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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0d0h | Let $(a_n)_{n \ge 1}$ be a sequence with the property that for every prime $p$ and for every positive integer $k$ the following relation holds:
$$
a_{kp+1} = pa_k - 3a_p + 13.
$$
Find $a_{2011}$. | [
"For $k = q$, prime, we have\n$$\na_{qp+1} = pa_q - 3a_p + 13 = qa_p - 3a_q + 13,\n$$\nso $(p+3)a_q = (q+3)a_p$. It follows that for every primes $p$ and $q$,\n$$\n\\frac{a_p}{p+3} = \\frac{a_q}{q+3}. \\qquad (1)\n$$\nThen, since $2011$ is a prime, we obtain\n$$\n\\frac{a_{2011}}{2014} = \\frac{a_7}{10}. \\qquad (2... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 2014 | |
0153 | Problem:
Does there exist an angle $\alpha \in (0, \pi / 2)$ such that $\sin \alpha$, $\cos \alpha$, $\tan \alpha$ and $\cot \alpha$, taken in some order, are consecutive terms of an arithmetic progression? | [
"Solution:\n\nSuppose that there is an $x$ such that $0 < x < \\frac{\\pi}{2}$ and $\\sin x$, $\\cos x$, $\\tan x$, $\\cot x$ in some order are consecutive terms of an arithmetic progression.\n\nSuppose $x \\leq \\frac{\\pi}{4}$. Then $\\sin x \\leq \\sin \\frac{\\pi}{4} = \\cos \\frac{\\pi}{4} \\leq \\cos x < 1 \\... | Baltic Way | Baltic Way 2008 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | No, such an angle does not exist. | |
0ezo | Problem:
Equally spaced perpendicular lines divide a large piece of paper into unit squares. $N$ squares are colored black. Show that you can always cut out a set of disjoint square pieces of paper, so that all the black squares are removed and the black area of each piece is between $1/5$ and $4/5$ of its total area. | [
"Solution:\n\nWe proceed by induction on $N$.\n\nBase case: $N = 1$. The single black square can be cut out as a $1 \\times 1$ square. The black area is $1$, which is between $1/5$ and $4/5$ of $1$ (since $1 = 1$). So the base case holds.\n\nInductive step: Suppose the statement is true for all numbers less than $N... | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
03ad | Let $X$ be an interior point of $\triangle ABC$ and $S_1 = S_{XBC}$, $S_2 = S_{XCA}$, $S_3 = S_{XAB}$. Find the minimal area of a convex polygon, containing three segments equal and parallel to $XA$, $XB$, $XC$. | [
"We may assume that $S_1 \\ge S_2 \\ge S_3$. We shall prove that the answer is $S_1$.\nIt is not difficult to see that the area of the convex hull of two segments $d_1$ and $d_2$ is not less than the area $S$ of a triangle with two sides equal and parallel to these segments. Indeed, if this hull is a triangle, two ... | Bulgaria | Team selection test for 50. IMO | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | max{S_{XBC}, S_{XCA}, S_{XAB}} | |
0f89 | Problem:
Find the minimum value of $\dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y}$ for positive reals $x$, $y$, $z$ with $x^2 + y^2 + z^2 = 1$. | [
"Solution:\nAnswer: min $\\sqrt{3}$ when all equal.\n\nLet us consider $z$ to be fixed and focus on $x$ and $y$. Put $f(x, y, z) = \\dfrac{xy}{z} + \\dfrac{yz}{x} + \\dfrac{zx}{y}$. We have $f(x, y, z) = \\dfrac{p}{z} + \\dfrac{z(1 - z^2)}{p} = \\dfrac{p + k^2 / p}{z}$, where $p = xy$, and $k = z\\sqrt{1 - z^2}$. N... | Soviet Union | 22nd ASU | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | sqrt(3) | |
04rc | Two circles $k_1(S_1, r_1)$ and $k_2(S_2, r_2)$ are externally tangent and both lie in a square $ABCD$ with side length $a$ so that $k_1$ touches the sides $AD$ and $CD$, while $k_2$ touches the sides $BC$ and $CD$. Prove that the area of at least one of the triangles $AS_1S_2$, $BS_1S_2$ is no more than $\frac{3}{16}a... | [
"The line segments $AS_2$ and $BS_1$ lie on the diagonals of the given square, so they are perpendicular to each other and intersect at the center $P$ of the square. We have\n$$\n\\begin{aligned}\n|DS_1| &= r_1 \\cdot \\sqrt{2}, & |BS_1| &= (a-r_1)\\sqrt{2}, & |PS_1| &= \\left(\\frac{a}{2} - r_1\\right)\\sqrt{2}, \... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
01hh | Let $a, b, c$ be real numbers, representing the side lengths of a triangle. Prove that
$$
\sqrt[3]{(a^2 + bc)(b^2 + ca)(c^2 + ab)} > \frac{a^2 + b^2 + c^2}{2}.
$$ | [
"We claim that\n$$\na^2 + bc > \\frac{a^2 + b^2 + c^2}{2},\n$$\nwhich will finish the proof. Note that the claimed inequality is equivalent to\n$$\n\\begin{aligned}\na^2 + bc > \\frac{a^2 + b^2 + c^2}{2} &\\iff 2a^2 + 2bc > a^2 + b^2 + c^2 \\\\\n&\\iff a^2 > (b-c)^2 \\iff a > |b-c|,\n\\end{aligned}\n$$\nwhich holds... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
0556 | Does there exist a positive integer $n$ which has exactly 9 positive divisors and whose all divisors can be placed in a 3-by-3 table such that the products of the 3 numbers in each row, each column and on each diagonal are all the same? | [
"The number $36$ has $9$ positive divisors $1$, $2$, $3$, $4$, $6$, $9$, $12$, $18$, $36$. Let the first row be $18$, $1$, $12$, second row $4$, $6$, $9$, and third row $3$, $36$, $2$. Then the product of each row, column and diagonal is $216$.",
"For each prime $p$ the number $p^8$ has exactly $9$ divisors $p^0,... | Estonia | Open Contests | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | Yes; for example, n = 36 works, and more generally any n = p^8 for a prime p. | |
0j94 | Problem:
Hugo plays a game: he places a chess piece on the top left square of a $20 \times 20$ chessboard and makes 10 moves with it. On each of these 10 moves, he moves the piece either one square horizontally (left or right) or one square vertically (up or down). After the last move, he draws an $\mathrm{X}$ on the ... | [
"Solution:\n\nIndex each square by its row number and column number, starting with $0$. For example, $(0,0)$ represents the top left square and $(2,5)$ represents the square in the third row down and the sixth column over. When the piece moves down or to the right, the sum of the indices of its square increases by ... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 36 | |
0749 | Let $x$, $y$, $z$ and $a$, $b$, $c$ be positive real numbers such that $x + y + z = a + b + c$ and $xyz = abc$. Suppose $\max\{x, y, z\} \ge \max\{a, b, c\}$. Prove that
$$
ab + bc + ca \ge xy + yz + zx.
$$ | [
"We may assume $x \\ge y \\ge z$ and $a \\ge b \\ge c$. We are given $x \\ge a$. Consider the cubics\n$$\nP(t) = (t - x)(t - y)(t - z), \\quad Q(t) = (t - a)(t - b)(t - c).\n$$\nSince $a$ is the largest root of $Q(t) = 0$, it follows that $Q(s) \\ge 0$ for $s \\ge a$. In particular, $Q(x) \\ge 0$. Observe that\n$$\... | India | Indija TS 2009 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0935 | Problem:
Let $ABC$ be an acute-angled triangle with $AB \neq AC$, circumcentre $O$ and circumcircle $\Gamma$. Let the tangents to $\Gamma$ through $B$ and $C$ meet each other at $D$, and let the line $AO$ intersect $BC$ at $E$. Denote the midpoint of $BC$ by $M$ and let $AM$ meet $\Gamma$ again at $N \neq A$. Finally,... | [
"Solution:\n\nWe may suppose $AB < AC$. It is known that $AD$ is a symmedian of triangle $ABC$. Let $Q$ be its second point of intersection with $k$. The triangles $ABM$ and $AQC$ are similar to each other, since their corresponding angles are equal, and it follows that\n$$\n\\angle AFQ = \\angle ACQ = \\angle AMB ... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Tri... | null | proof only | null | |
0cf5 | Let $A \in \mathcal{M}_n(\mathbb{R})$ be a matrix with the property that
$$
\det(A + X) \ge \det X, \text{ for all } X \in \mathcal{M}_n(\mathbb{R}).
$$
Prove that $A = O_n$. | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
027n | Problem:
Uma pilha de pedras está sobre uma mesa, Pedrinho joga o seguinte jogo: a cada momento, ele pode escolher uma pilha com pelo menos 3 pedras, retirar uma dessas pedras e dividir a pilha em duas pilhas não vazias. Por exemplo, se ele tem uma pilha com 15 pedras, ele pode dividir essa pilha em duas pilhas de 9 e... | [
"Solution:\n\na) Sim, Pedrinho consegue! Basta fazer o seguinte processo: primeiro dividimos a pilha original em duas, uma com 3 e outra com 15. Depois, dividimos a pilha com 15 em uma com 3, e outra com 11, ficando com duas pilhas de 3, e uma de 11. Continuando o processo, dividimos a de 11 em uma de 3 e uma de 7.... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) Yes. b) No. | |
047h | Find the maximum possible area of a right-angled triangle that can be covered by two closed disks of radius $1$. | [
"*Proof.* Let the right triangle be $\\triangle ABC$ with $\\angle C = 90^\\circ$. In this solution, the interior of a triangle or a circle includes its boundary.\nLet $\\omega_1$ and $\\omega_2$ be the circumferences of two closed disks with radius $1$, such that every point inside $\\triangle ABC$ lies in the int... | China | 2024 CGMO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 3*sqrt(3)/2 | |
07q5 | Let $ABC$ be a triangle and $D$, $E$, $F$ points on the circumcircle of triangle $ABC$ so that $AD$, $BE$ and $CF$ are diameters of the circumcircle. The lines $AE$, $BF$ and $CD$ form a triangle $A'B'C'$. Prove that $\triangle ABC$ is similar to $\triangle A'B'C'$. | [
"Let $A'$ be the intersection of $AE$ and $CD$. Because $BE$ and $AD$ are diameters, $\\angle BAA'$ and $\\angle ACA'$ are right angles. Hence, $\\angle BAC + \\angle CAA' = 90^\\circ = \\angle AA'C + \\angle CAA'$. This implies $\\angle BAC = \\angle AA'C$. Similarly, we obtain $\\angle CBA = \\angle BB'A$ and $\\... | Ireland | Ireland | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03mc | Let $x$, $y$ and $z$ be positive real numbers. Show that $x^2 + xy^2 + xyz^2 \ge 4xyz - 4$.
Soit $x$, $y$ et $z$ trois nombres réels positifs. Démontrez que $x^2 + xy^2 + xyz^2 \ge 4xyz - 4$. | [
"Note that\n$$\nx^2 \\ge 4x - 4, \\quad y^2 \\ge 4y - 4, \\quad \\text{and} \\quad z^2 \\ge 4z - 4,\n$$\nand therefore\n$$\nx^2 + xy^2 + xyz^2 \\ge (4x - 4) + x(4y - 4) + xy(4z - 4) = 4xyz - 4.\n$$"
] | Canada | Kanada 2012 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English, French | proof only | null | |
02jq | Problem:
Uma mesa quadrada tem 1 metro de lado. Qual o menor diâmetro de uma toalha redonda que cubra completamente o tampo da mesa?
(A) 1
(B) 1,5
(C) 2
(D) $\sqrt{2}$
(E) $\sqrt{3}$ | [
"Solution:\n\nPara que a toalha cubra inteiramente a mesa e que tenha o menor diâmetro possível, o quadrado deve estar inscrito no círculo. A diagonal do quadrado é o diâmetro do círculo, logo pelo Teorema de Pitágoras, temos:\n$$d^{2} = 1^{2} + 1^{2} \\Rightarrow d = \\sqrt{2}$$\n\n"
] | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | MCQ | D | |
0jo1 | Problem:
Let $ABC$ be a triangle with orthocenter $H$; suppose that $AB = 13$, $BC = 14$, $CA = 15$. Let $G_{A}$ be the centroid of triangle $HBC$, and define $G_{B}, G_{C}$ similarly. Determine the area of triangle $G_{A} G_{B} G_{C}$. | [
"Solution:\n\nAnswer: $\\quad 28 / 3$\n\nLet $D, E, F$ be the midpoints of $BC, CA$, and $AB$, respectively. Then $G_{A} G_{B} G_{C}$ is the $DEF$ about $H$ with a ratio of $\\frac{2}{3}$, and $DEF$ is the dilation of $ABC$ about $H$ with a ratio of $-\\frac{1}{2}$, so $G_{A} G_{B} G_{C}$ is the dilation of $ABC$ a... | United States | HMMT February 2015 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 28 / 3 | |
0d07 | Let $ABCD$ be a cyclic quadrilateral with $\hat{B} \neq 90^\circ$ and
$$
AB^2 + BC^2 + CD^2 + DA^2 = 2AC^2.
$$
Prove that the midpoint of diagonal $BD$ is on $AC$. | [
"Let $M$ be the intersection point of the diagonals $AC$ and $BD$.\n\n\n\nSince $\\hat{B} \\ne 90^\\circ$ it follows that $\\hat{D} \\ne 90^\\circ$, and hence we have $\\cos \\hat{B} \\ne 0$ and $\\cos \\hat{D} \\ne 0$. The quadrilateral $ABCD$ is cyclic, so therefore $\\hat{B} + \\hat{D} =... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
090c | Let $n \ge 2$ be an integer. Suppose that real numbers $a_1, a_2, \dots, a_{2n}$ satisfy $|a_k - a_{n+k}| \ge 1$ for every integer $k$ such that $1 \le k \le n$. Find the minimum possible value of
$$
(a_1 - a_2)^2 + (a_2 - a_3)^2 + \dots + (a_{2n-1} - a_{2n})^2 + (a_{2n} - a_1)^2.
$$ | [
"Set $a_{2n+1} = a_1$. Since $a_{n+1} - a_{2n+1} = -(a_1 - a_{n+1})$, the values\n$$\na_1 - a_{n+1}, \\ a_2 - a_{n+2}, \\ \\dots, \\ a_n - a_{2n}, \\ a_{n+1} - a_{2n+1}\n$$\nare all nonzero, and $a_1 - a_{n+1}$ and $a_{n+1} - a_{2n+1}$ have opposite signs. Therefore, there exists an integer $m$ with $1 \\le m \\le ... | Japan | The 35th Japanese Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | 2 | |
02n2 | Problem:
Calcule o valor da soma
$$
S = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \cdots + \frac{1}{2006 \cdot 2007} + \frac{1}{2007 \cdot 2008}
$$ | [] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2007/2008 | |
0bjb | Consider the function $f: \mathbb{N}^* \to \mathbb{N}^*$ which satisfies the properties:
a) $f(1) = 1$,
b) $f(p) = 1 + f(p-1)$ for any prime number $p$,
c) $f(p_1p_2\cdots p_n) = f(p_1) + f(p_2) + \cdots + f(p_n)$ for any prime numbers, not necessary distinct.
Prove that $2^{f(n)} \le n^3 \le 3^{f(n)}$, for any natura... | [
"As $2$ is a prime number, we have $f(2) = 1 + f(1) = 2$, whence $2^2 \\le 2^3 \\le 3^2$. Then $f(3) = 1 + f(2) = 3$, and thus $2^3 < 3^3 = 3^2$. We will prove the inequalities\n$$\n3 \\log_3 n \\le f(n) \\le 3 \\log_2 n\n$$\nby mathematical induction on $n$. Let $n$ be a natural number, $n \\ge 3$, and let $n = p_... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Other"
] | null | proof only | null | |
0ct6 | The diagonals $AC$ and $BD$ of a cyclic quadrilateral $ABCD$ meet at $P$. The point $Q$ is chosen on the segment $BC$ so that $PQ \perp AC$. Prove that the line of centers of the circumcircles of the triangles $APD$ and $BQD$ is parallel to $AD$. (A. Kuznetsov)
Диагонали $AC$ и $BD$ вписанного четырехугольника $ABCD$ ... | [
"Choose a point $T \\in QP$ so that $DT \\perp DA$ (see Fig. 16). The points $A$, $P$, $D$, and $T$ are concyclic, so the center of the circle $APD$ lies on the perpendicular bisector $\\ell$ of $DT$ (notice that $\\ell \\parallel AD$). Next, $\\angle QBD = \\angle PAD = \\angle QTD$, so the points $B$, $Q$, $D$, a... | Russia | Russian Mathematical Olympiad | [
"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 chasing",
"Geometry > Plane Geometry > Miscellaneou... | English; Russian | proof only | null | |
0f5q | Problem:
The positive reals $x$, $y$, $z$ satisfy $x^2 + x y + \dfrac{y^2}{3} = 25$, $\dfrac{y^2}{3} + z^2 = 9$, $z^2 + z x + x^2 = 16$. Find the value of $x y + 2 y z + 3 z x$. | [] | Soviet Union | 18th ASU | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 24√3 | |
03yg | Determine the least odd number $a > 5$ satisfying the following conditions: There are positive integers $m_1, m_2, n_1, n_2$ such that $a = m_1^2 + n_1^2$, $a^2 = m_2^2 + n_2^2$, and $m_1 - n_1 = m_2 - n_2$. | [
"The answer is $261$.\n\nNote that\n$$\n261 = 15^2 + 6^2, \\quad 261^2 = 189^2 + 180^2, \\quad 15 - 6 = 189 - 180.\n$$\nWe know that there is no number in between $5$ and $261$ that satisfies the condition of the problem. Assume on the contrary that $a$ is such a number. We may set $d = m_1 - n_1 > 0$. Because $a$ ... | China | China Girls' Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 261 | |
0emx | The real-valued function $f$ satisfies
$$
f(\tan 2x) = \tan^4 x + \cot^4 x
$$
for all real $x$. Prove that, for all real $x$,
$$
f(\sin x) + f(\cos x) \ge 196.
$$ | [
"Note that since $\\tan 2\\theta$ attains all real numbers in the interval $\\left(-\\frac{\\pi}{4}, \\frac{\\pi}{4}\\right)$, we may restrict our attention to this domain. Also,\n$$\nf(-\\tan 2\\theta) = f(\\tan 2(-\\theta))\n$$\n$$\n= \\tan^4(-\\theta) + \\cot^4(-\\theta) \\\\\n= \\tan^4 \\theta + \\cot^4 \\theta... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0e7u | Problem:
Za pozitivni realni števili $x$ in $y$ velja
$$
2013^{\log_{3} x} = y^{\log_{5} 2013} \quad \text{in} \quad \log_{\frac{1}{2}} x + \log_{\frac{1}{2}} y > 0
$$
Katero od števil $x$ in $y$ je večje? | [
"Solution:\n\nČe enakost logaritmiramo, dobimo\n$$\n\\frac{\\log x \\log 2013}{\\log 3} = \\frac{\\log y \\log 2013}{\\log 5}\n$$\noziroma\n$$\n\\log y = \\frac{\\log 5}{\\log 3} \\log x = \\log_{3} 5 \\log x.\n$$\nOd tod med drugim sledi, da sta $\\log y$ in $\\log x$ istega predznaka, torej sta $x$ in $y$ bodisi ... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | x | |
02op | Let $n$ be a positive integer and $A$ a subset of $\mathbb{Z}/(n)$, the set of the integers modulo $n$, define $f(A) = \min_{t \in \mathbb{Z}/(n)} |A \cap (A+t)|$, where $A+t = \{x+t, x \in A\} \subset \mathbb{Z}/(n)$. Define $g(n) = \max\{f(A); A \subset \mathbb{Z}/(n), |A| = \lfloor n/2 \rfloor\}$.
a. Prove that $g(... | [
"a.\nLet $A = \\{a_1, a_2, \\dots, a_{\\lfloor n/2 \\rfloor}\\}$. Consider the sum\n$$\n\\sum_{t=0}^{n-1} |A \\cap (A+t)|\n$$\nNow each element $a \\in A$ appears in a set $A + t_i$ $|A| = \\lfloor n/2 \\rfloor$ times: choose $t_i = a - a_i$ for each $i = 1, 2, \\dots, \\lfloor n/2 \\rfloor$. So\n$$\n\\sum_{t=0}^{n... | Brazil | Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof only | null | |
07zo | Problem:
In un trapezio isoscele, una diagonale è lunga $22~\mathrm{cm}$; si sa inoltre che tale diagonale forma con la base maggiore un angolo di $45^\circ$. Quanto vale l'area del trapezio?
(A) $121~\mathrm{cm}^2$
(B) $242~\mathrm{cm}^2$
(E) nessuna delle precedenti.
(C) $484~\mathrm{cm}^2$
(D) i dati sono insuffici... | [] | Italy | Italy Febbraio Contest | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | B | |
09k5 | In triangle $ABC$, the angle bisector of $\angle A$ intersects the angle bisector of $\angle C$ at point $E$. On the angle bisector of $\angle A$, a point $D$ is chosen such that $EC = ED$ and $E$ lies between points $A$ and $D$. If $ACDB$ is a cyclic quadrilateral, determine $\angle B$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 60° | |
0ama | Problem:
Find the value of $\sin \theta$ if the terminal side of $\theta$ lies on the line $5y - 3x = 0$ and $\theta$ is in the first quadrant.
(a) $\frac{3}{\sqrt{34}}$
(b) $\frac{3}{4}$
(c) $\frac{3}{5}$
(d) $\frac{4}{\sqrt{34}}$ | [] | Philippines | QUALIFYING STAGE | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | MCQ | a | |
0h12 | For any positive real numbers $x, y, z$ with $xyz=1$ prove the following inequality:
$$
(-x+y+z)(x-y+z) + (x-y+z)(x+y-z) + (x+y-z)(-x+y+z) \le 3.
$$ | [
"Without loss of generality, assume that $z = \\min\\{x, y, z\\}$. The inequality can be rewritten as\n$$\n4xy \\le 3 + (x+y-z)^2.\n$$\nSince $x+y-z \\ge 2\\sqrt{xy-z} > 0$, we have that $(x+y-z)^2 \\ge (2\\sqrt{xy-z})^2$. So, it is sufficient to prove that $3 + (2\\sqrt{xy-z})^2 \\ge 4xy$. From the problem conditi... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
06x7 | Let $ABC$ be a triangle, and let $\ell_{1}$ and $\ell_{2}$ be two parallel lines. For $i=1,2$, let $\ell_{i}$ meet the lines $BC$, $CA$, and $AB$ at $X_{i}$, $Y_{i}$, and $Z_{i}$, respectively. Suppose that the line through $X_{i}$ perpendicular to $BC$, the line through $Y_{i}$ perpendicular to $CA$, and finally the l... | [
"Throughout the solutions, $\\Varangle(p, q)$ will denote the directed angle between lines $p$ and $q$, taken modulo $180^{\\circ}$.\nLet the vertices of $\\Delta_{i}$ be $D_{i}, E_{i}, F_{i}$, such that lines $E_{i}F_{i}$, $F_{i}D_{i}$ and $D_{i}E_{i}$ are the perpendiculars through $X, Y$ and $Z$, respectively, a... | IMO | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Ta... | English | proof only | null | |
0gw9 | Prove that for any real numbers $x$ and $y$ the inequality
$$
|\cos x| + |\cos y| + |\cos(x+y)| \ge 1
$$
holds. | [
"$$\n\\begin{align*}\n& |\\cos x| + |\\cos y| + |\\cos(x+y)| \\ge |\\cos x \\sin y| + |\\cos y \\sin x| + \\cos^2(x+y) \\\\ \n& \\ge |\\sin(x+y)| + \\cos^2(x+y) \\ge \\sin^2(x+y) + \\cos^2(x+y) = 1.\n\\end{align*}\n$$"
] | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0gui | Let $ABC$ be a triangle. Let $D$, $E$, $F$ be points on the sides $BC$, $AC$, $AB$ respectively, such that
$$
DE \parallel AB,\ DF \parallel AC \text{ and } \frac{BD}{DC} = \frac{AB^2}{AC^2}.
$$
The circumcircle of the triangle $AEF$ meets $AD$ at $R$ again and meets the line passing through $A$ and tangent to the circ... | [
"\nFrom the condition $BD/DC = AB^2/AC^2$ we know that $AD$ is the $A$-symmedian of the triangle $ABC$. Let $a, b, c$ be the side lengths of the triangle, then we get $AF/FB = CD/DB = b^2/c^2$ hence $AF = b^2c/(b^2 + c^2)$ and similarly $AE = c^2b/(b^2 + c^2)$. Using these we get $AE \\cdot... | Turkey | 31st Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configur... | English | proof only | null | |
0j2d | Problem:
There are 111 StarCraft progamers. The StarCraft team SKT starts with a given set of eleven progamers on it, and at the end of each season, it drops a progamer and adds a progamer (possibly the same one). At the start of the second season, SKT has to field a team of five progamers to play the opening match. H... | [
"Solution:\n\nWe disregard the order of the players, multiplying our answer by $5! = 120$ at the end to account for it. Clearly, SKT will be able to field at most 1 player not in the original set of eleven players. If it does not field a new player, then it has $\\binom{11}{5} = 462$ choices. If it does field a new... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 4015440 | |
0571 | Does there exist an integer $n \ge 3$ such that some 3 diagonals of a regular $n$-gon meet in one point that is neither a vertex nor the center of the $n$-gon? If yes then find the least such $n$. | [
"If 3 diagonals of an $n$-gon meet in one point that is not a vertex of the $n$-gon then these diagonals have 6 endpoints in total, implying $n \\ge 6$. If $n = 6$ then the only way to leave the endpoints of every two diagonals to different sides of the third diagonal is connecting each vertex to the opposite one (... | Estonia | Final Round of National Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 8 | |
0bx9 | Let $I$ be the incenter of the scalene triangle $ABC$, with $AB < AC$, and let $I'$ be the reflection of point $I$ in the line $BC$. The angle bisector $AI$ meets the side $BC$ at $D$ and the circumcircle of $ABC$ at $E$. The line $EI'$ meets the circumcircle at $F$. Prove that $\frac{AI}{IE} = \frac{ID}{DE}$ and $IA =... | [
"We have $\\frac{AI}{ID} = \\frac{AB}{BD} = \\frac{EC}{DE} = \\frac{BE}{DE} = \\frac{IE}{DE}$, which proves the first relation.\n\nLet $G = EF \\cap BC$. We know that $GD$ is the bisector of the angle $\\angle EGI$, hence $\\frac{GI}{GE} = \\frac{ID}{DE} = \\frac{AI}{IE}$. (1)\n\nTriangles $FBE$ and $BGE$ are simil... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
059h | There are three consecutive positive integers on a blackboard. A move consists of erasing the smallest number on the blackboard and replacing it by the sum of itself and the greatest number on the blackboard. Is it possible that the sum of the numbers on the blackboard will be a power of 10:
a) after the 6th move?
b)... | [
"*Answer:* (a) Yes; (b) No.\n\nLet $x$ be the smallest of the three numbers initially on the blackboard. We will find the numbers on the blackboard after having made 0, 1, 2, 3, 4, 5, 6, 7 and 8 moves, and their sums:\n\n| Number of moves | Numbers on the blackboard | Sum of numbers |\n|-----------... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Other"
] | English | proof and answer | a) Yes; b) No | |
0e51 | Find the smallest positive integer $n$, such that the sum of its digits is equal to $27$ and the sum of the digits of $2n$ is greater than $27$. | [] | Slovenia | Selection Examinations for the IMO | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 14499 | |
0348 | Problem:
Find all triples $(x, y, z)$ of positive integers such that
$$
\sqrt{\frac{2005}{x+y}} + \sqrt{\frac{2005}{x+z}} + \sqrt{\frac{2005}{y+z}}
$$
is a positive integer. | [
"Solution:\n1. We first prove the following lemma.\n\nLEMMA. If $p, q, r$ and $\\sqrt{p} + \\sqrt{q} + \\sqrt{r}$ are rational numbers then $\\sqrt{p}$, $\\sqrt{q}$ and $\\sqrt{r}$ are also rational numbers.\n\nProof of the lemma. Let $\\sqrt{p} + \\sqrt{q} + \\sqrt{r} = s$, where $pqr \\neq 0$ and $s$ is a rationa... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | All permutations of (28070, 4010, 4010). | |
0fgq | Problem:
En un triángulo $ABC$ tenemos puntos $D$ y $E$ respectivamente sobre $AB$ y $AC$. Conocemos la medida de los ángulos indicados a continuación: $\widehat{ABE}=30^\circ$, $\widehat{EBC}=50^\circ$, $\widehat{ACD}=20^\circ$ y $\widehat{DCB}=60^\circ$. Hallar el valor del ángulo $\widehat{EDC}$. | [
"Solution:\n\n\n\nEn el triángulo de la figura se dibuja $DF$ la paralela a $BC$, el triángulo $BGC$ es equilátero, luego $GC=BC$, el triángulo $BCE$ es isósceles, puesto que tiene dos ángulos de $50^\\circ$, consecuentemente $BC=CE$. Luego $CE=GC$ y el triángulo $CGE$ también es isósceles;... | Spain | OME 23 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 30° | |
06wd | Let $S$ be an infinite set of positive integers, such that there exist four pairwise distinct $a, b, c, d \in S$ with $\operatorname{gcd}(a, b) \neq \operatorname{gcd}(c, d)$. Prove that there exist three pairwise distinct $x, y, z \in S$ such that $\operatorname{gcd}(x, y)=\operatorname{gcd}(y, z) \neq \operatorname{g... | [
"There exists $\\alpha \\in S$ so that $\\{\\operatorname{gcd}(\\alpha, s) \\mid s \\in S, s \\neq \\alpha\\}$ contains at least two elements. Since $\\alpha$ has only finitely many divisors, there is a $d \\mid \\alpha$ such that the set $B=\\{\\beta \\in S \\mid \\operatorname{gcd}(\\alpha, \\beta)=d\\}$ is infin... | IMO | IMO 2021 Shortlisted Problems | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
08oy | Problem:
Let $\triangle ABC$ be an acute triangle. The lines $(\varepsilon_{1}),(\varepsilon_{2})$ are perpendicular to $AB$ at the points $A$, $B$, respectively. The perpendicular lines from the midpoint $M$ of $AB$ to the sides of the triangle $AC; BC$ intersect the lines $(\varepsilon_{1}),(\varepsilon_{2})$ at the... | [
"Solution:\n\nLet $H, G$ be the points of intersection of $ME, MF$, with $AC, BC$ respectively. From the similarity of triangles $\\triangle MHA$ and $\\triangle MAE$ we get\n$$\n\\frac{MH}{MA} = \\frac{MA}{ME}\n$$\nthus, $MA^{2} = MH \\cdot ME$\n\nSimilarly, from the similarity of triangles $\\triangle MBG$ and $\... | JBMO | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fwm | Problem:
Sei $P(x) = x^{4} - 2 x^{3} + p x + q$ ein Polynom mit reellen Koeffizienten, dessen Nullstellen alle reell sind. Zeige, dass die grösste dieser Nullstellen im Intervall $[1,2]$ liegt. | [
"Solution:\n\nSeien $a, b, c, d$ diese Nullstellen. Nach Vieta gilt $a + b + c + d = 2$ und $ab + ac + ad + bc + bd + cd = 0$. Quadriert man die erste dieser Gleichungen und subtrahiert das Doppelte der zweiten, dann folgt $a^{2} + b^{2} + c^{2} + d^{2} = 4$.\n\nWir können $a \\geq b \\geq c \\geq d$ annehmen und m... | Switzerland | IMO Selektion 2008 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
00gs | Let $a$, $b$, $c$ be integers satisfying $0 < a < c - 1$ and $1 < b < c$. For each $k$, $0 \leq k \leq a$, let $r_{k}$, $0 \leq r_{k} < c$, be the remainder of $k b$ when divided by $c$. Prove that the two sets $\{r_{0}, r_{1}, r_{2}, \ldots, r_{a}\}$ and $\{0, 1, 2, \ldots, a\}$ are different. | [
"Suppose that two sets are equal. Then $\\operatorname{gcd}(b, c) = 1$ and the polynomial\n$$\nf(x) := \\left(1 + x^{b} + x^{2b} + \\cdots + x^{a b}\\right) - \\left(1 + x + x^{2} + \\cdots + x^{a-1} + x^{a}\\right)\n$$\nis divisible by $x^{c} - 1$. (This is because: $m = n + c q \\Longrightarrow x^{m} - x^{n} = x^... | Asia Pacific Mathematics Olympiad (APMO) | XX Asian Pacific Mathematics Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0l8k | In plane, let be given two circles intersecting at $A$ and $B$ and let $PQ$ be a common tangent ($P$ and $Q$ are touching points). Let $S$ be the point of intersection of the tangents at $P$ and at $Q$ of the circumcircle of triangle $APQ$ and let $H$ be the reflection of $B$ in the line $PQ$. Prove that the points $A$... | [] | Vietnam | VIETNAMESE MATHEMATICAL COMPETITION FOR TEAM SELECTION | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0c3p | Problem:
Determinaţi toate tripletele $(a, b, c)$ de numere reale pentru care au loc simultan relaţiile:
$$
\begin{aligned}
& a\left(b^{2}+c\right)=c(c+a b) \\
& b\left(c^{2}+a\right)=a(a+b c) \\
& c\left(a^{2}+b\right)=b(b+c a)
\end{aligned}
$$ | [
"Solution:\nFie $(a, b, c)$ o soluţie a sistemului. Dacă unul dintre numerele $a, b, c$ este egal cu $0$, de exemplu dacă $c=0$, atunci din $c\\left(a^{2}+b\\right)=a(b+c a)$ rezultă $a=0$, apoi, similar, că $b=0$. Aşadar, dacă $a b c=0$, atunci $a=b=c=0$.\n\nCăutăm în continuare soluţii cu $a b c \\neq 0$. Relaţii... | Romania | Al doilea test de selectie pentru OBMJ | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (x, x, x) for any real x | |
0jvg | Problem:
A contest has six problems worth seven points each. On any given problem, a contestant can score either $0$, $1$, or $7$ points. How many possible total scores can a contestant achieve over all six problems? | [
"Solution:\nFor $0 \\leq k \\leq 6$, to obtain a score that is $k \\pmod{6}$, exactly $k$ problems must get a score of $1$. The remaining $6-k$ problems can generate any multiple of $7$ from $0$ to $7(6-k)$, of which there are $7-k$. So the total number of possible scores is $\\sum_{k=0}^{6}(7-k)=28$."
] | United States | HMMT February | [
"Discrete Mathematics > Other"
] | null | proof and answer | 28 | |
0jvn | Problem:
For each positive integer $n$ and non-negative integer $k$, define $W(n, k)$ recursively by
$$
W(n, k)= \begin{cases}n^{n} & k=0 \\ W(W(n, k-1), k-1) & k>0\end{cases}
$$
Find the last three digits in the decimal representation of $W(555,2)$. | [
"Solution:\n\nFor any $n$, we have\n$$\nW(n, 1)=W(W(n, 0), 0)=\\left(n^{n}\\right)^{n^{n}}=n^{n^{n+1}}\n$$\nThus,\n$$\nW(555,1)=555^{555^{556}}\n$$\nLet $N=W(555,1)$ for brevity, and note that $N \\equiv 0 \\pmod{125}$, and $N \\equiv 3 \\pmod{8}$. Then,\n$$\nW(555,2)=W(N, 1)=N^{N^{N+1}}\n$$\nis $0 \\pmod{125}$ and... | United States | HMMT February | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 875 | |
05gh | Problem:
Trouver tous les triplets de nombres premiers $(p, q, r)$ tels que $(p+1)(q+2)(r+3)=4 p q r$. | [
"Solution:\n\nSi $p=2$ alors on a $3(q+2)(r+3)=8 q r$ ce qui implique $q=3$ ou $r=3$. On a alors la solution $(2,3,5)$.\n\nSi $q=2$ et $p, r>2$ alors on a $(p+1)(r+3)=2 p r$. Puisque $p$ et $r$ sont impairs, le membre de gauche est divisible par 4, alors que le membre de droite est divisible par 2, mais pas par 4, ... | France | Envoi 1 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2,3,5), (7,5,2), (5,3,3) | |
071a | Problem:
$ABC$ is a triangle. $BD$ is an angle bisector. $E$, $F$ are the feet of the perpendiculars from $A$, $C$ respectively to the line $BD$. $M$ is the foot of the perpendicular from $D$ to the line $BC$. Show that $\angle DME = \angle DMF$. | [
"Solution:\n\n\n\nLet $H$ be the foot of the perpendicular from $D$ to $AB$. $\\angle AHD = \\angle AED = 90^\\circ$, so $AHED$ is cyclic. Hence $\\angle DAE = \\angle DHE$. But $M$ is the reflection of $H$ in the line $BD$, so $\\angle DME = \\angle DAE$.\n\n$AE$ is parallel to $CD$, so $\... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0g22 | Problem:
Finde alle natürlichen Zahlen $n$, für die kein Tripel natürlicher Zahlen $(a, b, c)$ existiert, sodass die folgende Gleichung erfüllt ist:
$$
n=\frac{a \cdot \operatorname{kgV}(b, c)+b \cdot \operatorname{kgV}(c, a)+c \cdot \operatorname{kgV}(a, b)}{\operatorname{kgV}(a, b, c)}
$$ | [
"Solution:\n\nLes entiers $n$ recherchés sont les puissances de 2. On définit\n$$\nf(a, b, c):=\\frac{a \\cdot \\operatorname{kgV}(b, c)+b \\cdot \\operatorname{kgV}(c, a)+c \\cdot \\operatorname{kgV}(a, b)}{\\operatorname{kgV}(a, b, c)}\n$$\nNotons tout d'abord que\n$$\n\\frac{a \\cdot \\operatorname{kgV}(b, c)}{\... | Switzerland | SMO - Finalrunde | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | powers of 2 | |
0aiq | Let $\triangle ABC$ be an acute-angled triangle, and let $D$ be the foot of the altitude from $C$. The angle bisector of $\angle ABC$ intersects $CD$ at $E$ and meets the circumcircle $\omega$ of triangle $\triangle ADE$ again at $F$. If $\angle ADF = 45^\circ$, show that $CF$ is tangent to $\omega$. | [
"Since $\\angle CDF = 90^\\circ - 45^\\circ = 45^\\circ$, the line $DF$ bisects $\\angle CDA$, and so $F$ lies on the perpendicular bisector of segment $AE$, which meets $AB$ at $G$. Let $\\angle ABC = 2\\beta$. Since $ADEF$ is cyclic, $\\angle AFE = 90^\\circ$, and hence $\\angle FAE = 45^\\circ$. Further, as $BF$... | North Macedonia | Girls European Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate... | English | proof only | null | |
04z7 | In the buffet of the kitchen, there are three candy boxes, each containing the same number of candies. Every time when Juku goes into the kitchen, he takes either three candies from one box or one candy from every box. Prove that irrespectively of how Juku takes the candies, he always retains the possibility to complet... | [
"The difference of the numbers of candies in any two boxes can only be a multiple of $3$ because it is $0$ in the beginning and, with every move, it changes by either $0$ or $3$. Hence, starting from an arbitrary intermediate state, Juku can clean out the boxes as follows: he takes one candy from each box as many t... | Estonia | Estonija 2010 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
03c1 | Let $p$ and $4p+1$ be two primes such that $p > 10^9$. Prove that the decimal representation of the number $\frac{1}{4p+1}$ contains every digit from 0 to 9. | [
"Let $q = 4p+1$ and $a \\circ b$ be the remainder of $a$ upon division by $b$. It suffices to show that the last digits of the numbers $10^k \\circ q$ cover all digits from 0 to 9. Indeed, since $q$ and $10$ are coprime, that would mean that all digits from 0 to 9 occur among the last digits of the numbers $\\lfloo... | Bulgaria | 55th IMO Team Selection Test | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequ... | English | proof only | null | |
0l2y | Problem:
Let $ABC$ be an acute, scalene triangle with circumcenter $O$ and symmedian point $K$. Let $X$ be the point on the circumcircle of triangle $BOC$ such that $\angle AXO = 90^{\circ}$. Assume that $X \neq K$. The hyperbola passing through $B, C, O, K$, and $X$ intersects the circumcircle of triangle $ABC$ at po... | [
"Solution:\n\nLet $\\mathcal{H}$ denote the hyperbola, and also recall a well-known fact that $A, K, X$ are collinear. This solution is split into two independent parts.\n\nProof of $\\boldsymbol{U} \\boldsymbol{V} \\perp \\boldsymbol{AX}$.\n\nLet $T = OX \\cap BC$, and let $OX$ intersect $\\odot(ABC)$ at points $Y... | United States | HMIC 2024 | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane G... | null | proof only | null | |
081a | Problem:
Siano $x, y$ numeri reali positivi. Quale delle seguenti condizioni è sufficiente per garantire che $x^{y}>y^{x}$?
(A) $1 < x < y$
(B) $1 < y < x$
(C) $x < 1 < y$
(D) $x < y < 1$
(E) $y < x < 1$. | [] | Italy | Progetto Olimpiadi di Matematica 2000 GARA di SECONDO LIVELLO | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | MCQ | E | |
0cfk | Problem:
Let $p > 2$ be a prime number and $n \in \mathbb{N}^*$ such that $n \equiv 1 \pmod{p}$. If the ring $(A, +, \cdot)$ has the property that $x^{n+p} = x^n$ for any $x \in A$, prove that $x^{p+1} = x$ for any $x \in A$.
Mihai Opincariu | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Abstract Algebra > Ring Theory"
] | English | proof only | null | |
03br | Find the least positive real number $\alpha$ with the following property: if the weight of a finite number of pumpkins is 1 ton and the weight of every pumpkin is not more than $\alpha$ tons then the pumpkins can be distributed in 50 boxes (some of the boxes may remain empty) such that there are no more than $\alpha$ t... | [
"We prove that the desired value of $\\alpha$ is $\\alpha = \\frac{2}{51}$.\n\nAssume that some $\\alpha < \\frac{2}{51}$ satisfies the condition of the problem. Choose nonnegative integer $k \\ge 0$ such that $\\frac{1}{51 \\times 2^k} \\le \\alpha < \\frac{1}{51 \\times 2^{k-1}}$. Consider $51 \\times 2^k$ pumpki... | Bulgaria | 55th IMO Team Selection Test | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2/51 | |
037m | Problem:
Let $\mathbb{R}^{+}$ be the set of all positive real numbers and $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ be a function such that
$$
f(x+y)-f(x-y)=4 \sqrt{f(x) f(y)}
$$
for all $x>y>0$.
a) Prove that $f(2 x)=4 f(x)$ for all $x \in \mathbb{R}^{+}$.
b) Find all such functions. | [
"Solution:\n\na) It follows from $f(x+y)-f(x-y)>0$ that $f$ is an increasing function. Therefore the function $f(x)$ has a limit $l \\geq 0$ when $x \\rightarrow 0$, $x>0$ (prove!). Thus letting $x, y \\rightarrow 0$, $x>y>0$, we get $l-l=4 \\sqrt{l^{2}}$, i.e. $l=0$. Fixing $x$ and letting $y \\rightarrow 0$, $y>0... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All such functions are f(x) = c x^2 for a constant c > 0; in particular, f(2x) = 4 f(x) for all positive x. | |
01j7 | Let $\omega_1$ and $\omega_2$ be circles with no common points. Points $M$ and $N$ are chosen on the circles $\omega_1$ and $\omega_2$, respectively, such that the tangent to the circle $\omega_1$ at $M$ and the tangent to the circle $\omega_2$ at $N$ intersect at $P$ and $\triangle PMN$ is an isosceles triangle with a... | [
"Since $\\triangle MPN$ is an isosceles triangle, we have $\\angle PMA = \\angle PMN = \\angle MNP = \\angle BNP$. By tangent and chord theorem, $\\angle MCA = \\angle PMA = \\angle BNP = \\angle BDN$.\nSince $\\angle MCP = \\angle MNP$, the quadrilateral $CMPN$ is cyclic. Analogously, from $\\angle PDN = \\angle P... | Baltic Way | Baltic Way 2023 Shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0a4w | Problem:
Let $P(x) = x^{3} - 2x + 1$ and let $Q(x) = x^{3} - 4x^{2} + 4x - 1$. Show that
if $P(r) = 0$ then $Q(r^{2}) = 0$ | [
"Solution:\nNotice that $P(x) = x^{3} - 2x + 1 = (x - 1)(x^{2} + x - 1)$, and the roots of $x^{2} + x - 1$ are $\\frac{-1 \\pm \\sqrt{5}}{2}$. Therefore the roots of $P(x)$ are \n$$1, \\frac{\\sqrt{5} - 1}{2} \\text{ and } \\frac{-1 - \\sqrt{5}}{2}.$$ \nNotice also that $Q(x) = x^{3} - 4x^{2} + 4x - 1 = (x - 1)(x... | New Zealand | New Zealand Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0c0q | Let $a$, $b$, $c$ be positive real numbers such that $a^2 + b^2 + c^2 = 3$. Prove that
$$
\frac{1}{a} + \frac{3}{b} + \frac{5}{c} \geq 4a^2 + 3b^2 + 2c^2.
$$ | [
"*First solution.* The inequality can be written\n$$\n\\frac{1}{a} + \\frac{3}{b} + \\frac{5}{c} + b^2 + 2c^2 \\geq 4(a^2 + b^2 + c^2)\n$$\nor\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{2}{b} + b^2 + \\frac{4}{c} + 2c^2 \\geq 12.\n$$\nThis follows from the following inequalities by using the AM-GM ine... | Romania | 69th NMO Selection Tests for JBMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
08yu | As shown in the following figure, there are four types of blocks formed by four cubes of side one. How many ways are there for putting four blocks of these types in a $2 \times 2 \times 4$ rectangular box so that they do not stick out?
Note that we can use multiple blocks of the same type, and blocks can be rotated. Ad... | [
"$379$\nDenote a cube of side one as a *small cube*. The rectangular box consists of $16$ small cubes, and we denote a cube of side two (formed by eight small cubes) in the box as a *medium cube*.\nConsider the case where there are four blocks in the box. For each block in the box, there exists a medium cube contai... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof and answer | 379 | |
0eqg | As a decimal number $6.28\%$ is equal to
(A) $0.0628$ (B) $0.628$ (C) $6.28$ (D) $62.8$ (E) $628$ | [] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | English | MCQ | A | |
0i9g | Problem:
The sequence of real numbers $x_{1}, x_{2}, x_{3}, \ldots$ satisfies $\lim_{n \rightarrow \infty}\left(x_{2 n}+x_{2 n+1}\right)=315$ and $\lim_{n \rightarrow \infty}\left(x_{2 n}+x_{2 n-1}\right)=2003$. Evaluate $\lim_{n \rightarrow \infty}\left(x_{2 n} / x_{2 n+1}\right)$. | [
"Solution:\n\nWe have $\\lim_{n \\rightarrow \\infty}\\left(x_{2 n+1}-x_{2 n-1}\\right)=\\lim_{n \\rightarrow \\infty}\\left[\\left(x_{2 n}+x_{2 n+1}\\right)-\\left(x_{2 n}+x_{2 n-1}\\right)\\right]=315-2003=-1688$; it follows that $x_{2 n+1} \\rightarrow -\\infty$ as $n \\rightarrow \\infty$. Then\n\n$$\n\\lim_{n ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof and answer | -1 | |
0dda | Consider circles $(O_1)$, $(O_2)$, $(O_3)$ tangent to $d$ at $A$, $B$, $C$ and $(O_2)$ is the biggest circle, externally tangent to $(O_1)$, $(O_3)$. Let $BD$ be the diameter of $(O_2)$. The external tangent line (differs from $d$) of $(O_1)$, $(O_3)$ cuts $(O_2)$ at $X$, $Y$. Let $K$ be the midpoint of the arc $XBY$ o... | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gsw | In triangle $ABC$, the circle $\Gamma$ is tangent to the side $BC$ at $X$, and to the side $AC$ at $Y$. For an arbitrary point $P$ on the side $[AB]$, the second intersection of $XP$ with $\Gamma$ is $K$ and the second intersection of $YP$ with $\Gamma$ is $L$, the second intersection of $AK$ with $\Gamma$ is $R$ and t... | [
"Let $T$ be the intersection of $XY$ and $AB$ (since the following argument works on the projective plane, we need not distinguish the case $XY \\parallel AB$). Pascal's theorem applied to the degenerate cyclic hexagon $YYLRKX$ implies that the points\n$$\nA = YY \\cap KR, \\quad P = YL \\cap XK, \\quad LR \\cap XY... | Turkey | Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0a1w | Problem:
Zij $n \geq 3$ een vast natuurlijk getal. Er zijn $n$ dozen $A_{1}, A_{2}, \ldots, A_{n}$, elk met een aantal stenen erin $a_{1}, a_{2}, \ldots, a_{n}$ zo dat $a_{1}+a_{2}+\cdots+a_{n}=3 n$. Een zet bestaat uit de volgende handelingen:
kies een doos en verdeel alle stenen in de doos over de $n$ dozen (inclusi... | [
"Solution:\n\nAntwoord: $M=3 n-4$ en $f\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)=3 n-4$ dan en slechts dan als $a_{1}=a_{2}=\\ldots=a_{n}=3$.\n\nAllereerst merken we op dat er voor elke verdeling een zet bestaat zo dat $\\max \\left(a_{1}, \\ldots, a_{n}\\right)$ minstens een omhoog gaat, tenzij alle stenen in éé... | Netherlands | IMO-selectietoets II | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | M_n = 3n - 4; equality occurs exactly when every box initially contains three stones. | |
07s2 | Suppose $x$ is a real number and $\sin 3y = -\sin x$. Determine all possible values of $\sin y$. | [
"**Solution 1.** Note that\n$$\n\\sin 3\\theta = \\sin \\theta \\cos 2\\theta + \\cos \\theta \\sin 2\\theta = \\sin \\theta(1 - 2\\sin^2 \\theta) + \\cos \\theta (2\\sin \\theta \\cos \\theta) = 3 \\sin \\theta - 4 \\sin^3 \\theta.\n$$\nSo\n$$\n3 \\sin y - 4 \\sin^3 y = - \\sin x = -(3 \\sin z - 4 \\sin^3 z), \\qu... | Ireland | Irish | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | -sin(x/3), (1/2)(sin(x/3) + sqrt(3) cos(x/3)), (1/2)(sin(x/3) - sqrt(3) cos(x/3)) | |
0cf4 | Let $ABC$ be a triangle and let $M$ be the midpoint of the side $BC$. The parallels through $M$ to $AB$ and $AC$ cross the tangent at $A$ to circle $ABC$ at $X$ and $Y$, respectively. The circles $BMX$ and $CMY$ cross again at $S$. Prove that the circles $SXY$ and $SBC$ are tangent.
 | [
"The argument hinges on the four facts below:\n(1) $X'$ lies on circle $YMC$ and $Y'$ lies on circle $XMB$.\n(2) $S$, $A$ and $M$ are collinear.\n(3) $U$ and $V$ both lie on circle $SXY$.\n(4) $UV$ is parallel to $BC$.\nAssume these facts for the moment, to complete the solution as follows: By (3), the conclusion i... | Romania | Eighteenth STARS OF MATHEMATICS Competition | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geo... | English | proof only | null | |
0ftu | Problem:
Bestimme die grösste natürliche Zahl $n$, sodass
$$
4^{995}+4^{1500}+4^{n}
$$
eine Quadratzahl ist. | [
"Solution:\nAntwort: $n=2004$\nWir zeigen allgemeiner: Sind $a<b$ zwei natürliche Zahlen, dann ist $n=2b-a-1$ der grösste Wert, für den $4^{a}+4^{b}+4^{n}$ eine Quadratzahl ist.\nSei zunächst $n=2b-a-1$. Wir erhalten\n$$\n\\begin{aligned}\n4^{a}+4^{b}+4^{n} & =4^{a}\\left(1+4^{b-a}+4^{2(b-a)-1}\\right) \\\\\n& =\\l... | Switzerland | IMO Selektion | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2004 | |
04x3 | Determine all prime numbers $p$, such that the number
$$
\binom{p}{1}^2 + \binom{p}{2}^2 + \dots + \binom{p}{p-1}^2
$$
is divisible by $p^3$. | [
"We start from the observation that for $k = 1, 2, \\dots, p$ we have\n$$\n\\binom{p-1}{k-1} \\equiv \\pm 1 \\pmod{p}. \\qquad (1)\n$$\nTo see this, note that\n$$\np-1 \\equiv -1 \\pmod{p},\n$$\n$$\np-2 \\equiv -2 \\pmod{p},\n$$\n$$ \\vdots \\qquad \\vdots $$\n$$\nk \\equiv -(p-k) \\pmod{p},\n$$\nhence, multiplying... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | all primes p ≥ 5 | |
0f62 | Problem:
$C_1$, $C_2$, $C_3$ are circles with radii $r_1$, $r_2$, $r_3$ respectively. The circles do not intersect and no circle lies inside any other circle. $C_1$ is larger than the other two. The two outer common tangents to $C_1$ and $C_2$ meet at $A$ ("outer" means that the points where the tangent touches the tw... | [] | Soviet Union | 18th ASU | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | The quadrilateral is tangential, and its inscribed circle has radius r2*r3/r1. | |
02yp | Problem:
Em um determinado jogo, o número $1$ está escrito no quadro. Em qualquer momento, um movimento permitido consiste em trocar o número escrito no quadro pelo seu dobro ou por outro número que possui os mesmos dígitos que ele. Por exemplo, se estiver escrito no quadro o número $137$, um movimento permitido consi... | [
"Solution:\n\na) Sim, é possível. Após realizar o movimento de multiplicação por $2$ nove vezes, podemos trocar o $1$ original por $2^{9} = 512$. Em seguida, podemos trocá-lo por $125$. Multiplicando-o por $2$ três vezes, podemos trocá-lo por $125 \\cdot 2^{3} = 1000$.\n\nb) Sim, também é possível. Note que $1000 =... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | a) yes; b) yes; c) no | |
0cmu | On a party a company of 20 persons is to be arranged to sit around 4 tables. An arrangement is called *successful* if each two persons sharing a table are friends. It appears that there exists at least one successful arrangement, and for each successful arrangement exactly 5 persons sit around each table. Find the grea... | [
"**Answer.** $160$.\n\nIf two people are not friends, we will say that they are enemies. The total number of pairs of people in this company is $\\frac{20 \\cdot 19}{2} = 190$, so it is enough to prove that the minimal number of pairs of enemies is $30$.\n\nLet us prove that there cannot be fewer than $30$ pairs of... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 160 | |
0hzr | Problem:
A dart is thrown at a square dartboard of side length $2$ so that it hits completely randomly. What is the probability that it hits closer to the center than any corner, but within a distance $1$ of a corner? | [
"Solution:\n\nBy symmetry it will suffice to consider one quarter of the dartboard, which is a square of side length $1$. Therefore the probability is the area of the desired region in this square. The desired region is the part of the circle of radius $1$ centered at a corner that is closer to the opposite corner.... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | (π - 2)/4 | |
0dgd | Define $a_0 = 2$ and $a_{n+1} = a_n^2 + a_n - 1$ for $n \ge 0$. Prove that $a_n$ is coprime to $2n + 1$ for all $n \in \mathbb{N}$. | [
"Consider any prime divisor $p$ of $a_n$. Consider a directed graph with $p$ edges on $\\{0, 1, \\dots, p-1\\}$ (mod $p$) connecting $x \\to y$ if and only if\n$$\nx^2 + x - 1 \\equiv y \\pmod{p}.\n$$\nObserve that $\\frac{p-1}{2}$ is connected to $b = \\frac{p^2-5}{4}$ (mod $p$), every element has out-degree 1, an... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0fat | Problem:
$E$ is a point on the diagonal $BD$ of the square $ABCD$. Show that the points $A$, $E$ and the circumcenters of $ABE$ and $ADE$ form a square. | [
"Solution:\n\n\n\nLet $O$, $O'$ be the circumcenters of $ABE$, $ADE$ respectively. Then $OA = OE$ and $\\angle AOB = 2 \\angle ABE = 90^\\circ$. Similarly, $O'A = O'E$ and $\\angle AOE = 2\\angle ADE = 90^\\circ$. Hence $AOEO'$ is a square."
] | Soviet Union | 1st CIS | [
"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 | |
0khz | An organization has 30 employees, 20 of whom have a brand A computer while the other 10 have a brand B computer. For security, the computers can only be connected to each other and only by cables. The cables can only connect a brand A computer to a brand B computer. Employees can communicate with each other if their co... | [
"**Answer (B):** Let $P$ be one of the computers of brand A. If every brand B computer is connected to every brand A computer other than $P$, then there will be $10 \\cdot 19 = 190$ cables, but the employee using computer $P$ is completely isolated and cannot communicate with any other. Therefore the requested answ... | United States | AMC 12 A | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | MCQ | B | |
037s | Problem:
Let $k$ be the circumcircle of $\triangle ABC$ and $D$ be a point on the arc $\overparen{AB}$, which does not contain $C$. Denote by $I_{A}$ and $I_{B}$ the incenters of $\triangle ADC$ and $\triangle BDC$, respectively. Prove that the circumcircle of $\triangle I_{A} I_{B} C$ is tangent to $k$ if and only if
... | [
"Solution:\nLet $P = CI_{A} \\cap k$ and $Q = CI_{B} \\cap k$. First we prove that the circumcircle $k_{1}$ of $\\triangle I_{A} I_{B} C$ is tangent to $k$ if and only if $I_{A} I_{B} \\parallel PQ$.\nLet $T$ be the point on the tangent line to $k$ at $C$, for which $\\angle ACT = \\angle ABC$. If $k$ and $k_{1}$ a... | Bulgaria | Team selection test for 47. IMO | [
"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 only | null | |
0bxv | Determine the prime numbers $a$, $b$, $c$, such that $a = b^4 + c^3$ and $a \le 2017$. | [
"$a \\ge 2^4 + 2^3 > 2$ is a prime number, hence it is odd. Then precisely one of the primes $b$ or $c$ equals $2$.\n\nIf $b = 2$, then $a = 16 + c^3 \\le 2017$, hence $c \\le 11$, so $c \\in \\{3, 5, 7, 11\\}$.\n\nConsidering each of these values, we obtain:\n* for $c = 3$, $a = 43$, which is a prime;\n* for $c = ... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (a, b, c) = (43, 2, 3), (359, 2, 7), (89, 3, 2) | |
09lx | The internal bisector of angle $\angle B$ of triangle $ABC$ where $\angle A > 90^\circ$, intersects side $AC$ at point $D$. Points $A'$ and $F$ are chosen on side $BC$ such that $AB = A'B$ and $\angle AFB = 2\angle ADB$. Points $E$ and $G$ are on the line $BC$ such that $AD = DE$, $EF = FG$, $E \neq A'$ and $G \neq E$.... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0k5i | Let $n$ be a positive integer. A frog starts on the number line at $0$. Suppose it makes a finite sequence of hops, subject to two conditions:
* The frog visits only points in $\{1, 2, \dots, 2^n - 1\}$, each at most once.
* The length of each hop is in $\{2^0, 2^1, 2^2, \dots\}$. (The hops may be either direction, lef... | [
"We claim the answer is $\\frac{4^n-1}{3}$.\n\nWe first prove the bound. First notice that the hop sizes are in $\\{2^0, 2^1, \\dots, 2^{n-1}\\}$, since the frog must stay within bounds the whole time. Let $a_i$ be the number of hops of size $2^i$ the frog makes, for $0 \\le i \\le n-1$.\n\n**Claim.** For any $k = ... | United States | USA TSTST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Equations and Inequalities ... | null | proof and answer | (4^n - 1) / 3 | |
09ta | Problem:
Bepaal het product van alle positieve gehele getallen $n$ waarvoor $3(n!+1)$ deelbaar is door $2 n-5$. | [
"Solution:\n\nDe getallen $n=1$ tot en met $n=4$ voldoen, want $2 n-5$ is dan gelijk aan respectievelijk $-3,-1,1$ en $3$, dus altijd een deler dan $3(n!+1)$. Vanaf nu bekijken we alleen $n>4$ en dan is $2 n-5>3$.\n\nWe bewijzen eerst dat als $n$ voldoet, dan $2 n-5$ priem moet zijn. We onderscheiden twee gevallen.... | Netherlands | IMO-selectietoets III | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 2016 | |
05i7 | Problem:
Soit $ABCD$ un quadrilatère inscrit dans un cercle de centre $O$. On note $\Delta_{1}$ et $\Delta_{2}$ les images de la droite $(AB)$ par les symétries dont les axes sont respectivement les bissectrices intérieures de $\widehat{CAD}$ et $\widehat{CBD}$. Soit $P$ l'intersection de $\Delta_{1}$ et $\Delta_{2}$.... | [
"Solution:\n\nOn traite uniquement le cas où $O$ est à l'intérieur de $ABCD$ (les autres cas étant similaires). D'après les propriétés de l'angle inscrit, les deux bissectrices considérées s'intersectent en un point du cercle noté $S$, qui est le milieu de l'arc $CD$. En particulier, $(OS)$ et $(CD)$ sont perpendic... | France | Olympiades Françaises de Mathématiques - Épreuve en temps limité de Janvier | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0iwp | For $n \ge 2$ let $a_1, a_2, \dots, a_n$ be positive real numbers such that
$$
(a_1 + a_2 + \dots + a_n) \left( \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n} \right) \le \left( n + \frac{1}{2} \right)^2.
$$
Prove that $\max(a_1, a_2, \dots, a_n) \le 4 \min(a_1, a_2, \dots, a_n)$. | [
"**Solution 1.** The Cauchy-Schwarz Inequality gives\n$$\n\\begin{aligned} \\left(n + \\frac{1}{2}\\right)^2 &\\ge (a_1 + a_2 + \\dots + a_n) \\left(\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_n}\\right) \\\\ &= (m + a_2 + \\dots + a_{n-1} + M) \\left(\\frac{1}{M} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_... | United States | USAMO 2009 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
045m | Find all prime numbers $p$ and positive integers $a, b, c$ such that
$$
2^a p^b = (p + 2)^c + 1.
$$ | [
"Obviously, $p$ is odd, so $p \\ge 3$. If $c = 1$, then\n$$\np + 3 = 2^a p^b \\ge 2p \\ge p + 3,\n$$\nThis can only happen when $p = 3$, $a = b = 1$, giving one solution $(p, a, b, c) = (3, 1, 1, 1)$. In what follows, we assume that $c \\ge 2$.\n\nCase 1: $c$ is odd. Assume that $q$ is a prime factor of $c$. Since ... | China | 2022 China Team Selection Test | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | [(3, 1, 1, 1), (5, 1, 2, 2)] | |
00q4 | The cells of an $n \times n$ chessboard are coloured in several colours so that no $2 \times 2$ square contains four cells of the same colour. A proper path of length $m$ is a sequence $a_1, a_2, \dots, a_m$ of distinct cells in which the cells $a_i$ and $a_{i+1}$ have a common side and are coloured in different colour... | [
"(1) Consider the graph $G$ whose vertices are the board's cells and whose edges are all pairs of neighbouring (i.e., having a common side) cells coloured in different colours. We wish to show that $G$ contains a simple path of length at least $n$.\n\n(2) On a new $n \\times n$ chessboard $B$, colour each connected... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory"
] | English | proof only | null | |
074t | Problem:
How many 6-tuples $\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\right)$ are there such that each of $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ is from the set $\{1,2,3,4\}$ and the six expressions
$$
a_{j}^{2}-a_{j} a_{j+1}+a_{j+1}^{2}
$$
for $j=1,2,3,4,5,6$ (where $a_{7}$ is to be taken as $a_{1}$) are all... | [
"Solution:\n\nWithout loss of generality, we may assume that $a_{1}$ is the largest among $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$. Consider the relation\n$$\na_{1}^{2}-a_{1} a_{2}+a_{2}^{2}=a_{2}^{2}-a_{2} a_{3}+a_{3}^{2}\n$$\nThis leads to\n$$\n\\left(a_{1}-a_{3}\\right)\\left(a_{1}+a_{3}-a_{2}\\right)=0\n$$\nO... | India | INMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 40 | |
04yo | A ball bearing consists of two cylinders with the same axis and $n$ equal balls between them. The centers of all the balls are on the same plane perpendicular to the axis of the cylinders and each ball touches both cylinders and two adjacent balls. Let $r$ be the radius of the balls and let $R$ be the radius of the out... | [
"Consider the regular $n$-gon with vertices at the centers of the balls (Fig. 16). Its edges are of length $2r$ and its perimeter is $n \\cdot 2r$. The radius of the circumcircle of the $n$-gon is $R-r$ and the length of the circumcircle is $2\\pi(R-r)$. Since a chord of a circle is always shorter than the correspo... | Estonia | Estonija 2010 | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
07n2 | Find all integers $n$ for which $n^5 + n^4 + n^3 + n^2 + n + 1$ is divisible by 19. | [
"First note that $n \\equiv 1 \\pmod{19}$ implies $n^5 + n^4 + n^3 + n^2 + n + 1 \\equiv 6 \\pmod{19}$. Assume now that $n \\not\\equiv 1 \\pmod{19}$, i.e. $\\gcd(19, n-1) = 1$. Because $n^6 - 1 = (n-1)(n^5 + n^4 + n^3 + n^2 + n + 1)$ we then have\n$$\n19 \\mid n^5 + n^4 + n^3 + n^2 + n + 1 \\iff n^6 \\equiv 1 \\pm... | Ireland | Ireland | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All integers n with n ≡ 7, 8, 11, 12, or 18 (mod 19). | |
0k4d | Problem:
Call a polygon normal if it can be inscribed in a unit circle. How many non-congruent normal polygons are there such that the square of each side length is a positive integer? | [
"Solution:\n\nThe side lengths of the polygon can only be from the set $\\{1, \\sqrt{2}, \\sqrt{3}, 2\\}$, which take up $60^{\\circ}, 90^{\\circ}, 120^{\\circ}, 180^{\\circ}$ of the circle respectively. By working modulo $60$ degrees we see that $\\sqrt{2}$ must be used an even number of times. We now proceed to c... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 14 | |
00sw | Let $s \ge 2$ and $n \ge k \ge 2$ be integers, and let $\mathcal{A}$ be a subset of $\{1, 2, \dots, n\}^k$ of size at least $2sk^2n^{k-2}$ such that any two members of $\mathcal{A}$ share some entry. Prove that there are an integer $p \le k$ and $s+2$ members $A_1, A_2, \dots, A_{s+2}$ of $\mathcal{A}$ such that $A_i$ ... | [
"Fix a member $A$ of $\\mathcal{A}$. Note that there are at most $\\binom{k}{2}n^{k-2}$ $k$-tuples that share at least two entries with $A$. Indeed, there are $n^{k-2}$ $k$-tuples sharing any two given entries. The bound now follows, since the two entries can be chosen in $\\binom{k}{2}$ different ways.\n\nTherefor... | Balkan Mathematical Olympiad | BMO Short List | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
02jx | Problem:
O quadrado $STUV$ é formado de um quadrado limitado por 4 retângulos iguais. O perímetro de cada retângulo é $40~\mathrm{cm}$. Qual é a área, em $\mathrm{cm}^2$, do quadrado $STUV$?
(A) 400
(B) 200
(C) 160
(D) 100 | [
"Solution:\n\nDenotemos por $C$ e $L$, o comprimento e a largura respectivamente de cada um dos quatro retângulos. O perímetro de cada retângulo é $2(C+L)$. Então, $2 \\times (C+L) = 40 \\Rightarrow C+L = 20$. Observe na figura que o lado do quadrado $STUV$ é $C+L$, e portanto sua área é $A = (C+L)^2 = 20^2 = 400~\... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals"
] | null | MCQ | A | |
0l7b | Problem:
Let $\triangle ABC$ be a triangle with incenter $I$. The incircle of triangle $\triangle ABC$ touches $\overline{BC}$ at $D$. Let $M$ be the midpoint of $\overline{BC}$, and let line $AI$ meet the circumcircle of triangle $\triangle ABC$ again at $L \neq A$. Let $\omega$ be the circle centered at $L$ tangent ... | [
"Solution:\n\nLet $X$ and $Y$ be the bottom and top point on $\\omega$ (i.e., the tangents of $X$ and $Y$ to $\\omega$ are parallel to $BC$, and $Y$ and $A$ lie on the same side of $BC$). Note that $A$, $P$, $D$, and $X$ are collinear by homothety between the incircle and $\\omega$. The key claim is the following.\... | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tang... | null | proof only | null | |
03sh | The number of real solutions for equation
$$
(x^{2006} + 1)(1 + x^2 + x^4 + \cdots + x^{2004}) = 2006x^{2005}
$$
is \underline{\quad}.
| [
"We have\n$$\n\\begin{align*}\n& (x^{2006} + 1)(1 + x^2 + x^4 + \\cdots + x^{2004}) = 2006x^{2005} \\\\\n\\Leftrightarrow & \\left(x + \\frac{1}{x^{2005}}\\right)(1 + x^2 + x^4 + \\cdots + x^{2004}) = 2006 \\\\\n\\Leftrightarrow & x + x^3 + x^5 + \\cdots + x^{2005} + \\frac{1}{x^{2005}} + \\frac{1}{x^{2003}} + \\fr... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 1 | |
00nq | Determine all triples $(x, y, z)$ of positive integers satisfying
$$
x \mid (y+1), \quad y \mid (z+1) \quad \text{and} \quad z \mid (x+1).
$$ | [
"**Answer.** There are ten triples satisfying the three conditions. They are given by $(1, 1, 1)$, $(1, 1, 2)$, $(1, 3, 2)$, $(3, 5, 4)$ and their cyclic permutations.\n\nWithout loss of generality, let $x$ be the smallest of the three numbers (or one of the smallest), i.e. $x \\le y$ and $x \\le z$. From $z \\mid ... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All cyclic permutations of (1,1,1), (1,1,2), (1,3,2), (3,5,4). Equivalently, the ten triples: (1,1,1), (1,1,2), (1,2,1), (2,1,1), (1,3,2), (3,2,1), (2,1,3), (3,5,4), (5,4,3), (4,3,5). | |
00jp | Prove that if
$$
(n^2 + 1)^{2k} \cdot (44n^3 + 11n^2 + 10n + 2) = N^m
$$
holds for some non-negative integer values of $m$, $n$, $N$ and $k$, $m = 1$ must hold. | [
"Since the left side of the equation is certainly larger than 1, we first note that $m > 0$ must certainly hold.\n\nNow, we consider even values of $n$. Since $n^2+1 \\equiv 1 \\pmod 4$ and $44n^3+11n^2+10n+2 \\equiv 2 \\pmod 4$ are certainly true, we have $N^m \\equiv 2 \\pmod 4$. If $m > 1$, $N^m$ is odd for any ... | Austria | AustriaMO2013 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | m = 1 |
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