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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0clc | Find all integers $n \ge 2$ so that $n$ is divisible by each of the numbers $d_1, d_1+d_2, \dots, d_1+d_2+\dots+d_{k-1}$, where $1 = d_1 < d_2 < \dots < d_{k-1} < d_k = n$ are all the positive divisors of $n$. | [
"Answer: all primes and $n = 6$.\nThe primes have the required property, because the divisors of a prime $n$ are $1 = d_1 < d_2 = n$, and $n$ is divisible by $d_1$.\n\nTake now a positive integer $n$ with $k \\ge 3$ divisors and fulfilling the required property.\nIf $n$ is odd, then all its $k$ divisors are odd. Fr... | Romania | 75th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other"
] | English | proof and answer | all prime numbers and 6 | |
06pj | Given an isosceles triangle $A B C$ with $A B = A C$. The midpoint of side $B C$ is denoted by $M$. Let $X$ be a variable point on the shorter arc $M A$ of the circumcircle of triangle $A B M$. Let $T$ be the point in the angle domain $B M A$, for which $\angle T M X = 90^{\circ}$ and $T X = B X$. Prove that $\angle M ... | [
"Let $N$ be the midpoint of segment $B T$ (see Figure 1).\nLine $X N$ is the axis of symmetry in the isosceles triangle $B X T$, thus $\\angle T N X = 90^{\\circ}$ and $\\angle B X N = \\angle N X T$.\nMoreover, in triangle $B C T$, line $M N$ is the midline parallel to $C T$; hence $\\angle C T M = \\angle N M T$.... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0f3p | Problem:
35 containers of total weight 18 must be taken to a space station. One flight can take any collection of containers weighing 3 or less. It is possible to take any subset of 34 containers in 7 flights. Show that it must be possible to take all 35 containers in 7 flights. | [
"Solution:\n\nLet the lightest container be $S$ weighing $s$. The other containers can be taken in 7 flights. The smallest load on these flights must be $\\leq (18 - s) / 7$, and so that flight has spare capacity of at least $3 - (18 - s) / 7 = (3 + s) / 7$. Thus it can accommodate $S$ provided that $(3 + s) / 7 \\... | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0ef4 | Problem:
Natančno izračunaj vrednost izraza $\left(\sin \frac{38 \pi}{3}\right)^{5}: 4^{\log _{2} 3-\log _{16} 256}$. | [
"Solution:\n\n$$\n\\left(\\sin \\frac{38 \\pi}{3}\\right)^{5}=\\sin ^{5}\\left(12 \\pi+\\frac{2 \\pi}{3}\\right)=\\sin ^{5}\\left(\\frac{2 \\pi}{3}\\right)=\\left(\\frac{\\sqrt{3}}{2}\\right)^{5}=\\frac{9 \\sqrt{3}}{32}\n$$\nPoenostavimo in izračunamo drugi del izraza:\n$$\n4^{\\log _{2} 3-\\log _{16} 256}=4^{\\log... | Slovenia | 17. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Odbirno tekmovanje | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | sqrt(3)/2 | |
06tn | Let $\mathbb{Z}_{>0}$ denote the set of positive integers. For any positive integer $k$, a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ is called $k$-good if $\operatorname{gcd}(f(m)+n, f(n)+m) \leqslant k$ for all $m \neq n$. Find all $k$ such that there exists a $k$-good function. | [
"We first show that there is no $1$-good function. Suppose that there exists a function $f$ such that $G_{f}(m, n)=1$ for all $m \\neq n$. Now, if there are two distinct even numbers $m$ and $n$ such that $f(m)$ and $f(n)$ are both even, then $2 \\mid G_{f}(m, n)$, a contradiction. A similar argument holds if there... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English | proof and answer | All positive integers k greater than or equal to 2 | |
0ai1 | Let $x$, $y$ and $z$ be positive real numbers such that $xy + yz + zx = 3xyz$. Prove that
$$
x^2 y + y^2 z + z^2 x \geq 2(x + y + z) - 3
$$ | [
"The given condition can be rearranged to $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = 3$. Using this, we obtain:\n$$\n\\begin{aligned}\nx^2 y + y^2 z + z^2 x - 2(x + y + z) - 3 &= x^2 y - 2x + \\frac{1}{y} + y^2 z - 2y + \\frac{1}{z} + z^2 x - 2z + \\frac{1}{x} \\\\\n&= y \\left(x - \\frac{1}{y}\\right)^2 + z \\l... | North Macedonia | Balkan Mathematical Olympiads | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0aur | Problem:
Let $x$ and $y$ be real numbers that satisfy the following system of equations:
$$
\left\{\begin{array}{l}
\frac{x}{x^{2} y^{2}-1}-\frac{1}{x}=4 \\
\frac{x^{2} y}{x^{2} y^{2}-1}+y=2
\end{array}\right.
$$
Find all possible values of the product $x y$. | [
"Solution:\nThe given system can be expressed as follows:\n$$\n\\left\\{\\begin{array}{l}\n\\frac{x}{x^{2} y^{2}-1}-\\frac{1}{x}=4 \\\\\n\\frac{x^{2} y}{x^{2} y^{2}-1}+y=2\n\\end{array}\\right. \\Rightarrow \\left\\{\\begin{array}{l}\n\\frac{x^{2}}{x^{2} y^{2}-1}-1=4 x \\\\\n\\frac{x^{2}}{x^{2} y^{2}-1}+1=\\frac{2}... | Philippines | 19th Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | ± 1/√2 | |
0kuj | Problem:
Let $A$ and $B$ be points in space for which $AB = 1$. Let $\mathcal{R}$ be the region of points $P$ for which $AP \leq 1$ and $BP \leq 1$. Compute the largest possible side length of a cube contained within $\mathcal{R}$. | [
"Solution:\n\nLet $h$ be the distance between the center of one sphere and the center of the opposite face of the cube. Let $x$ be the side length of the cube. Then we can draw a right triangle by connecting the center of the sphere, the center of the opposite face of the cube, and one of the vertices that make up ... | United States | HMMT February 2023 | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (sqrt(10) - 1)/3 | |
01u1 | A point $P$ (distinct from $B$ and $C$) are marked on the side $BC$ of a triangle $ABC$. Let $I_1$ and $I_2$ be the centers of the inscribed circles of the triangles $ABP$ and $APC$, respectively. The circumcircle of the triangle $I_1PI_2$ intersects the segment $BC$ at $P$ and $Q$.
Prove that $AB + QC = AC + QB$.
(A. ... | [
"Without loss of generality we can assume that $Q$ belongs to the segment $PC$. First, note that if the incircle of the triangle $ABC$ touches the sides $AB$, $BC$, $CA$ at the points $C_1$, $A_1$, $B_1$, respectively, then\n$$\n\\begin{aligned}\nAB_1 &= AC_1 = \\frac{AB + AC - BC}{2}, \\\\\nBA_1 &= BC_1 = \\frac{B... | Belarus | 66th Belarusian 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 > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0fwc | Problem:
Bestimme alle Paare $(a, b)$ natürlicher Zahlen, sodass
$$
a^{2}+3 b \text{ und } b^{2}+3 a
$$
beides Quadratzahlen sind. | [
"Solution:\n\nWir setzen $A=a^{2}+3 b$ und $B=b^{2}+3 a$. Aus Symmetriegründen können wir $a \\geq b$ annehmen. Damit folgt nun\n$$\na^{2}<A \\leq a^{2}+3 a<a^{2}+4 a+4=(a+2)^{2}\n$$\nda $A$ selbst eine Quadratzahl ist, muss demnach $A=(a+1)^{2}$ sein. Daraus folgt $3 b=2 a+1$ und somit auch $B=b^{2}+\\frac{9}{2} b... | Switzerland | Vorrundenprüfung | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (1,1), (11,16), (16,11) | |
0fi2 | Problem:
En la figura, $AB$ es un segmento fijo y $C$ un punto variable dentro de él. Se construyen triángulos equiláteros $ACB'$
y $CBA'$ de lados $AC$ y $CB$ en el mismo semiplano definido por $AB$, y otro triángulo equilátero $ABC'$ de lado $AB$ en el semiplano opuesto. Demostrar:
a) Las rectas $AA'$, $BB'$ y $CC'... | [
"Solution:\n\nPara resolver este problema utilizaremos muchas veces la propiedad siguiente: si desde un punto $P$ de la circunferencia circunscrita a un triángulo equilátero trazamos segmentos que lo unan a los tres vértices, dichos segmentos forman ángulos de $60^{\\circ}, 60^{\\circ}$ y $120^{\\circ}$; y este últ... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Advanced Configurations > Napoleon and Fermat points",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Qu... | null | proof and answer | a) The three lines are concurrent at a point P.
b) As the interior point varies along the segment, P traces the arc of the circumcircle of the equilateral triangle on the whole segment lying on the side opposite to that triangle’s vertex; equivalently, the arc from which the segment is seen under an angle of one hundre... | |
0gjs | For any positive integer $n$, consider its binary representation. Denote by $f(n)$ the number we get after removing all the 0's in its binary representation, and $g(n)$ the number of 1's in the binary representation. For example, $f(19) = 7$ and $g(19) = 3$. Find all positive integers $n$ that satisfy
$$
n = f(n)^{g(n)... | [
"Let $g(n) = k$. Then $f(n) = 2^k - 1$, and so we just need to determine all $k$ such that $(2^k - 1)^k$ has $k$ 1's in its binary representation. To do this, we can simply expand\n$$\n(2^k - 1)^k = \\sum_{i=0}^{k} (-1)^i \\binom{k}{i} 2^{k^2 - ki} = \\sum_{i=0}^{\\lfloor \\frac{k-1}{2} \\rfloor} \\binom{k}{2i} 2^{... | Taiwan | IMO 2J, Independent Study 2 | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | Chinese; English | proof and answer | 1 and 9 | |
0ggn | 找出所有有下列性質的正整數 $n$: 存在一個 $n$ 的所有正因數 (共 $k$ 個) 的排列 $(d_1, d_2, \dots, d_k)$, 使得對於每個 $i = 1, 2, \dots, k$, 整數 $\sum_{j=1}^{i} d_j = d_1 + \dots + d_i$ 都是一個平方數。 | [
"The answer is $n = 1, 3$.\nFor $i = 1, 2, \\dots, k$ let $d_1 + \\dots + d_i = s_i^2$, and define $s_0 = 0$ as well. Obviously\n$0 = s_0 < s_1 < s_2 < \\dots < s_k$, so $s_i \\ge i$ and\n$$\nd_i = s_i^2 - s_{i-1}^2 = (s_i + s_{i-1})(s_i - s_{i-1}) \\ge s_i + s_{i-1} \\ge 2i - 1. \\quad (1)\n$$\nThe number 1 is one... | Taiwan | 2022 數學奧林匹亞競賽第一階段選訓營, 獨立研究(一) | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Chinese; English | proof and answer | 1, 3 | |
0l9e | Let be given two polynomials
$$
P(x) = 4x^3 - 2x^2 - 15x + 9
$$
and
$$ Q(x) = 12x^3 + 6x^2 - 7x + 1. $$
1/ Prove that each of these polynomials has three distinct real roots.
2/ Let $\alpha$ and $\beta$ be respectively the greatest roots of $P(x)$ and $Q(x)$. Prove that $\alpha^2 + 3\beta^2 = 4$. | [
"1. We have:\n$$\nP(-2) = -1;\\ P(-1) = 18;\\ P(3/2) = -9/2 \\text{ and } P(\\sqrt{33}/3) = (15 - \\sqrt{33})/9. \\quad (1)\n$$\n$$\nQ(-2) = -57;\\ Q(-1) = 2;\\ Q(1/3) = -2/9 \\text{ and } Q(1) = 12. \\quad (2)\n$$\nFrom (1) it follows that $P(x)$ has three distinct real roots, and from (2) it follows that $Q(x)$ h... | Vietnam | 2003 Vietnamese Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof only | null | |
0kr0 | Problem:
An ant starts at the point $(0,0)$ in the Cartesian plane. In the first minute, the ant faces towards $(1,0)$ and walks one unit. Each subsequent minute, the ant chooses an angle $\theta$ uniformly at random in the interval $\left[-90^\circ, 90^\circ\right]$, and then turns an angle of $\theta$ clockwise (neg... | [
"Solution:\n\nLet $\\alpha_{k}$ be a random variable that represents the turn made after step $k$, choosing $\\alpha_{k}$ uniformly at random on the complex plane among the arc of the unit circle containing $1$ from $-i$ to $i$. It is well known that $\\mathbb{E}\\left[\\alpha_{k}\\right]=\\frac{2}{\\pi}$. We have ... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 45 | |
046i | Let three circles $\odot O$, $\odot O_1$, $\odot O_2$, each externally tangent to the other two, lie on the same side of a line $\ell$. Let them be tangent to $\ell$ at points $A$, $A_1$, $A_2$ respectively, where point $A$ lies on segment $A_1A_2$. Denote the points of tangency of $\odot O$ with $\odot O_1$ and $\odot... | [
"*Proof* 1. Consider the antipodal point $P$ of $A$ on the circle $\\odot O$.\nWe notice that in the right trapezoid $A_1A_2O_1O_2$, we have $O_1A_1 = O_1B_1$ and $OB_1 = OA$.\nTherefore,\n$$\n\\begin{align*} \n\\angle A_1BA &= \\pi - \\angle O_1B_1A_1 - \\angle OB_1A \\\\ \n&= \\pi - \\frac{1}{2}(\\pi - \\angle A_... | China | China-TST-2023B | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05i3 | Problem:
Soit $a \in ] 0 ; 1[$ et $n>0$ un entier. On note $f_{n}$ la fonction définie sur $\mathbb{R}$ par $f_{n}(x)=x+\frac{x^{2}}{n}$, pour tout réel $x$. Prouver que
$$
\frac{a(1-a) n^{2}+2 a^{2} n+a^{3}}{(1-a)^{2} n^{2}+a(2-a) n+a^{2}}<\underbrace{\left(f_{n} \circ f_{n} \circ \cdots \circ f_{n}\right)}_{n}(a)<\fr... | [
"Solution:\nPour tout entier $k \\geq 0$, posons $a_{k}=\\underbrace{\\left(f_{n} \\circ f_{n} \\circ \\cdots \\circ f_{n}\\right)}_{k}(a)$. En particulier, $a_{0}=a$.\nOn remarque que, pour tout $k \\geq 0$, on a $a_{k+1}=a_{k}+\\frac{a_{k}^{2}}{n}>a_{k}$, donc la suite $\\left(a_{k}\\right)$ est strictement crois... | France | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
057c | Let $n$ be a positive integer. Real numbers $a_1, a_2, \dots, a_{2n}$ satisfy the following conditions:
(1) For every $i = 1, 2, \dots, 2n - 1$, one has $0 < a_{i+1} - a_i \le 1$;
(2) Rounding the numbers $a_1, a_2, \dots, a_{2n}$ to the closest integer (numbers equidistant from two closest integers are rounded up) g... | [
"Suppose that a numerator is greater than a denominator. Then interchanging these two numbers makes both fractions smaller. Thus we can assume w.l.o.g. that all numerators are less than all denominators.\n\nFor every $i = 1, 2, \\dots, 2n$, define $x_i = a_i - i + \\frac{1}{2}$. As $a_1 < a_2 < \\dots < a_{2n}$ and... | Estonia | Open Contests | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
07lu | A line drawn from the vertex $A$ of an equilateral triangle $ABC$ meets the side $BC$ at $D$ and the circumcircle at $P$. Show that
$$
\frac{1}{|PD|} = \frac{1}{|PB|} + \frac{1}{|PC|}
$$ | [
"Because $\\angle PAC = \\angle PBC$, $\\angle APC = \\angle ABC = 60^\\circ$ and $\\angle BPA = \\angle BCA = 60^\\circ$ the triangles $APC$ and $BPD$ are similar. Thus $\\frac{|PA|}{|PB|} = \\frac{|PC|}{|PD|}$, so\n$$\n|PA| \\cdot |PD| = |PB| \\cdot |PC| \\quad (1)\n$$\n\nSince $ABPC$ is ... | Ireland | Irska | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0158 | Problem:
Consider a set $A$ of positive integers such that the least element of $A$ equals $1001$ and the product of all elements of $A$ is a perfect square. What is the least possible value of the greatest element of $A$? | [
"Solution:\nWe first prove that $\\max A$ has to be at least $1040$.\nAs $1001 = 13 \\cdot 77$ and $13 \\nmid 77$, the set $A$ must contain a multiple of $13$ that is greater than $13 \\cdot 77$. Consider the following cases:\n\n- $13 \\cdot 78 \\in A$. But $13 \\cdot 78 = 13^{2} \\cdot 6$, hence $A$ must also cont... | Baltic Way | Baltic Way 2008 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1040 | |
0i2h | Let $P$ be a point in the plane of triangle $ABC$ such that the segments $PA$, $PB$, and $PC$ are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to $PA$. Prove that $\angle BAC$ is acute. | [
"By the **Cauchy-Schwarz Inequality**,\n$$\n\\sqrt{PB^2 + PC^2} \\sqrt{AC^2 + AB^2} \\ge PB \\cdot AC + PC \\cdot AB.\n$$\nApplying the **(Generalized) Ptolemy's Inequality** to quadrilateral $ABPC$ yields\n$$\nPB \\cdot AC + PC \\cdot AB \\geq PA \\cdot BC.\n$$\n\nBecause $PA$ is the longe... | United States | USA IMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"... | English | proof only | null | |
044h | Let $n$ be a positive integer, $a_1, a_2, \dots, a_{2n+1}$ be $2n+1$ positive real numbers. For $k = 1, 2, \dots, 2n+1$, define
$$
b_k = \max_{0 \le m \le n} \left( \frac{1}{2m+1} \sum_{i=k-m}^{k+m} a_i \right),
$$
where the subscript of $a_i$ is taken modulo $2n+1$. Prove: the number of subscripts $k$ satisfying that ... | [
"Define $I = \\{k \\mid b_k \\ge 1\\}$. For every $k \\in I$, assume that the maximum value $b_k$ of $\\frac{1}{2m+1} \\sum_{i=k-m}^{k+m} a_i$ is attained at $m = m_k$, and call\n$$\n[k - m_k, k + m_k] := \\{k - m_k, k - m_k + 1, \\dots, k + m_k\\}\n$$\na “nice segment”, where the subscripts are taken modulo $2n+1$... | China | China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof only | null | |
0dqb | Let $a_n, b_n$, $n = 1, 2, \ldots$ be two sequences of integers defined by $a_1 = 1$, $b_1 = 0$ and for $n \ge 1$,
$$
a_{n+1} = 7a_n + 12b_n + 6
$$
$$
b_{n+1} = 4a_n + 7b_n + 3.
$$
Prove that $a_n^2$ is the difference of two consecutive cubes. | [
"First we shall prove that $a_n^2$ is the difference of two consecutive cubes. To do so, we shall prove by induction that $a_n^2 = (b_n + 1)^3 - b_n^3$.\n\nWhen $n = 1$, this is true.\n\nSuppose for $n \\ge 1$, this is true. We have\n$$\n\\begin{aligned}\n(b_{n+1} + 1)^3 - b_{n+1}^3 &= 3b_{n+1}^2 + 3b_{n+1} + 1 \\\... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0ard | Problem:
Find the range of the function $f(x) = 2^{x^{2} - 4x + 1}$. | [
"Solution:\nLet $y = x^{2} - 4x + 1$.\n\nThe expression $x^{2} - 4x + 1$ is a quadratic in $x$.\n\nThe minimum value of $y$ occurs at $x = -\\frac{b}{2a} = \\frac{4}{2} = 2$.\n\nAt $x = 2$:\n$$\ny = (2)^{2} - 4 \\cdot 2 + 1 = 4 - 8 + 1 = -3\n$$\n\nAs $x \\to \\pm \\infty$, $y \\to +\\infty$.\n\nTherefore, $y \\in [... | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | [1/8, ∞) | |
0a23 | A rectangle $ABCD$ has a circumscribed circle with radius $5$. Points $P$, $Q$, $R$ and $S$ are the midpoints of the sides of rectangle $ABCD$.
What is the perimeter of quadrilateral $PQRS$?
A) $12$ B) $14$ C) $8\sqrt{5}$ D) $20$ E) not uniquely determined | [] | Netherlands | Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | D | |
09ij | Each face of a dodecahedral planet (see picture) is a country. In particular, there are 12 countries and each country is a regular pentagon. For any coloring of the countries in red, green, blue and yellow such that no two countries sharing a border has the same color, show that all colors are used the same number of t... | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
0k59 | Problem:
Farmer James has some strange animals. His hens have 2 heads and 8 legs, his peacocks have 3 heads and 9 legs, and his zombie hens have 6 heads and 12 legs. Farmer James counts 800 heads and 2018 legs on his farm. What is the number of animals that Farmer James has on his farm? | [
"Solution:\nNote that each animal has 6 more legs than heads. Thus, if there are $n$ animals, then there are $6n$ more legs than heads in total. There are $2018 - 800 = 1218$ more legs than heads in total, so there are $\\frac{1218}{6} = 203$ animals."
] | United States | HMMT November 2018 | [
"Math Word Problems"
] | null | final answer only | 203 | |
0i1t | Let $\triangle ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2 = BD_1$ and $CE_2 = AE_1$, and denote by $P$ the point of in... | [
"**First Solution.** The key observation is the following Lemma.\n**Lemma.** *Segment* $D_1Q$ *is a diameter of circle* $\\omega$.\n*Proof.* Let $I$ be the center of circle $\\omega$, that is, $I$ is the incenter of triangle $ABC$. Extend segment $D_1I$ through $I$ to intersect circle $\\omega$ again at $Q'$, and e... | United States | USA IMO | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nin... | English | proof only | null | |
0eee | Find all relatively prime integers $x$ and $y$ that solve the equation
$$
4x^3 + y^3 = 3xy^2.
$$ | [
"Adding $3y^3$ to the equation and factoring both sides we obtain $4(x + y)(x^2 - xy + y^2) = 3y^2(x + y)$. Moving all terms to the left side we can bring out $(x + y)$, so $(x + y)(4x^2 - 4xy + y^2) = 0$. By factoring the second expression we get $(x + y)(2x - y)^2 = 0$. So, $y = -x$ or $y = 2x$. Since $x$ and $y$... | Slovenia | Slovenija 2016 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (-1, 1), (1, -1), (1, 2), (-1, -2) | |
063y | Problem:
Es sei eine positive ganze Zahl $d$ und eine Folge $\left(a_{i}\right)_{i=1,2,3, \ldots}$ positiver ganzer Zahlen gegeben. Wir nehmen an, dass folgende zwei Bedingungen erfüllt sind:
- Jede positive ganze Zahl taucht genau einmal in der Folge auf.
- Für alle Indizes $i \geq 10^{100}$ gilt $\left|a_{i+1}-a_{i}... | [
"Solution:\n\nAngenommen, die Behauptung wäre nicht erfüllt. Dann existiert ein Index $N>10^{100}$, sodass für alle $i \\geq N$ entweder $a_{i} \\leq i-d$ oder $a_{i} \\geq i+d$ gilt. Im ersten Fall gilt wegen $i \\geq N>10^{100}$ laut der zweiten Voraussetzung an die Folge auch\n$$\na_{i+1} \\leq a_{i}+2 d \\leq i... | Germany | 1. Auswahlklausur | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0jq6 | Problem:
Consider a $5 \times 5$ grid of squares. Vladimir colors some of these squares red, such that the centers of any four red squares do not form an axis-parallel rectangle (i.e. a rectangle whose sides are parallel to those of the squares). What is the maximum number of squares he could have colored red? | [
"Solution:\n\nWe claim that the answer is 12. We first show that if 13 squares are colored red, then some four form an axis-parallel rectangle. Note that we can swap both columns and rows without affecting whether four squares form a rectangle, so we may assume without loss of generality that the top row has the mo... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 12 | |
02o9 | Problem:
Qual é o conjunto formado por todos os valores reais positivos de $x$ tais que $(x-1)(x-2)(x-3)<0$?
(a) $(1,2)$
(b) $(1,3)$
(c) $(0,1) \cup (2,3)$
(d) $(0,3)$
(e) $(0,2)$ | [
"Solution:\n\nPara que um produto de três fatores seja negativo, devemos ter dois fatores positivos e um fator negativo, ou os três negativos. As possibilidades são as seguintes.\n$(+)(+)(-) = -$\n$(+)(-)(+) = -$\n$(-)(+)(+) = -$\n$(-)(-)(-) = -$\n\n1) $\\underbrace{(x-1)}_{+} \\underbrace{(x-2)}_{+} \\underbrace{(... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | (c) | |
021r | Problem:
a) Uma minhoca anda sempre sobre uma linha reta. Todos os dias, ela avança $5\ \mathrm{m}$ e recua $3\ \mathrm{m}$. Ao final de $15$ dias, a minhoca estará a que distância do ponto de partida?
b) Ao final destes $15$ dias, quantos metros terá caminhado, no total, esta minhoca?
c) Uma outra minhoca anda també... | [
"Solution:\na) A cada dia, a minhoca avança $5$ metros e recua $3$ metros. Portanto, no final do dia a minhoca avança $5-3=2$ metros. Depois de $15$ dias, a minhoca estará a $15 \\times 2 = 30$ metros do ponto de partida.\n\nb) A cada dia, a minhoca anda $5+3=8$ metros. Portanto, após estes $15$ dias, a minhoca ter... | Brazil | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | a) 30 m; b) 120 m; c) 1000/1001 m; d) No | |
02e6 | Find all 10 digit numbers $a_0a_1\ldots a_9$ such that, for each $k$, $a_k$ is the number of times that the digit $k$ appears in the number. | [
"Note that $a_0 + a_1 + \\cdots + a_9$ is equal to the sum of number of 0s, 1s, ..., 9s, which is, in turn, equal to total number of digits, 10. In other words, the sum of the digits is 10. In particular, if $i > 5$, then $a_i = 0$ or 1 (because if $a_i = 2$, then we have at least 2 digits bigger than 5 with sum bi... | Brazil | VIII OBM | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 6210001000 | |
00oh | Let $ABC$ be an obtuse triangle with orthocenter $H$ and centroid $S$. Let $D$, $E$ and $F$ be the midpoints of segments $BC$, $AC$, $AB$, respectively.
Show that the circumcircle of triangle $ABC$, the circumcircle of triangle $DEF$ and the circle with diameter $HS$ have two distinct points in common. | [
"Let $M$ and $O$ denote the circumcenters of the triangles $DEF$ and $ABC$, respectively. We will use the well-known facts that the points $H$, $M$, $S$ and $O$ lie on the Euler line of triangle $ABC$ in this order, and that $HM : MS : SO = 3 : 1 : 2$.\n\nSince $S$ is in the interior of $ABC$, it is inside the circ... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0exz | Problem:
A graph has at least three vertices. Given any three vertices $A$, $B$, $C$ of the graph we can find a path from $A$ to $B$ which does not go through $C$. Prove that we can find two disjoint paths from $A$ to $B$.
[A graph is a finite set of vertices such that each pair of distinct vertices has either zero or... | [
"Solution:\nTake any path from $A$ to $B$. Suppose it is $A = A_0$, $A_1$, ..., $A_n = B$. We show by induction on $r$ that we can find two disjoint paths from $A$ to $A_r$.\n\nIf $r = 1$, then take any vertex $C$ distinct from $A$ and $A_1$. Take any path from $A_1$ to $C$ which does not go through $A$. Now take a... | Soviet Union | 6th ASU | [
"Discrete Mathematics > Graph Theory > Menger's theorem / max-flow, min-cut",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0dgn | Find all positive integers $n$ that have precisely $\sqrt{n+1}$ natural divisors. | [
"First observe that $n = k^2 - 1$ for some positive integer $k$. As $n$ is not a perfect square, $\\tau(n) = k$ must be even, ergo $n$ is odd.\nWe now establish a bound on $\\tau(n)$. Recall that, since $n$ is not a perfect square, there exists a bijective correspondence between factors of $n$ greater than $\\lfloo... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 3, 15 | |
0k1b | Problem:
Real numbers $x$, $y$, and $z$ are chosen from the interval $[-1,1]$ independently and uniformly at random. What is the probability that
$$
|x|+|y|+|z|+|x+y+z|=|x+y|+|y+z|+|z+x| ?
$$ | [
"Solution:\n\nWe assume that $x$, $y$, $z$ are all nonzero, since the other case contributes zero to the total probability.\n\nIf $x$, $y$, $z$ are all positive or all negative then the equation is obviously true. Otherwise, since flipping the signs of all three variables or permuting them does not change the equal... | United States | HMMT November 2018 | [
"Geometry > Solid Geometry > Volume",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3/8 | |
05kh | Problem:
On note $I$ le centre du cercle inscrit $\gamma$ dans le triangle $ABC$ et $O$ le centre de son cercle circonscrit. Soit $\omega_{A}$ le cercle qui passe par $B$ et $C$ et qui est tangent à $\gamma$. Les cercles $\omega_{B}$ et $\omega_{C}$ sont définis de manière analogue. On note $A'$ l'autre point commun d... | [
"Solution:\n\n\n\nOn note respectivement $A_{1}, B_{1}, C_{1}$ les points de contact de $\\gamma$ avec les côtés $[BC], [CA], [AB]$. Soit $X_{A}$ le point de commun des cercles $\\gamma$ et $\\omega_{A}$.\n\nLe cercle $\\omega_{A}$ est l'image de $\\gamma$ par une homothétie de centre $X_{A... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incente... | null | proof only | null | |
028i | Problem:
Amigos do século $XX$ - Dois amigos nasceram no século XX, com uma semana de intervalo e no mesmo mês e ano. Escrevendo da esquerda para a direita a data na forma o (ou os) algarismo(s) do dia, (ou os) algarismo(s) do mês, e os dois últimos algarismos do ano, obtemos dois números sendo um o sêxtuplo do outro.... | [
"Solution:\n\nOs dois amigos nasceram no mesmo mês e no mesmo ano, com uma diferença de 7 dias, assim um nasceu no dia $d/m/a$ e o outro no dia $(d+7)/m/a$. Com estas datas formamos os números $(d)(m)(a)$ e $(d+7)(m)(a)$. Sabemos que:\n$$\n(d+7)(m)(a) = (d)(m)(a) + 7 \\times 10^{k}\n$$\nAssim,\n$$\n(d+7)(m)(a) = 6 ... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 1 April 1900 | |
0ko5 | Problem:
Let $(a_{1}, a_{2}, \ldots, a_{8})$ be a permutation of $(1,2, \ldots, 8)$. Find, with proof, the maximum possible number of elements of the set
$$
\left\{a_{1}, a_{1}+a_{2}, \ldots, a_{1}+a_{2}+\cdots+a_{8}\right\}
$$
that can be perfect squares. | [
"Solution:\nWe claim the maximum is $5$, achieved by the sequence $(1,3,5,7,2,4,6,8)$. Now we prove that we cannot do better.\n\nSince $a_{1}+a_{2}+\\ldots+a_{8}=1+2+\\ldots+8=36$, then there are at most $6$ squares in\n$$\n\\left\\{a_{1}, a_{1}+a_{2}, \\ldots, a_{1}+a_{2}+\\cdots+a_{8}\\right\\}.\n$$\nNote that if... | United States | HMMT February 2022 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof and answer | 5 | |
08wk | How many quadruples $(a, b, c, d)$ of 1-digit positive integers are there which satisfy $a + bcd = ab + cd$? | [
"$$\n\\boxed{913}\n$$\nThe given equation can be transformed to $(a - cd)(b - 1) = 0$, so it is enough to determine quadruples $(a, b, c, d)$ satisfying either $a - cd = 0$ or $b - 1 = 0$.\n\nFor each $a = 1, 2, \\dots, 9$ the number of pairs $(c, d)$ satisfying $a = cd$ is the same as the number of positive factor... | Japan | Japan Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | English | proof and answer | 913 | |
0iys | Problem:
How many sequences of 5 positive integers $(a, b, c, d, e)$ satisfy $a b c d e \leq a+b+c+d+e \leq 10$? | [
"Solution:\n\nWe count based on how many $1$'s the sequence contains.\n\nIf $a = b = c = d = e = 1$ then this gives us $1$ possibility.\n\nIf $a = b = c = d = 1$ and $e \\neq 1$, $e$ can be $2, 3, 4, 5, 6$. Each such sequence $(1,1,1,1,e)$ can be arranged in $5$ different ways, for a total of $5 \\cdot 5 = 25$ ways... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 116 | |
0g8b | 設 $\omega$ 為三角形 $ABC$ 的外接圓。令 $AB$ 邊的中點為 $M$, $AC$ 邊的中點為 $N$, 並令 $\omega$ 上不含 $A$ 點的 $BC$ 弧的中點為 $T$。設三角形 $AMT$ 的外接圓與 $AC$ 邊的中垂線交於 $X$ 點, 三角形 $ANT$ 的外接圓與 $AB$ 邊的中垂線交於 $Y$ 點; 假設 $X, Y$ 兩點皆位於三角形 $ABC$ 的內部。設直線 $MN$ 與 $XY$ 交於 $K$ 點。證明 $KA = KT$。
Let $\omega$ be the circumcircle of a triangle $ABC$. Denote by $M$ and $N$ the... | [] | Taiwan | 14-2J-M1 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
... | null | proof only | null | |
0kav | Let $\mathbb{Z}/n\mathbb{Z}$ denote the set of integers considered modulo $n$ (hence $\mathbb{Z}/n\mathbb{Z}$ has $n$ elements). Find all positive integers $n$ for which there exists a bijective function $g: \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$, such that the 101 functions
$$
g(x),\ g(x) + x,\ g(x) + 2x,\... | [
"**First solution (original one)** The proof is split into two essentially orthogonal claims, which we state as lemmas.\n\n**Lemma (Lemma I: elimination of $g$)**\nAssume valiant $g: \\mathbb{Z}/n\\mathbb{Z} \\to \\mathbb{Z}/n\\mathbb{Z}$ exists. Then\n$$\nk! \\sum_{x \\in \\mathbb{Z}/n\\mathbb{Z}} x^k \\equiv 0 \\... | United States | USA IMO TST | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Algebra > Algebraic Expressi... | null | proof and answer | All positive integers whose prime divisors are greater than 101 (including 1). | |
03qj | In a plane rectangular coordinate system $xOy$ there are three points $A(0, \frac{4}{3})$, $B(-1, 0)$ and $C(1, 0)$. The distance from point $P$ to line $BC$ is the geometric mean of the distances from this point to lines $AB$ and $AC$.
(1) Find the locus equation of point $P$.
(2) If line $L$ passes through the ince... | [
"(1) The equations of lines $AB$, $AC$ and $BC$ are $y = \\frac{4}{3}(x+1)$, $y = -\\frac{4}{3}(x-1)$ and $y = 0$ respectively. The distances from point $P$ to $AB$, $AC$ and $BC$ are respectively\n$$\nd_1 = \\frac{1}{5} | 4x - 3y + 4 |,\n$$\n$$\nd_2 = \\frac{1}{5} | 4x + 3y - 4 |, \\quad d_3 = | y |.\n$$\nAccordin... | China | China Mathematical Competition (Hainan) | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscella... | English | proof and answer | Locus: 2x^2 + 2y^2 + 3y − 2 = 0 and 8x^2 − 17y^2 + 12y − 8 = 0. Slopes k: {0, ±1/2, ±(2√34)/17, ±√2/2}. | |
04p5 | Find all pairs $(x, y)$ of real numbers such that
$$
xy^3 = -135, \quad (x+y)y = -6. \quad (Matko Ljulj)
$$ | [] | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (-5, 3), (5, -3) | |
0e8m | Problem:
Tabelo velikosti $4 \times 4$ želimo pokriti z dominami oblike
(lahko tudi zrcaljena ali zasukana),
pri čemer se domine lahko prekrivajo ali segajo čez rob tabele. Najmanj koliko domin potrebujemo? | [
"Solution:\n\nPotrebujemo vsaj 5 domin. Če bi tabelo lahko pokrili s štirimi dominami, se te med sabo ne bi smele prekrivati in ne bi smele segati čez rob tabele, saj imajo štiri domine skupaj natanko 16 polj. Če to upoštevamo, potem lahko polje v levem zgornjem kotu pokrijemo le na dva načina, kot prikazuje skica,... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 5 | |
059a | A triangle with perimeter $P$ is divided into $k$ triangular pieces for some $k \ge 2$.
a. Show that there exists a piece with perimeter greater than $\frac{P}{k}$.
b. Show that if the initial triangle is equilateral, then there exists a piece with perimeter at least $\frac{P}{\sqrt{k}}$. | [
"a. The sides of the initial triangle are distributed between the pieces. As $k \\ge 2$, the sides of the pieces must also pass through the interior of the initial triangle. So the sum of the perimeters of the $k$ pieces is greater than $P$. Hence there must exist a piece with perimeter greater than $\\frac{P}{k}$.... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | English | proof only | null | |
04b2 | Determine all pairs $(x, y)$ of integers which satisfy
$$
6x^{2}y^{2} - 4y^{2} = 2012 - 3x^{2}.
$$ | [
"The given equality is equivalent to\n$$\n\\begin{aligned}\n6x^2y^2 + 3x^2 - 4y^2 &= 2012 \\\\\n3x^2(2y^2 + 1) - 2(2y^2 + 1) &= 2012 - 2 \\\\\n(3x^2 - 2)(2y^2 + 1) &= 2010\n\\end{aligned}\n$$\nSince $2010 = 2 \\cdot 3 \\cdot 5 \\cdot 67$ and the number $2y^2 + 1$ is odd, we conclude that\n$$\n\\begin{aligned}\n2y^2... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (-2, -10), (-2, 10), (2, -10), (2, 10) | |
0idt | Problem:
A freight train leaves the town of Jenkinsville at 1:00 PM traveling due east at constant speed. Jim, a hobo, sneaks onto the train and falls asleep. At the same time, Julie leaves Jenkinsville on her bicycle, traveling along a straight road in a northeasterly direction (but not due northeast) at 10 miles per ... | [
"Solution:\nJulie's distance is $(10\\ \\mathrm{mph}) \\cdot (6/5\\ \\mathrm{hrs}) = 12$ miles. Jim's walking distance, after falling off the train, is $(3.5\\ \\mathrm{mph}) \\cdot (1\\ \\mathrm{hr}) = 3.5$ miles at a right angle to the road. Therefore, Jim rode the train for $\\sqrt{12^{2} + 3.5^{2}} = \\frac{1}{... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 62.5 | |
0j2m | Problem:
What is the sum of the positive solutions to $2x^{2} - x\lfloor x\rfloor = 5$, where $\lfloor x\rfloor$ is the largest integer less than or equal to $x$? | [
"Solution:\nWe first note that $\\lfloor x\\rfloor \\leq x$, so $2x^{2} - x\\lfloor x\\rfloor \\geq 2x^{2} - x^{2} = x^{2}$. Since this function is increasing on the positive reals, all solutions must be at most $\\sqrt{5}$. This gives us 3 possible values of $\\lfloor x\\rfloor$: $0$, $1$, and $2$.\n\nIf $\\lfloor... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (3 + sqrt(41) + 2*sqrt(11))/4 | |
0jn7 | Problem:
Let $A$, $B$, $C$ be points in a line on this order, with $AB = BC$. A car travels from $A$ to $B$ at $40$ kilometers per hour and then from $B$ to $C$ at $60$ kilometers per hour, without stopping in between. What is the overall speed of the trip from $A$ to $C$? | [
"Solution:\n\nLet $d = AB = BC$. Then the total time taken is $\\frac{d}{40} + \\frac{d}{60}$, and the total distance is $2d$. Hence the overall speed is\n$$\n\\frac{2d}{\\frac{d}{40} + \\frac{d}{60}} = \\frac{2}{\\frac{1}{40} + \\frac{1}{60}} = 48\n$$"
] | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 48 | |
0dc1 | Find all integers $k > 1$ satisfy condition: there exists some nonnegative integer $n$ such that the number $A = 17^{18n} + 4 \cdot 17^{2n} + 7 \cdot 19^{5n}$ can be written as the product of $k$ consecutive positive integers. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 2 | |
0hu4 | Problem:
Let $a_{1}, a_{2}, a_{3}, \ldots$ be an infinite sequence of positive real numbers such that for all $n \geq 1$,
$$
a_{n} \leq a_{2 n}+a_{2 n+1}
$$
Prove that there exists an $N \geq 1$ such that
$$
\sum_{n=1}^{N} a_{n}>1
$$ | [
"Solution:\nLet us prove by induction that for any positive integer $k$, there is an $N$ for which\n$$\n\\sum_{n=1}^{N} a_{n} \\geq k a_{1}\n$$\nThe base case, $k=1$ and $N=1$, is trivial. Given that (1) is true for a given $k$, we have\n$$\nk a_{1} \\leq \\sum_{n=1}^{N} a_{n} \\leq \\sum_{n=1}^{N}\\left(a_{2 n}+a_... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
03te | Given complex numbers $a$, $b$, $c$, let $|a+b| = m$, $|a-b| = n$, and suppose $mn \neq 0$. Prove that
$$
\max\{|ac + b|, |a + bc|\} \ge \frac{mn}{\sqrt{m^2 + n^2}}
$$ | [
"**Proof I** We have\n$$\n\\begin{aligned}\n\\max\\{|ac + b|, |a + bc|\\} &\\ge \\frac{|b| \\cdot |ac + b| + |a| \\cdot |a + bc|}{|b| + |a|} \\\\\n&\\ge \\frac{|b(ac + b) - a(a + bc)|}{|a| + |b|} \\\\\n&= \\frac{|b^2 - a^2|}{|a| + |b|} \\\\\n&\\ge \\frac{|b + a| \\cdot |b - a|}{\\sqrt{2(|a|^2 + |b|^2)}}.\n\\end{ali... | China | China Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
072f | Problem:
All possible 6-digit numbers, in each of which the digits occur in non-increasing order (from left to right, e.g., $877550$) are written as a sequence in increasing order. Find the $2005$-th number in this sequence. | [
"Solution:\n\nConsider a 6-digit number whose digits from left to right are in non-increasing order. If $1$ is the first digit of such a number, then the subsequent digits cannot exceed $1$. The set of all such numbers with initial digit equal to $1$ is\n$$\n\\{100000, 110000, 111000, 111100, 111110, 111111\\}.\n$$... | India | INMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 864110 | |
0gju | Let $ABCD$ be a convex quadrilateral with $AC \perp BD$. Prove that there exist points $P, Q, R, S$ on the sides $AB, BC, CD, DA$, respectively, such that $PR \perp QS$ and the area of quadrilateral $PQRS$ is exactly half that of $ABCD$. | [
"Solution:\nDenote the area of $\\Gamma$ by $[\\Gamma]$. Let $E$ be the intersection of $AC$ and $BD$. From any point $P'$ on $AB$, we construct points $Q', R', S'$ as follows:\nLet the line $P'E$ intersect $CD$ at $R'$. Let $\\ell$ be the line perpendicular to $P'R'$ at $E$. Let $\\ell$ intersect $BC, DA$ at $Q', ... | Thailand | Selected Problems from Thailand Training Camp | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
044y | Let complex sequence $\{z_n\}$ satisfy
$$
z_1 = \frac{\sqrt{3}}{2}, \quad z_{n+1} = \overline{z_n}(1 + z_n i) \quad (n = 1, 2, \dots),
$$
where $i$ is the imaginary unit. Find the value of $z_{2021}$. | [
"For $n \\in \\mathbb{N}_+$, let $z_n = a_n + b_n i$ ($a_n, b_n \\in \\mathbb{R}$). Then\n$$\n\\begin{aligned}\na_{n+1} + b_{n+1}i &= z_{n+1} = \\overline{z_n}(1 + z_n i) \\\\\n&= \\overline{z_n} + |z_n|^2 \\cdot i \\\\\n&= a_n - b_n i + (a_n^2 + b_n^2)i,\n\\end{aligned}\n$$\nand hence $a_{n+1} = a_n$, $b_{n+1} = a... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | z_{2021} = \frac{\sqrt{3}}{2} + \left(\frac{1}{2} + \frac{1}{2^{2^{2020}}}\right) i | |
04on | Prove that positive real numbers $x$ and $y$ such that
$$(x^3 + y^3)(x^2 + y^2) = 2(x + y) = 2$$
do not exist. (Tamara Srnec) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0dh4 | Prove that not all zeros of a polynomial of the form $x^n + 2n x^{n-1} + 2n^2 x^{n-2} + \dots$ can be real. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof only | null | |
0ass | Problem:
The nonzero numbers $x$, $y$, and $z$ satisfy the equations
$$
x y = 2(x + y), \quad y z = 4(y + z), \quad \text{and} \quad x z = 8(x + z)
$$
Solve for $x$. | [
"Solution:\n\n$\\dfrac{16}{3}$"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 16/3 | |
0fh0 | Problem:
Los puntos $A'$, $B'$, $C'$ de los lados $BC$, $CA$, $AB$ del triángulo $ABC$ cumplen
$$
\frac{AC'}{C'B} = \frac{BA'}{A'C} = \frac{CB'}{B'A} = k
$$
Las rectas $AA'$, $BB'$, $CC'$ forman un triángulo $A_1B_1C_1$. Dados $k$ y el área $S$ del triángulo $ABC$, calcular el área del triángulo $A_1B_1C_1$. | [
"Solution:\n\nPara resolver el problema utilizaremos varias veces que si dos triángulos tienen la misma altura, entonces la relación de áreas es igual a la de las bases; o bien que si tienen la misma base, la relación de áreas es la de las alturas.\n\nPara empezar, llamaremos $s$ al área del triángulo pequeño $AB'C... | Spain | OME 25 | [
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | S * (1 - k)^2 / (1 + k + k^2) | |
0ag6 | Let the function $f: N \rightarrow R$ be such that for every natural number $n > 1$, there is a prime divisor $p$ of $n$ such that:
$$
f(n) = f\left(\frac{n}{p}\right) - f(p)
$$
If $f(2^{2007}) + f(3^{2008}) + f(5^{2009}) = 2006$, compute $f(2007^2) + f(2008^3) + f(2009^5)$
(adapted). | [
"If $n = p$ is a prime number then\n$$\nf(p) = f\\left(\\frac{p}{p}\\right) - f(p) = f(1) - f(p)\n$$\ni.e.\n$$\nf(p) = \\frac{f(1)}{2} \\qquad (1)\n$$\nIf $n = pq$ ($p$ and $q$ are prime numbers) then\n$$\nf(n) = f\\left(\\frac{n}{p}\\right) - f(p) = f(q) - f(p) \\text{ or } f(n) = f\\left(\\frac{n}{q}\\right) - f(... | North Macedonia | null | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 9 | |
0jdc | Problem:
Let $a$ and $b$ be real numbers such that $\frac{a b}{a^{2}+b^{2}}=\frac{1}{4}$. Find all possible values of $\frac{\left|a^{2}-b^{2}\right|}{a^{2}+b^{2}}$. | [
"Solution:\nAnswer: $\\frac{\\sqrt{3}}{2}$\nThe hypothesis statement is equivalent to\n$$\n\\begin{gathered}\na^{2}+b^{2}=4 a b \\\\\n1:\\ (a+b)^{2}=6 a b \\\\\n2:\\ (a-b)^{2}=2 a b\n\\end{gathered}\n$$\nMultiplying equations 1 and 2,\n$$\n\\begin{aligned}\n& \\left(a^{2}-b^{2}\\right)^{2}=12(a b)^{2} \\\\\n& \\lef... | United States | HMMT 2013 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | sqrt(3)/2 | |
0l29 | Problem:
Estimate the number of positive integers $n \leq 10^{6}$ such that $n^{2}+1$ has a prime factor greater than $n$.
Submit a positive integer $E$. If the correct answer is $A$, you will receive max $\left(0,\left\lfloor 20 \cdot \min \left(\frac{E}{A}, \frac{10^{6}-E}{10^{6}-A}\right)^{5}+0.5\right\rfloor\righ... | [
"Solution:\n\nLet $N$ denote $10^{6}$. We count by summing over potential prime factors $p$.\n\nFor any prime $p>2$, we have that $p \\mid n^{2}+1$ for two values of $n$ if $p \\equiv 1(\\bmod 4)$, and zero values otherwise. Pretending these values are equally likely to be any of $1, \\ldots, p$, we expect the numb... | United States | HMMT February 2024 Guts Round | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | null | |
05xu | Problem:
Trouver tous les couples d'entiers naturels non nuls $(k, n)$ pour lesquels
$$
1!+2!+\cdots+k!=1+2+\cdots+n
$$ | [
"Solution:\n\nDans toute la solution, on notera $A_{k}$ l'expression $1!+2!+\\cdots+k$ et $B_{n}$ l'expression $1+2+\\cdots+n$. Comme dans tout exercice de ce genre, on commence par rechercher d'éventuelles solutions pour de petites valeurs de $k$ et de $n$. À cette fin, il est pratique de reconnaître en $B_{n}$ la... | France | Préparation Olympique Française de Mathématiques | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (1,1), (2,2), (5,17) | |
0g99 | 假設你有以下兩種由四個正方形組成的骨牌, 左邊兩個稱為甲型骨牌, 右邊兩個稱為乙型骨牌。
現在有一個無限大的白色棋盤, 棋盤上每個格子的大小和骨牌的正方形一樣大。我們從棋盤上選一些格子塗黑。已知我們可以只用甲型骨牌蓋住所有黑色格子且不蓋到任何白色格子。證明: 如果我們改用甲和(或)乙型骨牌來蓋住所有黑色格子, 且不蓋到任何白色格子, 則我們必然使用了偶數個乙型骨牌。
Suppose you have two types of tetrominoes:
the two on the left are called “S-type”... | [
"令 $B$ 為黑色格子所成集合。將無限大棋盤用以下方式塗成紅綠兩色: 注意到甲型骨牌必然蓋住偶數個紅色格子, 而乙型骨牌必然蓋住奇數個紅色格子。\n\n由於 $B$ 可以由甲型骨牌組成，因此 $B$ 必然包含偶數個紅色格子。從而如果我们改用甲和 (或) 乙型骨牌，乙型骨牌的數量必然為偶數。證畢。"
] | Taiwan | 2015 Math Olympiad Second Stage Training Camp | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
02y8 | Problem:
Na figura abaixo, $D$ é o ponto médio do lado $AB$, $CE:DE=5:3$ e $BF:EF=1:3$. Se a área do triângulo $ABC$ é $192~\mathrm{cm}^2$, determine a área do triângulo $BDF$.
 | [
"Solution:\n\nDenote os comprimentos de $BD$, $BF$ e $DE$ por $x$, $z$ e $3y$, respectivamente. Em virtude das proporções dadas, segue que $EF=3z$, $CE=5y$ e $AD=x$. Portanto,\n$$\n\\begin{aligned}\n& \\frac{A_{BDF}}{A_{BDE}}=\\frac{z}{3z+z} \\\\\n& \\frac{A_{BDE}}{A_{CDB}}=\\frac{3y}{3y+5y} \\\\\n& \\frac{A_{CDB}}... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 9 | |
0fvk | Problem:
Die positiven Teiler der natürlichen Zahl $n$ seien $1=d_{1}<d_{2}<\ldots<d_{k}=n$. Bestimme alle $n$, für die gilt
$$
2 n=d_{5}^{2}+d_{6}^{2}-1
$$ | [
"Solution:\nFür jedes $1 \\leq i \\leq k$ ist $n / d_{i}$ ebenfalls ein positiver Teiler von $n$ und ein einfaches Abzählargument zeigt, dass dieser gleich $d_{k+1-i}$ sein muss. Folglich gilt $n=d_{i} d_{k+1-i}$ für alle $i$. Aus der gegebenen Gleichung folgen die beiden Abschätzungen\n$$\n\\begin{gathered}\n2 n=d... | Switzerland | IMO Selektion | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 272 | |
06m2 | Let $ABC$ be a triangle. Let $M$ be the midpoint of $BC$, and let $G$ be the centroid of $\triangle ABC$. Let $D$ be a point on the segment $GM$. A straight line passing through $D$ meets the sides $AB$ and $AC$ at $E$ and $F$ respectively (with $E, F \ne A$). Show that
$$
[BEGF] + [CFGE] \ge \frac{4}{9}[ABC].
$$
When ... | [
"Let $E'$ and $F'$ be the points on $AB$ and $AC$ such that $E'$, $G$, $F'$ are collinear and $E'F' \\parallel EF$. Since $\\triangle EFG$ and $\\triangle EFE'$ have the same base $EF$ and the same height, they have the same area. This yields\n$$\n[BEGF] = [BEF] + [EFG] = [BEF] + [EFE'] = [BFE'] \\ge [BF'E']\n$$\n\... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Mis... | null | proof and answer | Equality holds precisely when the point is the centroid and the line through it is parallel to the base, equivalently when the intersections with the sides are the points where the line through the centroid parallel to the base meets those sides. | |
050k | In his last research, professor $P$ was concentrating on natural numbers with a certain property. It is known that whenever a natural number $x$ has this property, all multiples of $x$ also have this property. Let $a_1, \dots, a_n$ be positive integers such that all their divisors that are greater than one have the pro... | [
"Let $k > 1$ be any divisor of the product $a_1 \\dots a_n$. Then $k$ has a prime divisor $p$, which is also a divisor of the product $a_1 \\dots a_n$. As $p$ is a prime, there exists $i$, such that $p$ is a divisor of $a_i$. As all the divisors of $a_i$ greater than $1$ have the property, $p$ also has this propert... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Yes | |
0i1w | Problem:
Find $x+y$, given that $x^{2}-y^{2}=10$ and $x-y=2$. | [
"Solution:\n$x^{2}-y^{2}=(x-y)(x+y)=2(x+y)=10$, so $x+y=5$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 5 | |
09p6 | Let $w_1 = a$, $w_2 = b$, and let the sequence of words ($w_n$) be defined recursively as follows:
$$
w_3 = ab, \quad w_4 = bab, \quad w_5 = abbab, \quad w_6 = bababbab, \dots,
$$
where each term is formed by concatenating the previous two terms, that is, $w_n = w_{n-2}w_{n-1}$ for all $n \ge 3$. Prove that for every $... | [] | Mongolia | MMO2025 Round 2 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0eid | Problem:
Poišči vsa realna števila $x$, ki rešijo enačbo
$$
6^{\log_{10} x} + x^{\log_{10} 6} = 72
$$ | [
"Solution:\nLogaritem je definiran le za pozitivna števila, zato mora biti $x > 0$. Število $x = 1$ očitno ni rešitev enačbe, zato lahko privzamemo, da je $x \\neq 1$. Za pozitivno realno število $x \\neq 1$ velja $\\log_{10} x = \\frac{1}{\\log_{x} 10}$, zato je $\\log_{10} 6 = \\frac{\\log_{x} 6}{\\log_{x} 10} = ... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 100 | |
02yb | Problem:
Qual é o maior inteiro positivo $n$ para o qual existe um único inteiro $k$, tal que
$$
\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13} ?
$$ | [
"Solution:\nPodemos escrever a desigualdade como\n$$\n\\frac{13}{7}<\\frac{n+k}{n}<\\frac{15}{8}\n$$\nAssim, multiplicando os membros da desigualdade por $56 n$, obtemos a desigualdade equivalente\n$$\n\\begin{aligned}\n104 n & <56 n+56 k & <105 n \\\\\n48 n< & 56 k & <49 n\n\\end{aligned}\n$$\nPara que exista um ú... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 112 | |
08sw | Does there exist a positive integer $n$ satisfying the following condition?
Condition: For an arbitrary rational number $r$, there exist an integer $b$ and nonzero integers $a_1, a_2, \dots, a_n$ such that
$$
r = b + \frac{1}{a_1} + \dots + \frac{1}{a_n}
$$ | [
"We will show that there exists no positive integer $n$ for which the statement of the problem is satisfied. Let us denote by $A_n$ the set of all rational numbers $r$ for which there exist an integer $b$ and nonzero integers $a_1, \\dots, a_n$ such that $r = b + \\sum_{k=1}^{n} \\frac{1}{a_k}$. In order to establi... | Japan | Japan Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Other"
] | English | proof and answer | No; there does not exist any positive integer satisfying the condition. | |
0jz2 | Problem:
Call a number $T$ persistent if the following holds: Whenever $a, b, c, d$ are real numbers different from $0$ and $1$ such that
$$
a + b + c + d = T
$$
and
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = T
$$
we also have
$$
\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} + \frac{1}{1-d} = T.
$$
a.... | [
"Solution:\n\na.\nSuppose $T$ is persistent. Observe that for any $u$ satisfying $|u| \\geq 2$, the equation $x + 1/x = u$ has real solutions (because it can be written in the form $x^2 - u x + 1 = 0$, which has discriminant $u^2 - 4$). Choose $u$ large enough so that both $u$ and $v = T - u$ have absolute value gr... | United States | 19th Bay Area Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
04da | Trapezium $ABCD$ with a longer base $AB$ is inscribed in the circle $k$. Let $A_0$, $B_0$ be respectively the midpoints of segments $BC$, $CA$. Let $N$ be the foot of the altitude from the point $C$ to $AB$, and $G$ the centroid of the triangle $ABC$. Circle $k_1$ goes through $A_0$ and $B_0$ and touches the circle $k$... | [
"Let $k_2$ be the circumcircle of the triangle $A_0B_0C$. Since $k_2$ is the image of the circle $k$ under homothety with centre $C$ (with the coefficient $\\frac{1}{2}$), circles $k_2$ and $k$ touch at point $C$. Tangents $t_X$ and $t_C$ of circle $k$ through points $X$ and $C$ are respectively radical axes of pai... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid... | English | proof only | null | |
02xt | Problem:
Em um estacionamento existem motos, carros, ônibus e caminhões, em um total de 80 veículos e 540 rodas. Cada moto tem 2 rodas, cada carro tem 4, cada ônibus tem 6 e cada caminhão tem 8. O número de carros é a soma do número de motos com o número de ônibus. Quantos são os caminhões neste estacionamento, se est... | [
"Solution:\n\nSejam as quantidades de motos, carros, ônibus e caminhões iguais a $m, k, o$ e $c$ respectivamente. Organizando as informações, temos:\n$$\n\\left\\{\\begin{array}{cccc}\nm+k+o+c & = & 80 & (I) \\\\\n2 m+4 k+6 o+8 c & = & 540 & (II) \\\\\nk & = & m+o & (III)\n\\end{array}\\right.\n$$\nSubstituindo (II... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 48 | |
0ey3 | Problem:
a. The digits of a natural number are rearranged and the resultant number is added to the original number. Prove that the answer cannot be $99\ldots9$ (1999 nines).
b. The digits of a natural number are rearranged and the resultant number is added to the original number to give $10^{10}$. Prove that the orig... | [
"Solution:\n\na. Let the digits of the original number be $a_1$, $a_2$, ... and the rearranged digits be $b_1$, $b_2$, ... Suppose that in the addition there is a carry, in other words $a_1 + b_1 > 9$ for some $i$. Take the largest such $i$. Then the resulting digit in that position cannot be a $9$. Contradiction. ... | Soviet Union | 1st ASU | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof only | null | |
07k7 | Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that for all positive real numbers $x$ and $y$, we have
$$
f(f(f(xy)) + x^2) = f(y)(f(x) - f(x + y))
$$
*Note. By $\mathbb{R}^+$, we mean the set of positive real numbers.* | [
"First, note that the function is clearly strictly decreasing. Assume the supremum of the function is finite. In this case, for any $\\epsilon > 0$, there exists $\\delta$ such that for any $x < \\delta$, we have $l - f(x) < \\frac{\\epsilon}{2}$. Now, in the given inequality, let $x, y$ tend to zero. We get $f(f(f... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 1/x | |
0f21 | Problem:
There are three straight roads. On each road a man is walking at constant speed. At time $t = 0$, the three men are not collinear. Prove that they will be collinear for $t > 0$ at most twice. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
00es | Let $n$ be a positive integer. Find the largest nonnegative real number $f(n)$ (depending on $n$) with the following property: whenever $a_{1}, a_{2}, \ldots, a_{n}$ are real numbers such that $a_{1}+a_{2}+\cdots+a_{n}$ is an integer, there exists some $i$ such that $\left|a_{i}-\frac{1}{2}\right| \geq f(n)$. | [
"The answer is\n$$\nf(n)= \\begin{cases}0 & \\text{ if } n \\text{ is even } \\\\ \\frac{1}{2 n} & \\text{ if } n \\text{ is odd }\\end{cases}\n$$\nFirst, assume that $n$ is even. If $a_{i}=\\frac{1}{2}$ for all $i$, then the sum $a_{1}+a_{2}+\\cdots+a_{n}$ is an integer. Since $\\left|a_{i}-\\frac{1}{2}\\right|=0$... | Asia Pacific Mathematics Olympiad (APMO) | null | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | f(n) = 0 if n is even; f(n) = 1/(2n) if n is odd | |
0irp | Problem:
Let $a$, $b$, $c$ be nonzero real numbers such that $a+b+c=0$ and $a^{3}+b^{3}+c^{3}=a^{5}+b^{5}+c^{5}$. Find the value of $a^{2}+b^{2}+c^{2}$. | [
"Solution:\nLet $\\sigma_{1}=a+b+c$, $\\sigma_{2}=ab+bc+ca$ and $\\sigma_{3}=abc$ be the three elementary symmetric polynomials. Since $a^{3}+b^{3}+c^{3}$ is a symmetric polynomial, it can be written as a polynomial in $\\sigma_{1}$, $\\sigma_{2}$ and $\\sigma_{3}$. Now, observe that $\\sigma_{1}=0$, and so we only... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 6/5 | |
0dhb | You plan to organize your birthday party, which will be attended either by exactly $m$ persons or by exactly $n$ persons (you are not sure at the moment). You have a big birthday cake and you want to divide it into several parts (not necessarily equal), so that you are able to distribute the whole cake among the people... | [
"We claim that the answer is $m + n - \\gcd(m, n)$. Firstly, note that if we consider the cake as the interval $[0, 1]$ and make cuts at points with coordinates $\\frac{k}{m}$ ($0 < k < m$) and $\\frac{l}{n}$ ($0 < l < n$), then we will be able to satisfy the condition of the problem. Moreover, there will be $m-1$ ... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | English | proof and answer | m + n - gcd(m, n) | |
09lz | Let $S$ be a set containing $11$ distinct positive integers, each not exceeding $60$. Prove that there exist two non-empty subsets $A$ and $B$ of $S$, such that $A \cap B = \emptyset$, $|A| = |B|$, and $\sum_{a \in A} a = \sum_{b \in B} b$.
For example, consider the set $\{1, 2, 3, \ldots, 11\}$. The subsets $\{1, 4\}... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
04tl | There is a figure of prince on a field of a $6 \times 6$ square chessboard. The prince can in one move jump either horizontally or vertically. The lengths of the jumps are alternately either one or two fields, and the jump on the next field is the first one. Decide, whether one can choose the initial field for the prin... | [
"Let us suppose the appropriate sequence exists and let us enumerate the fields of the chessboard as follows:\n\n| 1 | 2 | 3 | 4 | 1 | 2 |\n|---|---|---|---|---|---|\n| 2 | 3 | 4 | 1 | 2 | 3 |\n| 3 | 4 | 1 | 2 | 3 | 4 |\n| 4 | 1 | 2 | 3 | 4 | 1 |\n| 1 | 2 | 3 | 4 | 1 | 2 |\n| 2 | 3 | 4 | 1 | 2 | 3 |\n\nThe length o... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No, it is impossible to choose such a starting field. | |
0iby | Problem:
A regular hexagon has one side along the diameter of a semicircle, and the two opposite vertices on the semicircle. Find the area of the hexagon if the diameter of the semicircle is $1$. | [
"Solution:\nThe midpoint of the side of the hexagon on the diameter is the center of the circle. Draw the segment from this center to a vertex of the hexagon on the circle. This segment, whose length is $1/2$, is the hypotenuse of a right triangle whose legs have lengths $a/2$ and $a\\sqrt{3}$, where $a$ is a side ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3√3/26 | |
0ivx | Problem:
There are many sets of two different positive integers $a$ and $b$, both less than $50$, such that $a^{2}$ and $b^{2}$ end in the same last two digits. For example, $35^{2}=1225$ and $45^{2}=2025$ both end in $25$. What are all possible values for the average of $a$ and $b$?
For the purposes of this problem,... | [
"Solution:\n\nAssume that $b$ is the larger of the two numbers. Then $b^{2}$ and $a^{2}$ end in the same last two digits, so $b^{2}-a^{2}$ is a multiple of $100$. That is, using the difference of squares to factor, $(b+a) \\cdot (b-a)$ is a multiple of $100$, and hence also a multiple of both $4$ and $25$.\n\nFor t... | United States | Bay Area Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 10, 15, 20, 25, 30, 35, 40 | |
084b | Problem:
Siano $a$, $b$ interi positivi primi tra loro. Qual è il massimo valore che può assumere il massimo comun divisore fra $(a+b)^4$ e $a-b$?
(A) 3
(B) 4
(C) 16
(D) 32
(E) può essere grande a piacere. | [
"Solution:\n\nLa risposta è (C). Consideriamo anzitutto $d = MCD(a+b, a-b)$. Il MCD di due numeri divide anche la loro somma e la loro differenza, quindi $d$ divide $2a$ e $2b$. Dato che $a$ e $b$ sono primi fra loro, nessun primo dispari può comparire in $d$, e il $2$ compare al massimo con esponente $1$ (ciò avvi... | Italy | Progetto Olimpiadi di Matematica 2005 GARA di SECONDO LIVELLO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | MCQ | C | |
0dzu | Problem:
Od poljubno izbranega trimestnega števila odštejemo število, ki ima števke v obratnem vrstnem redu. Pokaži, da je razlika vedno deljiva z $11$. | [
"Solution:\n\nObe števili zapišemo z desetiškim mestnim zapisom in ju odštejemo. Izpostavimo skupni faktor in iz razstavljene oblike ugotovimo, da je izraz deljiv z $11$."
] | Slovenia | Državno tekmovanje | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
02zg | Problem:
Considere os números reais $p$ e $q$ e a equação cúbica:
$$
x^{3}+p x+q=0
$$
a) Se $x_{0}$ é uma raiz real da equação, então
$$
x^{3}+p x+q=\left(x-x_{0}\right)\left(x^{2}+a x+b\right)
$$
Verifique que $a=x_{0}$.
b) Verifique que $p^{2} \geq 4 x_{0} q$. | [
"Solution:\n\na) Expandindo a equação dada, temos\n$$\n\\begin{aligned}\nx^{3}+p x+q & =\\left(x-x_{0}\\right)\\left(x^{2}+a x+b\\right) \\\\\n& =x^{3}+\\left(a-x_{0}\\right) x^{2}+\\left(b-a x_{0}\\right) x-b x_{0}\n\\end{aligned}\n$$\nComo o coeficiente de $x^{2}$ no termo da esquerda é nulo, devemos ter $a-x_{0}... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0bj1 | a) Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that $g: \mathbb{R} \to \mathbb{R}$, $g(x) = f(x) + f(2x)$, and $h: \mathbb{R} \to \mathbb{R}$, $h(x) = f(x) + f(4x)$, are continuous functions. Prove that $f$ is also continuous.
b) Give an example of a discontinuous function $f: \mathbb{R} \to \mathbb{R}$, wit... | [
"a.\nSince $g$ and $h$ are continuous, and\n$$\nf(x) = (g(x) - g(2x) + h(x))/2,\n$$\n$x \\in \\mathbb{R}$, it follows that $f$ is continuous, as well.\n\nb.\nThe function $f: \\mathbb{R} \\to \\mathbb{R}$,\n$$\nf(x) = \\begin{cases} -1, & x < 0, \\\\ 0, & x = 0, \\\\ 1, & x > 0, \\end{cases}\n$$\nis discontinuous a... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
0hvt | Problem:
Let $ABCD$ be a convex quadrilateral. Suppose that the area of $ABCD$ is equal to
$$
\frac{AB+CD}{2} \cdot \frac{AD+BC}{2}
$$
Prove that $ABCD$ is a rectangle. | [
"Solution:\nIn the picture shown, we have replicated quadrilateral $ABCD$ four times, rotating by $180^{\\circ}$ successively about the midpoints of $BC$, $CF$, and $CH$. Because the angles of the quadrilateral sum to $360^{\\circ}$, the figure \"closes up\" so that the final quadrilateral has $CD$ as a side and wo... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
0eul | Given an arbitrary triangle $ABC$ with area $T$ and perimeter $L$. Let $P$, $Q$, $R$ be the points of tangency of sides $BC$, $CA$, $AB$ respectively and the inscribed circle. Prove the inequality
$$
\left(\frac{AB}{PQ}\right)^3 + \left(\frac{BC}{QR}\right)^3 + \left(\frac{CA}{RP}\right)^3 \geq \frac{2}{\sqrt{3}} \cdot... | [
"Put $BC = a$, $CA = b$, $AB = c$, $QR = p$, $RP = q$, $PQ = r$ and let $AP = x$, $BQ = y$, $CR = z$. For short,\nSince $x + y = c$, $y + z = a$, $z + x = b$, we have\n$$\nx = s - a, \\quad y = s - b, \\quad z = s - c \\quad \\left( s = \\frac{a+b+c}{2} \\right).\n$$\nThe Cosine Law applied to the triangles $ABC$ a... | South Korea | 23rd Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations and Inequalities >... | English | proof only | null | |
03ga | Problem:
Determine all real numbers $a$ such that the two polynomials $x^{2} + a x + 1$ and $x^{2} + x + a$ have at least one root in common. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a = 1 or a = -2 | |
0jkt | Problem:
Suppose we keep rolling a fair 2014-sided die (whose faces are labelled $1,2, \ldots, 2014$) until we obtain a value less than or equal to the previous roll. Let $E$ be the expected number of times we roll the die. Find the nearest integer to $100 E$. | [
"Solution:\nLet $n=2014$. Let $p_{k}$ denote the probability the sequence has length at least $k$. We observe that\n$$\np_{k} = \\frac{\\binom{n}{k}}{n^{k}}\n$$\nsince every sequence of $k$ rolls can be sorted in exactly one way. Now the answer is\n$$\n\\sum_{k \\geq 0} p_{k} = \\left(1 + \\frac{1}{n}\\right)^{n}\n... | United States | HMMT November 2014 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 272 | |
0gpg | Let $Z^+$ denote the set of all positive integers and $P$ denote the set of all prime numbers. For subsets $A$ and $S$ of $Z^+$, $A$ is called $S$-proper if there exists a positive integer $N$ such that for all $a \in A$ and integer $b$ with $0 \le b < a$ there exist not necessarily distinct elements $s_1, s_2, \dots, ... | [
"We will show that the set $S = \\{k \\cdot 5^{k-1} : k \\in \\mathbb{Z}^+\\}$ is an appropriate example.\n\nWe first prove that $\\mathbb{P}$ is $S$-proper for $N = 4$. Let $p \\neq 5$ be prime and $0 \\le n < p$. Then by Chinese remainder theorem there exists a positive integer $k$ satisfying the conditions $k \\... | Turkey | Team Selection Test | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | S = { k * 5^{k-1} : k ∈ Z^+ } | |
0bmf | Given an integer $k \ge 2$, determine the largest number of divisors the binomial coefficient $\binom{n}{k}$ may have in the range $n-k+1, \dots, n$, as $n$ runs through the integers greater than or equal to $k$. | [
"The required maximum is $k-1$ and is achieved, for instance, at $n = k!$. To complete the proof, we now show that at least one of the $k$ numbers $\\frac{1}{n-j}\\binom{n}{k}$, $j = 0, 1, \\dots, k-1$, is not an integer. To this end, we exhibit a $\\mathbb{Z}$-linear combination of these numbers which is not an in... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | proof and answer | k-1 |
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