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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
02yc | Problem:
a) Dado que a representação decimal de $5^{2018}$ possui 1411 algarismos e começa com 3 (o dígito não nulo mais à esquerda é 3), para quantos inteiros $1 \leq n \leq 2017$ o número $5^{n}$ começa com 1?
b) Os inteiros $4^{52}$ e $5^{52}$ ambos começam com o algarismo 2. Se as representações decimais das potênc... | [
"Solution:\na) Se $5^{k}$ começa com $a$ e possui $j$ algarismos, então\n$$\n10^{j} < 5^{k} < a \\cdot 10^{j+1}\n$$\ne assim\n$$\n\\begin{aligned}\n10^{j} & < 5 \\cdot 10^{j} \\\\\n& < 5 \\cdot 5^{k} \\\\\n& = 5^{k+1} \\\\\n& < 10 \\cdot 10^{j+1}\n\\end{aligned}\n$$\nIsso significa que a representação decimal de $5... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) 607; b) 2 and 4 | |
09rr | Problem:
Zij $H$ het hoogtepunt van een scherphoekige driehoek $A B C$. De lijn door $A$ loodrecht op $A C$ en de lijn door $B$ loodrecht op $B C$ snijden elkaar in $D$. De cirkel met middelpunt $C$ door $H$ snijdt de omgeschreven cirkel van driehoek $A B C$ in de punten $E$ en $F$. Bewijs dat $|D E|=|D F|=|A B|$. | [
"Solution:\n\nDe driehoek is scherphoekig, dus $H$ ligt binnen de driehoek. Dat betekent dat $E$ en $F$ op de korte bogen $A C$ en $B C$ liggen. Neem aan dat $E$ op de korte boog $A C$ ligt en $F$ op de korte boog $B C$.\nAls we $H$ spiegelen in $A C$, komt het spiegelbeeld $H'$ op de omgeschreven cirkel van $\\tri... | Netherlands | IMO-selectietoets II | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07uy | Consider the points $B$, $N$, $M$, $C$ in this order on a line, such that
$$
|BC| = 2|BM| = 4|BN|.
$$
The perpendiculars on $BC$ raised from $N$ and $M$ meet a line through $C$ at the points $A$ and $E$, respectively. Let $D$ be the intersection of $AM$ and $BE$. Prove the following statements:
a. Line $AC$ is tangent... | [
"Because $AN$ is the perpendicular bisector of $BM$, triangle $ABM$ is isosceles with $\\angle ABM = \\angle AMB$. Similarly, $\\angle ECB = \\angle EBC$ in the isosceles triangle $EBC$. We have $\\angle AMB = \\angle MAC + \\angle ECB$ (external angle) and $\\angle ABM = \\angle ABE + \\angle EBC$.\n\n![](attached... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | English | proof only | null | |
0gwp | Take the chequered board $10 \times 10$ and consider all the possible patterns of painting 10 sectors so that there is only one painted sector in every row and every column. For each pattern we find rectangle of maximal area the sides of which are on the lines of the grid. Besides, this rectangle must have no painted s... | [
"Let's consider rectangles with their sides along the grid lines.\nOn the board let there be a rectangle with dimensions $A \\times B$, which has no painted sectors ($A$ is its width, $B$ is its height). According to the problem statement $A$ columns which the rectangle occupies must have $A$ painted sectors. On th... | Ukraine | Ukrajina 2008 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 25 | |
01ck | A function $f: \mathbb{R} \to \mathbb{R}$ satisfies
$$
f(f(a)) = f(a) \quad \text{and} \quad f(a+b) = f(a) + f(b)
$$
for all real numbers $a, b$. Prove that, for all real $x$, there exists a unique $y$ such that $f(y) = 0$ and $x = y + f(z)$ for some real $z$. | [
"Let $x$ be given. First the uniqueness is proved. Assume that $x = y + f(z)$ with $f(y) = 0$. If $f$ is applied on both sides, then\n$$\nf(x) = f(y + f(z)) = f(y) + f(f(z)) = 0 + f(z) = f(z),\n$$\n\nNow we prove that $y = x - f(x)$ has the assumed property. Observe that $f(a - b) = f(a) - f(b)$, and hence\n$$\nf(y... | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
019a | Prove that there exist infinitely many natural numbers $n$ such that all prime factors of $n^2 + 1$ are less than $n$. | [
"It is true for all numbers $n = 2a^2$ where $a > 1$ and $a \\equiv 1 \\pmod{5}$.\nIf $n = 2a^2$ then $n^2 + 1 = 4a^4 + 1 = (2a^2 + 2a + 1)(2a^2 - 2a + 1)$.\nAs $2a^2 - 2a + 1 < 2a^2$ it remains to ensure that all prime factors of $2a^2 + 2a + 1$ are less than $2a^2$.\nIf $a \\equiv 1 \\pmod{5}$ then $2a^2 + 2a + 1... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof only | null | |
0c8c | The points $M \in (AB)$, $N \in (BC)$ and $P \in (CD)$ are chosen on three sides of the rhombus $ABCD$. Prove that the centroid of the triangle $MNP$ belongs to the line $AC$ if and only if $AM + DP = BN$. | [
"Let $a$ be the rhombus' side length. Then one can find $u, v, t \\in (0, 1)$ such that $AM = au$, $BN = av$, and $DP = at$.\n\nDenote by $G$ the centroid of $MNP$ and suppose that the diagonals of the rhombus intersect at $O$. Then $G \\in AC$ if and only if there exists some $k \\in \\mathbb{R}$ such that $\\over... | Romania | Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
0acu | Ilina ate $\frac{1}{5}$ plus three of the candies from the bag. From the remaining candies she ate $\frac{1}{5}$ plus five the next day. The third day she ate the remaining 15 candies. How many candies were there in the bag in the beginning? | [
"Let $x$ be the number of candies in the bag in the beginning. Then Ilina ate $\\frac{1}{5}x + 3$ during the first day and there were $\\frac{4}{5}x - 3$ candies left. During the second day Ilina ate $\\frac{1}{5}(\\frac{4}{5}x - 3) + 5$ and the third day she ate the remaining 15 candies. Hence\n$$\n\\frac{1}{5}x +... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 35 | |
0jzm | Problem:
Rthea, a distant planet, is home to creatures whose DNA consists of two (distinguishable) strands of bases with a fixed orientation. Each base is one of the letters $H$, $M$, $N$, $T$, and each strand consists of a sequence of five bases, thus forming five pairs. Due to the chemical properties of the bases, e... | [
"Solution:\n\nThere are $4 \\cdot 3 = 12$ ways to choose the first base pairs, and regardless of which base pair it is, there are $3$ possibilities for the next base on one strand and $3$ possibilities for the next base on the other strand. Among these possibilities, exactly $2$ of them have identical bases forming... | United States | HMMT November 2017 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 28812 | |
0guj | 33 balls are placed on unit squares of a $10 \times 10$ board such that no unit square contains more than one ball. For each empty unit square we calculate the total number of balls located on the same row with this unit square and the total number of balls located on the same column with this unit square and after tha... | [
"7. The radical axes of the circles (ABC), (BDE), (CDE) must be concurrent at T hence T, D, E are collinear. Moreover, $TD \\cdot TE = TB \\cdot TK$ hence the power of T with respect to the circles (ABC), (ADE) are equal and it lies on their radical axis. Since DE and BC are parallel, the radical axis of (ABC), (AD... | Turkey | Team Selection Test for JBMO 2023 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 438 | |
09t8 | Problem:
Zij $k > 2$ een geheel getal. Een positief geheel getal $\ell$ noemen we $k$-pabel als we de getallen $1, 3, 5, \ldots, 2k-1$ kunnen opdelen in twee verzamelingen $A$ en $B$ zodat de som van de elementen van $A$ precies $\ell$ keer zo groot is als de som van de elementen van $B$. Bewijs dat het kleinste $k$-p... | [
"Solution:\n\nWe gaan bewijzen dat als $p$ de kleinste priemdeler van $k$ is, dat dan $p-1$ het kleinste $k$-pabele getal is. Hieruit volgt het gevraagde, want er geldt dan $\\operatorname{ggd}(p-1, k) = 1$.\n\nEr geldt $1 + 3 + 5 + \\ldots + (2k-1) = k^2$. Als $\\ell$ een $k$-pabel getal is en $s$ is de bijbehoren... | Netherlands | IMO-selectietoets II | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0l9a | A triangle $ABC$ inscribed in a circle with center $O$ has three altitudes $AH$, $BK$, $CL$. Let $A_0$, $B_0$, $C_0$ respectively be the midpoints of $AH$, $BK$, $CL$.
The incircle with center $I$ of triangle $ABC$ touches the sides $BC$, $CA$, $AB$ respectively at $D$, $E$, $F$.
Prove that the four lines $A_0D$, $B_0... | [] | Vietnam | Vietnamese Team Selection Contest for the 44th IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Advanced Conf... | English | proof only | null | |
086o | Problem:
Un numero naturale $k$ si dice $n$-squadrato se, colorando comunque con $n$ colori diversi le caselle di una scacchiera $2n \times k$, esistono 4 caselle distinte dello stesso colore i cui centri sono vertici di un rettangolo avente i lati paralleli ai lati della scacchiera. Determinare, in funzione di $n$, i... | [] | Italy | XXV OLIMPIADE ITALIANA DI MATEMATICA | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n(2n−1)+1 | |
0hm5 | Problem:
Let $M$ be the midpoint of the side $AC$ of triangle $ABC$. If $N$ is the point on the side $AB$, $O$ intersection of the lines $BM$ and $CN$, and if the areas of triangles $BON$ and $COM$ are equal, prove that $N$ is the midpoint of $AB$. | [
"Solution:\n\nSince the areas of $\\triangle BON$ and $\\triangle COM$ are equal we see that the areas of triangles $\\triangle BCN$ and $\\triangle CBM$ are also equal. Since these two triangles share the side, they must have the corresponding altitudes equal. Hence the length of perpendiculars from $M$ and $N$ to... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0fpc | Si $n$ es un número natural, el $n$-ésimo número triangular es $T_n = 1 + 2 + \cdots + n$. Hallar todos los valores de $n$ para los que el producto de los 16 números triangulares consecutivos $T_n T_{n+1} \cdots T_{n+15}$ es un cuadrado perfecto. | [
"Como $T_n = \\dfrac{n(n+1)}{2}$, el producto de los 16 números triangulares es $P_n = nC_n(n+16)/2^{16}$, donde $C_n = (n+1)^2 \\cdots (n+15)^2$ es un cuadrado perfecto. Entonces $P_n$ es un cuadrado perfecto si y sólo si lo es $n(n+16)$. Como $n$ y $n+16$ no tienen divisores impares comunes, para que $n(n+16)$ se... | Spain | LII Olimpiada Matemática Española | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences an... | Spanish | proof and answer | [2, 9] | |
0e14 | Let $n \ge 4$ be a positive integer. Into every square of an $n \times n$ array we inscribe an integer so that the sum of the integers inside any $3 \times 3$ square is negative. Find all positive integers $n$ for which this can be done in such a way that the sum of all the numbers in the array is positive. | [
"When $n$ is divisible by $3$, $n = 3k$. The array can be divided into $k^2$ $3 \\times 3$ squares. The sum of the numbers inside each square is negative, so the sum of all these sums, which is the sum of all the numbers in the array, is negative as well.\n\nIf $n$ is not divisible by $3$, the numbers can be inscri... | Slovenia | Selection Examinations for the IMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All positive integers n with n ≡ 1 or 2 (mod 3). | |
047a | Given an integer $n \ge 3$. Let $\frac{n(n-1)}{2}$ non-negative real numbers $x_{i,j}$ ($1 \le i < j \le n$) satisfy: for any $1 \le i < j < k \le n$, we have $x_{i,j} + x_{j,k} \le x_{i,k}$. Prove that:
$$
\left\lfloor \frac{n^2}{4} \right\rfloor \cdot \sum_{1 \le i < j \le n} x_{i,j}^4 \ge \left( \sum_{1 \le i < j \l... | [
"*Proof.* First, let's point out a situation where the equality holds, which will help us understand the problem. Let $0 = y_1 = \\cdots = y_{\\lfloor \\frac{n}{2} \\rfloor} < y_{\\lfloor \\frac{n}{2} \\rfloor+1} = \\cdots = y_n = 1$, and take $x_{i,j} = y_j - y_i$ for $1 \\le i < j \\le n$. In this case, $x_{i,j} ... | China | The 65th IMO China National Team Selection Test | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
05sx | Problem:
Soit $ABC$ un triangle, $\Omega$ son cercle circonscrit et $O$ le centre de $\Omega$. Soit $S$ le centre du cercle tangent aux côtés $AB$ et $AC$ et tangent intérieurement au cercle $\Omega$ en un point $K$. Le cercle de diamètre $[AS]$ recoupe le cercle $\Omega$ en un point $T$. Soit $M$ le milieu du segment... | [
"Solution:\n\nDans ce problème, nous allons utiliser divers résultats autour du cercle tangent aux côtés $AB$ et $AC$ au cercle $\\Omega$. Ce cercle est appelé le cercle $A$-mixtilinéaire, on le notera $\\omega$. On note $U$ le point de contact de ce cercle avec le côté $AB$ et $V$ le point de contact avec le côté ... | France | Envoi 5: Pot Pourri | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: cen... | null | proof only | null | |
0e6n | In the first circle with diameter $4$ we inscribe a square. In the square we inscribe a new circle. In the last circle we again inscribe a square and repeat the process. What is the diameter of the fourth circle?
(A) $\frac{1}{\sqrt{2}}$ (B) $\frac{1}{2}$ (C) $1$ (D) $\sqrt{2}$ (E) $2\sqrt{2}$ | [
"The ratio of the diameters of two consecutive circles is equal to the ratio of the diagonal and the side of the square, hence $\\sqrt{2}$. The ratio of the diameters of the first and the fourth circle is thus equal to $(\\sqrt{2})^3 = 2\\sqrt{2}$. The diameter of the fourth circle is $\\frac{4}{2\\sqrt{2}} = \\sqr... | Slovenia | National Math Olympiad 2012 | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | MCQ | D | |
0f5o | Problem:
$\{x_1 < x_2 < x_3 < \ldots < x_n\}$. $\{y_i\}$ is a permutation of the $\{x_i\}$. We have that $x_1 + y_1 < x_2 + y_2 < \ldots < x_n + y_n$. Prove that $x_i = y_i$. | [] | Soviet Union | 18th ASU | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0j28 | Problem:
An icosahedron is a regular polyhedron with twenty faces, all of which are equilateral triangles. If an icosahedron is rotated by $\theta$ degrees around an axis that passes through two opposite vertices so that it occupies exactly the same region of space as before, what is the smallest possible positive val... | [
"Solution:\n\n$72^{\\circ}$\n\nBecause this polyhedron is regular, all vertices must look the same. Let's consider just one vertex. Each triangle has a vertex angle of $60^{\\circ}$, so we must have fewer than $6$ triangles; if we had $6$, there would be $360^{\\circ}$ at each vertex and you wouldn't be able to \"f... | United States | Harvard-MIT November Tournament | [
"Geometry > Solid Geometry > 3D Shapes",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 72° | |
08k2 | Problem:
Find all positive integers $n$, $n \geq 3$, such that $n \mid (n-2)!$. | [
"Solution:\nFor $n=3$ and $n=4$ we easily check that $n$ does not divide $(n-2)!$.\n\nIf $n$ is prime, $n \\geq 5$, then $(n-2)! = 1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot (n-2)$ is not divided by $n$, since $n$ is a prime not included in the set of factors of $(n-2)!$.\n\nIf $n$ is composite, $n \\geq 6$, then $n... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All composite integers greater than or equal to six. | |
06fm | Suppose $\{a_n\}$ is a sequence in which all the terms are integers, and $a_2$ is odd. For any natural number $n$, $n(a_{n+1} - a_n + 3) = a_{n+1} + a_n + 3$. Furthermore, $a_{2009}$ is divisible by $2010$. Find the smallest integer $n$, $n \ge 2$, such that $a_n$ is divisible by $2010$. | [
"The smallest possible integer is $269$.\n\nFor $n > 1$, we rewrite the recurrence relation as follows.\n$$\n\\begin{aligned}\nn(a_{n+1} - a_n + 3) &= a_{n+1} + a_n + 3 \\\\\n\\Rightarrow \\quad (n-1)a_{n+1} &= (n+1)a_n - 3(n-1) \\\\\n\\Rightarrow \\quad \\frac{a_{n+1}}{n(n+1)} &= \\frac{a_n}{n(n-1)} - \\frac{3}{n(... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 269 | |
0h6i | How many three-digit numbers are there with non-zero digits which have the following property: after any permutation of its digits one obtains three-digit number which is divisible by $4$? | [
"Obviously, all digits of such number are even because only even digit can be the last one. Also we cannot use $2$ or $6$ because if a number is divisible by $4$ then the last two digits of it are $12$, $32$, $...$, $92$ or $16$, $36$, $...$, $96$. So such number consists of the digits $4$ and $8$. So:\n\nThree dig... | Ukraine | 56th Ukrainian National Mathematical Olympiad, Third Round | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics"
] | English | proof and answer | 8 | |
099w | Find all natural $x$ such that for every natural $n$ with $10^n + n \mid x^n + n$? | [
"Only $x = 10$.\n\nAssume the contrary and a prime $p$ that does not divide $x - 10$. By the Chinese Remainder Theorem we can find a positive integer $n$ such that\n$$\n\\begin{cases}\nn \\equiv 1 \\pmod{p-1} \\\\\nn \\equiv -10 \\pmod{p}\n\\end{cases}.\n$$\nThen by Fermat's theorem,\n$$\n10^n + n \\equiv 10 + n \\... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 10 | |
068n | Let $ABC$ be an acute angled triangle inscribed in a circle $c(O, R)$ and $F$ a point on the side $AB$ such that $AF < \frac{AB}{2}$. The circle $c_1(F, FA)$ intersects the line $OA$ at point $A'$ and the circle $(c)$ at $K$. Prove that the quadrilateral $BKFA'$ is inscribed in a circle passing through $O$. | [
"The triangle $AFK$ is isosceles and hence $\\hat{F}_1 = 2\\hat{A}_1$. The angle $\\hat{A}_1$ is inscribed in the circle $(c)$ and $\\hat{O}_1 = 2\\hat{A}_1 = \\hat{F}_1$, and hence the quadrilateral $BKFO$ is cyclic.\n\nNext we will prove that the quadrilateral $OBKA'$ is cyclic. In fact, if $S$ be the counter poi... | Greece | SELECTION EXAMINATION | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0389 | Problem:
Find all real numbers $a$ for which the equation $x^{2}+a x+3 a^{2}-7 a-19=0$ has real roots $x_{1}$ and $x_{2}$ such that
$$
\frac{1}{x_{1}-2}+\frac{1}{x_{2}-2}=-\frac{2 a}{13}
$$ | [
"Solution:\nUsing Vieta's formulas we get\n$$\n\\frac{1}{x_{1}-2}+\\frac{1}{x_{2}-2}=\\frac{x_{1}+x_{2}-4}{\\left(x_{1}-2\\right)\\left(x_{2}-2\\right)}=-\\frac{a+4}{3 a^{2}-5 a-15}\n$$\nTherefore $3 a^{2}-5 a-15 \\neq 0$ and\n$$\n\\frac{a+4}{3 a^{2}-5 a-15}=\\frac{2 a}{13}\n$$\nHence $6 a^{3}-10 a^{2}-43 a-52=0 \\... | Bulgaria | Spring Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 4 | |
05x7 | Problem:
On place un certain nombre de segments ouverts dans le plan, aucun d'entre eux n'est parallèle aux axes $x$ et $y$. Ces segments sont disjoints. Thanima commence à se déplacer depuis $(0,0)$ parallèlement à l'axe $x$. À chaque fois qu'elle rencontre un mur, elle tourne de 90 degrés, et continue à se déplacer ... | [
"Solution:\n\nOn commence par démontrer qu'il existe un mur qui est plus bas que tous les autres (c'est à dire que l'ensemble des points sous ce segment n'intersecte aucun segment). Supposons par l'absurde que ce n'est pas le cas. On peut alors construire un cycle $A_{1}, B_{1}, A_{2}, B_{2}, \\ldots, A_{k}, B_{k}$... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0il5 | Problem:
Compute the sum of all real numbers $x$ such that
$$
2 x^{6}-3 x^{5}+3 x^{4}+x^{3}-3 x^{2}+3 x-1=0
$$ | [
"Solution:\nThe carefully worded problem statement suggests that repeated roots might be involved (not to be double counted), as well as complex roots (not to be counted). Let $P(x)=2 x^{6}-3 x^{5}+3 x^{4}+x^{3}-3 x^{2}+3 x-1$. Now, $a$ is a double root of the polynomial $P(x)$ if and only if $P(a)=P^{\\prime}(a)=0... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | -1/2 | |
068a | a. Examine if there is a real number $x$, such that both $x + \sqrt{3}$ and $x^2 + \sqrt{3}$ are rational numbers.
b. Examine if there is a real number $y$, such that both $y + \sqrt{3}$ and $y^3 + \sqrt{3}$ are rational numbers. | [
"a. Let $x + \\sqrt{3} = q$, $x^2 + \\sqrt{3} = p$ with $p, q \\in \\mathbb{Q}$. Then\n$$\nx = q - \\sqrt{3} \\Rightarrow x^2 = q^2 - 2q\\sqrt{3} + 3\n$$\nso substituting in the second one gives:\n$$\n(q^2 - 2q\\sqrt{3} + 3) + \\sqrt{3} = p \\Leftrightarrow -\\sqrt{3}(2q-1) = p - q^2 - 3\n$$\nIt follows that $2q-1=... | Greece | Hellenic Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | proof and answer | a) Yes: x = 1/2 − √3. b) No such y exists. | |
07tv | Let $n$ be a positive integer. Describe, in terms of the prime factorisation of $n$, the largest integer which is the side length of a square tile which can be used to completely tile a rectangle that is inscribed in a circle of radius $n$, if such a tiling is possible. | [
"This problem is the general version of Problem 16 and the first part of the solution is the same. Let $x$ be the side length of the square tile. If the rectangle is completely tiled with such square tiles, there exist integers $a, b$ such that the side lengths of the rectangle are $ax$ and $bx$. The diagonals of t... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson th... | null | proof and answer | If n has a prime factor congruent to 1 modulo 4, let p be the smallest such prime factor; then the maximal square tile side length is x = 2n / p. If n has no prime factor congruent to 1 modulo 4, no such tiling is possible. | |
0e6y | Find all natural numbers $n$ and prime numbers $p$ such that $\sqrt[3]{n} + \frac{p}{\sqrt[3]{n}}$ is the square of a natural number. | [
"Denote $\\sqrt[3]{n} + \\frac{p}{\\sqrt[3]{n}} = k^2$ where $k$ is a natural number. We raise the equation to the 3rd power and get $n + 3p\\sqrt[3]{n} + 3\\frac{p^2}{\\sqrt[3]{n}} + \\frac{p^3}{n} = k^6$, which is $n + 3pk^2 + \\frac{p^3}{n} = k^6$. From this we see that $n$ must divide $p^3$. Since $p$ is prime,... | Slovenia | National Math Olympiad 2012 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (n, p) = (1, 3) and (27, 3) | |
0k2t | Problem:
Let $n$ be a nonnegative integer. Prove that the numbers $n+2$ and $n^{2}+n+1$ cannot both be perfect cubes. | [
"Solution:\n\nIf both numbers are perfect cubes then so is their product. But\n$$\n(n+2)\\left(n^{2}+n+1\\right)=n^{3}+3 n^{2}+3 n+2=(n+1)^{3}+1,\n$$\nwhich cannot be a perfect cube, contradiction."
] | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
01bs | The quadrilateral $Q$ has a longest side of length $b$ and a shortest side of length $a$. Form a new quadrilateral $Q'$ by joining the successive midpoints of the edges of $Q$. Supposing that $Q$ and $Q'$ are similar, prove that $\frac{b}{a} < 1 + \sqrt{2}$. | [
"$Q'$ is a Varignon parallelogram, so that $Q$ is also a parallelogram. Let $v$ be the acute or right angle of $Q$. The diagonals $p$ and $q$ of $Q$, which are twice the sides of $Q'$, satisfy\n$$\np^2 = a^2 + b^2 - 2ab \\cos v \\quad \\text{and} \\quad q^2 = a^2 + b^2 + 2ab \\cos v.\n$$\nThe similarity of $Q$ and ... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0c3f | Determine all triples of real numbers $(a, b, c)$ that satisfy simultaneously the equations:
$$
\begin{align*}
a(b^2 + c) &= c(c + ab), \\
b(c^2 + a) &= a(a + bc), \\
c(a^2 + b) &= b(b + ca).
\end{align*}
$$ | [
"Let $(a, b, c)$ be a solution of the system. If one of the numbers $a, b, c$ is $0$, e.g. if $c = 0$, then $c(a^2 + b) = a(b + ca)$ leads to $a = 0$, and similarly one gets $b = 0$. Thus $abc = 0$ leads to $a = b = c = 0$ which is indeed a solution. We now look for solutions with $abc \\neq 0$. We rewrite the equa... | Romania | 69th NMO Selection Tests for JBMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | (x, x, x) for any real x | |
0gjg | 二、設 $x, y, z$ 為正實數。試求
$$
\frac{x}{3x+y+z} + \frac{y}{x+3y+z} + \frac{z}{x+y+3z}
$$
之值的範圍。
II. Let $x, y, z$ be three positive real numbers. Determine all possible values for the expression
$$
\frac{x}{3x+y+z} + \frac{y}{x+3y+z} + \frac{z}{x+y+3z}.
$$ | [
"二、首先注意到要求的式子是 $x, y, z$ 的齊次式, 故不失一般性可設 $x+y+z=2$, 使原式成為\n$$\n\\begin{aligned}\n& \\frac{x}{3x+y+z} + \\frac{y}{x+3y+z} + \\frac{z}{x+y+3z} \\\\\n&= \\frac{x}{2x+2} + \\frac{y}{2y+2} + \\frac{z}{2z+2} \\\\\n&= \\frac{1}{2} \\left( \\frac{x}{x+1} + \\frac{y}{y+1} + \\frac{z}{z+1} \\right) \\\\\n&= \\frac{3}{2} - \\f... | Taiwan | APMO Taiwan Preliminary Round 1 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | Chinese; English | proof and answer | (1/3, 3/5] | |
0bcs | a. Show that $x^4 - x^3 - x + 1 \ge 0$, for all real numbers $x$.
b. Find all real numbers $x_1, x_2$ and $x_3$ given that $x_1 + x_2 + x_3 = 3$ and $x_1^3 + x_2^3 + x_3^3 = x_1^4 + x_2^4 + x_3^4$. | [
"a. Write $x^4 - x^3 - x + 1 = (x-1)(x^3 - 1) = (x-1)^2(x^2 + x + 1)$ and notice that $x^2 + x + 1 > 0$ for all $x \\in \\mathbb{R}$ to get the claim.\n\nb. Notice that $\\sum_{k=1}^{3} (x_k^4 - x_k^3 - x_k + 1) = 0$ and use (a) to derive that $x_1^4 - x_1^3 - x_1 + 1 = 0$, $k = 1, 2, 3$. It follows that $x_1 = x_2... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x1 = x2 = x3 = 1 | |
0ix3 | Problem:
Prove that, for all $n > 3$, there exists a graph with chromatic number $n$ that does not contain any $n$-cliques. | [
"Solution:\n\nWe prove the claim by induction on $n$. The case $n = 3$ was addressed in (a).\n\nLet $n \\geq 3$ and suppose $G$ is a graph with chromatic number $n$ containing no $n$-cliques. We produce a graph $G'$ with chromatic number $n+1$ containing no $(n+1)$-cliques as follows. Add a vertex $v$ to $G$, and a... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
00l6 | Determine the largest constant $C$ such that
$$
(x_1 + x_2 + \dots + x_6)^2 \geq C \cdot (x_1(x_2 + x_3) + x_2(x_3 + x_4) + \dots + x_6(x_1 + x_2))
$$
holds for all real numbers $x_1, x_2, \dots, x_6$.
For this $C$, determine all $x_1, x_2, \dots, x_6$ such that equality holds. | [
"We rewrite the right-hand side\n\nExpanding yields\n$$\nX^2 + Y^2 + Z^2 \\geq XY + YZ + ZX\n$$\nThis is equivalent to\n$$\n(X - Y)^2 + (Y - Z)^2 + (Z - X)^2 \\geq 0\n$$\nwith equality for $X - Y = Y - Z = Z - X = 0$, i.e., $X = Y = Z$, thus $x_1 + x_4 = x_2 + x_5 = x_3 + x_6$."
] | Austria | National Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | C = 3; equality holds precisely when x1 + x4 = x2 + x5 = x3 + x6. | |
0bgn | Problem:
Fie $f:[0, \pi / 2] \rightarrow[0, \infty)$ o funcție crescătoare. Să se arate că:
a. $\int_{0}^{\pi / 2}(f(x)-f(\pi / 4))(\sin x-\cos x) \, \mathrm{d} x \geq 0$. | [] | Romania | Olimpiada Naţională de Matematică, Etapa judeţeană şi a municipiului Bucureşti | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Functions"
] | null | proof only | null | |
0j3r | Problem:
Let $f(x) = c x(x-1)$, where $c$ is a positive real number. We use $f^{n}(x)$ to denote the polynomial obtained by composing $f$ with itself $n$ times. For every positive integer $n$, all the roots of $f^{n}(x)$ are real. What is the smallest possible value of $c$? | [
"Solution:\nAnswer: $2$\n\nWe first prove that all roots of $f^{n}(x)$ are greater than or equal to $-\\frac{c}{4}$ and less than or equal to $1+\\frac{c}{4}$. Suppose that $r$ is a root of $f^{n}(x)$. If $r = -\\frac{c}{4}$, $f^{-1}(r) = \\left\\{ \\frac{1}{2} \\right\\}$ and $-\\frac{c}{4} < \\frac{1}{2} < 1+\\fr... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 2 | |
0e6z | Find all natural numbers $n$ and prime numbers $p$ such that $\sqrt[n]{n+\frac{2}{p}}$ is a natural number. | [
"Denote $\\sqrt[n]{n + \\frac{2}{p}} = k$ where $k$ is a natural number, hence $n + \\frac{2}{p} = k^n$. Thus $p$ is a divisor of $2$. Since $p$ is prime, we have $p = 2$ or $p = 1$ (but $1$ is not prime). So $p = 2$.\n\nNow, $n + \\frac{2}{2} = k^n \\implies n + 1 = k^n$.\n\nTry $n = 1$: $1 + 1 = 2 = k^1 \\implies... | Slovenia | National Math Olympiad 2012 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n = 1, p = 2 | |
0exx | Problem:
Given non-negative real numbers $a_1$, $a_2$, ..., $a_n$, such that $a_{i-1} \leq a_i \leq 2a_{i-1}$ for $i = 2, 3, \ldots, n$. Show that you can form a sum $s = b_1 a_1 + \ldots + b_n a_n$ with each $b_i = +1$ or $-1$, so that $0 \leq s \leq a_1$. | [
"Solution:\n\nWe show that you can pick $b_n, b_{n-1}, ..., b_r$ so that $s_r = b_n a_n + b_{n-1} a_{n-1} + \\ldots + b_r a_r$ satisfies $0 \\leq s_r \\leq a_r$. Induction on $r$.\n\nTrivial for $r = n$. Suppose true for $r$. Then $-a_{r-1} \\leq s_r - a_{r-1} \\leq a_r - a_{r-1} \\leq a_{r-1}$. So with $b_{r-1} = ... | Soviet Union | 6th ASU | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
01v8 | Given a convex quadrilateral $ABCD$. The point $A_1$ is on the boundary of $ABCD$ such that the segment $AA_1$ divides $ABCD$ into two parts with equal areas. In the same way we define points $B_1$, $C_1$ and $D_1$. It is known that the lengths of all segments $AA_1$, $BB_1$, $CC_1$, and $DD_1$ do not exceed $1$.
Prove... | [
"Let $O = AC \\cap BD$; and, without loss of generality, $BO \\geq DO$, $CO \\geq AO$. If $AB \\parallel DC$ then $1 \\leq \\frac{BO}{DO} = \\frac{AO}{CO} \\leq 1$, whence $BO = OD$ and $ABCD$ is a parallelogram. Then by the problem condition $AC \\leq 1$, $BD \\leq 1$, hence $S(ABCD) \\leq \\frac{1}{2} AC \\cdot B... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
086m | Problem:
Sia $ABC$ un triangolo scaleno e acutangolo, $\Gamma$ la sua circonferenza circoscritta, $K$ il piede della bisettrice relativa al vertice $A$. Sia $M$ il punto medio dell'arco $BC$ che contiene $A$. Detta $A'$ la seconda intersezione di $MK$ con $\Gamma$, sia $T$ l'intersezione delle tangenti a $\Gamma$ in $... | [] | Italy | XXV OLIMPIADE ITALIANA DI MATEMATICA | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08lg | Problem:
Let $\Gamma$ be a circle of center $O$, and $\delta$ be a line in the plane of $\Gamma$, not intersecting it. Denote by $A$ the foot of the perpendicular from $O$ onto $\delta$, and let $M$ be a (variable) point on $\Gamma$. Denote by $\gamma$ the circle of diameter $A M$, by $X$ the (other than $M$) intersect... | [
"Solution:\nConsider the line $\\rho$ tangent to $\\gamma$ at $A$, and take the points $K = A M \\cap X Y$, $L = \\rho \\cap X M$, and $F = O A \\cap X Y$.\n\n(Remark: Moving $M$ into its reflection with respect to the line $O A$ will move $X Y$ into its reflection with respect to $O A$. These old and the new $X Y$... | JBMO | 2008 Shortlist JBMO | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03wa | It is known that $p, q$ ($q \neq 0$) are real numbers; the equation $x^2 - px + q = 0$ has two real roots $\alpha, \beta$; the sequence $\{a_n\}$ satisfies $a_1 = p$, $a_2 = p^2 - q$, $a_n = p a_{n-1} - q a_{n-2}$ ($n = 3, 4, \dots$).
a. Find the general expression of $\{a_n\}$ in terms of $\alpha, \beta$.
b. If $p =... | [
"(1) By Vieta's theorem, we have $\\alpha \\times \\beta = q \\neq 0$, $\\alpha + \\beta = p$. Then\n$$\n\\begin{aligned}\na_n &= p a_{n-1} - q a_{n-2} \\\\\n&= (\\alpha + \\beta)a_{n-1} - \\alpha\\beta a_{n-2} \\quad (n = 3, 4, \\dots).\n\\end{aligned}\n$$\nThis can be rewritten as\n$$\na_n - \\beta a_{n-1} = \\al... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | a) If the roots are distinct, a_n = (β^{n+1} − α^{n+1})/(β − α). If the roots are equal, a_n = (n + 1) α^n. b) For p = 1 and q = 1/4, the sum of the first n terms is S_n = 3 − (n + 3)/2^n. | |
0e5u | Someone has written the numbers $1$, $2$, $\ldots$, $33$ on a chalkboard. In each step, we choose two (not necessarily different) numbers on the chalkboard such that one divides the other. We then erase these two numbers and write their quotient, which is a natural number, on the chalkboard. We repeat the process until... | [
"Let $P_k$ be the product of all the numbers on the chalkboard after the $k$\\text{th}$ step. If in the $k$\\text{th}$ step we choose the numbers $a > b$ such that $b$ divides $a$, then\n$$\nP_k = \\frac{P_{k-1}}{ab} \\cdot \\frac{a}{b} = \\frac{P_{k-1}}{b^2}.\n$$\nThus, after each step, the product of all the numb... | Slovenia | Selection Examinations for the IMO 2012 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 7 | |
04ux | Let $ABC$ be an isosceles triangle with base $AB$ and $P$ a point on its $C$-altitude. Ray $AP$ meets the circumcircle of the triangle $ABC$ again at $Q \neq A$. The line through $P$ parallel to $AB$ meets the side $BC$ at $R$. Prove that $QR$ bisects the angle $AQB$. (Jaroslav Švrček) | [] | Czech Republic | Second Round | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
020c | Problem:
Let $ABCD$ be a square. Consider a variable point $P$ inside the square for which $\angle BAP \geq 60^{\circ}$. Let $Q$ be the intersection of the line $AD$ and the perpendicular to $BP$ in $P$. Let $R$ be the intersection of the line $BQ$ and the perpendicular to $BP$ from $C$.
a. Prove that $|BP| \geq |BR|... | [
"Solution:\n\n\n\nWe claim that $\\triangle ABP$ and $\\triangle RCB$ are similar triangles. Indeed, if we denote the intersection of $BP$ and $CR$ by $S$, then $\\angle RCB = \\angle SCB = 90^{\\circ} - \\angle SBC = 90^{\\circ} - \\angle PBC = \\angle ABP$. Moreover, the right angles in $... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | Equality holds exactly when the interior point makes the triangle at that corner equilateral, that is, when the angle at the corner is sixty degrees and the adjacent side equals the segment to the point; there is a unique such point inside the square. | |
00ph | Determine the maximum possible number of distinct real roots of a polynomial $P(x)$ of degree $2012$ with real coefficients satisfying the condition
$$
P(a)^3 + P(b)^3 + P(c)^3 \geq 3P(a)P(b)P(c)
$$
for all real numbers $a, b, c$ with $a + b + c = 0$. | [
"We will prove that there exists a polynomial $P(x)$ which satisfies the given condition and has $2012$ distinct real roots.\nFirst we note that the given inequality is equivalent to\n$$\n(P(a) + P(b) + P(c))((P(a) - P(b))^2 + (P(b) - P(c))^2 + (P(c) - P(a))^2) \\geq 0,\n$$\nso it is enough to find a polynomial $P$... | Balkan Mathematical Olympiad | Balkan 2012 shortlist | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 2012 | |
0a67 | Problem:
Josie and Kevin are each thinking of a two digit positive integer. Josie's number is twice as big as Kevin's. One digit of Kevin's number is equal to the sum of digits of Josie's number. The other digit of Kevin's number is equal to the difference between the digits of Josie's number. What is the sum of Kevin... | [
"Solution:\n\nWe'll use $\\overline{AB}$ to denote a 2 digit number with $A$ in the tens digit and $B$ in the unit digit.\n\nLet Josie pick the number $\\overline{AB}$ and Kevin $\\overline{CD}$. Then we have\n\n$$\n\\overline{AB} = 2 \\times \\overline{CD}\n$$\n$$\n10A + B = 20C + 2D\n$$\n\nNow, $A \\geq 2C > C$ s... | New Zealand | NZMO Round One | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 51 | |
0hdy | Vasia wrote down all seven-digit numbers that contain every digit between $1$ and $7$ exactly once. Prove that neither of the numbers Vasia wrote down divides another such number. | [
"Suppose one of such numbers $a$ is divisible by $b$, that is, there is an integer $n > 1$ such that $a = n b$. Since both $a$ and $b$ have remainder $1$ when divided by $9$, then the number $n$ also has remainder $1$ when divided by $9$. Since $n \\ne 1$, then $n \\ge 10$, thus the number $a$ has at least one digi... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0j0z | Problem:
Suppose $a, b, c$ are real numbers such that $a+b \geq 0$, $b+c \geq 0$, and $c+a \geq 0$. Prove that
$$
a+b+c \geq \frac{|a|+|b|+|c|}{3} .
$$
(Note: $|x|$ is called the absolute value of $x$ and is defined as follows. If $x \geq 0$ then $|x|=x$; and if $x<0$ then $|x|=-x$. For example, $|6|=6$, $|0|=0$ and $... | [
"Solution:\nThe inequality $b+c \\geq 0$ gives $a+b+c \\geq a$. On the other hand, adding up the other two given inequalities yields $(a+b)+(c+a) \\geq 0$, resulting in $a+b+c \\geq -a$. Since $|a|=a$ or $-a$, we have in any case that\n$$\na+b+c \\geq |a| .\n$$\nSimilarly\n$$\n\\begin{aligned}\n& a+b+c \\geq |b| \\... | United States | Bay Area Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0615 | Problem:
In einem $10 \times 17$-Rechteck werden 74 Punkte markiert.
Man beweise, dass es dabei stets zwei markierte Punkte gibt, deren Abstand 2 nicht überschreitet. | [] | Germany | Auswahlwettbewerb zur IMO 2000 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Combinatorial Geometry > Minkowski's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0hbd | Points $B$ and $C$ are chosen on the circle with diameter $AD$ in such a way that $AB = AC$. Point $P$ is an arbitrary point of the segment $BC$, and points $M$ and $N$ are chosen on the segments $AB$ and $AC$ respectively in such a way that $PMAN$ is a parallelogram. Let $PL$ be a bisector in triangle $MPN$. Line $PD$... | [
"First, we are going to prove that $\\angle BDP = \\angle AMN$ and $\\angle PDC = ANM$. Indeed, $MP \\parallel AC$, $NP \\parallel AB$, because $PMAN$ is a parallelogram (fig. 38). Then $\\angle MPB = \\angle ABC = \\angle ACB = \\angle NPC$, meaning that $\\triangle BMP \\sim \\triangle PNC$. We also note that $BC... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0asb | Problem:
What is the smallest positive integral value of $n$ such that $n^{300} > 3^{500}$? | [
"Solution:\n\n$7$"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 7 | |
01y7 | Let $P(x)$ be a non-constant polynomial with integer coefficients such that $P(0) \neq 1$. Prove that there exist infinitely many primes $p$ such that $P(a) - a^{\frac{p-1}{2}}$ is divisible by $p$ for some positive integer $a$ (possibly, depending on $p$). | [
"Consider the polynomial $Q(x) = P(x^2) - 1 \\in \\mathbb{Z}[x]$. It's well-known that there exist infinitely many prime divisors of the numbers from the set $M = \\{Q(n) : n \\in \\mathbb{N} \\ \\&\\ Q(n) \\neq 0\\}$. Moreover, since $Q(0) = P(0) - 1 \\neq 0$, among these prime divisors there exist infinitely many... | Belarus | BY 2020-2021 tst for Navid | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
047v | Let point $P$ lie on the nine-point circle of triangle $ABC$. A line through $P$ perpendicular to $AP$ intersects $BC$ at $Q$. A line through $A$ perpendicular to $AQ$ intersects $PQ$ at $X$. Let $H$ be the orthocenter of triangle $ABC$, and let $D$ and $M$ be the midpoints of segments $BC$ and $AQ$, respectively. Prov... | [
"\n\n**Proof:** Let $N$ be the midpoint of $AH$. By the properties of the nine-point circle, $DN$ is its diameter, so $DP \\perp PN$. Since $AH \\perp BC$ and $AP \\perp PQ$, we have $\\triangle DPQ \\sim \\triangle NPA$. Therefore, $\\frac{DQ}{NA} = \\frac{PQ}{PA}$.\n\nSince $XA \\perp AQ$... | China | China-TST-2025A | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0c1f | Let $k > 2$ be a real number.
a) Prove that for all positive real numbers $x$, $y$ and $z$ the following inequality holds:
$$
\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} > 2\sqrt{\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}}
$$
b) Prove that there exist positive real numbers $x$, $y$ and $z$ such that
$$
\sqrt{x+y} + \sqrt{y+z} + \sqrt{z... | [
"a) We have $\\sqrt{x+y} + \\sqrt{y+z} + \\sqrt{z+x} > 2\\sqrt{\\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}} \\Leftrightarrow x+y+z + \\sqrt{x^2+xy+yz+zx} + \\sqrt{y^2+xy+yz+zx} + \\sqrt{z^2+xy+yz+zx} > 2 \\cdot \\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}$. But\n$\\sqrt{x^2+xy+yz+zx} > x$, $\\sqrt{y^2+xy+yz+zx} > y$ and $\\sqrt{z^2+xy+... | Romania | 69th NMO Selection Tests for JBMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
05sm | Problem:
Trouver toutes les fonctions $f$ de $\mathbb{R}$ dans $\mathbb{R}$ telles que pour tout couple $(x, y)$ de réels :
$$
f(f(x))+f(f(y))=2 y+f(x-y)
$$ | [
"Solution:\nAnalysons le problème : ici on est face à une équation fonctionnelle, avec deux variables. La première chose à faire est d'essayer les quelques substitutions classiques : $x=y=0$, $x=0$, $y=0$, $x=y$. Ici comme on a un $f(x-y)$, il est très tentant de regarder ce que ça donne pour $x=y$. Posons donc $C=... | France | Envoi 5: Pot Pourri | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x | |
03rh | Let $k$ be a real number such that the inequality $\sqrt{x-3} + \sqrt{6-x} \ge k$ has a solution. The maximum value of $k$ is ( ).
(A) $\sqrt{6}-\sqrt{3}$
(B) $\sqrt{3}$
(C) $\sqrt{6}+\sqrt{3}$
(D) $\sqrt{6}$ | [
"Set $y = \\sqrt{x-3} + \\sqrt{6-x}$, $3 \\le x \\le 6$.\nThen\n$$\n\\begin{aligned}\ny^2 &= (x-3) + (6-x) + 2\\sqrt{(x-3)(6-x)} \\\\\n&\\le 2[(x-3) + (6-x)] = 6.\n\\end{aligned}\n$$\nSo $0 < y \\le \\sqrt{6}$, and the maximum value of $k$ is $\\sqrt{6}$. Answer: D."
] | China | China Mathematical Competition (Jiangxi) | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | MCQ | D | |
07yq | Problem:
Siano date nel piano due circonferenze $\gamma_{1}$ e $\gamma_{2}$ di centri $A$ e $B$ rispettivamente, e intersecantesi in due punti $C$ e $D$. Si supponga che la circonferenza passante per $A$, $B$ e $C$ intersechi ulteriormente $\gamma_{1}$ e $\gamma_{2}$ in $E$ ed $F$ rispettivamente, e che l'arco $E F$ n... | [
"Solution:\n\nPer il teorema dell'angolo al centro, $\\widehat{C A D}=2 \\widehat{C E D}$. Poiché per simmetria $\\widehat{C A B}=\\widehat{D A B}$, si ha $\\widehat{C A B}=\\widehat{C E D}$. Inoltre, $\\widehat{C A B}=\\widehat{C E B}$, perché entrambi insistono sull'arco $C B$, e quindi $\\widehat{C E D}=\\wideha... | Italy | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
05l6 | Problem:
On note $\mathbb{N}^*$ l'ensemble des entiers naturels strictement positifs. Trouver toutes les fonctions $f: \mathbb{N}^* \rightarrow \mathbb{N}^*$ telles que
$$
m^2+f(n) \mid m f(m)+n
$$
pour tous entiers strictement positifs $m$ et $n$. | [
"Solution:\n\nSoit $f$ une solution éventuelle du problème.\nOn note $\\left( * \\right)$ la condition \"$m^2+f(n)$ divise $m f(m)+n$\".\n\nEn choisissant $m=n=2$ dans $\\left(*)\\right$, il vient que $4+f(2)$ divise $2 f(2)+2$. Or, on a $2 f(2)+2 < 2(f(2)+4)$, et il faut donc que $f(2)+4=2 f(2)+2$, d'où $f(2)=2$.\... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(n) = n for all positive integers n | |
09ab | Let $P(x)$ be a unitary polynomial with integer coefficients and $Q(x) = P(x^{2^{2010}})$. If $|P(0)| = 2010$ then prove that $Q(x)$ is irreducible over $\mathbb{Z}$. | [
"First, we will show that $Q_1 = P(x^2)$ is irreducible over $\\mathbb{Z}$. Let $\\deg P = n$. Then $\\deg Q_1 = 2n$. To the contrary, assume that there exist $R_1(x), R_2(x) \\in \\mathbb{Z}[x]$ such that $Q_1(x) = R_1(x)R_2(x)$. It is clear that $Q_1(x) = Q_1(-x) = R_1(-x) \\cdot R_2(-x)$. Let $F(x)$ be common fa... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | proof only | null | |
0i46 | Problem:
What is the maximum number of lattice points (i.e. points with integer coordinates) in the plane that can be contained strictly inside a circle of radius $1$? | [
"Solution:\n\n$4$. The circle centered at $(1/2, 1/2)$ shows that $4$ is achievable. On the other hand, no two points within the circle can be at a mutual distance of $2$ or greater. If there are more than four lattice points, classify all such points by the parity of their coordinates: (even, even), (even, odd), (... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4 | |
02ml | Problem:
Contando os zeros - Quantos zeros existem no final do número $9^{2007}+1$ ? | [
"Solution:\n\nInicialmente, verificamos como terminam as potências de $9$, ou seja, listamos os dois últimos algarismos, os da dezena e da unidade, das potências $9^{n}$, ordenadamente.\n\n| Se $n$ for | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: ... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | 1 | |
02r6 | Problem:
Um conjunto de inteiros consecutivos é equilibrado se ele pode ser dividido em dois subconjuntos com o mesmo número de elementos, de modo que:
1) os dois subconjuntos não tenham elementos em comum;
2) a soma dos elementos de um dos subconjuntos seja igual à soma dos elementos do outro;
3) a soma dos quadrados... | [
"Solution:\n\na) Dividimos o conjunto $\\{1,2,3,4,5,6,7,8\\}$ nos subconjuntos $\\{1,4,6,7\\}$ e $\\{2,3,5,8\\}$. Como\n$$\n1+4+6+7=18=2+3+5+8\n$$\n$$\n1^{2}+4^{2}+6^{2}+7^{2}=102=2^{2}+3^{2}+5^{2}+8^{2}\n$$\nvemos que $\\{1,2,3,4,5,6,7,8\\}$ é equilibrado.\n\nb) Seja $A=\\{a+1, a+2, a+3, \\ldots, a+8\\}$ um conjun... | Brazil | Nível 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
08et | Problem:
Sia $ABC$ un triangolo acutangolo, sia $M$ il punto medio di $BC$, e sia $H$ il piede dell'altezza uscente da $B$. Indichiamo con $Q$ il centro della circonferenza circoscritta al triangolo $ABM$, e con $X$ l'intersezione tra l'altezza $BH$ e l'asse di $BC$.
Dimostrare che i seguenti due fatti sono equivalent... | [
"Solution:\n\nSia $Y$ la proiezione di $X$ su $AB$. Dimostriamo che le circonferenze circoscritte ai triangoli $AMC$ e $AXH$ passano entrambe per $Y$. Questo è equivalente a dimostrare che $BY \\cdot BA = BM \\cdot BC$, e questo a sua volta è vero in quanto entrambi i prodotti sono uguali a $BX \\cdot BH$, dal mome... | Italy | XXXVII Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
002o | Hallar el mayor entero positivo no divisible por $10$ que es múltiplo de alguno de los números que se obtienen al suprimirle dos dígitos consecutivos de su escritura decimal, ninguno de ellos en la primera o en la última posición. | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | Español | proof and answer | 989901 | |
0asx | Problem:
Let $x$ and $y$ be positive real numbers such that $x + 2y = 8$. Determine the minimum value of
$$
x + y + \frac{3}{x} + \frac{9}{2y}
$$ | [
"Solution:\n\n8"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 8 | |
0dov | Find all functions $f : \mathbb{R} \to \mathbb{R}$, satisfying the identity
$$
f(x^2 + xy + f(y)) = (f(x))^2 + x f(y) + y
$$
for all $x, y \in \mathbb{R}$. | [
"Using the identity $x^2 + x y = (-x - y)^2 + (-x - y) y$ we obtain the following identity for the function $f$:\n$$\n(f(x))^2 + x f(y) = (f(-x - y))^2 - (x + y) f(y).\n$$\nLet's write down some particular cases of the last identity:\n$$\n(1) \\quad (f(x))^2 + x f(-x) = (f(0))^2 \\text{ (it is obtained by putting }... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x for all real x | |
00o8 | Determine all pairs $(x, y)$ of positive integers such that for $d = \gcd(x, y)$ the equation
$$
xyd = x + y + d^2
$$
holds. | [
"**Answer.** There are three such pairs, $(x, y) = (2, 2)$, $(x, y) = (2, 3)$ and $(x, y) = (3, 2)$.\n\nFor $x = 1$, we get $d = 1$ and the given equation becomes the contradiction $y = y + 2$. This works analogously for $y = 1$.\nTherefore, we can assume $x \\ge 2$ and $y \\ge 2$.\n\nWe start with the case $d = 1$... | Austria | AUT_ABooklet_2023 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (2, 2), (2, 3), (3, 2) | |
0jnc | Problem:
Consider a $10 \times 10$ grid of squares. One day, Daniel drops a burrito in the top left square, where a wingless pigeon happens to be looking for food. Every minute, if the pigeon and the burrito are in the same square, the pigeon will eat $10\%$ of the burrito's original size and accidentally throw it int... | [
"Solution:\n\nLabel the squares using coordinates, letting the top left corner be $(0,0)$. The burrito will end up in $10$ (not necessarily different) squares. Call them $p_{1} = (x_{1}, y_{1}) = (0,0), p_{2} = (x_{2}, y_{2}), \\ldots, p_{10} = (x_{10}, y_{10})$. $p_{2}$ through $p_{10}$ are uniformly distributed t... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Expected values",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 71.8 | |
04jc | Let $a$, $b$ and $c$ be real numbers and let $f: \mathbb{R} \to \mathbb{R}$ be a function given by
$$
f(x) = a x^5 + b x^3 + c \sin x - 1.
$$
If $f(-2015) = 2015$, determine $f(2015)$. | [] | Croatia | Croatia Mathematical Competitions | [
"Precalculus > Functions",
"Precalculus > Trigonometric functions"
] | null | proof and answer | -2017 | |
0ku7 | Problem:
Triangle $ABC$ has incenter $I$. Let $D$ be the foot of the perpendicular from $A$ to side $BC$. Let $X$ be a point such that segment $AX$ is a diameter of the circumcircle of triangle $ABC$. Given that $ID = 2$, $IA = 3$, and $IX = 4$, compute the inradius of triangle $ABC$. | [
"Solution:\n\nLet $R$ and $r$ be the circumradius and inradius of $ABC$, let $AI$ meet the circumcircle of $ABC$ again at $M$, and let $J$ be the $A$-excenter. We can show that $\\triangle AID \\sim \\triangle AXJ$ (e.g. by $\\sqrt{bc}$ inversion), and since $M$ is the midpoint of $IJ$ and $\\angle AMX = 90^\\circ$... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 11/12 | |
0a41 | Given is $\triangle ABC$ with circumcircle $\Gamma$. Let $M$ be the midpoint of the arc $BC$ of $\Gamma$ not containing $A$. The vertex $N$ on $\Gamma$ is the antipode of $A$. The line through $B$ perpendicular to $AM$ intersects $AM$ at $D$ and intersects $\Gamma$ a second time at $P \neq B$. The line through $D$ perp... | [
"We note the half of the angle at $A$ as $\\alpha = \\frac{1}{2} \\angle BAC = \\angle BAM = \\angle MAC$, as $M$ is the midpoint of arc $BC$. Then we note that $\\angle ABP = \\angle ABD = 90^\\circ - \\angle DAB = 90^\\circ - \\alpha$ and that $\\angle EDA = 90^\\circ - \\angle DAE = 90^\\circ - \\alpha$. Moreove... | Netherlands | BxMO/EGMO Team Selection Test | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dcm | The triangle $ABC$ ($AB > BC$) is inscribed in the circle $\Omega$. On the sides $AB$ and $BC$, the points $M$ and $N$ are chosen, respectively, so that $AM = CN$. The lines $MN$ and $AC$ intersect at point $K$. Let $P$ be the center of the inscribed circle of triangle $AMK$, and $Q$ the center of the excircle of the t... | [
"Let $T$ be the second intersection of two circles $(BMN)$ and $(O)$. We have\n$$\n\\angle TAB = \\angle TCB, \\quad \\angle TMB = \\angle TNB,\n$$\nand $AM = CN$, so $\\triangle TAM \\cong \\triangle TCN$. Then $TA = TB$, which means that $T$ is the midpoint of the arc $BAC$ of circle $(O)$.\n\n![](attached_image_... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscel... | English | proof only | null | |
0eav | There are 17 girls and 12 boys on the playground. How many children, at least, would have to join them, so that they could be divided into two groups of equal sizes both containing equal numbers of girls and boys?
(A) 1 (B) 3 (C) 5 (D) 7 (E) 9 | [
"If the children form 2 groups of equal sizes such that each group contains $n$ boys and $n$ girls, then there must be $2n$ girls on the playground at that time. This is an even number that must also be equal to the number of boys on the playground. In order for the number of girls to be even and equal to the numbe... | Slovenia | National Math Olympiad in Slovenia | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
0i0y | Problem:
Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive numbers, with $n \geq 2$. Prove that
$$
\left(x_{1}+\frac{1}{x_{1}}\right)\left(x_{2}+\frac{1}{x_{2}}\right) \cdots\left(x_{n}+\frac{1}{x_{n}}\right) \geq\left(x_{1}+\frac{1}{x_{2}}\right)\left(x_{2}+\frac{1}{x_{3}}\right) \cdots\left(x_{n-1}+\frac{1}{x_{n}}\right)... | [
"Solution:\nFirst we will prove a simple lemma involving only two variables: For all positive $a, b$,\n$$\n\\left(a^{2}+1\\right)\\left(b^{2}+1\\right) \\geq (a b+1)^{2}\n$$\nTo see why this is true, multiply out, and after simplifying, we have\n$$\na^{2}+b^{2} \\geq 2 a b\n$$\nThis is equivalent to\n$$\na^{2}-2 a ... | United States | 2nd Bay Area Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0gsp | For each real number $a$ let $\lfloor a \rfloor$ be the largest integer not exceeding $a$. Find all positive real numbers satisfying
$$
x \cdot \lfloor x \rfloor + 2022 = \lfloor x^2 \rfloor.
$$ | [
"Answer: $x = n + \\frac{2022}{n}$, where $n \\ge 2023$ is any integer.\n\nLet $x = n + \\alpha$, where $0 \\le \\alpha < 1$. Inserting it into the main equation\n$$\nx[x] + 2022 = [x^2]\n$$\nwe get\n$$\n(n + \\alpha)n + 2022 = \\lfloor(n + \\alpha)^2\\rfloor \\quad (1)\n$$\nwhich implies that $(n + \\alpha)n$ is a... | Turkey | Team Selection Test | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | x = n + 2022/n, where n is an integer with n ≥ 2023 | |
0kwu | Problem:
Compute the unique positive integer $n$ such that $\frac{n^{3}-1989}{n}$ is a perfect square. | [
"Solution:\nWe need $n^{2}-\\frac{1989}{n}$ to be a perfect square, so $n \\mid 1989$. Also, this perfect square would be less than $n^{2}$, so it would be at most $(n-1)^{2}=n^{2}-2 n+1$. Thus,\n\n$$\n\\frac{1989}{n} \\geq 2 n-1 \\Longrightarrow 1989 \\geq 2 n^{2}-n\n$$\n\nso $n \\leq 31$. Moreover, we need\n$$\nn... | United States | HMMT November | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 13 | |
03vb | Given three cubes with integer edge lengths, if the sum of their surface areas is $564\ \text{cm}^2$, then the sum of their volumes is ( ). | [
"Denote the edge lengths of the three cubes as $a$, $b$ and $c$, respectively. Then we have\n$$\n6(a^2 + b^2 + c^2) = 564,\n$$\ni.e. $a^2 + b^2 + c^2 = 94$. We may assume that\n$$\n1 \\le a \\le b \\le c < 10.\n$$\n\nThen\n$$\n3c^2 \\geq a^2 + b^2 + c^2 = 94.\n$$\nIt follows that $c^2 > 31$. So $6 \\leq c < 10$, an... | China | China Mathematical Competition | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 586 or 764 | |
06yf | Let $ABCD$ be a quadrilateral with $AB$ parallel to $CD$ and $AB < CD$. Lines $AD$ and $BC$ intersect at a point $P$. Point $X \neq C$ on the circumcircle of triangle $ABC$ is such that $PC = PX$. Point $Y \neq D$ on the circumcircle of triangle $ABD$ is such that $PD = PY$. Lines $AX$ and $BY$ intersect at $Q$.
Prove... | [
"Let $M$ and $N$ be the midpoints of $AD$ and $BC$, respectively and let the perpendicular bisector of $AB$ intersect the line through $P$ parallel to $AB$ at $R$.\n\nLemma. Triangles $QAB$ and $RNM$ are similar.\n\nProof. Let $O$ be the circumcentre of triangle $ABC$, and let $S$ be the midpoint of $CX$. Since $N,... | IMO | IMO2024 Shortlisted Problems | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geom... | English | proof only | null | |
0918 | Problem:
Let $n$ be a positive integer. Prove that if the sum of all positive divisors of $n$ is a perfect power of $2$, then the number of these divisors is also a perfect power of $2$. | [
"Solution:\n\nSuppose that $n = p_{1}^{s_{1}} p_{2}^{s_{2}} \\ldots p_{k}^{s_{k}}$, where $p_{1}, \\ldots, p_{k}$ are distinct primes and $s_{i} \\geqslant 1$ for each $i$, and that the sum of all positive divisors of $n$, which is given by\n$$\n\\left(1 + p_{1} + p_{1}^{2} + \\cdots + p_{1}^{s_{1}}\\right)\\left(1... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0jxz | Problem:
Consider a sequence $x_{n}$ such that $x_{1}=x_{2}=1$, $x_{3} = \frac{2}{3}$. Suppose that $x_{n} = \frac{x_{n-1}^{2} x_{n-2}}{2 x_{n-2}^{2} - x_{n-1} x_{n-3}}$ for all $n \geq 4$. Find the least $n$ such that $x_{n} \leq \frac{1}{10^{6}}$. | [
"Solution:\n\nThe recursion simplifies to $\\frac{x_{n-1}}{x_{n}} + \\frac{x_{n-3}}{x_{n-2}} = 2 \\frac{x_{n-2}}{x_{n-1}}$. So if we set $y_{n} = \\frac{x_{n-1}}{x_{n}}$ for $n \\geq 2$ then we have $y_{n} - y_{n-1} = y_{n-1} - y_{n-2}$ for $n \\geq 3$, which means that $\\{y_{n}\\}$ is an arithmetic sequence. From... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 13 | |
04zp | Ants has three pencils, each of a different color. In how many ways can he paint the faces of a regular octahedron in such a way that faces with a common edge always have different colors? Colorings that can be obtained from each other via rotations of the octahedron are considered the same. | [
"i) Case $4, 4, 0$. There are $3$ possibilities to choose two colors from the three. After that, there is only one possibility to paint the octahedron. Thus there are $3$ possibilities to paint.\n\nii) Case $4, 3, 1$. Ordering the $3$ colors can be done in $6$ ways. After that, there is only one possibility to pain... | Estonia | Selected Problems from the Final Round of National Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 15 | |
00kr | Let $x$, $y$ and $z$ be positive real numbers with $x + y + z = 3$. Prove that at least one of the three numbers
$$
x(x + y - z), \quad y(y + z - x) \quad \text{or} \quad z(z + x - y)
$$
is less or equal $1$. | [
"Since the three expressions are cyclic, we may w. l. o. g. assume that $x \\ge y, z$. Consequently we have $x \\ge \\frac{x+y+z}{3} = 1$. We now show that $a := y(y+z-x) = y(3-2x)$ satisfies $a \\le 1$.\n\n* Case a): For $\\frac{3}{2} \\le x < 3$ clearly $a \\le 0 < 1$.\n\n* Case b): For $1 \\le x < \\frac{3}{2}$ ... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0gfs | 設 $r \ge 2$ 為整數, 且 $m_1, n_1, m_2, n_2, \dots, m_r, n_r$ 為 $2r$ 個整數, 使得
$$
|m_i n_j - m_j n_i| = 1
$$
對所有的 $1 \le i < j \le r$ 皆成立。試求 $r$ 的最大可能值。 | [] | Taiwan | 2022 數學奧林匹亞競賽第二階段培訓營, 獨立研究 (三) | [
"Algebra > Linear Algebra > Determinants",
"Algebra > Linear Algebra > Vectors",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | Chinese; English | proof and answer | 3 | |
07xr | Two circles $\Gamma_1$ and $\Gamma_2$ intersect at two distinct points $A$ and $B$. A line $\ell$ through $A$ meets $\Gamma_1$ and $\Gamma_2$ at points $C$ and $D$ respectively, such that $D$ lies inside $\Gamma_1$. The line perpendicular to $\ell$ through $A$ meets $\Gamma_1$ and $\Gamma_2$ at points $E$ and $F$ respe... | [
"As $X$ lies on the perpendicular bisector of $EF$ we have $\\angle XEF = \\angle XFE$; call this angle $\\alpha$. Cyclicity of $AFBD$ implies $\\angle DBA = \\angle DFA = \\angle XFE = \\alpha$. Cyclicity of $EABC$ gives $\\angle CBA = 180^\\circ - \\angle CEA = 180^\\circ - \\angle XEF = 180^\\circ - \\alpha$. Th... | Ireland | IRL_ABooklet_2025 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0b8p | Consider a function $f : \mathbb{R} \to [0, \infty)$. Prove that $f$ satisfies the inequality $f(x+y) \ge (1+y)f(x)$ for any $x \in \mathbb{R}$ and any $y \ge 0$, if and only if the function $g : \mathbb{R} \to [0, \infty)$ defined by $g(x) = e^{-x}f(x)$, for $x \in \mathbb{R}$, is non-decreasing. | [
"The inequality $e^y \\ge 1+y$, holds for all $y \\in \\mathbb{R}$. Assume the function $g$ is monotonously increasing on $\\mathbb{R}$. Then, based on the inequality above, we get\n$$\nf(x + y) = e^{x+y}g(x + y) \\ge e^x(1 + y)g(x) = (1 + y)f(x),\n$$\nfor all $x \\in \\mathbb{R}$ and all $y \\ge 0$.\n\nConversely,... | Romania | Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0a5j | Problem:
$ABCD$ is a rectangle with side lengths $AB = CD = 1$ and $BC = DA = 2$. Let $M$ be the midpoint of $AD$. Point $P$ lies on the opposite side of line $MB$ to $A$, such that triangle $MBP$ is equilateral. Find the value of $\angle PCB$. | [
"Solution:\n\n$M$ is the midpoint of $AD$, by symmetry $MB = MC$. The side lengths of an equilateral triangle are all equal, so $MB = MP$.\n\n\n\nAs $MB = MC = MP$, $M$ is the circumcenter of triangle $BCP$. For any chord of any circle, the angle subtended at the center is always double the... | New Zealand | New Zealand Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 30° | |
0k5s | Problem:
Three friends wish to divide five different tasks among themselves, such that every friend must handle at least one task. In how many different ways can this be done? | [
"Solution:\n\nIf there were no restriction that all friends are assigned a task, the number of ways to assign would simply be $3^{5} = 243$. So we will use complementary counting.\n\nCall the friends $A$, $B$, and $C$. The number of ways to assign the tasks such that $A$ and $B$ have at least one task, but $C$ has ... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 150 | |
0bvo | A *special* set is a set of positive odd integers no element of which divides another, and each 3-element subset of which has a member dividing the sum of the other two. A *special* set is *maximal* if it is contained in no other *special* set. Determine the number of elements a maximal *special* set may have.
Yu. I. I... | [
"Leaving aside the trivial case $\\{1\\}$, a maximal special set may have only 3, 4 or 5 elements. Begin by noticing that if $a < b$ are positive odd integers, and $a$ does not divide $b$, then $a$, $b$ and $2b-a$ form a special set, so a maximal special set has at least three elements.\n\nAt the other extreme, a s... | Romania | Fifteenth IMAR Mathematical Competition | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1, 3, 4, or 5 | |
095z | Problem:
Se dă triunghiul ascuţitunghic $ABC$, înscris în cercul de centru $O$. Construim înălţimea $AD$ ($D \in BC$) şi bisectoarea $AE$ ($E \in BC$), care intersectează cercul în punctul $F$. Notăm cu $L$ intersecţia dreptei $AO$ cu cercul. Să se demonstreze că dreptele $FD$ şi $LE$ se intersectează pe cercul circum... | [
"Solution:\n\nUnghiul $ABL$ se sprijină pe diametrul $AL \\Rightarrow m(\\angle ABL) = 90^{\\circ}$. În $\\triangle CAD$ avem $m(\\angle CAD) = 90^{\\circ} - m(\\angle ACB)$. (1)\n\nÎn $\\triangle ALB$, $m(\\angle BAO) = 90^{\\circ} - m(\\angle ALB)$. Dar $m(\\angle ALB) = m(\\angle ACB)$ (se sprijină pe acelaşi ar... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0082 | 21 numbers are written in a row. If $u$, $v$, $w$ are three consecutive ones then $v = \frac{2uw}{u+w}$. The first number is $\frac{1}{100}$, the last one is $\frac{1}{101}$. Find the 15th number. | [
"Write $v = \\frac{2uw}{u+w}$ as $\\frac{1}{v} = \\frac{u+w}{2uw}$. This gives $\\frac{1}{v} = \\frac{1}{2} \\left( \\frac{1}{u} + \\frac{1}{w} \\right)$, or $\\frac{1}{v} - \\frac{1}{u} = \\frac{1}{w} - \\frac{1}{v}$. So look at the sequence of reciprocals of the given numbers: $\\frac{1}{v} - \\frac{1}{u} = \\fra... | Argentina | National Olympiad of Argentina | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | 10/1007 | |
0dim | Let $a_1, a_2, a_3, \dots, a_{10}$ and $b_1, b_2, \dots, b_{10}$ be real numbers such that the roots of these 10 polynomials
$$
x^2 + a_1x + b_1,\ x^2 + a_2x + b_2,\ \dots,\ x^2 + a_{10}x + b_{10}
$$
are all integer numbers $\pm 1, \pm 2, \dots, \pm 10$ (in some order).
a) What is the maximum amount of odd values among... | [
"a) Denote $x_i, y_i$ as the roots of the $i$-th polynomial, then by Vieta's theorem, $a_i = -(x_i + y_i)$ and $b_i = x_i y_i$. Thus $a_i b_i = -x_i y_i (x_i + y_i)$ which is always even, implying that at most 1 number among $a_i, b_i$ is odd. Hence, there are at most 10 odd values among their coefficients.\n\nThe ... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a) 10; b) minimum −385 and maximum 380 | |
0jd2 | Problem:
Let $x$ and $y$ be real numbers with $x > y$ such that $x^{2} y^{2} + x^{2} + y^{2} + 2 x y = 40$ and $x y + x + y = 8$. Find the value of $x$. | [
"Solution:\n\nWe have $(x y)^{2} + (x + y)^{2} = 40$ and $x y + (x + y) = 8$.\n\nSquaring the second equation and subtracting the first gives $x y (x + y) = 12$.\n\nSo $x y$, $x + y$ are the roots of the quadratic $a^{2} - 8a + 12 = 0$.\n\nIt follows that $\\{x y, x + y\\} = \\{2, 6\\}$.\n\nIf $x + y = 2$ and $x y ... | United States | HMMT | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 3 + sqrt(7) | |
0g41 | Problem:
Let $a$, $b$, $c$, $\lambda$ be positive real numbers with $\lambda \geq 1 / 4$. Show that
$$
\frac{a}{\sqrt{b^{2}+\lambda b c+c^{2}}}+\frac{b}{\sqrt{c^{2}+\lambda c a+a^{2}}}+\frac{c}{\sqrt{a^{2}+\lambda a b+b^{2}}} \geq \frac{3}{\sqrt{\lambda+2}}
$$ | [
"Solution:\nDenote the left side of the inequality by $LS$. By Hölder we have\n$$\n\\left(a\\left(b^{2}+\\lambda b c+c^{2}\\right)+b\\left(c^{2}+\\lambda c a+a^{2}\\right)+c\\left(a^{2}+\\lambda a b+b^{2}\\right)\\right)(LS)^{2} \\geq (a+b+c)^{3}\n$$\nSo now it is sufficient to prove\n$$\n\\frac{(a+b+c)^{3}}{a^{2} ... | Switzerland | IMO Selection | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0db8 | Find all pair of integers ($m$, $n$) and $m \geq n$ such that there exist a positive integer $s$ and
1. Product of all divisors of $s m$, $s n$ are equal.
2. Number of divisors of $s m$, $s n$ are equal. | [
"1) Denote $d(x)$, $\\pi(x)$ as the number of divisors, the product of divisors of positive integer $x$.\nFirstly, we can see that for any divisor $y_{k}$ of $x$, $1 \\leq k \\leq d(x)$ then $\\frac{x}{y_{i}}$ is also divisor of $x$, thus\n$$\n\\prod_{k=1}^{d(x)} y_{k} = \\prod_{k=1}^{d(x)} \\frac{x}{y_{k}} \\text{... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 1) The product-of-divisors condition holds if and only if m = n. 2) The equal-divisor-count condition holds if and only if either m = n or n does not divide m (equivalently, m and n are not comparable by divisibility). |
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