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
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values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0dd2 | Define sequence of positive integers $\left(a_{n}\right)$ as $a_{1}=a$ and $a_{n+1}=a_{n}^{2}+1$ for $n \geq 1$. Prove that there is no index $n$ for which
$$
\prod_{k=1}^{n}\left(a_{k}^{2}+a_{k}+1\right)
$$
is a perfect square. | [
"Denote $p$ as a prime of $a_{1}^{2}+a_{1}+1$, note that $a_{1}$ is odd (since $a_{1}^{2}+a_{1}+1= a_{1}\\left(a_{1}+1\\right)+1$ is an odd number) and $p \\mid a_{1}$. By induction, we can show that\n$$\na_{n} \\equiv a_{2} \\equiv -a_{1} \\pmod{p} \\text{ for any } n>1.\n$$\nThus $a_{n}^{2}+a_{n}+1 \\equiv a_{1}^... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
09zz | Problem:
Gegeven zijn twee cirkels $\Gamma_{1}$ en $\Gamma_{2}$ met middelpunten $O_{1}$ en $O_{2}$ en gemeenschappelijke uitwendige raaklijnen $\ell_{1}$ en $\ell_{2}$. De lijn $\ell_{1}$ raakt $\Gamma_{1}$ in $A$ en $\Gamma_{2}$ in $B$. Zij $X$ een punt op het lijnstuk $O_{1} O_{2}$, maar niet op $\Gamma_{1}$ of $\Ga... | [
"Solution:\nWe bekijken de configuratie waarbij $Y$ tussen $A$ en $X$ ligt; andere configuraties gaan analoog. Zij $C$ het raakpunt van $\\ell_{2}$ aan $\\Gamma_{1}$. Dan is $C$ de spiegeling van $A$ in $O_{1} O_{2}$. Er geldt\n$$\n\\begin{array}{rrr}\n\\angle O_{1} Y X & =180^{\\circ}-\\angle O_{1} Y A & \\text{ (... | Netherlands | MO-selectietoets | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09nb | Let $\omega$ be the circle passing through point $C$ and tangent to side $AB$ of triangle $ABC$ at point $B$. The interior bisector of angle $\angle A$ intersects circle $\omega$ at points $E$ and $F$, and side $BC$ at point $D$ such that $F$ lies inside triangle $ABC$. Point $S$ on segment $EC$ is chosen such that $2\... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0i5r | Problem:
Two fair coins are simultaneously flipped. This is done repeatedly until at least one of the coins comes up heads, at which point the process stops. What is the probability that the other coin also came up heads on this last flip? | [
"Solution:\n\n$\\frac{1}{3}$. Let the desired probability be $p$. There is a $\\frac{1}{4}$ chance that both coins will come up heads on the first toss. Otherwise, both can come up heads simultaneously only if both are tails on the first toss, and then the process restarts as if from the beginning; thus this situat... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1/3 | |
021c | Problem:
For each positive integer $n$, let $\operatorname{rad}(n)$ denote the product of the distinct prime factors of $n$. Show that there exist integers $a, b>1$ such that $\operatorname{gcd}(a, b)=1$ and
$$
\operatorname{rad}(a b(a+b))<\frac{a+b}{2024^{2024}}
$$
For example, $\operatorname{rad}(20)=\operatorname{ra... | [
"Solution:\nWe show that the pair $(a, b)$ of the form $a=2^{p(p-1)}, b=3^{p(p-1)}-2^{p(p-1)}$ for sufficiently large prime number $p$ satisfies the inequality. First, notice that $\\operatorname{gcd}(a, b)=\\operatorname{gcd}(a, a+b)=1$ indeed. In addition, see that $\\operatorname{rad}(a)=2$, and $\\operatorname{... | Benelux Mathematical Olympiad | 16th Benelux Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, in... | null | proof only | null | |
01sv | Find all pairs of positive integers $(a; b)$, $a \le b$, satisfying the equality $a^3 + b^3 = 1911ab$. | [
"Answer: $(756; 1008)$, $(600; 960)$.\n\nLet $d$ denote the greatest common divisor of $a$ and $b$, i.e., $a = da_1$, $b = db_1$, where $\\gcd(a_1, b_1) = 1$. Then the given equality can be presented in the form\n$$\nd(a_1^3 + b_1^3) = 1911a_1b_1.\n$$\nIt follows that $d a_1^3 : b_1$, and, since $a_1$ and $b_1$ are... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (756, 1008), (600, 960) | |
08dj | Problem:
Denotiamo con $\lfloor x\rfloor$ il più grande intero $\leq$ di $x$.
Siano $\lambda \geq 1$ un numero reale, e $n$ un intero positivo, tali che $\left\lfloor\lambda^{n+1}\right\rfloor,\left\lfloor\lambda^{n+2}\right\rfloor, \ldots,\left\lfloor\lambda^{4 n}\right\rfloor$ sono tutti quadrati perfetti. Dimostrar... | [
"Solution:\n\nDimostriamo, intanto, il caso $n=1$. Sapendo che $\\left\\lfloor\\lambda^{2}\\right\\rfloor,\\left\\lfloor\\lambda^{3}\\right\\rfloor$, e $\\left\\lfloor\\lambda^{4}\\right\\rfloor$ sono quadrati perfetti, vogliamo vedere che $\\lfloor\\lambda\\rfloor$ è un quadrato perfetto. Detto $a^{2}=\\left\\lflo... | Italy | XXXV Olimpiade Italiana di Matematica | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
055x | Let $ABCDE$ be a regular pentagon and let $c$ be the circle with diameter $AB$. Diagonals $AC$ and $AD$ intersect the circle $c$ at points $F$ and $G$, respectively. Line $FG$ intersects the side $AE$ at point $H$. Let $K$ be the midpoint of the side $DE$. Prove that points $F, H, E$, and $K$ are concyclic. | [
"As $AB$ is a diameter of $c$, $\\angle AFB = 90^\\circ$ (Fig. 27) and $BF$ is an altitude of triangle $ABC$. From $AB = BC$, $BF$ is a median and $F$ is the midpoint of $AC$. By symmetry, point $K$ lies on $BF$ and $\\angle FKE = 90^\\circ$. Notice that $\\angle BAC = \\angle CAD = \\angle DAE$, as $BAC, CAD$, and... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gwy | Points $A_0, B_0, C_0$ are feet of altitudes in an acute-angled triangle $ABC$. Points $A_1, B_1, C_1$ are placed inside the triangle so that $\angle A_1BC = \angle A_1AB$, $\angle A_1CB = \angle A_1AC$, $\angle B_1CA = \angle B_1BC$, $\angle B_1AC = \angle B_1BA$, $\angle C_1BA = \angle C_1CB$, $\angle C_1AB = \angle ... | [
"Let $H$ be the orthocenter of $\\triangle ABC$, lines $AA_1$ and $BC$ intersect at point $A_3$. Then $\\triangle A_3BA_1 \\sim \\triangle ABA_3$, $\\triangle A_1A_3C \\sim \\triangle ACA_3 \\Rightarrow CA_3^2 = A_3A_1 \\cdot AA_3 = BA_3^2 \\Rightarrow CA_3 = BA_3$ (fig. 6).\n\nSince $\\angle BA_1C = 180^\\circ - \... | Ukraine | Ukrajina 2008 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometr... | English | proof only | null | |
0l3j | What is the remainder when $7^{2024} + 7^{2025} + 7^{2026}$ is divided by 19?
(A) 0 (B) 1 (C) 7 (D) 11 (E) 18 | [
"The quantity in question is seen to be a multiple of 19 as follows:\n$$\n7^{2024} + 7^{2025} + 7^{2026} = 7^{2024} (1 + 7 + 7^2) = 7^{2024} \\cdot 57 = 7^{2024} \\cdot 3 \\cdot 19.\n$$\nTherefore the remainder when it is divided by 19 is 0.\n\nWorking modulo 19, $7^0 = 1$, $7^1 = 7 \\cdot 1 = 7$, $7^2 = 7 \\cdot 7... | United States | AMC 10 B | [
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | MCQ | A | |
0iax | Problem:
Let $f(x) = \sin (\sin x)$. Evaluate $\lim_{h \rightarrow 0} \frac{f(x+h) - f(h)}{x}$ at $x = \pi$. | [
"Solution:\nThe expression $\\frac{f(x+h) - f(h)}{x}$ is continuous at $h = 0$, so the limit is just $\\frac{f(x) - f(0)}{x}$. Letting $x = \\pi$ yields $\\frac{\\sin (\\sin \\pi) - \\sin (\\sin 0)}{\\pi} = 0$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Precalculus > Limits",
"Precalculus > Trigonometric functions"
] | null | final answer only | 0 | |
0eqt | Let $P$ be the product of any three consecutive positive odd integers. What is the highest common factor (greatest common divisor) of all such numbers $P$? | [
"For any three consecutive odd integers, exactly one is divisible by $3$, so $3$ is a common factor of all such numbers $P$. Even the first two non-overlapping values $P = 1 \\times 3 \\times 5$ and $P = 7 \\times 9 \\times 11$ have highest common factor $3$, which is therefore the highest common factor of all valu... | South Africa | South African Mathematics Olympiad Second Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 3 | |
04vc | In an acute-angled triangle $ABC$, let us denote $H$ its orthocenter and $I$ its incenter. Let $D$ be the perpendicular projection of $I$ on the line $BC$, and $E$ be the image of point $A$ in symmetry with center $I$. Furthermore, $F$ is the perpendicular projection of the point $H$ on the line $ED$. Prove that the po... | [
"Let $M$ be the midpoint of $BC$. In point reflection with the centre $M$, denote $L$ the image of $D$ and $J$ the image of $I$. It follows from this symmetry, that $J$ is the incenter of triangle $BCP$ and $L$ is the point where this incircle touches $BC$. Let $KL$ be the diameter of this incircle. Thus, $J$ is th... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Ge... | English | proof only | null | |
0fpw | ¿De cuántas formas se pueden colorear los vértices de un polígono con $n \ge 3$ lados usando tres colores de forma que haya exactamente $m$ lados, $2 \le m \le n$, con los extremos de colores diferentes? | [
"En el polígono señalamos los puntos medios de los $m$ lados cuyos extremos deben colorearse con colores diferentes. Esto puede hacerse de $\\binom{n}{m}$ formas. Los $n$ vértices del polígono quedan divididos así en $m$ grupos de vértices consecutivos en los que todos ellos tienen el mismo color pero los vértices ... | Spain | LII Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Spanish | proof and answer | binomial(n, m) * (2^m + 2(-1)^m) | |
03ai | Let $n \ge 3$ be a positive integer. Find all non-constant real polynomials $f_1(x), f_2(x), \dots, f_n(x)$ such that
$$
f_k(x)f_{k+1}(x) = f_{k+1}(f_{k+2}(x)), \quad 1 \le k \le n
$$
for every real $x$ (here $f_{n+1}(x) \equiv f_1(x)$ and $f_{n+2}(x) \equiv f_2(x)$). | [
"Let $\\deg(f_k) = \\alpha_k \\in \\mathbb{N}, 1 \\le k \\le n$. We consider all indexes modulo $n$. The condition implies the equalities $\\alpha_k + \\alpha_{k+1} = \\alpha_{k+1}\\alpha_{k+2}$, i.e. $\\alpha_{k+1}$ divides $\\alpha_k$ for every $k = 1, 2, \\dots, n$. Therefore $\\alpha_1 = \\alpha_2 = \\dots = \\... | Bulgaria | 58. National mathematical olympiad Final round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f_1(x)=f_2(x)=...=f_n(x)=x^2 | |
0epf | We call a divisor $d$ of a positive integer $n$ *special* if $d + 1$ is also a divisor of $n$. Prove: at most half the positive divisors of a positive integer can be special. Determine all positive integers for which exactly half the positive divisors are special. | [
"We prove that no positive divisor $d$ of $n$ that is greater or equal to $\\sqrt{n}$ can be special: if $d$ is special, then $d + 1$ is also a divisor, so $n/d$ and $n/(d + 1)$ are both integers, which means that their difference is at least 1. Thus\n$$\n\\frac{n}{d} \\geq \\frac{n}{d+1} + 1,\n$$\nwhich is equival... | South Africa | South African Mathematics Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | English | proof and answer | 2, 6, 12 | |
09sq | Problem:
Voor reële getallen $a_{1}, a_{2}, \ldots, a_{n}$, allemaal verschillend, berekenen we de $\frac{n(n-1)}{2}$ sommen $a_{i}+a_{j}$ met $1 \leq i<j \leq n$ en sorteren deze vervolgens van klein naar groot. Bepaal alle gehele $n \geq 3$ waarvoor er $a_{1}, a_{2}, \ldots, a_{n}$ bestaan zodat dit rijtje van $\fra... | [
"Solution:\n\nVoor $n=3$ bekijken we $(a_{1}, a_{2}, a_{3})=(1,2,3)$. De sommen van steeds twee elementen zijn gelijk aan $3,4$ en $5$, dus die vormen een rekenkundige rij. Voor $n=4$ bekijken we $(a_{1}, a_{2}, a_{3}, a_{4})=(1,3,4,5)$. De sommen van steeds twee elementen zijn gelijk aan $4,5,6,7,8$ en $9$, dus di... | Netherlands | IMO-selectietoets III | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n = 3 or n = 4 | |
0htv | Problem:
Prove that there are infinitely many primes $p$ with the following property: there exists a positive integer $k$ such that $2^{k}-3$ is divisible by $p$. | [
"Solution:\nSuppose that there were only finitely many such primes $p_{1}, p_{2}, p_{3}, \\ldots, p_{n}$. Consider the number\n$$\nN=2^{\\left(p_{1}-1\\right)\\left(p_{2}-1\\right) \\cdots\\left(p_{n}-1\\right)+2}-3 .\n$$\nClearly, none of the $p_{i}$'s is $2$. Using Fermat's little theorem, we can prove that $N$ i... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
08yg | Let $I$ be the in-center and $\Omega$ be the inscribed circle of a triangle $ABC$, and let $M$ be the mid-point of the side $BC$. Let $K$ be the point of intersection of the line, going through $A$ and perpendicular to line $BC$, and the line, going through $M$ and perpendicular to the line $AI$. Prove that the circle ... | [
"Let us write $XY$ to indicate the length of the line segment $XY$. If $AB = AC$, the points $K$ and $M$ coincide and the 2 circles become tangent to each other. So, in the sequel, we assume that $AB \\ne AC$.\n\nLet $\\Gamma$ be the ex-circle within $\\angle A$ of the triangle $ABC$. Let $D$ be the point of tangen... | Japan | 2019 Japanese Mathematical Olympiad, Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Spiral simila... | null | proof only | null | |
06j0 | Determine the maximum possible value of real number $k$, for which the inequality
$$
\frac{a}{1 + 9bc + k(b - c)^2} + \frac{b}{1 + 9ca + k(c - a)^2} + \frac{c}{1 + 9ab + k(a - b)^2} \ge \frac{1}{2}
$$
is satisfied for every choice of nonnegative real numbers $a, b, c$ satisfying $a + b + c = 1$. | [
"The maximum possible value of $k$ is $4$.\n\nFirstly, consider $a = 0$ and $b = c = \\frac{1}{2}$. The inequality becomes\n$$\n0 + \\frac{2}{4+k} + \\frac{2}{4+k} \\ge \\frac{1}{2}.\n$$\nHence, we need $k \\le 4$.\n\nIt remains to prove\n$$\n\\frac{a}{1 + 9bc + 4(b - c)^2} + \\frac{b}{1 + 9ca + 4(c - a)^2} + \\fra... | Hong Kong | 1997-2023 IMO HK TST | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 4 | |
0ffq | Problem:
Tenemos en el plano dos puntos diferentes, $A$ y $O$. Para cada punto $X$ del plano distinto de $O$, denotamos por $\alpha(X)$ la medida del ángulo entre $OA$ y $OX$, en radianes, y contado en sentido antihorario desde $OA$ $(0 \leq \alpha(X)<2\pi)$.
Sea $C(X)$ la circunferencia de centro $O$ y radio de long... | [] | Spain | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0ggz | 設 $a_1, a_2, a_3, \dots$ 為無窮正整數數列，且對所有正整數 $n, m$，都有 $a_{n+2m}$ 整除 $a_n + a_{n+m}$ 這個性質。證明這個數列最終有週期性，也就是說，存在正整數 $N$ 和 $d$，使得對於所有 $n > N$，都有 $a_n = a_{n+d}$。 | [
"We will make repeated use of the following simple observation:\n\n**Lemma 1.** If a positive integer $d$ divides $a_n$ and $a_{n-m}$ for some $m$ and $n > 2m$, it also divides $a_{n-2m}$.\n\n*Proof.* Both parts are obvious since $a_n$ divides $a_{n-2m} + a_{n-m}$\n\n**Claim.** The sequence $(a_n)$ is bounded.\n\n*... | Taiwan | 2022 數學奧林匹亞競賽第三階段選訓營, 獨立研究 (二) | [
"Number Theory > Other"
] | Chinese; English | proof only | null | |
0ayh | Problem:
Let $a$ and $b$ be integers for which $\frac{a}{2} + \frac{b}{1009} = \frac{1}{2018}$. Find the smallest possible value of $|a b|$. | [
"Solution:\n\nClear denominators to write this as $1009 a + 2 b = 1$. Clearly, $a = 1$, $b = -504$ is a solution, and so our solutions are of the form $a = 1 + 2k$, $b = -504 - 1009k$. Now, clearly $|a| \\geq 1$, and $|b| \\geq 504$, so $|a b| \\geq 504$, and equality is attained when $a = 1$ and $b = -504$."
] | Philippines | 20th Philippine Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 504 | |
0adf | Prove that if $(x+\sqrt{x^2+1}) \cdot (y+\sqrt{y^2+1}) = 1$ then $x+y=0$. | [
"Multiplying $(x+\\sqrt{x^2+1}) \\cdot (y+\\sqrt{y^2+1}) = 1$ with $x-\\sqrt{x^2+1}$ we have\n$$ (x-\\sqrt{x^2+1})(x+\\sqrt{x^2+1}) \\cdot (y+\\sqrt{y^2+1}) = x-\\sqrt{x^2+1}, \\text{ i.e. } -y-\\sqrt{y^2+1} = x-\\sqrt{x^2+1} \\quad (1). $$\nSimilar by multiplying with $y-\\sqrt{y^2+1}$ we have $-x-\\sqrt{x^2+1}=y-... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | null | proof only | null | |
07ia | a sequence $a_1, a_2, \dots$ of nonzero integer numbers is given such that for all $n \in \mathbb{N}$, if $a_n = 2^\alpha k$ where $k$ is an odd integer and $\alpha$ is a non-negative integer, then:
$$
a_{n+1} = 2^\alpha - k.
$$
Prove that if this sequence is periodic, then for all $n \in \mathbb{N}$ we have:
$$
a_{n+2... | [
"First we claim that:\n$$\n\\forall n \\in \\mathbb{N},\\ a_{n+1} \\equiv a_n + 1 \\pmod{2}\n$$\nAssume that $a_n = 2^\\alpha k$ where $k \\equiv 1 \\pmod{2}$ and $\\alpha \\in \\mathbb{Z}_{\\ge 0}$. So we have:\n$$\na_n + a_{n+1} = 2^\\alpha (k+1) - k \\equiv 1 \\pmod{2}\n$$\nFrom the claim we can get that if $\\{... | Iran | 40th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
012f | Problem:
Find all nonnegative integers $m$ such that
$$
a_{m} = \left(2^{2m+1}\right)^{2} + 1
$$
is divisible by at most two different primes. | [
"Solution:\nObviously $m=0,1,2$ are solutions as $a_{0}=5$, $a_{1}=65=5 \\cdot 13$, and $a_{2}=1025=25 \\cdot 41$. We show that these are the only solutions.\n\nAssume that $m \\geqslant 3$ and that $a_{m}$ contains at most two different prime factors. Clearly, $a_{m}=4^{2m+1}+1$ is divisible by $5$, and\n$$\na_{m}... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | m = 0, 1, 2 | |
09c5 | Натурал тоон дээр
1) Натурал зэрэг дэвшүүлэх
2) Сүүлийн 2 оронг 3-т үржүүлж өмнөх тоон дээр нэмэх үйлдлээр 243-аас 2010-г гарган авч болох уу? | [
"$n = 100a + b$-тоог авч үзье. Энд $b$ нь $n$-тооны сүүлчийн 2 цифрээс тогтох тоо. Хоёр дахь үйлдлээр $n$ тоо нь $n_1 = a + 3b$ болно. $3n - n_1 = 299a$-буюу 13-т хуваагдана гэж гарна. $n_1 \\equiv 3n \\pmod{13}$ буюу 2-р үйлдэл нь mod 13-аар 3-т үржүүлэх үйлдэл юм. $243 = 3^5$ тул бодлогын нөхцөлд өгсөн үйлдлээр г... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | Mongolian | proof and answer | No | |
09jy | Let $\Gamma$ and $\omega$ be two circles, intersecting at points $C$ and $D$. The center of circle $\omega$ is denoted as $P$, and it lies on circle $\Gamma$. A line passing through point $D$ intersects circle $\Gamma$ at point $A$ and circle $\omega$ at point $B$, with $D$ being located between points $A$ and $B$. Lin... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0b91 | Consider the set $\mathcal{F}$ of functions $f: \mathbb{N} \to \mathbb{N}$ having the property that
$$
f(a^2 - b^2) = f(a)^2 - f(b)^2, \text{ for all } a, b \in \mathbb{N}, a \ge b.
$$
a) Determine the set $\{f(1) \mid f \in \mathcal{F}\}$.
b) Prove that $\mathcal{F}$ has exactly two elements. | [
"a) For $a = b$ we have $f(0) = 0$. Then $f(a^2) = f(a)^2$ for all $a \\in \\mathbb{N}$, implying $f(1) = f(1)^2$ and then $f(1) \\in \\{0, 1\\}$.\nBoth following cases hold, $f_0(1) = 0$ for $f_0 \\equiv 0 \\in \\mathcal{F}$, and $f_1(1) = 1$ for $f_1 = 1_{\\mathbb{N}} \\in \\mathcal{F}$. The requested set is $\\{... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | Possible values of f at one are 0 and 1. The only functions are the zero function and the identity function. | |
014d | Problem:
Let $ABC$ be a triangle, let $B_1$ be the midpoint of the side $AB$ and $C_1$ the midpoint of the side $AC$. Let $P$ be the point of intersection, other than $A$, of the circumscribed circles around the triangles $ABC_1$ and $AB_1C$. Let $P_1$ be the point of intersection, other than $A$, of the line $AP$ wit... | [
"Solution:\n\nSince $\\angle PBB_1 = \\angle PBA = 180^{\\circ} - \\angle PC_1A = \\angle PC_1C$ and $\\angle PCC_1 = \\angle PCA = 180^{\\circ} - \\angle PB_1A = \\angle PB_1B$, it follows that $\\triangle PBB_1$ is similar to $\\triangle PC_1C$.\n\nLet $B_2$ and $C_2$ be the midpoints of $BB_1$ and $CC_1$ respect... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09g8 | Let $a$, $b$, $c$ be positive numbers such that $a + b + c = 16$. Prove that
$$
\frac{ab}{a + b} + \frac{bc}{b + c} + \frac{ca}{c + a} \leq 5 + \frac{13c}{16}.
$$ | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0fpa | Hallar todos los enteros positivos $n$ para los que en cada casilla de un tablero $n \times n$ se puede escribir una de las letras I, M y O de manera que:
* en cada fila y en cada columna, un tercio de las casillas tiene I, un tercio tiene M y un tercio tiene O; y
* en cualquier línea diagonal compuesta por un número d... | [] | Spain | LVII Olimpiada Internacional de Matemáticas | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | All positive integers divisible by 3 | |
07dx | Consider a triangle $ABC$ with incircle $\omega$ that is respectively tangent to sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$. Points $P$, $Q$ are inside of $\angle A$ so that $FP = FB$, $FP \parallel AC$ and $EQ = EC$, $EQ \parallel AB$. Prove that $P$, $Q$ and $D$ are collinear. | [
"It's obvious that\n$$\n\\left. \\begin{array}{l} FP \\parallel CE \\\\ FB \\parallel EQ \\\\ \\frac{FP}{FB} = \\frac{EC}{EQ} \\end{array} \\right\\} \\implies \\triangle PFB \\sim \\triangle CEQ.\n$$\n---\nSo $BP \\parallel CQ$, also\n$$\n\\left. \n\\begin{array}{l}\n\\frac{BP}{CQ} = \\frac{BF}{EQ} = \\frac{BF}{EC... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0arc | Problem:
Find all complex numbers $x$ satisfying $x^{3} + x^{2} + x + 1 = 0$. | [
"Solution:\nWe have the equation:\n$$\nx^{3} + x^{2} + x + 1 = 0.\n$$\n\nWe can factor the left side:\n$$\nx^{3} + x^{2} + x + 1 = (x^{3} + 1) + (x^{2} + 1) - x^{2} - 1 + x^{2} + 1 = (x^{3} + 1) + (x + 1).\n$$\nBut it's easier to factor by grouping:\n$$\nx^{3} + x^{2} + x + 1 = (x^{3} + x^{2}) + (x + 1) = x^{2}(x +... | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | x = -1, i, -i | |
039g | The real numbers $a_i, b_i$, $1 \le i \le n$, are such that
$$
\sum_{i=1}^{n} a_{i}^{2}=1, \sum_{i=1}^{n} b_{i}^{2}=1 \text{ and } \sum_{i=1}^{n} a_{i} b_{i}=0.
$$
Prove that
$$
\left(\sum_{i=1}^{n} a_{i}\right)^{2}+\left(\sum_{i=1}^{n} b_{i}\right)^{2} \leq n.
$$ | [] | Bulgaria | First selection test for IMO 2007, Vietnam | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Linear Algebra > Vectors"
] | English | proof only | null | |
054l | Real numbers $x$, $y$ and $z$ satisfy $x + y + z = 4$ and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{3}$. Find the largest and the smallest possible value of the expression $x^3 + y^3 + z^3 + xyz$. | [
"$$(x + y + z)^3 = x^3 + y^3 + z^3 + 3(x^2y + xy^2 + x^2z + y^2z + y^2x + xyz) + 6xyz,$$\nwhile\n$$(3(x + y + z) \\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) xyz = 3(x + y + z)(xy + xz + yz) \\\\ = 3(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) + 9xyz.$$ \nThus\n$$(x + y + z)^3 - 3(x + y + z) \\left(\\frac... | Estonia | National Olympiad Final Round | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | largest = 64, smallest = 64 | |
029i | Problem:
a) Observe as somas:
Verifique que vale:
$$
\begin{gathered}
\frac{(k+1) \cdot (k+2) \cdot (k+3) \cdot \ldots \cdot (k+901)}{901} + (k+2) \cdot (k+3) \cdot (k+4) \cdot \ldots \cdot (k+901) = \\
\frac{(k+2) \cdot (k+3) \cdot \ldots \cdot (k+901) \cdot (k+902)}{901}
\end{gathered}
$$
b... | [
"Solution:\na)\n$$\n\\begin{array}{r}\n\\frac{(k+1) \\cdot (k+2) \\cdot (k+3) \\cdot \\ldots \\cdot (k+901)}{901} + (k+2) \\cdot (k+3) \\cdot (k+4) \\cdot \\ldots \\cdot (k+901) \\\\\n= \\frac{(k+2) \\cdot (k+3) \\cdot \\ldots \\cdot (k+901) \\cdot [(k+1)+901]}{901} \\\\\n= \\frac{(k+2) \\cdot (k+3) \\cdot \\ldots ... | Brazil | NÍVEL 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0cje | Find all pairs of twice differentiable functions $f, g : \mathbb{R} \to \mathbb{R}$, such that $f''$ and $g''$ are continuous, such that
$$
(f(x) - g(y)) \cdot (f'(x) - g'(y)) \cdot (f''(x) - g''(y)) = 0,
$$
for all $x, y \in \mathbb{R}$. | [
"Let $(f, g)$ be a pair of functions satisfying the given condition. We shall show that $f''$ is constant. Suppose that $f''$ is not constant. Then, because $(f(x) - g(0)) \\cdot (f'(x) - g'(0)) \\cdot (f''(x) - g''(0)) = 0$, for all $x \\in \\mathbb{R}$, and $f''$ is continuous, there is $a \\in \\mathbb{R}$ and $... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All pairs with f(x) = m x^2 + n x + p and g(x) = m x^2 + n' x + p', where m, n, n', p, p' are real constants. | |
0g5z | 試求最小正整數 $n$ 滿足下列條件：
存在一組相異的正整數 $s_1, s_2, \dots, s_n$ 使得
$$
(1 - \frac{1}{s_1})(1 - \frac{1}{s_2})\cdots(1 - \frac{1}{s_n}) = \frac{17}{670}.
$$ | [
"答：$n = 39$.\n\n不妨假設 $s_1 < s_2 < \\cdots < s_n$. 顯然, $s_1 > 1$ 否則 $1 - \\frac{1}{s_1} = 0$. 因此, $2 \\le s_1 \\le s_2 - 1 \\le \\cdots \\le s_n - (n-1)$. 故 $s_i \\ge i + 1, \\forall i = 1, \\cdots, n$. 題設可得\n$$\n\\begin{aligned}\n\\frac{17}{670} &= (1 - \\frac{1}{s_1})(1 - \\frac{1}{s_2})\\cdots(1 - \\frac{1}{s_n})... | Taiwan | 二〇一一數學奧林匹亞競賽第一階段選訓營,模擬競賽(二) | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 39 | |
0dch | Let non-constant polynomial $f(x)$ with real coefficients is given with the following property: for any positive integer $n$ and $k$, the value of expression
$$
\frac{f(n+1) f(n+2) \ldots f(n+k)}{f(1) f(2) \ldots f(k)} \in \mathbb{Z} .
$$
Prove that $f(x)$ is divisible by $x$. | [
"Without loss of generality one may assume that $f(1) \\in \\mathbb{Z}$. Since for all positive $k$, we have $\\frac{f(k+1)}{f(1)}$ is integer, then we conclude that on all positive integer points our polynomial gets integer values. Assume that $\\operatorname{deg}(f)=d$ then, according to Lagrange interpolation fo... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other"
] | English | proof only | null | |
02bs | Problem:
Desigualdade racional - Quais são os valores de $x$ que satisfazem a desigualdade $\frac{1}{x-2}<4$ ?
(a) $x>\frac{9}{4}$
(c) $x<2$ ou $x>\frac{9}{4}$
(e) $x<2$
(b) $2<x$ e $x<\frac{9}{4}$
(d) $x<-2$ | [
"Solution:\n\nA opção correta é (c).\n\nTemos $\\frac{1}{x-2}<4 \\Longleftrightarrow \\frac{1}{x-2}-4<0 \\Longleftrightarrow \\frac{1-4(x-2)}{x-2}<0 \\Longleftrightarrow \\frac{9-4x}{x-2}<0$.\n\nPara que uma fração seja negativa, o numerador e o denominador devem ter sinais contrários.\n\n1o Caso: $9-4x>0$ e $x-2<0... | Brazil | Nível 2 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | (c) | |
0b4z | Problem:
Fixăm un număr întreg $n \geq 2$. Determinaţi valoarea minimă a expresiei
$$
\frac{x_{1}+x_{2}+\cdots+x_{n}}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}}
$$
când $x_{1}, x_{2}, \ldots, x_{n}$ parcurg mulţimea numerelor reale strict pozitive, supuse condiţiei
$$
\frac{1}{1+x_{1}^{2}}+\frac{1}{1+x_{2... | [] | Romania | TESTUL 1 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | Minimum value: n − 1, attained when x_1 = x_2 = ⋯ = x_n = √(n − 1). (For n = 2, the value is 1 for all feasible pairs.) | |
06qx | Let the real numbers $a, b, c, d$ satisfy the relations $a+b+c+d=6$ and $a^{2}+b^{2}+c^{2}+d^{2}=12$. Prove that
$$
36 \leq 4\left(a^{3}+b^{3}+c^{3}+d^{3}\right)-\left(a^{4}+b^{4}+c^{4}+d^{4}\right) \leq 48
$$ | [
"Observe that\n$$\n\\begin{gathered}\n4\\left(a^{3}+b^{3}+c^{3}+d^{3}\\right)-\\left(a^{4}+b^{4}+c^{4}+d^{4}\\right)=-\\left((a-1)^{4}+(b-1)^{4}+(c-1)^{4}+(d-1)^{4}\\right) \\\\\n+6\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)-4(a+b+c+d)+4 \\\\\n=-\\left((a-1)^{4}+(b-1)^{4}+(c-1)^{4}+(d-1)^{4}\\right)+52\n\\end{gathered}\... | IMO | 51st IMO Shortlisted Problems | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0hzy | Problem:
Evaluate $\sum_{n=0}^{\infty} \frac{\cos n \theta}{2^{n}}$, where $\cos \theta=\frac{1}{5}$. | [
"Solution:\n$\\cos n \\theta$ is the real part of $e^{i n \\theta}$, so the sum is the real part of $\\sum_{n=0}^{\\infty} \\frac{e^{i n \\theta}}{2^{n}}$.\n\nThis is a geometric series with initial term $1$ and ratio $\\frac{e^{i \\theta}}{2}$, so its sum is $\\frac{1}{1- e^{i \\theta}/2}$.\n\nWe are given $\\cos ... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | 6/7 | |
0iaa | Problem:
Find the real value of $x$ such that $x^{3}+3 x^{2}+3 x+7=0$. | [
"Solution:\nRewrite the equation as $(x+1)^{3}+6=0$ to get $(x+1)^{3}=-6 \\Rightarrow x+1=\\sqrt[3]{-6} \\Rightarrow x=-1-\\sqrt[3]{6}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | -1 - \sqrt[3]{6} | |
088l | Problem:
Sui lati di un triangolo $ABC$ rettangolo in $A$ vengono scelti tre punti $D$, $E$ ed $F$ (rispettivamente su $BC$, $AC$ e $AB$) in modo che il quadrilatero $AFDE$ sia un quadrato. Se $x$ è la lunghezza di un suo lato, dimostrare che
$$
\frac{1}{x} = \frac{1}{AB} + \frac{1}{AC}
$$ | [
"Solution:\n\nVisto che i denominatori non sono nulli, l'uguaglianza da dimostrare è equivalente a $AB \\cdot AC = AC \\cdot x + AB \\cdot x$. Considerando poi che $x$ è il lato del quadrato, possiamo ancora riscrivere la tesi come\n$$\nAB \\cdot AC = AC \\cdot DE + AB \\cdot DF.\n$$\nOra $AB \\cdot AC$ è il doppio... | Italy | Cesenatico | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
06zb | Problem:
The function $f$ is defined on the non-negative integers. $f\left(2^{n}-1\right)=0$ for $n=0,1,2, \ldots$. If $m$ is not of the form $2^{n}-1$, then $f(m)=f(m+1)+1$. Show that $f(n)+n=2^{k}-1$ for some $k$, and find $f\left(2^{1990}\right)$ | [
"Solution:\n\nWe claim that if $2^{m} \\leq n < 2^{m+1}$, then $f(n) = 2^{m+1} - n - 1$.\n\nPut $r = 2^{m+1} - n$. Then the claim follows by induction on $r$.\n\nHence $f\\left(2^{1990}\\right) = 2^{1990} - 1$."
] | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | f(2^{1990}) = 2^{1990} - 1 | |
0due | Problem:
Na stranici $BC$ ostrokotnega trikotnika $ABC$ leži taka točka $D$, da je $|AB| = |AD|$. Naj bo $E$ taka točka na višini iz $C$ trikotnika $ABC$, da se krožnica $\mathcal{K}_1$ s središčem v $E$ dotika premice $AD$ v točki $D$. Označimo s $\mathcal{K}_2$ krožnico skozi $C$, ki se dotika premice $AB$ v točki $... | [
"Solution:\n\nNarišimo dovolj veliko skico in privzemimo običajne oznake kotov trikotnika $ABC$. Označimo presečišče krožnice $\\mathcal{K}_2$ z daljico $AC$ z $F$. Kot med tetivo $FB$ in tangento $AB$ je enak obodnemu kotu nad to tetivo. Sledi $\\angle FBA = \\gamma$, kar nam da $\\angle AFB = \\beta$. Torej je $A... | Slovenia | 45. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07sl | Suppose $a$, $b$, $c$ are the side lengths of an isosceles triangle $ABC$ with area $\Delta$. Prove that
$$
\sqrt{a^2 + b^2 - 4\Delta} + \sqrt{a^2 + c^2 - 4\Delta} \ge \sqrt{b^2 + c^2 - 4\Delta}.
$$
Determine the cases of equality. | [
"Clearly, the desired inequality holds if either $a = b$ or $a = c$, in which case a term on the LHS is equal to that on the RHS. If, say, $a = c$, equality holds iff $a^2 + c^2 = 4\\Delta = 2ac \\sin B$, i.e., $\\sin B = 1$, which means that $ABC$ is right-angled with the right angle at $B$. If $a = b$, equality h... | Ireland | IRL_ABooklet_2020 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | Equality holds exactly in the following cases:
- If a = c, when the triangle is right-angled at B.
- If a = b, when the triangle is right-angled at C.
- If b = c, when 5a^2 = 2b^2 (equivalently, cos A = 4/5 and sin A = 3/5). | |
04ja | There are ten white, and one red, blue, green, yellow and purple card. White cards are identical. On exactly one side of each card is the sign $X$. In how many ways is it possible to put the cards one on another such that no two cards face each other with the side having the sign $X$? | [] | Croatia | First round – City competition | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 5765760 | |
0f0c | Problem:
$O$ is the point of intersection of the diagonals of the convex quadrilateral $ABCD$. Prove that the line joining the centroids of $ABO$ and $CDO$ is perpendicular to the line joining the orthocenters of $BCO$ and $ADO$. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0jef | Problem:
Let $\{a_n\}_{n \geq 1}$ be an arithmetic sequence and $\{g_n\}_{n \geq 1}$ be a geometric sequence such that the first four terms of $\{a_n+g_n\}$ are $0, 0, 1$, and $0$, in that order. What is the 10th term of $\{a_n+g_n\}$? | [
"Solution:\n\nAnswer: $-54$\n\nLet the terms of the geometric sequence be $a, r a, r^{2} a, r^{3} a$. Then, the terms of the arithmetic sequence are $-a, -r a, -r^{2} a + 1, -r^{3} a$. However, if the first two terms of this sequence are $-a, -r a$, the next two terms must also be $(-2 r + 1) a, (-3 r + 2) a$. It i... | United States | HMMT | [
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | final answer only | -54 | |
06lo | Is it possible to find a non-constant polynomial $P(x, y)$ such that $P([\alpha], [3\alpha]) = 0$ for every real number $\alpha$? (Here $[\mu]$ stands for the largest integer less than or equal to $\mu$.) | [
"Yes. We claim that $P(x, y) = (y - 3x)(y - 3x - 1)(y - 3x - 2)$ satisfies the conditions. Clearly, it is a non-constant polynomial. For any real number $\\alpha$, let $n = \\lfloor \\alpha \\rfloor$. Then we have $n \\le \\alpha < n + 1$ and hence $3n \\le 3\\alpha < 3n + 3$. This shows $\\lfloor 3\\alpha \\rfloor... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | P(x, y) = (y - 3x)(y - 3x - 1)(y - 3x - 2) | |
0ei0 | Problem:
Dolžine diagonal mejnih ploskev kvadra so enake $|XY| = 8~\mathrm{cm}$, $|YZ| = 9~\mathrm{cm}$ in $|ZX| = \sqrt{55}~\mathrm{cm}$ (glej sliko). Koliko centimetrov je dolga telesna diagonala tega kvadra?
(A) $\sqrt{90}$
(B) 10
(C) $\sqrt{120}$
(D) $\sqrt{200}$
(E) 20
 | [
"Solution:\n\nDolžino, širino in višino kvadra označimo z $a, b$ in $c$. Tedaj po Pitagorovem izreku velja $a^2 + b^2 = |XY|^2 = 64$, $b^2 + c^2 = |ZX|^2 = 55$ in $c^2 + a^2 = |YZ|^2 = 81$. Enačbe seštejemo, da dobimo $2\\left(a^2 + b^2 + c^2\\right) = 200$. Dolžina telesne diagonale kvadra je enaka $\\sqrt{a^2 + b... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Geometry > Solid Geometry > 3D Shapes"
] | null | MCQ | B | |
000q | Se consideran todos los números naturales de nueve dígitos que utilizan exclusivamente los dígitos $1$, $2$ y $3$ (el menor es el $111111111$ y el mayor es el $333333333$). Cada uno de estos números está escrito en una tarjeta; se tiene así un mazo de $19683$ tarjetas.
David, Juan y Pablo se repartieron las tarjetas d... | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | español | proof and answer | David | |
053u | Find the largest natural number $n$ for which $3^{2016} - 1$ is divisible by $2^n$. | [
"We have $3^{2016} - 1 = (3^{63} - 1)(3^{63} + 1)(3^{126} + 1)(3^{252} + 1) \\cdot (3^{504} + 1)(3^{1008} + 1)$.\n\nNumbers $3^{126}$, $3^{252}$, $3^{504}$ and $3^{1008}$ are squares of odd numbers, hence congruent to $1$ modulo $8$. Thus $3^{126} + 1$, $3^{252} + 1$, $3^{504} + 1$ and $3^{1008} + 1$ are congruent ... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | 7 | |
0e4y | A $7 \times 7$ board is divided into $49$ squares. Onto this board we place several tiles like the one in the picture
(the tiles can be rotated),
each tile covering two squares. At least how many tiles do we need to place onto the board, so that every uncovered square will be adjacent to at l... | [
"We can place nine tiles as shown in the first picture. Each uncovered square has at least one covered neighbour.\n\n\n\nNow, let us show that this cannot be the case if we use less than nine tiles. Put a tile onto the board and mark all the neighbouring squares. Each row contains at most t... | Slovenia | National Math Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 9 | |
0jgj | Given positive integers $m$ and $n$, prove that there is a positive integer $c$ such that the numbers $cm$ and $cn$ have the same number of occurrences of each non-zero digit when written in base ten. | [
"**Solution 1** (By Richard Stong). For a given positive integer $k$, write $10^k m - n = 2^r 5^s t$, where $\\text{gcd}(t, 10) = 1$. For large enough values of $k$, the number of times 2 and 5 divide the left-hand side is at most the number of times they divide $n$, hence by choosing $k$ large we can make $t$ arbi... | United States | USAMO | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
06ff | Let $n > 4$ be a positive integer such that $n$ is composite (not a prime) and divides $\varphi(n)\sigma(n) + 1$, where $\varphi(n)$ is the Euler's totient function of $n$ and $\sigma(n)$ is the sum of the positive divisors of $n$. Prove that $n$ has at least three distinct prime factors. | [
"If $p^2 \\mid n$ for some prime $p$, then $p \\mid \\varphi(p^2) \\mid \\varphi(n)$. This implies $p \\mid \\varphi(n)\\sigma(n) + 1$, and hence $p \\mid n + \\varphi(n)\\sigma(n) + 1$. Therefore, $n$ must be squarefree.\n\nIf $n = pq$ for some distinct primes $p$ and $q$, then\n$$\n\\varphi(pq)\\sigma(pq) + 1 = (... | Hong Kong | CHKMO | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebr... | null | proof only | null | |
09ov | Let triangle $ABC$ be equilateral with side length $a$ meters. Point $X$ moves from vertex $A$ toward $C$ at $3$ m/s, and point $Y$ moves from vertex $B$ toward $C$ at $4$ m/s, both starting at the same time. After how many seconds will the distance $XY$ be equal to the height of triangle $ABC$?
(Khulan Tumenbayar) | [] | Mongolia | MMO2025 Round 2 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | final answer only | a/26 seconds | |
08na | Problem:
Find the least positive integer such that the sum of its digits is $2011$ and the product of its digits is a power of $6$. | [
"Solution:\nDenote this number by $N$. Then $N$ can not contain the digits $0, 5, 7$ and its digits must be written in increasing order. Suppose that $N$ has $x_{1}$ ones, $x_{2}$ twos, $x_{3}$ threes, $x_{4}$ fours, $x_{6}$ sixes, $x_{8}$ eights and $x_{9}$ nines, then\n$$\nx_{1} + 2x_{2} + 3x_{3} + 4x_{4} + 6x_{6... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | The minimal number is the 235-digit integer consisting of 3, 4, followed by ninety-three 8s, and then one hundred forty 9s. | |
0dav | Suppose that $x, y, z$ are non-zero real numbers such that
$$
x = 2 - \frac{y}{z}, \quad y = 2 - \frac{z}{x}, \quad z = 2 - \frac{x}{y}
$$
Find all possible values of $T = x + y + z$. | [
"From the given conditions, we have\n$$\nxz = 2z - y, \\quad xy = 2x - z, \\quad yz = 2y - x.\n$$\nTaking the sum of these equations, side by side, we have\n$$\nxy + yz + zx = x + y + z = T.\n$$\nFrom $xz = 2z - y$, we also can get $2z - x - y = xz - x \\rightarrow 3z - T = x(z - 1)$. Make the similar equations and... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | 3 or 7 | |
09fg | Let $n \ge 2$ be a positive integer. A positive real number is written in each unit square of a $2 \times n$ grid, so that the sum of the two numbers in each column is $1$. Suppose that, regardless of the numbers written, we could always delete one number from each column so that the sum of the remaining numbers in eac... | [
"The minimum value of $a$ is\n$$\nc = \\frac{\\left[ \\frac{(n+1)^2}{4} \\right]}{n+1} = \\begin{cases} \\frac{k+1}{2}, & n = 2k+1 \\\\ \\frac{k(k+1)}{2k+1}, & n = 2k. \\end{cases}\n$$\n\nWe observe that if the $2 \\times n$ grid is given as below, the value of $a$ is not less than $c$.\n\n$n = 2k:$\n\n| $\\frac{k}... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | floor(((n+1)^2)/4)/(n+1) = { (k+1)/2 if n=2k+1; k(k+1)/(2k+1) if n=2k } | |
0ftq | Problem:
Sei $n>1$ eine ungerade natürliche Zahl. Die Felder eines $n \times n$ Schachbretts sind abwechselnd weiss und schwarz gefärbt, sodass die vier Eckfelder schwarz sind. Ein L-triomino ist eine L-förmige Figur, die genau drei Felder des Brettes bedeckt. Für welche Werte von $n$ ist es möglich, alle schwarzen Fe... | [
"Solution:\n\nSetze $n=2m+1$ mit $m \\geq 1$. Nummeriere die Zeilen und Spalten von $1$ bis $n$ und markiere alle Felder, die in einer ungeraden Zeile und einer ungeraden Spalte liegen. Diese $(m+1)^2$ markierten Felder sind alle schwarz. Ein L-triomino kann höchstens ein markiertes Feld bedecken, also ist die benö... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | All odd n with n ≥ 7; the minimal number of L-triominos is ((n+1)/2)^2. | |
061g | Problem:
Man ermittle die Anzahl aller Zahlen der Form $x^{2}+y^{2}$ ($x, y \in \{1,2,3, \ldots, 1000\}$), die durch 121 teilbar sind. | [
"Solution:\n\nDie Reste, die eine Quadratzahl bei der Division durch 11 haben kann, sind 0, 1, 4, 9, 5 und 3. Da aber, außer zur Null, keine komplementären Reste modulo 11 auftreten, müssen sowohl $x^{2}$ als auch $y^{2}$ und folglich auch $x$ und $y$ durch 11 teilbar sein.\n\nUnter den Zahlen von 1 bis 1000 gibt e... | Germany | Auswahlwettbewerb zur IMO 2002 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | null | |
0l2c | Problem:
Estimate the value of
$$
\frac{20!\cdot 40!\cdot 40!}{100!} \cdot \sum_{i=0}^{40} \sum_{j=0}^{40} \frac{(i+j+18)!}{i!j!18!}
$$
Submit a positive real number $E$ either in decimal or in a fraction of two positive integers written in decimal (such as $\frac{2024}{2025}$ ). If the correct answer is $A$, your will... | [
"Solution:\nNote that\n$$\n\\sum_{i=0}^{40} \\sum_{j=0}^{40} \\frac{(i+j+18)!}{i!j!18!} = \\sum_{i=0}^{40} \\sum_{j=0}^{40} \\frac{(98-i-j)!}{(40-i)!(40-j)!18!} = \\sum_{i=0}^{40} \\sum_{j=0}^{40}\\binom{98-i-j}{40-i, 40-j, 18}\n$$\n\nThe multinomial coefficient $\\binom{98-i-j}{40-i, 40-j, 18}$ counts the number o... | United States | HMMT November 2024 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 1159/10677 | |
0d8f | Find the greatest positive real number $M$ such that for all positive real sequence $\left(a_{n}\right)$ and for all real number $m < M$, it is possible to find some index $n \geq 1$ that satisfies the inequality
$$
a_{1} + a_{2} + a_{3} + \cdots + a_{n} + a_{n+1} > m a_{n}.
$$ | [
"Denote $S, T$ as the midpoints of $A M, A Q$ respectively. Hence, $K S$ is the perpendicular bisector of $A M$ and $T L$ is the perpendicular bisector of $A Q$.\n\nThese imply that $A I, S K, T L$ are concurrent at the circumcenter $G$ of triangle $A M Q$.\nNote that $B I$ is the perpendic... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 4 | |
0cf7 | The real numbers $a$, $x$, $y$ are such that $a > 0$, $x \neq y$ and $x + \sqrt{a - x^2} = y + \sqrt{a - y^2}$ (C).
a) Prove that, if $x$ and $y$ are rational, then $a$ is rational.
b) Prove that, if $a$ is a positive integer, then there exists a pair $(x, y)$, with $x$ and $y$ irrational, fulfilling conditions (C). | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | English | proof only | null | |
09bx | **АВС** гурвалжинд багтсан тойрог **ВС** ба **СА** талуудыг харгалзан **D** ба **Е** цэгүүдэд шүргэнэ. **DE**-ийн $\angle ABC$ ба $\angle BAC$-ийн биссектриссүүдтэй огтлолцсон цэгүүд нь харгалзан **F**, **G**. Мөн **А** ба **В** оройгоос **DE** руу буулгасан перпендикуляруудын сууриуд харгалзан **S**, **T** бол $SF = G... | [
"АВ-ийн дундаж $k$, багтсан О-ийн төв $I$ байг. $\\angle A = \\alpha$, $\\angle B = \\beta$, $\\angle C = \\gamma$ гэвэл\n$$\n\\angle AIF = \\angle BAI + \\angle ABI = \\frac{\\alpha}{2} + \\frac{\\beta}{2}, \\angle AFI = AEI = 90^\\circ\n$$\n---\nтул $AK = KF = KB$. Үүнтэй адилаар $\\angle BIG = \\frac{\\alpha}{2}... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | Mongolian | proof only | null | |
03bg | Let $M$ be convex 2011-gon. Consider 2011 points lying inside $M$ and such that no three of all 4022 points (the vertexes of $M$ and 2011 points inside $M$) are collinear. A coloring of all points in two colors is called *good* if it is possible to connect some of the points by segments such that the following conditio... | [
"Let the two colors be blue and red. We prove first the following\n\n*Lemma*. Consider $\\triangle ABC$ with vertices of both colors. Any coloring of $n$ points inside $\\triangle ABC$ is good.\n\n*Proof*. Without loss of generality assume that $A$ and $B$ are blue points and $C$ is a red point.\nWe proceed by indu... | Bulgaria | Bulgaria | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 2^{2011}(2011\cdot2010+2) | |
0haa | Let $x, y, z$ be positive real numbers such that $x + y + z = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$. Prove that $xy + yz + zx \ge 3$. | [
"Using this equation and the inequality between the arithmetic mean and the geometric mean, we have\n$$\nxy + yz + zx = xyz\\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) = \\frac{xyz\\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right)^2}{x + y + z} = \\frac{xyz\\left(\\frac{1}{x^2} + \\frac{1}{y^2} ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
005d | Problem:
Se tiene un tablero cuadriculado de $a$ filas y $b$ columnas ($a \ge 2, b \ge 2$) y piezas de dominó formadas por dos cuadrados de $1 \times 1$ que tienen escrito el número $1$ en uno de los cuadrados y el número $-1$ en el otro. Se debe cubrir el tablero con piezas de dominó sin huecos ni superposiciones y s... | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | Possible if and only if a is even and ab is divisible by 4; equivalently, either a is a multiple of 4 (any b) or b is even with a even. It is impossible precisely when a ≡ 2 mod 4 and b is odd. | |
03t8 | Let $f(x) = \sin^4 x - \sin x \cos x + \cos^4 x$, the range of $f(x)$ is ______. | [
"As\n$$\n\\begin{aligned} f(x) &= \\sin^4 x - \\sin x \\cos x + \\cos^4 x \\\\ &= 1 - \\frac{1}{2} \\sin 2x - \\frac{1}{2} \\sin^2 2x, \\end{aligned}\n$$\nwe define $t = \\sin 2x$, then\n$$\nf(x) = g(t) = 1 - \\frac{1}{2}t - \\frac{1}{2}t^2 = \\frac{9}{8} - \\frac{1}{2}\\left(t + \\frac{1}{2}\\right)^2.\n$$\nSo we ... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | [0, 9/8] | |
0d5d | Let $A B C$ be a triangle with orthocenter $H$. Let $P$ be any point of the plane of the triangle. Let $\Omega$ be the circle with the diameter $A P$. The circle $\Omega$ cuts $C A$ and $A B$ again at $E$ and $F$, respectively. The line $P H$ cuts $\Omega$ again at $G$. The tangent lines to $\Omega$ at $E, F$ intersect... | [
"Let $B Y, C Z$ be altitudes of $A B C$. Points $Y, Z$ lie on the circle of diameter $A H$. The line $H P$ cuts the circle $\\Omega$ again at $G$. Since $A P$ is a diameter in $\\Omega$, the lines $A G$ and $P H$ are perpendicular and therefore point $G$ lies on the circle of diameter $A H$.\n\n![](attached_image_1... | Saudi Arabia | SAMC 2015 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English, Arabic | proof only | null | |
0i3h | Problem:
Find $(x+1)\left(x^{2}+1\right)\left(x^{4}+1\right)\left(x^{8}+1\right) \cdots$, where $|x|<1$. | [
"Solution:\nLet $S = (x+1)\\left(x^{2}+1\\right)\\left(x^{4}+1\\right)\\left(x^{8}+1\\right) \\cdots = 1 + x + x^{2} + x^{3} + \\cdots$.\n\nSince $xS = x + x^{2} + x^{3} + x^{4} + \\cdots$, we have $(1-x)S = 1$, so $S = \\frac{1}{1-x}$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1/(1 - x) | |
0c5p | Let $m$ be a positive integer and let $n = m^2 + 1$. Determine all real numbers $x_1, x_2, \dots, x_n$ satisfying
$$
x_i = 1 + \frac{2m x_i^2}{x_1^2 + x_2^2 + \dots + x_n^2}, \quad i = 1, 2, \dots, n.
$$ | [
"The $x_i$ are either all equal to $1 + \\frac{2m}{n} = \\frac{(m+1)^2}{m^2+1}$ or exactly one is equal to $m+1$ and the other are all equal to $1 + \\frac{1}{m}$. The verification offers no difficulty and is hence omitted.\n\nLeaving aside the trivial case where the $x_i$ are all equal, consider a solution $x_1, x... | Romania | Stars of Mathematics Competition | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | Either all variables are equal to ((m+1)^2)/(m^2+1), or exactly one variable equals m+1 and each of the remaining variables equals 1 + 1/m. | |
0i6r | Problem:
A $5 \times 5$ square grid has the number $-3$ written in the upper-left square and the number $3$ written in the lower-right square. In how many ways can the remaining squares be filled in with integers so that any two adjacent numbers differ by $1$, where two squares are adjacent if they share a common edge... | [
"Solution:\n\nIf the square in row $i$, column $j$ contains the number $k$, let its \"index\" be $i + j - k$. The constraint on adjacent squares now says that if a square has index $r$, the squares to its right and below it each have index $r$ or $r + 2$. The upper-left square has index $5$, and the lower-right squ... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 250 | |
0bf8 | Problem:
Adott az $n \in \mathbb{N}^*$ szám. Igazold, hogy
$$
2 \sqrt{2^{n}} \cos \left(n \arccos \frac{\sqrt{2}}{4}\right)
$$
egy páratlan egész szám! | [] | Romania | Matematika tantárgyverseny Megyei szakasz | [
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
02uf | Problem:
A folha do caderno de desenho de João é um enorme plano cartesiano quadriculado. Um dos seus desenhos preferidos é a criação de cobrinhas cobrindo os lados dos quadradinhos com sua caneta. Basicamente uma cobrinha é uma sequência de $2n$ pontos distintos $P_{1}, P_{2}, \ldots, P_{2n}$ escolhidos nos vértices ... | [
"Solution:\n\na) A figura a seguir mostra como obter os valores $-1, 1, 3$ e $-3$ para o caso $n=3$.\n\n\nb) Sejam $S(P_{n})$ a soma das coordenadas do ponto $P_{n}$ e $S$ o número obtido por João. Queremos calcular os possíveis valores de $S$.\n$$\n\\begin{aligned}\nS & =\\left(S(P_{1})+S(... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) For six points, examples exist achieving totals −3, −1, 1, and 3 (as shown by suitable snakes). b) The possible values are exactly all integers between −n and n having the same parity as n, i.e., {−n, −(n−2), …, n−2, n}. | |
0hns | Problem:
The numbers $1, 8, 4, 0$ are the first four terms of the infinite sequence. Every subsequent term of the sequence is obtained as the last digit of the sum of previous four terms. Therefore the fifth term of the sequence is $3$, because $1+8+4+0=13$; the sixth term is $5$ because $8+4+0+3=15$, and so on.
a. W... | [
"Solution:\n\na. Yes, very soon, in fact the next four terms (from $7$ to $10$th) are $2, 0, 0, 7$.\n\nb. We will prove that $1, 8, 4, 0$ will be a subsequence again. Assume the contrary. Since there are only finitely many combinations of four digits (precisely $10^4$), and the sequence is infinite, some combinatio... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | null | proof and answer | a: yes; b: yes | |
0bce | Problem:
Fie $(a_{n})_{n \geq 1}$ un şir crescător şi mărginit. Calculaţi
$$
\lim_{n \rightarrow \infty} \left(2 a_{n} - a_{1} - a_{2}\right) \left(2 a_{n} - a_{2} - a_{3}\right) \cdots \left(2 a_{n} - a_{n-2} - a_{n-1}\right) \left(2 a_{n} - a_{n-1} - a_{1}\right)
$$ | [] | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 0 | |
07dz | $$
8abc \le \left(\sqrt{bc} + \frac{1}{2a + \sqrt{bc}}\right) + \left(\sqrt{ca} + \frac{1}{2b + \sqrt{ca}}\right) + \left(\sqrt{ab} + \frac{1}{2c + \sqrt{ab}}\right).
$$
where $a$, $b$, $c$ are positive real numbers such that $ab + bc + ca = 1$. | [
"$$\n\\frac{1}{2a + \\sqrt{bc}} = \\frac{ab + bc + ca}{2a + \\sqrt{bc}}.\n$$\nDue to LM-GM, the inequality\n$$\nab + ca = a(b + c) \\geq a \\times 2\\sqrt{bc}\n$$\nholds. Hence\n$$\n\\frac{1}{2a + \\sqrt{bc}} = \\frac{ab + bc + ca}{2a + \\sqrt{bc}} \\geq \\frac{2a\\sqrt{bc} + bc}{2a + \\sqrt{bc}} = \\sqrt{bc},\n$$\... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0a22 | Find all pairs of prime numbers $(p, q)$ for which there exist positive integers $(m, n)$ such that
$$
(p+q)^m = (p-q)^n.
$$ | [
"The only divisor $p-q$ and $p+q$ can have in common is $2$, because $p$ and $q$ are different prime numbers. Indeed, a divisor $d$ of $p+q$ and $p-q$ is also a divisor of $(p+q) + (p-q) = 2p$ and of $(p+q) - (p-q) = 2q$. And we know that $\\text{gcd}(2p, 2q) = 2$, so $d$ must be a divisor of $2$.\n\nSince each pri... | Netherlands | BxMO/EGMO Team Selection Test | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | [(3,5), (5,3)] | |
06zq | Problem:
Construct a cyclic trapezium $ABCD$ with $AB$ parallel to $CD$, perpendicular distance $h$ between $AB$ and $CD$, and $AB + CD = m$. | [] | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
028r | Problem:
O valor de $\frac{\sqrt[3]{-0,001} \times \sqrt{400}}{\sqrt{0,25}}-\frac{\sqrt{0,036}-\sqrt{0,4}}{\sqrt{0,4}}$ é:
(a) $-3,3$
(b) $-4,7$
(c) $-4,9$
(d) $-3,8$
(e) $-7,5$ | [
"Solution:\n\nTemos:\n$$\n\\frac{-0,1 \\times 20}{0,5}-\\frac{\\sqrt{0,4}(\\sqrt{0,09}-1)}{\\sqrt{0,4}}=-\\frac{20}{5}-(0,3-1)=-4-0,3+1=-3,3.\n$$"
] | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | a | |
0gtw | Find all prime numbers $p, q$ satisfying the equation
$$
p(p^4 + p^2 + 10q) = q(q^2 + 3).
$$ | [
"We are to find all prime numbers $p, q$ such that\n$$\np(p^4 + p^2 + 10q) = q(q^2 + 3).\n$$\n\nFirst, note that both sides are positive for positive primes $p, q$.\n\nLet us analyze the equation:\n$$\np(p^4 + p^2 + 10q) = q(q^2 + 3).\n$$\n\nExpand the left side:\n$$\np^5 + p^3 + 10pq = q^3 + 3q.\n$$\n\nBring all t... | Turkey | Team Selection Test for EGMO 2023 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | p = 2, q = 5 | |
0bie | Consider two integers $n \ge m \ge 4$ and $A = \{a_1, a_2, \dots, a_m\}$ a subset of the set $\{1, 2, \dots, n\}$ such that:
for all $a, b \in A$, $a \ne b$, if $a + b \le n$, then $a + b \in A$.
Prove that:
$$
\frac{a_1 + a_2 + \dots + a_m}{m} \ge \frac{n+1}{2}.
$$ | [
"Assume $1 \\le a_1 < a_2 < \\dots < a_m \\le n$.\n\nFor even $m$ we can group the elements of $A$ in pairs of the form $(a_i, a_{m+1-i})$, with $1 \\le i \\le \\frac{m}{2}$. We prove that the sum of the numbers in each pair is at least $n+1$. Assuming the contrary to be true, it would exist an $i$ for which $a_i +... | Romania | 65th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0i5o | Problem:
Dan is holding one end of a 26 inch long piece of light string that has a heavy bead on it with each hand (so that the string lies along two straight lines). If he starts with his hands together at the start and leaves his hands at the same height, how far does he need to pull his hands apart so that the bead... | [
"Solution:\n\nAfter he pulls the bead is 5 inches below his hands, and it is 13 inches from each hand. Using the Pythagorean theorem, his hands must be $2 \\cdot 12 = 24$ inches apart."
] | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 24 inches | |
0d90 | We put four numbers $1, 2, 3, 4$ around a circle in order. One starts at the number $1$ and every step, he moves to an adjacent number on either side. How many ways he can move such that the sum of the numbers he visits in his path (including the starting number) is equal to $21$? | [
"Let $a_{i}, b_{i}, c_{i}, d_{i}$ be the number of paths that end by $1, 2, 3, 4$ respectively and have the sum equal to $i$. These paths all start from $1$. So it is easy to check that\n$$\n\\begin{aligned}\n& a_{1}=1,\\ b_{1}=c_{1}=d_{1}=0,\\ a_{2}=b_{2}=c_{2}=d_{2}=0 \\\\\n& a_{3}=0,\\ b_{3}=1,\\ c_{3}=d_{3}=0,\... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 167 | |
0da5 | The $n \times n \times n$ cubic is constructed from $n^{3}$ unit cubic $1 \times 1 \times 1$ such that at least one of unit cubic is black. Show that we can always cut the $n \times n \times n$ cubic into rectangular boxes such that each box contains exactly one black unit cubic. | [
"We shall prove the problem also true for all brick $a \\times b \\times c$ for any positive integers $a, b, c$ by induction on the number of black boxes.\n\nFirst, if the brick contains only one black cubic then no any cut is needed. Suppose that it contains at least two black cubics.\n\nWe choose a plane that div... | Saudi Arabia | Team selection tests for JBMO 2018 | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0gul | Let $ABC$ be a triangle and $D$, $E$ be points on segments $AB$, $AC$ respectively, such that $DE \parallel BC$. Let the circumcircle of $ABC$ meet the circumcircles of $BDE$ and $CDE$ again at $K$, $L$ respectively. Let $T$ be the intersection of the lines $BK$ and $CL$. Prove that $TA$ is tangent to the circumcircle ... | [
"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 rad... | Turkey | Team Selection Test for JBMO 2023 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0khy | Problem:
A counter begins at $0$. Then, every second, the counter either increases by $1$ or resets back to $0$ with equal probability. The expected value of the counter after ten seconds can be written as $\frac{m}{n}$, where $m, n$ are positive integers and $\operatorname{gcd}(m, n)=1$. Find $100m+n$. | [
"Solution:\nThe probability that the counter is equal to $k$ corresponds to the last $k$ seconds all being increases by $1$ and the second before that being a reset to $0$, which happens with probability $2^{-k-1}$. The only contradiction to this is when $k=10$ and the counter gets there by only counting $1$'s. The... | United States | HMMT November 2021 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 103324 | |
099y | Ordered four points $A$, $B$, $C$, $D$ lie on a given circle. Let $AC$ and $BD$ segments meet at $M$. A line passing through the point $M$ and given circle meets at $M_1$ and $M_2$, which line and the circles $\omega(ABM)$ and $\omega(CDM)$ meets at $N_1$ and $N_2$ respectively. Show that $M_1N_1 = M_2N_2$.
(proposed ... | [
"Let $O_1$ and $O_2$ be circumcentres of $\\triangle ABM$ and $\\triangle CDM$ respectively. Then first, we shall prove that $OO_1MO_2$ is parallelogram. Denote by $\\angle ABM = \\angle DCM = \\varphi$; $\\Rightarrow \\angle AMO_1 = 90^\\circ - \\varphi$ and $\\angle MCD = \\varphi \\Rightarrow O_1M \\perp CD$; $O... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"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",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
02m8 | Problem:
Dízima periódica - Sabendo que $0,333\ldots=\frac{1}{3}$, qual é a fração irredutível equivalente a $0,1333\ldots$.
(a) $\frac{1}{13}$
(b) $\frac{1}{15}$
(c) $\frac{1}{30}$
(d) $\frac{2}{15}$
(e) $\frac{1333}{10000}$ | [
"Solution:\n\nComo $\\frac{1}{3}=0,333\\ldots$, segue que\n$$\n0,1333\\ldots=0,333\\ldots-0,2=\\frac{1}{3}-\\frac{2}{10}=\\frac{1}{3}-\\frac{1}{5}=\\frac{2}{15}\n$$\n\nSolução 2: Usando simplesmente a regra que fornece a geratriz de uma dízima periódica, também podemos obter\n$$\n0,1333\\ldots=\\frac{13-1}{90}=\\fr... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | d | |
09jf | Let $a$ and $b$ be distinct positive integers. If $n$ and $m$ are positive integers satisfying
$$
(a^n - b^n)^m = (a^m - b^m)^n,
$$
prove that $n = m$. | [] | Mongolia | Mongolian Mathematical Olympiad Round 1 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
06e6 | Let $\triangle ABC$ be an acute triangle, $D$ the foot of the perpendicular from $A$ to $BC$. With $AD$ as a diameter, draw a circle intersecting $AB$ and $AC$ at $E$ and $F$ respectively. Suppose $AD$ and $EF$ intersect at $G$, and $AD$ is extended to meet the circumcircle of $\triangle ABC$ at $H$. Show that $AD^2 = ... | [
"Firstly, since\n$$\n\\angle AED + \\angle AFD = 90^\\circ + 90^\\circ = 180^\\circ,\n$$\nthe points $A$, $E$, $D$, $F$ are concyclic.\n\nSecondly, since\n$$\n\\angle AHB = \\angle ACB = 90^\\circ - \\angle DAF = \\angle FDA = \\angle FEA,\n$$\nthe points $E$, $B$, $H$, $G$ are concyclic.\n\nThirdly, $AD$ is tangen... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03wu | Let $n$ be an integer greater than $3$. The points $V_1, V_2, \dots, V_n$, with no three collinear, lie on the plane. Some of the segments $V_iV_j$, with $1 \le i < j \le n$, are constructed. The points $V_i$ and $V_j$ are neighbors if $V_iV_j$ is constructed. Initially, the chess pieces $C_1, C_2, \dots, C_n$ are plac... | [
"The answer is $n+1$.\n\nFor a harmonic set, we consider a graph $G$ with $V_1, V_2, \\dots, V_n$ as its vertices and with the segments in the harmonic set as its edges.\n\nFirst, we show that there are at least $n$ edges in $G$. Note that $G$ must be connected. Also note that each vertex must have degree at least ... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof and answer | n+1 | |
0afc | За даден број ќе велиме дека е “шарен” ако е запишан со еднаков број парни и непарни цифри. Да се определи бројот на сите четирицифрени “шарени” броеви запишани со различни цифри? | [
"Имаме 5 парни цифри, $\\{0,2,4,6,8\\}$ и 5 непарни цифри, $\\{1,3,5,7,9\\}$. Два парни броја од 5 можеме да избереме на $C_5^2 = 10$ начини, т.е. ги имаме следниве можности\n$$\n\\{\\{0,2\\}, \\{0,4\\}, \\{0,6\\}, \\{0,8\\}, \\{2,4\\}, \\{2,6\\}, \\{2,8\\}, \\{4,6\\}, \\{4,8\\}, \\{6,8\\}\\}.\n$$\nНа исто толку на... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | Macedonian, English | proof and answer | 2160 | |
034n | Problem:
The sequence $\{a_{n}\}_{n=1}^{\infty}$ is defined by $a_{1}=0$ and $a_{n+1}= a_{n}+4 n+3$, $n \geq 1$.
a) Express $a_{n}$ as a function of $n$.
b) Find the limit
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{a_{n}}+\sqrt{a_{4 n}}+\sqrt{a_{4^{2} n}}+\cdots+\sqrt{a_{4^{10} n}}}{\sqrt{a_{n}}+\sqrt{a_{2 n}}+\sqr... | [
"Solution:\n\na.\nUsing the recurrence relation we easily get\n$$\n\\begin{aligned}\na_{k} & =a_{k-1}+4(k-1)+3=a_{k-2}+4(k-2)+4(k-1)+2 \\cdot 3=\\cdots \\\\\n& =a_{1}+4(1+2+\\cdots+k-1)+(k-1) \\cdot 3=2 k(k-1)+3(k-1) \\\\\n& =(2 k+3)(k-1)\n\\end{aligned}\n$$\n\nb.\nWe have $\\lim _{n \\rightarrow \\infty} \\frac{\\... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a_n = (2n+3)(n-1); the limit equals 683 | |
0juz | Problem:
Consider a $2 \times n$ grid of points and a path consisting of $2n-1$ straight line segments connecting all these $2n$ points, starting from the bottom left corner and ending at the upper right corner. Such a path is called efficient if each point is only passed through once and no two line segments intersec... | [
"Solution:\n\nThe general answer is $\\binom{2(n-1)}{n-1}$: Simply note that the points in each column must be taken in order, and anything satisfying this avoids intersections, so just choose the steps during which to be in the first column."
] | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | C(4030, 2015) |
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