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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0977 | Problem:
Să se arate că, pentru orice numere reale $x, y \in [0,1]$, este justă inegalitatea
$$
\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}} \leq \frac{2}{\sqrt{1+x y}}
$$ | [
"Solution:\nRidicând la pătrat, obținem $\\frac{1}{1+x^{2}}+\\frac{2}{\\sqrt{1+x^{2}} \\cdot \\sqrt{1+y^{2}}}+\\frac{1}{1+y^{2}} \\leq \\frac{4}{1+x y}$. Conform inegalității mediilor pentru 2 numere $u, v \\geq 0$, avem $2 \\sqrt{u v} \\leq u+v$. Atunci $\\frac{2}{\\sqrt{1+x^{2}} \\cdot \\sqrt{1+y^{2}}} \\leq \\fr... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
09p0 | Problem:
How many ways can three distinct bishops be placed on an $8 \times 8$ chessboard such that no two of them attack each other? | [] | Mongolia | MMO2025 Round 2 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 157152 | |
086q | Problem:
In quanti modi si possono ordinare le cifre $1$, $2$, $4$, $7$ e $9$ affinché formino un numero di cinque cifre divisibile per $11$?
(A) $0$
(B) $1$
(C) $10$
(D) $12$
(E) $24$. | [
"Solution:\n\nLa risposta è (D). Un numero è divisibile per $11$ se e solo se è divisibile per $11$ la differenza tra la somma delle cifre di posto dispari e la somma delle cifre di posto pari.\n\nSia $a$ la somma delle $3$ cifre di posto dispari e $b$ la somma delle $2$ cifre di posto pari. Siccome $a+b=1+2+4+7+9=... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Number Theory > Divisibility / Factorization"
] | null | MCQ | D | |
013s | Problem:
Let $m = 30030 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ and let $M$ be the set of its positive divisors which have exactly two prime factors. Determine the minimal integer $n$ with the following property: for any choice of numbers from $M$, there exist three numbers $a, b, c$ among them satisfying $a \c... | [
"Solution:\n\nTaking the 10 divisors without the prime $13$ shows that $n \\geq 11$. Consider the following partition of the 15 divisors into five groups of three each with the property that the product of the numbers in every group equals $m$.\n$$\n\\begin{array}{ll}\n\\{2 \\cdot 3, 5 \\cdot 13, 7 \\cdot 11\\}, & ... | Baltic Way | Baltic Way 2005 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 11 | |
0gkj | Let $a, b, c > 0$ and $a + b + c = 3$. Prove that
$$
\frac{1}{a\sqrt{2(a^2+bc)}} + \frac{1}{b\sqrt{2(b^2+ca)}} + \frac{1}{c\sqrt{2(c^2+ab)}} \ge \frac{1}{a+bc} + \frac{1}{b+ca} + \frac{1}{c+ab}.
$$ | [
"Without loss of generality, we may assume that $a \\ge b \\ge c$. Then\n$$\n\\frac{(c-a)(c-b)}{3(c+ab)} \\ge 0\n$$\nand\n$$\n\\frac{(a-b)(a-c)}{3(a+bc)} + \\frac{(b-a)(b-c)}{3(b+ca)} = \\frac{c(a-b)^2}{3} \\left( \\frac{1+a+b-c}{(a+bc)(b+ca)} \\right) \\ge 0\n$$\nTherefore\n$$\n\\sum_{\\text{cyc}} \\frac{(a-b)(a-c... | Thailand | Thai Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0ga0 | 已知 $n$ 是正整數。求多項式 $(x^2 - x + 1)^n$ 的奇係數的個數。
Let $n$ be a positive integer. Find the number of odd coefficients of the polynomial $(x^2 - x + 1)^n$. | [
"設多項式 $P(x), Q(x)$ 是整係數多項式。若 $P(x) - Q(x)$ 的所有係數都是偶數，則稱兩個多項式相似，記為 $P(x) \\sim Q(x)$。此時，多項式 $P(x), Q(x)$ 係數為奇數的項的個數相同，將多項式 $P(x)$ 的奇係數的個數記為 $\\beta(P)$。顯然，\n$$\n(x^2 - x + 1)^n \\sim (x^2 + x + 1)^n.\n$$\n底下討論 $P_n(x) = (x^2 + x + 1)^n$ 的奇係數的個數問題。\n\n利用數學歸納法證明：當 $n = 2^q$ ($q$ 為正整數) 時，$P_n(x) \\sim x^{2n} + x^n + 1$... | Taiwan | 二〇一六數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | Write n in binary as blocks of consecutive ones separated by blocks of zeros:
n = (1 repeated a_k times)(0 repeated b_k times)…(1 repeated a_1 times)(0 repeated b_1 times)_2,
where a_i, b_i are positive integers and b_1 ≥ 0. Then the number of odd coefficients of (x^2 − x + 1)^n equals
∏_{i=1}^k (2^{a_i+2} − (−1)^{a... | |
04in | Let $ABCD$ be a quadrilateral such that $|AB| = 6$, $|BC| = 9$, $|CD| = 18$ and $|AD| = 5$ hold. Determine the length of the diagonal $AC$ if it is known that it is a positive integer. (Andrea Aglić-Aljinović) | [] | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof and answer | 14 | |
0bp9 | Problem:
O funcţie $f$ de gradul al doilea are proprietatea: pentru orice interval $I$ de lungime 1, intervalul $f(I)$ are lungimea cel puţin 1.
Arătaţi că, pentru orice interval $J$ de lungime 2, intervalul $f(J)$ are lungimea cel puţin 4. | [
"Solution:\n\nDacă $f(x) = a x^{2} + b x + c$ şi $v$ este abscisa vârfului parabolei, atunci, pentru intervalul $I = [v - 1/2, v + 1/2]$, intervalul $f(I)$ are lungimea $\\frac{|a|}{4}$, deci $|a| \\geq 4$.\n\nPentru un interval $J$ de lungime 2 există $x, y \\in J$ astfel încât $x - y = 1$ şi $v \\notin (y, x)$.\n... | Romania | Olimpiada Naţională de Matematică, Etapa Naţională | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0gti | For a polynomial $Q$ with integer coefficient and prime $p$, we say that $Q$ excludes $p$ if there is no integer $n$ for which $p \mid Q(n)$. Does there exist a polynomial of degree $5$ with integer coefficients having no rational roots which excludes exactly one prime? | [
"Answer: Yes, for example $Q(x) = (2x^3 + 1)(x^2 - x + 1)$.\nClearly $Q(x)$ has no rational roots. We will show that the only prime excluded by $Q(x)$ is $p=2$.\n\n**Observation 1:** For any prime $p$ satisfying $p \\equiv 1 \\pmod 3$ there exists an integer $n$ such that $n^2 - n + 1 \\equiv 0 \\pmod p$.\n*Proof.*... | Turkey | Team Selection Test | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Yes; Q(x) = (2x^3 + 1)(x^2 - x + 1), which excludes only the prime 2. | |
03vk | Let $m, n \in \mathbb{N}^*, m, n > 1, a_{ij}$ ($i = 1, 2, \dots, n$, $j = 1, 2, \dots, m$) be non-negative real numbers (not all zero). Find the maximum and minimum values of
$$
f = \frac{n \sum_{i=1}^{n} (\sum_{j=1}^{m} a_{ij})^2 + m \sum_{j=1}^{m} (\sum_{i=1}^{n} a_{ij})^2}{(\sum_{i=1}^{n} \sum_{j=1}^{m} a_{ij})^2 + ... | [
"The maximum value of $f$ is $1$.\nFirstly, we prove that $f \\le 1$. It suffices to show that\n$$\nn \\sum_{i=1}^{n} (\\sum_{j=1}^{m} a_{ij})^2 + m \\sum_{j=1}^{m} (\\sum_{i=1}^{n} a_{ij})^2 \\le (\\sum_{i=1}^{n} \\sum_{j=1}^{m} a_{ij})^2 + mn \\sum_{i=1}^{n} \\sum_{j=1}^{m} a_{ij}^2,\n$$\n$$\n\\text{or} \\quad \\... | China | China National Team Selection Test | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | Maximum = 1; Minimum = (m + n) / (mn + min{m, n}) | |
0l4d | Problem:
Let $ABCDEF$ be a regular hexagon with center $O$ and side length $1$. Point $X$ is placed in the interior of the hexagon such that $\angle BXC = \angle AXE = 90^\circ$. Compute all possible values of $OX$. | [
"Solution:\n\nPoint $X$ is the intersection of circles with diameter $AE$ and $BC$. Thus, there are two possible intersection points. Since $AC \\perp BE$, the first point, $X_1$, is the intersection of $AC$ and $BE$, from which we can see $OX_1 = \\frac{1}{2}$ as our first answer. Let $X_2... | United States | HMMT November | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangl... | null | proof and answer | 1/2 and sqrt(7)/7 | |
0jiq | Let $a$, $b$, $c$, $d$ be real numbers such that $b - d \ge 5$ and all zeros $x_1$, $x_2$, $x_3$, and $x_4$ of the polynomial $P(x) = x^4 + a x^3 + b x^2 + c x + d$ are real. Find the smallest value the product $(x_1^2 + 1)(x_2^2 + 1)(x_3^2 + 1)(x_4^2 + 1)$ can take. | [
"**Solution 1** (by Titu Andreescu). Using Vieta's identities we have:\n$$\nx_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 - x_1x_2x_3x_4 \\ge 5,\n$$\nand so\n$$\nx_1(x_2 + x_3 + x_4 - x_2x_3x_4) + 1(x_2x_3 + x_2x_4 + x_3x_4 - 1) \\ge 4.\n$$\nIt follows that\n$$\n4^2 \\le [x_1(x_2 + x_3 + x_4 - x_2x_3x_4) + 1(... | United States | USAMO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | 16 | |
062d | Problem:
Man zeige, dass es in der Dezimaldarstellung von $\sqrt[3]{3}$ zwischen der 1000000. und 3141592. Nachkommastelle eine von 2 verschiedene Ziffer gibt. | [
"Solution:\n\nHätte $\\sqrt[3]{3}$ zwischen der 1000000. und 3141592. Nachkommastelle nur die Ziffer 2, wäre mit $a=\\left[10^{1000000} \\sqrt[3]{3}\\right]<10^{1000001}$ also $\\left|\\sqrt[3]{3}-(a+2 / 9) 10^{-1000000}\\right|<10^{-3141592}$ oder äquivalent\n\n$$\n\\left|\\sqrt[3]{3\\left(9 \\cdot 10^{1000000}\\r... | Germany | IMO-Auswahlklausur | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof only | null | |
09aw | Let $k, n$ be positive integers. Given $A_0, A_1, \dots, A_n$ points and edges that connect them are colored with $k$ colors. $S(n, k)$ is the maximum number of angles, whose sides are colored with different colors. Prove that
$$
S(n, k) > \binom{k}{2} \cdot \left[\frac{n}{k}\right]^2 \cdot n.
$$ | [
"We are going to give a construction of coloring, whose number of angles with different colored sides is greater than $\\binom{k}{2} \\cdot \\left[\\frac{n}{k}\\right]^2 \\cdot n$.\n\n**Case I.** Let $k$ be an odd number. Let $A_1, \\dots, A_k$ be subsets of $\\{A_0, A_1, \\dots, A_n\\}$ such that $A_1 \\cup \\dots... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
04au | Each side and each diagonal of a convex $n$-gon is colored in one of $k$ colors. It is known that there doesn't exist a closed monochrome broken line whose vertices are also the vertices of the given $n$-gon. Determine the largest possible value of $n$. (Russia 1990) | [
"It is known that a graph with $n$ vertices that contains no cycles has at most $n-1$ edges. That fact is easily proven by induction by the number of vertices $n$. Hence, at most $n-1$ edges (i.e. sides and diagonals) can be colored with the same color.\n\nSince the number of colors is $k$ and the total number of e... | Croatia | Hrvatska 2011 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2k | |
0del | Let us call a set of positive integers nice if the number of its elements equals to the average of its numbers. Call a positive integer $n$ an amazing number if the set $\{1, 2, \ldots, n\}$ can be partitioned into nice subsets.
a) Prove that every perfect square is amazing
b) Show that there are infinitely many posi... | [
"a) Let $A = \\{1, 2, \\dots, n\\}$ and $n = k^2$ for some positive integer $k$. Denote $B$ as the set of first $m$ odd positive integers. It's clear that $B$ is *nice*. Assume that $n \\ge 2m-1$, let $C = A \\setminus B$. $C$ is *nice* if and only if\n$$\n\\frac{(1 + 2 + \\cdots + n) - m^2}{n - m} = n - m \\iff n ... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
057i | Find all functions $f: \mathbb{R} \to \mathbb{R}$ that satisfy the equation
$$
f(xf(y) + y) = f(x^2 + y^2) + f(y)
$$
for all real numbers $x$ and $y$. | [
"Substituting $x = 0$ into the equation gives $f(y) = f(y^2) + f(y)$, implying $f(y^2) = 0$. Thus $f(x) = 0$ for all non-negative real numbers $x$. In particular $f(x^2 + y^2) = 0$, which allows to simplify the initial equation as\n$$\nf(xf(y) + y) = f(y).\n$$\nSuppose that $f(c) \\neq 0$ for some negative real num... | Estonia | Open Contests | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 for all real x | |
0dbd | Determine all arithmetic sequences $a_{1}, a_{2}, \ldots$ for which there exists integer $N>1$ such that for any positive integer $k$ the following divisibility holds
$$
a_{1} a_{2} \ldots a_{k} \mid a_{N+1} a_{N+2} \ldots a_{N+k} .
$$ | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | All such arithmetic sequences are exactly the constant sequences and the sequences of the form a_n = c·n with c an integer. For these, the divisibility holds for any shift greater than one. | |
0k3t | Problem:
Lil Wayne, the rain god, determines the weather. If Lil Wayne makes it rain on any given day, the probability that he makes it rain the next day is $75\%$. If Lil Wayne doesn't make it rain on one day, the probability that he makes it rain the next day is $25\%$. He decides not to make it rain today. Find the... | [
"Solution:\n\nLet $p_{n}$ denote the probability that Lil Wayne makes it rain $n$ days from today. We have $p_{0}=0$ and\n$$\np_{n+1}=\\frac{3}{4} p_{n}+\\frac{1}{4}\\left(1-p_{n}\\right)=\\frac{1}{4}+\\frac{1}{2} p_{n} .\n$$\nThis can be written as\n$$\np_{n+1}-\\frac{1}{2}=\\frac{1}{2}\\left(p_{n}-\\frac{1}{2}\\r... | United States | HMMT November 2018 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 9 | |
0dfr | Let convex $s$-gon is divided to $q$ quadrilaterals such that $b$ of them are not convex. Prove that
$$
q \geq b + \frac{s-2}{2}.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
01od | $N$ boys ($N \ge 3$), no two of them having the same height, are arranged along a circle. A boy in the given arrangement is said to be *tall* if he is taller than both of his neighbors; a boy is said to be *short* if he is shorter than both of his neighbors.
Prove that the number of tall boys is equal to the number of ... | [
"Solution:\n\nConsider arbitrary arrangement of the boys along the circle. We put the signs \"+\" or \"-\" before any boy in accordance with the following rule: we move clockwise along the circle and put the sign \"+\" before the boy if he is taller than the previous boy and we put the sign \"-\" if he is shorter t... | Belarus | Belorusija 2012 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
03m1 | Problem:
A set of points is marked on the plane, with the property that any three marked points can be covered with a disk of radius $1$. Prove that the set of all marked points can be covered with a disk of radius $1$. | [
"Solution:\n(For a finite set of points only.) Let $D$ be a disk of smallest radius that covers all marked points. Consider the marked points on the boundary $C$ of this disk. Note that if all marked points on $C$ lie on an arc smaller than the half circle (ASTTHC for short), then the disk can be moved a little tow... | Canada | CANADIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Combinatorial Geometry > Helly's theorem",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | null | proof only | null | |
05mh | Problem:
Soit $ABC$ un triangle dont tous les angles sont aigus, et $H$ son orthocentre. Les bissectrices de $\widehat{ABH}$ et $\widehat{ACH}$ se coupent en un point $I$. Montrer que $I$ est aligné avec les milieux de $[BC]$ et de $[AH]$. | [
"Solution:\n\n\n\nNotons $D, E, F$ les pieds des hauteurs, $O$ le centre du cercle circonscrit, $M$ le milieu de $[BC]$ et $N$ le milieu de $[AH]$. On a $2 \\overrightarrow{MN} = 2 \\overrightarrow{ON} - 2 \\overrightarrow{OM} = \\overrightarrow{OA} + \\overrightarrow{OH} - \\overrightarrow... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilater... | null | proof only | null | |
0ei5 | Problem:
David si je izmislil trimestno naravno število s tremi različnimi števkami. Vsako števko tega števila je nadomestil s črko, da je dobil besedo $E N A$, ki predstavlja njegovo število. Opazil je, da za njegovo število velja $c \cdot E N A = 2331$, kjer je $c$ neko naravno število. Koliko je vrednost števila $c... | [
"Solution:\n\nPozitivni delitelji števila $2331 = 3^{2} \\cdot 7 \\cdot 37$ so $1, 3, 7, 9, 21, 37, 63, 111, 259, 333, 777$ in $2331$. Med njimi je le $259$ trimestno naravno število s tremi različnimi števkami. Torej je $E N A = 259$ in zato $c = 2331 : 259 = 9$."
] | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | C | |
0j4q | Problem:
Joe has written 5 questions of different difficulties for a test with problems numbered 1 through 5. He wants to make sure that problem $i$ is harder than problem $j$ whenever $i-j \geq 3$. In how many ways can he order the problems for his test? | [
"Solution:\n\nAnswer: 25\n\nWe will write $p_{i} > p_{j}$ for integers $i, j$ when the $i$th problem is harder than the $j$th problem. For the problem conditions to be true, we must have $p_{4} > p_{1}$, $p_{5} > p_{2}$, and $p_{5} > p_{1}$.\n\nThen, out of $5! = 120$ total orderings, we see that half of them satis... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 25 | |
0ezc | Problem:
The vertices of a regular $n$-gon are colored (each vertex has only one color). Each color is applied to at least three vertices. The vertices of any given color form a regular polygon. Show that there are two colors which are applied to the same number of vertices. | [] | Soviet Union | 4th ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof only | null | |
0d9n | Let $ABC$ be an acute, non-isosceles triangle with $M, N, P$ as the midpoints of $BC, CA, AB$, respectively. Denote $d_{1}$ as the line passing through $M$ and perpendicular to the angle bisector of $\angle BAC$, and similarly define $d_{2}, d_{3}$. Suppose that $d_{2} \cap d_{3} = D$, $d_{3} \cap d_{1} = E$, $d_{1} \c... | [
"Denote $(I_{a}), (I_{b}), (I_{c})$ as the ex-circles with respect to angles $A, B, C$ of triangle $ABC$. It is easy to check that $\\mathscr{P}_{M/(I)} = \\mathscr{P}_{M/(I_{a})}$ and $\\mathscr{P}_{M/(I_{b})} = \\mathscr{P}_{M/(I_{c})}$. Let $O, K$ be the circumcenters of triangles $ABC$ and $DEF$.\n\nHence, $d_{... | Saudi Arabia | Team selection tests for IMO 2018 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle ... | English | proof only | null | |
08ua | Let $k$ be a positive integer and $m$ an odd integer. Show that there exists a positive integer $n$ for which the number $n^n - m$ is divisible by $2^k$. | [
"We prove the assertion of the problem by using the mathematical induction on $k$.\n\nIf $k=1$ let $n=1$. Then, $n^n - m = 1 - m$ is divisible by $2$ since $m$ is odd. So the assertion holds.\n\nSuppose the assertion holds when $k = t$, we will show that the assertion holds when $k = t + 1$.\nFrom the induction hyp... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0kmu | The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?
(A) 105 (B) 120 (C) 135 (D) 150 (E) 165 | [
"Let $a$, $b$, and $h$ be the length of the congruent sides, the base, and the height to the base of the obtuse isosceles triangle, respectively. Then the area of the triangle is $\\frac{1}{2} b h$, which by the stated condition equals $\\frac{1}{4} a^2$. The area is also $\\frac{1}{2} a^2 \\sin \\theta$, where $\\... | United States | AMC 12 B | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | D | |
06o2 | After expanding the polynomial $(1 + x + y)^{2023}$ and collecting like terms, we obtain the expression
$$
a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + \dots + a_ky^{2023}
$$
Find the total number of $a_i$'s which are divisible by $5$. | [
"Note that $(1+x+y)^5 \\equiv 1+x^5+y^5 \\pmod{5}$ since the coefficients of all other terms are of the form $\\frac{5!}{i!(5-i)!}$ for some $i, j$ satisfying $0 \\le i+j \\le 5$ and $(i, j) \\ne (0, 0), (0, 5), (5, 0)$, all of which are divisible by $5$ as the denominator is not divisible by $5$. Inductively, we f... | Hong Kong | Hong Kong Team Selection Test 1 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | 2044800 | |
06qi | Let $n$ be a positive integer. Given a sequence $\varepsilon_{1}, \ldots, \varepsilon_{n-1}$ with $\varepsilon_{i}=0$ or $\varepsilon_{i}=1$ for each $i=1, \ldots, n-1$, the sequences $a_{0}, \ldots, a_{n}$ and $b_{0}, \ldots, b_{n}$ are constructed by the following rules:
$$
\begin{gathered}
a_{0}=b_{0}=1, \quad a_{1}... | [
"For a binary word $w=\\sigma_{1} \\ldots \\sigma_{n}$ of length $n$ and a letter $\\sigma \\in\\{0,1\\}$ let $w \\sigma= \\sigma_{1} \\ldots \\sigma_{n} \\sigma$ and $\\sigma w=\\sigma \\sigma_{1} \\ldots \\sigma_{n}$. Moreover let $\\bar{w}=\\sigma_{n} \\ldots \\sigma_{1}$ and let $\\emptyset$ be the empty word (... | IMO | IMO Problem Shortlist | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Linear Algebra > Linear transformations"
] | English | proof only | null | |
0fke | Problem:
La igualdad $2008=1111+444+222+99+77+55$ es un ejemplo de descomposición del número $2008$ como suma de números distintos de más de una cifra, cuya representación (en el sistema decimal) utiliza un sólo dígito.
i) Encontrar una descomposición de este tipo para el número $2009$.
ii) Determinar para el número... | [
"Solution:\n\nAgrupando los números con igual cantidad de cifras tendremos la ecuación $2009=1111 a+111 b+11 c$\ndonde $a, b$ y $c$ son números enteros menores o iguales que $1+2+3+\\cdots+9=45$, puesto que los sumandos de la descomposición han de ser diferentes.\nSe tiene entonces $2009=182 \\cdot 11+7=11(101 a+10... | Spain | XLV Olimpiada Matemática Española | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | i) One example: 2009 = 1111 + 777 + 66 + 55.
ii) The minimal number of summands is four. All minimal decompositions are:
- 1111 + 777 + 99 + 22
- 1111 + 777 + 88 + 33
- 1111 + 777 + 77 + 44
- 1111 + 777 + 66 + 55
- 999 + 888 + 111 + 11
- 999 + 777 + 222 + 11
- 999 + 666 + 333 + 11
- 999 + 555 + 444 + 11
- 888 + 777 + ... | |
0lfv | Problem:
There are given 100 distinct positive integers. We call a pair of integers among them good if the ratio of its elements is either $2$ or $3$. What is the maximum number $g$ of good pairs that these 100 numbers can form? (A same number can be used in several pairs.) | [
"Solution:\nLike so often in Russian problems, numbers are used instead of generic symbols. Let us therefore denote $10 = n > 1$, $2 = k > 1$, $3 = \\ell > 1$, with the extra condition both $k$ and $\\ell$ aren't powers of a same number. Consider the digraph $G$ whose set of vertices $V(G)$ is made of $v = n^2$ dis... | Zhautykov Olympiad | International Zhautykov Olympiad in Sciences | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 180 | |
0gvp | Prove that for every positive integer $n$ the inequality
$$
\frac{3}{1!+ 2!+ 3!} + \frac{4}{2!+ 3!+ 4!} + \dots + \frac{n+2}{n!+ (n+1)!+ (n+2)!} < \frac{1}{2}
$$
holds. (Here $1!=1$, $k!=1 \cdot 2 \cdot \dots \cdot (k-1) \cdot k$, $k \ge 2$.) | [
"$$\n\\begin{aligned}\n& \\frac{3}{1!+ 2!+ 3!} + \\frac{4}{2!+ 3!+ 4!} + \\dots + \\frac{n+2}{n!+ (n+1)!+ (n+2)!} = \\\\\n&= \\left( \\frac{1}{(1+1)!} - \\frac{1}{(1+2)!} \\right) + \\left( \\frac{1}{(2+1)!} - \\frac{1}{(2+2)!} \\right) + \\dots + \\left( \\frac{1}{(n+1)!} - \\frac{1}{(n+2)!} \\right) = \\\\\n&= \\... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0igz | Problem:
Let $x$ be a positive real number. Find the maximum possible value of
$$
\frac{x^{2}+2-\sqrt{x^{4}+4}}{x} .
$$ | [
"Solution:\n$2 \\sqrt{2}-2$\n\nRationalizing the numerator, we get\n$$\n\\begin{aligned}\n\\frac{x^{2}+2-\\sqrt{x^{4}+4}}{x} \\cdot \\frac{x^{2}+2+\\sqrt{x^{4}+4}}{x^{2}+2+\\sqrt{x^{4}+4}} & =\\frac{\\left(x^{2}+2\\right)^{2}-\\left(x^{4}+4\\right)}{x\\left(x^{2}+2+\\sqrt{x^{4}+4}\\right)} \\\\\n& =\\frac{4 x^{2}}{... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 2 sqrt(2) - 2 | |
0i1i | Problem:
Show that there exist infinitely many natural numbers $n$ with the following property: the sum of all the positive divisors of $n$, excluding $n$ itself, equals $n+12$. | [
"Solution:\n\nLet $p$ be any prime number greater than $3$; we show that $n = 6p$ has the desired property. The positive divisors of $6p = 2 \\cdot 3 \\cdot p$ are $1, 2, 3, p, 2 \\cdot 3, 2 \\cdot p, 3 \\cdot p$, and $2 \\cdot 3 \\cdot p$. The sum of all the factors other than $6p$ is equal to $6p + 12$, as needed... | United States | Berkeley Math Circle Monthly Contest #7 | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0iem | Problem:
Let $S=\{s_{0}, \ldots, s_{n}\}$ be a finite set of integers, and define $S+k=\{s_{0}+k, \ldots, s_{n}+k\}$. We say that $S$ and $T$ are equivalent, written $S \sim T$, if $T=S+k$ for some $k$. Given a (possibly infinite) set of integers $A$, we say that $S$ tiles $A$ if $A$ can be partitioned into subsets eq... | [
"Solution:\n\nOne example is $\\{1,3,4,6\\}$. Since its elements are all distinct modulo $4$, it tiles $\\mathbf{Z}$ by translation by multiples of $4$. On the other hand, it is easy to see that it cannot tile $\\mathbf{N}$: $1$ is contained in $\\{1,3,4,6\\}$, but then there is no possible set for $2$ to be contai... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Other",
"Number Theory > Other"
] | null | proof and answer | {1,3,4,6} | |
086v | Problem:
Alberto, Barbara e Carlo stanno giocando a carte. Ad ogni mano, il vincitore guadagna 2 punti, mentre gli altri due giocatori perdono un punto a testa. Inizialmente, tutti hanno 0 punti. Qual è la probabilità che, dopo 10 mani, siano nuovamente tutti a zero punti?
(A) 0
(B) $\frac{1}{5}$
(C) $\frac{1}{3}$
(D... | [
"Solution:\n\nLa risposta è (A). Vogliamo dimostrare che è impossibile che tutti i giocatori, dopo 10 mani, abbiano nuovamente zero punti.\n\nSia $v$ il numero di mani vinte dal primo giocatore quando tutti si ritrovano ad avere 0 punti, e $p$ il numero di mani da lui perse.\n\nAllora, per la regola data nel testo ... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | MCQ | A | |
0h7k | Let $f(x) = ax^2 + bx + c$ be a polynomial with integer coefficients. For every integer $x$ $f(x)$ is divisible by $N$ where $N$ is a positive integer. Is it true that $N$ necessarily divides all the coefficients of $f(x)$ if
$$
\text{a) } N = 2016; \quad \text{b) } N = 2017?
$$ | [
"a) Note that for any integer $x$ the product $x(x+1)$ is even. This suggests the following example:\n$$\n1008x(x+1)+2016=1008x^2+1008x+2016.\n$$\nFor every integer $x$, $f(x)$ is divisible by $2016$, but $2016$ does not divide all the coefficients ($1008$ is not divisible by $2016$). Thus, the answer is no.\n\nb) ... | Ukraine | UkraineMO | [
"Number Theory > Modular Arithmetic",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | a) No. b) Yes. | |
0grh | Let $H$ be the orthocenter of an acute triangle $ABC$. The circumcircle of $ABC$ and the circle with a diameter $[AH]$ meet at a point $E$ which is different from $A$. Let $M$ be the midpoint of the smaller arc $BC$ of circumcircle of triangle $ABC$, and $N$ be the midpoint of the greater arc $BC$ of circumcircle of tr... | [
"Let the lines $EM$ and $BC$ meet at $K$, $HV$ and $BC$ meet at $P$. By the Power Rule in circumcircle of triangle $ABC$, we get $BK \\cdot KC = EK \\cdot KM$, and by the Power Rule in circumcircle of triangle $BHC$, we get $BP \\cdot PC = HP \\cdot PN$. If we show that $P = K$, then we obtain that $EK \\cdot KM = ... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0l25 | Problem:
Let $ABC$ be an acute triangle and $D$ be the foot of altitude from $A$ to $\overline{BC}$. Let $X$ and $Y$ be points on the segment $\overline{BC}$ such that $\angle BAX = \angle YAC$, $BX = 2$, $XY = 6$, and $YC = 3$. Given that $AD = 12$, compute $BD$.
Proposed by: Sarunyu Thongjarast
Answer: $12 \sqrt{2... | [
"Solution:\n\n\n\nLet the line tangent to $\\odot(ABC)$ at $A$ intersect line $BC$ at $P$.\n\nProof. Note that\n$$\n\\angle PAX = \\angle PAB + \\angle BAX = \\angle PCA + \\angle CAY = \\angle PYA\n$$\n\nThis proves the desired tangency.\nNow, by power of point, we have,\n$$\n\\begin{align... | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Misce... | null | proof and answer | 12 sqrt 2 - 16 | |
0j7p | Let $P$ be a given point inside quadrilateral $ABCD$. Points $Q_1$ and $Q_2$ are located within $ABCD$ such that
$$
\angle Q_1 BC = \angle ABP, \quad \angle Q_1 CB = \angle DCP, \quad \angle Q_2 AD = \angle BAP, \quad \angle Q_2 DA = \angle CDP.
$$
Prove that $\overline{Q_1Q_2} \parallel \overline{AB}$ if and only if $... | [
"We will prove that the lines $\\overline{AB}$, $\\overline{CD}$, and $\\overline{Q_1Q_2}$ are either concurrent or all parallel. Let $X$ and $Y$ denote the reflections of $P$ across the lines $\\overline{AB}$ and $\\overline{CD}$.\n\nWe first claim that $XQ_1 = YQ_1$ and $XQ_2 = YQ_2$. Indeed, let $Z$ be the refle... | United States | USAMO | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | null | proof only | null | |
0ddh | Let $ABC$ be an acute, non-isosceles triangle with circumcenter $O$. Tangent lines to $(O)$ at $B$, $C$ meet at $T$. A line passes through $T$ cuts segments $AB$ at $D$ and cuts ray $CA$ at $E$. Take $M$ as midpoint of $DE$ and suppose that $MA$ cuts $(O)$ again at $K$. Prove that $(MKT)$ is tangent to $(O)$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Inversio... | null | proof only | null | |
0a07 | Problem:
Bepaal alle positieve gehele getallen $d$ waarvoor er een $k \geq 3$ bestaat zodat je de getallen $d, 2d, 3d, \ldots, kd$ op een rij kan zetten op zo'n manier dat voor elk tweetal buurgetallen geldt dat de som van die twee getallen een kwadraat is. | [
"Solution:\nVoor $d=1$ nemen we $k=15$ en de rij\n$$\n8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9\n$$\nTwee buurgetallen in deze rij zijn samen altijd $9$, $16$ of $25$. Voor $d>1$ een kwadraat nemen we ook $k=15$ en dezelfde rij als hierboven, maar dan met alle getallen vermenigvuldigd met $d$. Twee buurgeta... | Netherlands | IMO-selectietoets II | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | null | proof and answer | Exactly the perfect squares | |
01mt | Points $M$ and $L$ are the midpoints of the sides $AB$ and $BC$ of an isosceles triangle $ABC$ ($BC = AC$). Point $N$ is marked on the side $AC$ so that $NA + AM = LN = LM$.
Find the value of the angle $NLM$.
(Slovenia, 2010) | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | 36 degrees | |
0915 | Problem:
Find all integers $k$ such that for every integer $n$, the numbers $4 n+1$ and $k n+1$ are relatively prime. | [
"Solution:\n\nSince $4 n+1$ is odd, the identity $k-4 = k(4 n+1) - 4(k n+1)$ shows that $4 n+1$ and $k n+1$ are relatively prime if $k-4$ has not any odd divisor $p > 1$, i.e. if $k-4 = \\pm 2^{k}$ with any nonnegative integer $k$.\n\nOn the other hand, if $k-4$ has got an odd divisor $p > 1$, then we can easily fi... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | All integers k with k − 4 = ± 2^t for some nonnegative integer t. | |
0ie9 | Problem:
Let $x$, $y$, and $z$ be three roots of unity. Prove that $x + y + z$ is also a root of unity if and only if $x + y = 0$, $y + z = 0$, or $z + x = 0$. | [
"Solution:\n\nAgain, we consider the geometric picture. Arrange the vectors $x$, $y$, $z$, and $-x-y-z$ so as to form a quadrilateral. If they are all roots of unity, they form a quadrilateral all of whose side lengths are $1$. If the quadrilateral is degenerate, then two of the vectors sum to $0$, which implies th... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0fzx | Problem:
Bestimme alle Paare $(m, p)$ natürlicher Zahlen, sodass $p$ eine Primzahl und
$$
2^{m} p^{2}+27
$$
die dritte Potenz einer natürlichen Zahl ist. | [
"Solution:\n\nDie dritte Potenz $n^{3}$ ist ungerade, also auch $n$. Umformen führt zu\n$$\n2^{m} p^{2}=n^{3}-3^{3}\n$$\nDie rechte Seite kann man ausklammern als\n$$\n2^{m} p^{2}=(n-3)\\left(n^{2}+3 n+9\\right)\n$$\nDer zweite Faktor ist ungerade, folglich gilt $2^{m} \\mid(n-3)$. Ausserdem gilt $p^{2} \\mid\\left... | Switzerland | SMO - Finalrunde | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (1,7) | |
05yj | Problem:
Soit $ABC$ un triangle dont tous les angles sont aigus. Soit $F$ le pied de la hauteur de $ABC$ issue de $A$, et soit $P$ un point situé sur le segment $[AF]$. On note $D$ le point de $(BC)$ tel que $(PD)$ soit parallèle à $(AC)$, puis $X$ le point où se recoupent le cercle circonscrit à $ABD$ et le cercle de... | [
"Solution:\n\n\n\nDans un tel problème, le plus simple est d'essayer d'éliminer au mieux les points qui pourraient s'avérer peu utiles pour la suite. Ici, c'est par exemple le cas du point $P$, qui ne sert qu'à construire les points $D$ et $E$ en tant qu'images de $B$ et $C$ par une même ho... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
09nr | Let $ABCD$ be a quadrilateral with $AD = BC$. Point $E$ lies such that $E$ and $A$ are on opposite sides of line $CD$ such that $BC = CE = DE$. If $\angle BAE + \angle CBE = 90^\circ$, prove that $\angle ABE + \angle DAE = 90^\circ$. | [] | Mongolia | MMO2025 Round 2 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0dvs | Problem:
V jami pod Krimom spi grozna pošast. Ko postane lačna, se zbudi in požre toliko ovc, kolikor je vsota števk tistega leta. Potem spet zaspi za toliko let, kolikor ovc je pojedla. Vemo, da se je zbudila 12. aprila leta 666. Ali je pošast lahko pred vrati? Ali se bo lahko zbudila leta 3003? | [
"Solution:\n\nKer je število $666$ deljivo z $9$, je tudi vsota njegovih števk in zato število požrtih ovc deljivo z $9$. Tudi vsako naslednje leto, ko se pošast zbudi, je torej deljivo z $9$. Pošast se ne more zbuditi ne leta $2003$ ne leta $3003$."
] | Slovenia | 47. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | No; No | |
0f1w | Problem:
The circles $C_1$, $C_2$, $C_3$ with equal radius all pass through the point $X$. $C_i$ and $C_j$ also intersect at the point $Y_{ij}$. Show that $\angle X O_1 Y_{12} + \angle X O_2 Y_{23} + \angle X O_3 Y_{31} = 180^\circ$ where $O_i$ is the center of circle $C_i$. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jkn | Let $n$ be an even positive integer, and let $G$ be an $n$-vertex graph with exactly $\frac{n^2}{4}$ edges, where there are no loops or multiple edges (each unordered pair of distinct vertices is joined by either 0 or 1 edge). An unordered pair of distinct vertices $\{x, y\}$ is said to be *amicable* if they have a com... | [
"The key idea is to rephrase amicable pairs in terms of more familiar concepts. Let $N^k(v)$ denote the $k$th neighbor set of $v$, and $S$ the set of vertices of nonzero degree. Note that $N^2(v)$ contains $v$ for all $v \\in S$, so a simple double-counting argument yields\n$$\n\\begin{align*} \n2(\\text{number of ... | United States | IMO Team Selection Test | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0cyq | Let $n$ be a positive integer such that $2011^{2011}$ divides $n!$. Prove that $2011^{2012}$ divides $n!$. | [
"Since $2011$ is a prime and $2011^{2011}$ divides $n!$, it follows that in $n!$ we have at least $2011$ multiples of $2011$. These are\n$$\n2011, 2 \\cdot 2011, 3 \\cdot 2011, \\ldots, 2011 \\cdot 2011.\n$$\n\nTherefore $2011! \\cdot 2011^{2011}$ divides $n!$, hence $2011^{2012}$ divides $n!$."
] | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0bym | If the positive integers $a, b, c$ satisfy the inequalities $a > b > c$ and $12b > 13c > 11a$, show that $a + b + c \ge 56$. | [
"Since $a, b, c \\in \\mathbb{N}$, from $a > b > c$ follows $a \\ge b + 1 \\ge c + 2$, hence $a - c \\ge 2$.\nIf $a - c \\ge 4$, then $13c > 11a \\ge 11(c + 4)$, so $c > 22$. It follows that $a + b + c > c + c + c > 66 > 56$.\nSo we still need to discuss the cases $a - c = 2$ and $a - c = 3$.\nIf $a - c = 2$, then ... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0gg7 | 平面上 $ABC$ 為銳角三角形，其外心為 $O$，外接圓為 $\Omega$。分別在線段 $AB, AC$ 上各取一點 $D, E$，並作過 $A$ 與 $DE$ 垂直的直線 $\ell$。設 $\ell$ 分別與三角形 $ADE$ 的外接圓及 $\Omega$ 再交於點 $P, Q$。令直線 $OQ$ 與 $BC$ 交於點 $N$，直線 $OP$ 與 $DE$ 交於點 $S$，且點 $W$ 為三角形 $AOS$ 的垂心。
試證：$S, N, O, W$ 四點共圓。 | [
"Let $D', E'$ be points on lines $AB, AC$, respectively, such that $D'E'$ is parallel to $DE$, and passes through $N$. From\n$$\n\\angle ND'B = 90^\\circ - \\angle BAQ = \\angle OQB = \\angle NQB,\n$$\nwe see that $B, D', N, Q$ are concyclic. Since\n$$\n\\angle QD'E' = \\angle QBC = \\angle QAE',\n$$\n$A, D', Q, E'... | Taiwan | 2022 數學奧林匹亞競賽第三階段選訓營, 國際競賽實作(二) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chas... | Chinese; English | proof only | null | |
0khq | Isosceles triangle $ABC$ has $AB = AC = 3\sqrt{6}$, and a circle with radius $5\sqrt{2}$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$. What is the area of the circle that passes through vertices $A$, $B$, and $C$?
(A) $24\pi$ (B) $25\pi$ (C) $26\pi$ (D) $27\pi$ (E) $28\pi$ | [
"Let $O$ be the center of the circle with radius $5\\sqrt{2}$. Consider the circle with diameter $\\overline{AO}$. Because $\\angle ABO$ and $\\angle ACO$ are right angles, the opposite angles of quadrilateral $ABOC$ are supplementary, and hence this quadrilateral is cyclic. Thus $O$ is also on the circle that pass... | United States | AMC 10 A | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | MCQ | C | |
000u | Sean $x$, $y$, $z$ números reales positivos tales que $x^2 + y^2 + z^2 = 1$. Pruebe que
$$
x^2 y + x y^2 + z^2 y + z^2 x + 2 x y z \le \frac{1}{3}
$$ | [] | Argentina | XII Olimpíada Matemática Rioplatense | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | español | proof only | null | |
0eoj | When a water-tank is $30\%$ empty it has $30$ litres more in it than when it is $30\%$ full. How many litres can the tank hold when it is full? | [
"$75$\n\nIf $W$ is the number of litres the tank can hold, then when it is $30\\%$ empty it is holding $0.7W$ litres, and so we have $0.7W = 0.3W + 30$, i.e. $0.4W = 30$ and then $W = 30 \\div 0.4 = 300 \\div 4 = 75$."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | English | final answer only | 75 | |
0bku | It is known that the equations $x^4 + y^4 + z^4 = t^4$ and $x^4 + y^4 + z^4 = 2t^4$ (with unknowns $x$, $y$, $z$, $t$) have solutions in $\mathbb{N}^*$.
a) Find the smallest natural number $k$ such that the equation
$$
x^4 + y^4 + z^4 = k t^4
$$
has no solutions in $\mathbb{N}^*$.
b) Prove that for every $n \in \math... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 4 | |
080w | Problem:
È data una piramide avente per base un quadrilatero $A B C D$ e vertice $V$, inscritta in una sfera. Si sa che $A D=2 B C$, e che le rette ottenute prolungando $A B$ e $C D$ si incontrano in un punto $E$ dalla parte del segmento $B C$. Calcolare il rapporto fra il volume della piramide avente per base il tria... | [
"Solution:\n\nL'intersezione del piano su cui giace la base della piramide con la sfera in cui essa è inscritta è una circonferenza alla quale appartengono i punti $A, B, C, D$. Di conseguenza il quadrilatero $A B C D$ è inscrivibile in una circonferenza ed ha quindi gli angoli opposti supplementari (perché a due a... | Italy | Cesenatico | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Solid Geometry > Volume"
] | null | proof and answer | 3/4 | |
0gz5 | Solve the inequality:
$$
2008^x + \log_{2008}(x+1) > \log_{x+2} 2008 + \sin^{2007} x + \cos^{2008} x.
$$ | [
"**Answer:** $x > 0$.\n\nEasy to understand, that range of a function in our expression is set $x > 0$. We will show that this is the solution. Really when $x > 0$ $2008^x > 1$, $\\log_{2008}(x+1) > 0$, because under such conditions, the logarithm base satisfies the conditions $0 < \\frac{x}{x+2} < 1$, $\\sin^{2007... | Ukraine | The Problems of Ukrainian Authors | [
"Algebra > Equations and Inequalities"
] | English | proof and answer | x > 0 | |
0dyb | Problem:
Poišči vsa realna števila $x$ in $y$, ki zadoščajo enačbama
$$
\begin{aligned}
x^{3}+8 y^{3} & =x+2 y \\
2 x^{2} y+4 x y^{2} & =x+2 y
\end{aligned}
$$ | [
"Solution:\n\nEnačbi lahko prepišemo v obliko\n$$\n\\begin{aligned}\n(x+2 y)\\left(x^{2}-2 x y+4 y^{2}\\right) & =x+2 y \\\\\n2 x y(x+2 y) & =x+2 y\n\\end{aligned}\n$$\nOčitno vsak par števil $x$ in $y$, ki zadošča zvezi $x+2 y=0$, reši enačbi. Naj bo sedaj $x+2 y \\neq 0$. Tedaj lahko delimo z $x+2 y$ in dobimo $x... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | {(x,y) : x + 2y = 0} ∪ {(1, 1/2), (-1, -1/2)} | |
0kic | Problem:
Triangle $A B C$ has side lengths $A B=19$, $B C=20$, and $C A=21$. Points $X$ and $Y$ are selected on sides $A B$ and $A C$, respectively, such that $A Y = X Y$ and $X Y$ is tangent to the incircle of $\triangle A B C$. If the length of segment $A X$ can be written as $\frac{a}{b}$, where $a$ and $b$ are rel... | [
"Solution:\n\nNote that the incircle of $\\triangle A B C$ is the $A$-excenter of $\\triangle A X Y$. Let $r$ be the radius of this circle. We can compute the area of $\\triangle A X Y$ in two ways:\n$$\n\\begin{aligned}\nK_{A X Y} & = \\frac{1}{2} \\cdot A X \\cdot A Y \\sin A \\\\\n& = r \\cdot (A X + A Y - X Y) ... | United States | HMMT Spring 2021 Guts Round | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | final answer only | 6710 | |
0gf7 | 給定一個大於 $1$ 的正整數 $N$。甲乙兩人進行下面遊戲：
(1) 甲說一個正整數 $A$。
(2) 乙說一個正整數 $B > 1$，其中 $B$ 必須是 $A$ 的因數或是 $A$ 的倍數。
(3) 甲從 $B-1$, $B$, $B+1$ 中說一個數作為新的 $A$。
遊戲以下輪流進行步驟 (2) 和 (3)。若前 $50$ 個被說到的數字中有 $N$，則乙勝；否則甲勝。試回答下列問題。
i. 證明：當 $N=10$ 時，甲有必勝策略。
ii. 證明：當 $N=24$ 時，乙有必勝策略。
iii. 找出讓乙有必勝策略的所有大於 $1$ 的正整數 $N$。 | [] | Taiwan | 2022 數學奧林匹亞競賽第一階段培訓營, 獨立研究(二) | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | Chinese; English | proof and answer | i: First player wins for ten. ii: Second player wins for twenty-four. iii: Second player wins precisely for all integers greater than one that are divisible by six. | |
0hej | Find all pairs of integers $(x, y)$ that satisfy the equality: $x - y = \frac{x}{y}$. | [
"Let us rewrite the equality as follows: $xy - y^2 = x$, or $x(y-1) = y^2$. Since numbers $y^2$ and $y-1$ are co-prime, i.e. don't have common divisors greater than 1, then from the last equality, $x \\mid y^2$ and $y^2 \\mid x \\Rightarrow x = \\pm y^2$. Note that $y \\neq 0$.\nIf $x = y^2$, then $y^3 - y^2 = y^2 ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (4, 2) | |
088x | Problem:
Normalmente Davide ha bisogno di dormire almeno 8 ore per notte. Se una notte dorme $k$ ore meno di quanto gli occorra, si ritrova ad aver bisogno di $k$ ore in più di sonno per le $k$ notti successive. Ogni notte dorme comunque un numero intero di ore minore o uguale al suo fabbisogno. Ad esempio, se lunedì ... | [
"Solution:\n\nDavide ha dormito come minimo 58 ore.\n\nNumeriamo i giorni della settimana da 1 a 7 e supponiamo che Davide dorma $k_{1}$ ore meno del suo fabbisogno il lunedì, $k_{2}$ ore il martedì, ..., $k_{7}$ ore la domenica. Sappiamo che $k_{1} + \\cdots + k_{7} = 7$.\n\nPoiché trascorsa una settimana Davide a... | Italy | Olimpiadi della Matematica - Gara di Febbraio | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 58 | |
0lgh | Problem:
In a set of 20 elements there are $2k+1$ different subsets of 7 elements such that each of these subsets intersects exactly $k$ other subsets. Find the maximum $k$ for which this is possible. | [
"Solution:\n\nLet $M$ be the set of residues mod $20$. An example is given by the sets $A_{i} = \\{4i+1, 4i+2, 4i+3, 4i+4, 4i+5, 4i+6, 4i+7\\} \\subset M$, $i = 0, 1, 2, 3, 4$.\n\nLet $k \\geq 2$. Obviously among any three 7-element subsets there are two intersecting subsets.\n\nLet $A$ be any of the $2k+1$ subsets... | Zhautykov Olympiad | XIV International Zhautykov Olympiad in Mathematics | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2 | |
0hn1 | Problem:
Let $p$ be a prime number. Find all possible values of the remainder when $p^{2}-1$ is divided by $12$. | [
"Solution:\nThe answers are $3$, $8$, and $0$.\n\nIt is clear that $p=2$ gives $3$, $p=3$ gives $8$, and $p=5$ gives $0$. We claim that all primes $p \\geq 5$ give the remainder $0$ as well, i.e. that $p^{2}-1$ is divisible by $12$ for these $p$.\n\nWe factor:\n$$\np^{2}-1=(p+1)(p-1)\n$$\nSince $p \\neq 2$, $p$ is ... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 3, 8, 0 | |
0abi | Prove that there is an angle $x$ such that
$$
sin x = \frac{\sin \beta \cdot \sin \gamma}{1 - \cos \alpha \cdot \cos \beta \cdot \cos \gamma},
$$
for every angle $\alpha$, and $\beta$ and $\gamma$ acute angles. | [
"We will prove that\n$$\n\\frac{\\sin \\beta \\cdot \\sin \\gamma}{1 - \\cos \\alpha \\cdot \\cos \\beta \\cdot \\cos \\gamma}\n$$\nbelongs in $[-1, 1]$.\n\nSince $\\beta$ and $\\gamma$ are acute angles we have $\\cos \\beta > 0$, $\\cos \\gamma > 0$, so $\\cos \\beta \\cdot \\cos \\gamma > 0$. Because $\\cos \\alp... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
019h | Determine all sequences $a_0, a_1, a_2, \dots$ of positive reals such that
$$
a_{n^2+m^2} = a_n a_m^n
$$
for all $n, m$. | [
"These sequences are given by $a_n = c^n$ for any positive real $c$.\n\n*Proof:* Evidently $a_n = c^n$ satisfies the equation. To prove that there are no other solutions we show that $a_1$ determines the entire sequence. This proves the claim because any $a_1$ may be obtained by a choice of $c$. To this end notice\... | Baltic Way | Baltic Way 2013 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | a_n = c^n for any positive real c | |
00ux | Let $n$, $k$ be positive integers. Julia and Florian play a game on a $2n \times 2n$ board. Julia has secretly tiled the entire board with invisible dominos. Florian now chooses $k$ cells. All dominos covering at least one of these cells then turn visible. Determine the minimal value of $k$ such that Florian has a stra... | [
"The minimal value of $k$ with this property is $n^2$.\nWe first show that in order for Florian to be able to deduce the entire tiling, we must have $k \\ge n^2$. If Julia picks an independent tiling on each of $n^2$ disjoint $2 \\times 2$ regions, then Florian needs to reveal at least one domino from each region t... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | n^2 | |
0frz | Problem:
Let $\mathbb{N}$ be the set of positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for every $m, n \in \mathbb{N}$
$$
f(m)+f(n) \mid m+n
$$ | [
"Solution:\nSoit $f$ une solution. Avec $m=n=1$, on obtient $f(1) \\mid 1$ et donc $f(1)=1$. Posons à présent $n=1$. On obtient\n$$\nf(m)+1 \\mid m+1\n$$\nIl serait intéressant de rendre le côté droit premier. Posons donc $m=p-1$ pour $p$ un nombre premier. On a donc $f(p-1)+1 \\mid p$ et comme $f(p-1)+1>1$, forcém... | Switzerland | null | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | f(n) = n for all n in N | |
027o | Problem:
Se $n$ e $k$ são inteiros positivos, então
$$
(n+1) \cdot(n+2) \cdot \ldots \cdot(n+k)<(n+k)^{k}
$$
Use isto para determinar qual dos dois números a seguir é maior que o outro:
$(100!)$ ! e $99!^{100!} \cdot 100!^{99!}$. | [
"Solution:\nSe $a$ é um inteiro arbitrário, usando a dica do enunciado e multiplicando as 100 desigualdades a seguir\n$$\n\\begin{aligned}\n1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot a & <a^{a} \\\\\n(a+1)(a+2)(a+3) \\cdot \\ldots \\cdot(2 a) & <(2 a)^{a} \\\\\n(2 a+1)(2 a+2)(2 a+3) \\cdot \\ldots \\cdot(3 a) & <(3 ... | Brazil | NÍVEL 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 99!^{100!} · 100!^{99!} is larger | |
02ye | Problem:
a) Verifique que para qualquer inteiro positivo $a$, com $a>1$, a equação
$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{a}
$$
possui pelo menos três soluções da forma $(x, y)$, com $x$ e $y$ inteiros positivos. Por exemplo, para $a=3$, os pares $(6,6),(4,12)$ e $(12,4)$ são soluções.
b) Encontre o número de pares de in... | [
"Solution:\na) Podemos encontrar uma equação equivalente:\n$$\n\\begin{aligned}\n\\frac{1}{x}+\\frac{1}{y} & =\\frac{1}{a} \\Leftrightarrow \\\\\n(x-a)(y-a) & =a^{2}\n\\end{aligned}\n$$\nComo $1 / x$ e $1 / y$ são menores que $1 / a$, segue que $x-a$ e $y-a$ são positivos. Para encontrarmos soluções dessa última eq... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 9 | |
0guz | Let $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$ be sequences of real numbers such that $a_0 = 40$, $b_0 = 41$ and for all $n \ge 0$ we have
$$
a_{n+1} = a_n + \frac{1}{b_n} \quad \text{and} \quad b_{n+1} = b_n + \frac{1}{a_n}
$$
Find the smallest positive integer $k$ such that $a_k > 80$. | [
"Answer: 2460.\nSince\n$$\n\\frac{a_{n+1}}{b_{n+1}} = \\frac{a_n + \\frac{1}{b_n}}{b_n + \\frac{1}{a_n}} = \\frac{a_n}{b_n}\n$$\nwe get that $\\frac{a_n}{b_n}$ is a constant and hence equals $\\frac{40}{41}$. Therefore, $a_k > 80$ is equivalent to $b_k > 82$ and $a_k b_k > 80 \\cdot 82 = 6560$. Multiplying both seq... | Turkey | Team Selection Test for JBMO 2024 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 2460 | |
051l | In the product
$$
\left(1 + \frac{1}{1}\right) \cdot \left(1 + \frac{1}{3}\right) \cdot \left(1 + \frac{1}{5}\right) \cdots \left(1 + \frac{1}{2n-1}\right)
$$
the denominators of the fractions are all odd numbers from $1$ to $(2n - 1)$. Is it possible to choose a natural number $n > 1$ such that this product would eval... | [
"By manipulating the given product we get\n$$\n\\begin{aligned} \\left(1 + \\frac{1}{1}\\right) \\cdot \\left(1 + \\frac{1}{3}\\right) \\cdot \\left(1 + \\frac{1}{5}\\right) \\cdots \\left(1 + \\frac{1}{2n-1}\\right) &= \\frac{2}{1} \\cdot \\frac{4}{3} \\cdot \\frac{6}{5} \\cdots \\frac{2n}{2n-1} \\\\ &= \\frac{2 \... | Estonia | Final Round of National Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No | |
0iij | For a given positive integer $k$ find, in terms of $k$, the minimum value of $N$ for which there is a set of $2k+1$ distinct positive integers that has sum greater than $N$ but every subset of size $k$ has sum at most $N/2$. | [
"The minimum is $N = 2k^3 + 3k^2 + 3k$. The set\n$$\n\\{k^2+1, k^2+2, \\dots, k^2+2k+1\\}\n$$\nhas sum $2k^3 + 3k^2 + 3k + 1 = N + 1$ which exceeds $N$, but the sum of the $k$ largest elements is only $(2k^3 + 3k^2 + 3k)/2 = N/2$. Thus this $N$ is such a value.\n\nSuppose $N < 2k^3 + 3k^2 + 3k$ and there are positi... | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | N = 2k^3 + 3k^2 + 3k | |
09l3 | Find all three-digit numbers that can be obtained by adding the product of their digits to seven times the sum of their digits. | [
"Let the three-digit number be $N = 100a + 10b + c$, where $a, b, c$ are digits and $a \\neq 0$.\n\nWe are told that $N = 7S + P$, where $S = a + b + c$ is the sum of the digits and $P = abc$ is the product of the digits.\n\nSo:\n$$\n100a + 10b + c = 7(a + b + c) + abc\n$$\n$$\n100a + 10b + c - 7a - 7b - 7c = abc\n... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | No three-digit numbers satisfy the condition. | |
0lck | The diagonals of a convex pentagon cut together divide the pentagon into one small pentagon and 10 sub-triangles. Find the maximum number of triangles among these sub-triangles such that they have the same area. | [
"\n\nIt is easy to see that the triples $(1,2,6)$, $(2,3,7)$, $(3,4,8)$, $(4,5,9)$, $(5,1,10)$ cannot have the same area because if we suppose that three triangles $1,2,6$ have the same area, then quadrilateral $ABNR$ is a parallelogram and $AR \\parallel BN$, a contradiction. The other cas... | Vietnam | Vietnamese Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 6 | |
0hmk | Problem:
In a certain kingdom, the only coin values are $3$ and $5$. Determine all possible amounts of money you can have using only these coins. | [
"Solution:\n\nThe amounts of money that you can have are all positive integers except $1, 2, 4$, and $7$. We can see that all these are impossible. To show that all other amounts are possible, it is clear that $3$ and $5$ are possible. Also, $6, 8, 9$, and $10$ are all possible, since\n$$\n\\begin{aligned}\n6 & = 3... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | All positive integers except 1, 2, 4, and 7. | |
06h2 | For any positive real numbers $a$, $b$, $c$ satisfying $a + b + c = 1$, prove that
$$
\sqrt{a^2 - 2bc + 2c} + \sqrt{b^2 - 2ca + 2a} + \sqrt{c^2 - 2ab + 2b} \ge \sqrt{5}.
$$ | [
"Using the condition $a + b + c = 1$, we obtain\n$$\na^2 - 2bc + 2c = a^2 + 2c(1 - b) = a^2 + 2c(a + c) = (c + a)^2 + c^2.\n$$\nSimilarly, the left-hand side of the inequality becomes\n$$\n\\sqrt{(c + a)^2 + c^2} + \\sqrt{(a + b)^2 + a^2} + \\sqrt{(b + c)^2 + b^2}.\n$$\nBy the triangle inequality, this is bounded b... | Hong Kong | IMO HK TST | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0cqg | A sequence $a_1, a_2, \dots$ is defined in the following way: $a_1 = 1$, $a_2 = 143$, and $a_{n+1} = 5 \cdot \frac{a_1 + a_2 + \dots + a_n}{n}$ for $n \ge 2$. Prove that $a_n$ is an integer for every $n \ge 1$.
Последовательность чисел $a_1, a_2, \dots$ задана условиями $a_1 = 1, a_2 = 143$ и $a_{n+1} = 5 \cdot \frac{... | [
"Show by induction that $a_n = (n+3)(n+2)(n+1)n$ for all $n \\ge 3$.\n\nFirst solution. The number $a_3 = 5 \\cdot 72 = 360$ is an integer. For $n \\ge 4$ the following equalities hold:\n$$\na_n = 5 \\cdot \\frac{a_1 + \\dots + a_{n-1}}{n-1} \\quad \\text{and} \\quad a_{n-1} = 5 \\cdot \\frac{a_1 + \\dots + a_{n-2}... | Russia | Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English; Russian | proof only | null | |
0fg4 | Problem:
Dado un polinomio $P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{k} x^{k}$ con coeficientes enteros, denotamos por $w(P)$ el número de coeficientes impares de $P$. Sea $Q_{i}(x)=(1+x)^{i}$, para $i=0,1, \ldots$ Demostrar que si $i_{1}, i_{2}, \ldots, i_{n}$ son enteros tales que $0 \leq i_{1}<i_{2}<\cdots<i_{n}$, ... | [] | Spain | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
062t | Problem:
Wir nennen eine natürliche Zahl $n$ ausgeglichen, wenn $n=1$ gilt oder wenn $n$ als Produkt einer geraden Anzahl von (nicht notwendigerweise verschiedenen) Primfaktoren geschrieben werden kann. Zu jedem Paar $(a, b)$ positiver ganzer Zahlen sei $P(x) = (x+a)(x+b)$.
a) Gibt es zwei verschiedene positive ganze... | [
"Solution:\n\na) Die Antwort lautet „Ja\". Damit $P(x)$ ausgeglichen ist, müssen $x+a$ und $x+b$ entweder beide eine gerade oder beide eine ungerade Anzahl von Primfaktoren besitzen. Hinsichtlich dieser Eigenschaft gibt es aber nur $2^{50}$, also endlich viele verschiedene Muster bei 50 aufeinanderfolgenden natürli... | Germany | IMO-Auswahlklausur | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | a) Yes. b) a = b. | |
004s | Se tienen dos tableros $A$ y $B$, ambos de $8 \times 8$.
Juan escribe un número en cada casilla del tablero $A$. Para cada casilla del tablero $A$, Juan suma el número escrito en dicha casilla con la suma de los números escritos en sus casillas vecinas, y luego escribe el resultado en la casilla que ocupa el mismo luga... | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Vectors"
] | Spanish | proof and answer | Exactly the four squares at the intersections of the third and sixth rows with the third and sixth columns. | |
0jda | Problem:
Evaluate $2+5+8+\cdots+101$. | [
"Solution:\n\nThere are $\\frac{102}{3}=34$ terms with average $\\frac{2+101}{2}$, so their sum is $17 \\cdot 103=1751$."
] | United States | HMMT November 2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 1751 | |
03mx | The integers $1$, $2$, $3$, $\ldots$, $2016$ are written on a board. You can choose any two numbers on the board and replace them with their average. For example, you can replace $1$ and $2$ with $1.5$, or you can replace $1$ and $3$ with a second copy of $2$. After $2015$ replacements of this kind, the board will have... | [
"a.\nFirst replace $2014$ and $2016$ with $2015$, and then replace the two copies of $2015$ with a single copy. This leaves us with $\\{1, 2, \\ldots, 2013, 2015\\}$. From here, we can replace $2013$ and $2015$ with $2014$ to get $\\{1, 2, \\ldots, 2012, 2014\\}$. We can then replace $2012$ and $2014$ with $2013$, ... | Canada | Kanada | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0afa | Реши го системот равенки
$$
\begin{cases}
x + y = z \\
x^2 + y^2 = z \\
x^3 + y^3 = z
\end{cases}
$$ | [
"Третата равенка на системот ќе ја трансформираме во облик $x^3 + y^3 = (x+y)^3 - 3xy(x+y)$, односно $z = z^3 - 3xyz$. Притоа од формулата за бином на квадрат имаме $xy = \\frac{1}{2}[(x+y)^2 - (x^2 + y^2)]$ и конечно третата равенка на системот добива облик $z = z^3 - \\frac{3}{2}z(z^2-z)$. Со средување истата ста... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | Macedonian, English | proof and answer | {(0,0,0), (0,1,1), (1,0,1), (1,1,2)} | |
09ss | Problem:
Vind alle positieve gehele $k$ waarvoor de vergelijking
$$
\operatorname{kgv}(m, n)-\operatorname{ggd}(m, n)=k(m-n)
$$
geen positieve gehele oplossingen $(m, n)$ met $m \neq n$ heeft. | [
"Solution:\nNoem $d=\\operatorname{ggd}(m, n)$ en schrijf $m=d a$ en $n=d b$. Er geldt $\\operatorname{kgv}(m, n) \\cdot \\operatorname{ggd}(m, n)=m n$, dus we kunnen de gegeven vergelijking schrijven als\n$$\n\\frac{d a \\cdot d b}{d}-d=k(d a-d b)\n$$\noftewel\n$$\na b-1=k(a-b) .\n$$\nWe kijken vanaf nu naar het e... | Netherlands | IMO-selectietoets | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2 | |
09ub | In the puzzle below, $a, b, c, d$, and $e$ are nonzero digits such that the two calculations are correct. The digits need not be distinct.
How many solutions are there for which $a < b$?
$ab \times ab = cde \quad \text{and} \quad ba \times ba = edc.$
A) 1 B) 2 C) 3 D) 4 E) 5 | [
"B) 2"
] | Netherlands | Junior Mathematical Olympiad, September 2019 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | B | |
03ug | In a round robin chess tournament each player plays with every other player exactly once. The winner of each game gets $1$ point and the loser gets $0$ point. If the game ends in a tie, each player gets $0.5$ point. Given a positive integer $m$, a tournament is said to have property $P(m)$ if the following holds: among... | [
"Note that if there are $2m-4$ players, we can label them\n$$\na_1, a_2, \\dots, a_{m-3}, A_{m-2}, B_{m-2}, a_{m-1}, \\dots, a_{2m-5},\n$$\nand assume that player $P_i$ beats player $P_j$ if and only if $i > j$, and $A_{m-2}$ and $B_{m-2}$ are in a tie. It is easy to see that in the group of $m$ players, there exis... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2m - 3 | |
09qk | Problem:
We bekijken betegelingen van een rechthoekig $m \times n$-bord met $1 \times 2$-tegels. De tegels mogen zowel horizontaal als verticaal liggen, maar ze mogen elkaar niet overlappen en niet buiten het bord uitsteken. Alle velden van het bord moeten bedekt worden door een tegel.
a) Bewijs dat bij elke betegeli... | [
"Solution:\n\na) Noem een rechte lijn die het bord in twee stukken verdeelt zodat elke tegel in zijn geheel binnen één van de stukken ligt, een scheidingslijn. Stel dat er een betegeling bestaat waar geen scheidingslijn te vinden is. Bekijk de kolommen $k$ en $k+1$, met $1 \\leq k \\leq 2009$. Er moet dan een horiz... | Netherlands | Dutch TST toets 8 juni | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0lfh | Find all polynomials $P(x)$, $Q(x)$ with real coefficients, such that for each real number $a$, $P(a)$ is the solution of the following equation
$$
x^{2023} + Q(a)x^2 + (a^{2024} + a)x + a^3 + 2025a = 0.
$$ | [
"From the hypothesis, we have\n$$\nP(x)^{2023} + Q(x)P(x)^2 + (x^{2024} + x)P(x) + x(x^2 + 2025) = 0, \\quad (1)\n$$\nfor all $x \\in \\mathbb{R}$. From that, we deduce that the polynomial $x(x^2 + 2025)$ is divisible by the polynomial $P(x)$. Thus, the polynomial $P(x)$ must have one of the following forms $k$, $k... | Vietnam | Vietnamese MO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All solutions are P(x) = k with k a nonzero real constant, and Q(x) = [-k(x^{2024} + x) - x^3 - 2025x - k^{2023}]/k^2. | |
0jpv | Problem:
Call a set of positive integers good if there is a partition of it into two sets $S$ and $T$, such that there do not exist three elements $a, b, c \in S$ such that $a^{b}=c$ and such that there do not exist three elements $a, b, c \in T$ such that $a^{b}=c$ ($a$ and $b$ need not be distinct). Find the smalles... | [
"Solution:\n\nFirst, we claim that the set $\\{2,4,8,256,65536\\}$ is not good. Assume the contrary and say $2 \\in S$. Then since $2^{2}=4$, we have $4 \\in T$. And since $4^{4}=256$, we have $256 \\in S$. Then since $256^{2}=65536$, we have $65536 \\in T$. Now, note that we cannot place $8$ in either $S$ or $T$, ... | United States | HMMT November 2015 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 65536 | |
083b | Problem:
Data una griglia $m \times n$ ($m, n \geq 1$) si dispone una pedina al centro di ogni casella e una in ogni vertice della griglia (nell'esempio mostrato a fianco con $n=4$ e $m=3$ si dispongono 32 pedine).
a. Trovare tutte le tabelle che hanno esattamente 500 pedine.
b. Dimostrare che esistono infiniti inte... | [
"Solution:\n\nData una griglia $m \\times n$, il numero di pedine poste sui vertici della griglia è $(m+1)(n+1)$, mentre il numero di pedine poste al centro di ogni casella è $m n$, sicché il totale di pedine è\n$$\nN = 2 m n + m + n + 1.\n$$\nOsserviamo che\n$$\n2N - 1 = 4 m n + 2 m + 2 n + 1 = (2m + 1)(2n + 1).\n... | Italy | XIX GARA NAZIONALE DI MATEMATICA | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All m-by-n grids with exactly 500 tokens are (m,n) = (1,166), (4,55), (13,18), (18,13), (55,4), (166,1). Moreover, there are no grids with exactly N tokens whenever 2N − 1 is prime, yielding infinitely many unattainable N. | |
0aq9 | Problem:
In $\triangle ABC$, let $D$, $E$, and $F$ be points on sides $BC$, $CA$, and $AB$, respectively, so that the segments $AD$, $BE$, and $CF$ are concurrent at point $P$. If $AF : FB = 4 : 5$ and the ratio of the area of $\triangle APB$ to that of $\triangle APC$ is $1 : 2$, determine $AE : AC$. | [] | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 4:11 | |
0a5h | Problem:
Is it possible to place a positive integer in every cell of a $10 \times 10$ array in such a way that both the following conditions are satisfied?
- Each number (not in the top row) is a proper divisor of the number immediately below.
- Each row consists of 10 consecutive positive integers (but not necessarily... | [
"Solution:\nAnswer: Yes. In fact it is even possible to achieve such an array where each row consists of ten consecutive positive integers in increasing order. We shall construct an example explicitly.\n\nInitially let the top row be $(1,2,3,\\ldots ,10)$ in this order. Then iteratively if the contents of a particu... | New Zealand | NZMO Round One | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Yes | |
06tv | In a convex pentagon $A B C D E$, let $F$ be a point on $A C$ such that $\angle F B C = 90^{\circ}$. Suppose triangles $A B F$, $A C D$ and $A D E$ are similar isosceles triangles with
$$
\angle F A B = \angle F B A = \angle D A C = \angle D C A = \angle E A D = \angle E D A.
$$
Let $M$ be the midpoint of $C F$. Point ... | [
"Denote the common angle in (1) by $\\theta$. As $\\triangle A B F \\sim \\triangle A C D$, we have $\\frac{A B}{A C} = \\frac{A F}{A D}$ so that $\\triangle A B C \\sim \\triangle A F D$. From $E A = E D$, we get\n$$\n\\angle A F D = \\angle A B C = 90^{\\circ} + \\theta = 180^{\\circ} - \\frac{1}{2} \\angle A E D... | IMO | IMO 2016 Shortlisted Problems | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triang... | English | proof only | null | |
0afo | Да се определат сите природни броеви $n$ за кои броевите $n+2$ и $n^2+n+1$ се точни кубови на природни броеви. | [
"Нека претпоставиме дека $n$ е природен број за кој $n+2$ и $n^2+n+1$ се точни кубови. Тогаш јасно бројот $(n+2)(n^2+n+1)$ е исто така точен куб на природен број. Но\n$$\n(n+2)(n^2+n+1) = n^3 + 3n^2 + 3n + 2 = (n+1)^3 + 1,\n$$\nи јасно не може да биде точен куб. Значи не постои природен број $n$ за кој се исполнети... | North Macedonia | Републички натпревар по математика за основно образование | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Macedonian, English | proof and answer | no such natural number |
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