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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
05lu | Problem:
Déterminer toutes les fonctions $f: \mathbb{R}_{+}^{*} \longmapsto \mathbb{R}_{+}^{*}$ telles que
$$
f\left(\frac{y}{f(x+1)}\right)+f\left(\frac{x+1}{x f(y)}\right)=f(y)
$$
pour tous $x, y \in \mathbb{R}_{+}^{*}$. | [
"Solution:\nEn cherchant un peu parmi les fonctions usuelles, on constate que la fonction $f: x \\longmapsto \\frac{1}{x}$ est une solution du problème. Nous allons maintenant prouver que c'est la seule.\nSoit $f$ une solution du problème. Si $y>0$ est tel que $y f(y)>1$, posons $x=\\frac{1}{y f(y)-1}$. La relation... | France | Iran | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1/x for all x > 0 | |
06cn | Find all polynomials $f(x)$ such that $f(f(x)) = (f(x))^m$, where $m > 1$ is a fixed integer. Substantiate your answer. | [
"The solutions are $f(x) = 0$, $f(x) = 1$, $f(x) = \\omega$ where $\\omega$ is an $(m-1)$st root of unity, and $f(x) = x^m$.\n\nIf $f(x) = c$ is a constant polynomial, then the relation holds if and only if $c = c^m$.\nClearly, the solutions are $c = 0, 1$ and all the $(m-1)$st roots of unity.\n\nIf $f$ is a non-co... | Hong Kong | Test 1 | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | All constant polynomials c with c = 0, c = 1, or c an (m−1)st root of unity; and the non-constant polynomial f(x) = x^m. | |
0hgu | Find the integer that is closest to the value of the expression:
$$
\left( (3 + \sqrt{1})^{2023} - \left( \frac{1}{3 - \sqrt{1}} \right)^{2023} \right) \cdot \left( (3 + \sqrt{2})^{2023} - \left( \frac{1}{3 - \sqrt{2}} \right)^{2023} \right) \cdot \left( (3 + \sqrt{3})^{2023} - \left( \frac{1}{3 - \sqrt{3}} \right)^{20... | [
"Let's consider the last factor:\n$$\n\\frac{1}{3 - \\sqrt{8}} = \\frac{3 + \\sqrt{8}}{9 - 8} = 3 + \\sqrt{8} \\Rightarrow (3 + \\sqrt{8})^{2023} = \\left(\\frac{1}{3 - \\sqrt{8}}\\right)^{2023},\n$$\nwhich means that the last factor equals $0$, and therefore the whole product equals $0$."
] | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | final answer only | 0 | |
0e3l | Let $ABCD$ be a square with the side of $20$ units. $Vid$ divides this square into $400$ unit squares. Eva then picks $4$ of the vertices of these unit squares. These vertices lie inside the square $ABCD$ and define a rectangle with the sides parallel to the sides of the square $ABCD$. There are exactly $24$ unit squar... | [
"Let $a$ and $b$ be the lengths of the sides of the rectangle. We may assume that $a \\ge b$. The unit squares that have at least one point in common with the rectangle are marked in the figure.\n\nThe number of the unit squares is equal to the area of this part. We can find this area by su... | Slovenia | National Math Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem"
] | English | proof and answer | 6, 8, 9 | |
0d8p | Let $ABC$ be a triangle inscribed in circle $(O)$ such that two points $B$, $C$ are fixed, $A$ moves on major arc $BC$ of $(O)$. Two tangents through $B$, $C$ of $(O)$ intersect at $P$. Circle with diameter $OP$ intersects $AC$, $AB$ at $D$, $E$, respectively. Prove that $DE$ is tangent to a fixed circle whose radius i... | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0eh4 | Problem:
Poišči vsa naravna števila $n$, za katera je
$$
\frac{100 n^{5}-n^{4}-50 n^{3}+2 n^{2}-290 n-2}{n^{2}-2}
$$
celo število.
(20 točk) | [
"Solution:\nČe polinom $100 n^{5}-n^{4}-50 n^{3}+2 n^{2}-290 n-2$ delimo s polinomom $n^{2}-2$, dobimo rezultat $100 n^{3}-n^{2}+150 n$ in ostanek $10 n-2$. Torej je\n$$\n\\frac{100 n^{5}-n^{4}-50 n^{3}+2 n^{2}-290 n-2}{n^{2}-2}=100 n^{3}-n^{2}+150 n+\\frac{10 n-2}{n^{2}-2}\n$$\nTo število bo celo natanko tedaj, ko... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | n = 1, 2, 3, 10 | |
07i9 | Given triangle $ABC$ and line $\ell$ passing through point $A$, point $X$ on $\ell$ is considered to be variable. Circles $\omega_b$ and $\omega_c$ pass through both of the points $A$ and $X$ and are tangent to sides $AB$ and $AC$, respectively. Tangents $BY$ and $CZ$ are drawn from vertices $B$ and $C$ to circles $\om... | [
"Let $D$ be intersection of $\\ell$ and $BC$. Let $P$ and $Q$ be the reflections of $A$ with respect to $BC$ and $D$, respectively. We claim that the circumcircle of $XYZ$ passes through $P$ and $Q$.\nSince $PQ \\parallel BC$, we have $\\angle XQP = \\angle XDB$, in line with $BX = BY = BP$, we have:\n$$\n\\begin{a... | Iran | 40th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0ijf | Problem:
A sequence is defined by $A_{0}=0$, $A_{1}=1$, $A_{2}=2$, and, for integers $n \geq 3$,
$$
A_{n}=\frac{A_{n-1}+A_{n-2}+A_{n-3}}{3}+\frac{1}{n^{4}-n^{2}}
$$
Compute $\lim _{N \rightarrow \infty} A_{N}$. | [
"Solution:\nIf we sum the given equation for $n=3,4,5, \\ldots, N$, we obtain\n$$\n\\sum_{n=3}^{N} A_{n}=\\sum_{n=3}^{N} \\frac{A_{n-1}+A_{n-2}+A_{n-3}}{3}+\\frac{1}{n^{4}-n^{2}}\n$$\nThis reduces dramatically to\n$$\nA_{N}+\\frac{2 A_{N-1}}{3}+\\frac{A_{N-2}}{3}=A_{2}+\\frac{2 A_{1}}{3}+\\frac{A_{0}}{3}+\\sum_{n=3... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 13/6 - π^2/12 | |
0fkm | Problem:
Halla las soluciones reales de la ecuación: $x\left(\frac{6-x}{x+1}\right)\left(\frac{6-x}{x+1}+x\right)=8$. | [
"Solution:\n\nSea $t = x + 1$. Entonces sustituyendo en la ecuación dada, se tiene $(t-1)\\left(\\frac{7}{t}-1\\right)\\left(\\frac{7}{t}+t-2\\right)=8$, que equivale a $\\left(7-t-\\frac{7}{t}+1\\right)\\left(\\frac{7}{t}+t-2\\right)=8$. Haciendo el cambio $u=\\frac{7}{t}+t$, se obtiene la ecuación $(8-u)(u-2)=8$ ... | Spain | FASE LOCAL DE LA XLIV OME | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 2 - sqrt(2), 2 + sqrt(2) | |
0gg4 | 愛莉拿到一個有理數 $r > 1$ 和一條直線, 直線有兩個點 $B \neq R$, 而 $R$ 上放著一個紅色的珠子, $B$ 上放著一個藍色的珠子。愛莉用這些東西來玩一個單人遊戲。每一回合, 她選定一個整數 $k$ (不一定為正) 和一個珠子來移動。如果這個珠子在位置 $X$, 而另外一個珠子在位置 $Y$, 那麼愛莉會把選中的珠子移動到位置 $X'$, 使得 $\overrightarrow{YX'} = r^k \overrightarrow{YX}$。愛莉的目標是把紅色的珠子移動到 $B$ 上。找出所有能讓愛莉在 2021 回合內達成目標的有理數 $r > 1$。 | [
"All $r = (b+1)/b$ with $b = 1, \\dots, 1010$.\n\nDenote the red and blue beads by $\\mathcal{R}$ and $\\mathcal{B}$, respectively. Introduce coordinates on the line and identify the points with their coordinates so that $R = 0$ and $B = 1$. Then, during the game, the coordinate $\\mathcal{R}$ is always smaller tha... | Taiwan | 2022 數學奧林匹亞競賽第二階段選訓營,國際競賽實作(二) | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Chinese; English | proof and answer | All rational numbers r = (b+1)/b for integers b = 1, 2, ..., 1010. | |
00tl | Let $\triangle ABC$ be an acute scalene triangle. Its $C$-excircle tangent to the segment $AB$ meets $AB$ at point $M$ and the extension of $BC$ beyond $B$ at point $N$. Analogously, its $B$-excircle tangent to the segment $AC$ meets $AC$ at point $P$ and the extension of $BC$ beyond $C$ at point $Q$. Denote by $A_1$ t... | [
"We shall use the standard notations for $ABC$, i.e. $\\angle ABC = \\beta$, $BC = a$ etc. We also write $s = \\frac{a+b+c}{2}$ for the semiperimeter and $r$ for the inradius.\nLet $MN$ intersect the altitude $AD$ ($D$ lies on $BC$) at the point $L$. We have that $\\angle BAD = 90^\\circ - \\beta$ and $\\angle AML ... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Co... | null | proof only | null | |
0cdc | Define $M = \{x \in \mathbb{R} \mid \lfloor nx \rfloor + \lfloor (n+1)x \rfloor = \lfloor (2n+1)x \rfloor, \forall n \in \mathbb{N}\}$.
a) Prove that, if $x \in M \cap [\frac{1}{2}, 1)$, then $\lfloor nx \rfloor = \lfloor \frac{n}{2} \rfloor, \forall n \in \mathbb{N}$.
b) Find the cardinal of the set $M \cap [0, 2023... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 4047 | |
08l6 | Problem:
Let $a$, $b$ and $c$ be positive real numbers such that $a b c = 1$. Prove the inequality
$$
\left(a b + b c + \frac{1}{c a}\right)\left(b c + c a + \frac{1}{a b}\right)\left(c a + a b + \frac{1}{b c}\right) \geq (1 + 2 a)(1 + 2 b)(1 + 2 c)
$$ | [
"Solution:\nBy Cauchy-Schwarz inequality and $a b c = 1$ we get\n$$\n\\begin{gathered}\n\\sqrt{\\left(b c + c a + \\frac{1}{a b}\\right)\\left(a b + b c + \\frac{1}{c a}\\right)} = \\sqrt{\\left(b c + c a + \\frac{1}{a b}\\right)\\left(\\frac{1}{c a} + a b + b c\\right)} \\geq \\\\\n\\left(\\sqrt{a b} \\cdot \\sqrt... | JBMO | 2008 Shortlist JBMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
06x3 | Let $\mathbb{Z}_{\geqslant 0}$ be the set of non-negative integers, and let $f: \mathbb{Z}_{\geqslant 0} \times \mathbb{Z}_{\geqslant 0} \rightarrow \mathbb{Z}_{\geqslant 0}$ be a bijection such that whenever $f\left(x_{1}, y_{1}\right)>f\left(x_{2}, y_{2}\right)$, we have $f\left(x_{1}+1, y_{1}\right)>f\left(x_{2}+1, ... | [
"We defer the constructions to the end of the solution. Instead, we begin by characterizing all such functions $f$, prove a formula and key property for such functions, and then solve the problem, providing constructions.\n\nCharacterization Suppose $f$ satisfies the given relation. The condition can be written mor... | IMO | International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | smallest 2500, largest 7500 | |
0gyo | On the side $AB$ of acute-angled triangle $ABC$ there is a point $K$, $M$ is the midpoint of $BC$, segments $AM$ and $CK$ intersect at a point $F$. It is known that $KF = AK$. Prove that $CF = AB$. | [
"On the ray $AM$ take a point $Q$, such that $MQ = AM$ (fig.19).\n\nThen $ACQB$ is a parallelogram and $\\angle KAF = \\angle FQC = \\angle CFQ \\Rightarrow CF = CQ = AB$, and we are done."
] | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
0esq | If the product of four consecutive integers is equal to the value of one of those integers, then what is the largest possible value for any of the integers? | [
"3\n\nWith a little trial and error it should soon be clear that this scenario is only possible if one of the integers is zero. The options are $-3$; $-2$; $-1$; $0$ or $-2$; $-1$; $0$; $1$ or $-1$; $0$; $1$; $2$ or $0$; $1$; $2$; $3$. The largest possible integer is thus $3$."
] | South Africa | South African Mathematics Olympiad Second Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 3 | |
0jsk | Problem:
Determine the number of integers $2 \leq n \leq 2016$ such that $n^{n}-1$ is divisible by $2,3,5,7$. | [
"Solution:\n\nOnly $n \\equiv 1 \\pmod{210}$ work. Proof: we require $\\gcd(n, 210) = 1$. Note that for all $p \\leq 7$ the order of $n$ $(\\bmod\\ p)$ divides $p-1$, hence is relatively prime to any $p \\leq 7$. So $n^{n} \\equiv 1 \\pmod{p} \\Longleftrightarrow n \\equiv 1 \\pmod{p}$ for each of these $p$."
] | United States | HMMT February | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 9 | |
0bau | Given an equilateral triangle $ABC$ and the points $M \in [BC]$, $N \in [AC]$, $P \in [AB]$ such that $BM = MC$, $3AN = NC$ and $2BP = AP$, find the measure of $\angle NMP$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 60° | |
002y | Sea $I$ el incentro de un triángulo $ABC$, y $D$ el punto de intersección de $AI$ con la circunferencia circunscrita a $ABC$. Sea $M$ el punto medio de $AD$, y $E$ el punto del segmento $BD$ tal que $IE$ es perpendicular a $BD$. Si $IB + IE = \frac{AD}{2}$, $ME$ es paralelo a $AB$, y el punto $M$ está en el interior de... | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety"
] | Español | proof and answer | ∠A = 60°, ∠B = 30°, ∠C = 90° | |
00ea | A set of points is called *antiparallelogram* if no four of them are the vertices of a parallelogram. Given a set $S$ of $2023$ points on the plane, no three of them on the same line, prove that there is a subset of $S$ containing $17$ points which is antiparallelogram. | [
"We run a greedy algorithm to find an antiparallelogram set $X$ contained in $S$. Set $X = \\emptyset$ to start. In each step, verify if there are points in $S \\setminus X$ that can be incorporated to $X$ while keeping it antiparallelogram. If so, choose any one of those points, add it to $X$, and repeat. Else, th... | Argentina | Rioplatense Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0eft | Problem:
Reši sistem enačb
$$
\begin{aligned}
\sqrt{2 x-3 y} &= \sqrt{x^{2}+4 y-1} \quad \text{in} \\
\sqrt{x-y+2}+2 &= x
\end{aligned}
$$ | [
"Solution:\n\nPrvo enačbo kvadriramo in preoblikujemo v $x^{2}-2 x+7 y-1=0$. V drugi enačbi koren osamimo, kvadriramo in izrazimo $y=-x^{2}+5 x-2$. To vstavimo v zgornjo enačbo in dobimo $2 x^{2}-11 x+5=0$. Za $x_{1}=5$ dobimo $y_{1}=-2$, rešitev ustreza prvotni enačbi. Za $x_{2}=\\frac{1}{2}$ dobimo $y_{2}=\\frac{... | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (5, -2) | |
0eqk | The last digit when $8\,045 - 4\,058$ is calculated is
(A) 1 (B) 3 (C) 5 (D) 7 (E) 9 | [
"The last digit will be found by subtracting $8$ from $15$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | D | |
01ya | Let $\mathbb{R}_+$ be the set of all positive integers. Find all functions $g: \mathbb{R}_+ \to \mathbb{R}_+$ such that numbers $x, y \in \mathbb{R}_+$ satisfy the equality
$$
xg(x + g(y)) = g(g(xy) + 1).
$$ | [
"Denote $xy = z$, then the variables $y$ and $z$ can take all positive real values independently. For these variables the equality has the form\n$$\n\\frac{z}{y}g\\left(\\frac{z}{y} + g(y)\\right) = g(g(z) + 1). \\qquad (1)\n$$\nFix any value of $z$ and vary $y$ over $\\mathbb{R}_+$, then $g(g(z) + 1)$ is fixed whi... | Belarus | BY 2020-2021 tst for Navid | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | g(x) = c/x for any positive constant c | |
0klb | Problem:
Aerith timed how long it took herself to solve a BMC monthly contest. She writes down the elapsed time as days:hours:minutes:seconds, and also simply as seconds. For example, if she spent $1,000,000$ seconds, she would write down $11:13:46:40$ and $1,000,000$.
Bob sees her numbers and subtracts them, ignorin... | [
"Solution:\n\nSay that Aerith took $d$ days, $h$ hours, $m$ minutes, and $s$ seconds. Bob would then get\n\n$$\n\\begin{aligned}\n\\Delta & =\\left(100^{3} d+100^{2} h+100 m+s\\right)-(24 \\cdot 60 \\cdot 60 d+60 \\cdot 60 h+60 m+s) \\\\\n& =\\left(100^{3}-24 \\cdot 60 \\cdot 60\\right) d+\\left(100^{2}-60 \\cdot 6... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 40 | |
046s | Given a prime number $p$ and a positive real number $\lambda$ less than $1$. Let $k$ be an integer, and let $S$ and $T$ be sets consisting of consecutive $s$ and $t$ integers, respectively, satisfying $1 \le s \le t < \frac{\lambda}{12}p$. Furthermore, assume that the number of elements in the set
$$
\{(x, y) \in S \ti... | [
"*Proof.* Let the solutions to the congruence equation $kx \\equiv y \\pmod{p}$ in $S \\times T$ be $(x_1, y_1), \\dots, (x_n, y_n)$, where $n \\ge 1 + \\lambda s \\ge 1$. Hence, we know that $S$ and $T$ are both non-empty, implying that $n \\ge 2$. Since $T$ is contained in a complete residue system modulo $p$, we... | China | China-TST-2023B | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
05n5 | Problem:
Soient $a$, $b$, $c > 0$. Montrer que
$$
\sqrt{\frac{a}{b+c}} + \sqrt{\frac{b}{c+a}} + \sqrt{\frac{c}{a+b}} > 2.
$$ | [
"Solution:\nL'idée est d'utiliser l'inégalité arithmético-géométrique pour minorer $\\frac{1}{\\sqrt{x+y}}$. Plus précisément, on a, par inégalité arithmético-géométrique :\n$$\n\\sqrt{\\frac{a}{b+c}} = \\frac{a}{\\sqrt{a(b+c)}} \\geqslant \\frac{a}{(a+b+c)/2} = \\frac{2a}{a+b+c}\n$$\nOn procède de même pour montre... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES - ENVOI No. 3 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0dc8 | In triangle $ABC$, such that $\angle ACB = 45^{\circ}$, let $O$ and $H$ be the circumcenter and orthocenter, respectively. The line passing through $O$ and perpendicular to $CO$ intersects $AC$ and $BC$ at $K$ and $L$, respectively. Prove that the perimeter of $KLH$ is equal to the diameter of the circumcircle of trian... | [
"Suppose that $AH$, $BH$ cut $(O)$ at the second points $E$, $F$. By angle chasing, we can see that $H$, $E$ are symmetric with respect to $BC$. Note that\n$$\n\\angle HBC = 90^{\\circ} - \\angle C = 45^{\\circ}\n$$\nso $\\angle CBE = \\angle CBH = 45^{\\circ}$.\nThus $\\angle CFE = 45^{\\circ}$, which implies that... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09so | Problem:
Zij $n$ een positief geheel getal en bekijk een vierkant met afmetingen $2^{n} \times 2^{n}$. We bedekken dit vierkant met een aantal (minstens 2) niet-overlappende rechthoeken, zodat elke rechthoek gehele afmetingen heeft en een tweemacht als oppervlakte. Bewijs dat twee van de rechthoeken in de bedekking de... | [
"Solution:\n\nMerk eerst op dat een rechthoek met gehele afmetingen een tweemacht als oppervlakte heeft precies dan als de afmetingen beide tweemachten zijn.\n\nBekijk een bedekking waarin geen twee rechthoeken met dezelfde afmetingen voorkomen. We bewijzen eerst dat er dan geen rechthoeken met breedte 1 voorkomen.... | Netherlands | IMO-selectietoets | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
01zr | Let $n \ge 3$ be a positive integer. Positive integers are written in two rows on the whiteboard $a_1\ a_2\ \dots\ a_n\ b_1\ b_2\ \dots\ b_n$, where the sets $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_n$ are some permutations of the numbers $1, 2, \dots, n$. The teacher allows the student Dima to use the following ... | [
"Answer: no. Each table $\\begin{array}{c} a_1 \\\\ b_1 \\end{array} \\begin{array}{c} a_2 \\\\ b_2 \\end{array} \\dots \\begin{array}{c} a_n \\\\ b_n \\end{array}$ corresponds to a pair of permutations $(s,t) \\in S_n \\times S_n$ such that\n$$\ns = \\begin{pmatrix} 1 & 2 & \\dots & n \\\\ a_1 & a_2 & \\dots & a_n... | Belarus | SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO | [
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | no | |
07rm | Let $M$ be the midpoint of side $BC$ of an equilateral triangle $ABC$. The point $D$ is on $CA$ extended such that $A$ is between $D$ and $C$. The point $E$ is on $AB$ extended such that $B$ is between $A$ and $E$, and $|MD| = |ME|$. The point $F$ is the intersection of $MD$ and $AB$. Prove that $\angle BFM = \angle BM... | [
"Let $N$ on $AC$ extended be such that $EN \\parallel BC$. Join $M$ to $N$.\n\n\n\nTriangles $\\triangle ABC$ and $\\triangle AEN$ are similar, hence $|AE| = |AN|$ and so $|BE| = |CN|$. Alternatively, we may define $N$ to be the point on $AG$ extended such that $|BE| = |CN|$ and $C$ is betw... | Ireland | Irish | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0l1t | Problem:
Suppose that $a$, $b$, and $c$ are distinct positive integers such that $a^{b} b^{c} = a^{c}$. Across all possible values of $a$, $b$, and $c$, compute the minimum value of $a + b + c$. | [
"Solution:\nWe claim that $(8, 2, 3)$ is the desired solution.\n\nObserve that $a^{c-b} = b^{c}$, so clearly $a \\neq 1$ and $b < a$. Furthermore, $a$ and $b$ must be distinct powers of the same integer.\n\nIf $a$ and $b$ were powers of an integer $n > 2$, then we would have $a + b + c \\geq 3^{2} + 3 + 1 = 13$. Th... | United States | HMMT November | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 13 | |
0c5i | Determine the numbers $x$, $y$, with $x$ integer and $y$ rational, for which the equality
$$
5(x^2 + xy + y^2) = 7(x + 2y)
$$
holds. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (-1, 3), (-1, 4/5), (0, 0), (0, 14/5), (1, 2), (1, -1/5) | |
074k | Problem:
Let $ABC$ be a triangle with circumcircle $\Gamma$. Let $M$ be a point in the interior of triangle $ABC$ which is also on the bisector of $\angle A$. Let $AM$, $BM$, $CM$ meet $\Gamma$ in $A_1$, $B_1$, $C_1$ respectively. Suppose $P$ is the point of intersection of $A_1C_1$ with $AB$; and $Q$ is the point of ... | [
"Solution:\n\nLet $A = 2\\alpha$. Then $\\angle A_1AC = \\angle BAA_1 = \\alpha$. Thus\n$$\n\\angle A_1B_1C = \\alpha = \\angle BB_1A_1 = \\angle A_1C_1C = \\angle BC_1A_1\n$$\nWe also have $\\angle B_1CQ = \\angle AA_1B_1 = \\beta$, say. It follows that triangles $MA_1B_1$ and $QCB_1$ are similar and hence\n$$\n\\... | India | INMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterials"
] | null | proof only | null | |
081j | Problem:
Nel quartiere di S. Maria ci sono $9897$ televisori. Solo tre famiglie del quartiere non possiedono televisori, mentre il $4\%$ ne ha due, il $2,5\%$ ne ha $3$ e lo $0,5\%$ ne ha addirittura $8$. Tutte le altre famiglie possiedono un solo televisore. Quante famiglie abitano nel quartiere di S. Maria?
(A) $99... | [
"Solution:\n\nLa risposta è (D). Se regaliamo un televisore alle famiglie che non ce l'hanno, la percentuale di famiglie che hanno solo un televisore diventa il $93\\%$, e il numero totale di televisori $9900$. Detto dunque $N$ il numero di famiglie del quartiere abbiamo la seguente relazione:\n$$\nN \\frac{93}{100... | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
04td | Mathematics clubs are very popular in a certain city. Any two of them have at least one common member. Prove that one can distribute rulers and compasses to the citizens in such a way that only one citizen gets both (compass and ruler) and any club has at its disposal both, compass and ruler, from its members. | [
"Let us consider the club $K$ with the least number of its members (in case there is more such clubs, we take any). We give to one of its members (let us call him Jacob) both a compass and a ruler. Each of the other members of the club will get a compass. Any other citizen will get a ruler. We show that this distri... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
03r0 | Let $n = \overline{\{abc\}}$ be a 3-digit number. If we can construct an isosceles triangle (including equilateral triangle) with $a$, $b$ and $c$ as the lengths of the sides. The number of such 3-digit integers $n$ is ( ).
(A) 45
(B) 81
(C) 165
(D) 216 | [
"If $a$, $b$ and $c$ are the lengths of the sides of a triangle, all of them are not zero, it follows that $a, b, c \\in \\{1, 2, \\dots, 9\\}$.\n\ni. If the triangle we construct is equilateral, let $n_1$ be the number of such 3-digit numbers. Since the three digits in such 3-digit number are equal, we have\n$$\nn... | China | China Mathematical Competition (Hainan) | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | MCQ | C | |
0kmg | Problem:
Milan has a bag of 2020 red balls and 2021 green balls. He repeatedly draws 2 balls out of the bag uniformly at random. If they are the same color, he changes them both to the opposite color and returns them to the bag. If they are different colors, he discards them. Eventually the bag has 1 ball left. Let $p... | [
"Solution:\n\nThe difference between the number of green balls and red balls in the bag is always $1$ modulo $4$. Thus the last ball must be green and $p=1$."
] | United States | HMMT Spring 2021 Guts Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2021 | |
0ja8 | Problem:
Let $n$ be the maximum number of bishops that can be placed on the squares of a $6 \times 6$ chessboard such that no two bishops are attacking each other. Let $k$ be the number of ways to put bishops on an $6 \times 6$ chessboard such that no two bishops are attacking each other. Find $n+k$. (Two bishops are c... | [
"Solution:\nColor the square with coordinates $(i, j)$ black if $i+j$ is odd and white otherwise, for all $1 \\leq i, j \\leq 6$. Looking at the black squares only, we note that there are six distinct diagonals which run upward and to the right, but that two of them consist only of a corner square; we cannot simult... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 74 | |
03gu | Problem:
Given the polynomial
$$
f(x) = x^{n} + a_{1} x^{n-1} + a_{2} x^{n-2} + \cdots + a_{n-1} x + a_{n}
$$
with integral coefficients $a_{1}, a_{2}, \ldots, a_{n}$, and given also that there exist four distinct integers $a, b, c$ and $d$ such that
$$
f(a) = f(b) = f(c) = f(d) = 5
$$
show that there is no integer $k$... | [
"Solution:\nLet $f(x)$ be as given, and suppose $f(a) = f(b) = f(c) = f(d) = 5$ for four distinct integers $a, b, c, d$.\n\nConsider the polynomial $g(x) = f(x) - 5$. Then $g(a) = g(b) = g(c) = g(d) = 0$, so $a, b, c, d$ are roots of $g(x)$.\n\nTherefore, $g(x)$ is divisible by $(x - a)(x - b)(x - c)(x - d)$, i.e.,... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
04ga | Let $p$ and $q$ be two parallel lines. Circle $k$ touches the line $p$ at $A$ and intersects $q$ at two different points, $B$ and $C$. Let $T$ be some point on $p$. Segments $\overline{TB}$ and $\overline{TC}$ intersect the shorter arc $\widehat{AC}$ at $K$ and $L$ respectively. Points $K$ and $L$ are both different fr... | [
"Let $P$ be the intersection of the lines $p$ and $KL$. Denote $\\angle TBC = x$.\n\nLine $BT$ is a transversal of the parallel lines $p$ and $q$, which implies that $\\angle BTA = \\angle TBC = x$.\nThe quadrilateral $BCLK$ is cyclic, hence $\\angle CLK = 180^\\circ - \\angle KBC = 180^\\c... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03uk | Find all positive integers $n$ such that there exist non-zero integers $x_1, x_2, \dots, x_n, y$, satisfying the following conditions
$$
\begin{cases} x_1 + \cdots + x_n = 0, \\ x_1^2 + \cdots + x_n^2 = n y^2. \end{cases}
$$ | [
"It is easy to see that $n > 1$.\n\nWhen $n = 2k$, $k \\in \\mathbb{N}$, let $x_{2i-1} = 1$, $x_{2i} = -1$, $i = 1, 2, \\dots, k$, and $y = 1$, then the condition is satisfied.\n\nWhen $n = 2k + 3$, $k \\in \\mathbb{N}$, let $y = 2$, $x_1 = 4$, $x_2 = x_3 = x_4 = x_5 = -1$, $x_{2i} = 2$, $x_{2i+1} = -2$, $i = 3, 4,... | China | China Western Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | All positive integers except 1 and 3 | |
0ds7 | Find the smallest positive integer $n$ so that $\sqrt{\frac{1^2+2^2+\dots+n^2}{n}}$ is an integer. | [
"Let $\\frac{1^2+2^2+\\dots+n^2}{n} = m^2$, $m \\in \\mathbb{Z}^+$. Then $(n+1)(2n+1) = 6m^2$. Thus $n$ is odd and $n = 6p \\pm 1$ or $n = 6p+3$ for some integer $p$.\n\nIf $n = 6p+3$, then $6m^2 = (6p+4)(12p+7)$ which gives no solution since $3$ does not divide the RHS.\n\nIf $n = 6p-1$, $m^2 = p(12p-1)$. Since $p... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 337 | |
0l4f | Problem:
Compute the number of triples $(f, g, h)$ of permutations on $\{1,2,3,4,5\}$ such that
$$
\begin{aligned}
& f(g(h(x)))=h(g(f(x)))=g(x), \\
& g(h(f(x)))=f(h(g(x)))=h(x), \text{ and } \\
& h(f(g(x)))=g(f(h(x)))=f(x)
\end{aligned}
$$
for all $x \in \{1,2,3,4,5\}$. | [
"Solution:\nLet $f g$ represent the composition of permutations $f$ and $g$, where $(f g)(x)=f(g(x))$ for all $x \\in \\{1,2,3,4,5\\}$.\n\nEvaluating $f g h f h$ in two ways, we get\n$$\nf = g f h = (f g h) f h = f g h f h = f(g h f) h = f h h,\n$$\nso $h h = 1$. Similarly, we get $f, g$, and $h$ are all involution... | United States | HMMT February 2024 | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | 146 | |
09p7 | A country consists of $n$ islands. Some pairs of islands are connected by bridges. For any two islands that are connected via some sequence of bridges, the *distance* between them is defined as the minimum number of bridges that must be crossed to travel from one island to the other.
Assume that each island is directly... | [
"*Answer:* If $m = 1$, then the maximum distance is $n-1$. If $m \\ge 2$, then the maximum distance is\n$$\n3 \\left\\lfloor \\frac{n}{m+1} \\right\\rfloor - \\varepsilon(n), \\quad \\text{where } \\varepsilon(n) = \\begin{cases} 3 & \\text{if } n \\equiv 0 \\pmod{m+1}, \\\\ 2 & \\text{if } n \\equiv 1 \\pmod{m+1},... | Mongolia | MMO2025 Round 4 | [
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | If m = 1, the maximum distance is n − 1. If m ≥ 2, the maximum distance is 3 floor(n/(m+1)) − ε(n), where ε(n) = 3 if n ≡ 0 mod (m+1), 2 if n ≡ 1 mod (m+1), and 1 otherwise. | |
0g3y | Problem:
Let $ABCD$ be a convex quadrilateral such that the circle with diameter $AB$ is tangent to the line $CD$, and the circle with diameter $CD$ is tangent to the line $AB$. Prove that the two intersection points of these circles and the point $AC \cap BD$ are collinear. | [
"Solution:\n\nLet $X$ be the tangency point of $CD$ with the first circle and $Y$ the tangency point of $AB$ with the second circle. Further, let $P$ be the intersection of $AC$ with $BD$. As we aim to use Pappus's theorem, we also introduce the points $Q = AX \\cap DY$ and $R = BX \\cap CY$.\n\nWe claim that $\\tr... | Switzerland | IMO Selection | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof only | null | |
06mc | $ABCD$ is a parallelogram with $\angle B$ acute. A circle is tangent to $BC$, $CD$ and $DA$. The circle intersects $AC$ at $M$ and $N$, where $M$ is closer to $A$ than $N$. If $AM = 9$, $MN = 16$ and $NC = 2$, find the area of $ABCD$. | [
"Let $P$, $Q$, $R$ be the points where the circle touches $BC$, $CD$ and $DA$ respectively. Using power, we have $AR = \\sqrt{AM \\cdot AN} = 15$ and $CP = CQ = \\sqrt{CN \\cdot CM} = 6$. Let $DR = DQ = x$ and $S$ be the foot of the perpendicular from $C$ to $AD$. Then we have $DS = x - 6$, $AS = 21$ and $CD = x + ... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 324*sqrt(2) | |
09z7 | On an $8 \times 8$-board there is a beetle on every square. At a certain moment the distribution of the beetles on the board changes: every beetle crawls either one square to the left or one square diagonally to the bottom right. If a beetle can make neither of the two movements without falling off the board, it stays ... | [] | Netherlands | First Round | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 32 | |
0hwr | Problem:
Let $ABC$ be an acute triangle with circumcenter $O$ and incenter $I$. Points $E, M$ lie on $AC$ and $F, N$ on $AB$ so that $BE \perp AC$, $CF \perp AB$, $\angle ABM = \angle CBM$ and $\angle ACN = \angle BCN$. Prove that $I$ lies on $EF$ if and only if $O$ lies on $MN$. | [
"Solution:\n\nLet $a = BC$, $b = CA$, $c = AB$. It is well-known (and follows from, say, Stewart's Theorem) that $AM = \\frac{bc}{a + c}$ and $AN = \\frac{bc}{a + b}$.\n\nNow, the distances from $O$ to $BC$, $AC$, $AB$ are $R \\cos \\alpha$, $R \\cos \\beta$, $R \\cos \\gamma$, respectively, where $\\alpha, \\beta,... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0avf | Problem:
Suppose $\frac{1}{2} \leq x \leq 2$ and $\frac{4}{3} \leq y \leq \frac{3}{2}$. Determine the minimum value of
$$
\frac{x^{3} y^{3}}{x^{6}+3 x^{4} y^{2}+3 x^{3} y^{3}+3 x^{2} y^{4}+y^{6}}
$$ | [
"Solution:\nNote that\n$$\n\\begin{aligned}\n\\frac{x^{3} y^{3}}{x^{6}+3 x^{4} y^{2}+3 x^{3} y^{3}+3 x^{2} y^{4}+y^{6}} & =\\frac{x^{3} y^{3}}{\\left(x^{2}+y^{2}\\right)^{3}+3 x^{3} y^{3}} \\\\\n& =\\frac{1}{\\frac{\\left(x^{2}+y^{2}\\right)^{3}}{x^{3} y^{3}}+3} \\\\\n& =\\frac{1}{\\left(\\frac{x^{2}+y^{2}}{x y}\\r... | Philippines | 18th PMO National Stage Oral Phase | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 27/1081 | |
0ecv | We inscribe a regular octagon in a square with side of length $a$, so that 4 sides of the octagon lie on the sides of the square. Express the side length of the inscribed octagon in terms of $a$. | [
"Denote by $x$ the side length of the inscribed octagon. The four triangles that are formed at the vertices of the square are isosceles right-angled triangles. Since their hypotenuse is of length $x$, their legs are of length $\\frac{x}{\\sqrt{2}}$. Thus\n$$\na = x + 2\\frac{x}{\\sqrt{2}} = x + x\\sqrt{2} = x(\\sqr... | Slovenia | National Math Olympiad 2015 – First Round | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | a(√2 - 1) | |
08z2 | Determine the number of ways to choose distinct $25$ integers from $1$ to $50$ such that for any two integers chosen, one is not a divisor of the other. | [
"For each odd integer $n$ from $1$ to $49$, define the group of $n$ as the set of integers from $1$ to $50$ which can be expressed as $n \\cdot 2^k$ for a non-negative integer $k$. Then each integer from $1$ to $50$ belongs to only one group.\nFor any two integers in the same group, one is a divisor of the other. T... | Japan | Japan 2022 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1632 | |
01ri | An $n \times n$ ($n \ge 4$) square is divided into $n^2$ unit cells. Find all possible values of $n$ such that this square can be covered with some layers of 4-cell figures of the following shape [ ] (i.e. each cell of the square must be covered with the same number of these figures).
(The sides of each figure must coi... | [
"Answer: $n = 4k$, $k \\in \\mathbb{N}$.\nIf $n = 4k$, $k \\in \\mathbb{N}$, then this square can be covered with one layer (and then with any number of layers) of the figures [ ] [ ] [ ] [ ].\n\nLet $n = 2m + 1$, $m \\ge 2$. In this case we use chess coloring of the square. Suppose that the square is covered with ... | Belarus | Final Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | n = 4k | |
0l17 | A disphenoid is a tetrahedron whose triangular faces are congruent to one another. What is the least total surface area of a disphenoid whose faces are scalene triangles with integer side lengths?
(A) $\sqrt{3}$ (B) $3\sqrt{15}$ (C) 15 (D) $15\sqrt{7}$ (E) $24\sqrt{6}$ | [
"**Answer (D):** First note that adjacent faces must be rotations of one another to create a disphenoid (see the figure below). If they were instead mirrored, then the other two faces would be isosceles (and be neither scalene nor congruent to the others, as required).\n\n\n\nNext, to deter... | United States | AMC 12 A | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | D | |
0fu7 | Problem:
Ein Bauklotz, bestehend aus 7 Einheitswürfeln, hat die Form eines $2 \times 2 \times 2$ Würfels mit einem fehlenden Eckeinheitswürfel. Aus einem Würfel der Kantenlänge $2^{n}$, $n \geq 2$, wird ein beliebiger Einheitswürfel entfernt. Zeige, dass sich der verbleibende Körper stets aus Bauklötzen aufbauen lässt... | [
"Solution:\n\nMan überlegt sich leicht, dass man aus 8 Bauklötzen einen doppelt so grossen Bauklotz herstellen kann. Induktiv folgt daraus, dass man für jedes $n \\geq 1$ einen Bauklotz der Kantenlänge $2^{n}$ herstellen kann. Wir nennen dies einen $n$-Klotz. Betrachte nun einen Würfel der Kantenlänge $2^{n}$ mit e... | Switzerland | IMO Selektion | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
09gm | Let $p$, $q$ and $r$ be different prime numbers. For a positive integer $n$, let $f(n)$ denote the greatest common divisor of $n$, $p$, $q$ and $r$. Find the number of triples $(a, b, c)$ such that $1 \le a, b, c \le pqr$ and $f(a)$, $f(b)$, $f(c)$, $f(a+b)$, $f(b+c)$, $f(c+a)$, $f(a+b+c)$ are mutually different from e... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 0 | |
05q5 | Problem:
Dans les carrés suivants, on s'autorise à remplacer tous les 0 par des 1 et réciproquement sur toute une ligne ou toute une colonne ou toute une diagonale. Dans chaque cas, peut-on n'obtenir que des 0 ?
$$
\left(\begin{array}{llll}
0 & 1 & 1 & 0 \\
1 & 1 & 0 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 1 & 1
\end{array}\r... | [
"Solution:\n\nDans tous les cas, on va dire que l'on agit sur une ligne, une colonne ou une diagonale si on y remplace les 0 par des 1 et réciproquement.\n\nDans le $1^{\\text{er}}$ cas, on remarque que l'on a un nombre impair de 1 : il y en a 9 dans la grille. Or, lors d'une action, la parité du nombre de 1 ne cha... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | First grid: no. Second grid: no. Third grid: yes. | |
02ok | Let $ABCD$ be a convex quadrilateral such that $AD = DC$, $AC = AB$ and $\angle ADC = \angle CAB$. Let $M$ and $N$ be the midpoints of $AD$ and $AB$. Prove that triangle $MNC$ is isosceles. | [
"Since $AD = CD$, $AB = AC$ and $\\angle ADC = \\angle BAC$, triangles $ADC$ and $BAC$ are similar by case SAS. Segments $CM$ and $CN$ are corresponding medians, so $\\frac{CM}{CN} = \\frac{CA}{CB}$ and $\\angle BCN = \\angle ACM \\iff \\angle BCN + \\angle NCA = \\angle ACM + \\angle NCA \\iff \\angle BCA = \\angl... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06jw | 4031 lines are drawn on the plane. No two lines are parallel or perpendicular, and no three lines meet at one point. Determine the maximum number of acute-angled triangles that may be formed. | [
"The maximum number of acute-angled triangles is $2729148240$.\nLet $n = 2015$, so that there are $2n + 1$ lines. We fix one of the lines $\\ell$ and place it as the $x$-axis of the coordinate plane. Then the other $2n$ lines can be partitioned into two groups, one consisting of those lines with positive slopes and... | Hong Kong | Year 2016 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 2729148240 | |
03uw | Suppose that a ball with radius $1$ moves freely inside a regular tetrahedron with edge length $4\sqrt{6}$. Then the area of the inner surface of the container, which the ball can never touch, is ______. | [
"As shown in Fig. 1, consider the situation where the ball is in a corner of the container. Draw the plane $A_1B_1C_1 \\parallel ABC$, tangent to the ball at point $D$. Then the ball center $O$ is also the center of the tetrahedron $P-A_1B_1C_1$, with $PO \\perp A_1B_1C_1$ and the foot point $D$ being the center of... | China | China Mathematical Competition | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | final answer only | 72√3 | |
0192 | Two disks are placed inside a square. What is the maximal proportion of the square that can be covered by the disks, if they are not permitted to overlap? Is it possible to cover more if overlap is allowed? | [
"Suppose the square has side length $1$ and centre $C$. Denote the radii of the two disks by $x$ and $y$ and the distance between their centres by $d$. The centres of the circles are then restricted within two squares centred at $C$ of sides $1 - 2x$ and $1 - 2y$, respectively. The farthest they can be from each ot... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | Maximum without overlap: pi*(9/2 - 3*sqrt(2)). Allowing overlap: yes, more can be covered. | |
0kvp | Problem:
Suppose $ABCD$ is a rectangle whose diagonals meet at $E$. The perimeter of triangle $ABE$ is $10\pi$ and the perimeter of triangle $ADE$ is $n$. Compute the number of possible integer values of $n$. | [
"Solution:\n\nFor each triangle $\\mathcal{T}$, we let $p(\\mathcal{T})$ denote the perimeter of $\\mathcal{T}$.\n\nFirst, we claim that $\\frac{1}{2} p(\\triangle ABE) < p(\\triangle ADE) < 2 p(\\triangle ABE)$. To see why, observe that\n$$\np(\\triangle ADE) = EA + ED + AD < 2(EA + ED) = 2(EA + EB) < 2 p(\\triang... | United States | HMMT February | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 47 | |
06fw | Let $a$, $b$, $c$ be the sides of a triangle, and $T$ its area. Prove that
$$
a^2 + b^2 + c^2 \ge 4\sqrt{3}T + (a-b)^2 + (b-c)^2 + (c-a)^2.
$$
When does the equality hold? | [
"Let $s$ be the semiperimeter of the triangle, and let $x = s - a$, $y = s - b$ and $z = s - c$. Note that $x$, $y$, $z > 0$ by the triangle inequality. Now,\n$$\n\\begin{align*}\n& a^2 + b^2 + c^2 \\ge 4\\sqrt{3T} + (a-b)^2 + (b-c)^2 + (c-a)^2 \\\\\n\\Leftrightarrow & \\quad 2ab + 2bc + 2ca \\ge 4\\sqrt{3s(s-a)(s-... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a = b = c | |
0ayl | Problem:
Factor $(a+1)(a+2)(a+3)(a+4)-120$ completely into factors with integer coefficients. | [
"Solution:\n\nWe have\n$$\n\\begin{aligned}\n(a+1)(a+2)(a+3)(a+4)-120 & =(a+1)(a+4)(a+2)(a+3)-120 \\\\\n& =\\left(a^{2}+5a+4\\right)\\left(a^{2}+5a+6\\right)-120 \\\\\n& =\\left(a^{2}+5a+5\\right)^{2}-1-120=\\left(a^{2}+5a+5\\right)^{2}-121 \\\\\n& =\\left(a^{2}+5a+5+11\\right)\\left(a^{2}+5a+5-11\\right) \\\\\n& =... | Philippines | 20th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | (a^2 + 5a + 16)(a - 1)(a + 6) | |
0fa9 | Problem:
Do there exist 4 vectors in the plane so that none is a multiple of another, but the sum of each pair is perpendicular to the sum of the other two? Do there exist 91 non-zero vectors in the plane such that the sum of any 19 is perpendicular to the sum of the others? | [
"## Problem 22"
] | Soviet Union | 25th ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
0l1v | Problem:
Compute the number of ways there are to assemble 2 red unit cubes and 25 white unit cubes into a $3 \times 3 \times 3$ cube such that red is visible on exactly 4 faces of the larger cube. (Rotations and reflections are considered distinct.) | [
"Solution:\n\nWe do casework on the two red unit cubes; they can either be in a corner, an edge, or the center of the face.\n\n- If they are both in a corner, they must be adjacent - for each configuration, this corresponds to an edge, of which there are 12.\n\n\n\n- If one is in the corner... | United States | HMMT February 2024 | [
"Geometry > Solid Geometry > 3D Shapes",
"Discrete Mathematics > Other"
] | null | final answer only | 114 | |
0ism | Problem:
Prove that 572 is not a juggling sequence. | [
"Solution:\n\nWe are given $j(0) = 5$, $j(1) = 7$ and $j(2) = 2$. So $f(3) = 3 + j(0) = 8$ and $f(1) = 1 + j(1) = 8$. Thus $f(3) = f(1)$ and so $f$ is not a permutation of $\\mathbb{Z}$, and hence 572 is not a juggling pattern. (In other words, there is a \"collision\" at times $t \\equiv 2 \\pmod{3}$.)"
] | United States | 11th Annual Harvard-MIT Mathematics Tournament - Team Round: B Division | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Other"
] | null | proof only | null | |
0bv4 | Problem:
Se consideră hexagonul inscriptibil $ABCDEF$ şi $H_{1}$, $H_{2}$, $H_{3}$, $H_{4}$ ortocentrele triunghiurilor $ABC$, $BCD$, $DEF$, $FAE$. Să se arate că pentru orice puncte $M$, $N$, $P$ şi $Q$ din plan care satisfac relaţia $\overrightarrow{H_{1}M} + \overrightarrow{H_{3}P} = \overrightarrow{H_{2}N} + \over... | [] | Romania | OLIMPIADA DE MATEMATICĂ ETAPA LOCALĂ | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0iml | Problem:
Compute the circumradius of cyclic hexagon $A B C D E F$, which has side lengths $A B = B C = 2$, $C D = D E = 9$, and $E F = F A = 12$. | [
"Solution:\n\nAnswer: 8. Construct point $E'$ on the circumcircle of $A B C D E F$ such that $D E' = E F = 12$ and $E' F = D E = 9$; then $\\overline{B E'}$ is a diameter. Let $B E' = d$. Then $C E' = \\sqrt{B E'^2 - B C^2} = \\sqrt{d^2 - 4}$ and $B D = \\sqrt{B E'^2 - D E'^2} = \\sqrt{d^2 - 144}$. Applying Ptolemy... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 8 | |
0bue | Problem:
a) $I_{1} = \int (\cos x \cdot \cos 2x \cdot \cos 3x) \, dx$
b) $I_{2} = \int_{1}^{\sqrt{3}} \frac{1}{x \sqrt{x^{4} + 1}} \, dx$ | [] | Romania | Olimpiada Națională de Matematică | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Trigonometric functions"
] | null | proof and answer | a) I1 = x/4 + (1/24) sin(6x) + (1/8) sin(2x) + (1/16) sin(4x) + C.
b) I2 = (1/2)[asinh(1) − asinh(1/3)] = (1/2) ln( 3(1 + sqrt(2)) / (1 + sqrt(10)) ). | |
01do | Determine all positive integers $a$ and all primes $p$ fulfilling the equation
$$
(a - p)^3 = a + p.
$$ | [
"Writing $n = a - p$ transforms the equation into\n$$\nn^3 = a + p = n + 2p,\n$$\nso that\n$$\n2p = n^3 - n = n(n + 1)(n - 1),\n$$\nwhich is divisible by $3$. Therefore $p = 3$ and $n = 2$, whence $a = 5$."
] | Baltic Way | Baltic Way 2016 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a = 5, p = 3 | |
0cun | Initially, Bazil thinks of eight cells on a chessboard, no two of which are in the same row or in the same column. Then Pete makes a series of guesses. By a guess, he places onto the chessboard 8 rooks of Pete's rooks which stand on positions he thinks of. If Bazil indicates an even number of rooks, Pete wins. Otherwis... | [
"2 guesses."
] | Russia | XLIII Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English; Russian | proof and answer | 2 | |
0452 | On a screen formed by $n \times n$ ($n \ge 2$) squares, every square displays initially one of the three colors: red, yellow, and blue. Every second, the screen changes the color of each square following the rules below:
* for each square $A$ whose current color is red, if there is a yellow square sharing a side with t... | [
"**Proof:** We first prove the claim: if the screen eventually turns into one color, say blue, then there must be one square which is constantly blue.\nSuppose not, namely, suppose that eventually all squares are blue but every square has changed color at some time. We construct an oriented graph $G$ as follows: th... | China | 2022 China Team Selection Test | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0gar | 求所有實係數多項式 $P$, 使得:
$$
P(x)P(x+1) = P(x^2 - x + 3) \quad \forall x \in \mathbb{R},
$$
其中 $\mathbb{R}$ 表所有實數所成的集合。 | [
"所有滿足題目要求之實係數多項式 $P$ 為零多項式與\n$$\nP(x) = (x^2 - 2x + 3)^n \\quad \\forall x \\in \\mathbb{R},\n$$\n其中 $n$ 是任一個非負整數。\n代入原式易知上述 $P(x)$ 都是此函數方程的解, 以下考慮 $P(x)$ 不是零多項式的情況。\n首先我們先證明 $P$ 沒有實根。使用反證法, 如果 $P$ 有實根 $\\alpha$, 原式中代 $x = \\alpha$ 可得\n$$\nP(\\alpha^2 - \\alpha + 3) = P(\\alpha)P(\\alpha + 1) = 0.\n$$\n設 $\\beta = ... | Taiwan | 二〇一七數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | Zero polynomial and P(x) = (x^2 - 2x + 3)^n for any nonnegative integer n. | |
0a1o | Problem:
Zij $\triangle ABC$ een driehoek met $|AB| < |AC| < |BC|$, omgeschreven cirkel $\Gamma$ met middelpunt $O$. Zij $\omega_1$ de cirkel met middelpunt $B$ en straal $|AC|$ en zij $\omega_2$ de cirkel met middelpunt $C$ en straal $|AB|$. De cirkels $\omega_1$ en $\omega_2$ snijden in een punt $E$ zodanig dat $A$ ... | [
"Solution:\n\nOmdat $|BE| = |AC|$ en $|CE| = |AB|$ is $ABEC$ een parallellogram. Dat betekent dat $BE \\parallel AC$. Aan de andere kant is $ABGC$ een koordenvierhoek met $|CG| = |AB|$. Hieruit volgt dat $ABGC$ een gelijkbenig trapezium is met $BG \\parallel AC$. In het bijzonder zijn $B$, $G$ en $E$ collineair. Ge... | Netherlands | IMO-selectietoets | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler lin... | null | proof only | null | |
0kxk | Problem:
If $a$, $b$, $c$, and $d$ are pairwise distinct positive integers that satisfy $\operatorname{lcm}(a, b, c, d) < 1000$ and $a + b = c + d$, compute the largest possible value of $a + b$. | [
"Solution:\nLet $a' = \\frac{\\operatorname{lcm}(a, b, c, d)}{a}$. Define $b'$, $c'$, and $d'$ similarly. We have that $a'$, $b'$, $c'$, and $d'$ are pairwise distinct positive integers that satisfy\n$$\n\\frac{1}{a'} + \\frac{1}{b'} = \\frac{1}{c'} + \\frac{1}{d'}\n$$\nLet $T$ be the above quantity. We have\n$$\na... | United States | HMMT February | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 581 | |
0abe | Prove that for every real root $x$ of $x^2 + p x + q = 0$, where $p, q \in \mathbb{R}$ and $a > 0$ we have $x \ge \frac{4q - (p + a)^2}{4a}$. | [
"The equation $x^2 + p x + q = 0$ has real roots, therefore $p^2 - 4q \\ge 0$. Let one of the roots be $\\frac{-p \\pm \\sqrt{p^2 - 4q}}{2}$.\nThen\n$$\n\\begin{aligned}\n\\frac{-p \\pm \\sqrt{p^2 - 4q}}{2} &\\ge \\frac{4q - (p + a)^2}{4a} \\\\\n&\\Leftrightarrow 2a(-p \\pm \\sqrt{p^2 - 4q}) \\ge 4q - (p + a)^2 \\\... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0jxr | Problem:
Find all real numbers $x$ satisfying the equation $x^{3}-8=16 \sqrt[3]{x+1}$. | [
"Solution:\nLet $f(x)=\\frac{x^{3}-8}{8}$. Then $f^{-1}(x)=\\sqrt[3]{8x+8}=2\\sqrt[3]{x+1}$, and so the given equation is equivalent to $f(x)=f^{-1}(x)$. This implies $f(f(x))=x$. However, as $f$ is monotonically increasing, this implies that $f(x)=x$.\n\nAs a result, we have\n$$\n\\frac{x^{3}-8}{8}=x \\Longrightar... | United States | February 2017 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | x = -2, 1 ± sqrt(5) | |
0aam | Let $ABCD$ be a cyclic quadrilateral such that $AB = AD + BC$ and $CD < AB$. The diagonals $AC$ and $BD$ intersect at $P$, while the lines $AD$ and $BC$ intersect at $Q$. The angle bisector of $\angle APB$ meets $AB$ at $T$. Show that the circumcenter of $\triangle CTD$ lies on the circumcircle of $\triangle CQD$. | [] | North Macedonia | Fourth Memorial Mathematical Contest | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | English | proof only | null | |
06re | Determine all pairs $(f, g)$ of functions from the set of positive integers to itself that satisfy
$$
f^{g(n)+1}(n) + g^{f(n)}(n) = f(n+1) - g(n+1) + 1
$$
for every positive integer $n$. Here, $f^{k}(n)$ means $\underbrace{f(f(\ldots f}_{k}(n) \ldots))$. | [
"The given relation implies\n$$\n\\begin{equation*}\nf\\left(f^{g(n)}(n)\\right) < f(n+1) \\quad \\text{ for all } n, \\tag{1}\n\\end{equation*}\n$$\nwhich will turn out to be sufficient to determine $f$.\nLet $y_1 < y_2 < \\ldots$ be all the values attained by $f$ (this sequence might be either finite or infinite)... | IMO | 52nd International Mathematical Olympiad 2011 Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Functional equations"
] | null | proof and answer | f(n) = n for all positive integers n, and g(n) = 1 for all positive integers n | |
0hor | Problem:
Let $A_{1} A_{2} \cdots A_{2 n}$ be a convex $2 n$-gon. Prove that there is an $i$ ($1 \leq i \leq n$) and a pair of parallel lines, each intersecting the $2 n$-gon only once, one at $A_{i}$ and one at $A_{n+i}$. | [
"Solution:\n\nConsider the \"highest\" and \"lowest\" points of the $2 n$-gon, with respect to any chosen orientation. Unless a side of the $2 n$-gon is perfectly horizontal, these points will be unique and thus a horizontal line through them will not meet the $2 n$-gon again. These two points will also be vertices... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0d1z | Prove that if $a$ is an integer relatively prime with $35$ then
$$
\left(a^{4}-1\right)\left(a^{4}+15 a^{2}+1\right) \equiv 0 \quad \bmod 35
$$ | [
"If $a$ is relatively prime with $35$ then it is relatively prime with both $5$, $7$.\nSince $a$ is relatively prime with $5$ then, by Fermat, $a^{4} \\equiv 1 \\bmod 5$ which implies\n$$\n\\left(a^{4}-1\\right)\\left(a^{4}+15 a^{2}+1\\right) \\equiv 0 \\quad \\bmod 5\n$$\nSince $a$ is relatively prime with $7$ the... | Saudi Arabia | Preselection tests for the full-time training | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English | proof only | null | |
06tq | Determine the largest real number $a$ such that for all $n \geqslant 1$ and for all real numbers $x_{0}, x_{1}, \ldots, x_{n}$ satisfying $0=x_{0}<x_{1}<x_{2}<\cdots<x_{n}$, we have
$$
\frac{1}{x_{1}-x_{0}}+\frac{1}{x_{2}-x_{1}}+\cdots+\frac{1}{x_{n}-x_{n-1}} \geqslant a\left(\frac{2}{x_{1}}+\frac{3}{x_{2}}+\cdots+\fra... | [
"We first show that $a=\\frac{4}{9}$ is admissible. For each $2 \\leqslant k \\leqslant n$, by the Cauchy-Schwarz Inequality, we have\n$$\n\\left(x_{k-1}+\\left(x_{k}-x_{k-1}\\right)\\right)\\left(\\frac{(k-1)^{2}}{x_{k-1}}+\\frac{3^{2}}{x_{k}-x_{k-1}}\\right) \\geqslant(k-1+3)^{2}\n$$\nwhich can be rewritten as\n$... | IMO | IMO 2016 Shortlisted Problems | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | 4/9 | |
0bk7 | Construct outside the square $ABCD$ the right isosceles triangle $ABE$, with hypotenuse $[AB]$. Denote $N$ the midpoint of the segment $[AD]$ and $\{M\} = CE \cap AB$, $\{P\} = CN \cap AB$, $\{F\} = PE \cap MN$. Take on the straight line $FP$ the point $Q$ so that $[CE$ is the bisector of the angle $\angle QCB$. Prove ... | [
"Clearly $PA = PB/2$, so $PA = AB = BC$. This leads to $\\triangle APE \\equiv \\triangle BCE$ (SAS), which implies $CE = PE$ and $\\overline{BEC} \\equiv \\overline{AEP}$. Now\n\n$m(\\widehat{BEC}) + m(\\widehat{AEC}) = 90^\\circ$ yields $m(\\widehat{AEC}) + m(\\widehat{AEP}) = 90^\\circ$, so $m(\\widehat{CEP}) = ... | Romania | 65th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0iww | Problem:
Simplify: $i^{0} + i^{1} + \cdots + i^{2009}$. | [
"Solution:\nBy the geometric series formula, the sum is equal to\n$$\n\\frac{i^{2010} - 1}{i - 1} = \\frac{-2}{i - 1} = 1 + i.\n$$"
] | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 1 + i | |
0ezj | Problem:
Five $n$-digit binary numbers have the property that every two numbers have the same digits in just $m$ places, but no place has the same digit in all five numbers. Show that $\frac{2}{5} \leq \frac{m}{n} \leq \frac{3}{5}$. | [] | Soviet Union | 4th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
05jc | Problem:
Un quadrilatère $ABCD$ est inscrit dans un cercle. Ses diagonales se coupent au point $K$. Le cercle passant par $A$, $B$, $K$ croise les droites $(BC)$ et $(AD)$ aux points $M$ et $N$ respectivement. Montrer que $KM = KN$. | [
"Solution:\n\n\nOn a $(NM, NK) = (BM, BK) = (BC, BD) = (AC, AD) = (AK, AN) = (MK, MN)$ donc $MKN$ est isocèle en $K$."
] | France | OFM | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
076e | Problem:
Let $ABCD$ be a convex quadrilateral. Let the diagonals $AC$ and $BD$ intersect in $P$. Let $PE$, $PF$, $PG$ and $PH$ be the altitudes from $P$ onto the sides $AB$, $BC$, $CD$ and $DA$ respectively. Show that $ABCD$ has an incircle if and only if
$$
\frac{1}{PE} + \frac{1}{PG} = \frac{1}{PF} + \frac{1}{PH}
$$ | [
"Solution:\nLet $AP = p$, $BP = q$, $CP = r$, $DP = s$; $AB = a$, $BC = b$, $CD = c$ and $DA = d$. Let $\\angle APB = \\angle CPD = \\theta$. Then $\\angle BPC = \\angle DPA = \\pi - \\theta$. Let us also write $PE = h_1$, $PF = h_2$, $PG = h_3$ and $PH = h_4$.\n\n\n\nObserve that\n$$\nh_1 ... | India | INMO | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0dbc | Chess horse attacks fields in distance $\sqrt{5}$. Let several horses are put on the board $12 \times 12$ such, that every square of size $2 \times 2$ contains at least one horse. Find the maximal possible number of cells that are not under attack (horse doesn't attack its own cell).
 | [
"Let's note, that if we put a horse in any green cell, then it will attack a grey cell. Since green cells form a square $2 \\times 2$, so one of them contains a horse, so at least one grey is under attack.\n\nNow let's split the board into 72 pairs like in the figure. According what we said above, at least 72 cells... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 72 | |
03jq | Problem:
Define $\{a_n\}_{n=1}$ as follows: $a_1 = 1989^{1989}$; $a_n$, $n > 1$, is the sum of the digits of $a_{n-1}$. What is the value of $a_5$? | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 9 | |
00ih | We are given a triangle $ABC$ and a point $D$ on the side $BC$. Let $U$ be the circumcenter of $\triangle BDA$ and $V$ the circumcenter of $\triangle CDA$. Prove that the triangles $AUV$ and $ABC$ are similar.
G. Baron, Vienna | [
"Let the point $C'$ be chosen in such a way that triangles $\\triangle ABC$ and $\\triangle ACC'$ are similar and have no common interior points. Furthermore, let $D'$ be chosen on $CC'$ such that triangles $\\triangle ABD$ and $\\triangle ACD'$ are also similar. This means that $\\triangle ACC'$ results from $\\tr... | Austria | Austria 2010 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations... | English | proof only | null | |
04c1 | Prove that there is no integer $n \ge 2$ such that
$$
f(x) = \cos(x\sqrt{1}) + \cos(x\sqrt{2}) + \dots + \cos(x\sqrt{n})
$$
is a periodic function. | [
"Suppose, on the contrary, that the function $f$ is periodic with the period $T$ for some integer $n \\ge 2$. Hence, $f(T) = f(0) = n$.\nNow we have\n$$\nf(T) = \\cos(T\\sqrt{1}) + \\cos(T\\sqrt{2}) + \\dots + \\cos(T\\sqrt{n}) = n,\n$$\nfrom which we conclude that $\\cos(T\\sqrt{1}) = \\cos(T\\sqrt{2}) = \\dots = ... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Other"
] | English | proof only | null | |
00u3 | Let $ABC$ be a triangle and let the tangent at $B$ to its circumcircle meet the internal bisector of angle $A$ at $P$. The line through $P$ parallel to $AC$ meets $AB$ at $Q$. Assume that $Q$ lies in the interior of segment $AB$ and let the line through $Q$ parallel to $BC$ meet $AC$ at $X$ and $PC$ at $Y$. Prove that ... | [
"\nSince $BP$ is tangent to the circumcircle and $BR \\parallel AC$, we have $\\angle PBZ = \\angle BAC = \\angle TBR$. It follows that the right-angled triangles $RTB$ and $PZB$ are similar and therefore $\\frac{PZ}{RT} = \\frac{PB}{RB}$.\n\nAnalogously the triangles $REB$ and $PDB$ are al... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0cff | Let $n \ge 2$ be an integer number. Determine all the values of $m \in \mathbb{N}$ such that there is $A \in \mathcal{M}_n(\mathbb{C})$ with the property that $\text{rank}(A^k) + \text{rank}(A^{n-k}) = m$, for any $k \in \{1, 2, \dots, n-1\}$. | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Linear transformations"
] | English | proof and answer | All m of the form 2r with 0 ≤ r ≤ n, together with m = n and m = n − 2. | |
034b | Problem:
Solve the system
$$
\left\lvert\, \begin{aligned}
& 3 \cdot 4^{x}+2^{x+1} \cdot 3^{y}-9^{y}=0 \\
& 2 \cdot 4^{x}-5 \cdot 2^{x} \cdot 3^{y}+9^{y}=-8
\end{aligned}\right.
$$ | [
"Solution:\nSet $u=2^{x}>0$ and $v=3^{y}>0$. Then the system becomes\n$$\n\\left\\lvert\\, \\begin{aligned}\n& 3 u^{2}+2 u v-v^{2}=0 \\\\\n& 2 u^{2}-5 u v+v^{2}=-8\n\\end{aligned}\\right.\n$$\nThe first equation can be written as $(u+v)(3 u-v)=0$, whence $u=-v$ or $3 u=v$. The first case is not possible since $u$ a... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | x = 1/2, y = 1 + (1/2) log_3 2 | |
0gb4 | 試證明: 存在一個整係數多項式 $f(x)$, 使得:
(1) $f(x) = 0$ 沒有有理實根。
(2) 對於任何正整數 $n$, 均存在整數 $m$, 使得 $f(m)$ 是 $n$ 的倍數。 | [
"(注意可能的 $f(x)$ 不唯一。) 以下證明\n$$\nf(x) = (x^2 + 1)(x^2 + 2)(x^2 - 2)(x^2 + 7)\n$$\n滿足條件。易見 $f(x) = 0$ 沒有有理實根。\n\n1. 首先證明：對於所有奇質數的幂次方 $p^\\alpha$，都存在整數 $m_{p^\\alpha}$ 使得\n$$\np^{\\alpha}|f(m_{p^{\\alpha}}).\n$$\n事實上，我們可以用歸納法證明以下引理：\n\n**引理一.** 對於所有奇質數 $p$，存在 $r_p \\in \\{-1, -2, 2\\}$，使得對於所有 $\\alpha \\ge 1$，必存在 $m_{p... | Taiwan | 二〇一七數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic re... | null | proof only | null | |
0i8g | Problem:
There are 100 houses in a row on a street. A painter comes and paints every house red. Then, another painter comes and paints every third house (starting with house number 3) blue. Another painter comes and paints every fifth house red (even if it is already red), then another painter paints every seventh hous... | [
"Solution:\nHouse $n$ ends up red if and only if the largest odd divisor of $n$ is of the form $4k+1$. We have 25 values of $n=4k+1$; 13 values of $n=2(4k+1)$ (given by $k=0,1,2,\\ldots,12$); 7 values of $n=4(4k+1)$ ($k=0,1,\\ldots,6$); 3 values of $n=8(4k+1)$ ($k=0,1,2$); 2 of the form $n=16(4k+1)$ (for $k=0,1$); ... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | null | proof and answer | 52 | |
052n | Find all pairs of positive rational numbers where the sum of the numbers in a pair is an integer and the sum of (multiplicative) inverses of the numbers in a pair is also an integer. | [
"Let the numbers in the pair be represented as reduced fractions $\\frac{a}{b}$ and $\\frac{c}{d}$. For\n$$\n\\frac{a}{b} + \\frac{c}{d} = \\frac{ad + bc}{bd}\n$$\nto be an integer, we must have\n$$\nad + bc = k \\cdot bd \\quad (2)\n$$\nwith $k$ being some integer. By writing the equality (2) as $bc = (kb - a) \\c... | Estonia | Open Contests | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | (1/2, 1/2), (1, 1), (2, 2) | |
0i7x | Problem:
A positive integer will be called "sparkly" if its smallest (positive) divisor, other than $1$, equals the total number of divisors (including $1$). How many of the numbers $2, 3, \ldots, 2003$ are sparkly? | [
"Solution:\nSuppose $n$ is sparkly; then its smallest divisor other than $1$ is some prime $p$. Hence, $n$ has $p$ divisors. However, if the full prime factorization of $n$ is $p_{1}^{e_{1}} p_{2}^{e_{2}} \\cdots p_{r}^{e_{r}}$, the number of divisors is $\\left(e_{1}+1\\right)\\left(e_{2}+1\\right) \\cdots\\left(e... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 3 | |
0i7i | Problem:
10 people are playing musical chairs with $n$ chairs in a circle. They can be seated in $7!$ ways (assuming only one person fits on each chair, of course), where different arrangements of the same people on chairs, even rotations, are considered different. Find $n$. | [
"Solution:\n\nThe number of ways 10 people can be seated on $n$ chairs is $n!$ multiplied by the number of ways one can choose $n$ people out of 10. Hence we must solve $7! = n! \\cdot \\dfrac{10!}{n! \\cdot (10-n)!}$. This is equivalent to $(10-n)! = \\dfrac{10!}{7!} = 8 \\cdot 9 \\cdot 10 = 720 = 6!$. We therefor... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 4 | |
06up | Let $p$ be an odd prime number and $\mathbb{Z}_{>0}$ be the set of positive integers. Suppose that a function $f: \mathbb{Z}_{>0} \times \mathbb{Z}_{>0} \rightarrow \{0,1\}$ satisfies the following properties:
- $f(1,1)=0$;
- $f(a, b)+f(b, a)=1$ for any pair of relatively prime positive integers $(a, b)$ not both equal... | [
"Denote by $\\mathbb{A}$ the set of all pairs of coprime positive integers. Notice that for every $(a, b) \\in \\mathbb{A}$ there exists a pair $(u, v) \\in \\mathbb{Z}^{2}$ with $u a+v b=1$. Moreover, if $(u_{0}, v_{0})$ is one such pair, then all such pairs are of the form $(u, v)=\\left(u_{0}+k b, v_{0}-k a\\rig... | IMO | International Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null |
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