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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0fht | Problem:
Sea $O \cdot X Y Z$ un triedro trirrectángulo de vértice $O$ y aristas $X$, $Y$ y $Z$. Sobre la arista $Z$ se fija un punto $C$ tal que $O C = c$. Sobre $X$ e $Y$ se consideran, respectivamente, puntos variables $P$ y $Q$ de manera que $O P + O Q$ sea una constante dada $k$. Para cada par de puntos $P$ y $Q$,... | [
"Solution:\n\nEn la figura se muestran con trazo discontinuo las circunferencias que resultan de intersecar la esfera con los planos coordenados. Las proyecciones del centro $W$ de la esfera sobre estos planos coinciden con los centros de estas circunferencias (denotados $F$, $G$ y $H$ en la figura) y al ser el tri... | Spain | OME 30 | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Projection locus: the line x + y = k/2 in the plane z = 0. Center locus: the line given by x + y = k/2 and z = c/2. | |
0iwv | Problem:
A ball inside a rectangular container of width $7$ and height $12$ is launched from the lower-left vertex of the container. It first strikes the right side of the container after traveling a distance of $\sqrt{53}$ (and strikes no other sides between its launch and its impact with the right side).
Find the h... | [
"Solution:\n\nLet $h$ be this height. Then, using the Pythagorean theorem, we see that $h^2 + 7^2 = 53$, so $h = 2$."
] | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2 | |
06d7 | Find all positive integers $n$ such that the equation $x^3 + y^3 + z^3 = nx^2y^2z^2$ has positive integer solutions. Be sure to give a proof. | [
"$n$ can be $1$ or $3$.\n\nWLOG assume $x \\ge y \\ge z$. Then we have\n$$\n3x^3 \\ge x^3 + y^3 + z^3 = nx^2y^2z^2,\n$$\nwhich gives $x \\ge \\frac{n}{3}y^2z^2$.\n\nNext, we rewrite the equation as $y^3 + z^3 = x^2(ny^2z^2 - x)$. Since the left-hand side is positive, we have $ny^2z^2 - x \\ge 1$. Thus,\n$$\n2y^3 \\... | Hong Kong | CHKMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1 and 3 | |
02ti | Problem:
Um tabuleiro de Xadrez tem suas linhas e colunas numeradas conforme a figura a seguir. Em cada casa é escrito o número que é a soma dos números da linha e da coluna dessa casa. Por exemplo, na casa que está na linha 4 e na coluna 5 é escrito o número $4+5=9$.
a) Qual a soma dos núme... | [
"Solution:\n\na) Veja que na linha 1 o número 1 é somado em cada uma das 8 casinhas, na linha 2 o número 2 também é somado oito vezes, e assim por diante. Desse modo, podemos contabilizar a contribuição das linhas por:\n$$\n\\begin{aligned}\n8 \\cdot (1+2+3+4+5+6+7+8) &= 8 \\cdot 36 \\\\\n&= 288\n\\end{aligned}\n$$... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a) 576; b) 0; c) 288 | |
00hc | Let $AB$ and $AC$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $AC$ at $E$ and ray $AB$ at $F$. Let $R$ be a point on segment $EF$. The line through $O$ parallel to $EF$ intersects line $AB$ at $P$. Let $N$ be the intersection of lines $PR$ and $AC... | [
"\n\nLet the line through $N$ tangent to $\\omega$ at point $X \\neq E$ intersect $AB$ at point $M'$. It suffices to show that $M'R \\parallel AC$, since this would yield $M' = M$.\nSuppose that the line $PO$ intersects $AC$ at $Q$ and the circumcircle of $AM'O$ at $Y$, respectively. Then\n... | Asia Pacific Mathematics Olympiad (APMO) | APMO 2016 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08bq | Problem:
Un poligono si dice convesso se tutti i suoi angoli interni hanno ampiezza strettamente minore di $180^{\circ}$. Quanti angoli di ampiezza minore di $150^{\circ}$ può avere al massimo un poligono convesso di 2016 lati?
(A) 11
(B) 12
(C) 13
(D) 17
(E) 2016 | [
"Solution:\n\nLa risposta è (A). Consideriamo un poligono convesso con 2016 lati, e supponiamo che abbia esattamente $a$ angoli di ampiezza inferiore a $150^{\\circ}$; la somma dei suoi angoli interni vale $(2016 - 2) \\cdot 180^{\\circ}$. Nel contempo, dal momento che ognuno dei $2016 - a$ angoli restanti ha ampie... | Italy | Progetto Olimpiadi della Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | MCQ | A | |
01nv | Prove that for every positive integer $n$, the set $\{2, \dots, 3n+1\}$ can be partitioned into $n$ triples in such a way that the numbers from each triple are the lengths of the sides of some obtuse triangle. | [] | Belarus | Belorusija 2012 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
08em | Problem:
Di un triangolo di vertici $A, B, C$ sappiamo che $AB = 5$, $BC = 4$ e $AC = AM$, dove $M$ è il punto medio del lato $BC$. Quanto vale la lunghezza del lato $AC$?
(A) 4
(B) $\sqrt{17}$
(C) $3\sqrt{2}$
(D) $2\sqrt{5}$
(E) $\sqrt{21}$ | [
"Solution:\n\nLa risposta è $\\mathbf{(B)}$.\n\n\n\nPoiché $M$ è punto medio otteniamo che $BM = MC = 2$. Sia $H$ il punto medio di $MC$, e in particolare $MH = HC = 1$; dato che il triangolo $AMC$ è isoscele su base $MC$, $H$ è anche il piede dell'altezza e perciò l'angolo $\\angle AHB$ è ... | Italy | Italian Mathematical Olympiad - February Round | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | MCQ | B | |
0k8x | Problem:
Let $p=2017$ be a prime and $\mathbb{F}_p$ be the integers modulo $p$. A function $f: \mathbb{Z} \rightarrow \mathbb{F}_p$ is called good if there is $\alpha \in \mathbb{F}_p$ with $\alpha \not \equiv 0 (\bmod p)$ such that
$$
f(x) f(y) = f(x+y) + \alpha^{y} f(x-y) \quad (\bmod p)
$$
for all $x, y \in \mathbb... | [
"Solution:\n\nWe will classify all good functions with parameter $\\alpha \\in \\mathbb{F}_p \\setminus \\{0\\}$. We will use $=$ to denote equality modulo $p$ when appropriate. Let the given statement be $P(x, y)$. Then $P(x, 0)$ gives $f(x) f(0) = 2 f(x)$ for all $x \\in \\mathbb{Z}$. $f$ cannot be the zero funct... | United States | HMIC | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Numbe... | null | proof and answer | 1327392 | |
04mc | Let $D_0, D_1, \dots, D_{2018}$ be points on the segment $\overline{AB}$ such that $D_0 = A$, $D_{2018} = B$ and
$$
|D_0 D_1| = |D_1 D_2| = \dots = |D_{2017} D_{2018}|.
$$
If $C$ is a point such that $\angle BCA = 90^\circ$, prove that
$$
|CD_0|^2 + |CD_1|^2 + \dots + |CD_{2018}|^2 = |AD_1|^2 + |AD_2|^2 + \dots + |AD_{... | [
"Let $E_i$ be the foot of the altitude from $D_i$ to the side $\\overline{AC}$ in the right-angled triangle $ABC$, for each $i = 0, 1, \\dots, 2018$. We have $E_0 = A$, $E_{2018} = C$.\n\n\n\nSince the lines $D_iE_i$ are parallel to $BC$, Thales' theorem asserts that\n$$\n|E_0E_1| = |E_1E_2... | Croatia | Croatia_2018 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0g43 | Problem:
Let $n \geq 3$ be an integer. Annalena has infinitely many cowbells in each of $n$ different colours. Given an integer $m \geq n+1$ and a group of $m$ cows standing in a circle, she is tasked with tying one cowbell around the neck of every cow so that every group of $n+1$ consecutive cows have cowbells of all... | [
"Solution:\n\nLet us first give a construction which works for all numbers $m \\geq n^{2}-n$:\n\nWrite $m = k n + r$ for some integer $k \\geq n-1$ and $r \\in \\{0,1, \\ldots, n-1\\}$. Note that $k - r \\geq 0$. If we call the colours $1,2, \\ldots, n$, we arrange them as follows:\n\n$$\n\\underbrace{1,1,2,3, \\ld... | Switzerland | Final round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | n^2 - n - 1 | |
02au | Problem:
As idades numa classe - Numa classe na escola, todos os alunos têm a mesma idade, exceto sete que têm 1 ano a menos e dois que têm 2 anos a mais.
A soma das idades de todos os alunos dessa classe é 330. Quantos alunos tem essa classe? | [
"Solution:\n\nDenotemos por $a$ a idade comum dos alunos e $n$ o número de alunos, assim temos 7 alunos com $a-1$ anos, 2 com $a+2$ anos e o resto, isto é, $n-9$ com $a$ anos. Assim a soma das idades é\n\n$$\n7(a-1) + 2(a+2) + (n-9)a = n a - 3 = 330\n$$\n\n$$\nn a = 333 = 9 \\times 37\n$$\n\nComo a classe tem mais ... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 37 | |
0899 | Problem:
Sia $x_{1}, x_{2}, x_{3}, \ldots$ la successione definita per ricorrenza come segue:
$$
\left\{\begin{array}{l}
x_{1}=4 \\
x_{n+1}=x_{1} x_{2} x_{3} \cdots x_{n}+5 \quad \text{ per } n \geq 1
\end{array}\right.
$$
(I primi termini della successione sono quindi $x_{1}=4, x_{2}=4+5=9, x_{3}=4 \cdot 9+5=41, \ldo... | [
"Solution:\n\nDimostreremo che l'unica coppia che soddisfa le condizioni del problema è $\\{1,2\\}$. (Chiaramente si ha una soluzione anche se $a=b=n$ per ogni $n$, ma in questo caso $\\{a, b\\}$ non è una coppia; aver o non aver considerato questo caso è comunque ininfluente ai fini della valutazione).\nPossiamo o... | Italy | Cesenatico | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | {1,2} | |
05iy | Problem:
Soient $A, B, C, D$ quatre points sur un même cercle. Notons $A'$, $C'$ les projetés orthogonaux de $A$ et $C$ sur $(BD)$, et $B'$, $D'$ les projetés orthogonaux de $B$ et $D$ sur $(AC)$. Montrer que $A'$, $B'$, $C'$, $D'$ sont cocycliques. | [
"Solution:\n\n\nComme $A, B, A', B'$ sont cocycliques (sur le cercle de diamètre $[AB]$), les angles orientés de droites $(BA', BA)$ et $(B'A', B'A)$ sont égaux, donc $(BD, BA) = (B'A', B'D')$. De même, $(CD, CA) = (C'A', C'D')$. Or, $(CD, CA) = (BD, BA)$ puisque $A, B, C, D$ sont cocycliqu... | France | OFM | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08pk | Problem:
Find the maximum positive integer $k$ such that for any positive integers $m, n$ such that $m^{3}+n^{3}>(m+n)^{2}$, we have
$$
m^{3}+n^{3} \geq (m+n)^{2}+k
$$ | [
"Solution:\nWe see that for $m=3$ and $n=2$ we have $m^{3}+n^{3}>(m+n)^{2}$, thus\n$$\n3^{3}+2^{3} \\geq (3+2)^{2}+k \\Rightarrow k \\leq 10\n$$\nWe will show that $k=10$ is the desired maximum. In other words, we have to prove that\n$$\nm^{3}+n^{3} \\geq (m+n)^{2}+10\n$$\nThe last inequality is equivalent to\n$$\n... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 10 | |
0fjx | Problem:
Sean $r$, $s$, $u$, $v$ números reales cualesquiera. Probar que:
$$
\min\left(r-s^{2},\ s-u^{2},\ u-v^{2},\ v-r^{2}\right) \leq \frac{1}{4}
$$ | [
"Solution:\nSupongamos que los cuatro números $r-s^{2}$, $s-u^{2}$, $u-v^{2}$ y $v-r^{2}$ son mayores estrictamente que $\\frac{1}{4}$. Entonces\n$$\nr-s^{2} + s-u^{2} + u-v^{2} + v-r^{2} > \\frac{1}{4} + \\frac{1}{4} + \\frac{1}{4} + \\frac{1}{4}\n$$\npero esta expresión es equivalente a\n$$\n0 > \\left(\\frac{1}{... | Spain | Spanish National Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0fps | $$(x - 1)(x - 2)\cdots(x - 2016) = (x - 1)(x - 2)\cdots(x - 2016)$$
que tiene $2016$ factores lineales en cada lado. Determinar el menor valor posible de $k$ para el cual pueden borrarse exactamente $k$ de estos $4032$ factores lineales, de modo que al menos quede un factor en cada lado y la ecuación que resulte no te... | [] | Spain | LVII Olimpiada Internacional de Matemáticas | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Spanish | proof and answer | 2016 | |
033h | Problem:
In any word with letters $a$ and $b$ the following changes are allowed: $a b a \rightarrow b$, $b \rightarrow a b a$, $b b a \rightarrow a$ and $a \rightarrow b b a$. Is it possible to obtain the word $b \underbrace{a a \ldots a}_{2003}$ from the word $\underbrace{a a \ldots a}_{2003} b$? | [
"Solution:\n\nThe answer is no.\n\nWe shall prove that applying any of the given changes the number of $a$'s in odd (even) positions does not change its parity. Indeed, let the replacement $a b a \\rightarrow b$ be applied to the word $w_{1} a b a w_{2}$. In the new word $w_{1} b w_{2}$ all $a$'s in $w_{1}$ do not ... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
0h5e | Let $BM$ be a median in an acute triangle $ABC$, whose sides $AB$ and $BC$ have different length. The extension of $BM$ intersects the circumcircle of $ABC$ at a point $N$. Let $D$ be a point on the circumcircle such that $\angle BDH = 90^\circ$, where $H$ is the orthocenter of $ABC$. Let $K$ be a point chosen so that ... | [
"Since $BC_1HA_1$ and $BANC$ are cyclic quadrilaterals, it follows that $\\angle A_1HC_1 = 180^\\circ - \\angle B$ and $\\angle ANC = 180^\\circ - \\angle B$. This implies $\\angle A_1HC_1 = \\angle ANC$. But $\\angle ANC = \\angle A_1HC_1$ (as vertical angles) and $\\angle AKC = \\angle ANC$ (as opposite angles in... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneo... | English | proof only | null | |
06cg | a. Given that $(1+x)^{p-2} = 1 + a_1x + a_2x^2 + \dots + a_{p-2}x^{p-2}$ where $p$ is an odd prime number, prove that $a_1 + 2, a_2 - 3, a_3 + 4, \dots, a_{p-3} - (p-2)$ and $a_{p-2} + (p-1)$ are all multiples of $p$.
b. How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 100$ where $-1 \le x... | [
"a.\nNote that the numbers are\n$$\n\\begin{aligned}\na_k + (-1)^{k+1}(k+1) &= \\binom{p-2}{k} + (-1)^{k+1}(k+1) \\\\\n&= \\frac{(p-2)(p-3)\\cdots(p-1-k)}{k!} + (-1)^{k+1}(k+1)\n\\end{aligned}\n$$\nwhere $1 \\le k \\le p-2$. Since\n$$\n(p-2)(p-3)\\cdots(p-1-k) \\equiv (-2)(-3)\\cdots(-1-k) = (-1)^k(k+1)! \\pmod{p}\... | Hong Kong | HKG TST | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 1728 | |
0gm8 | Find all continuous functions $f: [0,1] \to [0,1]$ satisfying the following condition: For any $x \in [0,1]$, there exists a positive integer $n$ such that $f^n(x) = x$.
(For $x \in [0,1]$, $f^n(x)$ is defined through $f^1(x) = f(x)$ and $f^{k+1}(x) = f(f^k(x))$ for all positive integers $k$.) | [] | Turkey | VIIth NATIONAL MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | All continuous functions f with f(f(x)) = x for all x in the interval. Equivalently, either f is the identity function, or f is a continuous strictly decreasing involution on the interval. | |
0apc | Problem:
Prove that the set $\{1,2, \ldots, 2007\}$ can be expressed as the union of disjoint subsets $A_{i}$ ($i=1,2, \ldots, 223$) such that
a. each $A_{i}$ contains 9 elements, and
b. the sum of all the elements in each $A_{i}$ is the same. | [
"Solution:\n\nWe first arrange the numbers $670, 671, \\ldots, 2007$ into 223 rows and 6 columns in the following way:\n\n\n\nLet $C_{i}$ represent the set containing the numbers in the $i$\\textsuperscript{th} row of the above arrangement. It is easy to check that the numbers in each $C_{i... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0075 | Determinar si es posible dividir un cuadrado de lado $11$ en las siguientes $5$ partes: un cuadrado de lado $1$ y cuatro rectángulos cuyas dimensiones son $8$ números enteros distintos y mayores que $1$. ¿Y si el cuadrado que se quiere dividir es de lado $10$? | [] | Argentina | Argentina 2009 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | Spanish | proof and answer | Side 11: possible. Side 10: impossible. | |
02je | Problem:
Se $x > 5$, então qual dos números abaixo é o menor?
(A) $\dfrac{5}{x}$
(B) $\dfrac{5}{x+1}$
(C) $\dfrac{5}{x-1}$
(D) $\dfrac{x}{5}$
(E) $\dfrac{x+1}{5}$ | [
"Solution:\n\nSolução 1:\nComo a questão tem uma única resposta, ela é válida para qualquer valor de $x$. Podemos então escolher um valor para $x$, por exemplo $x = 10$. Temos:\n\n$\\dfrac{5}{x} = \\dfrac{5}{10}$\n\n$\\dfrac{5}{x+1} = \\dfrac{5}{11}$\n\n$\\dfrac{5}{x-1} = \\dfrac{5}{9}$\n\n$\\dfrac{x}{5} = \\dfrac{... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | B | |
03bb | Every two of $n$ towns in a country are connected by one way or two way road. It is known that for every $k$ towns there exists round trip passing through each of these $k$ towns exactly once. Find the maximal possible number of one way roads. | [
"Suppose there exists a town $A$ with $k-1$ one way roads all of which are pointing in one and the same direction from $A$. Town $A$ with all $k-1$ end points of one way roads form a group of $k$ towns that violates the condition of the problem. We conclude that for any town there exist at most $k-2$ one way roads ... | Bulgaria | Bulgaria | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | English | proof and answer | If n ≤ 2k − 3, the maximum is n(n − 1)/2. If n > 2k − 3, the maximum is n(k − 2). | |
0l4b | Problem:
Suppose $a$, $b$, and $c$ are complex numbers satisfying
$$
\begin{aligned}
a^{2} & = b - c, \\
b^{2} & = c - a, \text{ and } \\
c^{2} & = a - b .
\end{aligned}
$$
Compute all possible values of $a + b + c$. | [
"Solution:\nSumming the equations gives $a^{2} + b^{2} + c^{2} = 0$ and summing $a$ times the first equation and etc. gives $a^{3} + b^{3} + c^{3} = 0$. Let $a + b + c = k$. Then $a^{2} + b^{2} + c^{2} = 0$ means $ab + bc + ca = k^{2} / 2$, and $a^{3} + b^{3} + c^{3} = 0 \\Longrightarrow -3abc = a^{3} + b^{3} + c^{... | United States | HMMT February 2024 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 0, i sqrt(6), -i sqrt(6) | |
0jf7 | Problem:
Let $\triangle X Y Z$ be a right triangle with $\angle X Y Z=90^{\circ}$. Suppose there exists an infinite sequence of equilateral triangles $X_{0} Y_{0} T_{0}, X_{1} Y_{1} T_{1}, \ldots$ such that $X_{0}=X$, $Y_{0}=Y$, $X_{i}$ lies on the segment $X Z$ for all $i \geq 0$, $Y_{i}$ lies on the segment $Y Z$ fo... | [
"Solution:\n\nLet $a = X Y$, $b = Y Z$, $r a = X_{1} Y_{1}$. Then $[\\mathcal{P}] = [X Y T_{0}](1 + r^{2} + r^{4} + \\cdots)$, $[X Y Z] = [X Y Y_{1} X_{1}](1 + r^{2} + r^{4} + \\cdots)$, $Y Y_{1} = r a \\sqrt{3}$, and $b = r a \\sqrt{3}(1 + r + r^{2} + \\cdots)$ (although we can also get this by similar triangles).... | United States | HMMT November 2013 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1 | |
01qv | $N$ cells are marked on an $n \times n$ table so that at least one marked cell is among any four cells of the table which form the figure $\Box\Box$ (the figure may be rotated). Find the smallest possible value of $N$. | [
"Answer: $(n - 3)([n/3] + 1) + 1$."
] | Belarus | Selection and Training Session | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | (n - 3)([n/3] + 1) + 1 | |
05hh | Problem:
Soit $ABC$ un triangle. On note $D$ et $E$ les milieux respectifs des côtés $AB$ et $AC$. Soit $O$ le centre du cercle circonscrit du triangle $ABC$. On note $K$ le point d'intersection des droites $(OE)$ et $(BC)$. On note $L$ le deuxième point d'intersection de la droite $(OD)$ avec le cercle circonscrit au... | [
"Solution:\n\n\n\nEn traçant la figure, on s'aperçoit que le point $A$ appartient également au cercle passant par $O$, $K$ et $B$. On commence par démontrer ce fait.\n\nD'une part, d'après le théorème de Thalès, les droites $(DE)$ et $(BC)$ sont parallèles, on a l'égalité des angles alterne... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"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 | |
0iie | Problem:
Begining at a vertex, an ant crawls between the vertices of a regular octahedron. After reaching a vertex, it randomly picks a neighboring vertex (sharing an edge) and walks to that vertex along the adjoining edge (with all possibilities equally likely.) What is the probability that after walking along $2006$... | [
"Solution:\n\nFor each nonnegative integer $n$, let $a_{n}, b_{n}$, and $c_{n}$ denote the respective probabilities that the ant is where it began, at a neighbor of where it began, or is opposite where it began after moving along $n$ edges. We seek $a_{2006}$. We have $a_{0}=1$ and $b_{0}=c_{0}=0$. We also have the... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | (2^2005 + 1) / (3 * 2^2006) | |
01jh | Let $\mathbb{Z}^+$ denote the set of positive integers, and $\text{lcm}(n, m)$ the least positive integer that is divisible by both $n$ and $m$.
Find all functions $f: \mathbb{Z}^+ \to \mathbb{Z}^+$ such that
$$
\text{lcm}(n \cdot f(n^2), f(m^2)) = n \cdot f(nm), \text{ for all positive integers } n, m
$$
and such that... | [
"First notice that $f(1)$ divides $1$, and hence $f(1) = 1$. For all primes $p$, we have\n$$\n\\begin{align*}\np \\cdot f(p^{2n}) &= \\text{lcm}(p \\cdot f(p^{2n}), 1) = \\text{lcm}(p \\cdot f(p^{2n}), f(1^2)) \\\\\n&= p \\cdot f(p^n \\cdot 1) = p \\cdot f(p^n),\n\\end{align*}\n$$\nwhich shows that $f(p^n) = f(p^{2... | Baltic Way | Baltic Way 2023 Shortlist | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Other"
] | English | proof and answer | All functions obtained by choosing an arbitrary subset S of primes and defining f(1)=1 and, for n=∏ p_i^{e_i}, f(n)=∏_{p_i | n, p_i ∈ S} p_i. Equivalently, for each prime p fix α_p ∈ {0,1} and set f(n)=∏_{p | n} p^{α_p}. | |
0j6j | Problem:
In how many ways can 13 bishops be placed on an $8 \times 8$ chessboard such that
(i) a bishop is placed on the second square in the second row,
(ii) at most one bishop is placed on each square,
(iii) no bishop is placed on the same diagonal as another bishop, and
(iv) every diagonal contains a bishop?
(For th... | [
"Solution:\nAnswer: 1152\nWe color the squares of the chessboard white and black such that B2 (the second square in the second row) is black. Note that at most 7 bishops can go on the white squares, and if there is a bishop on b2, at most 5 more can be on the white squares. So of the other 12 bishops, 7 go on white... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 1152 | |
0bo9 | Find the distinct primes $p, q, r$ and $s$ satisfying
$$
1 - \frac{1}{p} - \frac{1}{q} - \frac{1}{r} - \frac{1}{s} = \frac{1}{pqrs}.
$$ | [
"We will show the solution consists of all the permutations of the quadruple $2, 3, 7, 43$.\n\nWLOG we may suppose $p < q < r < s$. If $p \\ge 3$,\n$$\n1 - \\left(\\frac{1}{3} + \\frac{1}{5} + \\frac{1}{7} + \\frac{1}{11}\\right) = 1 - \\frac{886}{1155} > 0.2,\n$$\nin contradiction with $\\frac{1}{pqrs} < \\frac{1}... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All permutations of 2, 3, 7, and 43. | |
0l7j | Problem:
Let $\mathbb{Z}$ be the set of integers. Determine, with proof, all primes $p$ for which there exists a function $f\colon \mathbb{Z}\to \mathbb{Z}$ such that for any integer $x$,
$f(x + p) = f(x)$ and $p$ divides $f(x + f(x)) - x$. | [
"Solution:\nWe work in $\\mathbb{F}_p$, treating $f$ as a map from $\\mathbb{F}_p$ to itself. Clearly, $p = 2$ doesn't work. For $p > 2$ such that $5$ is a quadratic residue mod $p$, as well as $p = 5$ itself, there exists some $\\alpha$ such that $(2\\alpha + 1)^2\\equiv 5$ (mod $p$). Taking $f(x) = \\alpha x$ the... | United States | HMMT February 2025 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > F... | null | proof and answer | All primes p with 5 a quadratic residue modulo p (equivalently p ≡ 1 or 4 mod 5), together with p = 5; p = 2 does not work. | |
0hkz | Problem:
Let $ABCD$ be a square. We randomly select a point $P$ inside the square, uniformly and at random. What is the probability that $\angle BPC > 90^{\circ}$? | [
"Solution:\nLet $\\omega$ be the semicircle whose diameter is $\\overline{BC}$ and whose interior lies inside $ABCD$. Let $M$ denote the center of $\\omega$. We use the result that\n- $\\angle BPC < 90^{\\circ}$ if $P$ is inside $\\omega$,\n- $\\angle BPC = 90^{\\circ}$ if $P$ is on $\\omega$, and\n- $\\angle BPC >... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | π/8 | |
09ak | Given are $2n$ people and it is known that their heights are all different. They have to stand in two rows, each with $n$ people. How many different positions are there, if the front row person is always shorter than the back row person? | [
"In the first seats of the rows, we can choose 2 people $\\frac{2n(2n-1)}{2}$ different ways. Then in the second seats of the rows, we can choose 2 people $\\frac{(2n-2)(2n-3)}{2}$ different ways. Continuing this, we have $\\frac{2n(2n-1)}{2} \\cdot \\frac{(2n-2)(2n-3)}{2} \\cdots \\frac{2 \\cdot 1}{2} = \\frac{(2n... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | (2n)! / 2^n | |
04qd | On the side $\overline{AB}$ of the cyclic quadrilateral $ABCD$ there is a point $X$ such that the diagonals $\overline{BD}$ and $\overline{AC}$ bisect the segments $\overline{CX}$ and $\overline{DX}$, respectively.
Find the smallest possible value of $|AB| : |CD|$. (Belarus) | [
"Let $M$ and $N$ be the midpoints of the segments $\\overline{CX}$ and $\\overline{DX}$, respectively. Also, denote $\\alpha = \\angle BAC = \\angle BDC$ and $\\beta = \\angle CAD = \\angle CBD$.\n\n\n\nSince the triangles $AXN$ and $ADN$ have equal areas, we have $|AX| \\cdot |AN| \\sin \\... | Croatia | Croatian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 2 | |
05xq | Problem:
Soit $A, B, C, D$ et $E$ cinq points situés dans cet ordre sur un cercle $\Omega$, de sorte que $(C D)$ soit parallèle à $(B E)$ et que $(A B)$ soit parallèle à $(D E)$. Soit $X, Y$ et $Z$ les milieux respectifs des segments $[B D],[C E]$ et $[A E]$.
Démontrer que la droite $(A E)$ est tangente au cercle cir... | [
"Solution:\n\n\n\nSoit $O$ le centre du cercle $\\Omega$. Dès lors que l'on s'attache à tracer une figure incluant le cercle $\\Gamma$ circonscrit à $X Y Z$, on constate que celui-ci semble tangent aux trois droites $(B D),(C E)$ et $(A E)$, tandis que son centre semble coïncider avec $O$. ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0esa | Let $\triangle ABC$ be a triangle with circumcircle $\Gamma$. Let $D$ be a point on segment $BC$ such that $\angle BAD = \angle DAC$, and let $M$ and $N$ be points on segments $BD$ and $CD$, respectively, such that $\angle MAD = \angle DAN$. Let $S, P$ and $Q$ (all different from $A$) be the intersections of the rays $... | [
"\nLet $X$ be the intersection of $SM$ with the circumcircle. Note that $\\angle MXA = \\angle SXA = \\angle SBA = \\angle SBC + \\angle CBA$. Since $AS$ bisects the angle $\\angle BAC$, we have $\\angle SBC = \\angle SAC = \\angle BAS$. It follows that\n$$\n\\angle MXA = \\angle BAS + \\an... | South Africa | The South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09uo | In a class, $30$ students did a test. Every student got a grade that was an integer from $1$ to $10$. The grade $8$ was given more often than any of the other grades.
What is the smallest possible average grade of the students?
A) $3\frac{8}{15}$
B) $3\frac{2}{3}$
C) $3\frac{5}{6}$
D) $4\frac{11}{30}$
E) $4\frac{8}{15}... | [
"B) $3^{\\frac{2}{3}}$"
] | Netherlands | Junior Mathematical Olympiad, September 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | MCQ | B | |
0fa3 | Problem:
An $h \times k$ minor of an $n \times n$ table is the $hk$ cells which lie in $h$ rows and $k$ columns. The semiperimeter of the minor is $h + k$. A number of minors each with semiperimeter at least $n$ together include all the cells on the main diagonal. Show that they include at least half the cells in the ... | [] | Soviet Union | 25th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0ggs | 已知實數數列 $a_1, a_2, \dots$ 滿足下列性質：
i. 存在正整數 $N$ 使得 $a_n = 1$ 對所有 $n \ge N$ 都成立；
ii. 對所有 $n \ge 2$ 都有
$$
a_n \le a_{n-1} + 2^{-n} a_{2n}.
$$
證明：$a_n > 1 - 2^{-n}$ 對所有正整數 $n$ 均成立。 | [
"For all $k$, since there are only finitely many possible values for $a_n$ ($n \\ge k$) to take, there exists $n_{max} \\ge k$ so that $a_{n_{max}} \\ge a_n$ for all $n \\ge k$. Now, we can prove by induction that $a_m \\ge (1 - 2^{-n_{max}} - 2^{-n_{max}-1} - \\dots - 2^{-(m+1)})a_{n_{max}}$ for all $k \\le m \\le... | Taiwan | 2022 數學奧林匹亞競賽第一階段選訓營, 獨立研究 (二) | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | Chinese; English | proof only | null | |
06h8 | If $x-y=12$, find the value of $x^3-y^3-36xy$.
若 $x-y=12$, 求 $x^3-y^3-36xy$ 的值。 | [] | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English; Chinese | final answer only | 1728 | |
0euv | Let $ABC$ be a triangle with $AC < AB < BC$. Choose a point $D$ on the side $AB$. Suppose the circumcircle of $ABC$ meets the internal angle bisector of $\angle A$ and $CD$ at $E$ ($\neq A$) and $F$ ($\neq C$), respectively. Let $K$ be the intersection point of $BC$ and $DE$. Prove that $CK = AC$ if and only if $DK \cd... | [
"Let $T$ be the intersection point of $ED$ and the circumcircle of $ABC$ other than $E$. Since $\\angle CTD = \\angle EFD$ and $\\angle CDT = \\angle EDF$, $DCT$ and $DEF$ are similar triangles. Hence\n$$\n\\frac{EF}{DF} = \\frac{TC}{DT}\n$$\nNote that $AE$ is the perpendicular bisector of the line segment $CD$, as... | South Korea | 24th Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
064n | Problem:
Bestimmen Sie den kleinsten Wert, den der Ausdruck
$$
\left\lfloor\frac{a_{1}}{1}\right\rfloor+\left\lfloor\frac{a_{2}}{2}\right\rfloor+\cdots+\left\lfloor\frac{a_{2022}}{2022}\right\rfloor
$$
annehmen kann, wobei $a_{1}, a_{2}, \ldots, a_{2022} \geq 1$ reelle Zahlen sind, sodass $\left|a_{i}-a_{j}\right| \ge... | [
"Solution:\n\nAus der Voraussetzung folgt, dass die kleinste der Zahlen $a_{1}, a_{2}, \\ldots, a_{2022}$ mindestens $1$, die zweitkleinste mindestens $2$ ist, usw. bis zur $2022$-ten, die mindestens $2022$ sein muss. Wenn wir nun die Zahlen $a_{1}, \\ldots, a_{2022}$ sukzessive der Größe nach durch $1,2, \\ldots, ... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade 2022 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 11 | |
0fnj | The function $f : \mathbb{N}^* \rightarrow \mathbb{N}$ satisfies that $f(2) = 0$, $f(3) > 0$, $f(6042) = 2014$ and if $(m, n) \in \mathbb{N}^* \times \mathbb{N}^*$ then $f(m+n) - f(n) - f(m) \in \{0, 1\}$. Find the value of $f(2014)$. Here, $\mathbb{N}^* = \{1, 2, 3, \dots\}$. | [
"Since $f(2) \\ge f(1) + f(1) = 2 \\cdot f(1)$ and $f(2) = 0$ then $f(1) \\le 0$. Therefore, $f(1) = 0$. On the other hand, from $f(3) > 0$ and $f(3) - f(2) - f(1) \\in \\{0, 1\\}$ we have $f(3) = 1$. Putting $m = 1$ in the condition $f(m+n) - f(n) - f(m) \\in \\{0, 1\\}$, we obtain $f(n+1) - f(1) - f(n) = f(n+1) -... | Spain | International Mathematical Arhimede Contest | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Spanish | proof and answer | 671 | |
01hi | Given is a triangle with real side lengths $a$, $b$, $c$ and real area $A$. Prove that
$$ 3a^2 + 3b^2 - c^2 \ge 4\sqrt{3}A. $$ | [
"Let $\\gamma$ be the angle between the sides $a$ and $b$. We have\n$$\n\\begin{aligned}\na^2 + b^2 &\\ge 2ab \\\\\n&\\ge 2ab \\sin(\\gamma - 30^\\circ) \\\\\n&= 2ab [\\sin(\\gamma) \\cos(30^\\circ) - \\cos(\\gamma) \\sin(30^\\circ)] \\\\\n&= ab \\sin(\\gamma)\\sqrt{3} - ab \\cos(\\gamma) \\\\\n&= 2\\sqrt{3}A - \\f... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0kmc | Which of the following conditions is sufficient to guarantee that integers $x$, $y$, and $z$ satisfy the equation
$$
x(x - y) + y(y - z) + z(z - x) = 1?
$$
(A) $x > y$ and $y = z$
(B) $x = y - 1$ and $y = z - 1$
(C) $x = z + 1$ and $y = x + 1$
(D) $x = z$ and $y - 1 = x$
(E) $x + y + z = 1$ | [
"**Answer (D):** The given equation is equivalent to $2x(x - y) + 2y(y - z) + 2z(z - x) = 2$, which can be rewritten as $(x - y)^2 + (y - z)^2 + (z - x)^2 = 2$. This equation has an integer solution if and only if two of the squares are 1 and one is 0. This in turn means that two of the variables must be equal and ... | United States | AMC 12 B | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | MCQ | D | |
0gck | 對於任兩個由有限個正整數所成的集合 $X$ 和 $Y$, 令：
(1) $f_X(1) = \min\{n: n \in \mathbb{N}, n \notin X\}$，此處 $\mathbb{N}$ 表示所有正整數之集合；
(2) $\forall k > 1$, $f_X(k) = \min\{n: n > f_X(k-1), n \notin X\}$；
(3) $X \star Y = X \cup \{f_X(y) : y \in Y\}$。
令 $a, b$ 皆為正整數，$A$ 為由 $a$ 個正整數所成的集合，$B$ 為由 $b$ 個正整數所成的集合。
試證：若 $A \star B = B \star A$，則... | [
"(1) We first show that $\\star$ is associative. This can be done through the following lemma.\n\n**Lemma 1.** $f_{X\\star Y} = f_X \\circ f_Y$.\n\n*Proof.* Note that\n$$\n\\begin{align*}\nf_{X \\star Y}(\\mathbb{N}) &= \\mathbb{N} - (X \\star Y) = (\\mathbb{N} - X) - f_X(Y) \\\\\n&= f_X(\\mathbb{N}) - f_X(Y) = f_X... | Taiwan | 二〇一八數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof only | null | |
079q | Let $a$, $b$, $c$ be positive real numbers such that $ab + bc + ca = 1$. Show that
$$
\sqrt{3}(\sqrt{a} + \sqrt{b} + \sqrt{c}) \le \frac{a\sqrt{a}}{bc} + \frac{b\sqrt{b}}{ca} + \frac{c\sqrt{c}}{ab}.
$$ | [
"**Solution 1.** By Holder Inequality we have\n$$\n\\left(\\sum_{cyc} \\frac{a\\sqrt{a}}{bc}\\right)\\left(\\sum_{cyc} bc\\right)\\left(\\sum_{cyc} 1\\right) \\geq \\left(\\sum_{cyc} \\sqrt{a}\\right)^3 \\Rightarrow \\sum_{cyc} \\frac{a\\sqrt{a}}{bc} \\geq \\frac{1}{3}\\left(\\sum_{cyc} \\sqrt{a}\\right)^3.\n$$\nSo... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
01b6 | A function $f$ defined on the set of real numbers $\mathbb{R}$ and taking nonnegative real values satisfies the condition
$$
f(x + y) \le 2 \max\{f(x), f(y)\}
$$
for all $x, y \in \mathbb{R}$. Is it true that for each positive integer $k$ the inequality
$$
f(x_1 + \cdots + x_k) \le 2(f(x_1) + \cdots + f(x_k))
$$
holds ... | [
"**Answer:** Yes.\n\nThe required inequality obviously holds for $k = 1$. For $k \\ge 2$ we will prove the next (stronger) inequality\n$$\n\\frac{f(x_1 + \\cdots + x_k)}{2} \\le f(x_1) + \\cdots + f(x_k) - \\min\\{f(x_1), \\ldots, f(x_k)\\} \\quad (1)\n$$\nfor arbitrary $x_1, \\dots, x_k \\in \\mathbb{R}$.\n\nThe p... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | Yes | |
0kft | Two circles $\Gamma_1$ and $\Gamma_2$ have common external tangents $l_1$ and $l_2$ meeting at $T$. Suppose $l_1$ touches $\Gamma_1$ at $A$ and $l_2$ touches $\Gamma_2$ at $B$. A circle $\Omega$ through $A$ and $B$ intersects $\Gamma_1$ again at $C$ and $\Gamma_2$ again at $D$, such that quadrilateral $ABCD$ is convex.... | [
"**First solution, elementary (original)** We have $\\triangle YAC \\sim \\triangle YBD$, from which it follows\n$$\n\\frac{d(Y, AC)}{d(Y, BD)} = \\frac{AC}{BD}.\n$$\nMoreover, if we denote by $r_1$ and $r_2$ the radii of $\\Gamma_1$ and $\\Gamma_2$, then\n$$\n\\frac{d(T, AC)}{d(T, BD)} = \\frac{TA \\sin \\angle (A... | United States | USA IMO TST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry... | null | proof only | null | |
0hy2 | Problem:
A man named Juan has three rectangular solids, each having volume $128$. Two of the faces of one solid have areas $4$ and $32$. Two faces of another solid have areas $64$ and $16$. Finally, two faces of the last solid have areas $8$ and $32$. What is the minimum possible exposed surface area of the tallest tow... | [
"Solution:\nSuppose that $x$, $y$, $z$ are the sides of the following solids. Then Volume $= x y z = 128$.\n\nFor the first solid, without loss of generality (with respect to assigning lengths to $x$, $y$, $z$), $x y = 4$ and $y z = 32$. Then $x y^{2} z = 128$. Then $y = 1$. Solving the remaining equations yields $... | United States | HMMT 1998 | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 688 | |
0hc4 | For which of the sets $A = \mathbb{R}$ or $A = \mathbb{Q}$ is there a function $f : A \to (0, +\infty)$ such that $\forall x_1, x_2 \in A,\ x_1 \neq x_2$ the condition $\min\{f(x_1), f(x_2)\} \leq |x_1 - x_2|$ is satisfied?
(Andrii Anikushin) | [
"Suppose that the set $A$ is more than countable set.\nDivide half-plane $\\mathbb{R} \\times (0, +\\infty)$ to the counted number of rectangles:\n$$\nD_{n,m} = [n, n+1) \\times \\left[\\frac{1}{m+1}, \\frac{1}{m}\\right), \\ n \\in \\mathbb{Z},\\ m \\in \\mathbb{N},\\ D_{n,0} = [n, n+1) \\times [1, +\\infty).\n$$\... | Ukraine | Ukrainian Mathematical Competitions | [
"Discrete Mathematics > Other",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | Only for A = Q; no such function exists for A = R. | |
0jzo | Problem:
Let $a_{1}, a_{2}, \ldots, a_{1000}$ be real numbers such that
$$
\begin{array}{r}
a_{1} \cdot 1+a_{2} \cdot 2+a_{3} \cdot 3+\cdots+a_{1000} \cdot 1000=0 \\
a_{1} \cdot 1^{2}+a_{2} \cdot 2^{2}+a_{3} \cdot 3^{2}+\cdots+a_{1000} \cdot 1000^{2}=0 \\
a_{1} \cdot 1^{3}+a_{2} \cdot 2^{3}+a_{3} \cdot 3^{3}+\cdots+a_{... | [
"Solution:\nThe key idea is to multiply on weights which are polynomial coefficients. Specifically, let $P(n)=n(n-2)(n-3) \\ldots(n-1000)=\\sum_{k=0}^{1000} c_{k} n^{k}$, say. Note that $c_{1}=0$ and $c_{1000}=1$.\nNow take the $k$th equation and multiply it by $c_{k}$, then add all equations together. From this we... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | proof and answer | -1/999! | |
00xy | Problem:
Prove that for any odd positive integer $n$, $n^{12} - n^{8} - n^{4} + 1$ is divisible by $2^{9}$. | [
"Solution:\n\nFactorizing the expression, we get\n$$\nn^{12} - n^{8} - n^{4} + 1 = (n^{4} + 1)(n^{2} + 1)^{2}(n - 1)^{2}(n + 1)^{2}.\n$$\nNow note that one of the two even numbers $n - 1$ and $n + 1$ is divisible by $4$."
] | Baltic Way | Baltic Way 1993 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0iae | Problem:
Daniel and Scott are playing a game where a player wins as soon as he has two points more than his opponent. Both players start at par, and points are earned one at a time. If Daniel has a $60\%$ chance of winning each point, what is the probability that he will win the game? | [
"Solution:\n\nConsider the situation after two points. Daniel has a $9/25$ chance of winning, Scott, $4/25$, and there is a $12/25$ chance that the players will be tied. In the latter case, we revert to the original situation. In particular, after every two points, either the game returns to the original situation,... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 9/13 | |
0kzb | Problem:
Let $f$ and $g$ be two quadratic polynomials with real coefficients such that the equation $f(g(x))=0$ has four distinct real solutions: $112, 131, 146$, and $a$. Compute the sum of all possible values of $a$. | [
"Solution:\n\nClaim 1. If $a, b, c, d$ are roots of $f(g(x))$, then one can permute them so that $a+b=c+d$.\n\nProof. Let $v$ be the point for which $g(v)$ is the local minimum or maximum. Note that if $g(x)=g(y)$, then $x$ and $y$ are symmetric around $v$, or $x+y=2v$. Moreover, if $a, b, c, d$ are roots of $f(g(x... | United States | HMMT November 2024 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 389 | |
04g6 | Let $a$, $b$ and $c$ be integers and $a \neq 0$. Can the discriminant of the quadratic function
$$f(x) = ax^2 + bx + c$$
be equal to $51$? | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | No | |
0i06 | Problem:
Circles $C_{1}, C_{2}, C_{3}$ have radius $1$ and centers $O, P, Q$ respectively. $C_{1}$ and $C_{2}$ intersect at $A$, $C_{2}$ and $C_{3}$ intersect at $B$, $C_{3}$ and $C_{1}$ intersect at $C$, in such a way that $\angle A P B = 60^{\circ}$, $\angle B Q C = 36^{\circ}$, and $\angle C O A = 72^{\circ}$. Find... | [
"Solution:\n\nUsing a little trig, we have $BC = 2 \\sin 18$, $AC = 2 \\sin 36$, and $AB = 2 \\sin 30$ (see left diagram). Call these $a, b$, and $c$, respectively. By the law of cosines, $b^{2} = a^{2} + c^{2} - 2 a c \\cos ABC$, therefore\n$$\n\\cos ABC = \\frac{\\sin^{2} 18 + \\sin^{2} 30 - \\sin^{2} 36}{2 \\sin... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 90° | |
0jh9 | Problem:
A polynomial $f(x)=\sum_{i=0}^{n} a_{i} x^{i}$ of degree $n$ or less is called happy if
(i) Each coefficient $a_{i}$ satisfies $0 \leq a_{i}<1$;
(ii) $f(x)$ is an integer for all integers $x$.
Find the number of happy polynomials of degree $n$ or less. | [
"Solution:\nThe answer is the \"superfactorial\"\n$$\n1! \\cdot 2! \\cdots n! = 1^{n} 2^{n-1} \\cdots n^{1}\n$$\nFor $n=1$, the result is clear. We will prove as an induction step that there are $n!$ times as many happy polynomials of degree at most $n$ as of degree at most $n-1$.\n\nLet $f(x)$ be a happy polynomia... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 1!*2!*...*n! | |
0a1r | Problem:
Gegeven is een driehoek $ABC$ en een punt $D$ op het lijnstuk $AC$. Zij $M$ het midden van $CD$ en zij $\Omega$ de cirkel door $B$ en $D$ die raakt aan $AB$. Zij $E$ het punt zodat $\triangle MDB \sim \triangle MBE$ en zodat $D$ en $E$ aan weerszijden van de lijn $MB$ liggen.
Toon aan dat $E$ op $\Omega$ ligt... | [
"Solution:\n\nOplossing I. We bewijzen eerst dat $\\triangle CMB \\sim \\triangle DBE$. Omdat $D$ en $E$ aan weerszijden van $MB$ liggen, geldt er dat $\\angle DBE = \\angle DBM + \\angle MBE = \\angle DBM + \\angle MDB = \\angle CMB$ vanwege de gegeven gelijkvormigheid en de buitenhoekstelling. Ten tweede geldt er... | Netherlands | IMO-selectietoets II | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fjk | Problem:
Sea $x$ un número real tal que $x^{3} + 2x^{2} + 10x = 20$. Demostrar que tanto $x$ como $x^{2}$ son irracionales. | [
"Solution:\n\nPrimero veamos que $x$ no puede ser entero. Esto puede hacerse teniendo en cuenta que si lo fuese, sería un divisor de $20$ y basta probar los $8$ divisores para comprobar que ninguno cumple la ecuación.\n\nOtro modo de verlo es comprobar que $f(x) = x^{3} + 2x^{2} + 10x - 20$ es estrictamente crecien... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0ju9 | Problem:
Let $ABC$ be a triangle with $AB = 2$, $AC = 3$, $BC = 4$. The isogonal conjugate of a point $P$, denoted $P^*$, is the point obtained by intersecting the reflection of lines $PA$, $PB$, $PC$ across the angle bisectors of $\angle A$, $\angle B$, and $\angle C$, respectively.
Given a point $Q$, let $\mathfrak{... | [
"Solution:\nThe first main insight is that all the cubics pass through the points $A$, $B$, $C$, $H$ (orthocenter), $O$, and the incenter and three excenters. Since two cubics intersect in at most nine points, this is all the intersections of a cubic with a cubic.\n\nOn the other hand, it is easy to see that among ... | United States | HMMT February | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry >... | null | final answer only | 49 | |
00fi | Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that
(i) For all $x, y \in \mathbb{R}$,
$$
f(x)+f(y)+1 \geq f(x+y) \geq f(x)+f(y)
$$
(ii) For all $x \in [0,1)$, $f(0) \geq f(x)$,
(iii) $-f(-1)=f(1)=1$.
Find all such functions $f$. | [
"Plug $y \\rightarrow 1$ in (i):\n$$\nf(x)+f(1)+1 \\geq f(x+1) \\geq f(x)+f(1) \\Longleftrightarrow f(x)+1 \\leq f(x+1) \\leq f(x)+2 .\n$$\nNow plug $y \\rightarrow -1$ and $x \\rightarrow x+1$ in (i):\n$$\nf(x+1)+f(-1)+1 \\geq f(x) \\geq f(x+1)+f(-1) \\Longleftrightarrow f(x) \\leq f(x+1) \\leq f(x)+1 .\n$$\nHence... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1994 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | f(x) = ⌊x⌋ | |
0b7a | Prove that any continuous function $f : \mathbb{R} \to \mathbb{R}$ of the form
$$
f(x) = \begin{cases} a_1x + b_1, & \text{for } x \le 1 \\ a_2x + b_2, & \text{for } x > 1 \end{cases}
$$
with $a_1, a_2, b_1, b_2 \in \mathbb{R}$, can be represented as
$$
f(x) = m_1x + n_1 + \varepsilon|m_2x + n_2|, \text{ for } x \in \m... | [
"The continuity at $x = 1$ implies $a_1 + b_1 = a_2 + b_2$.\nIf $a_1 = a_2$, then $b_1 = b_2$ and we can take $m_1 = a_1 = a_2$, $n_1 = b_1 = b_2$, $m_2 = n_2 = 0$, $\\varepsilon = \\pm 1$.\nOtherwise, we search $A$ such that $m_2x + n_2 = A(x - 1)$. The required form can be obtained if $m_1 - \\varepsilon|A| = a_1... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
02aw | Problem:
Quais são os números? - Descubra quais números inteiros positivos $x$ e $y$ satisfazem a equação $x^{4}=y^{2}+71$. | [
"Solution:\n\nA equação pode ser escrita na forma $x^{4}-y^{2}=71$. Agora, fatorando $x^{4}-y^{2}$ temos:\n$$\n\\left(x^{2}-y\\right)\\left(x^{2}+y\\right)=71\\ (*)\n$$\nComo $x$ e $y$ são inteiros, então cada um dos fatores $\\left(x^{2}-y\\right)$ e $\\left(x^{2}+y\\right)$ também é um número inteiro. Logo em $\\... | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | x = 6, y = 35 | |
07kc | Suppose $a$, $b$ and $c$ are positive real numbers. Prove that
$$
\frac{a+b+c}{3} \le \sqrt{\frac{a^2+b^2+c^2}{3}} \le \frac{\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}}{3}.
$$
For each of the inequalities, find conditions on $a$, $b$ and $c$ such that equality holds. | [
"We first show that\n$$\n\\frac{a+b+c}{3} \\le \\sqrt{\\frac{a^2+b^2+c^2}{3}},\n$$\nwith equality iff $a = b = c$. The inequality is equivalent to saying that\n$$\n(a+b+c)^2 \\le 3(a^2+b^2+c^2) \\Leftrightarrow a^2+b^2+c^2+2(ab+bc+ca) \\le 3(a^2+b^2+c^2),\n$$\ni.e.,\n$$\n2(ab + bc + ca) \\le 2(a^2 + b^2 + c^2) \\Le... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds in both inequalities if and only if a = b = c. | |
09g2 | $$
S_n = \{(x, y) \in \mathbb{Z}^2 \mid 1 \le x, y \le 5n\}.
$$
A subset $A \subseteq S_n$ is said to be *sumfree* if
$$
(x_1, y_1), (x_2, y_2) \in A \implies (x_1 + x_2, y_1 + y_2) \notin A.
$$
Show that there exists a sumfree subset with at least $15n^2$ elements. | [
"Let $A := \\{(x, y) \\in S_n \\mid 4n+1 \\le x+y \\le 8n+1\\}$. Then it is clear that $A$ is sumfree.\nNow we show that $|A| \\ge 15n^2$.\n\nFor $k \\in \\mathbb{Z}_{\\ge 1}$, we let $T_k := \\{(x, y) \\in S_n \\mid x + y = k\\}$. Then it is easy to see that\n$$\n|T_k| = \\begin{cases} k-1 & k \\le 5n \\\\ 10n-k+1... | Mongolia | 2015 Mongolian IMO Team Selection Tests | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
06n4 | Let $(a_n)_{n \ge 1}$ be a sequence of positive real numbers with the property that
$$
(a_{n+1})^2 + a_n a_{n+2} \le a_n + a_{n+2}
$$
for all positive integers $n$. Show that $a_{2022} \le 1$. | [
"Let us denote $b_n = a_n$ for all $n$. The given inequality is:\n$$(a_{n+1})^2 + a_n a_{n+2} \\le a_n + a_{n+2}$$\nfor all $n \\ge 1$.\n\nWe can rearrange the inequality:\n$$(a_{n+1})^2 + a_n a_{n+2} - a_n - a_{n+2} \\le 0$$\n$$(a_{n+1})^2 - a_n + a_n a_{n+2} - a_{n+2} \\le 0$$\n$$(a_{n+1})^2 - a_n + (a_n - 1)a_{n... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0aup | Problem:
What is the smallest number that is greater than $2015$ and divisible by both $6$ and $35$? | [
"Solution:\nThe desired number is divisible by $2$, $3$, $5$ and $7$. Hence, it is the least multiple of $2 \\times 3 \\times 5 \\times 7 = 210$ that is greater than $2015$. The desired number is $2100$."
] | Philippines | 18th PMO National Stage Oral Phase | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 2100 | |
0ktd | Problem:
Consider parallelogram $ABCD$ with $AB > BC$. Point $E$ on $\overline{AB}$ and point $F$ on $\overline{CD}$ are marked such that there exists a circle $\omega_1$ passing through $A, D, E, F$ and a circle $\omega_2$ passing through $B, C, E, F$. If $\omega_1, \omega_2$ partition $\overline{BD}$ into segments $... | [
"Solution:\n\nWe want to find $AD = BC = EF$. So, let $EF$ intersect $BD$ at $O$. It is clear that $\\triangle BOE \\sim \\triangle DOF$. However, we can show by angle chase that $\\triangle BXE \\sim \\triangle DYF$:\n$$\n\\angle BEG = \\angle ADG = \\angle CBH = \\angle DFH\n$$\nThis means that $\\overline{EF}$ p... | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 51 | |
0gu0 | Find the smallest value of
$$
xy^3z^2 + \frac{4z}{x} - 8yz - \frac{4}{xy}
$$
where $x$, $y$, $z$ are positive real numbers satisfying at least one of the following inequalities:
$$
2xy > 1 \text{ and } yz > 1.
$$ | [
"4. For any $x > 0$, we have\n$$\nP(x) = x^n Q(1/x) \\ge Q(x) = P(1/x) x^n \\ge x^n Q(1/x) = P(x)\n$$\nhence we have equalities in each step so $P(x) = Q(x)$ for all positive values of $x$, hence for all $x$."
] | Turkey | Team Selection Test for EGMO 2023 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | -8 | |
0ek5 | Problem:
Dan je izraz $\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}$.
a) Racionaliziraj imenovalec in izraz poenostavi.
b) Izračunaj vrednost izraza za $x=\frac{5}{4}$. | [
"Solution:\n\na.\nDani izraz $\\frac{\\sqrt{x+1}+\\sqrt{x-1}}{\\sqrt{x+1}-\\sqrt{x-1}}$ v števcu in imenovalcu pomnožimo z $\\sqrt{x+1}+\\sqrt{x-1}$ ter poenostavimo:\n\n$$\n\\frac{\\sqrt{x+1}+\\sqrt{x-1}}{\\sqrt{x+1}-\\sqrt{x-1}} \\cdot \\frac{\\sqrt{x+1}+\\sqrt{x-1}}{\\sqrt{x+1}+\\sqrt{x-1}} = \\frac{(\\sqrt{x+1}... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | x + sqrt(x^2 - 1); at x = 5/4, the value is 2 | |
0ehz | Problem:
Naj bo $f: \mathbb{R} \rightarrow \mathbb{R}$ poljubna soda funkcija in $g: \mathbb{R} \rightarrow \mathbb{R}$ poljubna liha funkcija. Definirajmo funkcijo $\varphi: \mathbb{R} \rightarrow \mathbb{R}$ s predpisom $\varphi(x)=f(g(x))+g(f(x))$ za vse $x \in \mathbb{R}$. Katera od naslednjih enakosti je zagotovo... | [
"Solution:\n\nKer je $f(x)$ soda funkcija in $g(x)$ liha funkcija, velja $f(-x)=f(x)$ in $g(-x)=-g(x)$ za vsa realna števila $x$. Torej je $\\varphi(-x)=f(g(-x))+g(f(-x))=f(-g(x))+g(f(x))=f(g(x))+g(f(x))=\\varphi(x)$. Pravilen odgovor je (C).\n\nPrimer $f(x)=x^{2}$ in $g(x)=2 x$ pokaže, da nobena od ostalih enakost... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | MCQ | C | |
0d76 | Let $ABCDEF$ be a convex hexagon with $AB = CD = EF$, $BC = DE = FA$ and $\angle A + \angle B = \angle C + \angle D = \angle E + \angle F$. Prove that all angles of this hexagon are equal. | [
"Let $BC$, $DE$, $AF$ intersect and bound triangle $MNP$. Because the sum of angles in a hexagon is $720^\\circ$, we have $\\angle A + \\angle B = \\angle C + \\angle D = \\angle E + \\angle F = 240^\\circ$.\n\n\n\nTherefore, we easily see triangle $MNP$ is equilateral. To build the equilat... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0e9b | Let $x_1$ and $x_2$ be two different roots of the polynomial $p(x) = x^2 + a x + b$, and let $x_1^2 - \frac{1}{2}$ and $x_2^2 - \frac{1}{2}$ be the roots of $q(x) = x^2 + (a^2 - \frac{1}{2}) x + b^2 - \frac{1}{2}$. Find $a$ and $b$. | [
"The roots of the polynomial $p(x) = x^2 + a x + b$ are $x_1 = \\frac{-a + \\sqrt{a^2 - 4b}}{2}$ and $x_2 = \\frac{-a - \\sqrt{a^2 - 4b}}{2}$. They have to be different, so $a^2 - 4b \\neq 0$. From here we get\n$$\nx_1^2 - \\frac{1}{2} = \\frac{(a^2 - 2b - 1) - a \\sqrt{a^2 - 4b}}{2}\n$$\nand\n$$\nx_2^2 - \\frac{1}... | Slovenia | National Math Olympiad in Slovenia | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a = 0, b = -3/4 | |
0i98 | Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f : \mathbb{N} \to \mathbb{N}$ such that
$$
f(m+n)f(m-n) = f(m^2)
$$
for all $m, n \in \mathbb{N}$. | [
"Function $f(n) = 1$, for all $n \\in \\mathbb{N}$, is the only function satisfying the conditions of the problem.\n\nNote that\n$$\nf(1)f(2n-1) = f(n^2) \\quad \\text{and} \\quad f(3)f(2n-1) = f((n+1)^2)\n$$\nfor $n \\ge 3$. Thus\n$$\n\\frac{f(3)}{f(1)} = \\frac{f((n+1)^2)}{f(n^2)}.\n$$\nSetting $\\frac{f(3)}{f(1)... | United States | USA IMO 2003 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(n) = 1 for all positive integers n | |
0azo | Problem:
Two cards are chosen, without replacement, from a deck of 50 cards numbered $1,2,3, \ldots, 50$. What is the probability that the product of the numbers on these cards is divisible by 7? | [
"Solution:\nNote that there are $\\left\\lfloor\\frac{50}{7}\\right\\rfloor = 7$ multiples of $7$ from $1$ to $50$, so there are $43$ numbers which are not divisible by $7$. The probability that after choosing two cards, the product of these numbers is not divisible by $7$ is $\\binom{43}{2} / \\binom{50}{2} = \\fr... | Philippines | 20th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization"
] | null | final answer only | 46/175 | |
0eo6 | (a) Let $a$, $x$, $y$ be positive integers. Prove: if $x \neq y$, then also
$$
ax + \gcd(a, x) + \text{lcm}(a, x) \neq ay + \gcd(a, y) + \text{lcm}(a, y).
$$
(b) Show that there are no two positive integers $a$ and $b$ such that
$$
ab + \gcd(a, b) + \text{lcm}(a, b) = 2014.
$$ | [
"(a) Suppose that\n$$\nax + \\gcd(a, x) + \\text{lcm}(a, x) = ay + \\gcd(a, y) + \\text{lcm}(a, y)\n$$\nfor certain positive integers $a$, $x$, $y$. It follows that\n$$\n\\gcd(a, ax + \\gcd(a, x) + \\text{lcm}(a, x)) = \\gcd(a, ay + \\gcd(a, y) + \\text{lcm}(a, y)).\n$$\nSince $a$ divides both $ax$ and $\\text{lcm}... | South Africa | The South African Mathematical Olympiad Third Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof only | null | |
0app | Problem:
The length of one side of the square $ABCD$ is $4$ units. A circle is drawn tangent to $\overline{BC}$ and passing through the vertices $A$ and $D$. Find the area of the circle. | [
"Solution:\n\n$\\frac{25\\pi}{4}$ square units\n\nSee Figure 4. Let $R$ be the radius of the circle. Using the Extended Law of Sines, we have\n$$\n2R = \\frac{4}{\\sin 2A} = \\frac{4}{2 \\sin A \\cos A} = \\frac{2}{\\frac{2}{2\\sqrt{5}} \\cdot \\frac{4}{2\\sqrt{5}}} = 5\n$$\ngiving $R = \\frac{5}{2}$. Thus, we get\... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 25π/4 | |
0att | Problem:
What is the largest positive integer $k$ such that $27!$ is divisible by $2^{k}$? | [
"Solution:\n\nWe need to find the largest $k$ such that $2^{k}$ divides $27!$.\n\nThe exponent of $2$ in $27!$ is given by:\n$$\n\\left\\lfloor \\frac{27}{2} \\right\\rfloor + \\left\\lfloor \\frac{27}{4} \\right\\rfloor + \\left\\lfloor \\frac{27}{8} \\right\\rfloor + \\left\\lfloor \\frac{27}{16} \\right\\rfloor ... | Philippines | Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 23 | |
0l8g | Let $k$ and $d$ be positive integers. Prove that there exists a positive integer $N$ such that for every odd integer $n > N$, the digits in the base-$2n$ representation of $n^k$ are all greater than $d$. | [] | United States | USAMO | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0jgl | Problem:
How many positive integers less than $100$ are relatively prime to $200$? (Two numbers are relatively prime if their greatest common factor is $1$.) | [
"Solution:\n\nA positive integer $n < 100$ is relatively prime to $200$ if and only if it is relatively prime to $100$ (since $200$ and $100$ have the same prime factors). Thus, our answer is $\\varphi(100) = 100 \\times \\frac{1}{2} \\times \\frac{4}{5} = 40$."
] | United States | HMMT November 2013 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | final answer only | 40 | |
01d7 | All $n$-digit positive integers from $10^{n-1}$ to $10^n - 1$ are concatenated in the increasing order. What is the largest possible value of $k$ for which it is possible to find in this sequence of digits the same $k$-digit substring in at least two different places? | [
"Answer: $k = 2n - 1$.\n\nLet us look at the sequence\n$$\n\\underbrace{33\\ldots39}_{n-1} \\overline{\\phantom{33\\ldots3}} \\underbrace{33\\ldots34}_{n-2}\n$$\nIts length is $2n-1$ and it is a substring for two consecutive integers\n$$\n\\underbrace{33\\ldots39}_{n-1} \\text{ and } \\underbrace{33\\ldots340}_{n-2... | Baltic Way | Baltic Way 2016 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof and answer | 2n - 1 | |
0jhi | Problem:
Show that for all real numbers $a, b, c$,
$$
a^{6} + b^{6} + c^{6} - 3 a^{2} b^{2} c^{2} \geq \frac{1}{2}(a-b)^{2}(b-c)^{2}(c-a)^{2}
$$ | [
"Solution. A solution will be coming soon."
] | United States | Berkeley Math Circle | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0cjv | Let $n > 10$ be an integer, and let $A_1, A_2, \dots, A_n$ be distinct points in the plane such that the distances between the points are pairwise different. Define $f_{10}(j, k)$ to be the 10th smallest of the distances from $A_j$ to $A_1, A_2, \dots, A_k$, excluding $A_j$ if $k \ge j$. Suppose that for all $j$ and $k... | [
"To prove this, choose an arbitrary $i < n$, and let $A_i A_{j_1}, A_i A_{j_2}, \\dots, A_i A_{j_{10}}$ be the ten smallest numbers among the $A_i A_j$ with $j \\ne i$, ordered so that $j_1 < j_2 < \\dots < j_{10}$.\nIf $j_{10} < i$, then $i > 10$, and the problem condition yields\n$$\nb_i = \\max_{1 \\le k \\le 10... | Romania | Seventeenth ROMANIAN MASTER OF MATHEMATICS | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof only | null | |
0i5v | Problem:
Find all the integers $n>1$ with the following property: the numbers $1,2, \ldots, n$ can be arranged in a line so that, of any two adjacent numbers, one is divisible by the other. | [
"Solution:\n$2,3,4,6$\n\nThe values $n=2,3,4,6$ work, as shown by respective examples $1,2$; $2,1,3$; $2,4,1,3$; $3,6,2,4,1,5$. We shall show that there are no other possibilities.\n\nIf $n=2k+1$ is odd, then none of the numbers $k+1, k+2, \\ldots, 2k+1$ can divide any other, so no two of these numbers are adjacent... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2, 3, 4, 6 | |
0fza | Problem:
Finde alle surjektiven Funktionen $f: \mathbb{R} \rightarrow \mathbb{R}$, sodass für alle $x, y \in \mathbb{R}$ gilt:
$$
f(x+f(x)+2 f(y))=f(2 x)+f(2 y)
$$ | [
"Solution:\nSei $a \\in f^{-1}(0)$, solch ein $a$ existiert sicher, da $f$ surjektiv ist. Einsetzen von $x=y=a$ führt zu $f(2 a)=0$. Setzt man nun $x=a, 2 a$ ein so erhält man $f(a+2 f(y))= f(2 y)=f(2 a+2 f(y))$ für alle $y$, da $f$ surjektiv ist folgt, dass $f$ periodisch mit Periode $a$ ist, insbesondere gilt $f(... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = x for all real x | |
017q | Determine if there exists a set of $8$ consecutive positive integers that can be partitioned into two subsets with equal products of elements. | [
"We shall prove that no such set exists.\n\nSuppose that such a set exists and let $n$ be its smallest element. Consider any prime divisor $p$ of any of the numbers $n+i$, where $1 \\le i \\le 6$. If our set can be partitioned into two subsets with equal products of elements, then in each of these subsets there exi... | Baltic Way | BALTIC WAY | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | No, such a set does not exist. | |
0jt1 | Problem:
Let $ABC$ be an acute triangle with circumcenter $O$, orthocenter $H$, and circumcircle $\Omega$. Let $M$ be the midpoint of $AH$ and $N$ the midpoint of $BH$. Assume the points $M, N, O, H$ are distinct and lie on a circle $\omega$. Prove that the circles $\omega$ and $\Omega$ are internally tangent to each ... | [
"Solution:\n\nLet $R$ be the circumradius of $\\triangle ABC$. Recall that the circumcircle of $\\triangle HBC$ has radius equal to $R$. By a homothety at $H$ with factor $2$ at $H$, it follows that $\\triangle HMN$ has circumradius $R/2$. Thus $\\omega$ has circumradius $R/2$, and since $O$ lies on $\\omega$ the c... | United States | HMMT Invitational Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
0eai | The diagonals of a convex quadrilateral $ABCD$ intersect at $P$. Let $L, M$ and $N$ be the points on the segments $AB, BC$ and $CD$, respectively, such that
$$
\frac{|AL|}{|LB|} = \frac{|BM|}{|MC|} = \frac{|CN|}{|ND|}.
$$
Suppose that the line $LM$ is tangent to the circumcircle of the triangle $MCN$ and the line $NM$ ... | [
"Denote the circumcircle of $\\triangle MBL$ by $K_1$ and the circumcircle of $\\triangle MCN$ by $K_2$. The line $NM$ is tangent to $K_1$, so $\\angle NML = \\angle MBL$ (the tangent-chord angle theorem). Similarly, $LM$ is tangent to $K_2$, so $\\angle NML = \\angle NCM$. Hence,\n$$\n\\angle NCM = \\angle NML = \... | Slovenia | National Math Olympiad in Slovenia | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof only | null | |
0d42 | Define a domino to be an ordered pair of distinct positive integers. A proper sequence of dominoes is a list of distinct dominoes in which the first coordinate of each pair after the first equals the second coordinate of the immediately preceding pair, and in which $(i, j)$ and $(j, i)$ do not both appear for any $i$ a... | [
"Consider the complete graph $\\mathcal{G}_n$, over the set of vertices $\\{1,2, \\ldots, n\\}$. To each domino $(i, j)$ in $D_n$, we associate the edge joining the two vertices $i, j$. In this way, to each edge in $\\mathcal{G}_n$ we have associated exactly two dominoes $(i, j)$ and $(j, i)$, for some $1 \\leq i \... | Saudi Arabia | SAMC | [
"Discrete Mathematics > Graph Theory"
] | English, Arabic | proof and answer | If n is odd, the maximum length is n(n−1)/2. If n is even, the maximum length is (n^2 − 2n + 2)/2. | |
0d7v | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
x[f(x+y)-f(x-y)]=4 y f(x)
$$
for any real numbers $x, y$. | [
"Denote ($*$) as the given condition.\nLet $a = f(0)$. Taking $x = y$ in ($*$), we get\n$$\nf(2x) = a + 4 f(x)\n$$\nfor any $x \\neq 0$. Now, by replacing $x, y$ by $2x, 2y$ in ($*$) and using the above property (here we choose $x, y$ such that $x + y, x - y, x \\neq 0$), we get\n$$\n2x[4 f(x + y) - 4 f(x - y)] = 8... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = c x^2 for some real constant c | |
08hx | Problem:
Is it possible to cover a $2003 \times 2003$ board with $1 \times 2$ dominoes placed horizontally and $1 \times 3$ threeminoes placed vertically? | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No | |
02zf | Problem:
Em uma folha de papel estão desenhados 10 pontos, sendo 8 azuis e 2 vermelhos. Os dois pontos vermelhos são ligados a todos os pontos azuis, mas não são ligados um ao outro. Todos os pontos azuis são ligados um ao outro. Quando dois pontos são ligados, esta ligação é feita por um único segmento. De quantas ma... | [
"Solution:\n\nVamos chamar os dois pontos vermelhos de $A$ e $B$. Saindo de $A$, passando por exatamente um ponto azul, são 8 possibilidades; passando por dois pontos azuis, são $8 \\cdot 7 = 56$ possibilidades; passando por três pontos azuis, são $8 \\cdot 7 \\cdot 6 = 336$ possibilidades. Seguindo este raciocínio... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory"
] | null | final answer only | 219200 | |
0it9 | Problem:
Given that $x + \sin y = 2008$ and $x + 2008 \cos y = 2007$, where $0 \leq y \leq \pi / 2$, find the value of $x + y$. | [
"Solution:\nSubtracting the two equations gives\n$$\n(x + \\sin y) - (x + 2008 \\cos y) = 2008 - 2007\n$$\nso\n$$\n\\sin y - 2008 \\cos y = 1.\n$$\nBut since $0 \\leq y \\leq \\pi / 2$, the maximum of $\\sin y$ is $1$ and the minimum of $\\cos y$ is $0$, so we must have $\\sin y = 1$, so $y = \\frac{\\pi}{2}$.\n\nP... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 2007 + π/2 | |
0b7k | Find all primes $p$, $q$, $r$ such that $p(p-7) + q(q-7) = r(r-7)$. | [
"The given equality rewrites as $p^2 + q^2 - r^2 = 7(p + q - r)$. Since $(p+q-r)(p+q-r) = p^2 + q^2 - r^2 + 2pq$, it follows that $p+q-r$ divides $2pq$.\nIf $p$, $q$, $r > 2$, then $p+q-r$ is odd, so $p+q-r = p$, $q$ or $pq$. The first case gives $r = q$, then $p = 7$, so $(p, q, r) = (7, q, q)$, $q$ being an arbit... | Romania | NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (2, 7, 5), (7, 2, 5), (7, t, t), (t, 7, t) for any prime t | |
05cc | In how many ways can one choose 5 numbers from the list
$\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{1}{3}, \frac{1}{2}, 1, 2, 3, 4, 5, 6, 7, 8$
so that the product of the chosen numbers is 1?
*Remark:* Two choices are considered different if one choice contains a number that the other choic... | [
"Let's first count the choices that include the number $1$. If some three of the remaining numbers were $2$, $3$, and $4$, then the conditions of the problem would be violated, because the denominators of all the fractions in the list are smaller than $2 \\cdot 3 \\cdot 4 = 24$, and the product of the selected 5 nu... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | 41 | |
0gqn | Let $d(n)$ be the smallest prime divisor of integer $n \notin \{0, -1, +1\}$. Determine all polynomials $P(x)$ with integer coefficients satisfying
$$
P(n+d(n)) = n+d(P(n))
$$
for all integers $n > 2014$ for which $P(n) \notin \{0, -1, +1\}$. | [
"The answer: $P(x) = x, P(x) = 1, 0, -1$.\n\nWe start with the case when $\\deg(P(x)) \\ge 2$. Let us take $n = q$, where $q$ is prime: $P(q+d(q)) = q+d(P(q))$ yields $P(2q) = q+d(P(q))$. Therefore, $|P(2q)| \\le q + |P(q)|$ and $\\left|\\frac{P(2q)}{P(q)}\\right| \\le \\frac{q}{|P(q)|} + 1$ (1). Now when $q$ incre... | Turkey | Team Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | P(x) = x; P(x) = 0; P(x) = 1; P(x) = -1 |
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