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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
066i | Solve in the set of integers the equation: $y = 2x^2 + 5xy + 3y^2$. | [
"The equation can be written as\n$$\ny = 2x^2 + 2xy + 3xy + 3y^2 \\Leftrightarrow y = (x+y)(2x+3y). \\quad (1)\n$$\nBy putting $x+y=z \\in \\mathbb{Z}$, then equation (1) is written\n$$\ny = z(2z + y) \\Leftrightarrow y = 2z^2 + yz \\Leftrightarrow (z-1)y = -2z^2. \\quad (2)\n$$\nFor $z=1$ equation (2) becomes: $0 ... | Greece | Hellenic Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | [(0,0), (10,-8), (-2,1), (12,-9)] | |
0k7b | Problem:
Find the smallest positive integer $n$ such that
$$
\underbrace{2^{2^{2} \cdots^{2}}}_{n\ 2\text{\prime}s} > \underbrace{((\cdots((100!)!)!\cdots)!)!}_{100 \text{ factorials }}
$$ | [
"Solution:\nNote that $2^{2^{2^{2}}} > 100^{2}$. We claim that $a > b^{2} \\Longrightarrow 2^{a} > (b!)^{2}$, for $b > 2$. This is because\n$$\n2^{a} > b^{2b} \\Longleftrightarrow a > 2b \\log_{2}(b)\n$$\nand $\\log_{2}(b) < b^{2}/2$ for $b > 2$. Then since $b^{b} > b!$ this bound works. Then\n$$\n\\underbrace{\\le... | United States | HMMT February 2019 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | 104 | |
00mx | In the country of Oddland, there are stamps with values $1$ cent, $3$ cent, $5$ cent, etc., one type for each odd number. The rules of Oddland Postal Services stipulate the following: for any two distinct values, the number of stamps of the higher value on an envelope must never exceed the number of stamps of the lower... | [
"We construct a bijection between possible combinations in Oddland and possible combinations in Squareland. Suppose we have a combination of Squareland stamps that sum to $n$ cent, consisting of $a_1$ stamps of value $1$ cent, $a_2$ stamps of value $4$ cent, ..., $a_M$ stamps of value $M^2$ cent, so that\n$$\nn = \... | Austria | Austria2019 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | English | proof only | null | |
0024 | Consideramos los 2004 números enteros $n$, desde 1 hasta 2004. Determinar para cuántos de estos valores de $n$ se verifica que el número $n^3+3^n$ es múltiplo de 5. | [] | Argentina | XX OLIMPIADA MATEMÁTICA ARGENTINA | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | español | proof and answer | 401 | |
0ew7 | Problem:
Given unequal integers $x$, $y$, $z$ prove that $\left(x - y\right)^{5} + \left(y - z\right)^{5} + \left(z - x\right)^{5}$ is divisible by $5(x - y)(y - z)(z - x)$. | [
"Solution:\nPut $x - y = r$, $y - z = s$. Then $z - x = - (r + s)$, and\n$$\n\\left(x - y\\right)^{5} + \\left(y - z\\right)^{5} + \\left(z - x\\right)^{5} = r^{5} + s^{5} - (r + s)^{5}\n$$\nExpand $(r + s)^5$:\n$$\n(r + s)^5 = r^5 + 5r^4s + 10r^3s^2 + 10r^2s^3 + 5rs^4 + s^5\n$$\nSo,\n$$\nr^5 + s^5 - (r + s)^5 = r^... | Soviet Union | 2nd ASU | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0hub | Problem:
The point $K$ lies in the interior of the unit circle. Four lines are drawn through $K$ such that each two adjacent lines form an angle of $45^{\circ}$, as shown in the picture. In such a way the circle is divided into 8 regions. Four of these regions are colored such that no two colored regions share more th... | [
"Solution:\n\nWe will show that the total area of colored regions is equal to $\\frac{1}{2}$ the area of the circle no matter how the lines are drawn.\n\nFirst we will inscribe our circle in a square whose sides are parallel to 2 of these lines. If we color the regions of the square in the same way as the regions o... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Transformations > Translation"
] | null | proof and answer | Minimal = π/2, Maximal = π/2 | |
08hy | Problem:
Let $m > n$ be positive integers. For every positive integer $k$ we define the number $a_k = (\sqrt{5} + 2)^k + (\sqrt{5} - 2)^k$. Show that $a_{m+n} + a_{m-n} = a_m \cdot a_n$. | [
"Solution:\n\nLet $A = \\sqrt{5} + 2$ and $B = \\sqrt{5} - 2$. Then $a_k = A^k + B^k$.\n\nNote that $A \\cdot B = (\\sqrt{5} + 2)(\\sqrt{5} - 2) = (\\sqrt{5})^2 - 2^2 = 5 - 4 = 1$.\n\nAlso, $A + B = (\\sqrt{5} + 2) + (\\sqrt{5} - 2) = 2\\sqrt{5}$.\n\nWe want to show that $a_{m+n} + a_{m-n} = a_m a_n$.\n\nCompute $a... | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0dwc | Problem:
V trikotniku je $\beta=74^{\circ} 18^{\prime}$ in $\gamma=38^{\circ} 46^{\prime}$ ter $|AC|-|AB|=2,5~\mathrm{cm}$. Izračunaj dolžini stranic $|AB|$ in $|AC|$ ter rezultat zaokroži na dve mesti natančno. Nariši skico. | [
"Solution:\n\nZvezo $c = b - 2,5$ vstavimo v obrazec za sinusni izrek.\n\nIzrazimo $b = -\\frac{2,5 \\sin \\beta}{\\sin \\gamma - \\sin \\beta}$ in izračunamo $b = 7,1~\\mathrm{cm}$ ter $c = 4,6~\\mathrm{cm}$.\n\nSkica.\n\nZapis ali uporaba sinusnega izreka.\n\nZveza $c = b - 2,5$ vstavljena v sinusni izrek.\n\nIzr... | Slovenia | 4. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | AB = 4.6 cm, AC = 7.1 cm | |
0gfl | 試決定滿足下列性質的所有 $(1, 2, \dots, 2022)$ 的排列 $(a_1, a_2, \dots, a_{2022})$:
對差距大於 $20^{22}$ 的兩正整數 $m, n$, 以下不等式恆成立:
$$
\text{gcd}(m + 1, n + a_1) + \text{gcd}(m + 2, n + a_2) + \dots + \text{gcd}(m + 2021, n + a_{2021}) < 2|m - n|.
| [] | Taiwan | 2022 數學奧林匹亞競賽第二階段培訓營, 獨立研究(一) | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | Chinese; English | proof and answer | (1, 2, 3, ..., 2022) and (2, 3, 4, ..., 2022, 1) | |
0ehl | Problem:
Na celoštevilski mreži je označenih 16 točk (glej sliko). Največ koliko izmed teh točk lahko pobarvamo rdeče, tako da nobene tri rdeče točke ne bodo ležale na isti premici?
 | [
"Solution:\n\nKer izmed 4 točk, ki ležijo v isti vodoravni vrstici, lahko pobarvamo največ 2, lahko skupaj pobarvamo največ $4 \\cdot 2 = 8$ točk. Da 8 točk tudi res lahko pobarvamo, prikazuje spodnja slika.\n\n"
] | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 8 | |
06qe | Let $f$ be any function that maps the set of real numbers into the set of real numbers. Prove that there exist real numbers $x$ and $y$ such that
$$
f(x-f(y))>y f(x)+x .
$$ | [
"Assume that\n$$\nf(x-f(y)) \\leq y f(x)+x \\quad \\text{ for all real } x, y . \\tag{1}\n$$\nLet $a=f(0)$. Setting $y=0$ in (1) gives $f(x-a) \\leq x$ for all real $x$ and, equivalently,\n$$\nf(y) \\leq y+a \\quad \\text{ for all real } y . \\tag{2}\n$$\nSetting $x=f(y)$ in (1) yields in view of (2)\n$$\na=f(0) \\... | IMO | IMO Problem Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
02iu | Problem:
Quantas soluções inteiras e positivas satisfazem a dupla inequação $2000<\sqrt{n(n+1)}<2005$?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 | [
"Solution:\n\nComo os números que aparecem são todos positivos, podemos elevá-los ao quadrado mantendo os sinais, isto é: $2000^{2}<n(n+1)<2005^{2}$. Observe que $n$ e $n+1$ são inteiros consecutivos. Logo, temos as seguintes opções:\n$$\n\\begin{aligned}\n& 2000^{2}<2000 \\times 2001<2005^{2} \\\\\n& 2000^{2}<2001... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | E | |
09u8 | On a $2019 \times 2019$ chess board, there is a contagious disease. Each day some of the squares on the chess board are sick and the rest are healthy. A healthy square bordering a sick square (along a side), becomes sick itself the next day. A sick square will always be healthy the next day. A healthy square that has b... | [
"E) $10000$"
] | Netherlands | First Round, January 2019 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | MCQ | E | |
0atk | Problem:
Find the values of $x$ in $(0, \pi)$ that satisfy the equation
$$
(\sqrt{2014}-\sqrt{2013})^{\tan^{2} x}+(\sqrt{2014}+\sqrt{2013})^{-\tan^{2} x}=2(\sqrt{2014}-\sqrt{2013})^{3}
$$ | [] | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | x = \frac{\pi}{3},\ \frac{2\pi}{3} | |
0l7r | In a group of people, some are friends (friendship is mutual) and each person $p$ has a list $f_1(p), f_2(p), \dots, f_{d(p)}(p)$ of their friends, where $d(p)$ is the number of friends $p$ has. Additionally, any two people are connected by a series of friendships. Each person also has a *water balloon*. The following ... | [] | United States | USA TST Selection Test for 67th IMO and 15th EGMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0jgw | Given an integer $k \ge 2$, set $a_1 = 1$ and, for every integer $n \ge 2$, let $a_n$ be the smallest $x > a_{n-1}$ such that
$$
x = 1 + \sum_{i=1}^{n-1} \lfloor \sqrt[k]{\frac{x}{a_i}} \rfloor.
$$
Prove that every prime occurs in the sequence $a_1, a_2, \dots$. | [
"**Lemma 2.** For any positive integer $m$, we have\n$$\n\\sum_{x \\in X,\\ x \\le m} \\lfloor \\sqrt[k]{\\frac{m}{x}} \\rfloor = m.\n$$\n*Proof.* Observe that every positive integer may be written as the product of an element of $X$ and a perfect $k^{\\text{th}}$ power. Thus, we may divide the set $\\{1, 2, \\dots... | United States | RMM | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0gma | For all integers $x$, $y$, $z$, set
$$
S(x, y, z) = (xy - xz, yz - yx, zx - zy).
$$
Given any integers $a$, $b$, $c$ with $abc > 1$, show that there exist integers $n_0$ and $k$ with $0 < k \le abc$ such that
$$
S^{n+k}(a, b, c) \equiv S^n(a, b, c) \pmod{abc}
$$
for all integers $n \ge n_0$.
$(S^1 = S$ and, for each in... | [] | Turkey | TEAM SELECTION EXAMINATION FOR THE 42nd INTERNATIONAL MATH- EMATICAL OLYMPIAD. TURKEY. | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
05xe | Problem:
Soient $n \geqslant 2$ et $p$ un nombre premier impair. Soit $U$ l'ensemble des entiers positifs inférieurs ou égaux à $p^{n}$ et premiers avec $p$ et soit $N=|U|$. Montrer qu'il existe une permutation $a_{1}, \ldots, a_{N}$ des éléments de $U$ telle que $\sum_{k=1}^{N} a_{k} a_{k+1}$ (avec $a_{N+1}=a_{1}$) s... | [
"Solution:\n\nOn commence par traiter le cas $p=3$, dont la solution est différente. Soit $g \\in U$ un générateur modulo $p^{n}$, on pose $a_{i}=g^{i-1} \\bmod p^{n}$. Comme $g^{2}-1$ est divisible par $3$, la somme considérée $S$ vérifie $(g^{2}-1) S \\equiv \\sum_{i=1}^{N}(g^{2 i+1}-g^{2 i-1}) \\equiv g(g^{2 N}-... | France | Préparation Olympique Française de Mathématiques - Envoi 3: Arithmétique | [
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0es1 | A representation of $\frac{17}{20}$ as a sum of reciprocals
$$
\frac{17}{20} = \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_k}
$$
is called a *calm representation* with $k$ terms if the $a_i$ are distinct positive integers and at most one of them is not a power of two.
a. Find the smallest value of $k$ for whic... | [
"Note first that there is no calm representation with 2 terms: if either $a_1$ or $a_2$ is 1 or $a_1 = a_2 = \\frac{1}{2}$, then the sum is greater than $\\frac{17}{20}$. Otherwise, the sum is at most $\\frac{1}{2} + \\frac{1}{3} = \\frac{5}{6} < \\frac{17}{20}$, thus too small.\n\nOn the other hand, there is a rep... | South Africa | The South African Mathematical Olympiad Third Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | k = 3; infinitely many calm representations exist. | |
03ur | Let $\triangle ABC$ be a non-isosceles acute triangle, and point $O$ is the circumcenter. Let $A'$ be a point on the line $AO$ such that $\angle BA'A = \angle CA'A$. Construct $A'A_1 \perp AC$, $A'A_2 \perp AB$ with $A_1$ on $AC$, $A_2$ on $AB$ respectively. $AH_A$ is perpendicular to $BC$ at $H_A$. Write $R_A$ as the ... | [
"Firstly we claim that $A', B, O, C$ are concyclic points. Otherwise, extend $AO$ to intersect the circumcircle of $\\triangle BOC$ at point $P$ which is different from $A'$. We get\n$$\n\\angle BPA = \\angle BCO = \\angle CBO = \\angle CPA.\n$$\nThen $\\triangle PA'C \\cong \\triangle PA'B$, and $A'B = A'C$. So $A... | China | China Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscella... | English | proof only | null | |
0hjf | Problem:
Several weights are given, each of which is not heavier than $1 \mathrm{lb}$. It is known that they cannot be divided into two groups such that the weight of each group is greater than $1 \mathrm{lb}$. Find the maximum possible total weight of these weights. | [
"Solution:\n\nSuppose you start putting weights into two pans of a scale and stop when you cannot add another weight to either pan without it exceeding $1 \\mathrm{lb}$. At this point, you have put at most $2 \\mathrm{lb}$ of weight on the scale, and there can only be one weight left, otherwise you could add a weig... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 3 lb | |
038e | Problem:
Find all integers $a$ such that the equation
$$
x^{4} + 2x^{3} + (a^{2} - a - 9)x^{2} - 4x + 4 = 0
$$
has at least one real root. | [
"Solution:\nSet $u = x - \\frac{2}{x}$. Then the equation becomes\n$$\nu^{2} + 2u + a^{2} - a - 5 = 0$$\nSince the equation $x^{2} - u x - 2 = 0$ has real solutions for any real $u$, it suffices to find the integer values of $a$ for which the equation (1) has a real root. The last holds when $D = -a^{2} + a + 6 \\g... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | a = -2, -1, 0, 1, 2, 3 | |
08uf | A square $ABCD$ has a side length equal to $3$. Let $P$ be the point on the side $AB$ dividing $AB$ into $2 : 1$ ratio. Let $Q$, $R$ be the points on the side $CD$ dividing $CD$ into $1 : 2$, $2 : 1$ ratio, respectively. And let $S$, $T$ be the points on the side $DA$ dividing $DA$ into $1 : 2$, $2 : 1$ ratio, respecti... | [
"$\\frac{13}{5}$\nLet $E$, $F$ be the point of intersection of the line $PR$ and the lines $AD$, $BC$, respectively. Also, let $G$ be the point of intersection of the lines $TQ$ and $BC$. Then, it is clear that the triangles $APE$ and $DRE$ are similar so that we have $AE : DE = AP : DR = 2 : 1$, from which it foll... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 13/5 | |
0iwx | Problem:
You are given a $5 \times 6$ checkerboard with squares alternately shaded black and white. The bottom-left square is white. Each square has side length 1 unit. You can normally travel on this board at a speed of 2 units per second, but while you travel through the interior (not the boundary) of a black square... | [
"Solution:\n\nAnswer: $\\frac{1+5 \\sqrt{2}}{2}$\n\nIt is always faster to take a path around a black square than through it, since the length of the hypotenuse of any right triangle is greater than half the sum of the length of its legs. Therefore, an optimal path always stays on white squares or on boundaries, an... | United States | 2nd Annual Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (1+5√2)/2 | |
0fcj | Problem:
Dibuja un semicírculo con centro en $O$ y diámetro $A B$ y, en su interior, otro, con diámetro $O A$. Traza por un punto $C$ de $O A$ una recta perpendicular a dicho radio $O A$, que cortará al semicírculo pequeño en $D$ y al grande en $E$ y, finalmente, la recta $A D$ que cortará al semicírculo grande en $F$... | [
"Solution:\n\n-El triángulo $A E B$ es rectángulo.\nPor el Teorema del Cateto: $A E^{2}=A C . A B$\n-El cuadrilátero $B C D F$ es inscriptible, pues sus ángulos opuestos $C$ y $F$ son rectos.\n\nAsí, las rectas $A C B$ y $A D F$ son secantes a la circunferencia que lo circunscribe.\n- La po... | Spain | Fase Local | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dry | Let $P$ be a $2016$-sided polygon with all its adjacent sides perpendicular to each other, i.e., all its internal angles are either $90^\circ$ or $270^\circ$. If the lengths of its sides are odd integers, prove that its area is an even integer. | [
"We may assume the sides are either parallel to the $x$- or $y$-axes and that the lowest horizontal sides are at the $x$-axis. Start with a horizontal side that is on the $x$-axis, go around the polygon in clockwise direction, label the horizontal sides $1, 2, \\ldots, 1008$ and put an arrow on the side along the d... | Singapore | Singapur | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0c7t | Let $A, B \subset \mathbb{N}$ be two finite and nonempty sets. Denote by $\mathcal{F}$ the set of all functions $f : \mathcal{P}(A) \to B$ such that
$$
f(X \cap Y) = \min(f(X), f(Y)), \forall X, Y \subset A,
$$
and by $\mathcal{G}$ the set of all functions $g : \mathcal{P}(A) \to B$ such that
$$
g(X \cup Y) = \max(g(X)... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | sum_{t=1}^{|B|} t^{|A|} | |
00b9 | For every positive integer $m$, we write $S(m)$ for the sum of its digits. For example, $S(2018) = 2+0+1+8=11$.
We say a positive integer $n$ is *rioplatense* if there is a positive integer $m$ such that $m + 2S(m) = n$.
Find all positive integers that are *rioplatenses*. | [
"For every positive integer $m$, we know that $m$ and $S(m)$ have the same remainder $r$ modulo $3$. Then, $m + 2S(m)$ has the same remainder as $3r$ and so, it is a multiple of $3$. Therefore, if a number $n$ is rioplatense, then it is a multiple of $3$.\n\nNow we are going to show that every multiple of $3$ is ri... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | All positive integers divisible by 3. | |
0j3z | In an acute scalene triangle $ABC$, points $D$, $E$, $F$ lie on sides $BC$, $CA$, $AB$, respectively, such that $AD \perp BC$, $BE \perp CA$, $CF \perp AB$. Altitudes $AD$, $BE$, $CF$ meet at orthocenter $H$. Points $P$ and $Q$ lie on segment $EF$ such that $AP \perp EF$ and $HQ \perp EF$. Lines $DP$ and $QH$ intersect... | [
"$$\n\\frac{HQ}{PA} = \\frac{HS}{AS} \\quad \\text{and} \\quad \\frac{RH}{PA} = \\frac{HD}{AD}.\n$$\nIt suffices to show that\n$$\n\\frac{HS}{AS} = \\frac{HD}{AD}, \\qquad (29)\n$$\nwhich we remark is equivalent to showing that $(A, H)$ and $(S, D)$ are harmonic conjugates. Applying Ceva's theorem to triangle $AHC$... | United States | Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Pla... | null | proof and answer | 1 | |
0a9v | Problem:
Find all nonnegative integers $a, b, c$, such that
$$
\sqrt{a} + \sqrt{b} + \sqrt{c} = \sqrt{2014}
$$ | [
"Solution:\nWe start with a lemma:\n\nLemma. If $p, q$ are nonnegative integers and $\\sqrt{p} + \\sqrt{q} = r \\in \\mathbb{Q}$, then $p$ and $q$ are squares of integers.\n\nProof of lemma: If $r = 0$, then $p = q = 0$. For $r \\neq 0$, take the square of both sides to get $p + q + 2 \\sqrt{p q} = r^{2}$, which me... | Nordic Mathematical Olympiad | The 28th Nordic Mathematical Contest | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All permutations of (2014, 0, 0). | |
09l6 | Prove that there exists a sequence of 2023 consecutive positive integers in which exactly 2000 numbers are composite. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0c4q | Prove that for every real numbers $x_1, x_2, \dots, x_n$, where $n \ge 2$, it is possible to choose $+$ and $-$ so that
$$
\cos (x_1 \pm x_2 \pm \cdots \pm x_n) \le \cos x_1 \cos x_2 \cdots \cos x_n.
$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
0brs | Consider a cube $ABCD A'B'C'D'$ and two points $M \in (CD')$ and $N \in (DA')$. Prove that $MN$ is the common perpendicular of the lines $CD'$ and $DA'$ if and only if
$$
\frac{D'M}{D'C} = \frac{DN}{DA'} = \frac{1}{3}.
$$ | [
"If $\\frac{D'M}{D'C} = \\frac{DN}{DA'} = \\frac{1}{3}$, then $M$ and $N$ are the centroids of the triangles $DD'C$ and, respectively, $ADD'$, hence the points $C'$, $M$, $P$ and $A$, $N$, $P$ are collinear,\n\nwhere $P$ is the midpoint of the side $DD'$.\nFrom the triangle $APC'$, $\\frac{PM}{PC'} = \\frac{PN}{PA}... | Romania | 67th Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
096k | Problem:
Fie $A$ o matrice pătratică de ordinul $n$, elementele căreia sunt numerele $-2019$ sau $2019$.
Determinaţi valoarea maximă a determinantului matricei $A$, pentru:
a) $n=3$;
b) $n=4$. | [
"Solution:\n\na) Observăm că $A = 2019 \\cdot B$, unde $B$ este o matrice pătratică de ordinul $n$, care are ca elemente numerele $-1$ sau $1$. Atunci $\\det A = 2019^{n} \\det B$. Fie $n=3$. Adunăm elementele liniei a doua a matricei $B$ cu elementele respective ale primei linii. Adunăm elementele liniei a treia a... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof and answer | For n=3: 4 · 2019^3; For n=4: 16 · 2019^4 | |
00uc | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all real numbers $x$ and $y$,
$$
f(xy + f(x^2)) = x f(x + y).
$$ | [
"Answer: $f(x) = 0$ for all real numbers $x$ or $f(x) = x$ for all real numbers $x$.\nLet $P(x, y)$ be the assertion of the given equation. We will consider two cases:\n\n**Case 1.** Function is not periodic. This means that if $f(x + p) = f(x)$ for every real number $x$, then $p = 0$.\nConsidering $P(1, x - 1)$ we... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | Either f(x) = 0 for all real x, or f(x) = x for all real x. | |
0fr1 | Problem:
Determinar para qué valores de $n$ existe un polígono convexo de $n$ lados cuyos ángulos internos, expresados en grados, son todos enteros, están en progresión aritmética y no son todos iguales. | [
"Solution:\nSea $x$ la medida del ángulo mayor y $d$ la diferencia de la progresión aritmética. De esta manera,\n$$\nn x - \\frac{d n(n-1)}{2} = 180(n-2)\n$$\no alternativamente\n$$\n720 = n(360 - 2x + d(n-1))\n$$\nComo $360 - 2x > 0$ por tratarse de un polígono convexo, necesariamente sucede que\n$$\nn(n-1) < d n(... | Spain | FASE LOCAL DE LA OLIMPIADA MATEMÁTICA ESPAÑOLA. | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | {3,4,5,6,8,9,10,12,15,16,18} | |
0cpo | In Academy of Sciences, there are $999$ academicians. For each scientific topic there are exactly $3$ academicians interested in this topic. For each pair of academicians, there exists exactly one topic in which both these academicians are interested. Prove that there exists a set of $250$ topics such that each academi... | [
"Будем говорить, что две темы пересекаются по академику, если он интересуется обеими этими темами.\n\nВыберем наибольшее возможное количество непересекающихся тем; пусть это темы $T_1, \\dots, T_k$. Предположим, что $k \\le 249$. Обозначим через $S$ множество всех академиков, не интересующихся этими темами; тогда и... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English, Russian | proof only | null | |
0177 | Let $n$ be Tracey's favourite positive integer. Suppose that her opponent initially writes a certain positive integer $m$ on the blackboard and that whenever a positive integer $a$ is already written on the blackboard then Tracey is permitted to also write the numbers $17 \cdot a$ and $[\sqrt{a}]$ on it. Prove that whi... | [
"Fix $m$ and let $T$ refer to the set of all those positive integers whose presence on the blackboard Tracey can enforce. If $k$ denotes the least element of $T$, then clearly $k \\le [\\sqrt{k}] \\le \\sqrt{k}$, whence $k = 1$. Therefore $1 \\in T$, from which it inductively follows that all powers of $17$ likewis... | Baltic Way | BALTIC WAY | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0dr3 | Let $F$ be a finite nonempty set of integers and let $n$ be a positive integer. Suppose that
* Any $x \in F$ may be written as $x = y + z$ for some $y, z \in F$;
* If $1 \le k \le n$ and $x_1, \dots, x_k \in F$, then $x_1 + \dots + x_k \ne 0$.
Show that $F$ has at least $2n + 2$ distinct elements. | [
"Because of the second condition above, $0 \\notin F$. If $F$ contains only positive elements, let $x$ be the smallest element in $F$. But then $x = y+z$, and $y, z > 0$ imply that $y, z < x$, a contradiction. Hence $F$ contains negative elements. A similar argument shows that $F$ contains positive elements.\n\nPic... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0d55 | Let $a$, $b$, $c$, $d$ be positive integers such that $a c + b d$ is divisible by $a^{2} + b^{2}$. Prove that $\operatorname{gcd}\left(c^{2} + d^{2}, a^{2} + b^{2}\right) > 1$. | [
"Suppose for the sake of contradiction that $\\operatorname{gcd}\\left(c^{2} + d^{2}, a^{2} + b^{2}\\right) = 1$. We first notice that\n$$\n(a c + b d)^{2} + (a d - b c)^{2} = \\left(a^{2} + b^{2}\\right)\\left(c^{2} + d^{2}\\right).\n$$\nSince $a c + b d$ is divisible by $a^{2} + b^{2}$, it follows that $(a d - b ... | Saudi Arabia | SAMC 2015 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English, Arabic | proof only | null | |
0j03 | Problem:
A clue "$k$ digits, sum is $n$" gives a number $k$ and the sum of $k$ distinct, nonzero digits. An answer for that clue consists of $k$ digits with sum $n$. For example, the clue "Three digits, sum is 23" has only one answer: $6,8,9$. The clue "Three digits, sum is 8" has two answers: $1,3,4$ and $1,2,5$.
If... | [
"Solution:\n\nThe sum of $20$ has $12$ answers, and this is the largest number of answers for any four-digit clue.\n\nWe could simply list all the possible sets of four digits and then count. There are $126$ such sets.\n\nAlternatively, define $A(s, n, k)$ to be the number of options with sum $s$ using exactly $n$ ... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | n = 20, and it has 12 answers | |
0kcp | Problem:
Let $\omega_{1}$ be a circle of radius $5$, and let $\omega_{2}$ be a circle of radius $2$ whose center lies on $\omega_{1}$. Let the two circles intersect at $A$ and $B$, and let the tangents to $\omega_{2}$ at $A$ and $B$ intersect at $P$. If the area of $\triangle ABP$ can be expressed as $\frac{a \sqrt{b}}... | [
"Solution:\n\nLet $O_{1}$ and $O_{2}$ be the centers of $\\omega_{1}$ and $\\omega_{2}$, respectively. Because\n$$\n\\angle O_{2}AP + \\angle O_{2}BP = 90^{\\circ} + 90^{\\circ} = 180^{\\circ},\n$$\nquadrilateral $O_{2}APB$ is cyclic. But $O_{2}$, $A$, and $B$ lie on $\\omega_{1}$, so $P$ l... | United States | HMMO 2020 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 19285 | |
09ch | Хавтгайн $n$ цэгээс тогтох цэгүүдийн систем $N$-ийн хувьд $d = \max_{A \in N} |AB|$, $q = \min_{A \in N} |AB|$ бол
$$
\frac{d}{q} > \frac{\sqrt{3}}{2} (\sqrt{n} - 1) \text{ болохыг батал.}
$$ | [
"$d$-диаметртэй цэгүүдийн системийг $\\frac{d}{\\sqrt{3}}$ радиустай дугуй болно (Юнгийн теорем). Цэг болгон дээр төвтэй $\\frac{q}{2}$ радиустай дугуй байгуулбал $q$ нь $\\min_{A \\in N} |AB|$ учир аль ч 2 тойрог огт огтлолцохгүй, шүргэлцэх тохиолдол бий.\n\nНэгэнт бүх цэгийг $\\frac{d}{\\sqrt{3}}$ радиустай дугуй... | Mongolia | ОУМО-53 | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | Mongolian | proof only | null | |
0cjr | We say that a natural number $n \ge 3$ is almost square-free if there exists a prime number $p$, with $p \equiv 1 \pmod 3$, such that $n$ is divisible by $p^2$, and the number $\frac{n}{p}$ is square-free (i.e., not divisible by the square of any prime number).
Show that, for any natural number $n$ that is almost squa... | [
"Consider $n = p_1 \\cdots p_{k-1} \\cdot p_k^2 \\cdot p_{k+1} \\cdots p_s$, with $p_1 < p_2 < \\dots < p_s$ prime numbers, $1 \\le k \\le s$, and $p = p_k \\equiv 1 \\pmod 3$. Moreover, $p$ is odd and satisfies $p^2 + p + 1 \\equiv 0 \\pmod 3$.\n\nThe number of divisors of $n$, $\\tau(n)$, is\n$$\n\\tau(n) = \\und... | Romania | 75th NMO Selection Tests | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0j2g | Problem:
Show that, for every positive integer $n$, there exists a monic polynomial of degree $n$ with integer coefficients such that the coefficients are decreasing and the roots of the polynomial are all integers. | [
"Solution:\nWe claim we can find values $a$ and $b$ such that $p(x) = (x - a)(x + b)^n$ is a polynomial of degree $n+1$ that satisfies these constraints. We show that its coefficients are decreasing by finding a general formula for the coefficient of $x^k$.\n\nThe coefficient of $x^k$ is $b^k \\binom{n}{k} - a b^{k... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
05mz | Problem:
Déterminer tous les polynômes $P$ et $Q$ à coefficients entiers tels que, si l'on définit la suite $\left(x_{n}\right)$ par $x_{0}=2015$, $x_{2 n+1}=P\left(x_{2 n}\right)$ et $x_{2 n+2}=Q\left(x_{2 n+1}\right)$ pour tout $n \geqslant 0$, alors tout entier $m>0$ divise au moins un terme non nul de la suite. | [
"Solution:\n\nSoit $P$ et $Q$ deux tels polynômes.\nOn dira qu'une suite $\\left(y_{n}\\right)$ d'entiers possède la propriété D si tout entier $m>0$ divise au moins un terme non nul de cette suite. On dira qu'un polynôme $T$ à coefficients entiers possède la propriété D s'il existe un entier $a$ tel que la suite $... | France | TEST DU GROUPE A ET DES CANDIDATES À L'ÉPREUVE EGMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (P(X), Q(X)) = (ε X + b, ε X + d) with ε ∈ {+1, −1} and integers b, d, such that either b + ε d divides 2015 or d + ε b divides ε·2015 + b. | |
01jw | Given a trapezium $ABCD$ with $BC \parallel AD$, $AD > BC$. Let $M$ be the point of intersection of $AC$ and $BD$. A circle $\Gamma_1$ passes through $M$ and touches $AD$ at $A$. A circle $\Gamma_2$ passes through $M$ and touches $AD$ at $D$. Let $S$ be the point of intersection of the lines $AB$ and $DC$, $X$ be the p... | [
"Draw the line $LM$, where $L$ is the second (different from $M$) point of intersection of $\\Gamma_1$ and $\\Gamma_2$. (If $\\Gamma_1$ touches $\\Gamma_2$ at $M$, then $LM$ is the common tangent of these circles at $M$.) Let $K$ be the point of intersection of the lines $LM$ and $AD$. By the tangent-secant theorem... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, ... | English | proof only | null | |
0koa | Problem:
Sets $A$, $B$, and $C$ satisfy $|A| = 92$, $|B| = 35$, $|C| = 63$, $|A \cap B| = 16$, $|A \cap C| = 51$, $|B \cap C| = 19$. Compute the number of possible values of $|A \cap B \cap C|$. | [
"Solution:\n\nSuppose $|A \\cap B \\cap C| = n$. Then there are $16 - n$ elements in $A$ and $B$ but not $C$, $51 - n$ in $A$ and $C$ but not $B$, and $19 - n$ in $B$ and $C$ but not $A$. Furthermore, there are $25 + n$ elements that are only in $A$, $n$ only in $B$, and $n - 7$ that are only in $C$. Therefore, $7 ... | United States | HMMT February 2022 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 10 | |
02rz | Problem:
A figura a seguir mostra um cubo de aresta 1.
a) Qual o menor comprimento possível para um caminho formado por arestas do cubo que passa por todos os 8 vértices?
A figura abaixo mostra um cubo de aresta 1 no qual todas as 12 diagonais da face foram desenhadas. Assim, criou-se uma r... | [
"Solution:\n\na) A figura a seguir representa um pedaço de folha de papel, contendo oito quadrados de lado 1 posicionados lado a lado.\n\n\nObserve que, ao dobrarmos a folha de papel nas arestas que são comuns a dois quadrados, obteremos um cubo de lado 1 como aquele do enunciado do exercíc... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Other"
] | null | proof and answer | a) 7; b) 6*sqrt(2) + 1 | |
05cs | Let $ABCD$ be a rectangle. The bisector of the angle $CAD$ meets the side $CD$ at point $L$. Let $M$ be the midpoint of the line segment $AL$. The line $DM$ meets lines $AC$ and $AB$ at points $E$ and $F$, respectively. Given that line segments $AE$ and $AF$ are equal, prove that $ABCD$ is a square. | [
"Let $\\angle DAL = \\angle LAC = \\alpha$. Then $\\angle FAE = 90^\\circ - 2\\alpha$ (Fig. 1).\n\nAs $AE = AF$, we obtain $\\angle AFE = \\frac{180^\\circ - (90^\\circ - 2\\alpha)}{2} = 45^\\circ + \\alpha$.\n\nBut as $M$ bisects the hypotenuse $AL$ of the right triangle $ALD$, it follows that $M$ is the circumcen... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | English | proof only | null | |
0653 | Compute the sum
$$
\Sigma = \sqrt{1+\frac{8 \cdot 1^2-1}{1^2 \cdot 3^2}} + \sqrt{1+\frac{8 \cdot 2^2-1}{3^2 \cdot 5^2}} + \dots + \sqrt{1+\frac{8 \cdot 1003^2-1}{2005^2 \cdot 2007^2}}
$$ | [
"For $v = 1, 2, \\dots, 1003$, we have\n$$\n1 + \\frac{8v^2 - 1}{(2v-1)^2 (2v+1)^2} = \\frac{(4v^2 - 1)^2 + 8v^2 - 1}{(4v^2 - 1)^2} = \\frac{16v^4}{(4v^2 - 1)^2} = \\left( \\frac{4v^2}{4v^2 - 1} \\right)^2\n$$\nHence we can write\n$$\n\\begin{aligned}\n\\sqrt{1+\\frac{8v^2-1}{(2v-1)^2(2v+1)^2}} &= \\sqrt{\\left(\\f... | Greece | Selection Examination for Juniors | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | final answer only | 2014024/2007 | |
0lbs | Let $ABC$ be the isosceles triangle with $AB = AC$ and $M,N$ are the midpoints of $AB, AC$, respectively. The circle $(K)$ is tangent to line $AB$ at $B$ and tangent to $AC$ at $C$. An arbitrary circle $(I)$ tangents to $(K)$, passes through $A$ and cuts the line $MN$ at $D,E$. Prove that:
a. Point $I$ belongs to a fi... | [
"a. Suppose that two circles $(I)$ and $(K)$ tangent to each other at point $T$, then\n$$IK - IA = IK - IT = TK$$\nwhich is a constant.\nHence, point $I$ belongs to the hyperbola with two foci $A$ and $K$, which are fixed points.\nNext, we have to prove that the angle $\\angle MAN$ is constant.\n\n![](attached_imag... | Vietnam | Vietnamese Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasi... | null | proof only | null | |
0igi | Problem:
Ten positive integers are arranged around a circle. Each number is one more than the greatest common divisor of its two neighbors. What is the sum of the ten numbers? | [
"Solution:\n\nFirst note that all the integers must be at least $2$, because the greatest common divisor of any two positive integers is at least $1$. Let $n$ be the largest integer in the circle. The greatest common divisor of its two neighbors is $n-1$. Therefore, each of the two neighbors is at least $n-1$ but a... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 28 | |
0a71 | Problem:
Let $a$, $b$, and $c$ be non-zero real numbers and let $a \geq b \geq c$. Prove the inequality
$$
\frac{a^{3}-c^{3}}{3} \geq a b c\left(\frac{a-b}{c}+\frac{b-c}{a}\right)
$$
When does equality hold? | [
"Solution:\nSince $c-b \\leq 0 \\leq a-b$, we have $(a-b)^{3} \\geq (c-b)^{3}$, or\n$$\na^{3}-3 a^{2} b+3 a b^{2}-b^{3} \\geq c^{3}-3 b c^{2}+3 b^{2} c-b^{3}\n$$\nOn simplifying this, we immediately have\n$$\n\\frac{1}{3}\\left(a^{3}-c^{3}\\right) \\geq a^{2} b-a b^{2}+b^{2} c-b c^{2}=a b c\\left(\\frac{a-b}{c}+\\f... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Equality holds if and only if a = c. | |
0a8x | Problem:
There are 2005 young people sitting around a (large!) round table. Of these at most 668 are boys. We say that a girl $G$ is in a strong position, if, counting from $G$ to either direction at any length, the number of girls is always strictly larger than the number of boys. ($G$ herself is included in the coun... | [
"Solution:\n\nAssume the number of girls to be $g$ and the number of boys $b$. Call a position clockwise fairly strong, if, counting clockwise, the number of girls always exceeds the number of boys. No girl immediately followed by a boy has a fairly strong position. But no pair consisting of a girl and a boy follow... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 19 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
034u | Problem:
Let $M$ and $N$ be arbitrary points on the side $AB$ of a triangle $ABC$ such that $M$ lies between $A$ and $N$. The line through $M$ parallel to $AC$ meets the circumcircle of $\triangle MNC$ at point $P$, and the line through $M$ parallel to $NC$ meets the circumcircle of $\triangle AMC$ at point $Q$. Analo... | [
"Solution:\n\n$$\n\\begin{aligned}\n\\angle QCP &= \\angle QCA + \\angle ACN + \\angle NCP \\\\\n&= \\angle CNA + \\angle ACN + \\angle NAC \\\\\n&= 180^\\circ,\n\\end{aligned}\n$$\n\ni.e. the points $P$, $C$ and $Q$ are collinear.\n\nb) Set $\\angle ACM = \\varphi$, $\\angle NCM = \\psi$ a... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cdp | a) Prove that there exists irrational numbers $a$, $b$, and $c$ so that the numbers $a+bc$, $b+ac$ and $c+ab$ are rational.
b) Prove that, if the real numbers $a$, $b$, and $c$ are such that $a + b + c = 1$ and the numbers $a+bc$, $b+ac$, and $c+ab$ are non-nil rationals, then the numbers $a$, $b$, $c$ are rational. | [
"a) An example can be obtained if $a = b = c$, with irrational $a$ and rational $a^2 + a$. This happens if, for instance, $a = \\sqrt{2} - \\frac{1}{2}$.\n\nb) Since $b+ac$ and $c+ab$ are rational, so is $b+c+ac+ab = (b+c)(1+a) = (1-a)(1+a) = 1-a^2$, hence $a^2$ is rational.\nIn the same way, $b^2$ and $c^2$ are ra... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0en5 | Given a non-isosceles triangle $ABC$, let $D$, $E$ and $F$ denote the midpoints of the sides $BC$, $CA$ and $AB$ respectively. The circumcircle of triangle $BCF$ and the line $BE$ meet again at $P$, and the circumcircle of $ABE$ and the line $AD$ meet again in $Q$. Finally, the lines $DP$ and $FQ$ meet at $R$. Prove th... | [
"We will use the following lemma:\n\n**Lemma 1.** Let $AD$ be a median in triangle $ABC$. Then $\\cot \\angle BAD = 2 \\cot A + \\cot B$ and $\\cot \\angle ADC = \\frac{1}{2}(\\cot B - \\cot C)$.\n\n*Proof.* Let $CC_1$ and $DD_1$ be the perpendiculars from $C$ and $D$ to $AB$. Using signed lengths, we write\n$$\n\\... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry... | null | proof only | null | |
0b9t | Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that, for every $x, y \in \mathbb{R}$:
$$
|f(x + y) + \sin x + \sin y| \le 2.
$$
a. Prove that $|f(x)| \le 1 + \cos x$, for every $x \in \mathbb{R}$.
b. Give an example of such a function, which vanishes nowhere in the interval $(-\pi, \pi)$. | [
"a.\nLet $x = t - \\frac{\\pi}{2}$ and $y = \\frac{\\pi}{2}$. Then\n$$\n|f(t) + \\sin(t - \\tfrac{\\pi}{2}) + \\sin(\\tfrac{\\pi}{2})| \\le 2.\n$$\nBut $\\sin(t - \\tfrac{\\pi}{2}) = -\\cos t$ and $\\sin(\\tfrac{\\pi}{2}) = 1$, so\n$$\n|f(t) - \\cos t + 1| \\le 2.\n$$\nThis gives\n$$\nf(t) - \\cos t + 1 \\le 2 \\im... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities"
] | null | proof and answer | For all real x, |f(x)| ≤ 1 + cos x. An example is f(x) = 2 − 2|sin(x/2)|, which satisfies the condition and has no zeros in (−π, π). | |
0e63 | Let $BE$ and $CF$ be heights of an acute triangle $ABC$. Two circles through points $A$ and $F$ touch the line $BC$ in points $P$ and $Q$, respectively, where $B$ lies between the points $C$ and $Q$. Prove that the lines $PE$ and $QF$ intersect on the circumscribed circle of the triangle $AEF$. | [
"Let all the angles in the solution be directed. Given an acute triangle $ABC$, denote $G$ the foot (lying on the side $BC$) of the altitude from $A$. The powers of the point $B$ with respect to the circles through points $A$ and $F$ give us $|BP|^2 = |BA| \\cdot |BF| = |BQ|^2$, hence $|BP| = |BQ|$.\n\nAccording to... | Slovenia | Selection Examinations for the IMO 2012 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hh1 | Let $(x_n)$ be a sequence of real numbers from the interval $(0; 1)$. The sequence of positive integers $(a_n)$ is defined as follows: $a_1 = 1$, $a_{i+1} = m$, where $m$ is the smallest positive integer for which $[x_1 + x_2 + \cdots + x_m] = a_i$. Prove that for any indexes $i, j$ the inequality $a_{i+j} \ge a_i + a_... | [
"Note that we can put $a_0 = 0$, then we would get $a_1 = 1$. First, let's prove the following lemma:\n\n**Lemma.** For all $i \\in \\mathbb{N}$ the condition holds: $a_{i+1} - a_i \\ge a_i - a_{i-1}$.\n\n*Proof:*\n$$\nx_1 + x_2 + \\cdots + x_{2a_i - a_{i-1} - 1} = (x_1 + x_2 + \\cdots + x_{a_i}) + (x_{a_i + 1} + x... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0lbh | Cho dãy số $x_1, x_2, \ldots$ được xác định như sau: $x_1 = 1$, $x_2 = 2011$, $x_{n+2} = 4022x_{n+1} - x_n$ với mọi $n \ge 1$.
Chứng minh rằng $\frac{x_{2012} + 1}{2012}$ là số chính phương. | [] | Vietnam | Kì thi chọn học sinh vào Đội tuyển Quốc gia Dự thi IMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials"
] | Vietnamese | proof only | null | |
01ax | Show that
$$
\cos(56^\circ) \cdot \cos(2 \cdot 56^\circ) \cdot \dots \cdot \cos(2^{23} \cdot 56^\circ) = \frac{1}{2^{23}}
$$ | [
"We start by rewriting the expression as follows:\n$$\n\\cos(56^\\circ) \\cdot \\cos(2 \\cdot 56^\\circ) \\cdot \\dots \\cdot \\cos(2^{23} \\cdot 56^\\circ) = \\frac{\\sin(56^\\circ) \\cdot \\cos(56^\\circ) \\cdot \\cos(2 \\cdot 56^\\circ) \\cdot \\dots \\cdot \\cos(2^{23} \\cdot 56^\\circ)}{\\sin(56^\\circ)}\n$$\n... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 1/2^{23} | |
045e | Given positive integers $a, b, c, p, q, r$, where $p, q, r \ge 2$. Let $Q = \{(x, y, z) \in \mathbb{Z}^3 \mid 0 \le x \le a, 0 \le y \le b, 0 \le z \le c\}$ be the game board.
Initially, put $M$ game pieces on $Q$ (no restriction on the number of pieces in each position). There are three types of legal moves:
(1) Rem... | [
"To prove $M = p^a q^b r^c$ is necessary, notice that if one initially places fewer than $p^a q^b r^c$ pieces at $(a, b, c)$, then the goal cannot be achieved. Indeed, let the weight of a game piece $u$ at $(x, y, z)$ be $w(u) = \\frac{1}{p^x q^y r^z}$. According to the rule, before and after each move, the total w... | China | China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | p^a q^b r^c | |
0as6 | Problem:
Let $[|x|]$ be the integer part of $x$ and $\{x\}=x-[|x|]$, the decimal part of $x$. Solve $2[|x|]=x+2\{x\}$. | [
"Solution:\n(ans. $x=0, \\frac{4}{3}, \\frac{8}{3}$.\n$2[|x|]=x+2\\{x\\} \\Rightarrow 2[|x|]=[|x|]+\\{x\\}+2\\{x\\} \\Rightarrow [|x|]=3\\{x\\}<3 \\Rightarrow$ $\\left.[|x|]=0,1,2 \\Rightarrow x=0, \\frac{4}{3}, \\frac{8}{3}.\\right)$"
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | x = 0, 4/3, 8/3 | |
08mn | Problem:
Let $a$, $b$ be positive real numbers. Prove that
$$
\sqrt{\frac{a^{2}+a b+b^{2}}{3}}+\sqrt{a b} \leq a+b.
$$ | [
"Solution:\nApplying $x+y \\leq \\sqrt{2\\left(x^{2}+y^{2}\\right)}$ for $x=\\sqrt{\\frac{a^{2}+a b+b^{2}}{3}}$ and $y=\\sqrt{a b}$, we will obtain\n$$\n\\sqrt{\\frac{a^{2}+a b+b^{2}}{3}}+\\sqrt{a b} \\leq \\sqrt{\\frac{2 a^{2}+2 a b+2 b^{2}+6 a b}{3}} \\leq \\sqrt{\\frac{3\\left(a^{2}+b^{2}+2 a b\\right)}{3}}=a+b.... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
03xs | Let $a$, $b$, $c$ be complex numbers such that $|az^2 + bz + c| \le 1$ for all complex numbers $z$ with $|z| \le 1$. Find the maximum of $|bc|$. (Posed by Li Weigu) | [
"Write $f(z) = az^2 + bz + c$. We first prove that\n$$\n|f(z)| \\le 1 \\text{ for all } z, |z| \\le 1 \\Leftrightarrow |f(z)| \\le 1 \\text{ for all } z, |z| = 1.\n$$\n\nAssume that $f(z) = a(z - \\alpha)(z - \\beta)$. For any $z$, $|z| < 1$, if $\\alpha = \\beta$, one of the two intersection points of the line thr... | China | Chinese Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 3√3/16 | |
0h8o | Determine if there exist a function $f : (0; 1) \rightarrow (2018; +\infty)$, so that the following conditions hold:
* $f(x \cdot y) = f(x) \cdot f(y)$ for any $x, y \in (0; 1)$;
* for any $y \in (2018; +\infty)$ there exists $x \in (0; 1)$ such that $f(x) = y$? | [
"Suppose such a function exists. Since for any $x \\in (0; 1)$ equation $x = \\sqrt{x} \\cdot \\sqrt{x}$ holds, then $f(x) = f(\\sqrt{x} \\cdot \\sqrt{x}) = f(\\sqrt{x}) \\cdot f(\\sqrt{x}) > 2018^2$, that contradicts the fact that $E_f = (2018, +\\infty)$."
] | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | No such function exists. | |
030y | Problem:
Fie $n \geqslant 2$ un număr natural și $M = \mathbb{N} \times \mathbb{N} \times \ldots \times \mathbb{N}$ (de $n$ ori $\mathbb{N}$), mulțimea $n$-uplelor de numere naturale. Pentru fiecare $a = (a_{1}, a_{2}, \ldots, a_{n}) \in M$ notăm cu $d_{a}$ numărul perechilor $(i, j)$, $i, j \in \{1, 2, \ldots, n\}$, ... | [] | Brazil | Al treilea baraj de selecție pentru OBMJ | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | floor(n^2/4) | |
0hyq | Problem:
The real positive numbers $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ satisfy the relation $a_{n+1}^{2}=a_{n}+1$ for all $n=1,2, \ldots$ Prove that at least one of the $a_{i}$'s must be an irrational number. | [
"Solution:\n\nArgue by contradiction. Suppose that for all $i$, $a_{i}=p_{i} / q_{i}$ for some relatively prime positive integers $p_{i}, q_{i}$. Substitute in the given relation:\n$$\n\\frac{p_{n+1}^{2}}{q_{n+1}^{2}}=\\frac{p_{n}+q_{n}}{q_{n}} \\Rightarrow p_{n+1}^{2} q_{n}=q_{n+1}^{2}\\left(p_{n}+q_{n}\\right)\n$... | United States | BAMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Algebra > Equations and Inequalities > Linear and quadratic inequalitie... | null | proof only | null | |
0iea | Problem:
Find the sum of the absolute values of the roots of $x^{4}-4 x^{3}-4 x^{2}+16 x-8=0$. | [
"Solution: $2+2 \\sqrt{2}+2 \\sqrt{3}$\n$$\n\\begin{aligned}\nx^{4}-4 x^{3}-4 x^{2}+16 x-8 & =\\left(x^{4}-4 x^{3}+4 x^{2}\\right)-\\left(8 x^{2}-16 x+8\\right) \\\\\n& =x^{2}(x-2)^{2}-8(x-1)^{2} \\\\\n& =\\left(x^{2}-2 x\\right)^{2}-(2 \\sqrt{2} x-2 \\sqrt{2})^{2} \\\\\n& =\\left(x^{2}-(2+2 \\sqrt{2}) x+2 \\sqrt{2... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2+2√2+2√3 | |
0cms | Positive numbers $x_1, x_2, \dots, x_{2009}$ satisfy the relations $x_1^2 - x_1x_2 + x_2^2 = x_2^2 - x_2x_3 + x_3^2 = x_3^2 - x_3x_4 + x_4^2 = \dots = x_{2008}^2 - x_{2008}x_{2009} + x_{2009}^2 = x_{2009}^2 - x_{2009}x_1 + x_1^2$. Prove that $x_1 = x_2 = \dots = x_{2009}$. | [
"**Первое решение.** Без ограничения общности можно считать, что $x_1$ — наибольшее (или одно из наибольших) среди данных чисел. Тогда из равенства $x_1^2 - x_1x_2 + x_2^2 = x_2^2 - x_2x_3 + x_3^2$ следует $x_1^2 - x_3^2 - x_1x_2 + x_2x_3 = 0 \\Leftrightarrow (x_1 - x_3)(x_1 + x_3) - x_2(x_1 - x_3) = 0 \\Leftrighta... | Russia | Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English; Russian | proof only | null | |
07m7 | In a triangle $ABC$, $X$ and $Y$ are the mid-points of $AB$ and $BC$ respectively. On $BC$ there is a point $D$, which is not the mid-point of $BC$. Prove that $\angle XDY = \angle BAC$ implies $AD \perp BC$. | [
"Let $Z$ be the mid-point of $BC$. Because $X$, $Y$, $Z$ are mid-points, we have $XZ \\parallel AC$, $YZ \\parallel AB$ and $XY \\parallel BC$. Therefore $\\angle XBZ = \\angle XYZ$ and $\\angle XZY$. The latter, together with the assumption $\\angle XDY = \\angle BAC$ implies $\\angle XDY = \\angle XZY$, hence $X$... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fpt | Las circunferencias $C_1$ y $C_2$ se cortan en dos puntos distintos $A$ y $K$. La tangente común a $C_1$ y $C_2$ más cercana a $K$ toca a $C_1$ en $B$ y a $C_2$ en $C$. Sean $P$ el pie de la perpendicular desde $B$ sobre $AC$, y $Q$ el pie de la perpendicular desde $C$ sobre $AB$. Si $E$ y $F$ son los puntos simétricos... | [] | Spain | XXXI Olimpiada Iberoamericana de Matemáticas | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Spanish | proof only | null | |
040q | Find the smallest positive integer $k$ with the following property: for any $k$ element subset $A$ of the set $S = \{1, 2, \dots, 2012\}$, there exist three pairwise distinct elements $a, b, c$ of $S$ such that $a+b$, $b+c$, $c+a$ all belong to $A$. | [
"Without loss of generality, we may assume $a < b < c$. Write $x = a+b$, $z = b+c$, $y = a+c$, then $x < y < z$, $x+y > z$, and $x+y+z$ is even. On the other hand, if there exist $x, y, z \\in A$ such that $x < y < z$, $x+y > z$, and $x+y+z$ is even, set $a = \\frac{x+y-z}{2}$, $b = \\frac{x+z-y}{2}$, $c = \\frac{y... | China | China Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1008 | |
0h8b | Numbers $a$, $b$, $c$ satisfy the conditions:
$$
\frac{a+c}{a+1} = b, \quad \frac{c+b}{c+1} = a, \quad \frac{b+a}{b+1} = c.
$$
What values can the expression $(a+1)(b+1)(c+1)$ take? | [
"Let us subtract $1$ from the left and right parts of each equality and obtain:\n\n$$\nb - 1 = \\frac{a+c}{a+1} - 1 = \\frac{a+c - a - 1}{a+1} = \\frac{c-1}{a+1},\n$$\n\nand analogously from the other two equalities. Because of the conditions of existence of expressions, none of the variables equals $-1$, so we can... | Ukraine | UkraineMO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 1 and 8 | |
09xg | We consider sports tournaments with $n \ge 4$ participating teams and where every pair of teams plays against one another at most one time. We call such a tournament *balanced* if any four participating teams play exactly three matches between themselves. So, not all teams play against one another.
Determine the larges... | [
"We will show that $5$ is the largest value of $n$ for which a balanced tournament with $n$ teams exists. First we will show that in a balanced tournament with $n \\ge 5$ teams, there are no three teams that all play against one another in the tournament.\n\nSuppose towards a contradiction that we can find three te... | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | 5 | |
03dz | Given an acute triangle $ABC$ with circumcenter $O$. The point $P$ on $BC$ such that $BP < \frac{BC}{2}$ and the point $Q$ is on $BC$, such that $CQ = BP$. The line $AO$ meets $BC$ at $D$ and $N$ is the midpoint of $AP$. The circumcircle of $(ODQ)$ meets $(BOC)$ at $E$. The lines $NO$, $OE$ meet $BC$ at $K$, $F$. Show ... | [
"Let $A'$ be the antipode of $A$ and let $AO \\cap (BOC) = R$. Since $NO \\parallel PA'$, by Reim's theorem we need $APA'F$ being cyclic, or $DP \\cdot DF = DA \\cdot DA' = DB \\cdot DC = DO \\cdot DR$, so we need $OPRF$ being cyclic, or that $\\angle OFP = \\angle ORP$. By shooting lemma, $RDEF$ is cyclic, so $\\a... | Bulgaria | Autumn tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneo... | English | proof only | null | |
0dqd | Let $a_1, a_2, ..., a_n$ be positive integers, not necessarily distinct but with at least five distinct values. Suppose that for any $1 \le i < j \le n$, there exist $k, \ell$, both different from $i$ and $j$ such that $a_i + a_j = a_k + a_\ell$. What is the smallest possible value of $n$? | [
"$a_1 \\le a_2 \\le \\dots \\le a_n$. Suppose $x < y$ are the two smallest values. Then $a_1 = x$ and let $s$ be the smallest index such that $a_s = y$. Now there are two other terms whose sum is $x+y$. Thus we have $a_2 = x$ and $a_{s+1} = y$. Since $a_1 + a_2 = 2x$, we must have $a_3 = a_4 = x$. Similarly, by con... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 13 | |
01rj | $N$ numbers are marked in the set $\{1, 2, ..., 2000\}$ so that any pair of the numbers $(1, 2), (2, 4), ..., (1000, 2000)$ contains at least one marked number.
Find the least possible value of $N$. | [
"For any odd number $a$, $a < 2000$, define the set\n$$\nS_a = \\{a \\cdot 2^i \\mid i \\ge 0,\\ a \\cdot 2^i \\le 2000\\}.\n$$\nNote that $\\{1, 2, ..., 2000\\} = \\bigcup_a S_a$, where $a$ varies through all odd numbers less than $2000$. Both numbers of the pair $(k, 2k)$ belong to one and the same $S_a$. So we n... | Belarus | SELECTION and TRAINING SESSION | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 666 | |
0290 | Problem:
Dois triângulos retângulos congruentes possuem catetos de medidas $4~\mathrm{cm}$ e $7~\mathrm{cm}$. Na figura abaixo, à esquerda, os triângulos foram desenhados de modo a coincidirem os catetos de $7~\mathrm{cm}$. Assim, $AB=7~\mathrm{cm}$ e $AD=BC=4~\mathrm{cm}$. Já na figura à direita, eles foram desenhado... | [
"Solution:\n\nOs pontos $A, B, C$ e $D$ formam o retângulo $ABCD$.\n\nComo as diagonais de um retângulo o dividem em quatro triângulos de mesma área, a área sombreada é igual a três quartos da área do retângulo $ABCD$. Portanto, a área sombreada é igual a $\\frac{3}{4}(7 \\times 4)=21~\\mat... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | First figure: 21 cm^2; Second figure: 131/7 cm^2 | |
04nz | Let $a_1, a_2, \dots, a_{41}$ be an arithmetic sequence such that
$$
\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \dots + \frac{1}{\sqrt{a_{40}} + \sqrt{a_{41}}}
$$
is a positive integer. If $a_1 = 1$, and the common difference of the sequence is a positive integer, find the common difference... | [] | Croatia | Croatia_2018 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 2, 9, 38 | |
0gzv | Point $P$ lies inside triangle $ABC$. Let $I_A$, $I_B$, $I_C$ be incenters of triangles $PBC$, $PAC$, $PAB$ respectively. Let $I_P$ denote the incenter of triangle $I_A I_B I_C$. Prove that for point $P$ which satisfies the condition $I_P = P$, the following equalities hold:
$$
AP - BP = AC - BC, \quad BP - CP = BA - C... | [
"Let us denote the points $A' = AP \\cap I_B I_C$, $B' = BP \\cap I_A I_C$, $C' = CP \\cap I_B I_A$ (see Fig. 11).\n\n$I_P = P$ implies that $I_A P$ is a bisector of $\\angle BPC$ and $\\angle I_B I_A I_C$, therefore two couples of lines $I_A I_B$, $I_A I_C$, and also $BP$, $CP$ are symmetric with respect to $I_A P... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010) | [
"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 | |
0ij6 | Problem:
In triangle $A B C$, choose point $A_{1}$ on side $B C$, point $B_{1}$ on side $C A$, and point $C_{1}$ on side $A B$ in such a way that the three segments $A A_{1}$, $B B_{1}$, and $C C_{1}$ intersect in one point $P$. Prove that $P$ is the centroid of triangle $A B C$ if and only if $P$ is the centroid of t... | [
"Solution:\n\nAssume first that $P$ is the centroid of $\\triangle A B C$. This means that $A A_{1}$, $B B_{1}$ and $C C_{1}$ are the medians of $\\triangle A B C$, and hence $A_{1}$, $B_{1}$ and $C_{1}$ are the midpoints of $B C$, $C A$ and $A B$, respectively. As a midsegment of $\\triangle A B C$, segment $B_{1}... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Misc... | null | proof only | null | |
0cht | Let $n \ge 3$ be a positive integer, the set $S = \{1, 2, 3, \dots, n\}$, and the set $\mathcal{F}$ of functions from $S$ to $S$. We say that a set $\mathcal{G} \subset \mathcal{F}$ is a *generating set* for the set $\mathcal{H} \subset \mathcal{F}$ if every function in $\mathcal{H}$ can be represented as a composition... | [
"Let $fg$ denote the function composition $f \\circ g$ (where $f, g \\in \\mathcal{F}$), and let $(i_1, i_2, \\dots, i_p)$ denote the function $f : S \\to S$ defined by $f(i_j) = i_{j+1}$ for $j = \\overline{1, p-1}$, $f(i_p) = i_1$, and $f(x) = x$ for $x \\ne i_1, \\dots, i_p$ (where $i_1, \\dots, i_p$ are $p \\ge... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof only | null | |
05jj | Problem:
Dans le plan, on considère l'ensemble $S$ des points de coordonnées $(x, y)$, où $x, y \in \{1,2, \ldots, 2013\}$. Deux points de $S$ sont dits voisins s'ils sont à une distance $1$ l'un de l'autre. À chaque seconde, une mouche et des araignées se déplacent sur les points de $S$ de la façon suivante : tout d'... | [
"Solution:\n\nUne seule araignée ne peut attraper la mouche : en effet, la mouche peut attendre jusqu'à ce que l'araignée soit sur un point voisin de celui sur lequel elle se trouve. À partir de ce moment, elle se déplace à chaque fois sur un point diagonalement opposé à celui sur lequel est l'araignée si celle-ci ... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) 2; b) 2 | |
0i9l | Problem:
A partition of a number $n$ is a sequence of positive integers, arranged in descending order, whose sum is $n$. For example, $n=4$ has 5 partitions: $1+1+1+1=2+1+1=2+2=3+1=4$. Given two different partitions of the same number, $n=a_{1}+a_{2}+\cdots+a_{k}=b_{1}+b_{2}+\cdots+b_{l}$, where $k \leq l$, the first ... | [] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | final answer only | null | |
0bvh | Let $n \ge 3$ be a positive integer. Consider an $n \times n$ square. In each cell of the square, one of the numbers from the set $M = \{1, 2, \dots, 2n-1\}$ is to be written. One such filling is called "good" if, for every index $i$, $1 \le i \le n$, row no. $i$ and column no. $i$, together, contain all the elements o... | [
"a) For $n = 4$ the filling below is good.\n\n| 1 | 2 | 4 | 5 |\n|---|---|---|---|\n| 3 | 1 | 6 | 4 |\n| 7 | 5 | 1 | 2 |\n| 6 | 7 | 3 | 1 |\n\nb) We prove that there is no good filling of an $n \\times n$ square if $n$ is odd. Assume the contrary to be true. Then, for some odd number $n$, we have that, for every in... | Romania | The Danube Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0b5b | Problem:
Într-un cuvânt format din literele $a$ şi $b$ sunt posibile schimbările $a b a \rightarrow b$, $b \rightarrow a b a$, $b b a \rightarrow a$, $a \rightarrow b b a$. Dacă începem de la cuvântul $\underbrace{a a \ldots a}_{2003} b$, este posibil de obținem cuvântul $b \underbrace{a a \ldots a}_{2003}$ ? | [] | Romania | Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
07th | A convex quadrilateral $ABCD$ is such that $|AB| + |CD| = |AD| + |BC|$. Prove that a circle can be inscribed in $ABCD$ touching all four sides. | [
"There always exists a circle that touches three of the four sides. For example, the circle that touches $BC$, $CD$ and $DA$ is either the incircle or an excircle of the triangle that has $CD$ as one side and two other sides which are contained in the lines $BC$ and $DA$.\n\n\n\nLet $X$, $Y... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0251 | Problem:
Em certa loteria, existem 60 números distintos e 6 deles são sorteados sem reposição. Cada bilhete possui 6 números distintos entre os 60 possíveis. O prêmio máximo, conhecido como "gol-no-ângulo", é dado para o jogador que possuir o bilhete com os mesmos 6 números que foram sorteados. Nesta loteria, também e... | [
"Solution:\n\n(a) Tome os 6 menores números $\\{1,2,3,4,5,6\\}$. Neste caso, temos apenas 6 bilhetes bola-na-trave, que são formados por escolhas de 6 números do conjunto $\\{1,2,3,4,5,6,7\\}$. Em princípio, existiriam 7 bilhetes, mas como um bilhete gol-no-ângulo não é bola-na-trave, o número de bilhetes bola-na-t... | Brazil | NÍVEL 3 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | (a) Example: {1,2,3,4,5,6}; number of near-miss tickets: 6.
(b) Example: {3,6,9,12,15,18}; number of near-miss tickets: 728.
(c) For {2,3,8,11,14,17}: 485 near-miss tickets.
(d) For {8,10,12,14,16,18}: 376 near-miss tickets. | |
07ql | For each positive integer $n$, let
$$
C_n = \frac{1}{n+1} \binom{2n}{n},
$$
where, as usual,
$$
\binom{2n}{n} = \frac{(2n)!}{(n!)^2}.
$$
(i) Prove that $C_n$ is an integer for every positive integer $n$.
(ii) Prove that $C_n$ is not a prime number if $n > 3$. | [
"Observe first that\n$$\n\\binom{2n+1}{n} = \\frac{(2n+1)((2n)!)}{n!(n+1)!} = (2n+1)C_n,\n$$\nso\n$$\nC_n = (2n + 2 - (2n + 1))C_n = 2 \\binom{2n}{n} - \\binom{2n+1}{n},\n$$\nso $C_n$ is an integer, as all binomial coefficients $\\binom{m}{k}$ with positive integers $m, k$ with $k \\le m$ are integers. This establi... | Ireland | Irish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0jyc | Problem:
We define a chessboard polygon to be a polygon whose edges are situated along lines of the form $x = a$ and $y = b$, where $a$ and $b$ are integers. These lines divide the interior into unit squares, which we call cells.
Let $n$ and $k$ be positive integers. Assume that a square can be partitioned into $n$ c... | [
"Solution:\n\nNote that $n k = s^{2}$ for some $s$. By Factor Lemma, pick $n = a b$, $k = c d$, and $s = a c = b d$. Now we can tile the board with $a \\times b$ rectangles!"
] | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
04jf | Let $ABC$ be a right triangle with the right angle at $C$. Let $A'$, $B'$ and $C'$ be the pedals of the perpendiculars from the centroid of the triangle $ABC$ onto the lines $BC$, $CA$ and $AB$ respectively.
Determine the ratio of the areas of the triangles $A'B'C'$ and $ABC$. | [
"Let us denote by $T$ the centroid of the triangle $ABC$, by $a$, $b$ and $c$ the lengths of the sides $\\overline{BC}$, $\\overline{CA}$ and $\\overline{AB}$ respectively and by $v$ the length of the altitude to the side $\\overline{AB}$. Let us denote $\\alpha = \\angle BAC$ and $\\beta = \\angle CBA$.\nSince $T$... | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2:9 | |
09kp | Let $\mathbb{R}$ denote the set of all real numbers. Find all pairs of functions $f: \mathbb{R} \to \mathbb{R}$ and $h: \mathbb{R}^2 \to \mathbb{R}$ such that
$$
f(x + y - z)^2 = f(xy) + h(x + y + z, xy + yz + zx)
$$
for all $x, y, z \in \mathbb{R}$. | [
"Let us define two sets $D = \\{(x+y+z, xy+yz+zx) \\mid x, y, z \\in \\mathbb{R}\\}$ and $E = \\{(a, b) \\in \\mathbb{R}^2 \\mid a^2 \\ge 3b\\}$. It is obvious that $D \\subseteq E$. Now let us show that $E \\subseteq D$. Indeed, if $a^2 \\ge 3b$ there exists $c$ such that the polynomial $P(x) = x^3 - ax^2 + bx - c... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof and answer | All solutions are:
1) f(x) = c for some constant c, and h(a, b) = c^2 − c for all (a, b) with a^2 ≥ 3b; h can be arbitrary on pairs with a^2 < 3b.
2) f(x) = x/4, and h(a, b) = (a^2 − 4b)/16 for all (a, b) with a^2 ≥ 3b; h can be arbitrary on pairs with a^2 < 3b. | |
08uv | A point $P$ is located in the interior of a triangle $ABC$. If
$$
AP = \sqrt{3},\ BP = 5,\ CP = 2,\ AB : AC = 2 : 1,\ \angle BAC = 60^\circ,
$$
what is the value of the area of the triangle $ABC$?
Here by $XY$ we represent the length of the line segment $XY$. | [
"$$\n\\frac{6+7\\sqrt{3}}{2}\n$$\nTake a point $Q$ in the opposite side from the point $C$ with respect to the line $AB$ so as to make the triangles $ABQ$ and $ACP$ similar. The similarity ratio of the triangle $ABQ$ to the triangle $ACP$ is given by $AB : AC = 2 : 1$, and so we obtain $AQ = 2AP = 2\\sqrt{3}$, $BQ ... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (6+7√3)/2 | |
0evz | Problem:
Prove that any 39 successive natural numbers include at least one whose digit sum is divisible by 11. | [
"Solution:\nLet $n$ be the smallest number in the sequence and $m$ the smallest with last digit $0$. $m$ and $m+10$ have different digit sums unless (possibly) the penultimate digit of $m$ is $9$, but in that case $m+10$ and $m+20$ have different digit sums. So two of $m$, $m+10$, $m+20$ are sure to have different ... | Soviet Union | 1st ASU | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
00x3 | Problem:
Let $\mathbb{Q}^{+}$ denote the set of positive rational numbers. Show that there exists one and only one function $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$ satisfying the following conditions:
(i) If $0 < q < \frac{1}{2}$ then $f(q) = 1 + f\left(\frac{q}{1 - 2q}\right)$.
(ii) If $1 < q \leq 2$ then $f... | [
"Solution:\n\nBy condition (iii) we have $f(1) = 1$. Applying condition (iii) to each of (i) and (ii) gives two new conditions $(i')$ and $(ii')$ taking care of $q > 2$ and $\\frac{1}{2} \\leq q < 1$ respectively. Now, for any rational number $\\frac{a}{b} \\neq 1$ we can use (i), $(i')$, (ii) or $(ii')$ to express... | Baltic Way | Baltic Way 1992 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0dhw | Given that
$$
(10!)^3 = \overline{47\ 784\ 725\ 83a\ b72\ 000\ 000},
$$
determine the digits $a$ and $b$. | [
"As $(10!)^3$ is divisible by $9$, by the divisibility criterion for the number $9$ we get\n$$\n9|4+7+7+8+4+7+2+5+8+3+a+b+7+2+0+0+0+0+0+0=64+a+b.\n$$\nFurther, by Wilson's theorem we obtain $(10!)^3 \\equiv (-1)^3 \\equiv -1 \\pmod{11}$, and thus\n$$\n-1 \\equiv (7+8+7+5+3+b+2+0+0+0)-(4+7+4+2+8+a+7+0+0+0) = b-a \\p... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a = 9, b = 8 | |
072k | Find all triples $(a, b, c)$ integers such that $a^2 + b^2 = c^2$, $\text{gcd}(a, b, c) = 1$ and $2000 \le a, b, c \le 3000$. | [
"We use the known characterisation of primitive Pythagorean triples $(a, b, c)$: $a = 2pq$, $b = p^2 - q^2$, $c = p^2 + q^2$, where $p$ and $q$ are relatively prime integers, one even and $q < p$. The given condition entails $c^2 = a^2 + b^2 > 2 \\times 2000^2$. Hence $2000\\sqrt{2} < c = p^2 + q^2 < 3000$. Using t... | India | Indija TS 2006 | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2100, 2059, 2941) |
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