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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
08er | Problem:
Date due frazioni $a / b$ e $c / d$, definiamo la loro somma pirata come
$$
\frac{a}{b} \diamond \frac{c}{d}=\frac{a+c}{b+d}
$$
dove si intende che le due frazioni iniziali sono ridotte ai minimi termini (cioè semplificate il più possibile), ed anche il risultato viene poi ridotto ai minimi termini. Così, per... | [
"Solution:\n\nPoniamo $M_{n}=1 / 2$ e $m_{n}=1 /(n-1)$. Dimostriamo che $M_{n}$ è il massimo valore possibile per l'ultima frazione, mentre $m_{n}$ è il minimo valore possibile per l'ultima frazione. La dimostrazione consiste di due parti: una costruzione che mostra che tali valori sono effettivamente realizzabili,... | Italy | XXXVII Olimpiade Italiana di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Maximum = 1/2, Minimum = 1/(n−1) | |
0871 | Problem:
È dato un triangolo $A B C$, rettangolo in $A$ e con $A C$ cateto maggiore; sia $M$ il punto medio di $B C$, $N$ il simmetrico di $A$ rispetto a $B C$, $O$ l'intersezione fra la perpendicolare ad $M N$ passante per $N$ e la retta contenente $B C$.
a) Dimostrare che l'angolo $O M N$ è il doppio dell'angolo $A... | [
"Solution:\n\nIl punto medio $M$ dell'ipotenusa $B C$ è anche il centro della circonferenza circoscritta al triangolo $A B C$; l'angolo al centro $A \\widehat{M} B$ e l'angolo alla circonferenza $A \\widehat{C} B$ insistono sullo stesso arco, da cui $A \\widehat{M} B = 2 A \\widehat{C} B$. Inoltre $A \\widehat{M} B... | Italy | Progetto Olimpiadi di Matematica | [
"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 > Distance chasing"
] | null | proof only | null | |
008z | There are several positive integers smaller than $200$ written on a blackboard such that none of them divides the smallest common multiple of the rest of them. Determine the maximum amount of numbers that can be written on the blackboard. | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 46 | |
035v | Problem:
Solve the equation
$$
\left|\left| x-\frac{5}{2}\right|-\frac{3}{2}\right|=\left|x^{2}-5 x+4\right|
$$ | [
"Solution:\nSince both sides of the equation are non-negative, it is equivalent to\n$$\n\\left(\\left|x-\\frac{5}{2}\\right|-\\frac{3}{2}\\right)^{2}=\\left(x^{2}-5 x+4\\right)^{2}\n$$\nHence\n$$\n\\left(\\left|x-\\frac{5}{2}\\right|-\\frac{3}{2}-x^{2}+5 x-4\\right)\\left(\\left|x-\\frac{5}{2}\\right|-\\frac{3}{2}+... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 1 and 4 | |
02o7 | Prove that if $10^{2n} + 8 \cdot 10^n + 1$ has a prime factor of the form $60k + 7$, $k$ and $n$ both positive integers, then $n$ and $k$ are both even. | [
"Let $p = 60k + 7$ be such a prime. Then $10^{2n} + 8 \\cdot 10^n + 1 \\equiv 0 \\pmod{p} \\iff (10^n - 1)^2 \\equiv -10^{n+1} \\pmod{p}$. Now suppose $n$ is odd. Then $(10^n - 1)^2 \\equiv -1 \\cdot (10^{(n+1)/2})^2 \\pmod{p}$, and $-1$ is a quadratic residue. But by the Euler criterion, $\\left(\\frac{-1}{p}\\rig... | Brazil | Brazilian Math Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
05vp | Problem:
Soit $p$ et $q$ deux nombres premiers distincts, tels que $p < 2q$ et $q < 2p$. Démontrer qu'il existe deux entiers consécutifs dont l'un a $p$ pour plus grand facteur premier et l'autre a $q$ pour plus grand facteur premier. | [
"Solution:\n\nSupposons sans perte de généralité que $p < q$. Le théorème de Bézout indique qu'il existe des entiers $a$ et $b$ tels que $a p + b q = 1$.\n\nOn peut toujours remplacer un tel couple $(a, b)$ par $(a \\pm q, b \\mp p)$. Par conséquent, si on cherche un couple pour lequel la valeur de $|a|$ est minima... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0h3a | A function $f : \mathbb{R} \to \mathbb{R}$ is such that for all $x, y \in \mathbb{R}$
$$
f(x + 2xy) = f(x) + 2f(xy).
$$
Given that $f(2011) = 2012$, find $f(2012)$. | [
"Підставимо $x = 0$ і одержимо, що $f(0) = 0$. Підставляючи до нашого функціонального рівняння $y = -1$ та $y = -\\frac{1}{2}$, одержимо, що $f(-x) = -f(x)$ та $f(x) = 2f(\\frac{x}{2})$ для будь-яких дійсних $x$. Далі, підставимо $y = \\frac{z}{2x}$, де $x \\neq 0$, і одержимо, що $f(x+z) = f(x) + f(z)$. Оскільки $... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | 2012^2/2011 | |
0512 | The teacher drew a $3 \times 3$ table in Juku's exercise book and wrote a number in every position of the table. Then he gave Juku the following task.
1. Turn the next page and draw a similar table. Write in the first row the numbers obtained by subtracting the numbers in the third row of the corresponding column from... | [
"First note that after step 1 we get a table where the column sums of the table are $0$, and after step 2 we get a table where the row sums of the table are $0$. Suppose that after some steps we reach the table with all zeroes in it. By symmetry we can consider the case where we get this table after step 2. Then th... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Linear transformations"
] | null | proof only | null | |
0iuz | Problem:
Given that $\sin A + \sin B = 1$ and $\cos A + \cos B = 3/2$, what is the value of $\cos (A - B)$? | [
"Solution:\n\nSquaring both equations and adding them together, one obtains\n$$\n(\\sin A + \\sin B)^2 + (\\cos A + \\cos B)^2 = 1^2 + (3/2)^2 = 1 + 9/4 = 13/4.\n$$\nBut\n$$\n(\\sin A + \\sin B)^2 + (\\cos A + \\cos B)^2 = (\\sin^2 A + 2 \\sin A \\sin B + \\sin^2 B) + (\\cos^2 A + 2 \\cos A \\cos B + \\cos^2 B)\n$$... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 5/8 | |
02z1 | Problem:
Os triângulos $A B C$ e $A B D$ estão inscritos na mesma semicircunferência de diâmetro $A B$, que mede $15~\mathrm{cm}$. Se traça por $D$ a perpendicular a $A B$ que intersecta $A B$ em $P$, o segmento $A C$ em $Q$ e o prolongamento do lado $B C$ em $R$. Além disso, $P R=\frac{40}{3}~\mathrm{cm}$ e $P Q=\fra... | [
"Solution:\n\n\n\na) Como $A B$ é um diâmetro, segue que $\\angle A C B=90^{\\circ}$. Daí\n$$\n\\angle P R B=90^{\\circ}-\\angle P B R=\\angle B A C\n$$\nOs triângulos $A P Q$ e $P R B$ são semelhantes, pois possuem ângulos de mesmas medidas. Portanto,\n$$\n\\begin{aligned}\n\\frac{P Q}{A P... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 5*sqrt(2) | |
0hdl | Let $a_1 < a_2 < \dots < a_k$ be positive integers. For each $a_i$ Zarina has written down all the positive divisors of $a_i$ in the notebook (some numbers might be written several times). Then, Marina split all the numbers in the notebook into several groups. It turned out that the numbers in each of these groups form... | [
"Notice that number $a_k$ is written only once in the notebook. Then there exists some Marina's group that contains all divisors of $a_k$. Therefore $a_k$ is also on the desk. Then we cross out from the notebook all divisors of $a_k$ one time. Now, by similar thoughts it follows that $a_{k-1}$ is also on the desk, ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0exu | Problem:
Can both $x^2 + y$ and $x + y^2$ be squares for $x$ and $y$ natural numbers? | [
"Solution:\nNo. The smallest square greater than $x^2$ is $(x + 1)^2$, so we must have $y > 2x$. Similarly $x > 2y$. Contradiction."
] | Soviet Union | 6th ASU | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | No | |
0k5b | Problem:
Assume the quartic $x^{4}-a x^{3}+b x^{2}-a x+d=0$ has four real roots $\frac{1}{2} \leq x_{1}, x_{2}, x_{3}, x_{4} \leq 2$. Find the maximum possible value of $\frac{\left(x_{1}+x_{2}\right)\left(x_{1}+x_{3}\right) x_{4}}{\left(x_{4}+x_{2}\right)\left(x_{4}+x_{3}\right) x_{1}}$ (over all valid choices of $a, ... | [
"Solution:\nWe can rewrite the expression as\n$$\n\\begin{gathered}\n\\frac{x_{4}^{2}}{x_{1}^{2}} \\cdot \\frac{\\left(x_{1}+x_{1}\\right)\\left(x_{1}+x_{2}\\right)\\left(x_{1}+x_{3}\\right)\\left(x_{1}+x_{4}\\right)}{\\left(x_{4}+x_{1}\\right)\\left(x_{4}+x_{2}\\right)\\left(x_{4}+x_{3}\\right)\\left(x_{4}+x_{4}\\... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 5/4 | |
00jv | Determine all integers $x$ such that
$$
\left\lfloor \frac{x}{2} \right\rfloor \cdot \left\lfloor \frac{x}{3} \right\rfloor \cdot \left\lfloor \frac{x}{4} \right\rfloor = x^2
$$
holds. (Note that $\lfloor y \rfloor$ is the largest integer not greater than $y$.) | [
"Since $x^2 \\ge 0$ certainly holds, we must have $x \\ge 0$. $x = 0$ is obviously a solution. We now assume $x > 0$. Since $\\lfloor y \\rfloor \\le y$, we certainly have\n$$\nx^2 \\le \\frac{x}{2} \\cdot \\frac{x}{3} \\cdot \\frac{x}{4} = \\frac{x^3}{24} \\Rightarrow 24 \\le x.\n$$\nFurthermore, since $\\left\\lf... | Austria | AustriaMO2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 0 and 24 | |
0de0 | For a positive integer $k$, denote by $f(k)$ the number of positive integers $m$ such that the remainder of $k m$ modulo $2019^3$ is greater than $m$. Find the amount of different numbers among $f(1), f(2), \ldots, f(2019^3)$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 25 | |
0b5m | For real numbers $x, y, z \in (0, 1)$, with $xyz = (1-x)(1-y)(1-z)$, show that at least one of the numbers $(1-x)y, (1-y)z, (1-z)x$ is greater than or equal to $\frac{1}{4}$. | [
"From $x, y, z \\in (0, 1)$ it follows that also $1-x, 1-y, 1-z \\in (0, 1)$. We have\n$$\n\\sum (1-x)y = \\sum x - \\sum xy,\n$$\nwhile\n$$\n\\prod x = \\prod (1-x) \\text{ translates into } (\\sum x - \\sum xy) + 2xyz = 1, \\text{ hence } \\sum (1-x)y + 2\\sqrt{\\prod (1-x)y} = 1 = 3 \\cdot \\frac{1}{4} + 2\\sqrt... | Romania | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
00sk | Let $ABCD$ be a square of center $O$ and let $M$ be the symmetric of the point $B$ with respect to the point $A$. Let $E$ be the intersection of $CM$ and $BD$, and let $S$ be the intersection of $MO$ and $AE$. Show that $SO$ is the angle bisector of $\angle ESB$. | [
"We have\n$$\n\\begin{cases}\nDC \\equiv DA \\\\\n\\angle EDC \\equiv \\angle EDA \\\\\nDE \\equiv DE\n\\end{cases} \\Rightarrow \\triangle DEC \\equiv \\triangle DEA \\Rightarrow \\angle DAE \\equiv \\angle DCE (*).\n$$\nLet $CM \\cap AD = \\{P\\}$, then follows $\\triangle CDP \\equiv \\triangle BAP$ and $\\angle... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ixe | Problem:
In how many distinct ways can you color each of the vertices of a tetrahedron either red, blue, or green such that no face has all three vertices the same color? (Two colorings are considered the same if one coloring can be rotated in three dimensions to obtain the other.) | [
"Solution:\n\nIf only two colors are used, there is only one possible arrangement up to rotation, so this gives 3 possibilities. If all three colors are used, then one is used twice. There are 3 ways to choose the color that is used twice. Say this color is red. Then the red vertices are on a common edge, and the g... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Solid Geometry > Other 3D problems"
] | null | final answer only | 6 | |
0bo3 | Let $ABCDA'B'C'D'$ be a cuboid and $AB' \cap A'B = \{O\}$. Let $N$ be a point on the edge $[BC]$ so that $A'C \parallel (B'AN)$. It is known that $D'O \perp (B'AN)$. Prove that $ABCDA'B'C'D'$ is a cube.
Valeriu Bărbieru | [
"From $D'A' \\perp (ABB')$ and $D'O \\perp AB'$ follows $A'O \\perp AB'$, so $ABB'A'$ is a square. Then $A'C \\parallel (ANB')$ and $(ANB') \\cap (A'BC) = ON$, hence $A'C \\parallel ON$. Since $O$ is the midpoint of the segment $[A'B]$, $N$ is the midpoint of the segment $[BC]$. Rectangle $A'BCD'$ has $D'O \\perp O... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
0ae3 | The numbers $1$, $2$, $\ldots$, $2009$ are written on a board. Some of them are erased and the remainder of their sum divided with $13$ is written on the board. After a finite number of repetition of the above procedure only three numbers have left, two of which are $99$ and $999$. What is the third number? | [
"Let $x$ be the third number. After every procedure the remainder of the sum of the numbers on the board divided with $13$ is unchanged.\n\n$$1+2+3+\\ldots+2009 = \\frac{2009 \\cdot 2010}{2} = 1005 \\cdot 2009$$\n\ndivided with $13$ has remainder $2$.\n\nHence $99+999+x$ divided with $13$ has remainder $2$. Now $99... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | Macedonian, English | proof and answer | 9 | |
0efr | Problem:
Določi vse vrednosti, ki jih zavzame izraz $A = n \cdot \left\lfloor \frac{2017}{n} \right\rfloor$, ko $n$ preteče vsa naravna števila. Pri tem $\lfloor x \rfloor$ označuje največje celo število, ki ni večje od $x$. | [
"Solution:\n\nOznačimo z $m < n$ ostanek števila $2017$ pri deljenju z $n$. Tedaj lahko število $2017$ zapišemo v obliki $2017 = k n + m$, kjer je $k$ nenegativno celo število. Tedaj je\n$$\nA = n \\cdot \\left\\lfloor \\frac{2017}{n} \\right\\rfloor = n k\n$$\nkar je zagotovo celo število. Če je $k = 0$, je $A = 0... | Slovenia | Slovenian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Other"
] | null | proof and answer | 0 and all integers from 1009 through 2017 inclusive | |
0d6k | Let $ABC$ be an acute, non-isosceles triangle which is inscribed in a circle $(O)$. A point $I$ belongs to the segment $BC$. Denote by $H$ and $K$ the projections of $I$ on $AB$ and $AC$, respectively. Suppose that the line $HK$ intersects $(O)$ at $M, N$ ($H$ is between $M, K$ and $K$ is between $H, N$). Prove the fol... | [
"See the solution to Problem 1 in the test of level 4+. $\\square$"
] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
0e2j | Problem:
Na nogometni tekmi vratar brcne žogo. Pot žoge je opisana s funkcijo
$$
h(x) = -0,0126 x^{2} + 0,635 x
$$
kjer je $h$ višina žoge nad zemljo in $x$ vodoravna oddaljenost od mesta udarca (količini sta izraženi v merskih številih).
a) Na kolikšni višini je žoga, ko je njena vodoravna... | [
"Solution:\n\na) Vstavimo vrednost $x = 15$ in izračunamo:\n$$\nh(15) = -0,0126 \\cdot 15^{2} + 0,635 \\cdot 15 \\doteq 6,69~\\mathrm{m}\n$$\n\nb) Nastavimo enačbo $h(x) = 0$. Izračunamo rešitvi $x = 0$ in $x = 50,4~\\mathrm{m}$ ter izločimo $x = 0$.\n\nc) Ugotovimo, da je najvišja višina žoge enaka ordinati temena... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | a) 6.69 m; b) 50.4 m; c) 8 m | |
0d2s | Find the largest integer $k$ such that $k$ divides $n^{55} - n$ for all integer $n$. | [
"Let $p$ be a prime divisor of $n^{55} - n$ for all integer $n$. Whenever $n$ is not divisible with $p$, we have\n$$\nn^{54} \\equiv 1 \\quad \\bmod p\n$$\nIn this case, the order of $n$ modulo $p$ divides $54$. But there exists an integer $n$ of order $p-1$ modulo $p$. We deduce that $p-1$ divides $54$. But the on... | Saudi Arabia | Selection tests for the Gulf Mathematical Olympiad 2013 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 798 | |
04v3 | Consider any graph with $50$ vertices and $225$ edges. We say that a triplet of its (mutually distinct) vertices is *connected* if the three vertices determine at least two edges. Determine the smallest and the largest possible number of connected triples. | [] | Czech Republic | Final Round | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | minimum 600, maximum 5400 | |
06es | Let $x$, $y$, $z$ be positive real numbers such that $x + y + z = 1$. For positive integer $n$, define $S_n = x^n + y^n + z^n$. Furthermore, let $P = S_2 S_{2005}$ and $Q = S_3 S_{2004}$.
a. Find the smallest possible value of $Q$.
b. If $x$, $y$, $z$ are pairwise distinct, determine whether $P$ or $Q$ is larger. | [
"a. The smallest possible value of $Q$ is $\\frac{1}{3^{2005}}$.\nBy the power mean inequality, we have\n$$\nS_n \\ge 3 \\left(\\frac{S_1}{3}\\right)^n = \\frac{1}{3^{n-1}}\n$$\nfor any positive integer $n$. Therefore, we have\n$$\nQ = S_3 S_{2004} \\ge \\frac{1}{3^2} \\cdot \\frac{1}{3^{2003}} = \\frac{1}{3^{2005}... | Hong Kong | IMO HK TST | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof and answer | a) 1/3^{2005}. b) P > Q for pairwise distinct x, y, z. | |
06zw | Problem:
If $P$ and $Q$ are two points in the plane, let $m(PQ)$ be the perpendicular bisector of $PQ$. $S$ is a finite set of $n > 1$ points such that:
(1) if $P$ and $Q$ belong to $S$, then some point of $m(PQ)$ belongs to $S$,
(2) if $PQ$, $P'Q'$, $P''Q''$ are three distinct segments, whose endpoints are all in $... | [
"Solution:\n\nThere are $n(n-1)/2$ pairs of points. Each has a point of $S$ on its bisector. But each point of $S$ is on at most two bisectors, so $2n \\geq n(n-1)/2$. Hence $n \\leq 5$.\n\nThe equilateral triangle and regular pentagon show that $n = 3, 5$ are possible.\n\nConsider $n = 4$. There are 6 pairs of poi... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 3 and 5 | |
07fy | We call a monic polynomial $P(x) \in \mathbb{Z}[x]$ square-free mod $n$ if there are no polynomials $Q(x), R(x) \in \mathbb{Z}[x]$ such that $Q$ being non-constant and
$$
P(x) \equiv Q(x)^2 R(x) \pmod{n}.
$$
Given a prime $p$ and integer $m \ge 2$. Find the number of monic square-free mod $p$ polynomials $P(x)$ with de... | [
"The answer is $1$ for $m = 0$, $p$ for $m = 1$ and $\\varphi(p^m)$ for $m > 1$.\n\nNote that $\\mathbb{Z}_p[x]$ is a unique factorization domain. So any monic $P(x) \\in \\mathbb{Z}_p[x]$ can be uniquely expressed as $P(x) \\equiv A(x)^2 Q(x) \\pmod{p}$, where $A(x), Q(x)$ are monic polynomials and $Q(x)$ is squar... | Iran | 37th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Discrete Mathematics > Combinatorics > In... | English | proof and answer | φ(p^m) | |
05d3 | A triangle $ABC$ with $AB < BC$ and an obtuse angle at vertex $B$ is given. The incircle of the triangle $ABC$ with incentre $I$ touches the sides $BC$, $CA$ and $AB$ at points $D$, $E$ and $F$, respectively. The line $AI$ intersects the side $BC$ at point $K$. The ray $IB$ intersects the circumcircle of the triangle $... | [
"\nFig. 52\nSolution:\nLet $\\angle BAC = 2\\alpha$, $\\angle CBA = 2\\beta$, $\\angle ACB = 2\\gamma$; then $\\alpha + \\beta + \\gamma = 90^\\circ$. By the two tangents theorem, we have $AE = AF$ (Fig. 52), and because $\\angle FAK = \\alpha = \\angle KAE$, the triangles $AEK$ and $AFK$ a... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05g9 | Problem:
Soit $a, b, c$ des nombres réels. Montrer que:
$$
2 a^{2}+20 b^{2}+5 c^{2}+8 a b-4 b c-4 a c \geqslant 0
$$
et trouver les cas d'égalité. | [
"Solution:\n\nL'inégalité est équivalente à $4 b c+4 a c-8 a b \\leqslant 2 a^{2}+20 b^{2}+5 c^{2}$. Or on sait que $4 a c=2 \\times a \\times(2 c) \\leqslant a^{2}+(2 c)^{2}=a^{2}+4 c^{2}$ et que $4 b c=2 \\times c(2 b) \\leqslant 4 b^{2}+c^{2}$. On a également $-8 a b=2 \\times(-a) \\times(4 b) \\leqslant a^{2}+1... | France | ENVOI 2 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | Equality holds only for a = b = c = 0. | |
078o | Let $\triangle ABC$ be an acute angled triangle with $AC > AB$ and incircle $\omega$. Let $\omega$ touch the sides $BC, CA$, and $AB$ at $D, E$, and $F$ respectively. Let $X$ and $Y$ be points outside $\triangle ABC$ satisfying
$$
\angle BDX = \angle XEA = \angle YDC = \angle AFY = 45^\circ.
$$
Prove that the circumcir... | [
"\nWe have that $AB \\neq AC$, so $(AEF)$ and $(ABC)$ are not tangent at $A$, thus there is a point $S \\neq A$ which is the second intersection of $(AEF)$ and $(ABC)$.\n\nConsider an inversion about the incircle and let the inverse of a point $P$ be denoted by $P'$. Note that $A'$ is the m... | India | IMO TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Spiral s... | null | proof only | null | |
0fah | Problem:
An investigator works out that he needs to ask at most $91$ questions on the basis that all the answers will be yes or no and all will be true. The questions may depend upon the earlier answers. Show that he can make do with $105$ questions if at most one answer could be a lie. | [
"Solution:\n\nSuppose he asks $n$ questions as usual, and then asks \"did you lie to any of the last $n$ questions?\" If the reply is a truthful no, then the $n$ answers were correct. If the reply is a lying no, then the $n$ answers were still correct. On the other hand if the answer is yes, then the $n$ answers mi... | Soviet Union | 25th ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | 105 | |
0it6 | Problem:
Determine the number of juggling sequences of length $n$ with exactly $1$ ball. | [
"Solution:\nWith $1$ ball, we simply need to decide at which times the ball should land in our hand. That is, we need to choose a non-empty subset of $\\{0, 1, 2, \\ldots, n-1\\}$ where the ball lands. It follows that the answer is $2^{n} - 1$."
] | United States | 11th Annual Harvard-MIT Mathematics Tournament - Team Round: B Division | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2^n - 1 | |
0bpy | Problem:
Fie unghiurile adiacente suplementare $\Varangle A O B$ și $\Varangle B O C$ astfel încât raportul măsurilor să fie $\frac{1}{4}$. Fie $[O D$ semidreapta opusă bisectoarei unghiului $\Varangle B O C$. În interiorul unghiului $\Varangle C O D$ se consideră punctele $M$ și $N$ astfel încât $m(\Varangle C O N)=m... | [] | Romania | OLIMPIADA DE MATEMATICĂ - ETAPA LOCALĂ | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | m∠COD = 108°, and points B, O, M are collinear. | |
0iw4 | Problem:
Find all pairs of integer solutions $(n, m)$ to
$$
2^{3^{n}} = 3^{2^{m}} - 1
$$ | [
"Solution:\nWe find all solutions of $2^{x} = 3^{y} - 1$ for positive integers $x$ and $y$. If $x = 1$, we obtain the solution $x = 1, y = 1$, which corresponds to $(n, m) = (0, 0)$ in the original problem. If $x > 1$, consider the equation modulo $4$. The left hand side is $0$, and the right hand side is $(-1)^{y}... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | [(0, 0), (1, 1)] | |
0ks1 | Problem:
Aerith writes $100$ positive numbers on a blackboard. Every minute, Bob either replaces one number $x$ with $\frac{1}{x}$, or replaces two numbers $x, y$ with $\frac{x y+1}{x+y}$. Given that after $2021$ minutes, there is only one number left, show that this number only depends on Aerith's initial numbers. | [
"Solution:\n\nLet $x \\star y = \\frac{x y + 1}{x + y}$. One can check that $x \\star \\frac{1}{y} = \\frac{1}{x \\star y}$. Because of the latter property, Bob would get the same number by doing all $\\star$ operations first and doing all multiplicative inverses last. Every $\\star$ decreases the total number of n... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0bv5 | Problem:
Fie $G$ un grup și $x, y \in G$ cu proprietatea că ord $(x)=3$, $y^{4}=e$, $x y=y^{3} x$. Să se arate că dacă $y \in G \setminus \{e\}$ atunci ord $(y)=2$ și $x y=y x$. | [] | Romania | Olimpiada de Matematică Etapa Locală | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
0j8i | Problem:
Find all integers $x$ such that $2 x^{2}+x-6$ is a positive integral power of a prime positive integer. | [
"Solution:\n\nAnswer: $-3, 2, 5$\n\nLet $f(x) = 2 x^{2} + x - 6 = (2x - 3)(x + 2)$.\n\nSuppose a positive integer $a$ divides both $2x - 3$ and $x + 2$. Then $a$ must also divide $2(x + 2) - (2x - 3) = 7$. Hence, $a$ can either be $1$ or $7$.\n\nAs a result, $2x - 3 = 7^{n}$ or $-7^{n}$ for some positive integer $n... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | -3, 2, 5 | |
0270 | Problem:
Considere 2018 pontos em uma circunferência de raio 1. Verifique que existe um ponto $P$ da circunferência para o qual a soma das distâncias de $P$ aos 2018 pontos é pelo menos 2018. | [
"Solution:\nEscolha dois pontos $A$ e $B$ diametralmente opostos. Dado qualquer ponto $Q$ no círculo, temos $d(Q, A) + d(Q, B) > 2$, em que $d(X, Y)$ denota a distância entre $X$ e $Y$.\n\nConsidere as somas $S_{A}$ e $S_{B}$ das distâncias de todos os 2018 pontos até $A$ e $B$, respectivamente. Como $S_{A} + S_{B}... | Brazil | null | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ep4 | Let $f$ be a function satisfying
$$
f(xy) = \frac{f(x)}{y}
$$
for all positive real numbers $x$ and $y$. If $f(500) = 3$, what is the value of $f(100)$? | [
"We are given $3 = f(500) = f(100 \\times 5) = \\frac{f(100)}{5}$, so $f(100) = 3 \\times 5 = 15$."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | 15 | |
0gs6 | Let $n$ be a positive integer with $b$ digits and $l, r$ be non-negative integers satisfying $l + r < b$. We say that a positive integer number is a sub-divisor of $n$, if it divides the number obtained by erasing the first $l$ and last $r$ digits of $n$. (For example, sub-divisors of $143$ are $1$, $2$, $3$, $4$, $7$,... | [
"Answer: All positive integers which are coprime with $10$.\nIf a number is divisible by either $2$ or $5$ then it can not divide any number of the form $11\\ldots1$. Therefore, for all numbers which are not coprime with $10$ the set $A_d$ is not finite.\n\nNow let $d$ be a positive integer which is coprime with $1... | Turkey | Team Selection Test for IMO 2019 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | All positive integers coprime with 10. | |
0dyi | If $a(b + c) + b(c + a) + c(a + b) = ab + bc + ca$, then
$$
\frac{a^2(b+c) + b^2(a+c) + c^2(a+b)}{abc}
$$
is an integer. | [
"The given equation implies $ab + bc + ca = 0$, so we can write\n$$\n\\frac{a^2(b+c) + b^2(a+c) + c^2(a+b)}{abc} = \\frac{a(ab+ac) + b(ba+bc) + c(ca+cb)}{abc}\n$$\n\nSince $ab + ac = -bc$, $ba + bc = -ca$ and $ca + cb = -ab$ we have\n$$\n\\frac{a(ab + ac) + b(ba + bc) + c(ca + cb)}{abc} = \\frac{a(-bc) + b(-ca) + c... | Slovenia | Slovenija 2008 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0flg | Problem:
Sea $ABC$ un triángulo con $\angle B = 2 \angle C$ y $\angle A > 90^{\circ}$. Sean $D$ el punto de la recta $AB$ tal que $CD$ es perpendicular a $AC$, y $M$ el punto medio de $BC$. Demuestra que $\angle AMB = \angle DMC$. | [
"Solution:\n\nLa recta que pasa por $A$ y es paralela a $BC$ corta a $DM$ y a $DC$ en los puntos $N$ y $F$ respectivamente.\n\nSe sigue que $AN : BM = DN : DM = NF : MC$.\n\nPues $BM = MC$, resulta $AN = NF$. Como $\\angle ACF = 90^{\\circ}$, tenemos $AN = NC$, de manera que $\\angle NCA = \\angle NAC = \\angle ACB... | Spain | XLVII Olimpiada Matemática Española, Fase nacional (Pamplona) | [
"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"
] | null | proof only | null | |
0dr9 | Determine the largest odd positive integer $N$ such that every odd integer $k$ with $1 < k < N$ and $(k, N) = 1$ is a prime. | [
"The largest such integer $N$ is $105$. Let $p_i$ denote the $i$th prime, i.e. $p_1 = 2$, $p_2 = 3$, etc. We call a positive integer $N$ *admissible* if it is odd and has the property stated in the problem. Any admissible number exceeding $p_i^2$ must clearly contain the factor $p_i$ for $i \\ge 2$. By Bertrand's p... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 105 | |
0gh9 | Let $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. Points $D, E, F$ are the midpoints of the inferior arcs $BC, CA, AB$, respectively, on $\Omega$. Let $G$ be the point diametrically opposed to $D$ on $\Omega$. Let $X$ be the intersection of lines $GE$ and $AB$, while $Y$ the intersection of lines $FG$... | [
"令 $I$ 為 $\\triangle ABC$ 的內心, 我們有 $\\angle IFG = \\angle CBG = 90^\\circ - \\frac{1}{2}\\angle A = \\angle CIE$, 即 $FG \\parallel IE$。同理有 $GE \\parallel IF$。我們證明 $BY$ 與 $CX$ 的交點 $W$ 位於 $\\Omega$ 上: 考慮折線 $BWCFGE$, 由帕斯卡定理知 $W$ 位於 $\\Omega$ 上等價於 $X, Y, I$ 共線, 而這是因為\n$$\nA(X,Y; I, G) = (B,C; D, G) = -1 = G(E,F; I, \\i... | Taiwan | 2023 數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometr... | Chinese (Traditional) | proof only | null | |
04su | A strange calculator has only two buttons with positive integers, each consisting of two digits. It displays the number $1$ at the beginning. Whenever a button with number $N$ is pressed, the calculator replaces the displayed number $X$ with the number $X \cdot N$ or $X + N$. Multiplication and addition alternate, mult... | [
"Let $a, b$ be the numbers written on the buttons. Consider the sequence $(x_n)_{n=0}^{\\infty}$ such that $x_{n+1}$ is formed by the last four digits of $a(x_n + b)$ for each $n \\ge 0$, that is,\n$$\nx_{n+1} \\equiv a(x_n + b) \\pmod{10\\,000} \\quad \\text{and} \\quad 0 \\le x_{n+1} < 10\\,000.\n$$\nSince there ... | Czech Republic | 15th Czech-Polish-Slovak Mathematics Competition | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | (a) yes; (b) yes | |
0io0 | Problem:
Triangle $ABC$ has $\angle A = 90^{\circ}$, side $BC = 25$, $AB > AC$, and area $150$. Circle $\omega$ is inscribed in $ABC$, with $M$ its point of tangency on $AC$. Line $BM$ meets $\omega$ a second time at point $L$. Find the length of segment $BL$. | [
"Solution:\n\nLet $D$ be the foot of the altitude from $A$ to side $BC$. The length of $AD$ is $2 \\cdot 150 / 25 = 12$.\n\nTriangles $ADC$ and $BDA$ are similar, so $CD \\cdot DB = AD^2 = 144 \\Rightarrow BD = 16$ and $CD = 9 \\Rightarrow AB = 20$ and $AC = 15$.\n\nUsing equal tangents or the formula inradius as a... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 45*sqrt(17)/17 | |
08l7 | Problem:
Show that
$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq 4\left(\frac{x}{x y+1}+\frac{y}{y z+1}+\frac{z}{z x+1}\right)^{2}
$$
for all real positive numbers $x$, $y$ and $z$. | [
"Solution:\nThe idea is to split the inequality in two, showing that\n$$\n\\left(\\sqrt{\\frac{x}{y}}+\\sqrt{\\frac{y}{z}}+\\sqrt{\\frac{z}{x}}\\right)^{2}\n$$\ncan be intercalated between the left-hand side and the right-hand side. Indeed, using the Cauchy-Schwarz inequality one has\n$$\n(x+y+z)\\left(\\frac{1}{x}... | JBMO | 2008 Shortlist JBMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0882 | Problem:
Quante sono le coppie ordinate $(x, y)$ di interi relativi che verificano l'equazione $y^{4}-8 y^{2}+7=8 x^{2}-2 x^{2} y^{2}-x^{4}$? | [
"Solution:\n\nLa risposta è 4. L'equazione proposta è equivalente a:\n$$\n\\left(x^{2}+y^{2}-1\\right)\\left(x^{2}+y^{2}-7\\right)=0,\n$$\nle cui soluzioni sono date dalle coppie $(x, y)$ di interi che realizzano $x^{2}+y^{2}=1$ oppure $x^{2}+y^{2}=7$. L'ultima equazione, tuttavia, è impossibile, dato che un intero... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 4 | |
07rj | Suppose $x$ and $y$ are real numbers with $x \ge 0$ and $y^2 \ge x(x+1)$. Prove that $(y-1)^2 \ge x(x-1)$. | [
"Suppose $x \\ge 0$ and $y^2 \\ge x(x+1)$. If $0 \\le x \\le 1$, then $(y-1)^2 \\ge 0 \\ge x(x-1)$. If $x > 1$, either $y \\ge \\sqrt{x(x+1)}$ or $y \\le -\\sqrt{x(x+1)}$. If $y \\ge \\sqrt{x(x+1)}$, then $y > 1$, since $x > 1$. So $(y-1)^2 \\ge (\\sqrt{x(x+1)}-1)^2 = x^2+x+1-2\\sqrt{x(x+1)}$. It suffices to prove ... | Ireland | Irish | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0an2 | Problem:
Find all negative solutions to the equation $x = \sqrt[3]{20 + 21 \sqrt[3]{20 + 21 \sqrt[3]{20 + 21 x}}}$
(a) $-1, -2$
(b) $-5, -3$
(c) $-2, -4$
(d) $-4, -1$ | [] | Philippines | Qualifying Round | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Intermediate Algebra > Other"
] | null | MCQ | d | |
0fsl | Problem:
Beweise, dass es zu jedem Polynom $P(x)$ vom Grad $10$ mit ganzzahligen Koeffizienten eine (in beiden Richtungen) unendliche arithmetische Folge ganzer Zahlen gibt, die keinen der Werte $P(k)$, $k \in \mathbb{Z}$, enthält. | [
"Solution:\nWir bemerken zuerst, dass die Existenz einer solchen arithmetischen Folge äquivalent dazu ist, dass es natürliche Zahlen $a, m$ gibt, sodass $P(x) \\not\\equiv a \\pmod{m}$ gilt für alle $x \\in \\mathbb{Z}$. Wir müssen also zeigen, dass die Werte von $P$ bei den ganzen Zahlen eine gewisse Kongruenzklas... | Switzerland | IMO - Selektion | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
03bd | Isosceles $\triangle ABC$ ($AC = BC$) is inscribed in a circle $k$. A point $M$ lies on the side $BC$. A point $N$ from the ray $AM$ ($M$ lies between $A$ and $N$) is such that $AN = AC$. The circumcircle of $\triangle MCN$ intersects $k$ at $C$ and $P$, where $P$ is from the arc $BC$, not containing $A$. The lines $AB... | [
"Denote $\\angle MAC = \\alpha$ and $\\angle ACB = \\beta$. Let $J$ be the incenter of $\\triangle AMC$. It follows that $\\angle AJM = 90^\\circ + \\beta/2$ and $\\angle ABM = 90^\\circ - \\beta/2$, which implies that the quadrilateral $ABMJ$ is inscribed in a circle $k_1$. Analogously we have that the quadrilater... | Bulgaria | Bulgaria | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthoce... | English | proof only | null | |
0ewj | Problem:
Can we label each vertex of a 45-gon with one of the digits $0, 1, \ldots, 9$ so that for each pair of distinct digits $i, j$ one of the 45 sides has vertices labeled $i, j$? | [
"Solution:\n$10 \\times 5 > 45$, so some digit $i_0$ must appear less than $5$ times. But each occurrence can give at most $2$ edges $i_0, j$, so there are at most $8$ edges $i_0, j$, which is one too few."
] | Soviet Union | 3rd ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No | |
0eb3 | Find all functions $f: \mathbb{R} \to \mathbb{R}$, such that
$$
f(xy) = x f(y) + 3 f(x) + 3
$$
for all $x, y \in \mathbb{R}$. | [
"Plugging $x = 0$ into the equation, we get $f(0) = 3 f(0) + 3$, which implies that $f(0) = -\\frac{3}{2}$.\n\nNow plug in $y = 0$ to get $f(0) = x f(0) + 3 f(x) + 3$. From here we can obtain\n$$\nf(x) = -\\frac{1}{3} x f(0) + \\frac{1}{3} f(0) - 1 = \\frac{1}{2} x - \\frac{3}{2}.\n$$\nIt is easy to verify that the... | Slovenia | National Math Olympiad in Slovenia | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x/2 - 3/2 | |
0dji | Every positive integer greater than $1000$ is colored in red or blue in such a way that the product of any two distinct red numbers is blue. Can it happen that no two blue numbers have difference $1$? | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | No | |
0fur | Problem:
Seien $k_{1}$ und $k_{2}$ zwei Kreise, die sich im Punkt $P$ äusserlich berühren. Ein dritter Kreis $k$ berühre $k_{1}$ in $B$ und $k_{2}$ in $C$, so dass $k_{1}$ und $k_{2}$ im Innern von $k$ liegen. Sei $A$ einer der Schnittpunkte von $k$ mit der gemeinsamen Tangente von $k_{1}$ und $k_{2}$ durch $P$. Die G... | [
"Solution:\n\nWir untersuchen als erstes die Potenz von $A$ an den Kreisen $k_{1}$ und $k_{2}$. Es gilt\n$$\nA B \\cdot A R = A P^{2} = A C \\cdot A S\n$$\nDaraus folgt, dass $\\triangle A B C$ und $\\triangle A S R$ ähnlich sind.\nSei $T$ der Schnittpunkt der gemeinsamen Tangente von $k$ und $k_{1}$ mit der gemein... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a2d | Let $P(x)$ be a polynomial with integer coefficients of degree $n > 1$ for which it holds that $Q(x) = P(P(P(x))) - P(x)$ has exactly $n^3$ distinct real roots. Prove that the roots of $Q(x)$ can be partitioned into two groups with equal arithmetic means. | [
"Write $\\gamma_i$ with $1 \\le i \\le n^3$ for the distinct roots of $Q(x)$. For these $\\gamma_i$, we have that $P(P(P(\\gamma_i))) - P(\\gamma_i) = 0$, so $P(\\gamma_i)$ is a zero of the polynomial $P(P(x)) - x$. The degree of $P(P(x)) - x$ is $n^2$, so this polynomial has at most $n^2$ distinct roots. Denote th... | Netherlands | IMO Team Selection Test 3 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0gy4 | Let's call the filling of the square $2009 \times 2009$ which is divided into unit squares "correct" if it is filled with numbers $1,2,\ldots,2009$ in such a way that each of these numbers appears in every row and every column. Let's consider the distance from the central square to the nearest square with number «1» (b... | [
"The square with number «1» which is the nearest to the center is located on the longest diagonal. The distance to the border is $1004$, therefore the distance to the nearest square is $502$. Let us show that this distance is the shortest possible. By condition the distance between two squares is determined by the ... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 502 | |
08to | Let $\Gamma$ be the circum-circle of a triangle $\triangle ABC$. Suppose a circle with its center at a point $O$ is tangent to the line segment $BC$ at a point $P$, and is tangent to the arc $BC$ (not containing $A$) at a point $Q$. If $\angle BAO = \angle CAO$, show that $\angle PAO = \angle QAO$ must be satisfied. | [
"If $AB = AC$, then both of the points $P, Q$ lie on the bisector of the angle $\\angle BAC$, and we get $\\angle PAO = \\angle QAO = 0^\\circ$; so the assertion is valid in this case.\n\nLet us assume in the sequel that $AB \\ne AC$. Let $\\gamma$ be the circle with $O$ as its center and tangent to the line $BC$ a... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06y7 | Let $n$ be a positive integer. The integers $1, 2, 3, \ldots, n^{2}$ are to be written in the cells of an $n \times n$ board such that each integer is written in exactly one cell and each cell contains exactly one integer. For every integer $d$ with $d \mid n$, the $d$-division of the board is the division of the board... | [
"We first show by induction that $n = 2^{k}$ is a cool number. The base case of $n = 2$ is trivial as there is no such $d$.\n\nFor induction, assume that $2^{k}$ is a cool number. We construct a numbering of a $2^{k+1} \\times 2^{k+1}$ board that satisfies the conditions.\n\nTake the $2^{k+1} \\times 2^{k+1}$ board... | IMO | IMO2024 Shortlisted Problems | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | All even cool numbers are powers of two. | |
0bov | Let $x$, $y$, $n$ be positive integers such that
$$
n = \frac{x^2 - 1}{2} = \frac{y^2 - 1}{3}.
$$
Prove that $n = y^2 - x^2$ and $20$ divides $n$. | [
"a. As $x^2 = 2n + 1$ and $y^2 = 3n + 1$, one has $y^2 - x^2 = n$.\n\nb. Since $x$ is odd, both $x-1$ and $x+1$ are even and so $4$ divides $x^2 - 1 = 2n$, implying that $n$ is even. Then $y^2 = 3n + 1$ is odd, and so is $y$. Again, $4$ divides $y^2 - 1$ so $n = (y^2 - 1) - (x^2 - 1)$ is divisible by $4$; in fact, ... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
0i93 | Problem:
$O K R A$ is a trapezoid with $O K$ parallel to $R A$. If $O K=12$ and $R A$ is a positive integer, how many integer values can be taken on by the length of the segment in the trapezoid, parallel to $O K$, through the intersection of the diagonals? | [
"Solution:\n\nLet $R A = x$. If the diagonals intersect at $X$, and the segment is $P Q$ with $P$ on $K R$, then $\\triangle P K X \\sim \\triangle R K A$ and $\\triangle O K X \\sim \\triangle R A X$ (by equal angles), giving $R A / P X = A K / X K = 1 + A X / X K = 1 + A R / O K = (x+12)/12$, so $P X = 12x/(12+x)... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 10 | |
0hlc | Problem:
The set of positive integers is partitioned into finitely many subsets. Show that some subset $S$ has the following property: for every positive integer $n$, $S$ contains infinitely many multiples of $n$. | [
"Solution:\n\nLet the subsets be $S_{1}, S_{2}, \\ldots, S_{k}$. Suppose the statement is false and seek a contradiction. Then for each $S_{i}$ there exists some $n_{i}$ such that $S_{i}$ contains only finitely many multiples of $n_{i}$. Let $n = n_{1} n_{2} \\cdots n_{k}$; then every multiple of $n$ is a multiple ... | United States | Berkeley Math Circle Take-Home Contest #2 | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
09c0 | Гурвалжны талууд үүсгэх $a, b, c$ гурван натурал тооны их нь $2010$-аас хэтрэхгүй бол ийм бүх $(a, b, c)$ гуравтын тоог ол. | [
"$S_n$-ээр гурвалжны талууд үүсгэх ба тус бүр $n$-ээс хэтрэхгүй $(a, b, c)$ гурвалын тоог тэмдэглэе. Тэгвэл $S_n$-ийн хувьд дараах тэнцэтгэл үнэн байна.\n$$\nS_n = S_{n-1} + |A| + |B| + |C| - |A \\cup B| - |A \\cup C| - |B \\cup C| + |A \\cup B \\cup C| \\quad (*)\n$$\nЭнд\n$A = \\{(a, b, c) \\mid a = n, 1 \\leq b ... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | Mongolian | proof and answer | 2010*(2010^2 + 1)/2 | |
01li | Find all real $a$ such that there exists a function $f : \mathbb{R} \to \mathbb{R}$ satisfying the equality $f(\sin x) + a f(\cos x) = \cos 2x$ for all real $x$. (I. Voronovich) | [
"Answer: $a \\in \\mathbb{R} \\setminus \\{1\\}$.\n\nFirst, if $a \\neq 1$ then it is easy to verify that the function\n$$f(x) = \\frac{2x^2 - 1}{a - 1}$$\nsatisfies the given equation.\n\nOn the other hand, for $a = 1$ we have the functional equation\n$$f(\\sin x) + f(\\cos x) = \\cos 2x,$$\nwhich after changing $... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | all real numbers except 1 | |
069g | Let $a$, $b$ be two distinct real numbers and let $c$ be a positive real number such that
$$
a^4 - 2019a = b^4 - 2019b = c.
$$
Prove that $-\sqrt{c} < ab < 0$. | [
"Let $a^4 - 2019a = b^4 - 2019b = c$.\n\nSince $a \\neq b$, subtract the two equations:\n$$\na^4 - b^4 - 2019(a - b) = 0\n$$\n$$\n(a - b)(a^3 + a^2b + ab^2 + b^3) - 2019(a - b) = 0\n$$\nSince $a \\neq b$, divide both sides by $(a - b)$:\n$$\na^3 + a^2b + ab^2 + b^3 = 2019\n$$\n$$\n(a + b)^3 - 3ab(a + b) = 2019\n$$\... | Greece | 23rd Junior Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0kl2 | The six-digit number $20210A$ is prime for only one digit $A$. What is $A$?
(A) 1 (B) 3 (C) 5 (D) 7 (E) 9 | [] | United States | AMC 12 A | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | MCQ | E | |
035b | Problem:
Find all triangles $ABC$ with integer sidelengths such that the side $AC$ is equal to the bisector of $\angle BAC$ and the perimeter of $\triangle ABC$ is equal to $10p$, where $p$ is a prime number. | [
"Solution:\nUsing the standard notation for the elements of $\\triangle ABC$ we have $b^{2} = l_{a}^{2} = bc - \\frac{a^{2}bc}{(b+c)^{2}}$. Hence\n$$\na^{2}c = (c-b)(c+b)^{2}\n$$\nLet $\\frac{a}{c} = \\frac{m}{n},\\ (m, n) = 1$, and $\\frac{b}{c} = \\frac{r}{s},\\ (r, s) = 1$. Then (1) implies that\n$$\n\\frac{m^{2... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techn... | null | proof and answer | (a,b,c) = (28, 15, 27), (267, 280, 343), (62, 120, 128) | |
02vt | Problem:
O conteúdo multiplicativo de um conjunto é o produto de seus elementos. Caso o conjunto possua um único elemento, seu conteúdo multiplicativo é este único elemento e, caso o conjunto seja vazio, seu conteúdo multiplicativo é $1$. Por exemplo, o conteúdo multiplicativo de $\{1,2,3\}$ é $1 \cdot 2 \cdot 3=6$.
... | [
"Solution:\n\na) Observe que no produto $(1+a)(1+b)$, ao aplicarmos a propriedade distributiva da multiplicação dos números reais, aparecerão os produtos $1 \\cdot 1, 1 \\cdot b, a \\cdot 1$ e $a \\cdot b$. Esses produtos coincidem com os conteúdos multiplicativos dos conjuntos $\\varnothing, \\{b\\}, \\{a\\}$ e $\... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a) 120; b) 2017 | |
0crc | В республике математиков выбрали число $\alpha > 2$ и выпустили монеты достоинствами в 1 рубль, а также в $\alpha^k$ рублей при каждом натуральном $k$. При этом $\alpha$ было выбрано так, что достоинства всех монет, кроме самой мелкой, иррациональны. Могло ли оказаться, что любую сумму в натуральное число рублей можно ... | [
"Ответ. Могло."
] | Russia | XL Russian mathematical olympiad | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Могло. | |
0in8 | Problem:
Thomas and Michael are just two people in a large pool of well qualified candidates for appointment to a problem writing committee for a prestigious college math contest. It is 40 times more likely that both will serve if the size of the committee is increased from its traditional 3 members to a whopping $n$ ... | [
"Solution:\n\nAnswer: 16. Suppose there are $k$ candidates. Then the probability that both serve on a 3 membered committee is $(k-2)/\\binom{k}{3}$, and the odds that both serve on an $n$ membered committee are $\\binom{k-2}{n-2}/\\binom{k}{n}$. The ratio of the latter to the former is\n$$\n\\frac{\\binom{k}{3}\\bi... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 16 | |
0dby | Given positive integer $n$. Several cells of a $n \times n$ board were coloured green in a way that no two green cells share a common segment. Is it always true (i.e. no matter which cells were coloured green) that it is possible to place exactly $n$ rooks in some not green cells of board such that no rook captures any... | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | For n = 20: yes; for n = 19: no. | |
0g73 | 已知函數 $f$ 與 $g$ 為實數映至實數。試求滿足下列條件之所有函數序對 $(f, g)$
$$
g(f(x + y)) = f(x) + (2x + y)g(y), \text{對任意實數 } x, y.
$$ | [
"滿足題目的所有解為 $f(x) = x^2 + c, g(x) = x$, 其中 $c$ 是任意實數。\n\n觀察到原式左側 $x, y$ 對稱的, 因此若代入 $(y, x)$ 可得到\n$$\ng(f(x + y)) = f(y) + (2y + x)g(x)\n$$\n與原式相減即有\n$$\nf(x) + (2x + y)g(y) = f(y) + (2y + x)g(x)\n$$\n將 $(x, 0), (1, x)$ 代入上式中可得到\n$$\nf(x) + 2xg(0) = f(0) + xg(x)\n$$\n$$\nf(1) + (2 + x)g(x) = f(x) + (2x + 1)g(1)\n$$\n... | Taiwan | 二〇一二數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All pairs are f(x) = x^2 + c and g(x) = x, where c is any real constant. | |
0ees | Problem:
Naj bo $m$ realno število in $f(x) = (m-2)x^{2} - 2mx + 3m$.
a) Izračunaj vrednost parametra $m$ tako, da bo graf funkcije $f$ potekal skozi točko $T(-2,3)$. Za tako izračunano vrednost parametra $m$ zapiši predpis funkcije $f$.
b) Izračunaj, za katero vrednost parametra $m$ graf funkcije $f$ ni parabola. K... | [
"Solution:\n\na.\n\nVstavimo točko $T(-2, 3)$ v funkcijo $f(x)$:\n\n$$\nf(-2) = (m-2)(-2)^2 - 2m(-2) + 3m = 3\n$$\n\nIzračunamo:\n\n$$\n(m-2) \\cdot 4 + 4m + 3m = 3\n$$\n$$\n4m - 8 + 4m + 3m = 3\n$$\n$$\n(4m + 4m + 3m) - 8 = 3\n$$\n$$\n11m - 8 = 3\n$$\n$$\n11m = 11\n$$\n$$\nm = 1\n$$\n\nPredpis funkcije $f$ za $m =... | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških and strokovnih šol Državno tekmovanje | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) m = 1; f(x) = -x^2 - 2x + 3. b) m = 2; f(x) = -4x + 6; f^{-1}(x) = -1/4 x + 3/2. | |
0csq | Find all real $x$ such that exactly one of the four numbers $x - \sqrt{2}$, $x - 1/x$, $x + 1/x$, and $x^2 + 2\sqrt{2}$ is not an integer. (N. Agakhanov) | [
"Обозначим $a = x - \\sqrt{2}$, $b = x - 1/x$, $c = x + 1/x$, $d = x^2 + 2\\sqrt{2}$. Заметим, что $b$ и $c$ не могут одновременно быть целыми. Действительно, тогда число $b+c = 2x$ также целое, значит, $x$ рационально, поэтому как $a$, так и $d$ не будут целыми как суммы рационального и иррационального чисел. Итак... | Russia | XL Russian mathematical olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | sqrt(2) - 1 | |
0c77 | Let $n$ be an odd positive integer and let $A, B \in \mathcal{M}_n(\mathbb{C})$ be two matrices with complex entries, such that $(A - B)^2 = O_n$. Prove that
$$
\det(AB - BA) = 0.
$$ | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
09vs | Daan distributes the numbers $1$ to $9$ over the nine squares of a $3 \times 3$ table (each square receives exactly one number). Then, in each row, Daan circles the median number (the number that is neither the smallest nor the largest of the three). For example, if the numbers $8$, $1$, and $2$ are in one row, he circ... | [
"a. The smallest possible number of circled numbers is $3$. Fewer than $3$ is not possible since in each row at least one number is circled (and these are three different numbers).\n\nOn the right, a median table is shown in which only $3$ numbers are circled. In the rows, the numbers $7$, $5$, $3$ are circled, in ... | Netherlands | Final Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Minimum 3, maximum 7 | |
01pr | Find the smallest real number $x$ such that the inequality $x + c \le (x + a)(x + b)$ holds for any triangle, where $a \le b \le c$ are the sides of the triangle. | [
"$x = 1$.\n\nFirst, we prove that if $a$, $b$, $c$ are the sides of a triangle, then the inequality\n$$\nx + c \\le (x + a)(x + b) \\quad (*)\n$$\nholds for $x = 1$. Indeed, we can rewrite $(*)$ as $x + c \\le x^2 + (a + b)x + ab$. It is easy to see that this inequality holds for $x = 1$ since $x^2 = x = 1$ and, by... | Belarus | BelarusMO 2013_s | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1 | |
002f | Delante de la cueva de Alí Babá hay un dispositivo para abrir la puerta: es una calesita con forma de cuadrado que tiene cuatro cofres cerrados ubicados uno en cada vértice. En cada cofre hay una moneda que puede estar cara o ceca. La cueva se abre sólo si las cuatro monedas tienen la misma posición, todas cara o todas... | [] | Argentina | XX OLIMPIADA MATEMÁTICA ARGENTINA | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | español | proof only | null | |
04zh | In an acute triangle $ABC$ the angle $C$ is greater than the angle $A$. Let $AE$ be a diameter of the circumcircle of the triangle. Let the intersection point of the ray $AC$ and the tangent of the circumcircle through the vertex $B$ be $K$. The perpendicular to $AE$ through $K$ intersects the circumcircle of the trian... | [
"Since $AE$ is a diameter of the circumcircle of the triangle $ABC$, $\\angle ACE = \\angle ECK = 90^\\circ$. So it suffices to show that $\\angle ACB = \\angle DCK$ (Fig. 19).\n\nLet $L$ be the point of intersection of lines $AE$ and $DK$. Then $\\angle BAC = \\angle CBK = \\angle CDK$ by the inscribed angles theo... | Estonia | Estonija 2010 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0d86 | For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geq 2017$, the integer $P(n)$ is positive and $S(P(n)) = P(S(n))$. | [
"We consider the degree of polynomial $P$:\n\nCase 1: If $\\deg P = 0$ then $P(x) \\equiv c$ for some $c \\in \\mathbb{Z}$, the given condition becomes $S(c) = c$ which holds if and only if $1 \\leq c \\leq 9$.\n\nCase 2: If $\\deg P = 1$. We notice that $S(m+n) \\leq S(m) + S(n)$ for all positive integers $m, n$ a... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | P(x) = x or P(x) = c for c in {1, 2, 3, 4, 5, 6, 7, 8, 9} | |
0d8e | Let $ABC$ be a triangle inscribed in circle $(O)$, with its altitudes $BE$, $CF$ intersecting at orthocenter $H$ ($E \in AC$, $F \in AB$). Let $M$ be the midpoint of $BC$, $K$ be the orthogonal projection of $H$ on $AM$. $EF$ intersects $BC$ at $P$. Let $Q$ be the intersection of the tangent to $(O)$ which passes throu... | [
"Let $AD$ be the altitude of triangle $ABC$, $AA'$ be the diameter of $(O)$. $AO$ meets $EF$ at $N$. Let $L$ be the reflection of $A$ through $N$.\n\n\n\nWe have $AO \\perp EF$ then quadrilateral $NECA'$ is cyclic. Then\n$$\nAO \\cdot AL = \\frac{1}{2} AA' \\cdot 2AN = AA' \\cdot AN = AE \\... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Reflection",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid... | English | proof only | null | |
0ahi | A fixed circle $k$ is given, together with three collinear points $E$, $F$ and $G$ such that $E$ and $G$ lie outside the circle and $F$ lies inside the circle. Prove that if $ABCD$ is an arbitrary quadrangle, inscribed in the circle $k$ such that the extensions of the sides $AB$, $AD$ and $DC$ pass through $E$, $F$ and... | [
"Let $ABCD$ be such a quadrangle. Notice that, according to the conditions of the exercise, the line $EG$ intersects the side $BC$ in a point inside of $BC$. Let us denote that point by $H$. We will consider two cases:\n\n**Case 1: The lines $AB$ and $CD$ are not parallel.**\n\n\nLet them i... | North Macedonia | 19-th Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | English | proof only | null | |
08hc | Problem:
In the scalene triangle $ABC$ the points $A_1$ and $B_1$ are the bisector feet, drawn from the vertices $A$ and $B$ respectively. The straight line $A_1B_1$ intersects the line $AB$ at the point $D$. Prove that one of the angles $\angle ACD$ or $\angle BCD$ is obtuse and $m(\angle ACD) + m(\angle BCD) = 180^\... | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0g7t | 圓上有 $2^{2013}$ 個點, 任意編號為 $1, 2, \dots, 2^{2013}$, 每個數字恰編給一個點。試證: 從這些點中可以連出 500 條兩兩不相交的弦, 使得這些弦的兩端點的數字和皆相同。 | [
"本題的證明係基於以下的事實。\n\n**引理.** 在一個圖 $G$ 中, 設頂點 $v$ 的度數為 $d_v$。則 $G$ 包含一個由某些頂點所成的獨立集 $S$ 滿足 $|S| \\ge f(G)$, 其中\n$$\nf(G) = \\sum_{v \\in G} \\frac{1}{d_v + 1}.\n$$\n**證.** 對 $G$ 的頂點數 $|G| = n$ 進行數學歸納法。初始情形 $n=1$ 顯然成立。在歸納步驟中取 $G$ 中有最小度數 $d$ 的某頂點 $v_0$。將 $v_0$ 與它所有的鄰居 $v_1, \\dots, v_d$, 以及所有連到這些頂點的邊從圖 $G$ 中刪除, 將所得的圖稱為 $... | Taiwan | 二〇一三數學奧林匹亞競賽第三階段選訓營,模擬競賽(一) | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
0eyt | Problem:
In the quadrilateral $ABCD$, $BC$ is parallel to $AD$. The point $E$ lies on the segment $AD$ and the perimeters of $ABE$, $BCE$ and $CDE$ are equal. Prove that $BC = AD / 2$. | [
"Solution:\n\nTake $E_1$ on the line $AD$ so that $AE_1CB$ is a parallelogram. Then $AE_1 = BC$, $AB = CE_1$, so triangles $ABE_1$ and $BCE_1$ have equal perimeters. Moreover, $E_1$ is the only point on the line for which this is true. For if we move $E$ a distance $x$ from $E_1$, then we change $AE_1$ by $x$, and ... | Soviet Union | 3rd ASU | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
06fl | Let $f : \mathbb{Z} \to \mathbb{Z}$ ($\mathbb{Z}$ is the set of integers) be such that $f(1) = 1$, $f(2) = 20$, $f(-4) = -4$ and $f(x+y) = f(x)+f(y)+axy(x+y)+bxy+c(x+y)+4$ for all $x, y \in \mathbb{Z}$, where $a, b$ and $c$ are certain constants.
a. Find a formula for $f(x)$, where $x$ is any integer.
b. If $f(x) \ge... | [
"a.\nWe have $f(x) = x^3 + 4x^2 - 4$.\n\nLabel the equation as follows.\n$$\nf(x + y) = f(x) + f(y) + axy(x + y) + bxy + c(x + y) + 4 \\quad (1)\n$$\n\nPutting $x = y = 0$ in (1), we have\n$$\nf(0) = f(0) + f(0) + 4,\n$$\nwhich implies $f(0) = -4$. Next, putting $x = 1$ and $y = 0$ in (1), we have\n$$\nf(1) = f(1) ... | Hong Kong | null | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | f(x) = x^3 + 4x^2 - 4; m = -1 | |
0781 | Find all $f: \mathbb{N} \to \mathbb{N}$ such that $f(x) + y$ and $f(y) + x$ have the same number of 1's in their binary representation for any $x, y \in \mathbb{N}$. | [
"We claim that all such functions are of the form $f(n) = n + c$ for some natural number $c$. The verification is trivial; now consider some function $f$ satisfying the given condition. For $n \\in \\mathbb{N}$, let $d(n)$ denote the number of 1's in the binary representation of $n$. Let $P(x, y)$ denote the statem... | India | IMO TST Day 3 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | f(n) = n + c for some natural number c | |
0h7o | For positive numbers $a, b, c$ prove the inequality :
$$
\sqrt{a^2 + bc} + \sqrt{b^2 + ca} + \sqrt{c^2 + ab} \ge \sqrt{ab + bc} + \sqrt{bc + ca} + \sqrt{ca + ab}.
$$ | [
"It should be noted, that the inequality is symmetrical, that is, the interchange in places of any two variables does not change its appearance. So, without any loss of generality we can assume that $a \\ge b \\ge c$. If we move all summands to one side and group them accordingly, the inequality becomes like this:\... | Ukraine | UkraineMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0206 | Problem:
Let $n \geqslant 3$ be an integer. A frog is to jump along the real axis, starting at the point $0$ and making $n$ jumps: one of length $1$, one of length $2, \ldots$, one of length $n$. It may perform these $n$ jumps in any order. If at some point the frog is sitting on a number $a \leqslant 0$, its next jump... | [
"Solution:\nWe claim that the largest positive integer $k$ with the given property is $\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor$, where $\\lfloor x\\rfloor$ is by definition the largest integer not exceeding $x$.\n\nConsider a sequence of $n$ jumps of length $1,2, \\ldots, n$ such that the frog never lands on an... | Benelux Mathematical Olympiad | 5th Benelux Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | floor((n-1)/2) | |
0l62 | Problem:
Given that $x$, $y$, and $z$ are positive real numbers such that
$$x^{\log_{2}(yz)} = 2^{8} \cdot 3^{4}, \quad y^{\log_{2}(zx)} = 2^{9} \cdot 3^{6}, \quad \text{and} \quad z^{\log_{2}(xy)} = 2^{5} \cdot 3^{10},$$
compute the smallest possible value of $xyz$. | [
"Solution:\nLet $k = \\log_{2} 3$ for brevity. Taking the base-$2$ log of each equation gives\n$$(\\log_{2} x)(\\log_{2} y + \\log_{2} z) = 8 + 4k,$$\n$$(\\log_{2} y)(\\log_{2} z + \\log_{2} x) = 9 + 6k,$$\n$$(\\log_{2} z)(\\log_{2} x + \\log_{2} y) = 5 + 10k.$$\nAdding the first two equations and subtracting the t... | United States | HMMT February | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 1/576 | |
0gwf | Find all pairs of positive integers $m$ and $n$ such that
$$
14^n = 13 \cdot m^n + 1.
$$ | [
"Відповідь: $m = n = 1$. При $n \\geq 2$ ліва частина рівняння ділиться без остачі на $4$, а тому $m'' = 3 \\pmod{4}$, що можливо лише для непарних $m$ і $n$. При непарному $n$ маємо, що $14'' = 2 \\pmod{3}$, а тому $m = 1 \\pmod{3}$. Оскільки $m < 14$, то $m \\in \\{1; 7; 13\\}$. Але при $m = 7$ права частина рівн... | Ukraine | Ukrainian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | m=1, n=1 | |
09yn | A group of islands consists of a large, a medium and a small island. The total area of the three islands together is $23$ km². The difference between the areas of the large island and the medium island turns out to be exactly $1$ km² more than the area of the small island.
How many km² is the area of the large island?
... | [] | Netherlands | First Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | C | |
0b01 | Problem:
Basket $A$ contains two white balls and three black balls, while Basket $B$ contains a white ball and three black balls. Daniel randomly chooses one of the baskets and then randomly picks a ball from this basket. If he picked a white ball, what is the probability that his chosen basket was Basket $A$? | [
"Solution:\n\nLet $P(A)$, $P(B)$ be the probabilities of choosing Baskets $A$ and $B$, respectively, and let $P(W)$ be the probability of picking a white ball. The probability of picking a white ball from Basket $A$ is $P(W \\mid A) = 2/5$ and the probability of picking a white ball from Basket $B$ is $P(W \\mid B)... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 8/13 | |
0dx1 | Problem:
Na nogometnem turnirju so sodelovale le ekipe iz Malega mesta in Velikega mesta. Iz Velikega mesta je bilo 9 ekip več kot iz Malega mesta. Vsaki ekipi sta se srečali natanko enkrat, pri čemer je zmagovalna ekipa dobila 1 točko, poražena 0 točk, neodločen izid pa ni bil možen. Ekipe iz Velikega mesta so osvoji... | [
"Solution:\n\nOznačimo število ekip iz Malega mesta z $x$. Potem je število ekip iz Velikega mesta enako $x+9$. Ekipe iz Malega mesta so med seboj igrale $\\frac{x(x-1)}{2}$ tekem in so zato osvojile $\\frac{x(x-1)}{2}+k$ točk, kjer je $k$ število zmag, osvojenih nad ekipami iz Velikega mesta. Ekipe iz Velikega mes... | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 11 | |
0h62 | a) Determine whether there exist positive integer numbers $a_1, a_2, ..., a_{2015}$ such that any two of them are co-prime and $a_1a_2...a_{2015}-1$ is a product of two consequent odd numbers?
b) Determine whether there exist positive integer numbers $a_1, a_2, ..., a_{2015}$ such that: any two of them are co-prime an... | [
"a) Let $a_1 = p_1^2, a_2 = p_2^2, ..., a_{2015} = p_{2015}^2$, where $p_1 = 2, p_2, ..., p_{2015}$ are the first 2015 prime numbers. It is clear that every two of these numbers are co-prime and $a_1a_2...a_{2015}-1 = (p_1p_2...p_{2015}-1)(p_1p_2...p_{2015}+1)$ is a product of two consequent odd numbers.\n\nb) Let ... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | a) Yes. For example, take the squares of the first 2015 primes including two; then the product minus one equals the product of two consecutive odd numbers. b) Yes. For example, take the squares of the first 2015 odd primes; then the product minus one equals the product of two consecutive even numbers. | |
02rj | Each member of a population has two genes (possibly repeated) among the genes $G_1, G_2, \dots, G_n$. Suppose that, in generation 0 of this population, $p_{ij}(0)$ ($1 \le i, j \le n$) is the proportion of members of this population with genotype $G_i G_j$ (or $G_j G_i$, which is biologically identical to $G_i G_j$).
T... | [] | Brazil | Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
02kp | Problem:
Adriano, Bruno, César e Daniel são quatro bons amigos. Daniel não tinha dinheiro, mas os outros tinham. Adriano deu a Daniel um quinto do seu dinheiro, Bruno deu um quarto do seu dinheiro e César deu um terço do seu dinheiro. Cada um deu a Daniel a mesma quantia. A quantia que Daniel possui agora representa q... | [
"Solution:\n\nSuponha que Daniel tenha recebido $x$ reais de cada um de seus amigos. Então, Adriano tinha, inicialmente, $5x$ reais, Bruno tinha $4x$ reais e César tinha $3x$ reais. Segue que o total de dinheiro dos três no início era de $5x + 4x + 3x = 12x$ reais. Como cada um de seus três amigos lhe deu $x$ reais... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | B | |
0aln | Problem:
The remainders when the polynomial $p(x)$ is divided by $(x+1)$ and $(x-1)$ are $7$ and $5$, respectively. Find the sum of the coefficients of the odd powers of $x$.
(a) $-4$
(b) $2$
(c) $-1$
(d) $12$ | [] | Philippines | Qualifying Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | c |
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