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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0hia | A country held a one-round tennis tournament. Participants received $1$ point for winning the match, and $0$ points for losing. There are no draws in tennis. At the end of the tournament, Oleksii saw the number of points scored by each participant, as well as the schedule of all the matches in the tournament, which sho... | [
"Let's choose player $A$, who has the lowest number of points. He has at least as many defeats as wins, so by indicating a defeat in each his game, Oleksii will guess at least half of the results of player $A$. Then Oleksii subtracts the results of player $A$ from the results of each participant and again selects p... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
08dz | Problem:
Sia $ABC$ un triangolo scaleno con $BC > CA > AB$. Siano $\omega$ e $\gamma$ le circonferenze passanti per $A$ di centro, rispettivamente, $B$ e $C$. Esse intersecano il segmento $BC$ in $M$ e $N$, rispettivamente. Costruiamo $Z$ come il simmetrico di $A$ rispetto al punto medio di $MN$.
a. Chiamata $P$ l'in... | [
"Solution:\n\na. Il quadrilatero $ANZM$ è un parallelogramma perché le diagonali si intersecano nel loro punto medio per costruzione. Le rette $AN$ e $MZ$ sono dunque parallele. Su questa coppia di rette parallele, le trasversali $CN$ e $CA$ staccano angoli corrispondenti uguali ($\\angle CMP = \\angle CNA$ e $\\an... | Italy | Olimpiadi della Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0dls | Find $n \in \mathbb{N}$ for which $5^n + n^4 + 20$ is a perfect square. | [] | Saudi Arabia | Saudi Booklet | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | no such n | |
04c2 | Let $ABC$ be a right triangle with the right angle at vertex $C$. Let $D$ be a point on the segment $AC$ and let $E$ be a point on the segment $BD$ such that $\angle ABC = \angle DAE = \angle AED$. Prove that $|BE| = 2|CD|$. (Lithuania 2010) | [
"Let $\\triangle ABC = \\triangle DAE = \\triangle AED = \\beta$.\n\nThen\n$$\n\\begin{aligned}\n\\angle BAC &= 90^\\circ - \\beta, \\\\\n\\angle BAE &= (90^\\circ - \\beta) - \\beta = 90^\\circ - 2\\beta, \\\\\n\\angle BEA &= 180^\\circ - \\beta.\n\\end{aligned}\n$$\nApplying the law of si... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
04j8 | Let $ABC$ be a triangle in which $\tan \angle BAC = 1$ and $\tan \angle ABC = 2$ holds.
Determine the ratio $|BC| : |AB|$. (Ilko Brnetić) | [] | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | sqrt(5):3 | |
0j54 | Problem:
How many polynomials $P$ with integer coefficients and degree at most $5$ satisfy $0 \leq P(x) < 120$ for all $x \in \{0,1,2,3,4,5\}$? | [
"Solution:\n\nAnswer: $86400000$\n\nFor each nonnegative integer $i$, let $x^{\\underline{i}} = x(x-1) \\cdots (x-i+1)$. (Define $x^{0} = 1$.)\n\nLemma: Each polynomial with integer coefficients $f$ can be uniquely written in the form\n$$\nf(x) = a_{n} x^{\\underline{n}} + \\ldots + a_{1} x^{\\underline{1}} + a_{0}... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | proof and answer | 86400000 | |
0jul | Problem:
A positive integer is written on each corner of a square such that numbers on opposite vertices are relatively prime while numbers on adjacent vertices are not relatively prime. What is the smallest possible value of the sum of these 4 numbers?
Proposed by: Eshaan Nichani | [
"Solution:\nAnswer: 60\nTwo opposite vertices are relatively prime, but they both share a factor with their common neighbor. So that common neighbor must have two prime factors. So each of the 4 numbers has two prime factors, which are not shared with the opposite vertex. Moreover, it suffices to choose the vertice... | United States | HMMT November 2016 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 60 | |
0cv2 | Does there exist an infinite strictly increasing sequence $a_1, a_2, a_3, \ldots$ of positive integers such that the sum of any two distinct members of the sequence is coprime to the sum of any three its distinct members?
Существует ли бесконечно возрастающая последовательность $a_1, a_2, a_3, \ldots$ положительных це... | [
"One example of such sequence is determined by $a_1 = 1$, $a_2 = 7$, $a_{n+1} = (3a_n)! + 1$. Define a good pair/triple of elements of this sequence as such pair/triple where only number 1 may repeat. To show that the sequence fits, prove by induction that any good pair and any good triple have coprime sums.\n\nLet... | Russia | XLIII Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English; Russian | proof and answer | Yes; for example, define a1 = 1, a2 = 7, and a_{n+1} = (3 a_n)! + 1. | |
0jfi | Problem:
A square is inscribed in a circle of radius $1$. Find the perimeter of the square. | [
"Solution:\nThe square has diagonal length $2$, so the side length is $\\sqrt{2}$ and the perimeter is $4\\sqrt{2}$."
] | United States | HMMT November 2013 | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | null | final answer only | 4√2 | |
05b7 | Anu and Bert each play the following game with ChatGPT. Anu's game starts with the number $2023!$ on the board, Bert's game with $2024!$ on the board. Each move consists of two parts. First the active player divides the number on the board by one of its composite factors $d$. Then their opponent must do one of the foll... | [
"As $2024! = 2023! \\cdot 2024$ and $2024 = 2^3 \\cdot 11^1 \\cdot 23^1$, the sum of the exponents of all primes in the prime factorisation of the number on the board is odd in one of the games and even in the other. We will show that ChatGPT can win the game where the sum is even.\n\nOn any of its moves, ChatGPT c... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0kr9 | There is a positive real number $x$ not equal to either $\frac{1}{20}$ or $\frac{1}{2}$ such that
$$
\log_{20x}(22x) = \log_{2x}(202x).
$$
The value $\log_{20x}(22x)$ can be written as $\log_{10}(\frac{m}{n})$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | [] | United States | AIME II | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 112 | |
000e | En un triángulo escaleno $ABC$ se traza la bisectriz interior $BD$, con $D$ sobre $AC$. Sean $E$ y $F$, respectivamente, los pies de las perpendiculares trazadas desde $A$ y $C$ hacia la recta $BD$, y sea $M$ el punto sobre el lado $BC$ tal que $DM$ es perpendicular a $BC$. Demuestre que $\angle EMD = \angle DMF$. | [] | Argentina | XVII Olimpíada Iberoamericana de Matemática | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | español | proof only | null | |
0b5d | Problem:
Fie $p$ un număr prim şi $0 \leq a_{1}<a_{2}<\ldots<a_{m}<p$ şi $0 \leq b_{1}<b_{2}<\ldots<b_{n}<p$ numere întregi oarecare. Notăm cu $k$ numărul resturilor diferite pe care le dau $a_{i}+b_{j}, 1 \leq i \leq m, 1 \leq j \leq n$ la împărțirea cu $p$. Demonstrați că:
a) dacă $m+n>p$, atunci $k=p$;
b) dacă $m... | [] | Romania | Bulgarian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0b95 | On a circle are written several real numbers, of positive sum. Let $S$ be the largest and $s$ the least of the sums of consecutive numbers on the circle. Prove that $S + s > 0$. | [
"Denote by $T > 0$ the total sum of the numbers around the circle. Clearly $S \\ge T > 0$. If $s \\ge 0$, we are done. If $s < 0$, the sum of the numbers which are not terms of $s$ is equal to $T - s$. Since $S \\ge T - s$, then $S + s \\ge T > 0$, as needed."
] | Romania | NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
01kk | Given the set of $N$ four-digit numbers formed from the digits $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$. Find the smallest possible value of $N$ if the set satisfies the following condition:
a) For any two different digits from $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ there exists a number from the set which contains both... | [
"a.\nSee Problem B.4.\n\nb.\nLet it be possible to choose six four-digit numbers satisfying the problem condition. By the above, any of digits $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ must be exactly in three of these six numbers.\n\nFurther, it is not hard to prove that no two of the chosen numbers have the same thr... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | a) 6; b) 7 | |
03b1 | Find all positive integers $x$ and $y$ such that $3^x 7^y + 1$ is an odd degree (greater than 1) of a positive integer. | [] | Bulgaria | Selection test for 27. Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | x = 2, y = 1 | |
025g | Problem:
Arnaldo, Bráulio e Carlos participarão de uma corrida de rua. Depois de algumas semanas, eles estavam discutindo suas estratégias. Arnaldo corre a primeira metade da distância total da corrida a $9~\mathrm{km}/\mathrm{h}$ e a segunda metade a $11~\mathrm{km}/\mathrm{h}$. Já Bráulio corre um terço da distância... | [
"Solution:\n\nChamaremos os tempos, medidos em horas, de Arnaldo, Bráulio e Carlos de $t_{A}, t_{B}$ e $t_{C}$, respectivamente. Seja $6d$ a distância total da corrida, medida em quilômetros. Como Carlos corre metade do tempo a $9~\\mathrm{km}/\\mathrm{h}$ e metade a $11~\\mathrm{km}/\\mathrm{h}$, então\n$$\n6d = 9... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | t_C < t_B < t_A | |
0jfx | Problem:
On each side of a $6$ by $8$ rectangle, construct an equilateral triangle with that side as one edge such that the interior of the triangle intersects the interior of the rectangle. What is the total area of all regions that are contained in exactly $3$ of the $4$ equilateral triangles? | [
"Solution:\nAnswer: $\\frac{96 \\sqrt{3}-154}{\\sqrt{3}}$ OR $\\frac{288-154 \\sqrt{3}}{3}$ OR $96-\\frac{154}{\\sqrt{3}}$ OR $96-\\frac{154 \\sqrt{3}}{3}$\n\nLet the rectangle be $ABCD$ with $AB=8$ and $BC=6$. Let the four equilateral triangles be $ABP_1$, $BCP_2$, $CDP_3$, and $DAP_4$ (for convenience, call them ... | United States | HMMT November 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 96 - 154/√3 | |
064g | Problem:
Wir betrachten alle positiven reellen Zahlen $a, b, c, d$, für die $(a+c)(b+d)=a c+b d$ gilt.
Bestimmen Sie den kleinsten Wert, den $S=\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$ annehmen kann. | [
"Solution:\n\nDer kleinste mögliche Wert für $S$ ist $8$.\nUm zu zeigen, dass stets $S \\geq 8$ gilt, wenden wir zweimal die AM-GM-Ungleichung an:\n\n$\\left(\\frac{a}{b}+\\frac{c}{d}\\right)+\\left(\\frac{b}{c}+\\frac{d}{a}\\right) \\geq 2 \\sqrt{\\frac{a c}{b d}}+2 \\sqrt{\\frac{b d}{a c}}=\\frac{2(a c+b d)}{\\sq... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 8 | |
0367 | Problem:
The positive integers $M$ and $n$ are such that $M$ is divisible by all positive integers from $1$ to $n$ but it is not divisible by $n+1$, $n+2$ and $n+3$. Find all possible values of $n$. | [
"Solution:\n\nWe shall prove that $n+1$, $n+2$ and $n+3$ are prime powers. Assume the contrary and let some of them has the form $ab$, where $a \\geq 2$, $b \\geq 2$ and $(a, b) = 1$. Since $ab$ does not divide $M$, then $a$ or $b$ does not divide $M$. Let $a$ does not divide $M$. Then it follows that $a \\geq n+1$... | Bulgaria | Bulgarian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n = 1, 2, 6 | |
00kg | For a point $P$ in the interior of a triangle $ABC$, let $D$ be the intersection of $AP$ with $BC$, $E$ the intersection of $BP$ with $AC$ and $F$ the intersection of $CP$ with $AB$.
Furthermore, let $Q$ and $R$ be the intersections of the parallel to $AB$ through $P$ with the sides $AC$ and $BC$, respectively. Likewis... | [
"In the following, let $[XYZ]$ denote the area of the triangle $XYZ$.\n\n\nSince we are only dealing with ratios of areas and parallels to the sides of the triangle, we can assume without loss of generality that $ABC$ is equilateral for the following argument. (If i... | Austria | Austria 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | the centroid of triangle ABC | |
0gi0 | Let $ABC$ be a scalene triangle with circumcenter $O$ and orthocenter $H$. Let $AYZ$ be another triangle sharing the vertex $A$ such that its circumcenter is $H$ and its orthocenter is $O$. Show that if $Z$ is on $BC$, then $A, H, O, Y$ are concyclic.
設 $ABC$ 為不等邊三角形，其外心為 $O$，垂心為 $H$。有另一個三角形 $AYZ$ 與 $ABC$ 共用頂點 $A$，且三角... | [
"我們先證明一個重要的引理。\n\n**Lemma.** 若 $X$ 是 $BC$ 與 $YZ$ 的交點, 則 $X$ 落在 $OH$ 的中垂線上。\n\n*Proof of the lemma.* 令 $H'$ 為 $H$ 關於 $BC$ 的鏡射點, $O'$ 為 $O$ 關於 $YZ$ 的鏡射點。熟知有 $AO = H'O$ 且 $AH = O'H$。因此 $\\angle HH'O = \\angle OAH = \\angle HO'H$, 所以 $HH'OO'$ 共圓。故 $X$ 是圓 $HH'OO'$ 的圓心, 得證。 □\n\n*Another proof of the lemma.* 取點 $P$ 使得 $A... | Taiwan | 2023 數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | Chinese (Traditional) | proof only | null | |
0bdq | Suppose $a$ and $b$ are complex numbers. Prove that $|az + b\bar{z}| \le 1$ for all $z \in \mathbb{C}$ with $|z| = 1$ if and only if $|a| + |b| \le 1$.
Liviu Vlaicu | [
"Let $|a| + |b| \\le 1$ and let $z \\in \\mathbb{C}$ with $|z| = 1$. Then $|az + b\\bar{z}| \\le |az| + |b\\bar{z}| = |a| + |b| \\le 1$, as claimed.\n\nConversely, if $a = 0$ or $b = 0$ there is nothing to show. For $a, b \\neq 0$, write $\\frac{b}{a} = r(\\cos\\alpha + i\\sin\\alpha)$. Put $z = \\cos\\frac{\\alpha... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
05na | Problem:
On considère un échiquier $2015 \times 2015$, auquel on a retiré le carré $2 \times 2$ en bas à droite. Montrer qu'on peut le paver par des pièces composées de 3 cases en forme de L (les pièces peuvent être pivotées). | [
"Solution:\n\nComme l'échiquier peut être décomposé en trois rectangles $2010 \\times 2015$, $5 \\times 2004$ et $5 \\times 11$, et comme $2010$ et $2004$ sont divisibles par $6$, il suffit de montrer que\n\n1) Pour tout $n \\geqslant 2$ et tout entier $m$, un échiquier $6 m \\times n$ peut être pavé par des pièces... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Other"
] | null | proof only | null | |
0hen | In how many ways can a chessboard $n \times n$, $n \ge 3$, from which two diagonally opposite corner cells $1 \times 1$ were cut out, accommodate $n$ rooks, neither of which attacks one other? A rook is a chess piece that attacks all the cells adjacent horizontally or vertically to the cell it is located in. | [
"Without limitation of generality, suppose that the left lower cell $A$ and the right upper cell $Z$ were cut out. First, let us see in how many ways rooks can be placed on a $n \\times n$ chessboard. There are $n!$ ways, since for a rook, there are $n$ options in the first column, $n-1$ option in the second column... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | (n-2)!(n^2 - 3n + 3) | |
00jl | The two equilateral triangles $ABC$ and $ADB$ (with $C \neq D$) share the common side $AB$. The midpoints of $AC$ and $BC$ are denoted by $E$ and $F$, respectively. Show that $DE$ and $DF$ divide $AB$ into three parts of equal length.
G. Kirchner, Innsbruck
Figure 1: Problem 4. | [
"The intersections of $DE$, $DF$ and $DC$ with $AB$ are denoted by $S_1$, $S_2$ and $M$, respectively.\n\nWe consider the triangle $ACD$. In this triangle, $DE$ and $AM$ are medians. Therefore, their intersection $S_1$ is the centroid of this triangle and we have $\\overline{AS_1} = 2 \\cdot \\overline{S_1M}$. An a... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
00je | The equation $n! + A \cdot n = n^k$ has $(n, k) = (0, 0)$ as a solution for every non-negative integer $A$. Determine all non-negative integer solutions of this equation for $A = 7$ and $A = 2012$. | [
"We first note that, independently of the value of $A$, $n = 0$ implies $k = 0$ and vice versa. Furthermore, also independently of the value of $A > 0$, there can be no solution for either $n = 1$ or $k = 1$, since $n! + A \\cdot n > n^k$ certainly holds in either of these cases. In the following, we therefore limi... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | For A = 7: (n, k) = (0, 0), (2, 4), (3, 3). For A = 2012: (n, k) = (0, 0) only. | |
01qm | Let $ABCD$ be a cyclic quadrilateral, $P = AB \cap CD$, $Q = AD \cap BC$.
Prove that the distance between the orthocenters of the triangles $APD$ and $AQB$ is equal to that of the triangles $CQD$ and $BPC$. | [
"We refer to the following well-known fact: in any cyclic quadrilateral all perpendiculars drawn through the midpoints of the sides to the opposite sides meet at the same point $U$. It follows that the altitudes $DD_1$ and $CC_1$ of the triangle $ADB$ and $BCP$, respectively, are symmetric with respect to $U$. The ... | Belarus | SELECTION and TRAINING SESSION | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | English | proof only | null | |
0jql | Problem:
In triangle $ABC$, $AB = 2$, $AC = 1 + \sqrt{5}$, and $\angle CAB = 54^{\circ}$. Suppose $D$ lies on the extension of $AC$ through $C$ such that $CD = \sqrt{5} - 1$. If $M$ is the midpoint of $BD$, determine the measure of $\angle ACM$, in degrees. | [
"Solution:\n\nLet $E$ be the midpoint of $\\overline{AD}$. $EC = \\sqrt{5} + 1 - \\sqrt{5} = 1$, and $EM = 1$ by similar triangles ($ABD \\sim EMD$). $\\triangle ECM$ is isosceles, with $m \\angle CEM = 54^{\\circ}$. Thus $m \\angle ACM = m \\angle ECM = \\frac{180 - 54}{2} = 63^{\\circ}$."
] | United States | HMMT February 2015 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 63° | |
0igr | Problem:
The graph of $r=2+\cos 2\theta$ and its reflection over the line $y=x$ bound five regions in the plane. Find the area of the region containing the origin. | [
"Solution: $\\frac{9\\pi}{2} - 8$\n\nThe original graph is closer to the origin than its reflection for $\\theta \\in \\left(\\frac{\\pi}{4}, \\frac{3\\pi}{4}\\right) \\cup \\left(\\frac{5\\pi}{4}, \\frac{7\\pi}{4}\\right)$, and the region is symmetric about the origin. Therefore the area w... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 9π/2 − 8 | |
0d70 | Let $a$ be a positive integer. Find all prime numbers $p$ with the following property: there exist exactly $p$ ordered pairs of integers $(x, y)$, with $0 \leq x, y \leq p-1$, such that $p$ divides $y^{2}-x^{3}-a^{2} x$. | [
"The case $p=2$ works with solutions $(0,0)$ and $(1,1)$ if $a$ is even; $(0,0)$ and $(0,1)$ if $a$ is odd.\n\nWe claim that any odd prime divisor $p$ of $a$ also works. Of course for $0 \\leq x, y \\leq p-1$, the congruence $y^{2} \\equiv x^{3} \\pmod{p}$ implies that $x=0$ if and only if $y=0$. For any $0 < x \\l... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | p = 2, or p is an odd prime divisor of a, or p ≡ 3 (mod 4). | |
04i4 | For a positive integer $n$, we say that an array with three rows and $n$ columns is *magical* if there exists a positive integer $k$, $1 \le k \le n$, such that
* the first row is filled with numbers $1, 2, \ldots, n$ in that order
* the second row is filled with numbers $k, k+1, \ldots, n, 1, 2, \ldots, k-1$ in that o... | [] | Croatia | Croatia Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All positive integers n. The number of magical arrays is 1 for n = 1 and 2 for every n ≥ 2. Equivalently: for even n the unique valid shift is n/2 + 1 and there are two possible common sums; for odd n the two valid shifts are (n + 1)/2 and (n + 3)/2, each yielding exactly one common sum. | |
00fm | You are given three lists $A$, $B$, and $C$. List $A$ contains the numbers of the form $10^{k}$ in base 10, with $k$ any integer greater than or equal to $1$. Lists $B$ and $C$ contain the same numbers translated into base $2$ and $5$ respectively:
| $A$ | $B$ | $C$ |
| :--- | :--- | :--- |
| 10 | 1010 | 20 |
| 100 | ... | [
"Let $b_{k}$ and $c_{k}$ be the number of digits in the $k$th term in lists $B$ and $C$, respectively. Then\n$$\n2^{b_{k}-1} \\leq 10^{k} < 2^{b_{k}} \\Longleftrightarrow \\log_{2} 10^{k} < b_{k} \\leq \\log_{2} 10^{k} + 1 \\Longleftrightarrow b_{k} = \\left\\lfloor k \\cdot \\log_{2} 10 \\right\\rfloor + 1\n$$\nan... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1994 | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0b4r | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(2 f(x))=f(x-f(y))+f(x)+y
$$
for all $x, y \in \mathbb{R}$. | [
"Solution:\nLet $P(x, y)$ be the problem statement. Note that if $f(a)=f(b)$ where $a, b \\in \\mathbb{R}$, then $P(x, a)$ and $P(x, b)$ imply $a=b$. Thus $f$ is injective.\n\nThen $P(x,-f(x))$ implies $f(2 f(x))=f(x-f(-f(x)))$, so by injectivity we obtain $f(-f(x))=x-2 f(x)$.\n\nNow $P(x, 0)$ gives $f(2 f(x))=f(x-... | Philippines | 25th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x | |
05fa | Problem:
Soit $ABC$ un triangle non isocèle en $A$ et $I$ le centre de son cercle inscrit. On note $D$ le point de la droite $(BC)$ tel que $\widehat{DIA} = 90^{\circ}$. On note $E$ le pied de la hauteur issue de $I$ dans le triangle $ADI$. Montrer que le point $E$ est sur le cercle circonscrit du triangle $ABC$. | [
"Solution:\n\n\n\nOn introduit $\\Omega$ le cercle circonscrit du triangle $ABC$ ainsi que $\\omega$ le cercle antarctique de centre $S$, le milieu de l'arc $BC$ ne contenant pas $A$. $\\omega$ passe ainsi par $B$, $C$ et $I$. On note de plus $\\gamma$ le cercle de diamètre $[AI]$, ce cercl... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
082r | Problem:
(i) Si determinino tutte le coppie $(m, n)$ di interi positivi che soddisfano l'equazione $n^{2}-2^{m}=1$.
(ii) Si determinino tutte le coppie $(m, n)$ di interi positivi che soddisfano l'equazione $2^{m}-n^{2}=1$. | [
"Solution:\n\nAnzitutto $n$ deve essere dispari in entrambi i casi, perché $2^{m}$ è pari per $m \\geq 1$.\n\ni. $n = m = 3$. Infatti tra due numeri pari consecutivi uno non è divisibile per quattro, mentre da $2^{m} = n^{2} - 1 = (n-1)(n+1)$ si ottiene che $n-1$ e $n+1$ sono entrambi potenze di 2; rimane quindi so... | Italy | Progetto Olimpiadi di Matematica 2003 GARA di SECONDO LIVELLO BIENNIO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | i: (m, n) = (3, 3); ii: (m, n) = (1, 1) | |
094k | Problem:
Find the smallest integer $b$ with the following property: For each way of colouring exactly $b$ squares of an $8 \times 8$ chessboard green, one can place 7 bishops on 7 green squares so that no two bishops attack each other.
Remark. Two bishops attack each other if they are on the same diagonal. | [
"Solution:\nLet us place 40 bishops on 6 diagonals as shown in Figure 2. If we select any 7 of the placed bishops, by Pigeon hole principle, at least two of the selected bishops are on the same diagonal, so they attack each other. Thus, the number $b$ of selected bishops is at least 41.\n\n... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 41 | |
074u | Let $A = (a_{jk})$ be a $10 \times 10$ array of positive real numbers such that the sum of numbers in each row as well as in each column is $1$. Show that there exist $j < k$ and $l < m$ such that
$$
a_{jl}a_{km} + a_{jm}a_{kl} \geq \frac{1}{50}
$$ | [
"Consider the first column. Let $M$ be the maximum value of the entries in this column. Then $M \\ge 1/10$, by pigeonhole principle. Writing the first column $a_{1,1}, a_{2,1}, \\dots, a_{10,1}$, let $M = a_{j,1}$, for some $j$. Consider the sum\n$$\n\\sum_{l=2}^{10} (a_{j,1}a_{k,l} + a_{j,l}a_{k,1}),\n$$\nwhere $k... | India | Indija TS 2010 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
0f02 | Problem:
For positive reals $x$, $y$ let $f(x, y)$ be the smallest of $x$, $1/y$, $y + 1/x$. What is the maximum value of $f(x, y)$? What are the corresponding $x$, $y$? | [] | Soviet Union | ASU | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Maximum value: sqrt(2). Achieved at x = sqrt(2), y = 1/sqrt(2). | |
02u5 | Problem:
a) O Teorema do Bico diz que as distâncias de um ponto exterior a uma circunferência aos pontos onde suas tangentes tocam a circunferência são iguais. Na figura a seguir, $AP$ e $AQ$ são tangentes à circunferência. Mostre que $AP = AQ$.
b) Considere o hexágono da figura a seguir, no... | [
"Solution:\na) Trace $OA$. Observe que os triângulos $\\triangle OPA$ e $\\triangle OQA$ são congruentes pois são triângulos retângulos com a mesma hipotenusa e um dos catetos com a mesma medida. Desse modo, $AP = AQ$.\n\nb) Cada um dos lados é dividido pelo ponto de tangência em dois segmentos, conforme a figura. ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 4 | |
02p3 | Let $ABC$ be an acute triangle and $H$ is orthocenter. Let $D$ be the intersection of $BH$ and $AC$ and $E$ be the intersection of $CH$ and $AB$. The circumcircle of $ADE$ meets the circumcircle of $ABC$ at $F \neq A$. Prove that the angle bisectors of $\angle BFC$ and $\angle BHC$ concur at a point on line $BC$. | [
"\n\nBy the angle bisector theorem, it suffices to prove that $\\frac{BF}{FC} = \\frac{BH}{HC}$.\nWe have $\\angle EFB = 180^\\circ - \\angle FEA = 180^\\circ - \\angle FDA = \\angle FDC$ and $\\angle FBE = \\angle FBA = \\angle FCA = \\angle FCD$, so triangles $BEF$ and $CDF$ are similar. ... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles"... | English | proof only | null | |
0ftt | Problem:
Sei $M$ eine endliche Menge reeller Zahlen mit folgender Eigenschaft: Aus je drei verschiedenen Elementen von $M$ lassen sich stets zwei auswählen, deren Summe in $M$ liegt. Wieviele Elemente kann $M$ höchstens haben? | [
"Antwort: 7 .\n\nNehme an, $M$ enthalte mehr als drei positive Elemente und bezeichne die vier grössten mit $a > b > c > d > 0$. Nach Voraussetzung kann man aus den drei Elementen $a, b, c$ zwei auswählen, deren Summe in $M$ liegt. Wegen $a + b > a$ und $a + c > a$ und der Maximalität von $a$ liegen diese beiden Su... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 7 | |
03t1 | Let $S = \{n \mid n-1, n, n+1$ all can be expressed as the sum of the squares of two positive integers\}. Prove that, if $n \in S$, then $n^2 \in S$. | [
"$$\nx^2 + y^2 \\equiv 0, 1, 2 \\pmod{4}.\n$$\nLet $n \\in S$. By the above equality, we get $n \\equiv 1 \\pmod{4}$. Thus, we may assume that\n$$\nn - 1 = a^2 + b^2, \\ a \\geq b,\n$$\n$$\nn = c^2 + d^2, \\ c > d,\n$$\n$$\nn + 1 = e^2 + f^2, \\ e \\geq f,\n$$\nwhere $a, b, c, d, e, f$ are positive integers. Theref... | China | China Western Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | English | proof only | null | |
0i5p | Problem:
For $n \geq 1$, let $a_{n}$ be the largest odd divisor of $n$, and let $b_{n}=a_{1}+a_{2}+\cdots+a_{n}$. Prove that $b_{n} \geq \frac{n^{2}+2}{3}$, and determine for which $n$ equality holds. For example,
$$
a_{1}=1,\ a_{2}=1,\ a_{3}=3,\ a_{4}=1,\ a_{5}=5,\ a_{6}=3
$$
thus
$$
b_{6}=1+1+3+1+5+3=14 \geq \frac{6... | [
"Solution:\n\nInspecting a few values leads one to the correct guess that equality holds if and only if $n$ is a power of $2$. The idea is to use induction by \"doubling\" to prove the result. When $n=1$, we have $1=1$ as desired. Now suppose $n \\geq 2$. If $n=2k$ with $k \\geq 1$, then\n$$\n\\begin{aligned}\nb_{n... | United States | Bay Area Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Equality holds if and only if n is a power of two. | |
099i | Show that, for arbitrary positive integer $k$ there exist positive integer $n$ such that $\frac{2^{n^2} + 1}{n^3}$ is integer and $n$ has exactly $k$ prime divisors.
(proposed by G. Batzaya) | [
"To prove this problem we need following lemma.\n\n**Lemma.** Let $2 < p$-prime number.\n(i) If $\\alpha \\in \\mathbb{N}$ then $p^{\\alpha}|a+1$ if and only if $p^{\\alpha+1}|a+1$. From here, we get $p^2 \\nmid \\frac{a^p+1}{a+1}$.\n(ii) If $a \\ge 4$ integer then there exist $q$ prime number such that $q|\\frac{a... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof only | null | |
0hr4 | Problem:
Let $ABC$ be a triangle. The incircle, centered at $I$, touches side $BC$ at $D$. Let $E$ be the reflection of $D$ through $I$, and let $F$ be the reflection of $D$ through the midpoint of $BC$. Prove that $A$, $E$, and $F$ are collinear. | [
"Solution:\n\nLet the tangent line to the incircle at $E$ (which is of course parallel to $BC$) meet $AB$ and $AC$ at $X$ and $Y$, respectively.\n\n\n\nNote that\n$$\n\\angle IXE = \\frac{\\angle BXE}{2} = \\frac{180^\\circ - \\angle DBX}{2} = 90^\\circ - \\angle DBI = BID.\n$$\nTherefore t... | United States | Berkeley Math Circle | [
"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 | |
0jrg | Problem:
Let $z = e^{\frac{2\pi i}{101}}$ and let $\omega = e^{\frac{2\pi i}{10}}$. Prove that
$$
\prod_{a=0}^{9} \prod_{b=0}^{100} \prod_{c=0}^{100} \left(\omega^{a} + z^{b} + z^{c}\right)
$$
is an integer and find (with proof) its remainder upon division by $101$. | [
"Solution:\nLet $p = 101$ and $r = 10$. Note that $p \\nmid r$.\n\nIn the sequel, we will repeatedly use the polynomial identities $\\prod_{k (\\bmod p)} (x - z^{k}) = x^{p} - 1$, and $\\prod_{j (\\bmod r)} (x - \\omega^{j}) = x^{r} - 1$.\n\nThe product is an integer by standard symmetric sum theory (concrete precu... | United States | HMMT February 2015 | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Algebraic Number Theory... | null | proof and answer | 13 | |
08d8 | Problem:
Due scuole si scontrano in un torneo di scacchi a cui ciascuna fa partecipare 75 alunni: vengono organizzate 75 partite in cui far giocare tutti gli studenti uno contro uno (un membro della prima scuola contro uno della seconda) sotto il controllo di un arbitro esterno. Ogni scuola numera i propri studenti da... | [
"Solution:\n\nLa risposta è (D). Dividiamo i 75 alunni di ognuna delle due scuole in 3 gruppi, a seconda del resto del loro numero quando diviso per 3. In particolare, nella prima scuola, gli alunni sono divisi nei gruppi $A, B, C$ a seconda se il loro numero è del tipo $3k, 3k+1$ o $3k+2$ per qualche intero $k$. A... | Italy | Progetto Olimpiadi della Matematica | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | D | |
0ek2 | Problem:
V nekem domu za ostarele so pred enim tednom ugotovili prve okužene s covid-19. Od takrat se je njihovo število popeterilo, torej jih je sedaj petkrat toliko kot pred enim tednom. 21 od teh okuženih so zjutraj odpeljali v bolnišnico, tako je med oskrbovanci, ki so ostali v domu, $9 . \overline{09}\ \%$ okužen... | [
"Solution:\n\nNaj bo prvotno $x$ okuženih in v domu skupno $y$ oskrbovancev. Od vseh okuženih $5x$ so jih 21 odpeljali, okuženih je ostalo $5x-21$. Oskrbovancev je ostalo $y-21$. Delež okuženih je $\\frac{5x-21}{y-21}=9 . \\overline{09}\\ \\%$. Če periodično število $9 . \\overline{09}$ spremenimo v ulomek, dobimo ... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | 8 infected originally; 230 total residents originally | |
057h | Find all pairs $(x, y)$ of integers such that
$$
\sqrt{x+2019} - \sqrt{x} = \sqrt{y}.
$$ | [
"As $x$ and $y$ occur under square root, only non-negative solutions can exist. Bringing $\\sqrt{x}$ to the right, squaring both sides and collecting similar terms gives $2019 = y + 2\\sqrt{xy}$ which is equivalent to the initial equation. Thus $2\\sqrt{xy}$ is an integer. If $x = 0$ then $y = 2019$, the case $y = ... | Estonia | Open Contests | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (x, y) in {(0, 2019), (1009^2, 1), (336^2·3, 3), (335^2, 9), (673, 673)} | |
0h6c | There are $N$ points on the circle. Andriy and Olesya play the game as follows. Andriy goes first. They take turns and connect two of the given points with a chord, if it does not cross any of the prior chords. The one after whose move there is a triangle formed by the drawn chords wins. Who will win the fair game if:
... | [
"**Answer:** a) Andriy wins, b) Olesya wins.\n\nClearly, the one who is forced to draw a chord from the point from which there is another chord drawn already, will lose (Fig. 39). For example, one player draws $AC$ and chord $AB$ has already been drawn before (we assume that there are no other chords drawn from poi... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a) Andriy wins; b) Olesya wins. | |
0b18 | Problem:
Let $f(x) = x^{2} + 3$. How many positive integers $x$ are there such that $x$ divides $f(f(f(x)))$? | [] | Philippines | Philippines Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 6 | |
06ai | For each positive integer $n$, we consider a $3n \times 3n$ chessboard, and its squares are colored black and white as follows: Starting from the left up corner square every third diagonal is black and all other squares are white, as in the figure.
At each move we select a $2 \times 2$ square ... | [
"a) Firstly, observe that if we change the colour of one square 3 times, it goes back to its original colour. As the final configuration does not depend on the order of moves and placing a move on a square 3 times does not change the configuration, we can suppose that a move is placed on any square exactly 0, 1 or ... | Greece | Selection examinations | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0fx4 | Problem:
Seien $a, b$ natürliche Zahlen. Zeige, dass man die ganzen Zahlen mit drei Farben färben kann, sodass zwei ganze Zahlen mit Differenz $a$ oder $b$ stets verschieden gefärbt sind. | [
"Solution:\n\nWir färben $0$ beliebig und konstruieren die gesuchte Färbung zuerst induktiv für die positiven ganzen Zahlen. Nehme an, $0,1, \\ldots, n-1$ seien schon gefärbt. Wir können jetzt $n$ einfach so färben, dass keine der beiden Zahlen $n-a$ und $n-b$ (sofern diese überhaupt schon gefärbt sind) dieselbe Fa... | Switzerland | IMO Selektion 2008 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
00z6 | Problem:
Denote by $d(n)$ the number of distinct positive divisors of a positive integer $n$ (including $1$ and $n$). Let $a > 1$ and $n > 0$ be integers such that $a^{n} + 1$ is a prime. Prove that
$$
d\left(a^{n} - 1\right) \geq n.
$$ | [
"Solution:\n\nFirst we show that $n = 2^{s}$ for some integer $s \\geq 0$. Indeed, if $n = m p$ where $p$ is an odd prime, then $a^{n} + 1 = a^{m p} + 1 = \\left(a^{m} + 1\\right)\\left(a^{m(p-1)} - a^{m(p-2)} + \\cdots - a + 1\\right)$, a contradiction.\n\nNow we use induction on $s$ to prove that $d\\left(a^{2^{s... | Baltic Way | Baltic Way | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0gu6 | At the beginning Alice distributes $1000$ balls into $30$ boxes. After that Alice and Bob alternatively make moves, Alice begins. A person making move chooses a box and takes one ball from the chosen box. A person taking the very last ball from a box takes on that empty box. Find the maximal integer $k$ such that regar... | [
"**5.** For any positive real $x$ we have $(x-1)^2(3x^2+4x+3) \\ge 0$. Therefore,\n$$\n3(x^4 + 1) \\ge 2(x^3 + x^2 + x) \\quad (1)\n$$\nBy using (1) we get\n$$\n\\frac{a^4+1}{b^3+b^2+b} \\cdot \\frac{b^4+1}{c^3+c^2+c} \\cdot \\frac{c^4+1}{a^3+a^2+a} \\ge \\frac{8}{27}\n$$\nFinally, by using of by AM-GM inequality\n... | Turkey | Team Selection Test for JBMO 2023 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 15 | |
06wc | Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy
$$
(f(a)-f(b))(f(b)-f(c))(f(c)-f(a))=f\left(a b^{2}+b c^{2}+c a^{2}\right)-f\left(a^{2} b+b^{2} c+c^{2} a\right)
$$
for all real numbers $a, b, c$. | [
"Answer: $f(x)=\\alpha x+\\beta$ or $f(x)=\\alpha x^{3}+\\beta$ where $\\alpha \\in\\{-1,0,1\\}$ and $\\beta \\in \\mathbb{R}$.\n\nIt is straightforward to check that above functions satisfy the equation. Now let $f(x)$ satisfy the equation, which we denote $E(a, b, c)$. Then clearly $f(x)+C$ also does; therefore, ... | IMO | IMO 2021 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | f(x)=αx+β or f(x)=αx^3+β where α∈{−1,0,1} and β∈ℝ | |
01xv | The circles $\omega_1$ and $\omega_2$ centered at $O_1$ and $O_2$, respectively, intersect at two different points $E$ and $F$. The line $O_1E$ intersects for the second time $\omega_1$ at the point $A$, and $\omega_2$ at the point $C$. The line $O_2E$ intersects for the second time $\omega_2$ at the point $B$, and $\o... | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
007t | A cross is the shape obtained from a $3 \times 3$ grid square upon removing the 4 corner unit squares. Every unit square of a $2010 \times 2010$ table must be colored in one of 5 distinct colors so that the 5 unit squares of every cross contained in the table have different colors. In how many ways can this be done? Tw... | [
"For an admissible coloring let the 13-cell shape shown in the figure be entirely inside the table. Observe that the central row of 5 represents all 5 colors, as well as the central column of 5. Assume on the contrary that the central row misses color 1. Since the crosses centered at *a*, *b*, *c* contain color 1, ... | Argentina | National Olympiad of Argentina | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 2400000 | |
0ems | Prove that if $x \le y \le z$ are real numbers satisfying $xy + yz + zx = 1$, then $xz < \frac{1}{2}$. Is it possible to replace $\frac{1}{2}$ with a smaller number? | [
"If $x$ and $z$ have opposite signs, or if one of them is $0$, then the inequality is obviously true, so $x$, $y$ and $z$ are either all positive or all negative. If they are all negative, we may replace $x$, $y$ and $z$ with $-x$, $-y$ and $-z$ without changing the conditions of the problem, so we can assume that ... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | xz < 1/2, and 1/2 is optimal (cannot be replaced by a smaller number). | |
01cn | Let $B$, $A$, $L$, $T$, $I$, $C$ be positive numbers. Find all possible values of the expression
$$
\frac{BA}{(C+B)(A+L)} + \frac{LT}{(A+L)(T+I)} + \frac{IC}{(T+I)(C+B)}.
$$ | [
"The range of the expression is the interval $(0, 1)$. Writing $x = \\frac{A}{A+L}$, $y = \\frac{T}{T+I}$, $z = \\frac{C}{C+B}$, and observing that $\\frac{L}{A+L} = 1 - x$ &c., the expression transforms into\n$$\nx(1-z) + y(1-x) + z(1-y) = 1 - xyz - (1-x)(1-y)(1-z),\n$$\nwhere $0 < x, y, z < 1$. Clearly the expres... | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | (0,1) | |
0d11 | Let $ABC$ be a triangle with incenter $I$ and circumcenter $O$ and let $M$ be the midpoint of $BC$. The bisector of angle $A$ intersects lines $BC$ and $OM$ at $L$ and $Q$, respectively. Prove that
$$
AI \cdot LQ = IL \cdot IQ.
$$ | [
"The bisector of $\\widehat{BAC}$ and the perpendicular bisector of side $BC$ intersect at $Q$, the midpoint of arc $\\widehat{BC}$ not containing $A$.\nWe have $BQ = IQ$, since $\\widehat{QBI} = \\widehat{QIB} = 90^\\circ - \\frac{1}{2}\\widehat{C}$.\nOn the other hand, the triangles $BLQ$ and $ALC$ are similar, m... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"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 | |
07qh | Let $AE$ be a diameter of the circumcircle of triangle $ABC$. Join $E$ to the orthocentre, $H$, of $\triangle ABC$ and extend $EH$ to meet the circle again at $D$. Prove that the nine point circle of $\triangle ABC$ passes through the midpoint of $HD$. | [
"Let $F$ be the midpoint of $DH$, $K$ be the midpoint of $AH$, $L$ on $BC$ the foot of the altitude from $A$, and let $M$ be the intersection point of $BC$ and $DE$.\n\n\n\nBecause $AE$ is a diameter of the circumcircle, $\\angle ACE = \\angle ABE = 90^\\circ$, so $CE$ is perpendicular to $... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ao7 | Problem:
How many triangles can be formed if two sides have lengths $15$ and $19$ and the third side has even length?
(a) $13$
(b) $14$
(c) $15$
(d) $16$ | [] | Philippines | Qualifying Round | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | MCQ | b | |
09uv | In a cafe, each product costs at most $12$ ducats. Currently the cafe owner is only using coins worth one ducat. This is unpractical for the more expensive products, however. Therefore, the cafe owner has decided to introduce two types of coins next to the coins of one ducat. He is doing this in such a way that as many... | [
"$4$ and $6$"
] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 4 and 6 | |
0e6l | We have a cylinder with a radius of $1$ cm and a height of $4$ cm. While the point $P$ lies on the bottom circle of the cylinder, the point $Q$ lies on the top circle right above $P$. A string connecting the point $P$ and $Q$ makes a complete turn round the cylinder (see the picture). How long is the shortest such stri... | [
"If we cut the cylinder along the vertical line segment $PQ$ and unfold it, we get a rectangle with sides of lengths $4$ cm and $2\\pi$ cm. Now the problem asks for the shortest path between two opposite vertices of the rectangle. The length of such a path is the same as the length of the diagonal, hence $\\sqrt{(2... | Slovenia | National Math Olympiad 2012 | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | MCQ | D | |
05gt | Problem:
Trouver tous les couples de nombres premiers $\left(p, q\right)$ tels que :
$$
p^{2}\left(p^{3}-1\right)=q(q+1)
$$ | [
"Solution:\nSi $p = q$, alors $p^{4} - p = p + 1$. Or $p^{4} = p \\cdot p^{3} \\geqslant 8p > 2p + 1$ pour tout nombre premier $p$ car $p \\geqslant 2$. On a donc $p \\neq q$ et donc $p$ et $q$ sont premiers entre eux.\n\n$p^{2}$ est premier avec $q$ donc d'après le lemme de Gauss, $p^{2} \\mid q + 1$. On dispose d... | France | ENVOi 3 : ARITHMÉTIQUE | [
"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"
] | null | proof and answer | no solution | |
0kmx | Problem:
For each positive real number $\alpha$, define
$$
\lfloor \alpha \mathbb{N} \rfloor := \{ \lfloor \alpha m \rfloor \mid m \in \mathbb{N} \}
$$
Let $n$ be a positive integer. A set $S \subseteq \{1,2, \ldots, n\}$ has the property that: for each real $\beta > 0$,
$$
\text{if } S \subseteq \lfloor \beta \mathbb... | [
"Solution:\n\nFor each $k \\in \\{\\lceil n / 2 \\rceil, \\ldots, n\\}$, picking $\\beta = 1 + 1 / k$ gives\n$$\n\\lfloor \\beta \\mathbb{N} \\rfloor \\cap [n] = [n] \\setminus \\{k\\}\n$$\nso $S$ must contain $k$.\n\nNow we show that $S = \\{\\lceil n / 2 \\rceil, \\ldots, n\\}$ works; this set $S$ has $\\lfloor n... | United States | HMMT Spring 2021 Team Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | floor(n/2) + 1 | |
09f4 | Is it possible to divide set of numbers 55 digit numbers into parts with 4 numbers such that 54 digits of the numbers in every part are same and the remained digits are consecutive natural numbers? | [
"Denote $S(a)$ sum of digits of number $a$. Consider the polynomial $f(x) = \\sum_{a\\text{--}55\\ \\text{digit numbers}} x^{S(a)}$. It is obvious that $f(x) = (x + x^2 + \\ldots + x^9)$.\n\n$(1 + x + x^2 + \\ldots + x^9)^{54} = \\frac{x(x^9 - 1)(x^{10} - 1)^{54}}{(x - 1)^{55}}$.\n\nIf we suppose that one can divid... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | English | proof and answer | No, it is impossible. | |
02ni | Problem:
Clube de Matemática - Eu faço parte de um Clube de Matemática, onde tenho o mesmo número de colegas homens do que de colegas mulheres. Quando um garoto falta, três quartos da equipe são de meninas. Eu sou homem ou mulher? Quantas mulheres e quantos homens tem o clube? | [] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | Female; 3 women and 2 men | |
0c6m | Let $f : [0, \infty) \to \mathbb{R}$ be a continuous function which is constant on the set of nonnegative integers. Suppose that for any real numbers $0 \le a < b < c < d$, such that $f(a) = f(c)$ and $f(b) = f(d)$, we have $f(\frac{a+b}{2}) = f(\frac{c+d}{2})$. Prove that $f$ is a constant function.
Florin Stănescu | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof only | null | |
0ftf | Problem:
Betrachte ein $7 \times 7$ Feld, das in 49 Einheitsquadrate unterteilt ist. In dieses Feld wollen wir Kacheln der Form eines Schweizerkreuzes, bestehend aus 5 Einheitsquadraten, hineinlegen. Dabei sollen die Kanten der Kreuze auf den Linien des Feldes zu liegen kommen. Bestimme die kleinstmögliche Anzahl Quad... | [
"Solution:\n\nFühre Koordinaten ein. Es ist leicht zu sehen, dass jedes Kreuz eines der 7 Quadrate\n$$\n(2,5),(3,2),(3,3),(4,6),(5,4),(6,2),(6,5)\n$$\nbedeckt. Dies zeigt, dass 7 Markierungen genügen. Wir zeigen nun, dass 7 Markierungen nötig sind. Nehme an, nicht. Die Kreuze mit Mittelpunkten\n$$\n(2,2),(2,6),(3,4... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 7 | |
04tt | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for any real numbers $x, y$ we have
$$
f(y - xy) = f(x)y + (x - 1)^2 f(y).
$$ | [
"Setting $x = 1$ we learn $f(0) = f(1)y$ for all real $y$, hence $f(0) = f(1) = 0$. Setting $y = 1$ in the original equation we get\n$$\nf(1-x) = f(x)\n$$\nfor all real $x$. Let $t$ be an arbitrary real number. Setting $x = 1-t$ gives\n$$\nf(ty) = f(1-t)y + t^2 f(y) = f(t)y + t^2 f(y) \\tag{1}\n$$\nfor all real $y$... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = a x(x - 1) for any real constant a | |
0joz | Problem:
Let $\lfloor x\rfloor$ denote the largest integer less than or equal to $x$, and let $\{x\}$ denote the fractional part of $x$. For example, $\lfloor\pi\rfloor=3$, and $\{\pi\}=0.14159 \ldots$, while $\lfloor 100\rfloor=100$ and $\{100\}=0$. If $n$ is the largest solution to the equation $\frac{\lfloor n\rfloo... | [
"Solution:\nNote that $n=\\lfloor n\\rfloor+\\{n\\}$, so\n$$\n\\begin{aligned}\n\\frac{\\lfloor n\\rfloor}{n} & =\\frac{\\lfloor n\\rfloor}{\\lfloor n\\rfloor+\\{n\\}} \\\\\n& =\\frac{2015}{2016} \\\\\n\\Longrightarrow 2016\\lfloor n\\rfloor & =2015\\lfloor n\\rfloor+2015\\{n\\} \\\\\n\\Longrightarrow\\lfloor n\\rf... | United States | HMMT November 2015 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2014/2015 | |
0id6 | Problem:
Augustin has six $1 \times 2 \times \pi$ bricks. He stacks them, one on top of another, to form a tower six bricks high. Each brick can be in any orientation so long as it rests flat on top of the next brick below it (or on the floor). How many distinct heights of towers can he make? | [
"Solution:\n\nIf there are $k$ bricks which are placed so that they contribute either $1$ or $2$ height, then the height of these $k$ bricks can be any integer from $k$ to $2k$. Furthermore, towers with different values of $k$ cannot have the same height. Thus, for each $k$ there are $k+1$ possible tower heights, a... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics"
] | null | final answer only | 28 | |
0c9i | Problem:
Determinaţi numerele complexe $x, y, z$, de acelaşi modul, ştiind că numerele $x+y+z$ şi $x^{3}+y^{3}+z^{3}$ sunt reale. | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ Etapa a III-a | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | All solutions are those with a common modulus r, where after a permutation two of the numbers are complex conjugates and the third is real of absolute value r; equivalently, x = ± r, y = r e^{i t}, z = r e^{-i t} for some real r ≥ 0 and real t. | |
05vd | Problem:
Soit $n$ un entier tel que $n \geqslant 3$. Lucie dispose $n$ feuilles blanches en cercle, puis elle écrit un nombre réel sur chaque feuille. Trouver tous les entiers $k$ tels que $1 \leqslant k \leqslant n$ pour lesquels la propriété suivante est vraie :
Si au moins un des nombres qu'a écrits Lucie n'est pas... | [
"Solution:\n\nDans la suite, on numérote les feuilles de 1 à $n$, dans le sens des aiguilles d'une montre, et on note $x_{i}$ le nombre que Lucie écrit sur la feuille $i$. Les indices seront considérés modulo $n$. L'énoncé revient donc à trouver les entiers $k$ pour lesquels il est impossible que toute somme $\\mat... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | Exactly those integers k between 1 and n that are coprime to n | |
0iw2 | Problem:
In how many ways can you rearrange the letters of "HMMTHMMT" such that the consecutive substring "HMMT" does not appear? | [
"Solution:\n\nThere are $8!/(4!2!2!)=420$ ways to order the letters. If the permuted letters contain \"HMMT\", there are $5 \\cdot 4!/ 2!=60$ ways to order the other letters, so we subtract these. However, we have subtracted \"HMMTHMMT\" twice, so we add it back once to obtain 361 possibilities."
] | United States | $12^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 361 | |
07ta | A circle is drawn through points $A$ and $B$ of a triangle $ABC$ such that the angle in the segment external to the triangle is the complement of $\angle ACB$. Similarly a second circle is drawn through $A$ and $C$ with the angle in the segment external to the triangle equal to the complement of $\angle ABC$.
Prove tha... | [
"Let $\\beta = \\angle ABC$, $\\gamma = \\angle BCA$ and let $\\Omega_B$ denote the circle through $A$ and $B$ and $\\Omega_C$ the circle through $A$ and $C$. The assumptions about the two circles mean that for any point $D$ on $\\Omega_B$ which is on the arc $AB$ that is outside $\\triangle ABC$ we have $\\angle B... | Ireland | IRL_ABooklet_2020 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dfv | Do there exist an infinite sequence $p_1, p_2, p_3, \dots$ of prime numbers such that for any positive integer $n$ the following condition holds
$$
|p_{n+1} - 2p_n| = 1.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | Unknown; this reduces to the open question of the existence of infinite Cunningham chains (of either kind). | |
0foe | Sean $x$ e $y$ números reales entre $0$ y $1$. Probar que
$$
x^3 + xy^2 + 2xy \leq 2x^2y + x^2 + x + y
$$ | [
"La desigualdad equivale a\n$$\nP = 2x^2y + x^2 + x + y - x^3 - xy^2 - 2xy \\geq 0.\n$$\nEscribimos $P$ como un polinomio en la variable $x$,\n$$\nP = -x^3 + (2y + 1)x^2 + (1 - 2y - y^2)x + y.\n$$\nDividimos este polinomio entre $x - 1$, mediante el algoritmo de Ruffini.\n$$\n\\begin{array}{c|cccc}\n1 & -1 & 2y+1 &... | Spain | L Olimpiada Matemática Española | [
"Algebra > Equations and Inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Spanish | proof only | null | |
0f2e | Problem:
There are direct routes between every two cities in a country. The fare between each pair of cities is the same in both directions. Two travellers decide to visit all the cities. The first traveller starts at a city and travels to the city with the most expensive fare (or if there are several such, any one of... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0jqc | Problem:
Let $a, b, c$ be positive reals. Prove that
$$
\frac{a^{3}}{b c}+\frac{b^{3}}{c a}+\frac{c^{3}}{a b} \geq a+b+c
$$ | [
"Solution:\nBy the AM-GM inequality, we have\n$$\n\\frac{2 \\frac{a^{3}}{b c}+\\frac{b^{3}}{c a}+\\frac{c^{3}}{a b}}{4} \\geq \\sqrt[4]{\\frac{a^{3} \\cdot a^{3} \\cdot b^{3} \\cdot c^{3}}{b c \\cdot b c \\cdot c a \\cdot a b}}=a.\n$$\nAdding the analogous two inequalities yields the conclusion."
] | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0etx | Let $f: \mathbb{N} \to \mathbb{N}_0$ be a function that satisfies
$$
f(mn) = mf(n) + nf(m)
$$
for all positive integers $m, n$ and $f(2024) = 10120$.
Prove that there are two integers $m, n$ with $m \neq n$ such that $f(m) = f(n)$. | [
"By setting $n = 1$, we obtain $mf(1) = 0$ for all $m$, so $f(1) = 0$. Next we notice that since every positive integer can be written as a product of prime numbers, the function $f$ is determined completely by its value on the primes. If there exists $m \\ge 2$ such that $f(m) = 0$, then $f(m) = f(1)$ and we are d... | South Africa | The South African Mathematical Olympiad Third Round | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
00ck | En el triángulo $ABC$ sean $D$ y $E$ en los lados $AB$ y $AC$ respectivamente, tales que $BD = CE$. Sean $M$ y $N$ los puntos medios de $BC$ y $DE$ respectivamente.
Demostrar que la bisectriz del ángulo $BAC$ es paralela a la recta $MN$. | [] | Argentina | Nacional OMA 2019 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | Spanish | proof only | null | |
09hm | Let $A, B \subseteq \{0, 1, \dots, 2016\}$ be subsets with $|A| = 30$ and $|B| = 31$ elements. Show that there is an integer $n$ such that the set
$$
\{a + nb \pmod{2017} \mid a \in A, b \in B\}
$$
of residues has at least 730 elements. | [
"First note that for any finite family of finite sets $\\{X_i\\}_{i \\in I}$, we have\n$$\n|\\cup_i X_i| \\ge \\sum_i |X_i| - \\frac{1}{2} \\sum_{i \\ne j} |X_i \\cap X_j|.\n$$\nIndeed, if $x \\in \\cup_i X_i$ is contained in exactly $k$ of the sets $X_i$, then it is counted $k$ times in the first sum and $k(k-1)$ ... | Mongolia | Mongolian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
00ke | The sequence $\langle a_n \rangle$ is defined by the recursion
$$
a_{n+1} = 5a_n^6 + 3a_{n-1}^3 + a_{n-2}^2 \quad \text{for } n \ge 2
$$
and the set of starting values $\{a_0, a_1, a_2\} = \{2013, 2014, 2015\}$.
(i.e., the starting values are these three numbers in arbitrary order.)
Show that the sequence does not cont... | [
"We consider second, third and sixth powers modulo 7:\n\n| x | $x^2$ | $x^3$ | $x^6$ |\n|---|-------|-------|-------|\n| 0 | 0 | 0 | 0 |\n| 1 | 1 | 1 | 1 |\n| 2 | 4 | 1 | 1 |\n| 3 | 2 | 6 | 1 |\n| 4 | 2 | 1 | 1 |\n| 5 | 4 | 6 | 1 |\n| 6 | 1 ... | Austria | Austria 2014 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0e5h | Find all natural numbers $m$ such that for any odd natural number $k$ the number $2^m$ divides $k^m - 1$. | [
"If $m$ is odd, then $3^m \\equiv 3 \\pmod{4}$. Because $2^m$ divides $3^m - 1$, it must be $m = 1$. For $m = 1$, a natural number $k - 1$ is divisible by $2$ as soon as $k$ is odd.\n\nNow let $m$ be even, write $m = 2n$. Then $4^n$ divides $k^{2n} - 1$ for any odd natural number $k$. If $n$ is odd, then $3^{2n} \\... | Slovenia | Selection Examinations for the IMO 2012 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1, 2, 4 | |
0ckl | Let $a$ be a real number for which the number $a + a^2 + a^3$ is an integer. Show that, if one of the numbers $a$, $a^2$ or $a^3$ is rational, then $a$ is an integer.
Cristi Șăvescu | [
"Let $a + a^2 + a^3 = n$, with integer $n$.\nIf $a$ is rational, then $a = \\frac{p}{q}$, where $p, q$ are coprime integers. It follows from here that $\\frac{p}{q} + \\frac{p^2}{q^2} + \\frac{p^3}{q^3} = n$, so $q^2p + qp^2 + p^3 = q^3n$, which implies $q \\mid p^3$. This is possible only if $q = \\pm 1$, which sh... | Romania | 75th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0j2l | Problem:
Let $f(x) = x^{3} - x^{2}$. For a given value of $c$, the graph of $f(x)$, together with the graph of the line $c + x$, split the plane up into regions. Suppose that $c$ is such that exactly two of these regions have finite area. Find the value of $c$ that minimizes the sum of the areas of these two regions. | [
"Solution:\n\nAnswer: $-\\frac{11}{27}$\n\nObserve that $f(x)$ can be written as $\\left(x-\\frac{1}{3}\\right)^{3} - \\frac{1}{3}\\left(x-\\frac{1}{3}\\right) - \\frac{2}{27}$, which has $180^{\\circ}$ symmetry around the point $\\left(\\frac{1}{3}, -\\frac{2}{27}\\right)$. Suppose the graph of $f$ cuts the line $... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | -11/27 | |
0c2j | Consider a function $f : \mathbb{R} \to \mathbb{R}$. For each $a \in \mathbb{Z}$ denote by $f_a : \mathbb{R} \to \mathbb{R}$, the function defined by $f_a(x) = (x - a)f(x)$. Prove that if there are infinitely many values of $a \in \mathbb{Z}$ such that functions $f_a$ are non-decreasing, then $f$ is monotone. | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Precalculus > Functions"
] | null | proof only | null | |
07c3 | Let $f_1 : \mathbb{N} \to \mathbb{N}$ be a function. Function $f_2$ is constructed from $f_1$ such that for each $k \in \mathbb{N}$, $f_2(k)$ is the number of times $k$ has appeared in the range of $f_1$, i.e. $f_2(k)$ is the number of natural numbers $n$ for which $f_1(n) = k$.
Suppose that the range of $f_1$ is such ... | [
"a) Consider a table with $T+1$ rows and an infinite number of columns (towards right) such that the first column is filled by numbers $1, 2, \\dots, T$ and $1$. An algorithm is introduced for filling the cells of the table, then $f_i(j)$ is defined as the value of the intersection cell of the $i^{th}$ row and the ... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Algorithms",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | For every positive integer T, there exists a starting function whose iterates are all definable and form a cycle of period T. For the second question, yes, there exists a starting function such that all iterates are definable and pairwise distinct. | |
0bla | Let $n \in \mathbb{N}^*$. Prove that
$$
\frac{a}{(n+1)a+b+c} + \frac{b}{a+(n+1)b+c} + \frac{c}{a+b+(n+1)c} \le \frac{1}{2}\sqrt{\frac{3}{n}},
$$
for every $a, b, c \in (0, \infty)$. Find the cases of equality. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds if and only if n = 3 and a = b = c. | |
0985 | Problem:
La centrul „Anticovid” din localitatea $N$ au fost aduse vaccinuri de două tipuri: A și B. Vaccinurile A erau ambalate, în mod egal, în 9 cutii. Vaccinurile B erau ambalate, de asemenea în mod egal, în 14 cutii. În total au fost aduse 363 vaccinuri. Câte vaccinuri de fiecare tip au fost aduse în localitatea $N... | [
"Solution:\nFie $x \\in N^{*}$ numărul vaccinurilor $A$ dintr-o cutie, iar $y \\in N^{*}$ numărul vaccinurilor $B$ dintr-o cutie. Conform condiției, $9x + 14y = 363$. Evident, $y$ se divide la 3; fie $y = 3z$, $z \\in N^{*}$. Ecuația capătă forma\n$$\n3x + 14z = 121\n$$\nși urmează a fi rezolvată în numere naturale... | Moldova | Olimpiada Republicană la Matematică | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 279 vaccines of type A and 84 vaccines of type B | |
08t4 | (1) Find the value of $S(30)$ (you don't need to give a proof.)
(2) Determine all $n$ for which $S(n)$ is a prime.
Here, by a prime we mean an integer greater than or equal to 2, which has no factor other than 1 and itself.
For a positive integer $n$. Let $S(n)$ be the sum of all those integers $k$, $1 \le k \le n$,... | [
"(1): Since $30 = 2 \\times 3 \\times 5$, a positive integer relatively prime to $30$ is not a multiple of $2$, $3$, or $5$. Hence integers lying in between $1$ and $30$ and relatively prime to $30$ are $1, 7, 11, 13, 17, 19, 23, 29$. Therefore,\n$$\nS(30) = 1 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 120\n$$\n\n(2): Fir... | Japan | Japan Junior Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | S(30) = 120; and the only n with S(n) prime is n = 3. | |
0k9q | Problem:
Let $ABC$ be a triangle with $AB = 13$, $BC = 14$, $CA = 15$. Let $H$ be the orthocenter of $ABC$. Find the radius of the circle with nonzero radius tangent to the circumcircles of $AHB$, $BHC$, $CHA$. | [
"Solution:\nWe claim that the circle in question is the circumcircle of the anticomplementary triangle of $ABC$, the triangle for which $ABC$ is the medial triangle.\n\nLet $A'B'C'$ be the anticomplementary triangle of $ABC$, such that $A$ is the midpoint of $B'C'$, $B$ is the midpoint of $A'C'$, and $C$ is the mid... | United States | HMMT February 2019 February 16, 2019 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"... | null | proof and answer | 65/4 | |
0cl4 | Let $n \ge 3$ be a natural number. Ana chooses the non-zero natural numbers $a_1, a_2, \dots, a_n$, and for each non-empty subset $A \subset \{1, 2, \dots, n\}$, she computes the sum $s_A = \sum_{k \in A} a_k$, then arranges these sums in increasing order, obtaining the sequence $s_1 \le s_2 \le \dots \le s_{2^{n}-1}$.... | [
"Assume $a_1 \\le a_2 \\le \\dots \\le a_n$. If we choose the subset $B' = \\{1, 2, \\dots, 2^{n-2}\\}$, then among the sums $s_1, s_2, \\dots, s_{2^{n-2}}$, the values $a_1, a_2, \\dots, a_t$ will appear for some $t \\ge 1$. Clearly, $s_1 = s_{\\{1\\}} = a_1$, and if $a_i$ appears as a sum, then $a_{i-1}$ must hav... | Romania | 75th NMO Selection Tests | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
08ki | Problem:
A point $O$ and the circles $k_{1}$ with center $O$ and radius $3$, $k_{2}$ with center $O$ and radius $5$, are given. Let $A$ be a point on $k_{1}$ and $B$ be a point on $k_{2}$. If $ABC$ is an equilateral triangle, find the maximum value of the distance $OC$. | [
"Solution:\n\nIt is easy to see that the points $O$ and $C$ must be in different semi-planes with respect to the line $AB$.\n\nLet $OPB$ be an equilateral triangle ($P$ and $C$ on the same side of $OB$). Since $\\angle PBC = 60^{\\circ} - \\angle ABP$ and $\\angle OBA = 60^{\\circ} - \\angle ABP$, then $\\angle PBC... | JBMO | OJBM | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geo... | null | proof and answer | 8 | |
0ks7 | Problem:
Aerith writes a positive integer in each cell of a $2021 \times 2021$ grid. Every second, Bob will pick a cell with value at least five if one exists, decrease its value by four, and increment each of the cell's neighbors by $1$. Must this process always stop? | [
"Solution:\n\nYes. Let the sum of all numbers in the grid be $S$. Note that $S$ never increases, as each cell has at most $4$ neighbors; in fact, it decreases when Bob picks a cell on the boundary. There are finitely many grids with total sum $\\leq S$, so it suffices to show that Bob cannot encounter the same grid... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Yes | |
02hz | Problem:
Se $x + |x| + y = 5$ e $x + |y| - y = 6$ então $x + y$ é:
(A) -1
(B) 11
(C) $\frac{9}{5}$
(D) 1
(E) -11 | [
"Solution:\n\nSe $x < 0$, então $|x| = -x$ e da $1^{a}$ equação temos $x + (-x) + y = 5 \\Rightarrow y = 5$. Substituindo esse valor na $2^{a}$ equação obtemos $x = 6$, o que não é possível pois estamos supondo $x < 0$. Logo, não há solução para $x < 0$.\n\nSe $y \\geq 0$, então $|y| = y$ e da $2^{a}$ equação segue... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C |
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