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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
06w5 | In a regular 100-gon, 41 vertices are colored black and the remaining 59 vertices are colored white. Prove that there exist 24 convex quadrilaterals $Q_{1}, \ldots, Q_{24}$ whose corners are vertices of the 100-gon, so that
- the quadrilaterals $Q_{1}, \ldots, Q_{24}$ are pairwise disjoint, and
- every quadrilateral $Q... | [
"Call a quadrilateral skew-colored, if it has three corners of one color and one corner of the other color. We will prove the following\n\nClaim. If the vertices of a convex $(4k+1)$-gon $P$ are colored black and white such that each color is used at least $k$ times, then there exist $k$ pairwise disjoint skew-colo... | IMO | IMO 2020 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
00nc | Let $ABC$ denote a triangle. The point $X$ lies on the extension of $AC$ beyond $A$, such that $AX = AB$. Similarly, the point $Y$ lies on the extension of $BC$ beyond $B$ such that $BY = AB$.
Prove that the circumcircles of $ACY$ and $BCX$ intersect a second time in a point different from $C$ that lies on the bisector... | [
"\nFigure 2: Problem 10\n\nAs usual, we denote the angles of the triangle at $A$, $B$ and $C$ with $\\alpha$, $\\beta$ and $\\gamma$.\nIt is sufficient to show that the center $I_c$ of the excircle touching the line $AB$ lies on the two circles. To do this, we look at the respective inscrib... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
03yi | The number of positive integer solutions of equation $x + y + z = 2010$ with $x \le y \le z$ is ________. | [
"It is easy to find that the number of positive integer solutions of $x + y + z = 2010$ is $C_{2009}^2 = 2009 \\times 1004$.\nWe now classify these solutions into three categories:\n\n(1)\n$x = y = z$, the number in this category is obviously 1;\n\n(2) there are exactly two that are equal among $x, y, z$ —\nthe num... | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | final answer only | 336675 | |
069u | A $5 \times 100$ table is divided into $500$ unit square cells, where $n$ of them are coloured black and the rest are coloured white. Two unit square cells are called *adjacent* if they share a common side. Each of the unit square cells has at most two adjacent black unit square cells. Find the largest possible value o... | [] | Greece | 23rd Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 302 | |
0agz | Let $S$ be a finite set of positive integers which has the following property: if $x$ is a member of $S$, then so are all positive divisors of $x$. A non-empty subset $T$ of $S$ is a good if whenever $x, y \in T$ and $x < y$, the ratio $y/x$ is a power of a prime number. A non-empty subset $T$ of $S$ is bad if whenever... | [
"Notice first that a bad subset of $S$ contains at most one element from a good one, to deduce that a partition of $S$ into bad subset has at least as many members as a maximal good subset.\n\nNotice further the elements of a good subset of $S$ must be among the terms of a geometric sequence whose ratio is a prime:... | North Macedonia | XXVIII-th Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0amw | Problem:
Let $f$ be a quadratic function of $x$. If $2y$ is a root of $f(x-y)$, and $3y$ is a root of $f(x+y)$, what is the product of the roots of $f(x)$?
(a) $6y^{2}$
(b) $5y^{2}$
(c) $4y^{2}$
(d) $3y^{2}$ | [] | Philippines | Qualifying Round | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | MCQ | c | |
06ru | Players $A$ and $B$ play a game with $N \geq 2012$ coins and $2012$ boxes arranged around a circle. Initially $A$ distributes the coins among the boxes so that there is at least $1$ coin in each box. Then the two of them make moves in the order $B, A, B, A, \ldots$ by the following rules:
- On every move of his $B$ pas... | [
"We argue for a general $n \\geq 7$ instead of $2012$ and prove that the required minimum $N$ is $2n-2$. For $n=2012$ this gives $N_{\\text{min}}=4022$.\n\na. If $N=2n-2$ player $A$ can achieve her goal. Let her start the game with a regular distribution: $n-2$ boxes with $2$ coins and $2$ boxes with $1$ coin. Call... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4022 | |
03ss | Prove that for $i = 1, 2, 3$, there exist infinitely many integers $n$ satisfying the following condition: we can find $i$ integers in $\{n, n+2, n+28\}$ that can be expressed as the sum of the cubes of three positive integers. | [
"**Lemma** Let $m$ be the remainder of the sum of the cubes of three positive integers when divided by $9$, then $m \\neq 4$ or $5$.\n**Proof:** Since any integer can be expressed as $3k$ or $3k \\pm 1$ ($k \\in \\mathbb{Z}$),\n$$\n(3k)^3 = 9 \\times 3k^3, \\\\\n(3k \\pm 1)^3 = 9 \\times (3k^3 \\pm 3k^2 + k) \\pm 1... | China | China Girls' Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0b5t | The cells of a $(n^2 - n + 1) \times (n^2 - n + 1)$ matrix are coloured using $n$ colours. A colour is called *dominant* on a row (or a column) if there are at least $n$ cells of this colour on that row (or column). A cell is called *extremal* if its colour is dominant both on its row, and its column. Find all $n \ge 2... | [
"The answer is that there exist such colourings for $n = 2$ (almost trivial), and $n = 3$ (following a detailed and tedious case analysis). For $n = 4$ one must exhibit a colouring with no extremal cells (again a very tedious labor). The colourings with no extremal cells for $n \\ge 5$ are built inductively, using ... | Romania | Local Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | All integers n ≥ 2 | |
0k2r | Problem:
$H O W$, $B O W$, and $D A H$ are equilateral triangles in a plane such that $W O = 7$ and $A H = 2$. Given that $D$, $A$, $B$ are collinear in that order, find the length of $B A$. | [
"Solution:\n\nNote that $H \\neq B$ since otherwise $D A B$ is an equilateral triangle. Let $M$ be the midpoint of $D A$, so $H B = 7 \\sqrt{3}$ and $H M = \\sqrt{3}$, and $\\angle H M B = 90^{\\circ}$. By the Pythagorean theorem,\n$$\nB M = \\sqrt{(7 \\sqrt{3})^2 - (\\sqrt{3})^2} = 12\n$$\nThen $B A = B M - A M = ... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 11 | |
0e2d | Find all positive integers $n$ for which there exists a multiple of $11$ with the sum of its digits equal to $n$. | [
"Denote the sum of the digits of a positive integer $m$ by $S(m)$. Then $S(11) = 2$. We notice that for the first few multiples of $11$ the sum of the digits is always even. The first multiple for which the sum of the digits is odd is $209 = 19 \\cdot 11$ and this sum is $11$. Using these two examples we can constr... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | All even positive integers and all odd integers at least 11; no multiple of 11 has digit sum 1, 3, 5, 7, or 9. | |
0bh3 | Let $n$ be a positive integer and $M$ be the arithmetic mean of its positive divisors. Prove that $M \ge \sqrt{n}$. When does the equality hold? | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds only for n = 1. | |
0i4u | Problem:
The Antarctican language has an alphabet of just 16 letters. Interestingly, every word in the language has exactly 3 letters, and it is known that no word's first letter equals any word's last letter (for instance, if the alphabet were $\{a, b\}$ then $a a b$ and $a a a$ could not both be words in the languag... | [
"Solution:\n\n1024\n\nEvery letter can be the first letter of a word, or the last letter of a word, or possibly neither, but not both. If there are $a$ different first letters and $b$ different last letters, then we can form $a \\cdot 16 \\cdot b$ different words (and the desired conditions will be met). Given the ... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 1024 | |
04ky | Let $ABCD$ be a trapezium with parallel sides $\overline{AB}$ and $\overline{CD}$, and let $P$ be the intersection of its diagonals. If the area of the trapezium equals $25$, and the area of the triangle $CDP$ equals $9$, find the area of the triangle $ABP$. (Stipe Vidak) | [] | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4 | |
01g1 | For a positive integer $n$ it is known that the number $\sqrt{12n^2 + 1}$ also is a positive integer. Prove that the number
$$
\sqrt{\frac{\sqrt{12n^2 + 1} + 1}{2}}
$$
also is a positive integer. | [
"Let's denote\n$$\n\\sqrt{12n^2 + 1} = 2k + 1 \\qquad (\\text{Eq-6})\n$$\nbecause if the number $\\sqrt{12n^2 + 1}$ is integer then it is odd. We have to prove that\n$$\n\\sqrt{\\frac{\\sqrt{12n^2 + 1} + 1}{2}} = \\sqrt{\\frac{(2k + 1) + 1}{2}} = \\sqrt{k + 1}\n$$\nis an integer, that is, that $k + 1$ is a perfect ... | Baltic Way | Baltic Way 2019 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0ip6 | Problem:
Four students from Harvard, one of them named Jack, and five students from MIT, one of them named Jill, are going to see a Boston Celtics game. However, they found out that only 5 tickets remain, so 4 of them must go back. Suppose that at least one student from each school must go see the game, and at least o... | [
"Solution:\n\nLet us count the number of ways of distributing the tickets so that one of the conditions is violated. There is $1$ way to give all the tickets to MIT students, and $\\binom{7}{5}$ ways to give all the tickets to the $7$ students other than Jack and Jill. Therefore, the total number of valid ways is $... | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 104 | |
05mv | Problem:
Soit $A B C D$ un quadrilatère convexe. On note $P, Q, R$ et $S$ les milieux des côtés $[A B],[B C],[C D]$ et $[D A]$. Montrer que l'aire de $PQRS$ est égale à la moitié de l'aire de $A B C D$.
 | [
"Solution:\n\nOn note $X$ l'intersection des diagonales de $A B C D$. $[P Q]$ coupe $[B D]$ en $Y$, et $[Q R]$ coupe $[A C]$ en $Z$. D'après le théorème de la droite des milieux, $(P Q)$ est parallèle à $(A C)$, donc $(Q Z)$ est parallèle à $(C X)$ et, de nouveau d'après le théorème de la droite des milieux, $Z$ es... | France | French Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08fu | Problem:
Il soggiorno della casa di Filippo ha pianta rettangolare. Filippo ha notato che se attacca l'aspirapolvere alla presa vicino alla porta d'ingresso riesce a pulire tutto il pavimento: questo vuol dire che tutti i punti del pavimento sono a distanza minore di $5~\mathrm{m}$ dal punto della parete dove si trova... | [
"Solution:\n\nLa risposta è 25. Consideriamo il rettangolo che forma la pianta del soggiorno, e orientiamolo in modo che la presa $P$ si trovi sul lato orizzontale $AB$, suddividendolo in due segmenti di lunghezza $a$ e $b$. Chiamiamo inoltre $c$ la lunghezza dei due lati verticali $BC$ e $DA$, come in figura. In q... | Italy | Italian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 25 | |
03mo | Let $m$ and $n$ be odd positive integers. Each square of an $m$ by $n$ board is coloured red or blue. A row is said to be red-dominated if there are more red squares than blue squares in the row. A column is said to be blue-dominated if there are more blue squares than red squares in the column. Determine the maximum p... | [
"The answer is $m+n-2$ if $m, n \\ge 3$ and $\\max\\{m, n\\}$ if one of $m, n$ is equal to $1$.\n\nNote that it is not possible that all rows are red-dominated and all columns are blue-dominated. This is true, since the number of rows and columns are both odd, the number of squares is odd. Hence, there are more squ... | Canada | Kanada 2014 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | m + n − 2 if m, n ≥ 3; otherwise max{m, n} if one of m, n equals 1 | |
09b4 | Let $G$ be a graph, not containing $K_4$ as a subgraph and $|V(G)| = 3k$. What is the maximum number of triangles in $G$. | [
"Let us show that the maximum number of triangles is $k^3$ by induction. It is trivial for $k=1$.\nLet $G$ be a graph with the maximum number of triangles. Clearly, it contains a triangle. Say it $T_1T_2T_2$. $G - \\Delta T_1T_2T_2$ is a subgraph of $G$ with $3(k-1)$ vertices. $G - \\Delta T_1T_2T_2$ doesn't contai... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | k^3 | |
0fyq | Problem:
Seien $k$ und $k'$ zwei konzentrische Kreise mit Mittelpunkt $O$. Der Kreis $k'$ sei grösser als der Kreis $k$. Eine Gerade durch $O$ schneide $k$ in $A$ und $k'$ in $B$, sodass $O$ zwischen $A$ und $B$ liegt. Eine andere Gerade durch $O$ schneidet $k$ in $E$ und $k'$ in $F$, sodass $E$ zwischen $O$ und $F$ l... | [
"Solution:\n\nSei $S \\neq O$ der zweite Schnittpunkt der Umkreise von $O A E$ und $O B F$. Wir zeigen nun, dass $S$ auf den Kreisen mit Durchmesser $A B$ und $E F$ liegt. Sei $\\angle A E O=\\alpha$. Da die Dreiecke $A O E$ und $B O F$ gleichschenklig sind gilt $\\angle E A O=\\alpha$ und\n$$\n\\angle O B F=\\angl... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02xw | Problem:
Dizemos que uma tripla de inteiros $(x, y, z)$ é do tipo bacana se $x$, $y$ e $z$ são inteiros positivos, com $y \geq 2$, e $x^{2}-3 y^{2}=z^{2}-3$.
a) Encontre uma tripla $(x, y, z)$ do tipo bacana com $x=5$ e $x=7$.
b) Mostre que para todo $x \geq 5$ e ímpar existem pelo menos duas triplas distintas $\lef... | [
"Solution:\n\na) Para $x=5$ e $x=7$, temos alguns exemplos de triplas do tipo bacana: $(x, y, z)=(5,2,4)$, $(5,3,1)$, $(7,3,5)$ e $(7,4,2)$.\n\nb) Os casos particulares do item anterior permitem conjecturar as seguintes triplas para $x$ ímpar:\n$$\n(x, y, z)=(2 n+1, n, n+2) \\text{ e } (x, y, z)=(2 n+1, n+1, n-1)\n... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | null | |
0c4y | Find three integers so that their sum is $2^{93} + 2^{90} + 3^{51}$, the first is 4 times smaller than the second and the third is with $27^{17}$ larger than the second. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 70th NMO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (2^{90}, 2^{92}, 2^{92} + 27^{17}) | |
0kgd | Problem:
Compute the sum of all positive integers $n$ for which the expression
$$
\frac{n+7}{\sqrt{n-1}}
$$
is an integer. | [
"Solution:\nWe know $\\sqrt{n-1}$ must be a positive integer, because the numerator is a positive integer, and the square root of an integer cannot be a non-integer rational. From this,\n$$\n\\frac{n+7}{\\sqrt{n-1}} = \\sqrt{n-1} + \\frac{8}{\\sqrt{n-1}}\n$$\nis a positive integer, so $\\sqrt{n-1}$ must be a positi... | United States | HMMT Spring 2021 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 89 | |
0879 | Problem:
Il minuscolo, ma preziosissimo, Diamante Dodecaedrico si trova a 2 metri dalla parete sud e 3 metri dalla parete ovest di una stanza rettangolare le cui pareti nord e sud sono lunghe 4 metri e quelle est e ovest sono lunghe 3 metri. Un ladro si cala dal soffitto all'interno della stanza e tocca il pavimento a... | [
"Solution:\n\nLa risposta è (D). Sia $L$ il punto del pavimento in cui \"atterra\" il ladro, $D$ quello in cui si trova il diamante; siano $D_{E}, D_{W}, D_{N}, D_{S}$ i simmetrici di $D$ rispetto alle pareti est, ovest, nord, sud $(E, W, N, S$ in figura). Immaginiamo che il ladro tocchi il filo (per poterlo taglia... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | D | |
08my | Problem:
A set $S$ of natural numbers is called good, if for each element $x \in S$, $x$ does not divide the sum of the remaining numbers in $S$. Find the maximal possible number of elements of a good set which is a subset of the set $A=\{1,2,3, \ldots, 63\}$. | [
"Solution:\nLet set $B$ be the good subset of $A$ which has the maximum number of elements. We can easily see that the number $1$ does not belong to $B$ since $1$ divides all natural numbers. Based on the property of divisibility, we know that $x$ divides the sum of the remaining numbers if and only if $x$ divides ... | JBMO | JBMO Shortlist | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 61 | |
0l39 | Problem:
There are 5 people who start with $1, 2, 3, 4$, and $5$ cookies, respectively. Every minute, two different people are chosen uniformly at random. If they have $a$ and $b$ cookies and $a \neq b$, the person with more cookies eats $|a-b|$ of their own cookies. If $a = b$, the minute still passes with nothing ha... | [
"Solution:\n\nEach person's cookie count can never increase or go below $1$, so in order for everyone to have the same number of cookies, everyone must have exactly $1$ cookie. The only time the number of people with exactly $1$ cookie increases is when one person with exactly $1$ cookie and one person with more th... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 25/3 | |
01ww | The squares $AA_1DE$ and $AA_1FG$ are constructed on the bisector $AA_1$ of the non-isosceles triangle $ABC$ such that the points $B$ and $F$ lie in different half-planes with respect to the line $AA_1$.
Prove that the lines $BD$, $CF$ and $EG$ are concurrent. | [
"Denote the points of intersection of the line $EG$ with the lines $BD$ and $CF$ by\n\n$P_1$ and $P_2$, respectively. We will prove that $P_1 \\equiv P_2$. Denote the points of intersection of the lines $AB$ and $AC$ with the line $DF$ by $C_1$ and $B_1$, respectively. The segment $AA_1$ is... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0fyw | Problem:
Sei $n$ eine natürliche Zahl. In einem Affenkäfig mit $n$ Affen stehen $n$ Kletterstangen. Damit die Affen etwas Bewegung bekommen, platzieren die Wärter zur Fütterung jeweils eine Banane oben an jeder Stange. Zusätzlich verbinden sie die Stangen mit einer endlichen Anzahl Seile, sodass zwei verschiedene Seil... | [
"Solution:\n\nOn commence avec une observation: Si l'on connaît la position d'un singe à un certain instant, cela détermine de manière unique son parcours après mais aussi avant cet instant. (Plus précisément: pour trouver le parcours avant le moment observé, on peut simplement laisser le singe faire le chemin inve... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Other"
] | null | proof only | null | |
00jy | Let $a$ and $b$ be real numbers with $0 \le a, b \le 1$.
Prove that
$$
\frac{a}{b+1} + \frac{b}{a+1} \le 1
$$
and find the cases of equality. | [
"We clear denominators to get\n$$\n\\begin{align*}\n& a(a+1) + b(b+1) \\le (a+1)(b+1), \\\\\n\\Leftrightarrow \\quad & a^2 + a + b^2 + b \\le ab + a + b + 1, \\\\\n\\Leftrightarrow \\quad & a^2 - a + b^2 - b \\le ab - a - b + 1, \\\\\n\\Leftrightarrow \\quad & a(a-1) + b(b-1) \\le (a-1)(b-1), \\\\\n\\Leftrightarrow... | Austria | AustriaMO2013 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Equality holds exactly at (a, b) = (1, 0), (0, 1), and (1, 1). | |
09sh | Problem:
Vind alle functies $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ zodat $f(1)=2$ en zodat voor alle $m, n \in \mathbb{Z}_{>0}$ geldt dat $\min (2 m+2 n, f(m+n)+1)$ deelbaar is door $\max (f(m)+f(n), m+n)$. | [
"Solution:\n\nInvullen van $m=n$ geeft dat $\\min (4 n, f(2 n)+1)$ deelbaar is door $\\max (2 f(n), 2 n)$. Een getal hoogstens gelijk aan $4 n$ is dus deelbaar door een getal minstens gelijk aan $2 f(n)$. Daaruit volgt $4 n \\geq 2 f(n)$, dus $f(n) \\leq 2 n$ voor alle $n$.\n\nInvullen van $m=n=1$ geeft dat $\\min ... | Netherlands | IMO-selectietoets II | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Two functions: f(n) = n + 1 for all positive integers n, and f(n) = 2n for all positive integers n. | |
02gy | Consider the sequence $(a_n)_{n \in \mathbb{N}}$ with $a_0 = a_1 = a_2 = a_3 = 1$ and $a_n a_{n-4} = a_{n-1} a_{n-3} + a_{n-2}^2$, $n \ge 4$. Prove that all the terms of this sequence are integer numbers. | [
"Let's prove by induction that both $a_n$ is an integer and $\\gcd(a_n, a_{n-1}) = \\gcd(a_n, a_{n-2}) = \\gcd(a_n, a_{n-3}) = 1$. It is certainly true for $n = 3$ and direct substitutions show that it is true for $n \\le 7$. Suppose that it's true for $n \\le k$. First we prove that $a_{k+1}$ is an integer by show... | Brazil | XXVI OBM | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0bdc | Let $n$ be a positive integer. Determine all positive integers $p$ for which there exist positive integers $x_1 < x_2 < \dots < x_n$ such that
$$
\frac{1}{x_1} + \frac{2}{x_2} + \dots + \frac{n}{x_n} = p.
$$ | [
"Call good a number $p$ for which there exist positive integers $x_1 < x_2 < \\dots < x_n$ such that $\\frac{1}{x_1} + \\frac{2}{x_2} + \\dots + \\frac{n}{x_n} = p$.\nSince $x_1, x_2, \\dots, x_n$ are integers and $x_1 < x_2 < \\dots < x_n$, we have $x_k \\ge k$, so $\\frac{k}{x_k} \\le 1$, for all $k = 1, 2, \\dot... | Romania | 64th NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | {1, 2, ..., n} | |
03jx | Problem:
Suppose that a function $f$ defined on the positive integers satisfies
$$
\begin{gathered}
f(1)=1, \quad f(2)=2 \\
f(n+2)=f(n+2-f(n+1))+f(n+1-f(n)) \quad(n \geq 1)
\end{gathered}
$$
a) Show that
(i) $0 \leq f(n+1)-f(n) \leq 1$
(ii) if $f(n)$ is odd, then $f(n+1)=f(n)+1$.
b) Determine, with justification, all... | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 2048 | |
0js0 | Problem:
Carl computes the number
$$
N=5^{555}+6^{666}+7^{777}
$$
and writes it in decimal notation. What is the last digit of $N$ that Carl writes? | [
"Solution:\n\nWe look at the last digit of each term.\n- The last digit of $5^{\\bullet}$ is always $5$.\n- The last digit of $6^{\\bullet}$ is always $6$.\n- The last digit of $7^{\\bullet}$ cycles $7, 9, 3, 1, 7, 9, 3, \\ldots$.\n\nSo the last digits are $5, 6, 7$ in that order. Since $5+6+7=18$, the answer is $8... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Number Theory > Modular Arithmetic"
] | null | final answer only | 8 | |
09hk | Let $p \ge 5$ be a prime number. Prove that there are positive integers $n$ and $m$ with $n + m \le (p+1)/2$ such that $p$ divides $2^n \cdot 3^m - 1$. | [] | Mongolia | Mongolian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0fq0 | Problem:
Determina el número de valores distintos de la expresión
$$
\frac{n^{2}-2}{n^{2}-n+2}
$$
donde $n \in\{1,2, \ldots, 100\}$. | [
"Solution:\n\nSumando y restando $2-n$ al numerador se obtiene\n$$\na_{n}=\\frac{n^{2}-2}{n^{2}-n+2}=\\frac{n^{2}-2-n+2+n-2}{n^{2}-n+2}=1+\\frac{n-4}{n^{2}-n+2}\n$$\nAhora vamos a ver si hay dos términos iguales, es decir, cuando es $a_{p}=a_{q}$ para $p \\neq q$. Esto es equivalente a encontrar los enteros $p \\ne... | Spain | LIII Olimpiada matemática Española | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 98 | |
01za | Vitya and Masha play the game. First, Vitya thinks of three different integers. Then Masha can ask one of the following quantities: either the sum of the numbers, or the sum of pairwise products of the numbers, or the product of the numbers suggested by Vitya. Masha asks questions sequentially, and Vitya gives an answe... | [
"a) If Masha asks all three different questions, then for the numbers $a$, $b$ and $c$ thought by Vitya, she will know the coefficients of the polynomial\n$$\np(x) = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + bc + ac)x - abc.\n$$\nSolving the cubic equation $p(x) = 0$ (for example, using Cardano's formula... | Belarus | Belarus2022 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 3 | |
05lv | Problem:
On dispose de $n$ jetons portant chacun un numéro entier (qui peut être négatif). Si on trouve parmi eux deux jetons portant le même numéro $m$, on les enlève et on met à leur place un jeton portant le numéro $m-1$, et un autre portant le numéro $m+1$. Montrer qu'au bout d'un nombre fini de tels changements, ... | [
"Solution:\n\nOn commence par montrer le lemme suivant:\nSoit $m_{0}$ le plus petit des numéros des jetons au début. On note $m_{1}^{(k)}, \\ldots, m_{j_{k}}^{(k)}$ les numéros inférieurs ou égaux à $m_{0}$ qui sont portés par des jetons après $k$ changements, comptés avec multiplicité et rangés par ordre décroissa... | France | Envoi de combinatoire | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
03zo | Suppose $n$ small balls have been placed into $n$ numbered boxes $B_1, B_2, \dots, B_n$. Each time we can select a box $B_k$ and do the following operations:
(1) If $k=1$ and there is at least one ball in $B_1$, move one ball from $B_1$ into $B_2$.
(2) If $k=n$ and there is at least one ball in $B_n$, move one ball fro... | [
"For any two vectors $\\mathbf{x} = (x_1, x_2, \\dots, x_n)$ and $\\mathbf{y} = (y_1, y_2, \\dots, y_n)$, if there exists $1 \\leq k \\leq n$ such that\n$$\nx_1 = y_1, \\dots, x_{k-1} = y_{k-1}, x_k > y_k,\n$$\nwe then denote it as $\\mathbf{x} > \\mathbf{y}$. Let $\\mathbf{x} = (x_1, x_2, \\dots, x_n)$ represent t... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
00x4 | Problem:
Let $\mathbb{N}$ denote the set of positive integers. Let $\varphi: \mathbb{N} \rightarrow \mathbb{N}$ be a bijective function and assume that there exists a finite limit
$$
\lim _{n \rightarrow \infty} \frac{\varphi(n)}{n}=L
$$
What are the possible values of $L$? | [
"Solution:\nIn this solution we allow $L$ to be $\\infty$ as well. We show that $L=1$ is the only possible value. Assume that $L>1$. Then there exists a number $N$ such that for any $n \\geq N$ we have $\\frac{\\varphi(n)}{n}>1$ and thus $\\varphi(n) \\geq n+1 \\geq N+1$. But then $\\varphi$ cannot be bijective, si... | Baltic Way | Baltic Way 1992 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | 1 | |
04yl | Find all functions $f: (0, \infty) \to [0, \infty)$ such that for all $x, y \in (0, \infty)$ it holds that
$$
f(x + y f(x)) = f(x) f(x + y).
$$ | [
"Any $f$ such that $f(x) \\in \\{0, 1\\}$ for all $x \\in \\mathbb{R}^{+}$ works. Furthermore, any $f$ such that\n$$\nf(x) = \\begin{cases} 0 & \\text{or } 1 \\\\ c & x = x_0 \\\\ 0 & x \\in (x_0, \\infty) \\end{cases}\n$$\nworks as well, where $x_0 > 0$, $c \\ge 0$ are arbitrary constants. We now show that these a... | Czech-Polish-Slovak Mathematical Match | CAPS Match 2025 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All solutions are exactly the following:
1) Any function that takes only the values zero and one on the entire positive real line.
2) For some fixed positive threshold and a fixed nonnegative constant, the function is arbitrary zero or one on inputs below the threshold, equals the fixed constant at the threshold, and i... | |
0i81 | Problem:
Find the sum of the reciprocals of all the (positive) divisors of $144$. | [
"Solution:\nAs $d$ ranges over the divisors of $144$, so does $144 / d$, so the sum of $1 / d$ is $1 / 144$ times the sum of the divisors of $144$. Using the formula for the sum of the divisors of a number (or just counting them out by hand), we get that this sum is $403$, so the answer is $403/144$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 403/144 | |
0bw6 | Let $ABC$ be a triangle, let $O$ and $\gamma$ be its circumcenter and circumcircle, respectively, and let $P$ and $Q$ be distinct points interior to $\gamma$ such that $O, P$ and $Q$ are not collinear. Reflect $O$ in the midpoint of the segment $PQ$ to obtain $R$, then reflect $R$ in the center of the nine-point circle... | [
"\n\nSolution:\n\nUse complex coordinates and the standard convention where $O$ is the origin of the complex plane, and $\\gamma$ is the unit circle; as usual, a lower case Roman letter denotes the complex coordinate of the point denoted by the corresponding upper case Roman letter.\n\nBegi... | Romania | Eleventh STARS OF MATHEMATICS Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Eule... | null | proof only | null | |
0jsm | Problem:
Compute the sum of all integers $1 \leq a \leq 10$ with the following property: there exist integers $p$ and $q$ such that $p$, $q$, $p^{2}+a$ and $q^{2}+a$ are all distinct prime numbers. | [
"Solution:\n\nOdd $a$ fail for parity reasons and $a \\equiv 2(\\bmod 3)$ fail for $\\bmod 3$ reasons. This leaves $a \\in\\{4,6,10\\}$. It is easy to construct $p$ and $q$ for each of these, take $(p, q)=(3,5),(5,11),(3,7)$, respectively."
] | United States | HMMT February | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 20 | |
0h7y | A sequence of real numbers: $a_0, a_1, a_2, \dots, a_{2012}$ satisfies the following conditions:
$$
|a_0 - a_1| = \frac{3^1}{2^0} |a_1 - a_2| = \frac{3^2}{2^1} |a_2 - a_3| = \dots = \frac{3^{2011}}{2^{2010}} |a_{2011} - a_{2012}| = \frac{3^{2012}}{2^{2011}} |a_{2012} - a_0|.
$$
Which values the subtraction: $a_0 - a_{1... | [
"Let $|a_0 - a_1| = k$. We can write down the equations:\n$$\n\\begin{align*}\n a_0 - a_1 &= \\pm k, \\\\\n a_1 - a_2 &= \\pm \\frac{2^0}{3^1} k, \\\\\n a_2 - a_3 &= \\pm \\frac{2^1}{3^2} k, \\\\\n \\dots, \\\\\n a_{2011} - a_{2012} &= \\pm \\frac{2^{2010}}{3^{2011}} k, \\\\\n a_{2012} - a_0 &= \\pm \\frac{2^{2011}... | Ukraine | UkraineMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 0 | |
03n0 | Problem:
Find all polynomials $P(x)$ with integer coefficients such that $P(P(n)+n)$ is a prime number for infinitely many integers $n$. | [
"Solution:\nNote that if $P(n)=0$ then $P(P(n)+n)=P(n)=0$ which is not prime. Let $P(x)$ be a degree $k$ polynomial of the form $P(x)=a_{k} x^{k}+a_{k-1} x^{k-1}+\\cdots+a_{0}$ and note that if $P(n) \\neq 0$ then\n$$\n\\begin{aligned}\n& P(P(n)+n)-P(n)= \\\\\n& \\quad a_{k}\\left[(P(n)+n)^{k}-n^{k}\\right]+a_{k-1}... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | Exactly the constant polynomials equal to a prime, and the linear polynomials of the form −2x + b with b an odd integer. | |
0kzm | A set of 12 tokens—3 red, 2 white, 1 blue, and 6 black—is to be distributed at random to 3 game players, 4 tokens per player. The probability that some player gets all the red tokens, another player gets all the white tokens, and the remaining player gets the blue token can be written as $\frac{m}{n}$, where $m$ and $n... | [
"The situation can be modeled by arranging the 12 tokens in a row, with the first player receiving the first 4 tokens, the second player receiving the next 4 tokens, and the third player receiving the last 4 tokens. There are $\\frac{12!}{3! \\cdot 2! \\cdot 1! \\cdot 6!}$ such arrangements. The given conditions ar... | United States | AMC 12 A | [
"Statistics > Probability > Counting Methods > Combinations",
"Statistics > Probability > Counting Methods > Permutations"
] | null | MCQ | C | |
0b80 | Let $ABC$ be an isosceles triangle with $AB = AC$ and let $n > 1$ be an integer. Point $M$ lies on the line segment $AB$ such that $nAM = AB$. Consider the points $P_1, P_2, \dots, P_{n-1}$ on the side $BC$ with $BP_1 = P_1P_2 = P_2P_3 = \dots = P_{n-1}C = \frac{1}{n}BC$. Prove that
$$
\angle MP_1A + \angle MP_2A + \do... | [
"Consider the point $N$ on the side $AC$ such that $nAN = AC$. The configuration is symmetric with respect to the perpendicular bisector of the segment $BC$, implying $\\angle MP_iA = \\angle NP_{n-i}A$, $i = 1, 2, \\dots, n-1$. The claim is equivalent to $\\angle MP_1N + \\angle MP_2N + \\dots + \\angle MP_{n-1}N ... | Romania | NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
04aa | In how many ways can the number $\frac{2011}{2010}$ be represented as a product of two fractions of the form $\frac{n+1}{n}$, where $n$ is a positive integer? (Order of the factors is not important.) | [
"Let $p$ and $q$ be positive integers such that $\\frac{2011}{2010} = \\frac{p+1}{p} \\cdot \\frac{q+1}{q}$.\nThen $2011pq = 2010(pq + p + q + 1)$, i.e. $pq = 2010(p + q + 1)$.\n\nFrom the last equation we find\n$$\np = \\frac{2010(q+1)}{q-2010} = \\frac{2010(q-2010)+2010 \\cdot 2011}{q-2010} = 2010 + \\frac{2010 \... | Croatia | Hrvatska 2011 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 16 | |
09nj | Let a four digit number $ABCD$ be called a *good number* if all its digits are distinct and nonzero, and the following are all integers: $\frac{CD}{AB}$, $\frac{C}{A}$, $\frac{D}{B}$. For example, the number $1284$ is good because all its digits are distinct, nonzero, and $\frac{84}{12} = 7$, $\frac{8}{1} = 8$, $\frac{... | [] | Mongolia | MMO2025 Round 2 | [
"Number Theory > Divisibility / Factorization"
] | English | final answer only | 1236, 1248, 1284, 1296, 1326, 1428, 1498, 2163, 2184, 2346, 2369, 3162, 3264, 3296, 3468, 4182, 4386 | |
0c8e | a) Let $a$ be a positive integer. Prove that none of the numbers $a^2 + 1, a^2 + 2, \dots, a^2 + 2a$ is a square.
b) Are there positive integers $m, n, p$ such that
$$
m^2 + n + p, \ n^2 + p + m, \ p^2 + m + n
$$
are all squares? | [
"a) The next square after $a^2$ is $(a+1)^2 = a^2 + 2a + 1$, therefore $a^2 + 1, a^2 + 2, \\dots, a^2 + 2a$ fit between two consecutive squares, hence they are not squares.\n\nb) Suppose, by way of contradiction, that such numbers exist. Then $m^2 + n + p > m^2$, hence $m^2 + n + p \\ge m^2 + 2m + 1$. Writing the o... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | No | |
09ja | Let $n \ge 3$. Let a sequence $a_1, a_2, \dots, a_n$ of integers, the following operations are admitted. For any $1 \le k \le n-2$, we can
1) increase $a_k$ and $a_{k+2}$ by 1 each, and decrease $a_{k+1}$ by 1, or
2) decrease $a_k$ and $a_{k+2}$ by 1 each, and increase $a_{k+1}$ by 1.
Starting from the sequence $1, 2, ... | [
"Answer: No.\nFor a sequence $a_1, a_2, \\dots, a_n$, we associate the remainder\n$$\nS = S(a_1, \\dots, a_n) = \\sum_{k=1}^{n} 3^k a_k \\pmod{7}.\n$$\nFor $c = \\pm 1$, the operation $(a_k, a_{k+1}, a_{k+2}) \\to (a_k+c, a_{k+1}-c, a_{k+2}+c)$ transforms $S$ to $S+3^k 7c \\equiv S \\pmod{7}$, thus $S$ is an invari... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | No | |
035n | Problem:
Let $a_{1}, a_{2}, \ldots, a_{2005}, b_{1}, b_{2}, \ldots, b_{2005}$ be real numbers such that the inequality
$$
\left(a_{i} x-b_{i}\right)^{2} \geq \sum_{j=1, j \neq i}^{2005}\left(a_{j} x-b_{j}\right)
$$
holds true for every real number $x$ and all $i=1,2, \ldots, 2005$. Find the maximum possible number of t... | [
"Solution:\nWe first prove that at least one of the numbers $a_{1}, a_{2}, \\ldots, a_{2005}$ is not positive. To do this we assume the contrary and choose $i$ such that\n$$\n\\frac{b_{i}}{a_{i}}=M=\\max _{1 \\leq j \\leq 2005}\\left(\\frac{b_{j}}{a_{j}}\\right)\n$$\nThen we can find $\\varepsilon>0$ such that\n$$\... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 4009 | |
09x2 | Problem:
Op een cirkel met middelpunt $M$ liggen drie verschillende punten $A$, $B$ en $C$ zodat $|AB| = |BC|$. Punt $D$ ligt binnen de cirkel op zo'n manier dat $\triangle BCD$ gelijkzijdig is. Het tweede snijpunt van $AD$ met de cirkel noemen we $F$. Bewijs dat $|FD| = |FM|$.
 | [
"Solution:\nWe gaan bewijzen dat $|FD| = |FC|$ en dat $|FC| = |FM|$, waaruit het gevraagde volgt.\n\nIn koordenvierhoek $ABCF$ is $\\angle BCF = 180^{\\circ} - \\angle BAF$. Wegens $|AB| = |BC| = |BD|$ geldt verder $\\angle BAF = \\angle BAD = \\angle ADB$, dus $\\angle BDF = 180^{\\circ} - \\angle ADB = 180^{\\cir... | Netherlands | Selectietoets | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kh2 | Problem:
Let $n$ be a positive integer. Given that $n^{n}$ has 861 positive divisors, find $n$. | [
"Solution:\nIf $n = p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\ldots p_{k}^{\\alpha_{k}}$, we must have $\\left(n \\alpha_{1} + 1\\right)\\left(n \\alpha_{2} + 1\\right) \\ldots\\left(n \\alpha_{k} + 1\\right) = 861 = 3 \\cdot 7 \\cdot 41$.\n\nIf $k = 1$, we have $n \\mid 860$, and the only prime powers dividing 860... | United States | HMMT November 2021 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 20 | |
09ga | Тус бүр $1$, $2$, $\ldots$, $99$, $100$ грам жинтэй $100$ ширхэг туухайг жинлүүрийн $2$ таваг дээр жин тэнцүү байхаар хувааж тавив. Үлдсэн туухайнуудын жин тэнцүү байхаар жинлүүрийн таваг тус бүрээс хоёр, хоёр туухайг авч болохыг батал. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
088i | Problem:
Da un punto $L$ partono due strade rettilinee che formano un angolo acuto $\alpha$. Lungo una delle due strade ci sono due lampioni, posizionati in $P$ e $Q$, tali che $L P=40~\mathrm{m}$ e $L Q=90~\mathrm{m}$. Eva si trova in $E$ sull'altra strada, e vede i due lampioni sotto un angolo $P \widehat{E} Q$. A c... | [
"Solution:\n\nLa risposta è (B). Sia $f$ la strada dove si trova Eva ed $s$ quella dove si trovano i lampioni. Consideriamo la circonferenza passante per $P, Q, E$, che esiste in quanto $P, Q$ appartengono a $s$ mentre $E$ non vi appartiene se no l'angolo $P \\hat{E} Q$ sarebbe 0, e supponiamo per assurdo essa non ... | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | MCQ | B | |
0byq | Prove that, if a ring $R$ is not a skew-field, and $x^2 = x$ for every non-invertible element $x$ of $R$, then $x^2 = x$ for every element $x$ of $R$. | [
"*First solution.* An element $x$ of $R$ such that $x^2 = x$ is called *idempotent*. We show that $1$ is the only unit in $R$. The proof is based on the remark below:\n(*) Let $x$ and $y$ be elements of a ring $R$. If $x \\neq 1$ is idempotent, then $xyz \\neq 1$ and $zyx \\neq 1$ for all $z$ in $R$; in particular,... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Abstract Algebra > Ring Theory"
] | English | proof only | null | |
02om | We have a red cube with sidelength $2$ cm. What is the minimum number of identical cubes that must be adjoined to the red cube in order to obtain a cube with volume $\left(\frac{12}{5}\right)^3$ cm? | [
"The bigger cube has sidelength $\\frac{12}{5}$ cm, so the difference between the side-lengths is $\\frac{12}{5} - 2 = \\frac{2}{5}$ cm, that is, the red cubes should not have sidelength greater than this length. Cubes with sidelength $\\frac{2}{5}$ cm are the natural candidates, so we set a new unit $u = \\frac{2}... | Brazil | Brazilian Math Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Volume"
] | English | proof and answer | 91 | |
0gkx | Let $P_1, P_2, \dots, P_{2556}$ be distinct interior points of a regular hexagon $ABCDEF$ with side length $1$. Assume that no three points of the set
$$
S = \{A, B, C, D, E, F, P_1, P_2, \dots, P_{2556}\}
$$
are collinear. Show that there is a triangle with area less than $\frac{1}{1700}$ all of whose vertices belong ... | [
"Draw line segments from $P_1$ to the points $A, B, C, D, E, F$, we get six smaller triangles. Since any three points in $S$ are not collinear, $P_2$ is inside one of the six triangles. Draw line segments joining $P_2$ with the vertices of this triangle to divide it into three triangles. The total number of triangl... | Thailand | The 10th Thailand Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0j91 | Problem:
Let $ABC$ be a triangle with $AB = 9$, $BC = 10$, and $CA = 17$. Let $B'$ be the reflection of the point $B$ over the line $CA$. Let $G$ be the centroid of triangle $ABC$, and let $G'$ be the centroid of triangle $AB'C$. Determine the length of segment $GG'$. | [
"Solution:\nAnswer: $\\frac{48}{17}$\n\nLet $M$ be the midpoint of $AC$. For any triangle, we know that the centroid is located $2/3$ of the way from the vertex, so we have $MG / MB = MG' / MB' = 1/3$, and it follows that $MGG' \\sim MBB'$. Thus, $GG' = BB'/3$. However, note that $BB'$ is t... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 48/17 | |
0b7w | How many four digit numbers $abcd$ simultaneously satisfy the equalities $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$? | [
"From $a + b = c + d$ we get $(a + b)^2 = (c + d)^2$, hence $ab = cd$ and furthermore $a^2 - 2ab + b^2 = c^2 - 2cd + d^2$. As $(a - b)^2 = (c - d)^2$, we have $|a - b| = |c - d|$, which implies $a - b = c - d$ or $a - b = d - c$. Recall that $a + b = c + d$, so either $a = c, b = d$ or $a = d, b = c$.\nThe numbers ... | Romania | Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | 171 | |
0f15 | Problem:
Find all real $a$, $b$, $c$ such that $a x + b y + c z + b x + c y + a z + c x + a y + b z = x + y + z$ for all real $x$, $y$, $z$. | [] | Soviet Union | ASU | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a + b + c = 1 | |
0ezm | Problem:
Given any set $S$ of $25$ positive integers, show that you can always find two such that none of the other numbers equals their sum or difference. | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
07li | Denote by $m_a$, $m_b$, $m_c$, respectively, the lengths of the medians from the vertices $A$, $B$, $C$ of a triangle $ABC$ to the opposite sides, respectively. Prove that
$$
m_a + m_b + m_c \geq a \sin A + b \sin B + c \sin C
$$
with equality iff the triangle is equilateral. | [
"Let the median from $A$ meet the side $BC$ at $D$, so that $|BD| = |DC| = a/2$. We begin by deriving an expression for $m_a$. Let $\\theta = \\angle BDA$.\n\n\n\nBy the Cosine Rule, $2|BD||AD| \\cos \\theta = |AD|^2 + |BD|^2 - |AB|^2$, i.e.\n$$\nam m_a \\cos \\theta = m_a^2 + \\frac{a^2}{4... | Ireland | Irska | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
0j08 | Problem:
Let $f$ be a function such that $f(0)=1$, $f'(0)=2$, and
$$
f''(t)=4 f'(t)-3 f(t)+1
$$
for all $t$. Compute the 4th derivative of $f$, evaluated at $0$. | [
"Solution:\nPutting $t=0$ gives $f''(0)=6$.\n\nBy differentiating both sides, we get\n$$\nf^{(3)}(t)=4 f''(t)-3 f'(t)\n$$\nand\n$$\nf^{(3)}(0)=4 \\cdot 6 - 3 \\cdot 2 = 18.\n$$\nSimilarly,\n$$\nf^{(4)}(t)=4 f^{(3)}(t)-3 f''(t)\n$$\nand\n$$\nf^{(4)}(0)=4 \\cdot 18 - 3 \\cdot 6 = 54.\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Equations > ODEs"
] | null | final answer only | 54 | |
0h7a | In acute-angled triangle $ABC$ bisector $AL$, height $BH$ and the perpendicular bisector of line $AB$ intersect at one point. Find the angle $BAC$.
**Answer:** $\angle BAC = 60^\circ$. | [
"Let the angle $BAC$ be $2\\alpha$ (fig. 15). Hence $\\triangle APB$ is isosceles, thus $\\angle PBA = \\alpha$. Since $\\triangle AHB$ is right triangle, $3\\alpha = 90^\\circ$, thus $\\angle BAC = 2\\alpha = 60^\\circ$."
] | Ukraine | UkraineMO | [
"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 and answer | 60 degrees | |
0524 | Find the greatest natural number $n$ for which it is possible to choose $n$ vertices of a cube such that no three of them form a right triangle.
Figure 9 | [
"Let some vertex of a cube be $A$ and let $B$, $C$ and $D$ be the opposite vertices of the faces of the cube that $A$ belongs to (see fig. 9). Then of the vertices $B$, $C$ and $D$ any two are also the opposite vertices of some face of the cube. Therefore any two of the chosen four vertices are at the distance of a... | Estonia | Final Round of National Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 4 | |
0059 | Sea $ABCDE$ un pentágono convexo que cumple las siguientes condiciones:
* Existe una circunferencia $\Gamma$ tangente a cada uno de sus lados.
* Las longitudes de todos sus lados son números enteros.
* Por lo menos uno de los lados del pentágono mide $1$.
* El lado $AB$ mide $2$.
Sea $P$ el punto de tangencia de $\Gamm... | [] | Argentina | XVIII Olimpiada Matemática del Cono Sur | [
"Geometry > Plane Geometry > Circles > Tangents"
] | Español | proof and answer | a) AP and BP are 1/2 and 3/2 (in some order). b) One example is a tangential pentagon with consecutive side lengths 2, 2, 1, 2, 2 (for instance, with tangent lengths around the pentagon 1/2, 3/2, 1/2, 1/2, 3/2). | |
0j6d | Problem:
Let $S$ be the set of points $(x, y, z)$ in $\mathbb{R}^3$ such that $x, y$, and $z$ are positive integers less than or equal to $100$. Let $f$ be a bijective map between $S$ and the $\{1,2, \ldots, 1000000\}$ that satisfies the following property: if $x_1 \leq x_2, y_1 \leq y_2$, and $z_1 \leq z_2$, then $f\l... | [
"Solution:\nWe examine the $6$ planes, their intersections and the lines between $2$ points in one of the three pairs of parallel planes. The expression is equivalent to summing differences in values along all these lines. We examine the planes intersections. There is one cube, $3 \\cdot 98$ squares and $3 \\cdot 9... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 30604 | |
082s | Problem:
Si vogliono regalare sette pacchi dono a sette bambini, uno a ciascuno. Si vuol fare in modo che in ciascun pacco ci siano tre giochi diversi e che, comunque si scelgano due bambini, essi ricevano al più un gioco in comune. Qual è il minimo numero di tipi di giochi distinti che è necessario usare? | [
"Solution:\n\nLa risposta è $7$. Che sia possibile realizzare sette pacchi con le proprietà richieste con sette tipi di giochi è dimostrato dalla seguente griglia, in cui $A$, $B$, $C$, $D$, $E$, $F$ e $G$ indicano i tipi di giochi e le colonne danno la composizione dei pacchi.\n\n| $\\mathrm{A}$ | $\\mathrm{A}$ | ... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 7 | |
0jg5 | Problem:
Let $\omega_{1}$ and $\omega_{2}$ be circles with centers $O_{1}$ and $O_{2}$, respectively, and radii $r_{1}$ and $r_{2}$, respectively. Suppose that $O_{2}$ is on $\omega_{1}$. Let $A$ be one of the intersections of $\omega_{1}$ and $\omega_{2}$, and $B$ be one of the two intersections of line $O_{1} O_{2}$... | [
"Solution:\n\nAnswer: $\\frac{-1+\\sqrt{5}}{2}, \\frac{1+\\sqrt{5}}{2}$\n\nThere are two configurations to this problem, namely, $B$ in between the segment $O_{1} O_{2}$ and $B$ on the ray $O_{1} O_{2}$ passing through the side of $O_{2}$.\n\nCase 1:\n\nLet us only consider the triangle $A B O_{2}$. $A B = A O_{1} ... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | (-1 + sqrt(5)) / 2, (1 + sqrt(5)) / 2 | |
02kn | Problem:
São dadas 4 moedas aparentemente iguais, das quais 3 são verdadeiras e por isso têm o mesmo peso; uma é falsa e por isso tem peso diferente. Não se sabe se a moeda falsa é mais leve ou mais pesada que as demais. Mostre que é possível determinar a moeda diferente empregando somente duas pesagens em uma balança... | [
"Solution:\n\nSejam $A$, $B$, $C$ e $D$ as quatro moedas. Comparamos as moedas $A$ e $B$ na balança, colocando uma em cada prato. Dois casos podem ocorrer: a balança fica em equilíbrio ou a balança não fica em equilíbrio. Vamos analisar separadamente cada caso.\n\n$1^{\\circ}$ Caso: A balança fica equilibrada. Pode... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Logic"
] | null | proof only | null | |
01f6 | Let $F_n$ be a sequence defined recursively by $F_1 = F_2 = 1$ and $F_{n+1} = F_n + F_{n-1}$ for $n \ge 2$. Find all pairs of positive integers $(x, y)$ such that
$$
5F_x - 3F_y = 1
$$ | [
"From the equation $5F_x = 3F_y + 1$ we have\n$$\n3F_y + 1 = 5F_x > 3F_x + 1 \\implies y > x\n$$\nOn the other hand, if $y \\ge x + 2$ and $x > 1$ then\n$$\n3F_y + 1 \\ge 3F_{x+2} + 1 = 3(F_{x+1} + 3F_x) + 1 = 6F_x + 3F_{x-1} + 1 > 5F_x\n$$\nwe have a contradiction. Therefore $y = x + 1$ and we have to solve the eq... | Baltic Way | Baltic Way 2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (3,4), (5,6), (6,7) | |
0cab | Problem:
Determinaţi numerele întregi nenule $a$ pentru care există funcţiile $f, g: \mathbb{Q} \rightarrow \mathbb{Q}$ care verifică ecuatia funcţională:
$$
f(x+g(y))=g(x)+f(y)+a y, \text{ oricare ar fi } x, y \in \mathbb{Q}
$$
Determinaţi toate aceste funcţii. | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ Etapa a III-a | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | a must be of the form k(k−1) with k ∈ Z and k ≠ 0,1. For each such a, let p be either integer root of p^2 − p = a (equivalently p = (1 ± √(1+4a))/2 ∈ Z). Then all solutions are g(x) = p x and f(x) = p x + r, where r ∈ Q is arbitrary. | |
0gfm | 在一個 $100 \times 100$ 格西洋棋盤的每一格內填入一個非負實數。我們稱這個棋盤是平衡的, 若且唯若該棋盤上每一直排的數字總和都是 1, 且每一橫列的數字總和也都是 1。求最大的正實數 $x$, 使得在任何一個平衡棋盤中, 我們都能挑出 100 個格子, 滿足任兩個格子都在不同直排且不同橫列, 並且這些格子裡的數字都不小於 $x$。 | [
"最大的 $x$ 為 $\\frac{1}{50 \\times 51}$。一般性的，對於 $n \\times n$ 的棋盤，令 $a = \\lfloor \\frac{n+1}{2} \\rfloor$，$b = \\lfloor \\frac{n+2}{2} \\rfloor$，則最大的 $x$ 為 $\\frac{1}{ab}$。\n\n首先構造達到此 $x$ 的棋盤。讓我們將棋盤分為四區：\n- 左上角的 $a \\times b$ 格都填 $\\frac{1}{ab}$;\n- 右上角的 $(n-a) \\times b$ 格都填 $\\frac{1}{b}$;\n- 左下角的 $a \\times (n-b)... | Taiwan | 2022 數學奧林匹亞競賽第二階段選訓營, 國際競賽實作(一) | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Chinese; English | proof and answer | 1/2550 | |
0isd | Three nonnegative real numbers $r_1$, $r_2$, $r_3$ are written on a blackboard. These numbers have the property that there exist integers $a_1$, $a_2$, $a_3$, not all zero, satisfying $a_1 r_1 + a_2 r_2 + a_3 r_3 = 0$. We are permitted to perform the following operation: find two numbers $x$, $y$ on the blackboard with... | [
"If two of the $a_i$ vanish, say $a_2$ and $a_3$, then $r_1$ must be zero and we are done. Assume at most one $a_i$ vanishes. If any one $a_i$ vanishes, say $a_3$, then $r_2/r_1 = -a_1/a_2$ is a nonnegative rational number. Write this number in lowest terms as $p/q$, and put $r = r_2/p = r_1/q$. We can then write $... | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0i13 | Problem:
Find the minimum of the function
$$
f(x, y)=\sqrt{(x+1)^{2}+(2 y+1)^{2}}+\sqrt{(2 x+1)^{2}+(3 y+1)^{2}}+\sqrt{(3 x-4)^{2}+(5 y-6)^{2}},
$$
defined for all real $x, y>0$. | [
"Solution:\nNote that $\\sqrt{(x+1)^{2}+(2 y+1)^{2}}$ is the distance in the coordinate plane from $(0,0)$ to $(x+1,2 y+1)$; $\\sqrt{(2 x+1)^{2}+(3 y+1)^{2}}$ is the distance from $(x+1,2 y+1)$ to $(3 x+2,5 y+2)$; $\\sqrt{(3 x-4)^{2}+(5 y-6)^{2}}$ is the distance from $(3 x+2,5 y+2)$ to $(6,8)$. Thus, $f(x, y)$ is ... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 10 | |
0d1n | Determine all positive integers $k$ such that there is a function $f: \mathbb{N} \to \mathbb{N}$ satisfying the following condition: for every positive integer $n$, we have both $f(f(n)) = kn$ and $f(n) < f(n+1)$. | [
"The answer is all $k \\neq 2$. If $k = 2$, note that $f(f(n)) = 2n$ implies that $f(n) \\neq n$ for all $n$. Therefore $f(1) \\neq 1$. We also have $f(1) \\neq 2$ since then $f(f(1)) = f(2) = 2$, so $f(1) > 2$. But then $2 = f(f(1)) < f(1)$ while $f(1) > 1$, so we have a contradiction of the second condition.\n\nF... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Algorithms"
] | English | proof and answer | all positive integers except 2 | |
06k1 | The sequence $\{x_n\}$ is defined by $x_1 = 5$ and $x_{k+1} = x_k^2 - 3x_k + 3$ for $k = 1, 2, 3, \dots$. Prove that $x_k > 3^{2^{k-1}}$ for any positive integer $k$. | [
"The recurrence relation can be rewritten as $x_{k+1} - 3 = x_k(x_k - 3)$. By repeating the process, we obtain\n$$\nx_{k+1} - 3 = x_k(x_k - 3) = x_k x_{k-1}(x_{k-1} - 3) = \\cdots = x_k x_{k-1} \\cdots x_1(x_1 - 3).\n$$\nChanging the index, this gives $x_k = 10x_2 \\cdots x_{k-1} + 3$. We now prove the assertion by... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0178 | For a positive integer $k$, let $d(k)$ denote the number of divisors of $k$ (e.g. $d(12) = 6$) and let $s(k)$ denote the digit sum of $k$ (e.g. $s(12) = 3$). A positive integer $n$ is said to be *amusing* if there exists a positive integer $k$ such that $d(k) = s(k) = n$. What is the smallest amusing odd integer greate... | [
"The answer is $9$. For every $k$ we have $s(k) \\equiv k \\pmod{9}$. Calculating remainders modulo $9$ we have the following table\n\n| $m$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |\n|-----|---|---|---|---|---|---|---|---|---|\n| $m^2$ | 0 | 1 | 4 | 0 | 7 | 7 | 0 | 4 | 1 |\n| $m^6$ | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 ... | Baltic Way | BALTIC WAY | [
"Number Theory > Modular Arithmetic",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 9 | |
00pj | Let $N$ be a nonnegative integer and let $f, g: \mathbb{Z} \to [0, \infty)$ be functions such that $f(n) = g(n) = 0$ for all $|n| \geq N$ where $\mathbb{Z}$ is the set of all integers. Define $h: \mathbb{Z} \to [0, \infty)$ by
$$
h(n) = \max \{f(k)g(n-k) : k \in \mathbb{Z}\}
$$
for all $n \in \mathbb{Z}$. Prove that
$$... | [
"Let $m_0$ be an integer at which $f$ achieves its maximum. Then $h(n) \\geq f(m_0)g(n - m_0)$ for all $n \\in \\mathbb{Z}$ and\n$$\n\\sum_{n \\in \\mathbb{Z}} h(n) \\geq f(m_0) \\sum_{n \\in \\mathbb{Z}} g(n - m_0) = f(m_0) \\sum_{n \\in \\mathbb{Z}} g(n).\n$$\nSimilarly, if $n_0$ is an integer at which $g$ achiev... | Balkan Mathematical Olympiad | Balkan 2012 shortlist | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0e1w | For a real number $t$ and positive real numbers $a$ and $b$ we have
$$
2a^2 - 3abt + b^2 = 2a^2 + abt - b^2 = 0.
$$
Find $t$. | [
"From $2a^2 - 3ab t + b^2 = 0$ we get $t = \\frac{2a^2 + b^2}{3ab}$ and from $2a^2 + ab t - b^2 = 0$ we get $t = \\frac{b^2 - 2a^2}{ab}$. So, $\\frac{2a^2 + b^2}{3ab} = \\frac{b^2 - 2a^2}{ab}$. Eliminating the fractions we obtain $8a^2 = 2b^2$ or $4a^2 = b^2$. Since $a$ and $b$ are positive, we conclude that $b = 2... | Slovenia | National Math Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 1 | |
01id | Find all $k \in \mathbb{Z}$ such that there exists a function $f : \mathbb{Z} \to \mathbb{Z}$ satisfying
$$
f(f(n)) = n + k
$$
for all $n \in \mathbb{Z}$. | [
"If $k \\in \\mathbb{Z}$ is even then for $f : \\mathbb{Z} \\to \\mathbb{Z}$, $f(x) = x + \\frac{k}{2}$ we get:\n$$\nf(f(n)) = \\left(n + \\frac{k}{2}\\right) + \\frac{k}{2} = n + k\n$$\nFor $k$ even it is therefore possible to find function $f : \\mathbb{Z} \\to \\mathbb{Z}$ with the property that $f(f(n)) = n + k... | Baltic Way | Baltic Way 2021 Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All even integers k | |
08y5 | Let $l$ be a straight line lying on the $xy$-plane. For given $20 \times 15$ points $(m, n) : (m = 1, 2, \dots, 20, n = 1, 2, \dots, 15)$ on the $xy$-plane, there are $222$ straight lines parallel to the line $l$ ($l$ itself may be considered as one of those parallel lines) and going through at least one of these point... | [
"212\nLet us call a point $(m, n)$ a lattice point if both $m$ and $n$ are integers. Call a lattice point $(m, n)$ a good point if it lies in the portion of the $xy$-plane given by $1 \\le x \\le 20$, $1 \\le y \\le 15$.\nIf a straight line $\\ell$ lying on the $xy$-plane is parallel either to the $x$-axis or to th... | Japan | Japan 2015 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 212 | |
06i6 | $2n + 1$ distinct points are chosen on a circle and each two of them are connected with a vector going in one of the two possible directions. Let $R$ be the number of triangles with the vertices at the given points such that the sum of the vectors going along the sides of the triangle is equal to zero, (i.e. starting f... | [
"The smallest and the largest possible values of $R$ are $0$ and $\\frac{n(n+1)(2n+1)}{6}$ respectively.\nLet the points be $A_1, A_2, \\dots, A_{2n+1}$. If we draw the vector $\\overrightarrow{A_iA_j}$ whenever $i < j$, then for any $\\triangle A_iA_jA_k$ with $i < j < k$, we have\n$$\n\\overrightarrow{A_iA_j} + \... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | minimum 0, maximum n(n+1)(2n+1)/6 | |
00zi | Problem:
$ABCD$ is a trapezium ($AD \parallel BC$). $P$ is the point on the line $AB$ such that $\angle CPD$ is maximal. $Q$ is the point on the line $CD$ such that $\angle BQA$ is maximal. Given that $P$ lies on the segment $AB$, prove that $\angle CPD = \angle BQA$. | [
"Solution:\n\nThe property that $\\angle CPD$ is maximal is equivalent to the property that the circle $CPD$ touches the line $AB$ (at $P$). Let $O$ be the intersection point of the lines $AB$ and $CD$, and let $\\ell$ be the bisector of $\\angle AOD$. Let $A'$, $B'$ and $Q'$ be the points symmetrical to $A$, $B$ a... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
04dk | Determine all ordered triplets of real numbers $(x, y, z)$ such that
$$
\begin{cases}
x + y = \sqrt{28} \\
xy - 2z^2 = 7.
\end{cases}
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (sqrt(7), sqrt(7), 0) | |
05mx | Problem:
Soit $ABC$ un triangle. Soit $P$ appartenant au cercle circonscrit. On sait que les projetés de $P$ sur $(BC)$, $(CA)$ et $(AB)$ sont alignés sur la droite dite de Simson. On suppose que cette droite passe par le point diamétralement opposé à $P$. Montrer qu'elle passe également par le centre de gravité de $A... | [
"Solution:\n\n\n\nNotons $P'$ le point diamétralement opposé à $P$. Soit $\\Delta$ la droite de Simson. Soit $h$ l'homothétie de centre $P$ et de rapport $2$. Alors $h(\\Delta)$ est la droite de Steiner. On sait qu'elle passe par l'orthocentre $H$, donc $\\Delta$ passe par le milieu de $[PH... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0jor | Problem:
Let $f: \mathbb{R}^{+} \rightarrow \mathbb{R}$ be a continuous function satisfying $f(x y)=f(x)+f(y)+1$ for all positive reals $x, y$. If $f(2)=0$, compute $f(2015)$. | [
"Solution:\nAnswer: $\\log_{2} 2015-1$\nLet $g(x)=f(x)+1$. Substituting $g$ into the functional equation, we get that\n$$\n\\begin{gathered}\ng(x y)-1=g(x)-1+g(y)-1+1 \\\\\ng(x y)=g(x)+g(y)\n\\end{gathered}\n$$\nAlso, $g(2)=1$. Now substitute $x=e^{x'}$, $y=e^{y'}$, which is possible because $x, y \\in \\mathbb{R}^... | United States | HMMT November 2015 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | log_2 2015 - 1 | |
05z6 | Problem:
Un jardinier et un pivert jouent au jeu suivant, dans leur jardin dont la forme est celle d'une grille $2022 \times 2022$ formée de $2022^{2}$ cases. Deux cases sont considérées comme voisines si elles ont un sommet ou une arête en commun. Initialement, chaque case abrite un arbre de taille 0. Puis, à chaque ... | [
"Solution:\n\nNous allons démontrer que l'entier recherché est $A=5 n=2271380$, où l'on a posé $n=674^{2}=(2022 / 3)^{2}$.\n\nTout d'abord, voici une stratégie pour le pivert. Il numérote les lignes et les colonnes de 1 à 2022, puis colorie en noir chaque case située dans une colonne $c$ et une ligne $\\ell$ pour l... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combi... | null | proof and answer | 2271380 | |
09fl | Let $ABCD$ be a circumscribed quadrilateral with circumcenter $O$ and let $K$, $L$, $M$ and $N$ denote the midpoints of the sides $AB$, $BC$, $CD$ and $DA$ respectively. Suppose that the lines $KM$ and $LN$ do not pass through $O$. Let $E := KM \cap LN$. Point $P$ is chosen on the interval $KM$ to satisfy $\angle KOE =... | [
"It suffices to prove that $\\angle POQ = 180^\\circ$.\n\n\n\nSince $O$ is the circumcenter, the points $K$, $L$, $M$, $N$ are the feet of the perpendiculars from $O$ to the sides of $ABCD$. Hence $OMDN$ and $OKBL$ are circumscribed. Hence\n$$\n\\begin{align*}\n\\angle POQ &= \\angle POM + ... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0426 | In a plane with cartesian coordinates, let $P$ and $Q$ be two regions of convex polygon (including boundary and interior) whose vertices are all integer points (i.e., their coordinates are all integers) and $T = P \cap Q$. Prove that if $T$ is not empty and does not contain integer point, then $T$ is a non-degenerate c... | [
"Since the non-empty intersection $T$ of two convex closed polygons is a closed convex polygon or degenerated polygon, there are three possible cases.\n\n(1) $T$ is a point. Then $T$ must be the vertex of $P$ or $Q$, contradicting the fact that $T$ contains no integer point.\n\n(2) $T$ is a segment. Then $T$ must b... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
0bec | Triangle $ABC$ is inscribed in the circle $C$ and $AA'$, $BB'$, $CC'$ are diameters of this circle. Denote $H_1, H_2, H_3$ the orthocenters of the triangles $A'BC$, $AB'C$, $ABC'$ respectively.
a) Prove that there exists a triangle $PQR$ with sides equal to $[AH_1]$, $[BH_2]$, $[CH_3]$ and compute the ratio $S_{PQR}/S... | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | 3 | |
04n5 | The rows of a $50 \times 50$ table are labelled with numbers $a_1, \dots, a_{50}$, while the columns are labelled with numbers $b_1, \dots, b_{50}$. These 100 numbers are mutually distinct, and exactly 50 of them are rational. The table is filled so that the number $a_i + b_j$ is written in the $(i, j)$ cell, for $i, j... | [] | Croatia | Croatia_2018 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Number Theory > Other"
] | English | proof and answer | 1250 | |
09zu | We call a positive integer *sunny* if it has four digits and if moreover each of the two digits on the outside is exactly 1 larger than the digit next to it. The numbers $8723$ and $1001$ for example are sunny, but $1234$ and $87245$ are not.
a) How many sunny numbers are there such that twice the number is again a su... | [
"a. First we look at the last two digits of a sunny number. There are nine possibilities for these: $01$, $12$, $23$, $34$, $45$, $56$, $67$, $78$, and $89$. If we then look at twice a sunny number, we get the following nine possibilities, respectively, for the last two digits: $02$, $24$, $46$, $68$, $90$, $12$, $... | Netherlands | Second Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 16 | |
0cvg | Let $a_1, a_2, a_3, \dots$ be an infinite sequence of real numbers such that for each $n \ge 2018$ the number $a_{n+1}$ is the smallest root of the polynomial
$$
P_n(x) = x^{2n} + a_1 x^{2n-2} + a_2 x^{2n-4} + \dots + a_n.
$$
Prove that there exists a positive integer $N$ such that the sequence $a_N, a_{N+1}, a_{N+2}, ... | [
"Пусть $n \\ge 2018$. Заметим, что $P_n(a) = P_n(-a)$ при всех $a$. Значит, поскольку $P_n(x)$ имеет ненулевой корень, он имеет и отрицательный корень, откуда $a_{n+1} < 0$.\n\nДалее, поскольку $P_{n+1}(x) = x^2 P_n(x) + a_{n+1}$, имеем\n$$P_{n+1}(a_{n+1}) = a_{n+1}^2 P_n(a_{n+1}) + a_{n+1} = 0 + a_{n+1} < 0. \\ (*... | Russia | Regional round | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English; Russian | proof only | null | |
0h0p | Olesya was given a homework to add two canonical fractions $\frac{a}{b}$ and $\frac{c}{d}$. Her classmate Andriy who missed the class asked her by phone about the homework, and due to bad connection he heard that they need to add $\frac{b}{a}$ and $\frac{d}{c}$. After he added them, Andriy asked Olesya for the answer. ... | [
"Suppose his calculations are correct, then the following equality holds: $\\frac{a}{b} + \\frac{c}{d} = \\frac{b}{a} + \\frac{d}{c}$ or $\\frac{ad+bc}{bd} = \\frac{bc+ad}{ac}$. Therefore, we have $bd = ac$ or $\\frac{b}{a} = \\frac{c}{d}$, which contradicts to the assumption that all fractions are distinct."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | No | |
02ae | Problem:
Diogo e Helen jogam o Jogo do Tira, que consiste no seguinte. Dado um quadriculado de quadrados $1 \times 1$, cada jogador, em sua vez, tem o direito de escolher um quadrado e então retirar do quadriculado todos os quadrados abaixo dele, todos os quadrados à esquerda dele, e todos os outros que estejam abaixo... | [
"Solution:\n\na) Helen começa tirando o quadrado abaixo:\n\nDiogo tem duas opções agora: se tira o quadrado abaixo, então Helen tira o quadrado seguinte e ganha a partida, pois Diogo terá que tirar o último quadrado.\n\nE se Diogo tira o quadrado a seguir, então Hel... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null |
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