id stringlengths 4 4 | problem_markdown stringlengths 36 3.59k | solutions_markdown listlengths 0 10 | images images listlengths 0 15 | country stringclasses 58
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|---|---|---|---|---|---|---|---|---|---|
039j | For a finite set $S$ denote by $P(S)$ the set of all subsets of $S$ (including the empty set and the set $S$). The function $f : P(S) \to \mathbb{R}$ is such that
$$
f(X \cap Y) = \min(f(X), f(Y))
$$
for any two $X, Y \in P(S)$. Find the largest number of distinct values that $f$ can take. | [] | Bulgaria | First selection test for IMO 2007, Vietnam | [
"Discrete Mathematics > Combinatorics > Functional equations"
] | English | proof and answer | |S| + 1 | |
0b9l | An infinite set $A$ of real numbers contains at least an irrational number. Prove that for each positive integer $n$ it is possible to find $n$ elements of $A$ whose sum is irrational. | [
"Let $a_1$ be an irrational number in $A$. If $A$ contains infinitely many irrational numbers, then for any $n$ we can pick $n$ distinct irrational numbers from $A$; their sum is irrational (since the sum of irrational numbers is irrational unless they sum to a rational, but with infinitely many choices, we can avo... | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof only | null | |
0f7k | Problem:
Players take turns in choosing numbers from the set $\{1, 2, 3, \ldots, n\}$. Once $m$ has been chosen, no divisor of $m$ may be chosen. The first player unable to choose a number loses. Who has a winning strategy for $n = 10$? For $n = 1000$? | [] | Soviet Union | 21st ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | First player for n=10; First player for n=1000. | |
0ec1 | Problem:
Določi parameter $a \in \mathbb{R}$ tako, da bo graf funkcije $f(x)=\frac{3-2a}{a+5} x+\frac{2a-1}{3-a}$ sekal ordinatno os nad koordinatnim izhodiščem in da bo funkcija $f$ padajoča. | [] | Slovenia | 15. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (3/2, 3) | |
0g84 | 已知 $a, b, c$ 為正實數, 試證不等式
$$
3(a + b + c) \geq 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3 + b^3 + c^3}{3}}
$$
Let $a, b, c$ be positive real numbers. Prove that
$$
3(a + b + c) \geq 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3 + b^3 + c^3}{3}}
$$ | [
"由算幾不等式得\n$$\n8\\sqrt[3]{abc} + \\sqrt[3]{\\frac{a^3 + b^3 + c^3}{3}} \\le 9\\sqrt[3]{\\frac{8abc + \\frac{a^3+b^3+c^3}{3}}{9}} = 3\\sqrt[3]{a^3 + b^3 + c^3 + 24abc}\n$$\n故只須證明\n$$\n3(a + b + c) \\ge 3\\sqrt[3]{a^3 + b^3 + c^3 + 24abc}\n$$\n或\n$$\n(a + b + c)^3 \\ge a^3 + b^3 + c^3 + 24abc.\n$$\n將上式展開, 等價於\n$$\na^2... | Taiwan | 二〇一四數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
05k9 | Problem:
Soient $\omega_{1}$ et $\omega_{2}$ deux cercles sécants en deux points $X$ et $Y$. Un cercle $\omega$ est tangent intérieurement à $\omega_{1}$ et $\omega_{2}$ en $P$ et $Q$ respectivement. Le segment $[X, Y]$ coupe $\omega$ en $M$ et $N$. Les demi-droites $[P, M)$ et $[P, N)$ coupent $\omega_{1}$ en $A$ et ... | [
"Solution:\n\nSoit $\\ell$ la tangente en $M$ à $\\omega$ et $\\ell_{1}$ la tangente en $A$ à $\\omega_{1}$. Comme $\\omega$ est tangent en $P$ à $\\omega_{1}$, il existe une homothétie $h$ de centre $P$ qui envoie $\\omega$ sur $\\omega_{1}$. Or, $P, M, A$ sont alignés donc $h(M)=A$. On en déduit que $h(\\ell)=\\e... | France | null | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03ao | Find all positive integers $n$ and all primes $p$ for which the following equality holds $n^3 - p^5 = p^6 + 3n - 2$. | [] | Bulgaria | Selection test for 27. Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials ... | English | proof and answer | (n, p) = (10, 3) | |
0gbb | 令 $N$ 表示所有正整數所成之集合。試求所有函數 $f: N \rightarrow N$, 滿足: 對於所有 $m, n \in N$, $f(m) + f(n) - mn \neq 0$ 且 $(f(m) + f(n) - mn)$ 能整除 $(mf(m) + nf(n))$. | [
"$f(n) = n^2$.\n\nIt is given that\n$$\n(f(m) + f(n) - mn) \\mid (mf(m) + nf(n)). \\quad (1)\n$$\n\n1. Take $m = n = 1$ in (1), we have $2f(1) - 1 \\mid 2f(1) \\Rightarrow f(1) = 1$.\n\n2. Let $p$ be a prime number, and take $(m, n) = (p, 1)$, then we have $f(p) - 1 + p \\mid pf(p) + 1$, which is\n$$\nf(p) - p + 1 ... | Taiwan | 二〇一七數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | f(n) = n^2 | |
0fv8 | Problem:
Finde alle natürlichen Zahlen $k$, sodass $3^{k}+5^{k}$ eine Potenz einer natürlichen Zahl mit Exponent $\geq 2$ ist. | [
"Solution:\n\nNehme an, es sei $3^{k}+5^{k}=n^{t}$ mit $t \\geq 2$. Ist $k$ gerade, dann gilt $3^{k}+5^{k} \\equiv 1+1=2$ $(\\bmod 4)$, also ist die linke Seite gerade aber nicht durch 4 teilbar, ein Widerspruch zu $t \\geq 2$. Ist $k$ ungerade, dann gilt\n$$\n3^{k}+5^{k}=(3+5)\\left(3^{k-1}-3^{k-2} \\cdot 5+\\ldot... | Switzerland | IMO Selektion | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | 1 | |
05va | Problem:
Un graphe $G$ fini simple à $n$ sommets est dit divisible s'il est possible d'attribuer à chaque sommet $s$ de $G$ un numéro $n_{s}$ de sorte que deux sommets distincts possèdent toujours deux numéros distincts et deux sommets quelconques $s$ et $s^{\prime}$ sont reliés par une arête si et seulement si $n_{s}... | [
"Solution:\n\nUn graphe est de divisibilité ssi on peut l'orienter (dans le sens des divisibilités) de manière à ce que s'il y a un arc (arête orientée) de $A$ vers $B$ et un de $B$ vers $C$, il y en a un de $A$ vers $C$. En effet, on peut alors attribuer à chaque sommet un nombre premier différent, et le nombre at... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Algorithms",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0h1t | a) Rectangle $ABCD$ is partitioned into squares, each of which has integer perimeter. Is it true that $ABCD$ has integer perimeter?
b) Square $ABCD$ is partitioned into squares, each of which has integer perimeter. Is it true that $ABCD$ has integer perimeter? | [
"a)\nWe construct a counterexample. Consider two squares $ABMN$ and $NMCD$ with side length $\\frac{1}{4}$. Then, perimeter of the rectangle $ABCD$ is $2 \\cdot (\\frac{1}{2} + \\frac{1}{4}) = \\frac{3}{2}$ — non-integer, while both squares have integer perimeter.\n\nb)\nConsider the side of the external square, an... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | a) No. b) Yes. | |
03ay | Given a $\triangle ABC$ with $\angle ACB = 90^\circ$ and altitude $CH$ ($H \in AB$). A circle $k$ touches the segments $BH$ and $CH$, and the circumcircle of $\triangle ABC$. If $D = k \cap BH$, prove that $CD$ bisects $\angle BCH$. | [] | Bulgaria | Selection test for 27. Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ch1 | Let $x_0$ be a real number. Prove that there exists a useful function $f$ so that $f(x_0) = 0$ if and only if $|x_0| \ge 4$.
We will call an affine function $f : \mathbb{R} \to \mathbb{R}$ *useful* if it has the properties:
(i) $|f(x)| \le 2$, for every real number $x$ such that $|x| \le 2$;
(ii) $|f(x)| \ge 1$, for e... | [
"\"$\\Rightarrow$\" If $f$ is useful, then $-f$ has properties (i) and (ii), hence it is useful. Consequently, we may look only at the useful functions $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = a x + b$, with $a > 0$, $b \\in \\mathbb{R}$.\n\nSince $a > 0$, the inequality (i) is equivalent to $f(2) \\le 2$ and $f... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0c5v | Let $n$ be a positive integer, and let $G$ be a finite group of order $n$. A function $f: G \to G$ is a *pseudoendomorphism* if $f(xyz) = f(x)f(y)f(z)$, for all $x, y, z$ in $G$.
a) If $n$ is odd, show that every pseudoendomorphism of $G$ is an endomorphism.
b) If $n$ is even, is every pseudoendomorphism of $G$ an en... | [
"a) Let $e$ denote the unit of $G$. Let $x = y = z = e$ to write $f(e)^3 = f(e)$, so $f(e)^2 = e$. Since $n$ is odd, it follows that $f(e) = e$.\n\nIf $x$ and $y$ are members of $G$, write $f(xy) = f(xye) = f(x)f(y)f(e) = f(x)f(y)$, to conclude that $f$ is indeed an endomorphism of $G$.\n\n\nb) The answer is negati... | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Abstract Algebra > Group Theory"
] | English | proof and answer | a) Yes. b) No; a constant map to an element of order two is a counterexample. | |
04pr | Rudi and Miljen take turns playing a game on a board. One move consists of selecting two relatively prime positive integers that are already written on the board, erasing them and replacing them with their sum. The player who cannot make a move loses. Rudi plays first. Prove that Miljen has a winning strategy if the in... | [
"a)\nWe claim that Miljen can play so that he leaves some odd number $n$ and $2019 - n$ ones on the board after each of his moves, i.e. before Rudi's move. In that case, after $1009$ of Miljen's moves only the number $2019$ is written on the board, so Rudi cannot make a move and Miljen wins.\n\nIn the beginning, th... | Croatia | Croatian Mathematical Society Competitions | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0090 | Determine, in each case, all real numbers $x$ such that:
$$
a) \lfloor x \rfloor + \lfloor 2x \rfloor + \dots + \lfloor 2012x \rfloor = 2013;
$$
$$
b) \lfloor x \rfloor + \lfloor 2x \rfloor + \dots + \lfloor 2013x \rfloor = 2014.
$$ | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | a) x in [1/671, 3/2012). b) No real solutions. | |
0gne | Let $ABC$ be a triangle with $m(\hat{B}) > m(\hat{C})$. Interior and exterior angle bisectors at vertex $A$ intersect $BC$ at $D$ and $E$ respectively. A variable point $P$ lies on the ray $[EA$ such that $A$ is closer to $E$ than $P$. Lines $DP$ and $AC$ intersect at point $M$ and lines $ME$ and $AD$ intersect at poin... | [
"Let $PQ \\cap BC = N$. Menelaus' theorem applied to $\\triangle (AEC)$ and collinear points $P, M, D$ yields:\n$$\n\\frac{|PA|}{|PE|} \\cdot \\frac{|ED|}{|DC|} \\cdot \\frac{|CM|}{|MA|} = 1.\n$$\nMenelaus' theorem applied to $\\triangle (ADC)$ and collinear points $E, Q, M$ yields:\n$$\n\\frac{|ED|}{|EC|} \\cdot \... | Turkey | Team Selection Test for IMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
03a5 | Find all primes $p < q < r$ such that $p + q = r$ and the number $(r - p)(q - p) - 27p$ is a perfect square. | [
"It is obvious that $p = 2$ and then $r - 2 = q$. Hence $q(q-2) - 54 = u^2$ which can be written as $(q-1)^2 - u^2 = 55 \\iff (q-1-u)(q-1+u) = 55 = 1 \\cdot 55 = 5 \\cdot 11$. Hence we have the following two cases:\n\nCase 1. $q-1-u = 1$ and $q-1+u = 55$. Then $q = u+2$, whence $q = 29$ and $r = 29+2 = 31$.\n\nCase... | Bulgaria | Fall Mathematical Competition | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | p = 2, q = 29, r = 31 | |
0ccb | Find the largest possible value of the expression
$$
\frac{a+b-c}{a^3+b^3+abc} + \frac{b+c-a}{b^3+c^3+abc} + \frac{c+a-b}{c^3+a^3+abc},
$$
where $a$, $b$, $c$ are positive real numbers such that $a + b + c \ge \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$. | [
"Denote by $E(a, b, c)$ the given expression. First, we find the maximum value of $E(a, b, c)$ under the assumption that $a + b - c$, $b + c - a$, $c + a - b$ are non-negative.\nSince $a^2 + b^2 - ab \\ge ab$, it follows that $a^3 + b^3 = (a + b)(a^2 + b^2 - ab) \\ge ab(a + b)$, so $a^3 + b^3 + abc \\ge ab(a + b + ... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - FOURTH SELECTION TEST | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1 | |
0adv | Find all positive integers $x$, $y$ and $z$ such that $1 + 2^x 3^y = z^2$. | [
"It is easily seen that for $z = 1, 2, 3$ the given equation has no solution. Let $z \\ge 4$. Then $2^x 3^y = (z-1)(z+1)$. Both of the numbers $z-1$ and $z+1$ can't be divisible by $3$ (because if they are then $3 | (z+1)-(z-1) = 2$, which is impossible). From $2 | (z-1)(z+1)$ it follows that the numbers $z-1$ and ... | North Macedonia | Sixteenth Macedonian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Macedonian, English | proof and answer | (x, y, z) = (3, 1, 5), (5, 2, 17), (4, 1, 7) | |
09dt | In a convex quadrilateral $ABCD$, $\angle B = \angle C = 120^\circ$ and $AB^2 + BC^2 + CD^2 = AD^2$. Prove that there is a circle inscribed in $ABCD$. | [
"$$\n(AB + x)^2 + (CD + x)^2 - (AB + x)(CD + x) = AD^2 \\iff \\\\\n\\Rightarrow AB \\cdot x + CD \\cdot x - AB \\cdot CD = 0.\n$$\nOn the other hand, $(AB+CD-x)^2 = AB^2+CD^2+x^2-2(AB \\cdot x+CD \\cdot x-AB \\cdot CD) = AB^2+CD^2+x^2 = AD^2$ and we get $AB+CD = AD+x = AD+BC$. It means in the quadrilateral $ABCD$ c... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
0b4t | Problem:
Suppose that $P(x)$ and $Q(x)$ are both quadratic polynomials with leading coefficient $1$ such that $P(P(x)-x) = (Q(x))^2$ for all real numbers $x$ and $P(2) = 0$. Find the sum of all possible values of $P(10)$. | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 64 | |
0gi3 | 令 $n$ 與 $m$ 為正整數。托兒所的老師用 $n \times m$ 塊正方形巧拼，排成一個 $n \times m$ 的長方形區域。每塊巧拼上有一個嬰兒，各自面向正方形的某一個邊。老師一拍手，所有嬰兒就同時往其面對的方向前進一塊巧拼，之後所有嬰兒原地順時鐘轉 90 度。若一個嬰兒爬出 $n \times m$ 的長方形區域，則該嬰兒大哭。若兩個嬰兒前進到相同的巧拼上，他們會對撞然後一起大哭。
已知老師可以適當安排一開始每個嬰兒的面向，使得不論老師拍多少次手，都沒有嬰兒大哭。
試求 $n$ 與 $m$ 的所有可能值。 | [
"答案是所有偶數 $n$ 與 $m$。顯然所有 $2k \\times 2h$ 的情形都可以用 $k \\times h$ 個如下的 $2 \\times 2$ 嬰兒陣拼出來：\n\n\n現在證明只有 $n$ 與 $m$ 皆為偶數時可滿足題意。讓我們將直排依序編號為 1, 2, ..., $n$，橫列依序編號為 1, 2, ..., $m$，並將第 $i$ 排 $j$ 列的巧拼標記為 $(i, j)$。讓我們考慮所有 $i$ 和 $j$ 都是奇數的“奇”格，以及所有 $i$ 和 $j$ 都是偶數的“偶”格。注意到不論起始嬰兒的面向如何，在老師拍完兩次手後，所有奇格的嬰兒都會出... | Taiwan | 2023 年台灣數學奧林匹亞考試 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Chinese (Traditional) | proof and answer | Both dimensions are even. | |
0fuc | Problem:
Sei $m$ eine natürliche Zahl grösser als $1$. Die Folge $x_{0}, x_{1}, x_{2}, \ldots$ ist definiert durch
$$
x_{i}= \begin{cases}2^{i}, & \text{ für } \quad 0 \leq i \leq m-1 \\ \sum_{j=1}^{m} x_{i-j} & \text{ für } \quad i \geq m\end{cases}
$$
Finde das grösste $k$, sodass es $k$ aufeinanderfolgende Folgegli... | [
"Solution:\n\nSei $r_{i}$ der Rest von $x_{i}$ bei Division durch $m$. Wir betrachten statt der Folge $\\left(x_{i}\\right)$ die Folge $\\left(r_{i}\\right)$. Mit Hilfe der Rekursionsformel lassen sich aus $m$ aufeinanderfolgenden Folgegliedern das nächste, aber auch das vorhergehende ausrechnen. Solche $m$ aufeina... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | m - 1 | |
0640 | Problem:
Man bestimme alle ganzen Zahlen $n \geq 2$ mit der folgenden Eigenschaft:
Für beliebige, nicht notwendigerweise verschiedene ganze Zahlen $m_{1}, m_{2}, \ldots, m_{n}$, deren Summe nicht durch $n$ teilbar ist, existiert ein Index $i$ $(1 \leq i \leq n)$, so dass keine der Zahlen
$$
m_{i},\ m_{i}+m_{i+1},\ m_{... | [
"Solution:\n\nBei den gesuchten Zahlen handelt es sich genau um die Primzahlen.\n\nTeilbeweis 1: Keine Nichtprimzahl erfüllt alle Voraussetzungen.\n\nEs sei $n=a \\cdot b$ mit $1<a, b<n$ eine Zerlegung von $n$ in zwei echte Teiler. Wir wählen $m_{i}=a$ für $1 \\leq i<n$ sowie $m_{n}=0$. Dann ist die Summe $m_{1}+m_... | Germany | 2. Auswahlklausur | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | all prime numbers | |
0355 | Problem:
Find all values of the real parameters $a$ and $b$ such that the graph of the function $y = x^{3} + a x + b$ has exactly three common points with the coordinate axes and they are vertices of a right triangle. | [
"Solution:\nThe first condition of the problem is equivalent to the assertion that the equation $x^{3} + a x + b = 0$ has a double real root $x_{1} \\neq 0$ and a simple real root $x_{2} \\neq 0$, where $x_{2} \\neq x_{1}$. Therefore\n$$\nx^{3} + a x + b = (x - x_{1})^{2}(x - x_{2})\n$$\nWe also have $\\angle ACB =... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | (a,b)=(-3/√2, 2^{1/4}) or (a,b)=(-3/√2, -2^{1/4}) | |
0efs | Problem:
Naj bo $ABC$ tak ostrokotni trikotnik, da oglišči $A$ in $B$ ter središči trikotniku očrtane in včrtane krožnice ležijo na isti krožnici. Dokaži, da na tej krožnici leži tudi višinska točka trikotnika $ABC$. | [
"Solution:\n\nOznačimo z $O$, $I$ in $H$ zaporedoma središče očrtane krožnice, središče včrtane krožnice in višinsko točko trikotnika $ABC$. Kote trikotnika označimo kot običajno z $\\alpha$, $\\beta$ in $\\gamma$. Ker je $\\angle BAI = \\frac{\\alpha}{2}$ in $\\angle IBA = \\frac{\\beta}{2}$, je $\\angle AIB = 180... | Slovenia | 61. matematično tekmovanje srednješolcev Slovenije, Odbirno tekmovanje | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fb6 | Problem:
Sea $m$ un entero positivo. Demuestra que no existen números primos de la forma $2^{5m} + 2^{m} + 1$. | [
"Solution:\n\nSumando y restando $2^{2m}$ resulta, $2^{5m} + 2^{m} + 1 = 2^{5m} + 2^{m} + 1 - 2^{2m} + 2^{2m} = 2^{2m}(2^{3m} - 1) + 2^{2m} + 2^{m} + 1$.\n\nTeniendo en cuenta que $2^{3m} - 1 = (2^{m} - 1)(2^{2m} + 2^{m} + 1)$, resulta que,\n\n$2^{5m} + 2^{m} + 1 = (2^{3m} - 2^{2m} + 1)(2^{2m} + 2^{m} + 1)$, que es... | Spain | null | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0k6i | Problem:
Equilateral $\triangle ABC$ has side length $6$. Let $\omega$ be the circle through $A$ and $B$ such that $CA$ and $CB$ are both tangent to $\omega$. A point $D$ on $\omega$ satisfies $CD = 4$. Let $E$ be the intersection of line $CD$ with segment $AB$. What is the length of segment $DE$? | [
"Solution:\n\nLet $F$ be the second intersection of line $CD$ with $\\omega$. By power of a point, we have $CF = 9$, so $DF = 5$. This means that $\\frac{[ADB]}{[AFB]} = \\frac{DE}{EF} = \\frac{DE}{5 - DE}$.\n\nNow, note that triangle $CAD$ is similar to triangle $CFA$, so $\\frac{FA}{AD} = \\frac{CA}{CD} = \\frac{... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 20/13 | |
0lew | Consider 2 non-constant polynomials $P(x), Q(x)$, with non-negative coefficients. The coefficients of $P(x)$ is not larger than $2021$ and $Q(x)$ has at least one coefficient larger than $2021$. Assume that $P(2022) = Q(2022)$ and $P(x), Q(x)$ has a common rational root $\frac{p}{q} \neq 0$ for some $p, q \in \mathbb{Z... | [
"Since the coefficients of $P(x)$ are non-negative, the root $x = p/q$ must be negative. Without loss of generality, we assume that $p < 0, q > 0$ then $|p| + n|q| = nq - p$. Let $R(x) = Q(x) - P(x)$, then $R(x)$ is an integer polynomial where $x = p/q$ is a root. This implies that\n$$\nR(x) = (qx - p)T(x)\n$$\nwhe... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factoriz... | English | proof only | null | |
0bzh | The incircle of triangle $ABC$ touches the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$ respectively. On the line segments $EF$, $FD$, and $DE$, consider the points $M$, $N$, and $P$ respectively such that the sums $BM + MC$, $CN + NA$, and $AP + PB$ are minimum.
a) Prove that the lines $AM$, $BN$, and $CP$ are con... | [
"a) Let $S$ be the reflection of $C$ across the line $EF$. As $AE = AF$, it follows that $\\angle AFE = \\angle AEF$, hence $\\angle BFE = \\angle FEC = \\angle FES$, therefore $FB$ and $ES$ are parallel. $FBES$ is a trapezoid (or a parallelogram); let $M$ be the intersection point of its diagonals. According to th... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscel... | English | proof only | null | |
0adk | Prove that the sum of six consecutive positive integers, such that each of them is not divisible with $7$, is divisible with $21$ but it's not divisible with $42$. Find six such numbers which sum is a four-digit number that is a square of a positive integer. | [
"Because none of the six consecutive positive integers is divisible with $7$ they are of this kind: $7n+1$, $7n+2$, $7n+3$, $7n+4$, $7n+5$, $7n+6$, $n \\in \\mathbb{N}_0$. Their sum is $S=42n+21=21(2n+1)$, from where it follows that $S$ is divisible with $21$ but it's not divisible with $42$. In order $S$ to be a s... | North Macedonia | Macedonian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 659, 660, 661, 662, 663, 664 | |
0j9t | Problem:
Mark and William are playing a game. Two walls are placed 1 meter apart, with Mark and William each starting an orb at one of the walls. Simultaneously, they release their orbs directly toward the other. Both orbs are enchanted such that, upon colliding with each other, they instantly reverse direction and go... | [
"Solution:\n\n$2^{17} / 125$ If the two orbs leave their respective walls at the same time, then they will return to their walls at the same time (because colliding affects both their speeds). After returning to the wall $n$ times, Mark's orb will travel at $\\frac{4^{n}}{1000}$ meter/s and William's will travel at... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 2^17 / 125 | |
0gbc | 試求所有的整數 $c \in \{0, 1, \dots, 2016\}$ 使得滿足下列兩條件之函數 $f: Z \rightarrow \{0, 1, \dots, 2016\}$ 的個數最小:
(1) 函數 $f$ 之週期為 $2017$;
(2) $f(f(x) + f(y) + 1) - f(f(x) + f(y)) \equiv c \pmod{2017}$.
註: 此處 $Z$ 表示所有整數所成的集合。 | [
"What we want are $c = 1, 1008, 1009, 2016$, and the number of $f$ is $2017$.\nLet $a\\%b$ denote the remainder of $a$ divided by $b$, where $0 \\le a\\%b < b$.\nNote that $(cx + k)\\%2017$ satisfies the condition (where $k = 0, 1, \\dots, 2016$).\n\nIn the following, we'll prove there is another solution when $c \... | Taiwan | 二〇一七數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Other"
] | null | proof and answer | c = 1, 1008, 1009, 2016; minimal number of functions = 2017 | |
048r | Prove that no odd integers $x, y, z$ can satisfy $(x-z)^2 + (y-z)^2 = (x+y)^2$. | [] | Croatia | CroatianCompetitions2011 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0inm | Problem:
Find the smallest positive integer that is twice a perfect square and three times a perfect cube. | [
"Solution:\n\nLet $n$ be such a number. If $n$ is divisible by $2$ and $3$ exactly $e_{2}$ and $e_{3}$ times, then $e_{2}$ is odd and a multiple of three, and $e_{3}$ is even and one more than a multiple of three. The smallest possible exponents are $n_{2}=3$ and $n_{3}=4$. The answer is then $2^{3} \\cdot 3^{4}=64... | United States | $10^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 648 | |
05un | Problem:
Trouver tous les entiers naturels non nuls $a$, $b$ et $c$ pour lesquels il existe des entiers naturels non nuls $x$, $y$ et $z$ tels que $x! = a b + 1$, $y! = b c + 1$ et $z! = c a + 1$.
Remarque : Pour tout entier naturel non nul $n$, l'entier $n!$ désigne le produit $1 \times 2 \times \cdots \times n$. | [
"Solution:\n\nSoit $(a, b, c)$ un triplet solution éventuel. Sans perte de généralité, et puisque $a$, $b$ et $c$ jouent des rôles symétriques, on suppose que $a \\leqslant b \\leqslant c$, de sorte que $x \\leqslant z \\leqslant y$ et que $x!$ divise à la fois $y!$ et $z!$.\n\nSoit $d$ un diviseur de $x!$, c'est-à... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | All permutations of (1, 1, n! - 1) with integer n ≥ 2. | |
0ddf | Let $ABCD$ be a convex quadrilateral and $X$ lying inside it such that $XA \cdot XC^2 = XB \cdot XD^2$ and $\angle AXD + \angle BXC = \angle CXD$. Prove that $\angle XAD + \angle XCD = \angle XBC + \angle XDC$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kgc | Problem:
Let $a$ and $b$ be positive integers with $a > b$. Suppose that
$$
\sqrt{\sqrt{a} + \sqrt{b}} + \sqrt{\sqrt{a} - \sqrt{b}}
$$
is an integer.
a. Must $\sqrt{a}$ be an integer?
b. Must $\sqrt{b}$ be an integer? | [
"Solution:\n\nLet $r = \\sqrt{\\sqrt{a} + \\sqrt{b}}$ and $s = \\sqrt{\\sqrt{a} - \\sqrt{b}}$. We know $r^{2} + s^{2} = 2 \\sqrt{a}$ and $r^{2} - s^{2} = 2 \\sqrt{b}$. If $r + s$ is an integer $k$, then\n$$\n\\sqrt{a} = \\frac{r^{2} + s^{2}}{2} = \\frac{(r + s)^{2} + (r - s)^{2}}{4} = \\frac{k^{2} + 4b / k^{2}}{4},... | United States | HMMT Spring 2021 Team Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | a: Yes. b: No; for example, a = 36 and b = 32. | |
02xu | Problem:
Considere as soluções do sistema
$$
\left\{\begin{array}{l}
2019=a+b-c \\
2019=a^{2}+b^{2}-c^{2}
\end{array}\right.
$$
em que $a$, $b$ e $c$ são inteiros.
a) Encontre pelo menos uma solução do sistema.
b) Verifique que o número de soluções é finito. | [
"Solution:\na) Da primeira equação, segue que $c=a+b-2019$. Substituindo na segunda equação, obtemos\n$$\n2019=a^{2}+b^{2}-(a+b-2019)^{2}=-2 a b+4038 a+4038 b-2019^{2}\n$$\nDaí,\n$$\n\\begin{aligned}\n2019-2019^{2} & =-2 a b+4038 a+4038 b-2 \\cdot 2019^{2} \\\\\n-2019 \\cdot 2018 & =-2(a-2019)(b-2019) \\\\\n2019 \\... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | One solution is (a, b, c) = (2020, 2039190, 2039191); the number of solutions is finite. | |
01yj | Given $n \ge 2$ pairwise distinct integers bigger than $-a$, such that among them the amount of odd numbers equals to the largest even number and the amount of even numbers equals to the largest odd number.
a) Find the minimal possible $n$ over all positive integers $a \ge 2$.
b) For each integer $a \ge 2$ find the m... | [
"**a)** The largest odd and even numbers are positive integers since there are numbers of both parities. This means that the largest even number is not less than two, the largest even number is not less than one, and the total number of numbers is not less than three. Note that $n = 3$ could be, for example, if $a ... | Belarus | Belarus2022 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Other"
] | English | proof and answer | a) 3; b) 2a - 1 | |
0hdb | Out of three expressions $\frac{x}{y}$, $\frac{x^2+4}{y^2+4}$ and $\frac{x^3+8}{y^3+8}$, for some integer $x, y$, two take the same value, and the remaining one takes a different value. For which pairs of integers $x, y$ is this possible?
(Bogdan Rublyov) | [
"Obviously, $x \\neq y$, because otherwise, all three expressions would take the same value, which contradicts problem statement.\nIf the first two expressions are equal, then we have the equality:\n$$\n\\frac{x}{y} = \\frac{x^2+4}{y^2+4} \\Rightarrow xy^2 + 4x = yx^2 + 4y \\Rightarrow xy(y-x) = 4(y-x).\n$$\nSince ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | [(1, 4), (4, 1)] | |
0fd2 | Problem:
Los años recientes se han podido expresar como sumas, restas y multiplicaciones de números con un mismo y único dígito; por ejemplo:
$$
2009=7 \times 7 \times 7 \times 7-7 \times 7 \times 7-7 \times 7, \quad 2010=66 \times 6 \times 6-66 \times 6+6 \times 6-6
$$
¿Se puede hacer lo mismo con el 2011, sin repeti... | [
"Solution:\n\nSi $2011$ fuera expresable como sumas, restas y multiplicaciones de números con el mismo dígito $a$, como cada uno de estos números es divisible por $a$, se tiene que $a$ es divisor de $2011$. Ahora bien, $2011$ es un número primo, por tanto $a=1$.\n\nEs sencillo observar que\n$$\n\\begin{aligned}\n10... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 2011=1111*1111-111*11111+1111-111+11 | |
02mh | Problem:
Um professor propõe 80 problemas a um aluno, informando que lhe atribuirá cinco pontos por problema resolvido corretamente e lhe descontará três pontos por problema não resolvido ou resolvido incorretamente. No final, o aluno fica com oito pontos. Quantos problemas ele resolveu corretamente? | [
"Solution:\n\nSeja $c$ o número de problemas resolvidos corretamente e seja $e$ a soma do número de problemas resolvidos incorretamente e de problemas não resolvidos. Logo, $c + e = 80$ e $5c - 3e$ é o número de pontos do aluno na avaliação. No caso presente,\n$$\n\\left\\{\\begin{aligned}\nc + e & = 80 \\\\\n5c - ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 31 | |
055s | Let $BE$ be an altitude of an acute triangle $ABC$ and let $P$ be the point on side $AB$ such that $AP = AE$. Let $N$ be the point for which $BCEN$ is a parallelogram. The areas of the triangles $AEP$ and $BNP$ are equal. Lines $NE$ and $AB$ intersect at $Q$.
a) Prove that the median of triangle $ABC$ drawn from the v... | [
"\nFig. 40\n\nDenote $AC = b$, $AB = c$, $\\angle BAC = \\alpha$, and $AE = AP = u$. As $BCEN$ is a parallelogram, $BN = CE = b-u$ and $BN \\parallel CE$. The latter implies $\\angle NBP = \\alpha$ (Fig. 40). As $\\angle QAE = \\angle QBN$ and $\\angle AQE = \\angle BQN$, triangles $AQE$ an... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
037i | Problem:
Find all complex numbers $a \neq 0$ and $b$ such that for every complex root $w$ of the equation $z^{4}-a z^{3}-b z-1=0$ the inequality $|a-w| \geq|w|$ holds. | [
"Solution:\nLet $z_{k}, 1 \\leq k \\leq 4$, be the roots of the given equation. Using Vieta's formulas we obtain\n$$\nz_{1}+z_{2}+z_{3}+z_{4}=a \\text{ and } z_{1}^{2}+z_{2}^{2}+z_{3}^{2}+z_{4}^{2}=a^{2}\n$$\nSet $u_{k}=\\frac{2 z_{k}}{a}=x_{k}+i y_{k}, 1 \\leq k \\leq 4$, where $x_{k}, y_{k} \\in \\mathbb{R}$. The... | Bulgaria | Spring Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | (2, -2), (-2, 2), (2i, 2i), (-2i, -2i) | |
0hzk | Problem:
If a right triangle is drawn in a semicircle of radius $1/2$ with one leg (not the hypotenuse) along the diameter, what is the triangle's maximum possible area? | [
"Solution:\n\nIt is easy to see that we will want one vertex of the triangle to be where the diameter meets the semicircle, so the diameter is divided into segments of length $x$ and $1 - x$, where $x$ is the length of the leg on the diameter. The other leg of the triangle will be the geometric mean of these two nu... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 3√3/32 | |
05uj | Problem:
Morgane a écrit les trois entiers $3$, $4$ et $12$ au tableau. Elle effectue ensuite des changements successifs en procédant comme suit : elle choisit deux nombres $a$ et $b$ écrits au tableau, les efface, et les remplace par $(3a+4b)/5$ et $(4a-3b)/5$.
Morgane peut-elle, après un nombre fini de tels changem... | [
"Solution:\n\nNous allons démontrer que la réponse est négative dans les deux cas. Soit $\\mathcal{S}$ la somme des carrés des nombres écrits au tableau, et soit $f$ et $g$ les fonctions définies par $f:(x, y) \\rightarrow (3x+4y)/5$ et $g:(x, y) \\rightarrow (4x-3y)/5$. La relation\n$$\nf(x, y)^2 + g(x, y)^2 = x^2... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Linear transformations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | a) no; b) no | |
04xm | In a convex quadrilateral $ABCD$ let $M$ and $N$ be midpoints of sides $AD$ and $BC$, respectively. Points $K$ and $L$ are chosen on sides $AB$ and $CD$, respectively, in such a manner that $\angle MKA = \angle NLC$. Prove that if lines $BD$, $KM$ and $LN$ meet at one point, then
$$
\angle KMN = \angle BDC, \quad \angl... | [
"Let $P$ be the midpoint of $BD$ and $Q$ be the common point of lines $BD$, $KM$ and $LN$. Without losing generality assume that point $B$ lies between $Q$ and $D$. By Tales theorem, $PM \\parallel AB$ and $PN \\parallel CD$. Therefore, $\\angle PNL = \\angle NLC = \\angle MKA = \\angle KMP$. Note that this implies... | Czech-Polish-Slovak Mathematical Match | 11-th Czech-Slovak-Polish Match, 2011 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
091n | Problem:
Let $k$ and $m$, with $k > m$, be positive integers such that the number $k m (k^{2} - m^{2})$ is divisible by $k^{3} - m^{3}$. Prove that $(k - m)^{3} > 3 k m$. | [
"Solution:\nLet $d$ be the greatest common divisor of $k$ and $m$. Write $k = d a$, $m = d b$. Then $a$ and $b$ are relatively prime. Moreover, $a > b$.\n\nThe number $k m (k^{2} - m^{2}) = d^{4} a b (a^{2} - b^{2}) = d^{4} a b (a - b)(a + b)$ is divisible by $k^{3} - m^{3} = d^{3} (a^{3} - b^{3}) = d^{3} (a - b)(a... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0fnr | Encontrar las tres últimas cifras del número $7^{2014}$. | [
"Usaremos el teorema de Euler-Fermat: si $m \\mid a, m = 1$, entonces\n$$\na^{\\varphi(m)} \\equiv 1 \\pmod{m}.\n$$\nEn nuestro caso, queremos calcular $7^{2014} \\pmod{1000}$. Por ser $1000 = 2^3 5^3$, se tiene que $\\varphi(1000) = 2^2(2-1) \\cdot 5^2(5-1) = 400$. Entonces,\n$$\n7^{2014} = 7^{5 \\cdot 400 + 14} =... | Spain | L Olimpiada Matemática Española | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | Spanish | proof and answer | 849 | |
0car | Problem:
Fixăm două numere întregi $m \geq 2$ şi $n \geq 2$. Fie $S$ o mulţime de puncte laticiale situate în dreptunghiul cartezian $[1, m] \times [1, n]$; un punct laticial este un punct care are ambele coordonate întregi. Arătaţi că, dacă $|S| \geq m+n+\left\lfloor \frac{1}{4} m+\frac{1}{4} n-\frac{1}{2} \right\rfl... | [] | Romania | Olimpiada Nationala de Matematica 2022 baraj 2 de selectie seniori | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0fmw | Sean $k$ y $n$ enteros, con $n \ge k \ge 3$. Se consideran $n + 1$ puntos en el plano, no alineados entre sí tres a tres. A cada segmento que une entre sí dos de esos puntos se le asigna un color de entre $k$ colores dados. Se dice que un ángulo es **bicolor** si tiene por vértice uno de los $n + 1$ puntos, y por lados... | [] | Spain | Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof only | null | |
0kfe | Problem:
Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let the internal angle bisector of $\angle BAC$ intersect $BC$ and $\Gamma$ at $E$ and $N$, respectively. Let $A'$ be the antipode of $A$ on $\Gamma$ and let $V$ be the point where $AA'$ intersects $BC$. Given that $EV=6$, $VA'=7$, and $A'N=9$, comput... | [
"Solution:\n\nLet $H_a$ be the foot of the altitude from $A$ to $BC$. Since $AE$ bisects $\\angle H_a AV$, by the angle bisector theorem $\\frac{AH_a}{H_aE}=\\frac{AV}{VE}$. Note that $\\triangle AH_aE \\sim \\triangle ANA'$ are similar right triangles, so $\\frac{AN}{NA'}=\\frac{AH_a}{H_aE}$.\n\nLet $R$ be the rad... | United States | HMMT February | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 15/2 | |
0cc1 | Inside a square of side length $1$ there are $n > 1$ squares, whose side lengths add up to $1$, and each parallel to a side of the large square intersects the interior of at most one of the interior squares. Prove that the sides of the small squares are parallel to the sides of the large square.
Cristian Șăvescu | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
086x | Problem:
Francesco vuole scrivere il polinomio $x^{16}+x$ come prodotto di più polinomi a coefficienti interi, ognuno di grado almeno 1. Quanti fattori potrà ottenere al massimo?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 . | [
"Solution:\n\nLa risposta è (E). Ricordiamo la fattorizzazione notevole $a^{k}+b^{k}=(a+b)\\left(a^{k-1}-a^{k-2} b+a^{k-3} b^{2}-\\cdots+b^{k-1}\\right)$, valida per $k$ dispari. Utilizzando due volte detta fattorizzazione (una volta con $k=3$, una con $k=5$) abbiamo\n$$\n\\begin{aligned}\nx^{16}+x & =x\\left(x^{15... | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | MCQ | E | |
01f8 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(xf(y) - y^2) = (y + 1)f(x - y)
$$
holds for all $x, y \in \mathbb{R}$. | [
"Putting $y = -1$ and $x = 0$, we obtain $f(-1) = 0$.\nSuppose that $f$ has another zero, i.e., there is an $a \\neq -1$ with $f(a) = 0$. Putting $x = 0$ and $y = -a$ gives $f(-a^2) = 0$. Now choosing $y = a$, we obtain $f(x - a) = 0$. Hence, $f(x) = 0$ for all $x$ which clearly is a solution.\n\nLet us assume now ... | Baltic Way | Baltic Way 2019 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = 0 for all x; f(x) = x + 1 for all x | |
05kf | Problem:
Soient $E$ un ensemble de cardinal $n$ et $\mathscr{F}$ un ensemble de parties de $E$ avec $|\mathscr{F}| = 2^{n-1}$ tel que pour tous $A$, $B$ et $C$ dans $\mathscr{F}$, $A \cap B \cap C$ est non vide.
Montrer que l'intersection de tous les éléments de $\mathscr{F}$ est non vide. | [
"Solution:\n\nSi $A \\subset E$, on notera $A^{c}$ son complémentaire : l'ensemble des parties de $E$ est partitionné en $2^{n-1}$ doubletons de la forme $\\{A, A^{c}\\}$ où $A \\subset E$. Or, il est impossible que les deux éléments d'un doubleton soient dans $\\mathscr{F}$, car leur intersection est vide, donc $\... | France | OFM 2013-2014 Envoi 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0bd2 | Let $(a_n)_{n \ge 1}$ be an increasing bounded sequence of real numbers. Evaluate
$$
\lim_{n \to \infty} (2a_n - a_1 - a_2)(2a_n - a_2 - a_3) \cdots (2a_n - a_{n-2} - a_{n-1})(2a_n - a_{n-1} - a_1).
$$ | [
"The limit is equal to $0$.\nLet $x_n = (2a_n - a_1 - a_2)(2a_n - a_2 - a_3) \\cdots (2a_n - a_{n-2} - a_{n-1})(2a_n - a_{n-1} - a_1)$. The sequence $(a_n)_{n \\ge 1}$ is convergent; let $L = \\lim_{n \\to \\infty} a_n$. Since $a_n \\le L$ for all $n \\ge 1$ we have $2a_n - a_k - a_{k+1} \\le 2L - a_k - a_{k+1} \\l... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 0 | |
06lp | Find all real triples $(a, b, c)$ satisfying
$$(2^{2a} + 1)(2^{2b} + 2)(2^{2c} + 8) = 2^{a+b+c+5}.$$ | [
"The only solution is $(a, b, c) = \\left(0, \\frac{1}{2}, \\frac{3}{2}\\right)$.\nNote that $(2^a - 1)^2 \\ge 0$, $(2^b - \\sqrt{2})^2 \\ge 0$ and $(2^c - \\sqrt{8})^2 \\ge 0$. These yield\n$$\n2^{2a} + 1 \\ge 2^{a+1},\n$$\n$$\n2^{2b} + 2 \\ge 2^{b+\\frac{3}{2}},\n$$\n$$\n2^{2c} + 8 \\ge 2^{c+\\frac{5}{2}}.\n$$\nM... | Hong Kong | Year 2021 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | (0, 1/2, 3/2) | |
0eg5 | Problem:
Dano je zaporedje $a_{1}, a_{2}, a_{3}, \ldots$ z začetnima členoma $a_{1}=4$ in $a_{2}=16$, za katerega je $\log _{2}\left(\log _{2} a_{1}\right), \log _{2}\left(\log _{2} a_{2}\right), \log _{2}\left(\log _{2} a_{3}\right), \ldots$ aritmetično zaporedje. Dokaži, da je
$$
\log _{2}\left(\log _{2}\left(4 a_{1... | [
"Solution:\n\nZaporedje $\\log _{2}\\left(\\log _{2}\\left(a_{n}\\right)\\right)$ je aritmetično z začetnima členoma $\\log _{2}\\left(\\log _{2} 4\\right)=\\log _{2} 2=1$ in $\\log _{2}\\left(\\log _{2} 16\\right)=\\log _{2} 4=2$, torej je $\\log _{2}\\left(\\log _{2} a_{n}\\right)=n$. Od tod sledi $\\log _{2} a_{... | Slovenia | 61. matematično tekmovanje srednješolcev Slovenije, Odbirno tekmovanje | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof only | null | |
02g9 | Let $f: \mathbb{Z}_{+}^{*} \rightarrow \mathbb{Z}_{+}^{*}$ be a function defined as follows: given $n$, we write $n = 2^a \cdot (2b+1)$, with $a$ and $b$ non-negative integers, and define $f(n) = a^2 + a + 1$.
Determine the least positive integer $n$ such that $f(1) + f(2) + \cdots + f(n) \ge 123456$. | [
"Let $S_k = f(1) + f(2) + f(3) + \\dots + f(2^k)$. Then\n$$\nS_k = f(1) + f(2) + \\dots + f(2^{k-1}) + f(1) + f(2) + \\dots + f(2^{k-1} - 1) + f(2^k)\n$$\n$$\n\\Leftrightarrow S_k = 2S_{k-1} - f(2^{k-1}) + f(2^k)\n$$\n$$\n\\Leftrightarrow S_k = 2(S_{k-1} + k)\n$$\nIf $n = (a_t a_{t-1} \\dots a_0)_2$ is the binary e... | Brazil | XXII OBM | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 24710 | |
0b9u | Points $M, N, P, Q$ are given on the sides $AB, BC, CD, DA$ of a parallelogram $ABCD$ such that $MN + QP = AC$. Prove that $PN + QM = DB$. | [
"Denote $m = \\frac{AM}{AB}$, $n = \\frac{BN}{BC}$, $p = \\frac{DP}{DC}$, $q = \\frac{AQ}{AD}$, to get $MN = AN - AM = AB + BN - AM = (1-m)AB + nAD$, $QP = pAB + (1-q)AD$. The given condition becomes $(1-m+p)AB + (1-q+n)AD = AC = AB + AD$.\n\nNotice that vectors $AB$ and $AD$ have distinct directions to deduce that... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0g7k | 是否存在一個由 10 個相異正整數所形成的遞增等差數列 $a_1, a_2, \dots, a_{10}$, 使得 $\prod_{i=1}^{10} a_i$ 整除一個形如 $n^2 + 1$ 的整數 (其中 $n$ 為正整數)? 若存在, 試求出滿足此條件的數列使其公差為最小。 | [
"滿足題設的數列存在。其中公差最小為 84, 其中一數列為\n29, 113, 197, 281, 365, 449, 533, 617, 701, 785.\n我們需要以下引理。\n**Lemma 1.** 形如 $n^2 + 1$ 的整數沒有 $4k + 3$ 型的質因數。\n*Proof.* 利用費馬小定理即可得證。\n\n**Lemma 2.** 若 $a_1, b_1, a_2, b_2$ 為非零整數且 $a_i$ 與 $b_i$ 互質 ($i = 1, 2$), 則 $(a_1^2 + b_1^2)(a_2^2 + b_2^2)$ 亦可表為兩個互質整數的平方和。\n*Proof.* 注意到 $(a_1^2 + b... | Taiwan | 二〇一三數學奧林匹亞競賽第三階段選訓營, 獨立研究 (二) | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | null | proof and answer | Yes. The minimal common difference is 84. One example sequence is 29, 113, 197, 281, 365, 449, 533, 617, 701, 785. | |
09rg | Problem:
Gegeven is een driehoek $ABC$. Zij $\Gamma_1$ de cirkel door $B$ die raakt aan zijde $AC$ in $A$. Zij $\Gamma_2$ de cirkel door $C$ die raakt aan zijde $AB$ in $A$. Het tweede snijpunt van $\Gamma_1$ en $\Gamma_2$ noemen we $D$. De lijn $AD$ snijdt de omgeschreven cirkel van $\triangle ABC$ nog een keer in $E... | [
"Solution:\n\nWe bekijken de configuratie waar $D$ binnen $\\triangle ABC$ ligt. De andere configuratie gaat analoog. Vanwege de raaklijnomtrekshoekstelling in $\\Gamma_1$ geldt $\\angle DAC = \\angle DBA$. Vanwege de raaklijnomtrekshoekstelling in $\\Gamma_2$ geldt $\\angle DAB = \\angle DCA$. Er geldt dus $\\tria... | Netherlands | MO-selectietoets | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
0k2j | Problem:
Let $ABC$ be a triangle with $\angle A = 18^{\circ}$, $\angle B = 36^{\circ}$. Let $M$ be the midpoint of $AB$, $D$ a point on ray $CM$ such that $AB = AD$; $E$ a point on ray $BC$ such that $AB = BE$, and $F$ a point on ray $AC$ such that $AB = AF$. Find $\angle FDE$. | [
"Solution:\nLet $\\angle ABD = \\angle ADB = x$, and $\\angle DAB = 180^{\\circ} - 2x$. In triangle $ACD$, by the law of sines,\n$$\nCD = \\frac{AD}{\\sin \\angle ACM} \\cdot \\sin(198^{\\circ} - 2x),\n$$\nand by the law of sines in triangle $BCD$,\n$$\nCD = \\frac{BD}{\\sin \\angle BCM} \\cdot \\sin(x + 36^{\\circ... | United States | HMMT February 2018 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 27° | |
0foy | En la primera fila de un tablero $5 \times 5$ se colocan $5$ fichas que tienen una cara blanca y otra negra, mostrando todas la cara blanca. Cada ficha se puede mover de una casilla a cualquiera de las contiguas (horizontal o verticalmente) dándole la vuelta en cada movimiento. Además, varias fichas pueden ocupar una m... | [
"Si pintamos las casillas del tablero alternativamente de blanco y negro como en un tablero de ajedrez, sucede que una ficha cuyo color visible coincida con el de la casilla, al moverse seguirá teniendo el mismo color que la nueva casilla (puesto que tanto el color de la ficha como el de la casilla cambian). Supues... | Spain | LII Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof only | null | |
0h6e | On the side $BC$ of acute triangle $ABC$ one chooses an arbitrary point $D$. Let $O$ be the circumcenter of $\triangle ABC$, $Z$ the point on this circle which is diametrically opposite to $A$. Let $X$, $Y$ be points on segments $BO$, $CO$ such that:
$$
\angle BXD + \angle ABC = 180^{\circ} = \angle CYD + \angle ACB.
$... | [
"Here is the solution for the location of points showed in the picture. For other cases, the location of the solution will be similar. Prove that for any selected point $D$ quadrilateral $XOYZ$ is inscribed. Then $\\angle XZY = 180^{\\circ} - 2\\angle A$, so doesn't depend on $D$.\n\nFrom the statement of the probl... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ... | English | proof only | null | |
0lau | Given a positive integer $n$. Show that for any positive real number $x$, we have the inequality:
$$
\frac{x^n (x^{n+1} + 1)}{x^n + 1} \le \left( \frac{x+1}{2} \right)^{2n+1}
$$
When does the equality take place? | [
"The inequality is easily proved by induction on $n$. The equality occurs if and only if $x = 1$."
] | Vietnam | Vijetnam 2011 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | x = 1 | |
01je | In a triangle $\triangle ABC$ let the incircle be tangent to $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively, and let the excircle opposite to $A$ be tangent to $BC$, $CA$, $AB$ at $P$, $Q$, $R$, respectively. Let $EX$ and $FY$ be altitudes in triangle $\triangle DEF$, and $QZ$ and $RW$ altitudes in triangle $\triangle... | [
"Let $K$ be the orthogonal projection of $B$ onto the angle bisector of $\\angle BAC$. Now by the Iran lemma and its analogue for the excircle (both easy angle chases), $K$ lies on lines $DE$ and $PQ$. Now let $L$ be the orthogonal projection of $K$ onto line $AB$. Our main claim is that $L$ lies on both $XY$ and $... | Baltic Way | Baltic Way 2023 Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homot... | English | proof only | null | |
008f | For each natural number $x$, let $S(x)$ be the sum of its digits. Find the smallest natural number $n$ such that $9S(n) = 16S(2n)$. | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | 55555555555555569999999 | |
03ms | Let $\mathbb{N} = \{1, 2, 3, \dots\}$ be the set of positive integers. Find all functions $f$, defined on $\mathbb{N}$ and taking values in $\mathbb{N}$, such that $(n - 1)^2 < f(n)f(f(n)) < n^2 + n$ for every positive integer $n$. | [
"The only such function is $f(n) = n$.\nAssume that $f$ satisfies the given condition. It will be shown by induction that $f(n) = n$ for all $n \\in \\mathbb{N}$. Substituting $n = 1$ yields that $0 < f(1)f(f(1)) < 2$ which implies the base case $f(1) = 1$. Now assume that $f(k) = k$ for all $k < n$ and assume for ... | Canada | Kanada 2015 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | f(n) = n | |
0cdg | Find the strictly increasing functions $f : \mathbb{N} \to \mathbb{N}$ with the following property: the number $f(x) \cdot f(y)$ divides the number $(1+2x) \cdot f(y) + (1+2y) \cdot f(x)$, for any natural numbers $x$ and $y$. | [
"For $x = y = 0$ we deduce $f^2(0) \\mid 2f(0)$, whence $f(0) \\in \\{0, 1, 2\\}$.\n\nIf $f(0) = 0$, for $y = 0$ we have $0 \\mid f(x)$, $\\forall x \\in \\mathbb{N}$, so $f(x) = 0$, $\\forall x \\in \\mathbb{N}$, which contradicts the fact that $f$ is strictly increasing.\n\nIf $f(0) = 1$, for $y = 0$, we have $f(... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | f(x) = 2x + 1 or f(x) = 4x + 2 | |
0h9i | For which positive integers $n$, square $n \times n$ can be completely covered (without overlaps) by rectangles $k \times 1$ and one square $1 \times 1$, where:
a) $k=4$;
b) $k=8$? | [
"For both a) and b) it is obvious that $n$ has to be odd and greater than $k$.\n\na) Hence, $n$ is odd and greater than $4$. Let us show that any such odd $n$ satisfies the condition. It is clear that any stripe $4 \\times l$ can be covered by rectangles $4 \\times 1$. For $5 \\times 5$ and $7 \\times 7$ the requir... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a) Exactly the odd n with n ≥ 5.
b) Exactly the odd n with n > 8 and n ≡ 1 or 7 (mod 8), equivalently n = 8m + 9 or n = 8m + 15 for integers m ≥ 0. | |
07oa | Find the smallest positive integer $m$ such that $5m$ is an exact 5th power, $6m$ is an exact 6th power, and $7m$ is an exact 7th power. | [
"Let $m = 2^a \\cdot 3^b \\cdot 5^c \\cdot 7^d$. $N$ where $N$ is a product of primes different from $2$, $3$, $5$ and $7$. For a number $n$ to be an exact 5th power, all the exponents in the prime factorisation of $n$ must be divisible by $5$. Similarly for 6th and 7th powers.\nFor primes other than $2$, $3$, $5$ ... | Ireland | Ireland | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 2^35 * 3^35 * 5^84 * 7^90 | |
03b9 | Given a $\triangle ABC$ and a function $f : \mathbb{R}^+ \to \mathbb{R}$ with the following property: for any segment $DE$ of the interior of the triangle and its midpoint $M$ one has that
$$
f(d(D)) + f(d(E)) \le 2f(d(M)).
$$
where $d(X)$ denotes the distance from $X$ to the boundary of $\triangle ABC$. Prove that for... | [
"Denote by $k(I, r)$ the incircle of $\\triangle ABC$. It is clear that the image of $d$ is the interval $\\Delta = (0, r]$.\n\nVarying $D$ and $E$ on $AI$, it follows that $f$ is a mid-point concave function on $\\Delta$, i.e. $f(2x) + f(2y) \\le f(x + y)$.\n\nNote that the points on given distance (from the bound... | Bulgaria | Bulgaria | [
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line,... | English | proof only | null | |
0byj | a) Compare the numbers $2^{53}$ and $3^{35}$.
b) Show that, if $5b \ge 3a > 0$, then $2^{ab} < 3^{ba}$. | [
"a) Since $3^7 = 2187 > 2048 = 2^{11}$, we find $3^{35} = (3^7)^5 > (2^{11})^5 > 2^{53}$.\n\nb) The inequality $2^{53} < 3^{35}$ can be written $\\left(\\frac{2^{10}}{3}\\right)^5 < \\left(\\frac{3^{10}}{2}\\right)^3$. It follows that $\\left(\\frac{2^{10}}{3}\\right)^{5a} < \\left(\\frac{3^{10}}{2}\\right)^{3a}$. ... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof only | null | |
0c6e | Let $H$ be the orthocenter of the acute triangle $ABC$. Let $X$ be a point in the plane such that the triangle $XAH$ is an isosceles right triangle, with the hypotenuse $AH$, and the points $B$ and $X$ are separated by the line $AH$. Prove that $\overrightarrow{XA} + \overrightarrow{XC} + \overrightarrow{XH} = \overrig... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"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 > Transformations > Rotation"
] | English | proof only | null | |
0gnw | In a group of $2009$ people, any pair of persons have exactly one common friend. Determine the smallest possible value of the difference between the numbers of friends of the person with the most friends and the person with the least friends in such a group. | [
"Suppose that $a$ and $b$ are not friends. If $x$ is a friend of $a$, then by assumption, $x$ and $b$ have exactly one common friend $y$. The function that takes each $x$ to the corresponding $y$ is a bijection between the friends of $x$ and the friends of $y$. Therefore, any two persons who are not friends have th... | Turkey | Team Selection Test for IMO 2009 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 2006 | |
092b | Problem:
Find all pairs of positive integers $(m, n)$ for which there exist relatively prime integers $a$ and $b$ greater than $1$ such that
$$
\frac{a^{m}+b^{m}}{a^{n}+b^{n}}
$$
is an integer. | [
"Solution:\n\nIf $\\frac{m}{n}=q$ is an odd integer, we have\n$$\n\\frac{a^{m}+b^{m}}{a^{n}+b^{n}}=\\frac{\\left(a^{n}\\right)^{q}+\\left(b^{n}\\right)^{q}}{a^{n}+b^{n}}=\\left(a^{n}\\right)^{q-1}-\\left(a^{n}\\right)^{q-2} \\cdot b^{n}+\\cdots-a^{n} \\cdot\\left(b^{n}\\right)^{q-2}+\\left(b^{n}\\right)^{q-1}\n$$\n... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (m, n) = (q n, n) where q is an odd positive integer | |
01ge | Find the smallest positive integer $k$ with a property: for any three points with integer coordinates on a plane
$$
L_{\max} - L_{\min} > \frac{1}{\sqrt{k} \cdot L_{\max}},
$$
where $L_{\max}$ and $L_{\min}$ are respectively maximum and minimum distances among them. | [
"Answer: $k = 4$.\nAt first we note that $L_{\\max} > L_{\\min}$ because the triangle made by these three points is not equilateral. Assume the contrary: then the area of a triangle with vertices in integral points is a rational number, but the area of an equilateral triangle can be computed by formula $S = \\frac{... | Baltic Way | Baltic Way 2020 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 4 | |
0h8a | Let $x_1, x_2, \dots, x_n$ be positive numbers such that:
$$
x_1^{n-1} + x_2^{n-1} + \dots + x_n^{n-1} = x_1 x_2 \dots x_n
$$
Prove the inequality:
$$
(x_1 - n + 1)(x_2 - n + 1)\dots(x_n - n + 1) \ge 1. \quad (\text{Serdiuk Nazar})
$$ | [
"Using Cauchy inequality we get $\\forall i=1, n$:\n$$\n\\begin{align*}\nx_1 x_2 \\dots x_n &= x_1^{n-1} + x_2^{n-1} + \\dots + x_n^{n-1} \\ge x_1^{n-1} + (n-1) x_1 x_2 \\dots x_{n-1} x_{n+1} \\dots x_n \\\\\n&\\Rightarrow x_1 x_2 \\dots x_{n-1} x_{n+1} \\dots x_n (x_i - n + 1) \\ge x_i^{n-1} \\Rightarrow x_i - n +... | Ukraine | UkraineMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
04ji | Let $n$ be a positive integer and $a_0, a_1, \dots, a_{2n} \in \left\langle -\frac{\pi}{2}, \frac{\pi}{2} \right\rangle$ real numbers such that
$$
\tan a_k = 2^{k-n} \quad \text{for} \quad k = 0, 1, \dots, 2n.
$$
Determine the sum $a_0 + a_1 + \dots + a_{2n}$. | [] | Croatia | Croatia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (2n+1)π/4 | |
0iyd | Problem:
If $a$, $b$, $x$, and $y$ are real numbers such that $a x + b y = 3$, $a x^{2} + b y^{2} = 7$, $a x^{3} + b y^{3} = 16$, and $a x^{4} + b y^{4} = 42$, find $a x^{5} + b y^{5}$. | [
"Solution:\nWe have $a x^{3} + b y^{3} = 16$, so $(a x^{3} + b y^{3})(x + y) = 16(x + y)$ and thus\n$$\na x^{4} + b y^{4} + x y (a x^{2} + b y^{2}) = 16(x + y)\n$$\nIt follows that\n$$\n42 + 7 x y = 16(x + y)\n$$\nFrom $a x^{2} + b y^{2} = 7$, we have $(a x^{2} + b y^{2})(x + y) = 7(x + y)$ so $a x^{3} + b y^{3} + ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 20 | |
011j | Problem:
Given a triangle $A B C$ with $\angle A=120^\circ$. The points $K$ and $L$ lie on the sides $A B$ and $A C$, respectively. Let $B K P$ and $C L Q$ be equilateral triangles constructed outside the triangle $A B C$. Prove that
$$
|P Q| \geqslant \frac{\sqrt{3}}{2} \cdot (|A B|+|A C|) .
$$ | [
"Solution:\n\nSince $\\angle A B C+\\angle A C B=60^\\circ$, the lines $B P$ and $C Q$ are parallel. Let $X$ and $Y$ be the feet of perpendiculars from $A$ to $B P$ and $C Q$, respectively (see Figure 4). Then $|A X|=\\frac{\\sqrt{3}}{2}|A B|$ and $|A Y|=\\frac{\\sqrt{3}}{2}|A C|$. Since the points $X, A$ and $Y$ a... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0efu | Problem:
Dano je število $n=100\ldots 001$, zapisano z 2017 ničlami in 2 enkama.
a. Ali je število $n$ deljivo z 11?
b. Ali je število $n$ deljivo s 101?
c. Ali je število $n$ deljivo s 1001? | [
"Solution:\n\na.\nUporabimo pravilo za deljivost z 11. Naravno število $\\overline{a_{k} \\ldots a_{3} a_{2} a_{1}}$, kjer so $a_{i}$ števke, je deljivo z 11 natanko tedaj, ko je število $a_{1}-a_{2}+a_{3}-\\ldots+(-1)^{k+1} a_{k}$ deljivo z 11. Ker\n$$\n1-\\underbrace{0+0-0+\\ldots+0-0}_{2017 \\text{ ničel}}+1=2\n... | Slovenia | 61. matematično tekmovanje srednješolcev Slovenije, Odbirno tekmovanje | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | a: no; b: yes; c: no | |
09h5 | Two circles $\omega_1$ and $\omega_2$ intersected at points $A$ and $B$. Line through $B$ is intersect the $\omega_1$ at point $C$ and intersect the $\omega_2$ at point $D$. The line $AD$ intersect the $\omega_1$ at point $E$ different from $A$ and the line $AC$ intersect the $\omega_2$ at point $F$ different from $A$.... | [
"We draw the circumcircle of $AEF$, and denote $\\angle EAC = \\alpha$. So $\\angle EAF = 180^\\circ - \\alpha$ and from here we get $\\angle EOF = 2\\alpha$. Other hand $\\alpha = \\angle EBC = \\angle EAC = \\angle FAD = \\angle FBD$, so $\\angle EBF = 180^\\circ - 2\\alpha$. From here we have $\\angle EOF + \\an... | Mongolia | Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ksg | Problem:
Show that the product of any two side lengths of a triangle is greater than the product of the diameters of the inscribed and circumscribed circles. | [
"Solution:\nLet our triangle be $ABC$, with side lengths $a$, $b$, $c$, and let $r$, $R$ be the diameters of the circumscribed and inscribed circles, respectively. We want to show $ab > 4 R r$.\n\nThe triangle inequality tells us that $a + b > c$.\n\nHeron's Formula tells us that the area of the triangle is $S = s ... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > T... | null | proof only | null | |
0bel | On the side $AB$ of the rectangle $ABCD$ are taken the points $S$ and $T$ so that $AS = ST = TB$. Denote $M, N$ and $P$ the orthogonal projections of $A, S$ and $T$ onto the straight lines $DS, DT$ respectively $DB$. Prove that the points $M, N$ and $P$ are collinear if and only if $15AD^2 = 2AB^2$.
Gabriel Popa
![](a... | [
"Let $M', N', P'$ be the orthogonal projections onto $AB$ of $M, N$ and $P$ respectively. Then $M, N, P$ are collinear if and only if the intersection $N_1$ of $MP$ and $NN'$ coincides with $N$. Since $MM' > PP'$ and $N'$ is between $M'$ and $P'$, from similar triangles (Fig. 2) we get $\\frac{P'N'}{P'M'} = \\frac{... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
04cy | There are $1234$ pebbles on a table. Ratko and Rudi play the following game: first Ratko takes an even number of pebbles, at least two but no more than $100$, and then Rudi takes an odd number of pebbles, at least one but no more than $99$. The players take turns taking pebbles, always following the same conditions. A ... | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Ratko | |
03k6 | Problem:
Show that the number $x$ is rational if and only if three distinct terms that form a geometric progression can be chosen from the sequence
$$
x, x+1, x+2, x+3, \ldots
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
03dt | Consider a $\triangle ABC$ with integer sides, a median $CM$ ($M \in AB$), and a circumscribed center $O$. If the circumcircle of $AMO$ passes through the midpoint of $CM$, find the smallest possible value for the perimeter of $ABC$. | [
"Let $N$ and $P$ be the midpoints of $AC$ and $CM$, respectively. Then the pentagon $ANPOM$ is cyclic with $AM \\parallel PN$, meaning that $\\triangle CAM = 180^\\circ - \\angle ANP = 90^\\circ - \\angle PNO = 90^\\circ - \\angle PMO = \\angle AMC$,\n\nwhich is equivalent to $AC = CM$. Let $Q$ be the midpoint of $... | Bulgaria | Bulgaria 2022 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > D... | null | proof and answer | 24 | |
06ln | For a sequence with some ones and zeros, we count the number of continuous runs of equal digits in it. (For example the sequence $011001010$ has $7$ continuous runs: $0$, $11$, $00$, $1$, $0$, $1$, $0$.) Find the sum of the number of all continuous runs for all possible sequences with $2019$ ones and $2019$ zeros. | [
"The answer is $2020\\binom{4038}{2019}$.\nWe show that for all sequences with $n$ zeros and $n$ ones, the sum of all continuous runs is given by $(n+1)\\binom{2n}{n}$.\n\nWe first count the number of changes, which means a pair of consecutive digits which is $01$ or $10$. There are $2n-1$ positions for which chang... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 2020*binom(4038,2019) | |
03g0 | Given is a triangle $ABC$ and a circle $\omega$ with center $I$ that touches $AB$, $AC$ and meets $BC$ at $X$, $Y$. The line through $I$ perpendicular to $BC$ meets the line through $A$ parallel to $BC$ at $Z$. Show that the circumcircles of $\triangle XYZ$ and $\triangle ABC$ are tangent to each other. | [
"Let $W$ be the midpoint of the major arc $BAC$, $A' \\in (ABC)$ be such that $AA' \\parallel BC$, $T$ be the intersection of the circle with diameter $AI$ and $(ABC)$, and let $\\omega$ touch $AC$, $AB$ at $E$, $F$. We claim the two circles touch at $T$.\n\nFirstly, observe that $Z \\in (AEF)$, so\n\n$$\n\\begin{a... | Bulgaria | Bulgarian National Olympiad - Final Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
08xj | Suppose we assign one of positive integers $n$ satisfying $1 \le n \le 8$ to each of the 8 vertices of a regular octagon. How many ways of assigning these numbers are there, which satisfy the following 2 conditions?:
* assigned numbers are all distinct.
* any pair of numbers assigned to adjacent vertices are relatively... | [
"576 ways\nSince pair of even integers are not relatively prime, any pair of adjacent vertices cannot both be occupied by even numbers. Hence even and odd numbers must be assigned to vertices alternately. So, the assignment of numbers must be carried out as in the figures 1 or 2 below.\n\n\... | Japan | Japan Mathematical Olympiad Initial Round | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | final answer only | 576 | |
0c6g | Let $n \ge 2$, be an integer. Find in terms of $n$ the maximum value of $m \le n$ such that we can select $m$ numbers from the set $1, 2, 3, \dots, n$ in such a way that for any two of them, say $a, b$, with $a > b$, the number $a-b$ does not divide $a+b$.
Vasile Pop | [] | Romania | 70th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | ceil(n/3) | |
0fy7 | Problem:
Sei $ABC$ ein Dreieck mit $AB \neq AC$ und Inkreismittelpunkt $I$. Der Inkreis berühre die Seiten $BC$, $CA$ bzw. $AB$ bei $D$, $E$ bzw. $F$. Sei $M$ der Mittelpunkt von $EF$. Die Gerade $AD$ schneide den Inkreis bei $P \neq D$. Beweise, dass $PMID$ ein Sehnenviereck ist. | [
"Solution:\n\nDa das Dreieck $AFE$ gleichschenklig und der Punkt $M$ der Mittelpunkt von der Strecke $EF$ ist, liegt $M$ auf der Winkelhalbierenden von $\\angle FAE$, also auf $AI$. Wir zeigen nun, dass $AP \\cdot AD = AM \\cdot AI$. Aus dem Potenzsatz folgt dann die Behauptung. Da $AE$ die Tangente an den Inkreis ... | Switzerland | SMO Finalrunde | [
"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 > Miscellaneous > Angle chasing"
... | null | proof only | null | |
0bsq | Let $M$ be the set of the natural numbers $k$ for which it exists $n \in \mathbb{N}$ such that the remainder of $3^n$ when divided by $n$ is $k$. Show that $M$ is infinite. | [
"Let $j$ be a fixed positive integer and $p > 2$ a prime so that $2^j p > 3^{2^j}$. Then $3^{2^j}(3^{2^j(p-1)} - 1) \\equiv 0 \\pmod{2^j p}$, because $2\\varphi(2^j p) = 2^j(p-1)$. Hence $3^{2^j p} \\equiv 3^{2^j} \\pmod{2^j p}$, so, for $n = 2^j p$, one has $r_n = 3^{2^j}$. It follows that $3^{2^j} \\in M$, for al... | Romania | 67th NMO Selection Tests for JBMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | English | proof only | null | |
0jjn | Problem:
Let $a_{1}, a_{2}, \ldots$ be an infinite sequence of integers such that $a_{i}$ divides $a_{i+1}$ for all $i \geq 1$, and let $b_{i}$ be the remainder when $a_{i}$ is divided by $210$. What is the maximal number of distinct terms in the sequence $b_{1}, b_{2}, \ldots$? | [
"Solution:\n\nAnswer: $127$\n\nIt is clear that the sequence $\\{a_{i}\\}$ will be a concatenation of sequences of the form $\\{v_{i}\\}_{i=1}^{N_{0}},\\{w_{i} \\cdot p_{1}\\}_{i=1}^{N_{1}},\\{x_{i} \\cdot p_{1} p_{2}\\}_{i=1}^{N_{2}},\\{y_{i} \\cdot p_{1} p_{2} p_{3}\\}_{i=1}^{N_{3}}$, and $\\{z_{i} \\cdot p_{1} p... | United States | HMMT 2014 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Other"
] | null | proof and answer | 127 | |
0azw | Problem:
Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}-11x+24=0$. Let $f$ be the polynomial of least degree, with integer coefficients and leading coefficient $1$, such that $\sqrt{\alpha}+\sqrt{\beta}$ and $\sqrt{\alpha\beta}$ are zeros of $f$. Find $f(1)$. | [] | Philippines | 21st PMO Area Stage | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | -92 |
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