the arcs may overlap). Find all $d$ for which such a configuration exists. | [
"Consider the problem for a circle of length $n$ divided into unit arcs by $n$ black points, and $d$ arcs of lengths $1$, $2$, ..., $d$ with black endpoints such that none of them contains another. We show that the maximal admissible $d$ equals $\\lfloor \\frac{n+1}{2} \\rfloor$; here $\\lfloor \\cdot \\rfloor$ den... | Argentina | Argentina_2017 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | All integers d with 1 ≤ d ≤ 500 | |
0dp3 | Let $a$, $b$, $c$, $d$ be positive integers such that $d$ divides $a^{2b} + c$ and $d \ge a + c$. Prove that $d \ge a + \sqrt[2b]{a}$. | [
"We have $a^{2b} + c \\equiv (d - a)^{2b} + c \\pmod d$, since\n$$\na^{2b} - (d - a)^{2b} = (a^2 - (d - a)^2) \\times \\\\\n\\times (a^{2(b-1)} + a^{2(b-2)}(d - a)^2 + \\dots + a^2(d - a)^{2(b-2)} + (d - a)^{2(b-1)})\n$$\n$$\na^{2b} - (d - a)^{2b} \\vdots (a + (d - a)) = d.\n$$\nWe deduce consequently $(d - a)^{2b}... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0fyv | Problem:
Betrachte ein Spielbrett mit ungeraden Seitenlängen, das in Einheitsquadrate aufgeteilt ist. Das Brett ohne ein Eckfeld wird irgendwie mit Dominos bedeckt. Man kann nun in einem Zug ein Domino in Längsrichtung um eins verschieben, sodass das vorher leere Feld bedeckt wird, dafür ein neues (zwei Felder davon e... | [
"Solution:\n\nBetrachte ein Eckfeld $E$, welches von einem Dominostein $D_{1}$ bedeckt ist. Dieser grenzt an eine weiteres Feld (zwei Felder vom Eckfeld entfernt), welches entweder ein freies Eckfeld ist oder von einem weiteren Domino $D_{2}$ bedeckt ist. So erhält man eine Folge von verschiedenen Dominosteinen $D_... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0987 | Problem:
În triunghiul scalen $ABC$ notăm cu $I$ punctul de intersecție al bisectoarelor. Demonstrați că dreapta, care include linia mijlocie a triunghiului, paralelă cu $BC$, intersectează dreptele $BI$ și $CI$ în puncte, situate pe cercul de diametru $[AI]$. | [
"Solution:\n\nFie $N$ mijlocul laturii $[AB]$ și $M$ mijlocul laturii $[AC]$. Fie $MN \\cap BI = \\{F\\}, \\quad MN \\cap CI = \\{E\\}$, $AE \\cap BC = \\{P\\}, \\quad AF \\cap BI = \\{Q\\}$. Este clar că în $\\triangle PAB$, $NE$ este linie mijlocie $\\Rightarrow [AE] = [PE]$, deci în $\\triangle APC$, $CE$ este b... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01pl | $N$ segments are arranged inside a unit circle $\Gamma$. The sum of the lengths of all these segments is equal to $2\sqrt{N}$.
Prove that there exists a concentric with $\Gamma$ circumference intersecting at least two of these segments. | [
"Consider $360^\\circ$ rotation of $\\Gamma$ (together with all segments) about its center. Under this rotation each of the segments covers some ring with the center at the center of $\\Gamma$. If we show that the sum of the areas of all these rings is not less than $\\pi$, then the statement will be proved.\n\nWe ... | Belarus | BelarusMO 2013_s | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
09uw | Agatha, Isa and Nick each have a different kind of bike. One of them has an electric bike, one has a racing bike, and one has a mountain bike. The bikes have different colours: green, blue and black. The three owners make two statements each, of which one is true and the other is false:
* Agatha says: "I have an electr... | [
"D) Isa has a mountain bike."
] | Netherlands | First Round, January 2019 | [
"Discrete Mathematics > Logic"
] | English | MCQ | D | |
09f7 | Let $H$ be the intersection point of the altitudes $AD$ and $BE$ of an acute triangle $ABC$. The circumcircle of the triangle $ABC$ intersects the circle with diameter $CH$ at the point $K$ other than $C$. Prove that
$$
\frac{DK}{KE} = \frac{DH}{HE}.
$$ | [
"\nSince $\\angle BDH = \\angle AEH$ and $\\angle BHD = \\angle AHE$, we have $\\triangle BHD \\sim \\triangle AHE$. Therefore\n$$\n\\frac{DH}{HE} = \\frac{BD}{AE}. \\qquad (1)\n$$\n\nAlso it is easy to observe that $\\angle CDK = \\angle CEK$, and moreover $\\angle BOK = \\angle AEK$, $\\a... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00fd | Let $ABCD$ be a quadrilateral such that all sides have equal length and angle $\angle ABC$ is $60$ degrees. Let $\ell$ be a line passing through $D$ and not intersecting the quadrilateral (except at $D$). Let $E$ and $F$ be the points of intersection of $\ell$ with $AB$ and $BC$ respectively. Let $M$ be the point of in... | [
"\nTriangles $AED$ and $CDF$ are similar, because $AD \\parallel CF$ and $AE \\parallel CD$. Thus, since $ABC$ and $ACD$ are equilateral triangles,\n$$\n\\frac{AE}{CD} = \\frac{AD}{CF} \\Longleftrightarrow \\frac{AE}{AC} = \\frac{AC}{CF} .\n$$\nThe last equality combined with\n$$\n\\angle E... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1993 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06mj | Let $n$ be a positive integer. Show that if $p$ is a prime dividing $5^{4n} - 5^{3n} + 5^{2n} - 5^n + 1$, then $p \equiv 1 \pmod 4$. | [
"Clearly, $p \\neq 2, 5$. Let $m = 5^{4n} - 5^{3n} + 5^{2n} - 5^n + 1$. Then\n$$\n(2 \\cdot 5^{2n} - 5^n + 2)^2 - 5 \\cdot 5^{2n} = 4m \\equiv 0 \\pmod{p}.\n$$\nThis gives $5 \\equiv (5^{-n}(2 \\cdot 5^{2n} - 5^n + 2))^2 \\pmod{p}$. Using the Legendre symbol, we have $\\left(\\frac{5}{p}\\right) = 1$. On the other ... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
0ed8 | Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$, such that
$$
f\left(\frac{x}{f(y)}\right) = \frac{(f(x))^2}{y f(f(x))}
$$
for all $x, y > 0$. | [
"Let us show that the function $f$ is surjective. Substituting $y \\mapsto \\frac{(f(x))^2}{y f(f(x))}$ in the initial equation we get\n$$\nf\\left(\\frac{x}{f\\left(\\frac{(f(x))^2}{y f(f(x))}\\right)}\\right) = y,\n$$\nwhich means that for any $y$ there exists a number which is mapped into $y$ by $f$. Hence, $f$ ... | Slovenia | Slovenija 2016 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = k x for any constant k > 0 | |
0idi | Problem:
For $x > 0$, let $f(x) = x^{x}$. Find all values of $x$ for which $f(x) = f'(x)$. | [
"Solution:\n\nLet $g(x) = \\log f(x) = x \\log x$. Then $f'(x)/f(x) = g'(x) = 1 + \\log x$. Therefore $f(x) = f'(x)$ when $1 + \\log x = 1$, that is, when $x = 1$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives"
] | null | proof and answer | x = 1 | |
0fhu | Problem:
Una Oficina de Turismo va a realizar una encuesta sobre el número de días soleados y de días lluviosos a lo largo de un año. Para ello recurre a seis regiones, que le transmiten los datos de la tabla siguiente:
| Región | Sol o lluvia | Inclasificable |
| :---: | :---: | :---: |
| A | 336 | 29 |
| B | 321 | ... | [
"Solution:\n\nAl suprimir una región, la suma de días soleados o lluviosos de las restantes ha de ser múltiplo de $4$. Esta suma vale $1994$ para las seis regiones, valor que dividido entre $4$ da resto $2$. El único dato de esta columna que da resto $2$ al dividirlo entre $4$ es $330$ correspondiente a la región $... | Spain | OME 30 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | F | |
0jha | Problem:
Values $a_{1}, \ldots, a_{2013}$ are chosen independently and at random from the set $\{1, \ldots, 2013\}$. What is expected number of distinct values in the set $\left\{a_{1}, \ldots, a_{2013}\right\}$? | [
"Solution:\n\nAnswer: $\\frac{2013^{2013}-2012^{2013}}{2013^{2012}}$\n\nFor each $n \\in \\{1,2, \\ldots, 2013\\}$, let $X_{n}=1$ if $n$ appears in $\\left\\{a_{1}, a_{2}, \\ldots, a_{2013}\\right\\}$ and $0$ otherwise. Defined this way, $\\mathrm{E}\\left[X_{n}\\right]$ is the probability that $n$ appears in $\\le... | United States | HMMT | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | (2013^{2013}-2012^{2013})/2013^{2012} | |
06gz | Let $S = \{1, 2, 3, \dots, 2n\}$, where $n$ is a positive integer greater than or equal to $1$. For any subset $T$ of $S$, $T$ is called a *good* subset if in $T$ the number of even elements is greater than the number of odd elements.
a. Find the total number of good subsets of $S$.
b. Find the sum of all the element... | [
"a.\nThe answer is $2^{2n-1} - \\frac{1}{2}\\binom{2n}{n}$.\n\nA subset $T$ of $S$ is called a *bad* subset if in $T$ the number of odd elements is greater than the number of even elements. A subset of $S$ is neither good nor bad if it has exactly $k$ odd elements and $k$ even elements for some $k = 0, 1, \\dots, n... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | a: 2^{2n-1} - \tfrac{1}{2}\binom{2n}{n}; b: 2^{2n-2}(2n^2 + n) - n^2\binom{2n-1}{n} | |
08gf | Problem:
Alberto ha davanti a sé 13 caselle disposte una sopra l'altra, e vuole inserirvi i numeri da 1 a 10, uno per casella (tre caselle rimarranno vuote). Vuole inoltre che, se due numeri sono scritti in caselle che si toccano, quello più in alto sia maggiore. In quanti modi può farlo?
(A) $3^{10}$
(B) $2^{11} \cd... | [
"Solution:\n\nLa risposta è (D). Le quattro caselle lasciate bianche partizionano i 10 numeri in quattro sottoinsiemi (eventualmente vuoti). All'interno di ciascuno di questi sottoinsiemi, l'ordine dei numeri inseriti nelle caselle è determinato. Poiché ciascuno dei 10 numeri può finire in uno qualsiasi dei quattro... | Italy | Italian Mathematical Olympiad - February Round | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | MCQ | D | |
02yh | Problem:
a) Verifique que se $a \in \{1,2,4\}$, então $n(a+n)$ não é um quadrado perfeito para qualquer inteiro positivo $n$.
b) Verifique que se $a=2^{k}$, com $k \geq 3$, então existe um inteiro positivo $n$ tal que $n(a+n)$ é um quadrado perfeito.
c) Verifique que se $a \notin \{1,2,4\}$, então sempre existe um inte... | [
"Solution:\na) Para $a \\in \\{1,2,4\\}$ e $n$ inteiro positivo, em virtude das desigualdades\n$$\nn^{2}<n(n+1)<n(n+2)<(n+1)^{2}\n$$\ne\n$$\n(n+1)^{2}<n(n+4)<(n+2)^{2}\n$$\npodemos concluir que $n(n+a)$ está entre dois quadrados perfeitos consecutivos e, consequentemente, não pode ser um quadrado perfeito.\n\nb) Se... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0c4e | Problem:
Fie $n$ un număr întreg, $n \geq 2$, și fie $\mathbf{A}$ o matrice din $\mathcal{M}_{n}(\mathbb{C})$, astfel încât $\mathbf{A}$ și $\mathbf{A}^{2}$ să aibă ranguri diferite. Arătați că există o matrice nenulă $\mathbf{B}$ în $\mathcal{M}_{n}(\mathbb{C})$, astfel încât $\mathrm{AB}=\mathrm{BA}=\mathrm{B}^{2}=\... | [
"Solution:\n\nÎntrucât $\\mathbf{A}$ și $\\mathbf{A}^{2}$ au ranguri diferite, $\\mathbf{A}$ este o matrice singulară nenulă.\n\nDacă $n=2$, atunci $\\mathbf{A}^{2}=(\\operatorname{tr} \\mathbf{A}) \\mathbf{A}$, conform teoremei Hamilton-Cayley. Deoarece $\\mathbf{A}$ și $\\mathbf{A}^{2}$ au ranguri diferite, rezul... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
089k | Problem:
Un modellino di automobile viene testato su alcuni circuiti chiusi lunghi 600 metri, composti da tratti piani e tratti in salita o discesa. Tutti i tratti in salita e in discesa hanno la stessa pendenza. I test mettono in risalto alcuni fatti curiosi:
a. la velocità del modellino dipende solo dal fatto che l... | [
"Solution:\n\nIndichiamo tutte le lunghezze in metri e tutti i tempi in secondi, omettendo le unità di misura.\nLe terne di velocità possibili sono $\\left(v_{s}, v_{p}, v_{d}\\right)=(10,12,15),(9,12,18),(8,12,24)$ e $(7,12,42)$.\n\nConsideriamo innanzitutto un circuito completamente in piano lungo 600. L'ipotesi ... | Italy | Cesenatico | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (10,12,15), (9,12,18), (8,12,24), (7,12,42) | |
0dio | An acute scalene triangle $ABC$ is inscribed in a circle $k$. The bisector of angle $\angle ABC$ meets side $BC$ at point $D$. Let $I$ be an arbitrary point on the segment $AD$, and let $H$ be the orthogonal projection of $I$ onto $BC$. Circle $\omega$ is centered at $I$ and passes through $H$, and $U$ is the internal ... | [
"Let $M$, $N$ be the midpoint of minor arc and major arc $BC$ of circle $(O)$, respectively, then points $A$, $D$, $M$ are collinear. Let $T$ be the intersection of the two tangent lines at $B$ and $C$ of $(O)$ and $J$ the midpoint of $BC$. First, by the angle chasing, we have $\\angle TBM = \\angle BAM = \\angle M... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous >... | English | proof only | null | |
0dyd | Problem:
Poišči vsa praštevila $p$ in $q$, za katera je število $2 p^{2} q + 45 p q^{2}$ popoln kvadrat. | [
"Solution:\n\nNajprej denimo, da je $p = q$. Potem mora biti število $47 p^{3}$ popoln kvadrat. Ker je deljivo s $47$ in je $47$ praštevilo, mora biti deljivo tudi s $47^{2}$, od koder sledi, da $47$ deli $p^{3}$ oziroma $47$ deli $p$. Toda $p$ je praštevilo, torej mora biti enako $47$. Res, pri $p = q = 47$ je šte... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | p = q = 47 and p = 3, q = 2 | |
08w6 | Suppose the points $A$, $B$, $C$, $D$ are located on the circumference of a circle in this order as indicated in the figure below.
Suppose the angle formed by the line tangent to the circle at $B$ and the line $AB$ is $30^\circ$, and that formed by the line tangent to the circle at $C$ and the line $CD$ is $10^\circ$. ... | [
"Since $AB \\parallel DC$ and since $\\angle DCA$ and $\\angle DBA$ are angles subtended by the same arc $\\widearc{AD}$ at the points $C$ and $B$ on the circle, we have $\\angle BDC = \\angle DBA = \\angle DCA$.\n\nSince the angle subtended by an arc $\\widearc{AB}$ at the point $C$ on the circle equals the angle ... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 70° | |
00i7 | Consider an infinite sequence $a_{1}, a_{2}, \ldots$ of positive integers such that
$$
100!\left(a_{m}+a_{m+1}+\cdots+a_{n}\right) \quad \text{ is a multiple of } a_{n-m+1} a_{n+m}
$$
for all positive integers $m, n$ such that $m \leq n$.
Prove that the sequence is either bounded or linear.
Observation: A sequence of ... | [
"Let $c=100!$. Suppose that $n \\geq m+2$. Then $a_{m+n}=a_{(m+1)+(n-1)}$ divides both $c\\left(a_{m}+a_{m+1}+\\cdots+a_{n-1}+a_{n}\\right)$ and $c\\left(a_{m+1}+\\cdots+a_{n-1}\\right)$, so it also divides the difference $c\\left(a_{m}+a_{n}\\right)$. Notice that if $n=m+1$ then $a_{m+n}$ divides $c\\left(a_{m}+a_... | Asia Pacific Mathematics Olympiad (APMO) | APMO 2025 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequence... | null | proof only | null | |
0d94 | Let $S = \{-17, -16, \ldots, 16, 17\}$. We call a subset $T$ of $S$ a good set if $-x \in T$ for any $x \in T$ and if $x, y, z \in T$ ($x, y, z$ may be equal) then $x + y + z \neq 0$. Find the largest number of elements in a good set. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 18 | |
098q | Problem:
Fie paralelipipedul dreptunghic $A B C D A_{1} B_{1} C_{1} D_{1}$, în care $A B=a, B C=2 a, A A_{1}=3 a$. Pe muchiile $C C_{1}$ și $A D$ se consideră punctele $M$ și $N$ respectiv, astfel încât $A N=C_{1} M=a$. Determinați măsura unghiului dintre dreptele $A M$ și $N B_{1}$. | [
"Solution:\n\nPe dreapta suport a muchiei $B C$ considerăm punctul $Q$, astfel încât $N Q \\| A C$.\nConsiderăm punctul $Q_{1} \\in B_{1} C_{1}$, astfel încât $Q_{1} Q \\| C_{1} C, Q_{1} Q=C_{1} C$, și punctul $M_{1} \\in Q_{1} Q$, astfel încât $Q_{1} M_{1}=a$.\nAtunci $A N\\|C Q, C Q\\| M ... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | arccos(5√11/33) | |
0hov | Problem:
$p$ is a prime number such that the period of its decimal reciprocal is $200$. That is,
$$
\frac{1}{p}=0 . X X X X \ldots
$$
for some block of $200$ digits $X$, but
$$
\frac{1}{p} \neq 0 . Y Y Y Y \ldots
$$
for all blocks $Y$ with less than $200$ digits. Find the $101$st digit, counting from the left, of ... | [
"Solution:\n\nLet $X$ be a block of $n$ digits and let $a=0 . X \\ldots$ Then $10^{n} a=X . X \\ldots$. Subtracting the previous two equalities gives us $\\left(10^{n}-1\\right) a=X$, i.e. $a=\\frac{X}{10^{n}-1}$.\n\nThen the condition that $a=\\frac{1}{p}$ reduces to $\\frac{1}{p}=\\frac{X}{10^{n}-1}$ or $p X=10^{... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 9 | |
0cli | From a point $O$ inside the square $ABCD$ the perpendicular line $OS$ is raised to the plane of the square. Let $M, N, P, Q$ be projections of point $O$ onto the planes $(SAB), (SBC), (SCD)$, respectively $(SDA)$. Prove that the points $M, N, P, Q$ are coplanar if and only if $O$ lies on one of the diagonals of the squ... | [
"We assume that $O$ lies, for example, on the diagonal $AC$. Let $OE \\perp AB$, $E \\in AB$ and $OF \\perp AD$, $F \\in AD$. Then we have successively $OE = OF$, $\\triangle SOE \\equiv \\triangle SOF$ (C.C.), $SE = SF$. Then $M \\in SE$ and $OM \\perp SF$, $Q \\in SF$ and $OQ \\perp SF$, $\\triangle SOM \\equiv \... | Romania | 75th Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
0b7h | Let $ABC$ be a triangle such that $AB \neq AC$. The internal bisector lines of the angles $ABC$ and $ACB$ meet the opposite sides of the triangle at points $B_0$ and $C_0$, respectively, and the circumcircle $ABC$ at points $B_1$ and $C_1$, respectively. Further, let $I$ be the incenter of the triangle $ABC$. Prove tha... | [
"Let the internal bisector of the angle $BAC$ meet again the circumcircle $ABC$ at point $A_1$. The lines $A_1B_1$ and $AC$ meet at point $B_2$, and the lines $AB$ and $A_1C_1$ meet at point $C_2$. Apply Pascal's theorem to the hexagon $AC_1BA_1CB_1$ to deduce that the points $B_2$, $I$ and $C_2$ are collinear; mor... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0c6v | Let $ABCDA'B'C'D'$ be a rectangular parallelepiped, and $M$, $N$, $P$ the projections of the points $A$, $C$, respectively $B'$, on the diagonal $BD'$.
a)
Show that $BM + BN + BP = BD'$.
b) Show that $3(AM^2 + B'P^2 + CN^2) \ge 2D'B^2$ if and only if the rectangular parallelepiped $ABCDA'B'C'D'$ is cube. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
087v | Problem:
I rossi e i verdi stanno facendo una battaglia a gavettoni. La base dei rossi è un'area a forma di triangolo equilatero di lato 8 metri. I verdi non possono entrare nella base dei rossi, ma possono lanciare i loro proiettili nella base stando comunque fuori dal perimetro. Sapendo che i verdi riescono a colpir... | [
"Solution:\n\nLa risposta è $\\mathbf{(A)}$. Detto $ABC$ il triangolo che forma la base, la zona di sicurezza è un triangolo $A'B'C'$ (con $A'$ appartenente alla bisettrice dell'angolo in $A$ e cicliche) interno al triangolo $ABC$. Dette $H$ e $K$ le proiezioni di $A'$ e $B'$ rispettivamente sul lato $AB$, si ha $A... | Italy | Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | A | |
0fnx | De un prisma recto de base cuadrada, con lado de longitud $L_1$, y altura $H$, extraemos un tronco de pirámide, no necesariamente recto, de bases cuadradas, con lados de longitud $L_1$ (para la inferior) y $L_2$ (para la superior), y altura $H$. Las dos piezas obtenidas aparecen en la imagen siguiente.
![](attached_ima... | [
"Si prolongamos una altura $h$ el tronco de pirámide hasta obtener una pirámide completa de altura $H + h$ tendrá una sección como la que se muestra en la figura anterior.\n\nUn argumento de semejanza de triángulos permite comprobar que\n$$\n\\frac{h+H}{L_1} = \\frac{h}{L_2}\n$$\ny, por tanto,\n$$\nh = \\frac{H L_2... | Spain | L Olimpiada Matemática Española | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Transformations > Homothety",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | Spanish | proof and answer | (1 + sqrt(5)) / 2 | |
00ol | Determine the maximal number of consecutive positive integers such that each of these integers has a common divisor with $2024$ greater than $1$. | [
"We observe that $2024 = 2^3 \\cdot 11 \\cdot 23$. An integer has a common divisor greater than $1$ with $2024$ if and only if it is divisible by $2$, $11$ or $23$.\nLet $N$ be the desired maximal number. Only each $11$th integer is divisible by $11$. That means that if $z$ is divisible by $11$, then $z+1, z+2, \\d... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 5 | |
03gh | Problem:
Let $ABC$ be the right-angled isosceles triangle whose equal sides have length $1$. $P$ is a point on the hypotenuse, and the feet of the perpendiculars from $P$ to the other sides are $Q$ and $R$. Consider the areas of the triangles $APQ$ and $PBR$, and the area of the rectangle $QCRP$. Prove that regardless... | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 2/9 | |
02xq | Problem:
a) Encontre o valor da soma
$$
\frac{1}{1+1/x}+\frac{1}{1+x}
$$
b) Encontre o valor da soma
$$
\frac{1}{2019^{-2019}+1}+\ldots+\frac{1}{2019^{-1}+1}+\frac{1}{2019^{0}+1}+\frac{1}{2019^{1}+1}+\ldots+\frac{1}{2019^{2019}+1}
$$ | [
"Solution:\n\na) Temos\n$$\n\\begin{aligned}\n\\frac{1}{1+1/x}+\\frac{1}{1+x} & =\\frac{1}{(x+1)/x}+\\frac{1}{1+x} \\\\\n& =\\frac{x}{1+x}+\\frac{1}{1+x} \\\\\n& =1\n\\end{aligned}\n$$\n\nb) Em virtude do item anterior, considerando $x=a^{b}$, podemos agrupar as frações $\\frac{1}{a^{-b}+1}$ e $\\frac{1}{a^{b}+1}$ ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | a) 1; b) 4039/2 | |
079i | There are $11$ men sitting around a circular table with equal distances and $11$ cards with numbers $1,2,\ldots,11$ on them are dealt among them. It is possible that one has no cards and the other has more than one. In each step one can give one of his cards to his adjacent individual if the card number $i$ has the fol... | [
"First divide the table into $11$ equal arcs. Now if the cards $i, j$ be on the points $A, B$ on the table, we define the distance between these two cards as the number of arcs between $A, B$ on the table (the smaller one). For example, the distance between $i, j$ is $5$ in the following figure:\n\n![](attached_ima... | Iran | 27th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0ite | Problem:
Trodgor the dragon is burning down a village consisting of 90 cottages. At time $t=0$ an angry peasant arises from each cottage, and every 8 minutes (480 seconds) thereafter another angry peasant spontaneously generates from each non-burned cottage. It takes Trodgor 5 seconds to either burn a peasant or to bu... | [
"Solution:\n\nAnswer: 1920\n\nWe look at the number of cottages after each wave of peasants. Let $A_{n}$ be the number of cottages remaining after $8n$ minutes. During each 8 minute interval, Trodgor burns a total of $480 / 5 = 96$ peasants and cottages. Trodgor first burns $A_{n}$ peasants and spends the remaining... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 1920 | |
0a8h | Problem:
Determine the number of real roots of the equation
$$
x^{8}-x^{7}+2 x^{6}-2 x^{5}+3 x^{4}-3 x^{3}+4 x^{2}-4 x+\frac{5}{2}=0
$$ | [
"Solution:\nWrite\n$$\n\\begin{gathered}\nx^{8}-x^{7}+2 x^{6}-2 x^{5}+3 x^{4}-3 x^{3}+4 x^{2}-4 x+\\frac{5}{2} \\\\\n=x(x-1)\\left(x^{6}+2 x^{4}+3 x^{2}+4\\right)+\\frac{5}{2}\n\\end{gathered}\n$$\nIf $x(x-1) \\geq 0$, i.e. $x \\leq 0$ or $x \\geq 1$, the equation has no roots. If $0<x<1$, then $0>x(x-1)=\\left(x-\... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 15 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 0 | |
0e11 | Let $E$ and $F$ be the points on the sides $AB$ and $AD$ of a convex quadrilateral $ABCD$, such that $EF$ is parallel to $BD$. The segment $CE$ intersects the diagonal $BD$ at $G$, while the segment $CF$ intersects the diagonal $BD$ at $H$. Prove: if $AGCH$ is a parallelogram, then $ABCD$ is a parallelogram as well. | [
"Denote the intersection of the lines $EF$ and $AG$ by $I$ and the intersection of the lines $EF$ and $AH$ by $J$.\nNow, $EF$ is parallel to $BD$ and $AH$ is parallel to $CE$, so the quadrilateral $EGHJ$ is a parallelogram. Furthermore, $AG$ and $CF$ are parallel, so $FIGH$ is also a parallelogram. Hence, $|FH| = |... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0dyq | Given a sequence of integers $a_1, a_2, a_3, \dots$ such that
$$
0 \le a_k \le k-1 \quad \text{and} \quad a_1 + \dots + a_k \equiv 0 \pmod{k}
$$
for all $k > 1$. Prove that the sequence is constant from some point on. For example, when $a_1 = 9$ the sequence is $9, 1, 2, 0, 3, 3, 3, \dots$. | [
"Taking a look at the sequences we obtain for different values of $a_1$, we notice the following: Assume there is an index $k$ such that $a_1 + a_2 + \\dots + a_k = d \\cdot k$ and $0 \\le d < k$. Then $a_1 + a_2 + \\dots + a_k + d = d \\cdot (k+1)$ and since $a_{k+1}$ is a uniquely determined number between $0$ an... | Slovenia | Slovenija 2008 | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
03z6 | Suppose $\cos^5\theta - \sin^5\theta < 7(\sin^3\theta - \cos^3\theta)$, $\theta \in [0, 2\pi)$. Then the range of $\theta$ is ______. | [
"From the inequality\n$$\n\\cos^5\\theta - \\sin^5\\theta < 7(\\sin^3\\theta - \\cos^3\\theta),\n$$\nwe have\n$$\n\\sin^3\\theta + \\frac{1}{7}\\sin^5\\theta > \\cos^3\\theta + \\frac{1}{7}\\cos^5\\theta.\n$$\nSince $f(x) = x^3 + \\frac{1}{7}x^5$ is increasing over $(-\\infty, +\\infty)$, then $\\sin \\theta > \\co... | China | China Mathematical Competition | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | English | proof and answer | (π/4, 5π/4) | |
08n9 | Problem:
Find all prime positive integers $p, q$ such that $2 p^{3}-q^{2}=2(p+q)^{2}$. | [
"Solution:\nThe given equation can be rewritten as $2 p^{2}(p-1)=q(3 q+4 p)$.\nHence $p\\mid 3 q^{2}+4 p q \\Rightarrow p\\mid 3 q^{2} \\Rightarrow p \\mid 3 q$ (since $p$ is a prime number) $\\Rightarrow p \\mid 3$ or $p \\mid q$. If $p \\mid q$, then $p=q$. The equation becomes $2 p^{3}-9 p^{2}=0$ which has no pr... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (3, 2) | |
0ine | Problem:
Three brothers Abel, Banach, and Gauss each have portable music players that can share music with each other. Initially, Abel has $9$ songs, Banach has $6$ songs, and Gauss has $3$ songs, and none of these songs are the same. One day, Abel flips a coin to randomly choose one of his brothers and he adds all of... | [
"Solution:\n\nIf Abel copies Banach's songs, this can never happen. Therefore, we consider only the cases where Abel copies Gauss's songs. Since all brothers have Gauss's set of songs, the probability that they play the same song is equivalent to the probability that they independently match whichever song Gauss ch... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 1/288 | |
0eit | Problem:
Dan je trikotnik $A B C$. Naj bosta $D$ in $E$ taki točki, ki ležita zaporedoma na poltrakih $C A$ in $C B$, a ne na stranicah trikotnika $A B C$, da velja $|A D|=|B E|=|A B|$. Naj bo $F$ presečišče vzporednice $k A C$ skozi točko $E$ in vzporednice $k B C$ skozi točko $D$. Presečišče premic $A E$ in $B D$ oz... | [
"Solution:\n\nOznačimo s $P$ presečišče daljice $B C$ in simetrale kota $\\Varangle B A C$, z $Q$ pa presečišče daljice $A C$ in simetrale kota $\\Varangle C B A$. Presečišče premic $A P$ in $B Q$ je torej središče trikotniku $A B C$ včrtane krožnice, označimo ga z $I$.\n\nKer je trikotnik $B A D$ enakokrak z vrhom... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"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 | |
0ezt | Problem:
(1) Player $A$ writes down two rows of $10$ positive integers, one under the other. The numbers must be chosen so that if $a$ is under $b$ and $c$ is under $d$, then $a + d = b + c$. Player $B$ is allowed to ask for the identity of the number in row $i$, column $j$. How many questions must he ask to be sure o... | [
"Solution:\n\n(1) is trivial. We can write the condition as $b - a = d - c$, so the $10$ numbers in the first row and $1$ in the second row can all be chosen arbitrarily. Hence at least $11$ questions are needed. But they are also sufficient. Having determined those numbers, the others immediately follow.\n\n(2). T... | Soviet Union | ASU | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Functional equations",
"Algebra > Linear Algebra > Matrices"
] | null | proof and answer | (1) 11; (2) m + n - 1 | |
08ay | Problem:
Camilla ha una scatola che contiene 2015 graffette. Ne prende un numero positivo $n$ e le mette sul banco di Federica, sfidandola al seguente gioco. Federica ha a disposizione due tipi di mosse: può togliere 3 graffette dal mucchio che ha sul proprio banco (se il mucchio contiene almeno 3 graffette), oppure t... | [
"Solution:\n\na. Federica vince se e solo se $n$ è multiplo di $3$.\nSe $n$ è multiplo di $3$ Federica può vincere: le basta effettuare la mossa con la quale toglie tre graffette dal banco esattamente $n / 3$ volte.\nD'altra parte, se ad un certo punto sul banco di Federica c'è un numero di graffette non multiplo d... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | a) 671; b) 1344 | |
0a50 | Problem:
You have an unlimited supply of square tiles with side length $1$ and equilateral triangle tiles with side length $1$. For which $n$ can you use these tiles to create a convex $n$-sided polygon? The tiles must fit together without gaps and may not overlap. | [
"Solution:\nAll the angles in squares and equilateral triangles are multiples of $30^{\\circ}$. So all the external angles of the $n$-sided polygon are multiples of $30^{\\circ}$. Since the polygon is convex, this implies that all external angles are greater than or equal to $30^{\\circ}$. However, the sum of the e... | New Zealand | NZMO Round One | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | All integers n with 3 ≤ n ≤ 12 | |
0jg9 | Problem:
Let $p$ be a prime number that has the form $a^{3}-b^{3}$ for some positive integers $a$ and $b$. Prove that $p$ also has the form $c^{2}+3 d^{2}$ for some positive integers $c$ and $d$. | [
"Solution:\n\nWe can factor\n$$\np = a^{3} - b^{3} = (a-b)\\left(a^{2} + a b + b^{2}\\right).\n$$\nSince $a$ and $b$ are positive integers, the only way this can happen is if $a-b=1$.\nEither $a$ or $b$ is even. If $a$ is even, let $a=2u$, so $b=2u-1$. Then\n$$\n\\begin{aligned}\np & = (2u)^{2} + (2u)(2u-1) + (2u-1... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0fx8 | Problem:
Betrachte sieben verschiedene Geraden in der Ebene. Ein Punkt heisst gut, falls er auf mindestens drei dieser Geraden liegt. Bestimme die grösstmögliche Anzahl guter Punkte. | [
"Solution:\n\nMan überlegt sich leicht, dass 6 gute Punkte möglich sind. Wir zeigen nun, dass dies die grösstmögliche Anzahl guter Punkte ist. Wir nennen die sieben Geraden aus der Aufgabenstellung gut, um sie von irgendwelchen anderen Geraden zu unterscheiden.\nFür $n \\geq 2$ sei $a_{n}$ die Anzahl guter Punkte, ... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Combinatorial Geometry > Sylvester's theorem",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 6 | |
03p1 | An acute triangle $ABC$ has three heights $AD$, $BE$ and $CF$ respectively. Prove that the perimeter of triangle $DEF$ is not over half of the perimeter of triangle $ABC$. (posed by Qi Jianxin) | [
"**Proof** Since $\\angle ADB = \\angle AEB = 90^\\circ$, so four points $A$, $B$, $D$ and $E$ are concyclic, and furthermore, $AB$ is the diameter. Hence, by the sine rule, we can get\n$$\n\\frac{DE}{\\sin \\angle DAE} = AB = c,\n$$\nso\n$$\nDE = c \\sin \\angle DAE.\n$$\nIn addition, $\\angle DAC + \\angle DCA = ... | China | China Girls' 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 > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometri... | English | proof only | null | |
0kdl | Problem:
Anastasia is taking a walk in the plane, starting from $(1,0)$. Each second, if she is at $(x, y)$, she moves to one of the points $(x-1, y)$, $(x+1, y)$, $(x, y-1)$, and $(x, y+1)$, each with $\frac{1}{4}$ probability. She stops as soon as she hits a point of the form $(k, k)$. What is the probability that $k... | [
"Solution:\nThe key idea is to consider $(a+b, a-b)$, where $(a, b)$ is where Anastasia walks on. Then, the first and second coordinates are independent random walks starting at $1$, and we want to find the probability that the first is divisible by $3$ when the second reaches $0$ for the first time. Let $C_{n}$ be... | United States | HMMT February 2020 | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | (3 - sqrt(3))/3 | |
0e35 | For real numbers $a$, $b$ and $c$ we have
$$
(2b - a)^2 + (2b - c)^2 = 2(2b^2 - ac).
$$
Prove that the numbers $a$, $b$ and $c$ are three consecutive terms in some arithmetic sequence. | [
"The given equation is equivalent to $8b^2 - 4ab - 4bc + a^2 + c^2 = 4b^2 - 2ac$ or\n$$\n4b^2 - 4ab - 4bc + a^2 + 2ac + c^2 = 0.\n$$\nThis can be further rewritten as $(a + c)^2 - 4b(a + c) + 4b^2 = 0$ and finally as\n$$\n(a + c - 2b)^2 = 0.\n$$\nFrom here we conclude that $a + c = 2b$ or, equivalently, $b - a = c ... | Slovenia | National Math Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0k8w | Problem:
Alison is eating 2401 grains of rice for lunch. She eats the rice in a very peculiar manner: every step, if she has only one grain of rice remaining, she eats it. Otherwise, she finds the smallest positive integer $d > 1$ for which she can group the rice into equal groups of size $d$ with none left over. She ... | [
"Solution:\n\nNote that $2401 = 7^{4}$. Also, note that the operation is equivalent to replacing $n$ grains of rice with $n \\cdot \\frac{p-1}{p}$ grains of rice, where $p$ is the smallest prime factor of $n$.\n\nNow, suppose that at some moment Alison has $7^{k}$ grains of rice. After each of the next four steps, ... | United States | HMMT November 2019 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 17 | |
0hjj | Problem:
Suppose you have an unlimited number pennies, nickels, dimes, and quarters. Determine the number of ways to make 30 cents using these coins. | [
"Solution:\n\nWe use cases to organize our work, based first on the number of quarters and then the number of dimes. First note that the number of quarters must be $0$ or $1$, since $2$ quarters would be too much. This gives $2$ cases:\n\n- $1$ quarter: There are $2$ possibilities: a quarter and a nickel or a quart... | United States | Berkeley Math Circle: Monthly Contest 8 | [
"Discrete Mathematics > Combinatorics"
] | null | final answer only | 17 | |
0fr0 | Problem:
Dado un número natural $n>1$, realizamos la siguiente operación: si $n$ es par, lo dividimos entre dos; si $n$ es impar, le sumamos $5$. Si el número obtenido tras esta operación es $1$, paramos el proceso; en caso contrario, volvemos a aplicar la misma operación, y así sucesivamente. Determinar todos los val... | [
"Solution:\n\nEn primer lugar, es inmediato comprobar que siempre que empezamos por $2$, $3$ o $4$ el proceso termina y que si empezamos por $5$ entramos en el bucle $(5,10,5,10,\\ldots)$ y nunca acabamos.\n\nSi el número por el que se empieza es mayor que $5$, en uno o dos pasos siempre pasamos a un número más peq... | Spain | FASE LOCAL DE LA OLIMPIADA MATEMÁTICA ESPAÑOLA. | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | The process is finite if and only if n is not a multiple of 5. | |
0azy | Problem:
Among all victims of zombie bites, $10\%$ are prescribed the experimental drug Undetenin to treat them. Overall, $4\%$ of the human population suffer an adverse reaction to Undetenin. Out of all the patients being treated with Undetenin, $2\%$ suffer an adverse reaction to the drug. What is the probability th... | [
"Solution:\n\nThis is an application of Bayes' theorem. If $A$ is the event of being prescribed Undetenin and $B$ is being allergic to it, then we are looking for $P(A \\mid B)$. We have\n$$\nP(A \\mid B) = \\frac{P(B \\mid A) P(A)}{P(B)} = \\frac{(0.02)(0.1)}{0.04} = 0.05.\n$$"
] | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 5% | |
0lc4 | Find the number of ordered 6-tuples satisfying the following system of modular equations
$$
\begin{cases}
ab + a'b' \equiv 1 \pmod{15} \\
bc + b'c' \equiv 1 \pmod{15} \\
ca + c'a' \equiv 1 \pmod{15}
\end{cases}
$$
with $a, b, c, a', b', c' \in \{0, 1, \dots, 14\}$. | [
"For any integer $k$, let $N_k$ be the number of ordered 6-tuple $(a, b, c, a', b', c')$ that satisfy\n$$\nab + a'b' \\equiv bc + b'c' \\equiv ca + c'a' \\equiv 1 \\pmod{k}\n$$\nand $a, b, c, a', b', c' \\in \\{0, 1, \\dots, k-1\\}$. By the Chinese Remainder Theorem,\n$$\nN_{mn} = N_n \\times N_m \\text{ if } \\gcd... | Vietnam | VMO | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | 3472 | |
0g0c | Problem:
Sei $p$ eine ungerade Primzahl. Bestimme die Anzahl Tupel $\left(a_{1}, a_{2}, \ldots, a_{p}\right)$ natürlicher Zahlen mit folgenden Eigenschaften:
1) $1 \leq a_{i} \leq p$ für alle $i=1, \ldots, p$.
2) $a_{1}+a_{2}+\cdots+a_{p}$ ist nicht durch $p$ teilbar.
3) $a_{1} a_{2}+a_{2} a_{3}+\ldots a_{p-1} a_{p... | [
"Solution:\n\nFür ein Tupel $t=\\left(a_{1}, \\ldots, a_{p}\\right)$ wie in 1) definieren wir $S(t)=a_{1}+\\cdots+a_{p}$ und $P(t)=a_{1} a_{2}+a_{2} a_{3}+\\cdots+a_{p} a_{1}$. Wir halten fest\n$$\n|\\{t \\mid t \\text{ erfüllt } 1 \\text{ und } 2\\}| = p^{p-1}(p-1)\n$$\nda wir die ersten $p-1$ Einträge frei wählen... | Switzerland | SMO - Finalrunde | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | p^{p-2}(p-1) | |
0i6g | Problem:
$x$, $y$ are positive real numbers such that $x + y^{2} = x y$. What is the smallest possible value of $x$? | [
"Solution:\n\nNotice that $x = \\dfrac{y^{2}}{y-1} = 2 + (y-1) + \\dfrac{1}{y-1} \\geq 2 + 2 = 4$. Conversely, $x = 4$ is achievable, by taking $y = 2$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 4 | |
090f | The JMO Cluster initially consists of five stars $O$, $A$, $B$, $C$, and $D$. Each star is assigned a value called its **importance**. The importance of $O$ is $0$, and the importance of each of $A$, $B$, $C$, and $D$ is $1$. Moreover, there are one-way direct flights from $O$ to $A$ and $C$; from $A$ to $B$ and $D$; f... | [
"Let the initial configuration be called state $0$, and for each integer $n \\ge 1$, let state $n$ denote the configuration after the $n$-th operation.\n\n**Lemma 1.** There exists an assignment of every star into exactly one of groups $0$, $1$, or $2$ such that:\n(1) Star $O$ is assigned to group $0$.\n(2) For eve... | Japan | The 35th Japanese Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relat... | English | proof and answer | 2^68 * (2^102 - 1) / 3 |
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