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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
066m | Let $ABC$ be an acute angle scalene triangle with $AB < AC$, inscribed in the circle $c(O, R)$. The circle $c_1(B, AB)$ intersects the side $AC$ at point $K$ and the circle $c$ at point $E$. The line $KE$ intersects the circle $c$ also at point $F$. The line $BO$ intersects $KE$ at point $L$ and $AC$ at point $M$. Fina... | [
"In the circle $c$, the chords $AB$ and $BE$ are equal and hence\n$$\n\\hat{A}_1 = \\hat{F}_1 = \\hat{F}_2 = \\hat{E}_1 = \\hat{C},\n$$\n\nFigure 1\nbecause in the circle $c$ the above angles go to the equal arches $AB$ and $BE$. $OB$ is the line of the centers of the circles $c$ and $c_1$,... | Greece | Selection Examination A | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line,... | English | proof only | null | |
07x3 | Let $A$, $B$, $C$ be three points on a circle $\Gamma$ and consider points $M$, $N$ on the side $BC$, point $P$ on the side $AC$ and $Q$ on the side $BC$, such that $MP$ is parallel to $BC$ and $NQ$ is parallel to $AC$. Let $T$ be a point on line $MP$, so that $TC$ is tangent to $\Gamma$. The circumcircle of $\triangle... | [
"The quadrilateral $TCMA$ is cyclic. Indeed, the Alternate Segment Theorem applied to the tangent $TC$, and $MP \\parallel BC$ implies\n$$\n\\angle TCA = \\angle CBA = \\angle TMA.\n$$\nLet $S$ be the intersection point of the lines $TC$ and $NQ$. It then follows in a similar way that $SCNB$ is cyclic.\n\nMoreover,... | Ireland | IRL_ABooklet_2024 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0lf4 | A school has two classes $A$ and $B$ which have $m$ and $n$ students each. The students of the two classes sit in a circle. Each student is then given a number of candies equal to the number of consecutive students sitting to the left of him that are from his same class. After distributing the candies, the teacher deci... | [
"Arrange $m+n$ students on a circle forming the arcs on which any two students are in the same class, and as few arcs as possible. Obviously with this division, two adjacent arcs contain students who are not in the same class. We claim that the number of students have candies is exactly the same as the number of ar... | Vietnam | Team selection tests | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | a) 2·min(m, n). b) min(m, n), except when m = n is odd, in which case it is m − 1. | |
033i | Problem:
Let $a$, $b$, $c$ and $d$ be positive integers such that there are exactly $2004$ ordered pairs $(x, y)$, $x, y \in (0,1)$, for which $a x + b y$ and $c x + d y$ are integers. If $(a, c) = 6$, find $(b, d)$. | [
"Solution:\nSuppose first that $a d \\neq b c$. The set of points $(a x + b y, c x + d y)$, $x, y \\in (0,1)$, coincides with the interior of the parallelogram with vertices $A = (0, 0)$, $B = (a, c)$, $C = (b, d)$ and $D = (a + b, c + d)$. Its area $S$ equals $|a d - b c|$. The Pick formula implies that $S = n + \... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Linear Algebra > Determinants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | {1, 7, 49} | |
0iz2 | Problem:
Travis is hopping around on the vertices of a cube. Each minute he hops from the vertex he's currently on to the other vertex of an edge that he is next to. After four minutes, what is the probability that he is back where he started? | [
"Solution:\n\nAnswer: $\\boxed{\\frac{7}{27}}$\n\nLet the cube have vertices all $0$ or $1$ in the $x, y, z$ coordinate system. Travis starts at $(0,0,0)$. If after $3$ moves he is at $(1,1,1)$ he cannot get back to $(0,0,0)$. From any other vertex he has a $\\frac{1}{3}$ chance of getting back on the final move. T... | United States | Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | 7/27 | |
0dhn | From a point $O$ lying outside the line $d$, draw the projection $A$ of $O$ onto $d$. Take some point $M$ on $d$ different from $A$ and $H$ is the projection of $A$ onto $OM$. Denote $D$ as the midpoint of $HM$ and take $N$ on $OA$ such that $NH \perp AD$. Suppose that two circumcircles of triangles $HMN$ and $OAH$ are... | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 2 | |
07cn | Let $n > 1$ be an integer. Prove that there exists an integer $n-1 \ge m \ge \lfloor \frac{n}{2} \rfloor$ such that the following equation has integer solutions with $a_m > 0$
$$
\frac{a_m}{m+1} + \frac{a_{m+1}}{m+2} + \dots + \frac{a_{n-1}}{n} = \frac{1}{\text{lcm}(1, 2, \dots, n)}.
$$ | [
"Two simple lemmas are needed to prove the problem.\n\n**Lemma.** For all integers $k > 1$, if $x_1, x_2, \\dots, x_k$ are integers with\n$$\n\\gcd(x_1, \\dots, x_k) = 1,\n$$\nthen there are integers $a_1, \\dots, a_k$ such that\n$$\na_1x_1 + \\dots + a_kx_k = 1.\n$$\n\n*Proof.* For $k = 2$ the statement of the lem... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof only | null | |
01ea | An invisible hare occupies one of $N$ vertices of a graph $G$. Several hunters try to kill the hare. Each minute all of them simultaneously shoot: each hunter shoots to a single vertex, they choose the target vertices cooperatively. If the hare was in the target vertex during a shoot, the hunting is finished. Otherwise... | [
"Let hunters apply optimal (fastest) algorithm. Let say that a vertex has a smell of a hare, if there exists an initial vertex and a sequence of moves of the hare for which the hare is still alive and now occupies this vertex. After every shoot mark the set of all the vertices that have a smell of a hare. In the be... | Baltic Way | Baltic Way shortlist | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | English | proof only | null | |
0gu5 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(x + f(x)) = f(-x)
$$
for all real numbers $x$ and $f(x) \le f(y)$ for all real numbers $x \le y$. | [
"All constant functions.\n\nLet $a < b$ be two arbitrary numbers. Consider a sufficiently large $x$ so that $x > -a$ and $x > b - f(-a)$, thus $-x < a$ and $x + f(x) > b - f(-a) + f(-a) = b$ (here we used $f(x) \\ge f(-a)$ since $x > -a$). Now $-x < a < b < x + f(x)$ while $f(-x) = f(x+f(x))$, hence $f(a) = f(b)$, ... | Turkey | Team Selection Test for JBMO 2023 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All constant functions | |
0fbv | Problem:
Sea $O$ el circuncentro del triángulo acutángulo $ABC$ y sea $M$ un punto arbitrario del lado $AB$. La circunferencia circunscrita al triángulo $AMO$ interseca por segunda vez a la recta $AC$ en el punto $K$ y la circunferencia circunscrita al triángulo $BOM$ interseca por segunda vez a la recta $BC$ en el pu... | [
"Solution:\n\nUtilizaremos las notaciones habituales en Geometría del triángulo ($R$, radio de la circunferencia circunscrita al $\\triangle ABC$, $O$ su incentro; $a$ longitud del lado $BC$ del triángulo $ABC$, etc.). Sea $\\frac{MA}{MB}=k>0$, entonces\n$$\nk+1=\\frac{MA+MB}{MB}=\\frac{c}{MB} \\Rightarrow MB=\\fra... | Spain | LIV Olimpiada matemática Española (Concurso Final) | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequal... | null | proof and answer | Area(MNK) ≥ 1/4 Area(ABC), with equality when M is the midpoint of AB. | |
062h | Problem:
Ein Quadrat wird so in $n>1$ Rechtecke zerlegt, dass die Seiten der Rechtecke parallel zu den Seiten des gegebenen Quadrats verlaufen. Jede Gerade, die parallel zu einer der Seiten des Quadrats verläuft und das Innere des Quadrats schneidet, soll dabei auch im Inneren wenigstens eines der Rechtecke verlaufen.... | [
"Solution:\n\nWir beweisen die Kontraposition und nehmen dazu an, dass jedes Rechteck der Zerlegung mindestens einen Punkt mit dem Rand des Quadrates gemeinsam hat. Nach den Voraussetzungen hat es dann wenigstens eine seiner Seiten mit dem Rand des Quadrates gemeinsam. Daher lässt sich der Rand des Quadrates abschn... | Germany | 2. IMO-Auswahlklausur | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0895 | Problem:
Alice, Berto e Carlo stanno cercando un tesoro. Sapendo che i tre amici si trovano sui vertici di un triangolo equilatero e che il tesoro si trova in un punto al di fuori del triangolo, a 1 metro di distanza da Alice e da Berto e 2 metri di distanza da Carlo, quanti metri misura il lato del triangolo?
(A) $\... | [
"Solution:\n\nLa risposta è $(\\mathbf{E})$. Siano $A, B, C$ i tre amici e $T$ il tesoro. Per simmetria $C T$ è perpendicolare ad $A B$.\n\nConsideriamo il triangolo $A C T$. $\\widehat{A C T}=30^\\circ$ e $C T=2 A T$; questo è sufficiente per concludere che $A C T$ è emiequilatero (ovvero è la metà di un triangolo... | Italy | Olimpiadi della Matematica - Gara di Febbraio | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | E | |
0f0b | Problem:
Prove that a collection of squares with total area $1$ can always be arranged inside a square of area $2$ without overlapping. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0enc | Let $ABC$ be a triangle with orthocentre $H$ and let $P$ be a point on its circumcircle. The line through $A$ parallel to $BP$ meets $CH$ at $Q$ and the line through $A$ parallel to $CP$ meets $BH$ at $R$. Prove that $QR$ is parallel to $AP$. | [
"\n\nLet $T$ be the intersection of $BH$ and $PC$ and $F$ be the intersection of $AB$ and $CH$. Then\n$$\n\\begin{aligned}\n\\angle HQA &= 90^\\circ - \\angle BAQ \\\\\n&= 90^\\circ - \\angle ABP \\quad (\\text{since } PB \\parallel AQ) \\\\\n&= 90^\\circ - \\angle ACP \\\\\n&= \\angle RTC ... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gte | Let $c$ be a given real number. Suppose that for all real numbers $a$ and $b$ the function $f(x) = x^2 - 2a x + b$ satisfies the following inequality
$$
f(c) \cdot f(-c) \geq f(a).
$$
Find all possible values of $c$. | [
"Answer: $c = \\pm\\frac{1}{2}$.\n\nNote that\n$$\n\\begin{align*}\nf(c) \\cdot f(-c) - f(a) &= (c^2 - 2a c + b)(c^2 + 2a c + b) + a^2 - b \\\\\n&= (c^2 + b)^2 - 4a^2 c^2 + a^2 - b.\n\\end{align*}\n$$\nPutting $d = c^2$ we get\n$$\n\\begin{align*}\nf(c) \\cdot f(-c) - f(a) &= (d + b)^2 - 4a^2 d + a^2 - b \\\\\n&= a... | Turkey | Team Selection Test | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | c = ±1/2 | |
0er5 | $1 \div 13$ is a recurring decimal that begins $0.076923076923076923\ldots$. The 100th digit after the decimal comma is
(A) $0$
(B) $2$
(C) $6$
(D) $7$
(E) $9$ | [
"There are $6$ digits in the repeating part of the decimal form. $100 = 6 \\times 16 + 4$, so the $100$th digit will be the $4$th in the repeating part $076923$, which is $9$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | MCQ | E | |
04ic | Determine all triples $(a, b, c)$ of real numbers such that
$$
a^2 + b^2 + c^2 = 1 \quad \text{and} \quad (2b - 2a - c)a \ge \frac{1}{2}. \quad \text{(Ivan Kokan)}
$$ | [] | Croatia | Croatia Mathematical Competitions | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | (a, b, c) = (1/√6, 2/√6, −1/√6) or (−1/√6, −2/√6, 1/√6). | |
0126 | Problem:
Let $ABCD$ be a convex quadrilateral, and let $N$ be the midpoint of $BC$. Suppose further that $\angle AND = 135^\circ$. Prove that
$$
|AB| + |CD| + \frac{1}{\sqrt{2}} \cdot |BC| \geqslant |AD|
$$ | [
"Solution:\n\nLet $X$ be the point symmetric to $B$ with respect to $AN$, and let $Y$ be the point symmetric to $C$ with respect to $DN$ (see Figure 3).\nThen\n$$\n\\angle XNY = 180^\\circ - 2 \\cdot (180^\\circ - 135^\\circ) = 90^\\circ\n$$\nand $|NX| = |NY| = \\frac{|BC|}{2}$. Therefore, $|XY| = \\frac{|BC|}{\\sq... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0dqk | Let $n$ be a positive integer. Find the smallest positive integer $k$ with the property that for any colouring of the squares of a $2n \times k$ chessboard with $n$ colours, there are 2 columns and 2 rows such that the 4 squares in their intersections have the same colour. | [
"The answer is $2n^2 - n + 1$.\n\nConsider an $n$-colouring of the $2n \\times k$ chessboard. A vertical-pair is a pair of squares in the same column that are coloured the same. In every column there are at least $n$ vertical-pairs. Let $P$ be the total number of vertical-pairs and $P_i$ be the number of vertical-p... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof and answer | 2n^2 - n + 1 | |
0bgi | Problem:
Un grup $(G, \cdot)$ are proprietatea $(P)$, dacă, pentru orice automorfism $f$ al lui $G$, există două automorfisme $g$ şi $h$ ale lui $G$, astfel încât $f(x)=g(x) \cdot h(x)$, oricare ar fi $x \in G$. Să se arate că:
a) Orice grup care are proprietatea $(P)$ este comutativ.
b) Orice grup comutativ finit d... | [] | Romania | Olimpiada Naţională de Matematică, Etapa judeţeană şi a municipiului Bucureşti | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
0l3k | In $\triangle ABC$, $\angle ABC = 90^\circ$ and $BA = BC = \sqrt{2}$. Points $P_1, P_2, \dots, P_{2024}$ lie on hypotenuse $\overline{AC}$ so that $AP_1 = P_1P_2 = P_2P_3 = \dots = P_{2023}P_{2024} = P_{2024}C$. What is the length of the vector sum
$$
\overrightarrow{BP_1} + \overrightarrow{BP_2} + \overrightarrow{BP_3... | [
"**Answer (D):** For $1 \\le i \\le 2024$, the vector sum $\\overrightarrow{BP_i} + \\overrightarrow{BP_{2025-i}}$ is the vector pointing from the apex of isosceles right triangle $\\triangle ABC$ to the reflection of the apex across the hypotenuse, as seen in the figure below.\n\nIts lengt... | United States | AMC 12 A | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | final answer only | 2024 | |
0btb | Two circles, $\omega_1$ and $\omega_2$, of equal radius intersect at different points $X_1$ and $X_2$. Consider a circle $\omega$ externally tangent to $\omega_1$ at a point $T_1$, and internally tangent to $\omega_2$ at a point $T_2$. Prove that lines $X_1T_1$ and $X_2T_2$ intersect at a point lying on $\omega$.
Luxem... | [
"Let the line $X_kT_k$ and $\\omega$ meet again at $X'_k$, $k = 1,2$, and notice that the tangent $t_k$ to $\\omega_k$ at $X_k$ and the tangent $t'_k$ to $\\omega$ at $X'_k$ are parallel. Since the $\\omega_k$ have equal radii, the $t_k$ are parallel, so the $t'_k$ are parallel, and consequently the points $X'_1$ a... | Romania | 2016 European Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0cqt | Назовём натуральное число хорошим, если среди его делителей есть ровно два простых числа. Могут ли 18 подряд идущих натуральных чисел быть хорошими?
(О. Подлипский) | [
"Ответ. Не могут.\n\nПредположим, что нашлись 18 хороших чисел подряд. Среди них найдутся три числа, делящихся на $6$. Пусть это числа $6n$, $6(n+1)$ и $6(n+2)$. Поскольку эти числа — хорошие, и в разложение каждого из них на простые множители входят двойка и тройка, других простых делителей у них быть не может.\n\... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0hko | Problem:
A triangle, two of whose sides are $3$ and $4$, is inscribed in a circle. Find the minimal possible radius of the circle. | [
"Solution:\nSince the circle has a chord of length $4$, its diameter is at least $4$ and so its radius is at least $2$. To achieve equality, choose a right triangle with hypotenuse $4$ and one leg $3$ (the other leg will, by the Pythagorean theorem, have length $\\sqrt{7}$). Then the midpoint of the hypotenuse is t... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2 | |
04dp | Prove that the equation
$$
x^2 = 2y^2 - 75y + 5
$$
has no integer solutions. | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
04du | Determine all real numbers $a$ such that there exists a complex number $z$ which satisfies
$$
|z| = 1 \quad \text{and} \quad |az - 1| = a|z + 1|.
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a >= 1/3 or a = -1 | |
06sw | Consider $n \geqslant 3$ lines in the plane such that no two lines are parallel and no three have a common point. These lines divide the plane into polygonal regions; let $\mathcal{F}$ be the set of regions having finite area. Prove that it is possible to colour $\lceil\sqrt{n / 2}\rceil$ of the lines blue in such a wa... | [
"Let $L$ be the given set of lines. Choose a maximal (by inclusion) subset $B \\subseteq L$ such that when we colour the lines of $B$ blue, no region in $\\mathcal{F}$ has a completely blue boundary. Let $|B|=k$. We claim that $k \\geqslant\\lceil\\sqrt{n / 2}\\rceil$.\n\nLet us colour all the lines of $L \\backsla... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
08mu | Problem:
Inside of a square whose side length is $1$ there are a few circles such that the sum of their circumferences is equal to $10$. Show that there exists a line that meets at least four of these circles. | [
"Solution:\n\nFind projections of all given circles on one of the sides of the square. The projection of each circle is a segment whose length is equal to the length of a diameter of this circle. Since the sum of the lengths of all circles' diameters is equal to $10 / \\pi$, it follows that the sum of the lengths o... | JBMO | JBMO Shortlist | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Circles"
] | null | proof only | null | |
075s | For a positive integer $n$, a sum-friendly odd partition of $n$ is a sequence $(a_1, a_2, \dots, a_k)$ of odd positive integers with $a_1 \le a_2 \le \dots \le a_k$ and $a_1 + a_2 + \dots + a_k = n$ such that for all positive integers $m \le n$, $m$ can be uniquely written as a subsum $m = a_{i_1} + a_{i_2} + \dots + a... | [
"We consider the sum-friendly odd partitions of a positive integer $n$. Clearly $(1, 1, \\dots, 1)$ is a sum-friendly odd partition. On the other hand, if $a_i > 1$ for some $i$, then let $r$ be the smallest such that $a_r > 1$. It follows that $a_r = r$ and that $r$ divides $a_i$ for all $i \\ge r$. Therefore $r$ ... | India | Indija TS 2013 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 16 | |
055p | A beetle is creeping on the coordinate plane, starting from point $(0; -1)$, along a straight line until reaching the $x$-axis at point $(-x; 0)$ where $x$ is a positive real number. After that it turns $90^\circ$ to the right and creeps again along a straight line until reaching the $y$-axis. Then it again turns right... | [
"Let $O$ be the origin of coordinates. Let $A_0$ be the starting point of the beetle's journey, $A_1$ the first turning point, $A_2$ the second turning point, $A_3$ the third turning point and $A_4$ the endpoint (see figure below).\n\n\n\nThe right triangles $OA_0A_1$ and $OA_1A_2$ are simi... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | a) Yes, for example taking x equal to four thirds. b) No, it is impossible. | |
0bc4 | Find all the perfect squares whose product of their decimal digits is a prime. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Number Theory > Modular Arithmetic",
"Number Theory > Other"
] | null | proof and answer | 121 | |
03ts | In a mathematical competition some competitors are friends. Friendship is always mutual. Call a group of competitors a clique if each two of them are friends. (In particular, any group of fewer than two competitors is a clique.) The number of members of a clique is called its size.
Given that, in this competition, the ... | [
"**Proof** We provide an algorithm to distribute the competitors.\nDenote the rooms $A$ and $B$. At some initial stage, we move one person at a time from one room to the other one. We achieve the target by going through several adjustments. In every step of the algorithm, let $A$ and $B$ be the sets of competitors ... | China | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | English | proof only | null | |
0a2r | The uppercase letters $A$, $E$, $F$, $H$, $I$, $K$, $L$, $M$, $N$, $T$, $V$, $W$, $X$, $Y$, $Z$ can be written using only straight line segments. For the $I$, one line segment is enough; for the $E$, four line segments are needed. We call a sequence of at least two of these letters a word; so it does not have to be an ... | [] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Other"
] | null | proof and answer | 43 | |
0iqm | Problem:
For a positive integer $n$, let $\theta(n)$ denote the number of integers $0 \leq x < 2010$ such that $x^{2} - n$ is divisible by $2010$. Determine the remainder when $\sum_{n=0}^{2009} n \cdot \theta(n)$ is divided by $2010$. | [
"Solution:\n\nAnswer: $335$\n\nLet us consider the sum $\\sum_{n=0}^{2009} n \\cdot \\theta(n) \\pmod{2010}$ in another way. Consider the sum $0^{2} + 1^{2} + 2^{2} + \\cdots + 2009^{2} \\pmod{2010}$. For each $0 \\leq n < 2010$, in the latter sum, the term $n$ appears $\\theta(n)$ times, so the sum is congruent to... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 335 | |
0520 | A positive integer $n$ is written on the board once, then $n-1$ is written on the board twice, etc.; on every step the number smaller by $1$ from the previous number is written twice as many times as the previous number. When reaching zeros this process stops. Prove that in the end the sum of the numbers on the board i... | [
"The sum of the numbers on the board is\n$$\ns_n = 1 \\cdot n + 2 \\cdot (n-1) + 4 \\cdot (n-2) + \\dots + 2^{n-1} \\cdot 1.\n$$\nLet us also define\n$$\nr_n = \\frac{s_n}{2^n} = \\frac{1}{2} \\cdot 1 + \\frac{1}{4} \\cdot 2 + \\frac{1}{8} \\cdot 3 + \\dots + \\frac{1}{2^n} \\cdot n.\n$$\n\n$\\frac{1}{2} + \\frac{1... | Estonia | Final Round of National Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0fm2 | Points $A_1, A_2, \dots, A_{2n}$ are vertices of a regular $2n$-polygon. Find the number of triples $A_i, A_j, A_k$ such that triangle $A_iA_jA_k$ is right-angled and the number of triples such that triangle $A_iA_jA_k$ is acute-angled. | [] | Spain | Spanija 2012 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | Right-angled: 2n(n−1); Acute-angled: n(n−1)(n−2)/3 | |
08c6 | Problem:
Maga Magò ha un mazzo di 52 carte, disposte in pila, con il dorso in alto. Magò separa il mazzetto costituito dalle sette carte in cima alla pila, lo capovolge, e lo mette sotto alla pila. Ora tutte le carte sono nuovamente in pila, ma non tutte hanno ancora il dorso in alto: le sette in fondo sono girate al ... | [
"Solution:\n\nColoriamo le carte del mazzo di blu e di rosso, a gruppetti alternati di tre e quattro rispettivamente. Le tre in cima sono blu, le quattro successive rosse, le tre ancora dopo blu, e via dicendo. Osserviamo che, quando la maga Magò gira un mazzetto, questa colorazione a gruppetti alternati si mantien... | Italy | Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 112 | |
0fd0 | Problem:
Considérese la sucesión definida como $a_{1}=3$, y $a_{n+1}=a_{n}+a_{n}^{2}$.
Determínense las dos últimas cifras de $a_{2000}$. | [
"Solution:\nSe tiene $a_{1}=3$ y $a_{n+1}=a_{n}+a_{n}^{2}=a_{n}(1+a_{n})$.\nEscribimos los primeros términos de la sucesión:\n$$\n3, 12, 156, 156157 = 24492, 24492 \\cdot 24493 = \\ldots 56, \\ldots\n$$\nSupongamos que $a_{n}$ termina en 56. Entonces, $a_{n}=100a+56$, y tenemos\n$$\na_{n+1} = (100a+56)(100a+57) = 1... | Spain | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 92 | |
0f9i | Problem:
Given $1990$ piles of stones, containing $1, 2, 3, \ldots, 1990$ stones. A move is to take an equal number of stones from one or more piles. How many moves are needed to take all the stones? | [] | Soviet Union | 24th ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | 11 | |
0f7i | Problem:
A real number with absolute value at most $1$ is put in each square of a $1987 \times 1987$ board. The sum of the numbers in each $2 \times 2$ square is $0$. Show that the sum of all the numbers does not exceed $1987$. | [] | Soviet Union | 21st ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof only | null | |
0kk1 | Let $m \ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1 \le a_i \le m$ for all $i$ such that $m$ divides $a_1 + a_2 + a_3 + a_4$. There is a polynomial $q(x) = c_3x^3 + c_2x^2 + c_1x + c_0$ such that $D(m) = q(m)$ for all odd integers $m \ge 5$... | [
"Let $s(m,r)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct residue classes modulo $m$ such that $a_1 + a_2 + a_3 + a_4 \\equiv r \\pmod{m}$. Because the total number of quadruples of distinct residue classes is $m(m-1)(m-2)(m-3)$, it follows that\n$$\n\\sum_{r=1}^{m} s(m, r) = m(m-1)(m-2)(m-3)... | United States | AMC 12 A | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | E | |
05rf | Problem:
Soit $x$ et $y$ deux entiers tels que $5x + 6y$ et $6x + 5y$ soient des carrés parfaits. Montrer que $x$ et $y$ sont tous deux divisibles par $11$.
Note : on dit qu'un entier $n$ est un carré parfait si c'est le carré d'un entier. | [
"Solution:\n\nSoit $a$ et $b$ deux entiers tels que $5x + 6y = a^{2}$ et $6x + 5y = b^{2}$. On note que $a^{2} + b^{2} = 11(x + y)$ est divisible par $11$. Or, modulo $11$, les carrés sont $0, 1, 3, 4, 5$ et $9$ : ainsi, la somme de deux carrés est nulle (mod $11$) si et seulement si les deux carrés en question son... | France | Préparation Olympique Française de Mathématiques | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0jae | Problem:
Suppose $ABC$ is a triangle with circumcenter $O$ and orthocenter $H$ such that $A$, $B$, $C$, $O$, and $H$ are all on distinct points with integer coordinates. What is the second smallest possible value of the circumradius of $ABC$? | [
"Solution:\nAnswer: $\\sqrt{10}$\n\nAssume without loss of generality that the circumcenter is at the origin. By well known properties of the Euler line, the centroid $G$ is such that $O$, $G$, and $H$ are collinear, with $G$ in between $O$ and $H$, such that $GH = 2GO$. Thus, since $G = \\frac{1}{3}(A+B+C)$, and w... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Translation"
] | null | proof and answer | sqrt(10) | |
0fzc | Problem:
Bestimme die grösste natürliche Zahl $k$ mit der folgenden Eigenschaft: Die Menge der natürlichen Zahlen kann so in $k$ disjunkte Teilmengen $A_{1}, \ldots, A_{k}$ aufgeteilt werden, dass sich jede natürliche Zahl $n \geq 15$ für jedes $i \in\{1, \ldots, k\}$ als Summe zweier verschiedener Elemente aus $A_{i}... | [
"Solution:\n\nFür $k=3$ kann man die natürlichen Zahlen wie folgt aufteilen:\n$$\n\\begin{aligned}\n& A_{1}=\\{1,2,3\\} \\cup \\{3m \\mid m \\geq 4\\} \\\\\n& A_{2}=\\{4,5,6\\} \\cup \\{3m-1 \\mid m \\geq 4\\} \\\\\n& A_{3}=\\{7,8,9\\} \\cup \\{3m-2 \\mid m \\geq 4\\}\n\\end{aligned}\n$$\nIn $A_{1}$ kann man alle Z... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 3 | |
0bk8 | Let $n$ be an integer with $n \ge 2$, and let $a_0, a_1, a_2, \dots, a_n$ be complex numbers with $a_n \ne 0$. Prove that the following statements are equivalent:
(P) $|a_n z^n + a_{n-1} z^{n-1} + \dots + a_1 z + a_0| \le |a_n + a_0|$, for every complex number $z$ of modulus 1,
(Q) $a_1 = a_2 = \dots = a_{n-1} = 0$ a... | [
"(Q) $\\Rightarrow$ (P). If $a_1 = a_2 = \\dots = a_{n-1} = 0$ and $\\frac{a_0}{a_n} \\in [0, \\infty)$, then\n$$\n\\begin{aligned}\n|a_n z^n + a_{n-1} z^{n-1} + \\dots + a_1 z + a_0| &= |a_n z^n + a_0| \\\\ &\\le |a_n z^n| + |a_0| = |a_n| + |a_0| = |a_n + a_0|,\n\\end{aligned}\n$$\nfor every complex number $z$ of ... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof only | null | |
00fu | Determine all pairs $(a, b)$ of integers with the property that the numbers $a^{2}+4b$ and $b^{2}+4a$ are both perfect squares. | [
"Without loss of generality, assume that $|b| \\leq |a|$. If $b=0$, then $a$ must be a perfect square. So $(a = k^{2}, b = 0)$ for each $k \\in \\mathbb{Z}$ is a solution.\n\nNow we consider the case $b \\neq 0$. Because $a^{2}+4b$ is a perfect square, the quadratic equation\n\n$$\nx^{2} + a x - b = 0 \\tag{*}\n$$\... | Asia Pacific Mathematics Olympiad (APMO) | XI APMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | (k^2, 0), (0, k^2), (k, 1 - k), (-4, -4), (-6, -5), (-5, -6) for any integer k | |
0d4z | Find all strictly increasing functions $f: \mathbb{Z} \rightarrow \mathbb{R}$ such that for any $m, n \in \mathbb{Z}$ there exists a $k \in \mathbb{Z}$ such that $f(k)=f(m)-f(n)$. | [
"Let $f: \\mathbb{Z} \\rightarrow \\mathbb{R}$ be such a function. We prove that $f(n+1)-f(n)$ is a positive constant $a$ independent of the integer $n$.\n\nIndeed, assume that there exist two integers $n_{0}, n_{1}$ such that\n$$\nf\\left(n_{0}+1\\right)-f\\left(n_{0}\\right)<f\\left(n_{1}+1\\right)-f\\left(n_{1}\... | Saudi Arabia | SAMC 2015 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English, Arabic | proof and answer | All functions of the form f(n) = a(n − n0) with a > 0 real and n0 an integer. | |
0ju6 | Problem:
Camille the snail lives on the surface of a regular dodecahedron. Right now he is on vertex $P_{1}$ of the face with vertices $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$. This face has a perimeter of $5$. Camille wants to get to the point on the dodecahedron farthest away from $P_{1}$. To do so, he must travel along ... | [
"Solution:\n\nAnswer: $\\frac{17+7 \\sqrt{5}}{2}$\n\nConsider the net of the dodecahedron. It suffices to look at three pentagons $ABCDE$, $EDF GH$, and $GFIJK$, where $AJ = L$. This can be found by the law of cosines on triangle $AEJ$. We have $AE = 1$, $EJ = \\tan 72^{\\circ}$, and $\\angle AEJ = 162^{\\circ}$. T... | United States | HMMT November | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | (17+7*sqrt(5))/2 | |
04h1 | Prove that for every positive integer $n$
$$
\sqrt{n + \sqrt{(n-1) + \sqrt{(n-2) + \dots + \sqrt{2 + \sqrt{1}}}}} < \sqrt{n} + 1.
$$ | [
"Let $S_n = \\sqrt{n + \\sqrt{(n-1) + \\sqrt{(n-2) + \\dots + \\sqrt{2 + \\sqrt{1}}}}}$.\n\nWe will prove by induction on $n$ that $S_n < \\sqrt{n} + 1$ for all positive integers $n$.\n\nBase case ($n = 1$):\n$S_1 = \\sqrt{1} = 1 < \\sqrt{1} + 1 = 2$.\n\nInductive step:\nAssume $S_{n-1} < \\sqrt{n-1} + 1$.\nConside... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0b26 | Problem:
A $20 \times 19$ rectangle is plotted on the Cartesian plane with one corner at the origin and with sides parallel to the coordinate axes. How many unit squares do the two diagonals of this rectangle pass through? | [
"Solution:\n\nSuppose that one corner of the rectangle is on $(20,19)$. First of all, note that $20$ and $19$ are relatively prime. This means that the line does not intersect any vertex of a unit square in the interior of the grid.\n\nNow, consider the diagonal from $(0,0)$ to $(20,19)$. This diagonal intersects e... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 74 | |
014c | Problem:
The altitudes of a triangle are $12$, $15$ and $20$. What is the area of the triangle? | [
"Solution:\n\nDenote the sides of the triangle by $a$, $b$ and $c$ and its altitudes by $h_{a}$, $h_{b}$ and $h_{c}$. Then we know that $h_{a}=12$, $h_{b}=15$ and $h_{c}=20$. By the well known relation $a : b = h_{b} : h_{a}$ it follows $b = \\frac{h_{a}}{h_{b}} a = \\frac{12}{15} a = \\frac{4}{5} a$. Analogously, ... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Triangles"
] | null | final answer only | 150 | |
0cwh | Find all positive integers $n$ for which there exists an *even* positive integer $a$ such that $(a-1)(a^2-1)\dots(a^n-1)$ is a perfect square. | [
"For $n=1$ and $n=2$.\n\nFor $n=1$, any even number $a$ of the form $m^2 + 1$ works, for example, $a = 2$.\n\nFor $n = 2$, any even number $a$ of the form $m^2 - 1$ works, for example, $a = 8$.\n\nAssume that for $n = 3$, such a number $a$ exists. Then the number $(a-1)(a^2-1)(a^3-1) = (a-1)^3(a+1)(a^2+a+1)$ must b... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Russian | proof and answer | 1 and 2 | |
092c | Problem:
Prove that for all positive real numbers $a, b, c$ such that $a b c=1$ the following inequality holds:
$$
\frac{a}{2 b+c^{2}}+\frac{b}{2 c+a^{2}}+\frac{c}{2 a+b^{2}} \leqslant \frac{a^{2}+b^{2}+c^{2}}{3}
$$ | [
"Solution:\nUsing the given condition $a b c=1$ we get the following:\n$$\n\\begin{aligned}\n\\sum_{\\text {cyc }} \\frac{a}{2 b+c^{2}} & =\\sum_{\\text {cyc }} \\frac{a}{b+b+c^{2}} \\\\\n& \\stackrel{\\text { AM-GM }}{\\leqslant} \\sum_{\\text {cyc }} \\frac{a}{3 \\sqrt[3]{b^{2} c^{2}}}=\\sum_{\\text {cyc }}\\left... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof only | null | |
057f | Find all triples $(x, y, z)$ of real numbers that satisfy the system of equations
$$
\begin{cases} xy + x + y = z, \\ yz + y + z = x, \\ zx + z + x = y. \end{cases}
$$ | [
"Denote $x+1 = a$, $y+1 = b$ and $z+1 = c$. Adding 1 to the sides of all equations and factorizing in the left gives\n$$\n\\begin{cases} ab = c, \\\\ bc = a, \\\\ ca = b. \\end{cases} \\qquad (2)\n$$\nIf $a=0$ then the first and the third equation of (2) imply $b=0$ and $c=0$. Analogously, if $b=0$ or $c=0$ then $a... | Estonia | Open Contests | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | [(-1, -1, -1), (0, 0, 0), (-2, -2, 0), (-2, 0, -2), (0, -2, -2)] | |
0547 | Angles $\alpha$ and $\beta$ are such that $\frac{\tan \alpha}{\tan \beta} = k \neq 1$. Express $\frac{\sin(\alpha+\beta)}{\sin(\alpha-\beta)}$ in terms of $k$. | [
"We have $k = \\frac{\\tan \\alpha}{\\tan \\beta} = \\frac{\\sin \\alpha \\cdot \\cos \\beta}{\\cos \\alpha \\cdot \\sin \\beta}$, or $\\sin \\alpha \\cos \\beta = k \\cdot \\cos \\alpha \\sin \\beta$. Therefore\n$$\n\\frac{\\sin(\\alpha + \\beta)}{\\sin(\\alpha - \\beta)} = \\frac{\\sin \\alpha \\cos \\beta + \\co... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | (k+1)/(k-1) | |
0gmb | Find all pairs $(x, y)$ of integers satisfying
$$
5^x = 1 + 4y + y^4
$$ | [
"Let us consider the equation:\n$$\n5^x = 1 + 4y + y^4\n$$\nWe seek integer solutions $(x, y)$.\n\nFirst, note that $y^4 + 4y + 1$ grows rapidly for large $|y|$, so $5^x$ must also be a perfect power of $5$.\n\nLet us try small integer values for $y$:\n\nFor $y = 0$:\n$$\n1 + 4 \\cdot 0 + 0^4 = 1\n$$\nSo $5^x = 1 \... | Turkey | TEAM SELECTION EXAMINATION FOR THE 42nd INTERNATIONAL MATH- EMATICAL OLYMPIAD. TURKEY. | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (0, 0) and (2, 2) | |
0197 | For any natural number $n$, denote by $N(n)$ the number of digits of $n$ and by $S(n)$ the sum of digits of $n$. (Assume that numbers do not start with zero.)
Which digits can occur in a natural number $n$ if $\frac{n}{S(n)} < \frac{m}{S(m)}$ for all other $m$ such that $N(m) = N(n)$? | [
"Let $n$ be fixed. Consider the number that is obtained by increasing or decreasing one of its digits by $i$, i.e., the number $n \\pm bi$ where $b = 10^k$ for some $k$. Then\n$$\n\\begin{align*} \n\\frac{n \\pm bi}{S(n \\pm bi)} > \\frac{n}{S(n)} &\\iff \\frac{n \\pm bi}{S(n) \\pm i} > \\frac{n}{S(n)} \\\\ \n&\\if... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 0, 1, and 9 | |
08s1 | On a plane, the band with width $d$ is the set of all points whose distance from a line is less than or equal to $\frac{d}{2}$. There are four points $A$, $B$, $C$, $D$ on the plane. If you chose any three points among them, there exists a band with width $1$ containing them. Prove that there exists a band with width $... | [
"When a group $X$ consisting of points in the plane is included in band $B$, we say that band $B$ covers $X$. First, we prove the following lemma.\n\n**Lemma.** For triangle $XYZ$, let $H_X$ be the foot of the perpendicular from $X$ to $YZ$, $H_Y$ be the foot of the perpendicular from $Y$ to $ZX$, and $H_Z$ be the ... | Japan | Japan 2007 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | English | proof only | null | |
0ehq | Problem:
Dan je izraz
$$
X=\left(a+a^{-1}\right)^{-1}\left(a^{2}+3 a+2\right)\left(a^{2}-3 a+2\right)\left(a^{2}-4\right)^{-1}
$$
a) Izraz $X$ poenostavi in zapiši v obliki produkta.
(8 točk)
b) Izračunaj vrednost izraza $X$ za $a=-\frac{1}{3}$. | [
"Solution:\n\nPrvi faktor preoblikujemo v $\\left(a+a^{-1}\\right)^{-1}=\\frac{a}{a^{2}+1}$.\n\nV produktu $\\left(a^{2}+3 a+2\\right)\\left(a^{2}-3 a+2\\right)$ lahko vsak člen iz prvega oklepaja pomnožimo z vsakim členom iz drugega oklepaja. Hitrejša možnost pa je, da to preoblikujemo v produkt vsote in razlike i... | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Odbirno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | Simplified: X = a(a − 1)(a + 1)/(a^2 + 1). For a = −1/3: X = 4/15. | |
0ezi | Problem:
$ABC$ is an acute-angled triangle. The angle bisector $AD$, the median $BM$ and the altitude $CH$ are concurrent. Prove that angle $A$ is more than $45$ degrees. | [
"Solution:\n\nWe use Ceva's theorem. Since $AD$, $BM$, $CH$ are concurrent, we have $(BD/DC) \\cdot (CM/MA) \\cdot (AH/BH) = 1$. But $CM = MA$ and since $AD$ is the angle bisector $BD/DC = AB/AC$, so $(AB/AC) \\cdot (AH/BH) = 1$. Hence $AH/AC = BH/AB < 1$. So angle $HAC >$ angle $HCA$. But angle $AHC = 90^\\circ$, ... | Soviet Union | 4th ASU | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fvi | Problem:
Ein Kreis mit Umfang $6 n$ wird durch $3 n$ Punkte in je $n$ Intervalle der Länge $1$, $2$ und $3$ zerlegt. Zeige, dass es stets zwei dieser Punkte gibt, welche auf dem Kreis diametral gegenüber liegen. | [
"Solution:\n\nWir betrachten neben den $3 n$ Intervallendpunkten auch die Mittelpunkte aller Intervalle der Länge $2$ und die Drittelpunkte aller Intervalle der Länge $3$. Diese $6 n$ Punkte bilden ein reguläres Polygon. Wir färben die Intervallendpunkte schwarz, die anderen Punkte weiss. Nehme an, es gäbe keine zw... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0fcd | Problem:
Sea $ABCD$ un cuadrilátero convexo y $P$ un punto interior. Determina cuáles son las condiciones que deben cumplir el cuadrilátero y el punto $P$ para que los cuatro triángulos $PAB$, $PBC$, $PCD$ y $PDA$ tengan la misma área. | [
"Solution:\n\nConsideremos, primero, los triángulos $PCD$ y $PCB$. Tienen la base común $PC$ y alturas correspondientes $DX$ y $BY$. Si queremos que tengan la misma área, las alturas deben ser iguales. Por lo tanto, el punto $Q$ tiene que ser el punto medio de la diagonal $BD$. La recta $CP$ debe pasar por $Q$.\n\n... | Spain | null | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | The diagonals must intersect at the midpoint of one of them, and the interior point must be the midpoint of the other diagonal. | |
0jbb | Problem:
During the weekends, Eli delivers milk in the complex plane. On Saturday, he begins at $z$ and delivers milk to houses located at $z^{3}, z^{5}, z^{7}, \ldots, z^{2013}$, in that order; on Sunday, he begins at $1$ and delivers milk to houses located at $z^{2}, z^{4}, z^{6}, \ldots, z^{2012}$, in that order. E... | [
"Solution:\n\nAnswer: $\\frac{1005}{1006}$\n\nNote that the distance between two points in the complex plane, $m$ and $n$, is $|m-n|$. We have that\n$$\n\\sum_{k=1}^{1006}\\left|z^{2 k+1}-z^{2 k-1}\\right|=\\sum_{k=1}^{1006}\\left|z^{2 k}-z^{2 k-2}\\right|=\\sqrt{2012}\n$$\nHowever, noting that\n$$\n|z| \\cdot \\su... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 1005/1006 | |
0do6 | Problem:
За низ ненегативних реалних бројева $a_{1}, a_{2}, \ldots, a_{k}$ кажемо да је уложив у интервал $[b, c]$ ако постоје бројеви $x_{0}, x_{1}, \ldots, x_{k}$ из интервала $[b, c]$ такви да важи $\left|x_{i}-x_{i-1}\right|=a_{i}$ за $i=1,2, \ldots, k$. Низ је нормиран ако су сви његови чланови не већи од 1. За з... | [
"Solution:\n\na) Довољно је доказати да је сваки нормиран низ $a_{1}, a_{2}, \\ldots, a_{2 n+1}$ уложив у неки интервал дужине $2-\\frac{1}{2^{n}}$. Тврђење доказујемо индукцијом по $n$. Оно је тачно за $n=0$; нека је $n \\geqslant 1$. По индуктивној претпоставци постоји низ $x_{0}, x_{1}, \\ldots, x_{2 n-1} \\in\\... | Serbia | 13. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
042i | Let $A_1A_2\cdots A_{101}$ be a regular 101-gon, and color every vertex red or blue. Let $N$ be the number of obtuse triangles satisfying the following: The three vertices of the triangle must be vertices of the 101-gon, both the vertices with acute angles have the same color, and the vertex with obtuse angle have diff... | [
"Define $x_i = 0$ or $1$ depending on whether $A_i$ is red or blue. For obtuse triangle $A_{i-a}A_iA_{i+b}$ (vertex $A_i$ is the vertex of the obtuse angle, i.e. $a + b \\le 50$), these three vertices satisfy the conditions of the problem if and only if\n$$\n(x_i - x_{i-a})(x_i - x_{i+b}) = 1, \\qquad \\textcircled... | China | China Team Selection Test | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | Largest N = 32175; number of colorings achieving this maximum = 2 * (C(75, 24) + C(76, 25)) | |
0bw9 | Determine the numbers written with three distinct nonzero even digits which are divisible by the product of their digits. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 624 | |
0gvl | Find all functions $f : (0; +\infty) \to \mathbf{R}$ such that the equality
$$
f(x)f(y) = f(xy) + 2005 \left( \frac{1}{x} + \frac{1}{y} + 2004 \right)
$$
holds for all positive real $x$ and $y$. | [
"При $y=1$ із вихідного співвідношення одержимо, що для всіх $x > 0$\n$$\nf(x) \\cdot f(1) = f(x) + 2005 \\left( \\frac{1}{x} + 2005 \\right).\n$$\nЗвідси дістаємо рівність $f(1)^2 = f(1) + 2005 \\cdot 2006$. Таким чином, $f(1) = -2005$ або $f(1) = 2006$.\nЯкщо $f(1) = -2005$, то при всіх $x > 0$\n$$\nf(x) = -\\fra... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1/x + 2005 or f(x) = -(2005/2006) (1/x + 2005), for all x > 0 | |
0ee8 | Non-zero real numbers $a$ and $b$ satisfy $\frac{a}{b+1} + \frac{b}{a+1} = 1$. Which of the statements about the expression $\frac{a}{b} + \frac{b}{a} - \frac{1}{ab}$ is correct?
(A) The expression can take any value in the interval $(0, 1]$.
(B) The expression can take any value in the interval $[1, 2)$.
(C) The value... | [
"Remove the fractions and simplify the equation to get $a^2 + b^2 = ab + 1$. Change the fractions in the given expression to the common denominator to obtain $\\frac{a^2 + b^2 - 1}{ab}$. The equality now implies that the value of the expression is $\\frac{ab}{ab} = 1$. The correct answer is (C)."
] | Slovenia | Slovenija 2016 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
0eez | Problem:
Naj bo $n \geq 3$ naravno število. Na vsako polje tabele velikosti $n \times n$ želimo zapisati eno izmed števil $1,2$ ali $3$, tako da bodo na poljubnih treh poljih, ki jih lahko prekrijemo z domino oblike $\square \square$, pri čemer lahko domino tudi zavrtimo, zapisana različna števila. Na koliko načinov l... | [
"Solution:\n\nNajprej opazimo, da, če zapišemo števili v dve sosednji polji vrstice oziroma stolpca, potem so števila, ki jih moramo zapisati v preostala polja te vrstice oziroma stolpca, enolično določena.\nČe znamo tabelo izpolniti na pravilen način, potem bo ostala pravilno izpolnjena tudi, če števila $1$, $2$ i... | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 12 | |
097b | Problem:
a)
$$
\int_{1}^{2}\left(\frac{1}{x}-\arcsin \frac{1}{x}\right) d x
$$
b) Comparați numerele: $\frac{\pi}{6}$ și $\ln \frac{2+\sqrt{3}}{2}$. | [
"Solution:\na)\n$$\n\\begin{aligned}\n& I=\\int_{1}^{2}\\left(\\frac{1}{x}-\\arcsin \\frac{1}{x}\\right) d x=\\left.\\ln x\\right|_{1} ^{2}-\\left.x \\arcsin \\frac{1}{x}\\right|_{1} ^{2}+\\int_{1}^{2} x \\cdot \\frac{1}{\\sqrt{1-\\frac{1}{x^{2}}}} \\cdot\\left(-\\frac{1}{x^{2}}\\right) d x= \\\\\n& \\ln 2+\\frac{\... | Moldova | Olimpiada Republicană la Matematică | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Integral Calculus > Applications"
] | null | proof and answer | I = π/6 − ln((2+√3)/2); and π/6 < ln((2+√3)/2) | |
06by | Let $x_1$, $x_2$, $x_3$, $\dots$ be a sequence of nonzero real numbers satisfying
$$
x_n = \frac{x_{n-2} x_{n-1}}{2x_{n-2} - x_{n-1}} \quad \text{for } n = 3, 4, 5, \dots
$$
Find all pairs $(x_1, x_2)$ such that $x_n$ is an integer for infinitely many $n$. | [
"The only possibilities are $(x_1, x_2) = (c, c)$ for some nonzero integer $c$.\n\nRewrite the recurrence relation as\n$$\n\\frac{2}{x_{n-1}} = \\frac{1}{x_n} + \\frac{1}{x_{n-2}}.\n$$\nLet $y_n = \\frac{1}{x_n}$ for all $n$. This implies\n$$\ny_{n-1} - y_{n-2} = y_n - y_{n-1}.\n$$\nTherefore, $\\{y_n\\}$ is an ari... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | All pairs where both entries are equal to the same nonzero integer, i.e., (c, c) with c an integer and c not zero. | |
07de | Consider quadrilateral $ABCD$ inscribed in circle $\omega$. $P$ is the intersection point of $AC$, $BD$. Points $E$, $F$ lie on sides $AB$, $CD$ respectively such that $\widehat{APE} = \widehat{DPF}$. Circles $\omega_1$, $\omega_2$ are tangent to $\omega$ at $X$, $Y$ respectively and also both tangent to the circumcirc... | [
"Consider an inversion with center $P$ and radius $-PA \\cdot PC$. Let $Z'$ be the inversion of $Z$. We know that $A$, $C$ and $B$, $D$ are inverse points.\n\nAlso we have\n$$\n\\frac{EY}{E'Y'} = \\frac{PE}{PY'}, \\quad \\frac{EX}{E'X'} = \\frac{PE}{PX'}\n$$\nWhich implies\n$$\n\\frac{EX}{E... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cl6 | Let $p$ be a prime, $p \geq 3$, and $k$ an odd number, not a multiple of $p$. Let $K$ be a finite field with $kp+1$ elements and $A = \{x_1, x_2, \dots, x_t\}$, the set of elements $K^* = K \setminus \{0\}$ of order different of $k$ in the multiplicative group $(K^*, \cdot)$. Prove that the polynomial $P(X) = (X + x_1)... | [
"As $k$ and $p$ are odd, $|K|$ is even, in fact a power of $2$, the characteristic of $K$ being $2$. It follows that $x_1, x_2, \\dots, x_t$ are the roots of $P$. Consider $P = \\sum_{i=0}^{t} a_j X^j$.\n\nThe multiplicative group $(K^*, \\cdot)$ is cyclic of order $kp$. Consider $a \\in K^*$ one of its generators.... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Field Theory",
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | English | proof only | null | |
0f77 | Problem:
$ABC$ is acute-angled. What point $P$ on the segment $BC$ gives the minimal area for the intersection of the circumcircles of $ABP$ and $ACP$? | [] | Soviet Union | 20th ASU | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | The foot of the perpendicular from A to BC. | |
03pb | Let $a$, $b$, $c$, $d$ be positive integers and $\log_a b = \frac{3}{2}$, $\log_c d = \frac{5}{4}$. If $a-c=9$, then $b-d=$ ________. | [
"We have $b = a^{\\frac{3}{2}}$, $d = c^{\\frac{5}{4}}$ from the assumption. We may assume that $a = x^2$, $c = y^4$ with $x$ and $y$ being positive integers, since $a$, $b$, $c$, $d$ are all positive integers. Then $a-c = x^2 - y^4 = (x-y^2)(x+y^2) = 9$. It follows that $(x-y^2, x+y^2) = (1, 9)$. So we obtain the ... | China | China Mathematical Competition (Shaanxi) | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | final answer only | 93 | |
0lf9 | Given are two coprime positive integers $a, b$ with $b$ odd and $a > 2$. The sequence $(x_n)$ is defined by $x_0 = 2$, $x_1 = a$ and $x_{n+2} = a x_{n+1} + b x_n$ for $n \ge 1$. Prove that
a) If $a$ is even then there do not exist positive integers $m, n, p$ such that $\frac{x_m}{x_n x_p}$ is a positive integer.
b) I... | [
"The general formula for the sequence is $x_n = \\alpha^n + \\beta^n$ with $\\alpha$ and $\\beta$ such that\n$$\n\\alpha + \\beta = a, \\quad \\alpha\\beta = -b.\n$$\n\nWe will now prove $\\gcd(b, x_m) = 1$ for all $m$. Indeed, suppose there are $m$ and a prime $p$ such that $x_m$ and $b$ are divisible by $p$. Sinc... | Vietnam | Team selection tests | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof only | null | |
0ebn | Problem:
Poišči vse pare naravnih števil $a$ in $b$, za katere je $a-b=101$ in je $a b$ popoln kvadrat. | [
"Solution:\n\n1. način. Naj bo $d$ največji skupni delitelj števil $a$ in $b$. Torej je $a=d m$ in $b=d n$, kjer sta $m$ in $n$ tuji naravni števili. Ker je $a b=d^{2} m n$ popoln kvadrat in sta $m$ in $n$ tuji števili, sta tudi $m$ in $n$ popolna kvadrata. Pišimo $m=x^{2}$ in $n=y^{2}$, kjer sta $x$ in $y$ naravni... | Slovenia | 59. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis... | null | proof and answer | a = 2601, b = 2500 | |
08ii | Problem:
Find all real solutions of the equation $x^{4} + 7x^{3} + 6x^{2} + 5\sqrt{2003}x - 2003 = 0$. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | x = -3 + sqrt(9 + sqrt(2003)) and x = -3 - sqrt(9 + sqrt(2003)) | |
0a0a | Problem:
Gegeven zijn positieve, reële getallen $a_{1}, a_{2}, \ldots, a_{n}$ met $n \geq 2$ waarvoor geldt dat $a_{1} a_{2} \cdots a_{n}=1$. Bewijs dat
$$
\left(\frac{a_{1}}{a_{2}}\right)^{n-1}+\left(\frac{a_{2}}{a_{3}}\right)^{n-1}+\ldots+\left(\frac{a_{n-1}}{a_{n}}\right)^{n-1}+\left(\frac{a_{n}}{a_{1}}\right)^{n-1... | [
"Solution:\n\nWe nemen het rekenkundig-meetkundig gemiddelde op $\\frac{1}{2} n(n-1)$ termen:\n$$\n\\begin{aligned}\n\\frac{(n-1)\\left(\\frac{a_{1}}{a_{2}}\\right)^{n-1}+(n-2)\\left(\\frac{a_{2}}{a_{3}}\\right)^{n-1}+\\ldots+\\left(\\frac{a_{n-1}}{a_{n}}\\right)^{n-1}}{\\frac{1}{2} n(n-1)} & \\geq\\left(\\left(\\f... | Netherlands | Selectietoets | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds for all positive pairs with product one when the number of variables is two; for more than two variables, equality holds only when all variables are equal to one. | |
04hw | Positive integers $a$, $b$ and prime number $p$ satisfy the equation $a^2 + p^2 = b^2$.
Prove that $2(b + p)$ is a perfect square. | [
"We are given $a^2 + p^2 = b^2$, where $a$, $b$ are positive integers and $p$ is a prime number.\n\nRewrite the equation:\n$$\n b^2 - a^2 = p^2 \n$$\n$$\n (b - a)(b + a) = p^2 \n$$\n\nSince $p$ is prime, $p^2$ has only three positive divisors: $1$, $p$, $p^2$.\nSo, the pairs $(b - a, b + a)$ must be $(1, p^2)$, $(p... | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0cz0 | For each positive integer $n$ let the set $A_{n}$ consist of all numbers $\pm 1 \pm 2 \pm \ldots \pm n$. For example,
$$
\begin{gathered}
A_{1}=\{-1,1\}, \quad A_{2}=\{-3,-1,1,3\} \\
A_{3}=\{-6,-4,-2,0,2,4,6\}
\end{gathered}
$$
Find the number of elements in $A_{n}$. | [
"The greatest element of set $A_{n}$ is\n$$\n1+2+\\ldots+n=\\frac{n(n+1)}{2}\n$$\nand the smallest element of $A_{n}$ is\n$$\n-1-2-\\ldots-n=-\\frac{n(n+1)}{2} .\n$$\nAlso, the difference of any two elements of $A_{n}$ is even, hence all elements of $A_{n}$ are of the same parity.\nLet us prove that all integers be... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | n(n+1)/2 + 1 | |
084x | Problem:
Sia $ABCD$ un parallelogramma. Si sa che il lato $AB$ misura $6$, l'angolo $\angle BAD$ misura $60^\circ$ e l'angolo $\angle ADB$ è retto. Sia $P$ il baricentro del triangolo $ACD$. Calcolare il valore del prodotto delle aree del triangolo $ABP$ e del quadrilatero $ACPD$. | [
"Solution:\n\nLa risposta è $27$. Sia $H$ la proiezione ortogonale di $D$ su $AB$. Dai dati del problema segue subito che $AD = 3$ e $DH = \\frac{3 \\sqrt{3}}{2}$. Siano $Q, R$ le proiezioni ortogonali di $P$ su $CD$ e $AB$, rispettivamente.\n\n\n\nPoiché $P$ è il baricentro di $ACD$, e poi... | Italy | Progetto Olimpiadi di Matematica 2006 GARA di SECONDO LIVELLO | [
"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 > Distance chasing"
] | null | proof and answer | 27 | |
0fqz | Problem:
Demostrar que la suma de los divisores positivos de un número de la forma $3k+2$ siempre es un múltiplo de $3$. | [
"Solution:\nSea $n$ el número en cuestión y consideremos un divisor suyo cualquiera, $d$. Como $n=3k+2$, $d$ no puede ser múltiplo de $3$. Si $d=3\\ell+1$, entonces $n/d$, que es otro divisor distinto de $n$ (dado que $n$ no es cuadrado perfecto), será de la forma $3\\ell'+2$, y viceversa. Por tanto, podemos agrupa... | Spain | FASE LOCAL DE LA OLIMPIADA MATEMÁTICA ESPAÑOLA. | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
04qn | We say that a positive integer $n$ is fantastic, if there exist positive rational numbers $a$ and $b$ such that
$$
n = a + \frac{1}{a} + b + \frac{1}{b}.
$$
a. Prove that there exist infinitely many prime numbers $p$ such that no multiple of $p$ is fantastic.
b. Prove that there exist infinitely many prime numbers $p... | [
"Note that\n$$\nr(a,b) := a + \\frac{1}{a} + b + \\frac{1}{b} = \\frac{(a+b)(ab+1)}{ab}.\n$$\nWe put $a = \\frac{t}{u}$ and $b = \\frac{v}{w}$, where $t, u, v$ and $w$ are positive integers such that both $t$ and $u$ and also $v$ and $w$ are coprime. Then we get $r(a,b) = \\frac{(tv+uw)(tw+uv)}{tuvw}$, whence the D... | Czech Republic | null | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Seri... | English | proof only | null | |
0cdh | Let $a$ be a positive real number. Show that there are no real numbers $b$ and $c$, with $b < c$, such that $\left|\frac{x+y}{x-y}\right| \le a$, for every $x, y \in (b, c)$, $x \ne y$. | [
"Assume, by the sake of contradiction, that there exist $b, c \\in \\mathbb{R}$, $b < c$, such that $\\left|\\frac{x+y}{x-y}\\right| \\le a$, for every $x, y \\in (b, c)$, $x \\ne y$, and consider $x \\in (b, c)$, $x \\ne 0$.\n\nIf $x > 0$, there exist infinitely many positive integers $n$ such that $n > \\frac{1}{... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - THIRD SELECTION TEST | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0kul | Problem:
The graph of the equation $x+y=\left\lfloor x^{2}+y^{2}\right\rfloor$ consists of several line segments. Compute the sum of their lengths. | [
"Solution:\nWe split into cases on the integer $k=\\left\\lfloor x^{2}+y^{2}\\right\\rfloor$. Note that $x+y=k$ but $x^{2}+y^{2} \\geq \\frac{1}{2}(x+y)^{2}=\\frac{1}{2} k^{2}$ and $x^{2}+y^{2}<k+1$, which forces $k \\leq 2$.\n\nIf $k=0$, the region defined by $0 \\leq x^{2}+y^{2}<1$ and $x+y=0$ is the diameter fro... | United States | HMMT February 2023 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | 4 + sqrt(6) - sqrt(2) | |
05qz | Problem:
Soit $q$ un nombre réel. Margaret a écrit 10 nombres réels, deux à deux distincts, sur une ligne. Puis elle ajoute trois lignes comme suit :
$\triangleright$ sur la $2^{\text{nde}}$ ligne, elle écrit tous les nombres de la forme $a-b$, où $a$ et $b$ sont deux réels (non nécessairement distincts) de la $1^{\t... | [
"Solution:\n\nOn va dire qu'un réel $q$ est bon s'il a la propriété demandée dans l'énoncé. Tout d'abord, il est clair que, si $q$ est bon, alors $-q$ l'est aussi, et réciproquement. D'autre part, $q=0$ est manifestement bon. On cherche donc les bons réels $q>0$, s'il y en a.\n\nSoit $\\lambda$ un grand nombre réel... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Intermediate Algebra > Other",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | -2, 0, 2 | |
0jy9 | Problem:
Victoria paints every positive integer either pink or blue. Is it possible that both conditions below are satisfied?
- For every positive integer $n$, the numbers $n$ and $n+5$ are different colors.
- For every positive integer $n$, the numbers $n$ and $2 n$ are different colors. | [
"Solution:\nThe answer is no.\nAssume for contradiction that such a coloring exists. Let's say $10$ was colored pink. Then $10+5=15$ must be blue, and $15+5=20$ must be pink. But now $20=10 \\cdot 2$, violating the second condition.\n\nNow if $10$ was colored blue, the same argument works with \"pink\" and \"blue\"... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No | |
08qi | Problem:
Viktor and Natalia bought $2020$ buckets of ice-cream and want to organize a degustation schedule with $2020$ rounds such that:
- In every round, each one of them tries $1$ ice-cream, and those $2$ ice-creams tried in a single round are different from each other.
- At the end of the $2020$ rounds, each one of... | [
"Solution:\n\nIf we fix the order in which Natalia tries the ice-creams, we may consider 2 types of fair schedules:\n\n1) Her last $1010$ ice-creams get assigned as Viktor's first $1010$ ice-creams, and vice versa: Viktor's first $1010$ ice-creams are assigned as Natalia's last $1010$ ice-creams. This generates $(1... | JBMO | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof only | null | |
0b05 | Problem:
In square $ABCD$ with side length $1$, $E$ is the midpoint of $AB$ and $F$ is the midpoint of $BC$. The line segment $EC$ intersects $AF$ and $DF$ at $G$ and $H$, respectively. Find the area of quadrilateral $AGHD$. | [] | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 7/15 | |
0jm5 | Problem:
The Evil League of Evil is plotting to poison the city's water supply. They plan to set out from their headquarters at $(5,1)$ and put poison in two pipes, one along the line $y=x$ and one along the line $x=7$. However, they need to get the job done quickly before Captain Hammer catches them. What's the short... | [
"Solution:\n\n$4 \\sqrt{5}$\n\nAfter they go to $y=x$, we reflect the remainder of their path in $y=x$, along with the second pipe and their headquarters. Now, they must go from $(5,1)$ to $x=7$ crossing $y=x$, and then go to $(1,5)$. When they reach $x=7$, we reflect the remainder of their path again, so now their... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 4√5 | |
0412 | Suppose $f(x) = a \sin x - \frac{1}{2} \cos 2x + a - \frac{3}{a} + \frac{1}{2}$, $a \in \mathbf{R}$, $a \neq 0$.
(1)
If $f(x) \le 0$ for any $x \in \mathbf{R}$, find the range of $a$.
(2)
If $a \ge 2$ and there exists $x \in \mathbf{R}$ such that $f(x) \le 0$, find the range of $a$. | [
"(1) We have $f(x) = \\sin^2 x + a \\sin x + a - \\frac{3}{a}$. Let $t = \\sin x$ ($-1 \\le t \\le 1$). Then\n$$\ng(t) = t^2 + a t + a - \\frac{3}{a}.\n$$\nThe sufficient and necessary condition for $f(x) \\le 0, \\forall x \\in \\mathbf{R}$ is\n$$\n\\begin{cases} g(-1) = 1 - \\frac{3}{a} \\le 0, \\\\ g(1) = 1 + 2a... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (1) a in (0, 1]. (2) a in [2, 3]. | |
03el | $$
A_n = 1 \cdot 2 + 3 \cdot 4 + 5 \cdot 8 + \dots + (2n - 1) \cdot 2^n.
$$
(Nedyalka Dimitrova) | [
"Since\n$$\n2A_n = 1 \\cdot 4 + 3 \\cdot 8 + \\dots + (2n - 3) \\cdot 2^n + (2n - 1) \\cdot 2^{n+1},\n$$\nit follows\n$$\n\\begin{align*}\nA_n = 2A_n - A_n &= (2n - 1) \\cdot 2^{n+1} - (1 \\cdot 2 + 2 \\cdot 4 + 2 \\cdot 8 + \\dots + 2 \\cdot 2^n) \\\\\n&= (2n - 1) \\cdot 2^{n+1} - 2 \\cdot (2 + 4 + 8 + \\dots + 2^... | Bulgaria | 2 Bulgarian Winter Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | 4045 * 2^{2025} + 6 | |
0ceq | Let $a$ and $b$ be positive integers such that $b-a$ is a prime. Prove that
$$(a^n + a + 1)(b^n + b + 1)$$
is not the square of an integer for infinitely many positive integers $n$. | [
"Suppose, if possible, that $(a^n + a + 1)(b^n + b + 1)$ is a square for all but finitely many positive integers $n$. Then\n$$\nab \\equiv \\frac{(a^{p+1} + a + 1)(b^{p+1} + b + 1)}{(a^{p-2} + a + 1)(b^{p-2} + b + 1)}\n$$\nand\n$$\n(a + 2)(b + 2) \\equiv (a^{p-1} + a + 1)(b^{p-1} + b + 1) \\pmod{p}\n$$\nare both qu... | Romania | Seventeenth Stars of Mathematics Competition | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
01cx | Find all real solutions of the equation
$$
\frac{(x + y)(2 - \sin(x + y))}{4 \sin^2(x + y)} = \frac{xy}{x + y}.
$$ | [
"Under the condition $\\sin(x + y) \\neq 0$, the equation will be equivalent to\n$$\n(x+y)^2(2 - \\sin(x+y)) = 4xy \\sin^2(x+y).\n$$\nThe left-hand side is non-negative; hence $xy \\ge 0$. We then have\n$$\n(x+y)^2(2 - \\sin(x+y)) \\ge (x+y)^2 \\ge 4xy \\ge 4xy \\sin^2(x+y).\n$$\nEquality holds in the second inequa... | Baltic Way | Baltic Way 2016 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x = y = π/4 + nπ for any integer n | |
0c00 | The orthocenter $H$, the centroid $G$ and the incenter $I$ of a triangle are collinear points. Prove that the triangle is isosceles. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"... | null | proof only | null | |
00h3 | Let $ABC$ be an acute triangle with altitudes $AD$, $BE$ and $CF$, and let $O$ be the center of its circumcircle. Show that the segments $OA$, $OF$, $OB$, $OD$, $OC$, $OE$ dissect the triangle $ABC$ into three pairs of triangles that have equal areas. | [
"Let $M$ and $N$ be midpoints of sides $BC$ and $AC$, respectively. Notice that $\\angle MOC = \\frac{1}{2} \\angle BOC = \\angle EAB$, $\\angle OMC = 90^{\\circ} = \\angle AEB$, so triangles $OMC$ and $AEB$ are similar and we get $\\frac{OM}{AE} = \\frac{OC}{AB}$. For triangles $ONA$ and $BDA$ we also have $\\frac... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01cy | Find all quadruples $(a, b, c, d)$ of real numbers that simultaneously satisfy the following equations:
$$
\begin{cases} a^3 + c^3 = 2 \\ a^2b + c^2d = 0 \\ b^3 + d^3 = 1 \\ ab^2 + cd^2 = -6. \end{cases}
$$ | [
"Consider the polynomial $P(x) = (a x + b)^3 + (c x + d)^3 = 2 x^3 - 18 x + 1$. By $P(0) > 0$, $P(1) < 0$, $P(3) > 0$, it has two distinct real zeros $x_1$ and $x_2$. Since $P(x) = 0$ implies that $(a + c)x + (b + d) = 0$, it follows that $a + c = b + d = 0$. This contradicts the first equation $a^3 + c^3 = 2$. Hen... | Baltic Way | Baltic Way 2016 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | no real solution | |
0km1 | Problem:
A light pulse starts at a corner of a reflective square. It bounces around inside the square, reflecting off of the square's perimeter $n$ times before ending in a different corner. The path of the light pulse, when traced, divides the square into exactly 2021 regions. Compute the smallest possible value of $n... | [
"Solution:\nThe main claim is that if the light pulse reflects vertically (on the left/right edges) $a$ times and horizontally $b$ times, then $\\gcd(a+1, b+1)=1$, and the number of regions is $\\frac{(a+2)(b+2)}{2}$. This claim can be conjectured by looking at small values of $a$ and $b$; we give a full proof at t... | United States | HMMT Spring 2021 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 129 | |
0etr | Let $ABIH$, $BDEC$, and $ACFG$ be arbitrary rectangles constructed (externally) on the sides of triangle $ABC$. Choose point $S$ outside rectangle $ABIH$ (on the opposite side as triangle $ABC$) such that $\angle SHI = \angle FAC$ and $\angle HIS = \angle EBC$. Prove that the lines $FI$, $EH$, and $CS$ are concurrent. | [
"Let $T = EH \\cap FI$. Let $K$ and $L$ be the feet of the perpendiculars from $A$ to $FI$ and from $B$ to $EH$, respectively. Then $K$ and $L$ are on the circumcircle $\\Gamma$ of $ABIH$. Note that $K$ is also on the circumcircle of $ACFG$ and $L$ is also on the circumcircle of $BDEC$. Let $CK$ extended intersect ... | South Africa | The South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null |
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