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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0hmq | Problem:
As usual, let $n!$ denote the product of the integers from $1$ to $n$ inclusive. Determine the largest integer $m$ such that $m!$ divides $100! + 99! + 98!$. | [
"Solution:\n\nThe answer is $m = 98$. Set\n$$\nN = 98! + 99! + 100! = 98! (1 + 99 + 99 \\cdot 100)\n$$\nHence $N$ is divisible by $98!$. But\n$$\n\\frac{N}{98!} = 1 + 99 \\cdot 101\n$$\nis not divisible by $99$. Hence $N$ is not divisible by $99!$."
] | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 98 | |
09vu | Francisca has a square piece of paper whose sides have length $10$ cm. She also has a rectangular piece of paper having the exact same area as the square piece of paper. She puts the rectangle right on top of the square, putting the left bottom corner of both pieces of paper in the same spot. Exactly one quarter of the... | [
"E) $13\\frac{1}{3}$"
] | Netherlands | First Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | E | |
0g4x | Problem:
Sei $\omega_{1}$ ein Kreis mit Durchmesser $J K$. Sei $t$ die Tangente an $\omega_{1}$ bei $J$ und sei $U \neq J$ ein weiterer Punkt auf $t$. Sei $\omega_{2}$ der kleinere Kreis mit Mittelpunkt $U$, welcher $\omega_{1}$ an einem einzigen Punkt $Y$ berührt. Sei $I$ der zweite Schnittpunkt von $J K$ mit dem Umk... | [
"Solution:\n\nSei $O$ der Mittelpunkt von $\\omega_{1}$ und sei $\\omega_{3}$ der Umkreis des Dreiecks $J Y U$. Wir behaupten, dass $F$ auf $\\omega_{3}$ liegt. Um dies zu zeigen, bemerken wir zuerst, dass $U, Y$, und $O$ kollinear sind, da $\\omega_{1}$ und $\\omega_{2}$ sich im Punkt $Y$ berühren. Da die Dreiecke... | Switzerland | Zweite Runde 2023 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00gy | Let $ABC$ be a triangle with $\angle BAC \neq 90^{\circ}$. Let $O$ be the circumcenter of the triangle $ABC$ and let $\Gamma$ be the circumcircle of the triangle $BOC$. Suppose that $\Gamma$ intersects the line segment $AB$ at $P$ different from $B$, and the line segment $AC$ at $Q$ different from $C$. Let $ON$ be a di... | [
"From the assumption that the circle $\\Gamma$ intersects both of the line segments $AB$ and $AC$, it follows that the 4 points $N, C, Q, O$ are located on $\\Gamma$ in the order of $N, C, Q, O$ or in the order of $N, C, O, Q$. The following argument for the proof of the assertion of the problem is valid in either ... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"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 | |
046e | Let $a, b, c, d$ be nonnegative real numbers not exceeding $1$. Prove that
$$
\frac{1}{1+a+b} + \frac{1}{1+b+c} + \frac{1}{1+c+d} + \frac{1}{1+d+a} \le \frac{4}{1+2\sqrt[4]{abcd}}.
$$ | [
"Notice that when $\\sqrt{ac} \\le x$, we have\n$$\n\\frac{1}{x+a} + \\frac{1}{x+c} - \\frac{2}{x+\\sqrt{ac}} = \\frac{(\\sqrt{a}-\\sqrt{c})^2(\\sqrt{ac}-x)}{(x+a)(x+c)(x+\\sqrt{ac})} \\le 0. \\quad (*)\n$$\nGiven the conditions, $\\sqrt{ac} \\le 1 \\le 1+b$, and $\\sqrt{ac} \\le 1+d$. Substituting $x = 1+b$ and $x... | China | 22nd Chinese Girls' Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0a31 | Tomorrow, the Janssen family will be travelling by car and they have a nice route in mind. The youngest of the family notes that their planned stopover in Germany is exactly halfway along the route in terms of distance. Father responds: "When we cross the border after 150 kilometres tomorrow, our stopover will only be ... | [] | Netherlands | Dutch Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 400 | |
0iyq | Problem:
Let $f(x) = x^{4} + 14 x^{3} + 52 x^{2} + 56 x + 16$. Let $z_{1}, z_{2}, z_{3}, z_{4}$ be the four roots of $f$. Find the smallest possible value of $\left|z_{a} z_{b} + z_{c} z_{d}\right|$ where $\{a, b, c, d\} = \{1,2,3,4\}$. | [
"Solution:\nNote that $\\frac{1}{16} f(2x) = x^{4} + 7x^{3} + 13x^{2} + 7x + 1$. Because the coefficients of this polynomial are symmetric, if $r$ is a root of $f(x)$ then $\\frac{4}{r}$ is as well. Further, $f(-1) = -1$ and $f(-2) = 16$ so $f(x)$ has two distinct roots on $(-2,0)$ and two more roots on $(-\\infty,... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Equatio... | null | proof and answer | 8 | |
0bqu | Let $f : [0, 2] \to \mathbb{R}$ be a continuous function. Compute
$$
\lim_{t \to 0} \int_{0}^{1} \frac{f(x+t) - f(x)}{t} \, dx.
$$ | [] | Romania | 67th NMO Shortlisted Problems | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Applications"
] | English | proof and answer | f(1) - f(0) | |
0ek3 | Problem:
V eksplicitni, implicitni in odsekovni obliki zapiši enačbo premice s pozitivnim smernim koeficientom, ki s koordinatnima osema oblikuje pravokotni trikotnik s ploščino $9$ kvadratnih enot in abscisno os seka pri $x = -3$. | [
"Solution:\n\nPremica s koordinatnima osema oblikuje pravokotni trikotnik, katerega ploščino izračunamo kot polovični produkt med katetama $p = \\frac{k_1 \\cdot k_2}{2}$. Iz naloge je razvidno, da je dolžina ene katete $3$, dolžino druge katete pa izračunamo s pomočjo ploščine in dobimo rezultat $6$. Ker je smerni... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | Intercept form: x/(-3) + y/6 = 1; Implicit form: -2x + y - 6 = 0; Explicit form: y = 2x + 6 | |
0gzu | Acute-angled triangle $ABC$ is given. On the perpendicular bisectors to sides $AB$ and $BC$ respectively, points $P$ and $Q$ are chosen. Let $M$ and $N$ be the projections of $P$ and $Q$ onto $AC$ (Fig.09). It turns out, that $2MN = AC$. Prove, that circumcircle of triangle $PBQ$ passes through the circumcenter of tria... | [
"Let $O$ be the circumcenter of $ABC$. Let our perpendicular bisectors meet sides $AB$ and $BC$ at points $K$ and $L$ respectively. Then $KL$ is a midline of triangle $ABC$. Therefore, $KL \\parallel AC$ and $2KL = AC$. Thus, $MKLN$ is parallelogram, because $KL \\parallel MN$ and $KL = MN$. Due to the fact, that $... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03qt | The solution set of the inequality $\sqrt{\log_2 x - 1} + \frac{1}{2} \log_{\frac{1}{2}} x^3 + 2 > 0$ is ( ).
(A) [2, 3]
(B) (2, 3]
(C) [2, 4)
(D) (2, 4] | [
"$$\n\\begin{cases} \\sqrt{\\log_2 x - 1} - \\frac{3}{2} \\log_2 x + \\frac{3}{2} + \\frac{1}{2} > 0, \\\\ \\log_2 x - 1 \\ge 0. \\end{cases}\n$$\nLet $t = \\sqrt{\\log_2 x - 1}$, we have\n$$\n\\begin{cases} t - \\frac{3}{2}t^2 + \\frac{1}{2} > 0, \\\\ t \\ge 0. \\end{cases}\n$$\n\nThe solution of the above inequal... | China | China Mathematical Competition (Hainan) | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | MCQ | C | |
05rs | Problem:
Soit $P_{1}, P_{2}, \ldots, P_{2019}$ des polynômes non constants à coefficients réels tels que
$$
P_{1}\left(P_{2}(x)\right)=P_{2}\left(P_{3}(x)\right)=\ldots=P_{2019}\left(P_{1}(x)\right)
$$
pour tout réel $x$. Démontrer que $P_{1}=P_{2}=\ldots=P_{2019}$. | [
"Solution:\n\nNous allons en fait démontrer le résultat plus général suivant : si $P_{1}, \\ldots, P_{2 n+1}$ sont des polynômes non constants à coefficients réels et tels que $P_{1} \\circ P_{2}= P_{2} \\circ P_{3}=\\ldots=P_{2 n+1} \\circ P_{1}$, alors $P_{1}=\\ldots=P_{2 n+1}$. Par commodité, dans la suite, on p... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
05s5 | Problem:
Soit $n \geqslant 2$ et soient $x_{1}, x_{2}, \ldots, x_{n}$ des nombres réels tels que $x_{1}+x_{2}+\cdots+x_{n}=0$ et $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$.
Montrer qu'il existe $i$ tel que $x_{i} \geqslant \frac{1}{\sqrt{n(n-1)}}$. | [
"Solution:\n\nOn note $N^{+}$ le nombre d'indices $i$ tels que $x_{i}>0$ et $N^{-}$ le nombre d'indices $i$ tels que $x_{i} \\leqslant 0$, de sorte que $N^{+}+N^{-}=n$. On note également\n$$\nS_{1}^{+}=\\sum_{i \\text{ tel que } x_{i}>0} x_{i} \\quad \\text{et} \\quad S_{1}^{-}=\\sum_{i \\text{ tel que } x_{i} \\le... | France | Préparation Olympique Française de Mathématiques - ENVOI 4 : POT-POURRI | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0f1g | Problem:
A triangle $ABC$ has unit area. The first player chooses a point $X$ on side $AB$, then the second player chooses a point $Y$ on side $BC$, and finally the first player chooses a point $Z$ on side $CA$. The first player tries to arrange for the area of $XYZ$ to be as large as possible, the second player tries... | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Choose the midpoint of side AB first; this guarantees the area of the resulting triangle is one quarter of the original, and the second player’s optimal reply makes the last choice irrelevant. | |
0drl | Find all positive integers $k$ such that $k^k + 1$ is divisible by $30$. Justify your answer. | [
"An integer is divisible by $30$ iff it is divisible by $2$, $3$ and $5$.\n\nNote that $2 \\mid k^k + 1$ iff $k$ is odd. Thus we may assume that $k$ is odd. Write $k = 2t + 1$.\n\nIf $k \\equiv 0$ or $1 \\pmod{3}$, then $3 \\nmid k^k + 1$. If $k \\equiv 2 \\equiv -1 \\pmod{3}$, then $3 \\mid k^k + 1$ iff $k$ is odd... | Singapore | Singapur 2015 | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | All positive integers k with k ≡ 29 mod 30, i.e., k = 30n + 29 for n = 0, 1, 2, … | |
03hd | Problem:
Let $n$ be a fixed positive integer. To any choice of $n$ real numbers satisfying
$$
0 \leq x_{i} \leq 1, \quad i=1,2, \ldots, n
$$
there corresponds the sum
(*)
$$
\begin{aligned}
& \sum_{1 \leq i<j \leq n}\left|x_{i}-x_{j}\right| \\
& \qquad \begin{aligned}
= & \left|x_{1}-x_{2}\right|+\left|x_{1}-x_{3}\righ... | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | floor(n^2/4) | |
0abd | Find the value of $y$ so that $\sqrt{y^2 + 2y + 1}$, $\frac{y^2 + 3y - 1}{3}$, $y-1$ are consecutive terms in the arithmetic progression. | [
"If $p, q, r$ are consecutive terms in the arithmetic progression, then $p + r = 2q$. Because of this, we have\n$$\n\\sqrt{y^2 + 2y + 1} + y - 1 = 2 \\frac{y^2 + 3y - 1}{3}\n$$\n$$\n3(\\sqrt{(y+1)^2} + y - 1) = 2(y^2 + 3y - 1)\n$$\n\nSince $\\sqrt{(y+1)^2} = |y+1|$, we have\n$$\n\\begin{aligned}\n3(|y+1| + y - 1) &... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof and answer | {-2, -1, 1} | |
033a | Problem:
Find all non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients such that $P(x) Q(x+1) = P(x+2004) Q(x)$ for any $x$. | [
"Solution:\n\nSet $R(x) = P(x) P(x+1) \\ldots P(x+2003)$. It follows by the given condition that if $x$ is greater than the largest real root of $P(x)$, then\n$$\n\\frac{Q(x)}{R(x)} = \\frac{Q(x+1)}{R(x+1)}\n$$\nWe get by induction that $\\frac{Q(x)}{R(x)} = \\frac{Q(x+n)}{R(x+n)}$ for any positive integer $n$. Not... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All solutions are given by choosing any nonconstant real polynomial P, and setting Q(x) = c · P(x) P(x+1) … P(x+2003) for some nonzero real constant c. | |
033q | Problem:
Let $a$, $b$ and $n$ be positive integers. Denote by $K(n)$ the number of the representations of $1$ as a sum of $n$ numbers of the form $\frac{1}{k}$, where $k$ is a positive integer. Let $L(a, b)$ be the least positive integer $m$ such that the equation $\sum_{i=1}^{m} \frac{1}{x_{i}}=\frac{a}{b}$ has a solu... | [
"Solution:\nThe function $K(n)$ is increasing for $n \\geq 3$, since $\\sum_{i=1}^{n} \\frac{1}{x_{i}}=1$ implies that\n$$\n\\sum_{i=1}^{n-1} \\frac{1}{x_{i}}+\\frac{1}{x_{n}+1}+\\frac{1}{x_{n}(x_{n}+1)}=1\n$$\nThus it is enough to find $t \\leq L(b)$ such that $K(t+2)+2 L(b) \\geq d(b)$, where $d(b)$ is the number... | Bulgaria | Bulgarian Mathematical Competitions | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Other"
] | null | proof only | null | |
02st | Problem:
Pedro escreveu a lista de todos os números inteiros positivos menores que $10000$ nos quais cada um dos algarismos $1$ e $2$ aparecem uma única vez. Por exemplo, $1234$, $231$, $102$ foram escritos na lista, mas $1102$ e $235$ não estão na lista. Quantos números há na lista escrita por Pedro? | [
"Solution:\nNotemos que um número natural menor do que $10000$ pode ser representado por exatamente quatro algarismos escolhidos em $\\{0,1,2,3,4,5,6,7,8,9\\}$, possivelmente com repetições. Assim, temos quatro posições para serem preenchidas com esses algarismos. Por exemplo, o número $12$ seria representado por $... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 768 | |
073y | Problem:
Let $ABC$ be a triangle and let $P$ be an interior point such that $\angle BPC = 90^{\circ}$, $\angle BAP = \angle BCP$. Let $M, N$ be the mid-points of $AC, BC$ respectively. Suppose $BP = 2PM$. Prove that $A, P, N$ are collinear. | [
"Solution:\n\nExtend $CP$ to $D$ such that $CP = PD$. Let $\\angle BCP = \\alpha = \\angle BAP$. Observe that $BP$ is the perpendicular bisector of $CD$. Hence $BC = BD$ and $BCD$ is an isosceles triangle. Thus $\\angle BDP = \\alpha$. But then $\\angle BDP = \\alpha = \\angle BAP$. This implies that $B, P, A, D$ a... | India | Indian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0cq6 | For some $2011$ positive integers, all $2011 \cdot 2010/2$ of their pairwise sums are written onto a blackboard. Is it possible that exactly a third part of the numbers written are divisible by $3$, and another third part of them has residue $1$ modulo $3$? (I. Bogdanov)
Для некоторых $2011$ натуральных чисел выписали... | [
"Могло.\n\nПусть в нашем наборе $a$, $b$ и $c$ чисел, дающих соответственно остатки $0$, $1$ и $2$ при делении на $3$. Тогда условие переписывается в виде\n$$\n\\frac{a(a-1)}{2} + bc = \\frac{b(b-1)}{2} + ac = \\frac{c(c-1)}{2} + ab,\n$$\n$a+b+c = 2011$.\n\nНесложно видеть, что этим условиям удовлетворяет, например... | Russia | Russian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English, Russian | proof and answer | Yes; for example, take the integers from one to two thousand eleven. | |
0ew2 | Problem:
Given $n > 1$ points, some pairs joined by an edge (an edge never joins a point to itself). Given any two distinct points you can reach one from the other in just one way by moving along edges. Prove that there are $n - 1$ edges. | [
"Solution:\nEvery point must have at least one edge. We show that there is a point with just one edge. Suppose the contrary, that every point has at least two edges. We now construct a path in which the same edge or point never appears twice. Starting from any point $b$, move along an edge to $c$. $c$ is not alread... | Soviet Union | 1st ASU | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | n - 1 edges | |
0cjl | Let $r$ be a positive rational number such that the numbers $r$ and $\sqrt{r+1}$ have the same fractional part. Show that $r$ is an integer.
Andrei Bâra | [
"The hypothesis implies that $\\sqrt{r+1} - r = a$, where $a \\in \\mathbb{Z}$.\n\nWe get: $r + 1 = r^2 + 2ra + a^2$.\n\nLet $r = \\frac{m}{n}$, with $m, n \\in \\mathbb{N}^*$ and $\\text{gcd}(m, n) = 1$. Then,\n$$\nmn + n^2 = m^2 + 2mna + a^2n^2,\n$$\nand therefore\n$$\nm^2 = n(m + n - 2ma - a^2n).\n$$\nHence, $m^... | Romania | 75th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0bte | Let $x$ and $y$ be positive real numbers such that $x + y^{2016} \ge 1$. Prove that $x^{2016} + y > 1 - 1/100$. | [
"If $x \\ge 1 - 1/(100 \\cdot 2016)$, then\n$$\nx^{2016} \\ge \\left(1 - \\frac{1}{100 \\cdot 2016}\\right)^{2016} > 1 - 2016 \\cdot \\frac{1}{100 \\cdot 2016} = 1 - \\frac{1}{100}\n$$\nby Bernoulli's inequality, whence the conclusion.\n\nTo establish the latter, refer again to Bernoulli's inequality to write\n$$\n... | Romania | 2016 Eighth Romanian Master of Mathematics | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0k8t | Problem:
Let $a, b, c$ be positive real numbers such that $a \leq b \leq c \leq 2 a$. Find the maximum possible value of
$$
\frac{b}{a}+\frac{c}{b}+\frac{a}{c}
$$ | [
"Solution:\nFix the values of $b, c$. By inspecting the graph of\n$$\nf(x)=\\frac{b}{x}+\\frac{x}{c}\n$$\nwe see that on any interval the graph attains its maximum at an endpoint. This argument applies when we fix any two variables, so it suffices to check boundary cases in which $b=a$ or $b=c$, and $c=b$ or $c=2 a... | United States | HMMT November 2019 | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 7/2 | |
06xw | A sequence of integers $a_{0}, a_{1}, a_{2}, \ldots$ is called kawaii, if $a_{0}=0, a_{1}=1$, and, for any positive integer $n$, we have
$$
\left(a_{n+1}-3 a_{n}+2 a_{n-1}\right)\left(a_{n+1}-4 a_{n}+3 a_{n-1}\right)=0 .
$$
An integer is called kawaii if it belongs to a kawaii sequence.
Suppose that two consecutive pos... | [
"We start by rewriting the condition in the problem as:\n$$\na_{n+1}=3 a_{n}-2 a_{n-1}, \\text{ or } a_{n+1}=4 a_{n}-3 a_{n-1}\n$$\nWe have $a_{n+1} \\equiv a_{n}$ or $a_{n-1}(\\bmod 2)$ and $a_{n+1} \\equiv a_{n-1}$ or $a_{n}(\\bmod 3)$ for all $n \\geqslant 1$. Now, since $a_{0}=0$ and $a_{1}=1$, we have that $a_... | IMO | International Mathematical Olympiad Shortlist | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
015j | Determine all functions $f: \mathbf{R} \to \mathbf{R}^+$ satisfying
$$
f(x + y) = f(x)f(y) (1 + (f(y) - 1)^{2009})
$$
for all $x, y \in \mathbf{R}$. (Here $\mathbf{R}^+$ is the set of all positive real numbers.) | [
"Let first $y = 0$:\n$$\nf(x) = f(x)f(0)(1 + (f(0) - 1)^{2009}).\n$$\n$f(x) > 0$ may be divided away:\n$$\n1 = f(0)(1 + (f(0) - 1)^{2009}).\n$$\nThe right-hand side is strictly increasing in $f(0)$, so there is a unique solution $f(0) = 1$.\n\nNow let $x = 0$:\n$$\nf(y) = f(0)f(y)(1 + (f(y) - 1)^{2009}).\n$$\nDivid... | Baltic Way | Baltic Way SHL | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1 for all real x | |
04js | A magical triangulation is a partition of a triangle on smaller triangles by a finite number of segments whose endpoints are vertices of the triangle or points in its interior, such that in every point (including the vertices of the triangle) meets the same number of segments.
What is the maximal number of smaller tria... | [
"Let $n$ be the number of smaller triangles, $t$ the number of points in the triangulation (including the vertices of the triangle), $d$ the number of segments (including the sides of the triangle) and $k$ the number of segments meeting in each point of the triangulation.\n\nObviously $t \\cdot k = 2 \\cdot d$ hold... | Croatia | Croatian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 19 | |
0i7w | Problem:
If $x y = 5$ and $x^{2} + y^{2} = 21$, compute $x^{4} + y^{4}$. | [
"Solution:\nWe have $441 = (x^{2} + y^{2})^{2} = x^{4} + y^{4} + 2(x y)^{2} = x^{4} + y^{4} + 50$, yielding $x^{4} + y^{4} = 391$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 391 | |
02e5 | The Poincaré plane is a half-plane bounded by a line $R$. The lines are taken to be (1) the half-lines perpendicular to $R$, and (2) the semicircles with center on $R$. Show that given any line $L$ and any point $P$ not on $L$, there are infinitely many lines through $P$ which do not intersect $L$. Show that if $ABC$ i... | [
"\nFor the first part there are three cases to consider: $L$ a semicircle and $P$ inside it, $L$ a semicircle and $P$ outside it, and $L$ a perpendicular half-line. The diagram above shows the first case. Take $L$ to have radius $R$. The semicircle through $P$ with the same center $O$ as $L... | Brazil | VIII OBM | [
"Geometry > Non-Euclidean Geometry > Hyperbolic Geometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08wy | Suppose the least common multiple of three positive integers $x$, $y$, $z$ is $2100$.
What is the minimum possible value that the sum $x + y + z$ can take? | [
"Since $2100 = 2^2 \\cdot 3 \\cdot 5^2 \\cdot 7$, if we take $x = 5^2 = 25$, $y = 7$, $z = 2^2 \\cdot 3 = 12$, then the least common multiple of $x$, $y$, $z$ is $2100$ and we get $x+y+z = 44$.\n\nNow let us show that $44$ is the desired minimum value. Assume that the least common multiple of $x$, $y$, $z$ is $2100... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 44 | |
02ma | Problem:
O valor absoluto $|a|$ de um número $a$ qualquer é definido por
$$
|a|=\left\{\begin{array}{cl}
a & \text{ se } a>0 \\
0 & \text{ se } a=0 \\
-a & \text{ se } a<0
\end{array}\right.
$$
Por exemplo, $|6|=6$, $|-4|=4$ e $|0|=0$. Quanto vale $N=|5|+|3-8|-|-4|$ ?
(a) 4
(b) -4
(c) 14
(d) -14
(e) 6 | [] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | e | |
0bak | Consider $n \in \mathbb{N}$, $n \ge 2$, and $A$ a unitary ring with $n$ elements, such that the equation $x^{n+1} + x = 0$ has in $A \setminus \{0\}$ the unique solution $x = 1$. Prove that $A$ is a field.
Ioan Băetu | [
"Because $1$ satisfies $x^{n+1} + x = 0$, we have $1 + 1 = 0$, so all non-zero elements of the additive group $(A, +)$ are of order $2$. By Cauchy's theorem $n = 2^m$, $m \\in \\mathbb{N}^*$.\n\nLet $a \\in A$, $a \\neq 0$. The ring $A$ being finite, there exists $p < q$, such that $a^p = a^q$. By successive multip... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Abstract Algebra > Field Theory"
] | null | proof only | null | |
0hur | Problem:
Given a polynomial $P(x)$ with integer coefficients, assume that for every positive integer $n$ we have $P(n) > n$. Consider the sequence
$$
x_{1} = 1,\quad x_{2} = P(x_{1}),\quad \ldots,\quad x_{n} = P(x_{n-1}),\quad \ldots
$$
If for every positive integer $N$ there exists a member of the sequence divisible ... | [
"Solution:\n\nAssume the contrary. The polynomial $Q(x) = P(x) - x$ is non-decreasing for all $x$ greater than some $M$, as otherwise it wouldn't be positive, and it has to be bigger than $1$ for each $n$.\n\nIf there are infinitely many integers $n$ such that $P(n) = n + 1$ then we would have $P(x) = x + 1$ (nonze... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | P(x) = x + 1 | |
0624 | Problem:
Lassen sich für jede positive ganze Zahl $n$ nicht-negative ganze Zahlen $a, b, c, d, e, f, g, h$ mit
$$
n=\frac{2^{a}-2^{b}}{2^{c}-2^{d}} \cdot \frac{2^{e}-2^{f}}{2^{g}-2^{h}}
$$
finden? Die Antwort ist zu begründen. | [
"Solution:\nVorbemerkung: Im Folgenden sei $n$ ungerade. Ohne Einschränkung ist $a>b, c>d, e>f, g>h$. Die Bedingungsgleichung ist dann äquivalent zu\n$$\nn 2^{d+h-b-f}\\left(2^{c-d}-1\\right)\\left(2^{g-h}-1\\right)=\\left(2^{a-b}-1\\right)\\left(2^{e-f}-1\\right).\n$$\nDa $n$ ungerade ist, gilt $d+h-b-f=0$, und ma... | Germany | 1. IMO-Auswahlklausur | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | null | proof and answer | No; for example, n = 19 (and more generally n ≡ 19 mod 64) cannot be represented in the given form. | |
0daf | In each of the cells of a $13 \times 13$ board is written an integer such that the integers in adjacent cells differ by $1$. If there are two $2$s and two $24$s on this board, how many $13$s can there be? | [
"Let us define the distance between any two cells of the board to be the minimum number of steps required to get from one of the cells to the other one provided that one moves between adjacent cells in each step. Consequently, the distance of any cell to itself is $0$, the distance between adjacent cells is $1$ and... | Saudi Arabia | Team selection tests for GMO 2018 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 13 | |
0bsk | Let $A \in \mathcal{M}_2(\mathbb{R})$ be a matrix satisfying the conditions:
$$
\det (A^{2014} - I_2) = \det (A^{2014} + I_2) \quad \text{and} \quad \det (A^{2016} - I_2) = \det (A^{2016} + I_2).
$$
Prove that $\det (A^n - I_2) = \det (A^n + I_2)$, for any positive integer $n$. | [
"If $M \\in \\mathcal{M}_2(\\mathbb{R})$, we have\n$$\n\\det (M - xI_2) = x^2 - \\operatorname{tr}(M)x + \\det(M), \\quad \\forall x \\in \\mathbb{R}. \\quad (1)\n$$\nLet $A \\in \\mathcal{M}_2(\\mathbb{R})$ be a matrix satisfying the hypothesis. From (1) we obtain\n$$\n\\operatorname{tr}(A^{2014}) = \\operatorname... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
0enf | a sequence $(a_n)_{n=1}^{\infty}$ of natural numbers has the property that for any $n \ge 1$, $a_{n+1} = a_n + b_n$, where $b_n$ is the number having the same digits as $a_n$, but in the reverse order (unlike $a_n$, the number $b_n$ may start with one or more zeroes in the decimal notation). For instance, if $a_1 = 170... | [
"Let $a_n = a_{n,1}a_{n,2}\\dots a_{n,k_n}$ be the decimal representation of $a_n$. Note that if $k_n$ is even, then $a_n \\equiv_{11} a_{n,1} - a_{n,2} + \\dots - a_{n,k_n}$ and $b_n \\equiv_{11} a_{n,k} - a_{n,k-1} + \\dots - a_{n,1} \\equiv -a_n$. Hence $a_{n+1} = a_n + b_n \\equiv_{11} 0$, so $a_{n+1}$ is divis... | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | a. No; b. No | |
0g6l | 已知四邊形 $A_1A_2A_3A_4$ 不是圓內接四邊形。令 $O_1$ 與 $r_1$ 分別為三角形 $A_2A_3A_4$ 的外接圓圓心與半徑。類似地定義 $O_2, O_3, O_4$ 與 $r_2, r_3, r_4$。試證:
$$
\frac{1}{O_1A_1^2 - r_1^2} + \frac{1}{O_2A_2^2 - r_2^2} + \frac{1}{O_3A_3^2 - r_3^2} + \frac{1}{O_4A_4^2 - r_4^2} = 0.
$$ | [
"將平面直角坐標化，則每個圓皆可表示成 $p(x, y) = x^2 + y^2 + l(x, y) = 0$，其中 $l(x, y)$ 為至多一次的多項式。又注意到對於平面上的每一點 $A = (x_A, y_A)$，$p(x_A, y_A) = d^2 - r^2$，其中 $d$ 為 $A$ 到圓心的距離，而 $r$ 為圓的半徑。\n\n現在，對於每個 $i \\in \\{1, 2, 3, 4\\}$，令 $p_i(x, y) = x^2 + y^2 + l_i(x, y) = 0$ 表示所對應之圓其圓心為 $O_i$，半徑為 $r_i$ 的圓方程式，並令 $d_i$ 為 $A_i$ 到 $O_i$ 的距離。則 $A_... | Taiwan | 二〇一二數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0kcb | Problem:
Let $n$ be a fixed positive integer. An $n$-staircase is a polyomino with $\frac{n(n+1)}{2}$ cells arranged in the shape of a staircase, with arbitrary size. Here are two examples of 5-staircases:
Prove that an $n$-staircase can be dissected into strictly smaller $n$-staircases. | [
"Solution:\n\nViewing the problem in reverse, it is equivalent to show that we can use multiple $n$-staircases to make a single, larger $n$-staircase, because that larger $n$-staircase is made up of strictly smaller $n$-staircases, and is the example we need.\n\nFor the construction, we first attach two $n$-stairca... | United States | HMMT February 2020 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
00g7 | Let $a$, $b$, $c$ be the sides of a triangle, with $a + b + c = 1$, and let $n \geq 2$ be an integer. Show that
$$
\sqrt[n]{a^{n} + b^{n}} + \sqrt[n]{b^{n} + c^{n}} + \sqrt[n]{c^{n} + a^{n}} < 1 + \frac{\sqrt[n]{2}}{2}
$$ | [
"Without loss of generality, assume $a \\leq b \\leq c$. As $a + b > c$, we have\n$$\n\\frac{\\sqrt[n]{2}}{2} = \\frac{\\sqrt[n]{2}}{2}(a + b + c) > \\frac{\\sqrt[n]{2}}{2}(c + c) = \\sqrt[n]{2 c^{n}} \\geq \\sqrt[n]{b^{n} + c^{n}}. \\quad [2 \\text{ marks}] \\tag{1}\n$$\nAs $a \\leq c$ and $n \\geq 2$, we have\n$$... | Asia Pacific Mathematics Olympiad (APMO) | XV APMO | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0k0q | Problem:
A square can be divided into four congruent figures as shown:
If each of the congruent figures has area $1$, what is the area of the square? | [
"Solution:\nThere are four congruent figures with area $1$, so the area of the square is $4$."
] | United States | HMMT February 2018 | [
"Geometry > Plane Geometry > Quadrilaterals"
] | null | final answer only | 4 | |
04yk | We are given an acute triangle $ABC$. Point $D$ lies in the halfplane $AB$ containing $C$ and satisfies $DB \perp AB$ and $\angle ADB = 45^\circ + \frac{1}{2}\angle ACB$. Similarly, $E$ lies in the halfplane $AC$ containing $B$ and satisfies $AC \perp EC$ and $\angle AEC = 45^\circ + \frac{1}{2}\angle ABC$. Let $F$ be ... | [
"Denote $\\angle ABC = \\beta$ and $\\angle ACB = \\gamma$. The conditions translate as $\\angle BAD = 45^\\circ - \\gamma$ and $\\angle EAC = 45^\\circ - \\beta$. Denote by $G$ the intersection point of $BD$ and $CE$. Clearly $\\angle BAG = 90^\\circ - \\gamma = 2\\angle BAD$, and so $AD$ is the angle bisector of ... | Czech-Polish-Slovak Mathematical Match | CAPS Match 2025 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles"
] | null | proof only | null | |
0kp5 | Problem:
Given an angle $\theta$, consider the polynomial
$$
P(x) = \sin (\theta) x^{2} + (\cos (\theta) + \tan (\theta)) x + 1
$$
Given that $P$ only has one real root, find all possible values of $\sin (\theta)$. | [
"Solution:\nNote that if $\\sin (\\theta) = 0$, then the polynomial has 1 root. Now assume this is not the case then the polynomial is a quadratic in $x$.\n\nFactor the polynomial as $(\\tan (\\theta) x + 1)(x + \\sec (\\theta))$. Then the condition is equivalent to $\\sec (\\theta) = \\frac{1}{\\tan (\\theta)}$, w... | United States | HMMT November 2022 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | sin(theta) ∈ {0, (sqrt(5) - 1)/2} | |
0hrd | Problem:
Evaluate the sum
$$
\frac{1}{1+\tan 1^{\circ}}+\frac{1}{1+\tan 2^{\circ}}+\frac{1}{1+\tan 3^{\circ}}+\cdots+\frac{1}{1+\tan 89^{\circ}}
$$
(The tangent $(\tan)$ of an angle $\alpha$ is the ratio $BC/AC$ in a right triangle $ABC$ with $\angle C=90^{\circ}$ and $\angle A=\alpha$, and its value does not depend on... | [
"Solution:\nBy examining a triangle with angles $x$, $90-x$, and $90$, it is not hard to see the trigonometric identity\n$$\n\\tan \\left(90^{\\circ}-x\\right)=\\frac{1}{\\tan x} .\n$$\nLet us pair up the terms\n$$\n\\frac{1}{1+\\tan x} \\text{ and } \\frac{1}{1+\\tan \\left(90^{\\circ}-x\\right)}\n$$\nfor $x=1,2,3... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 89/2 | |
02gh | One has a long row of glasses and $n$ stones in the central glass (glass $0$). The following movements are allowed:
• Movement $A$
If there is at least one stone in glass $i$ and at least one in glass $i+1$, one may make one stone in glass $i+1$ jump to glass $i-1$, eliminating one stone in g... | [
"Assign $x'$ to a stone in the $i$th glass. Let's consider what happens to the sum of the number assigned to all stones with each movement.\n\nMovement *A* replaces $x^i + x^{i+1}$ with $x^{i-1}$; movement *B* replaces $2x^i$ with $x^{i-1} + x^{i+2}$. So the difference in the sum is either $x^{i-1}(1 - x - x^2)$ or... | Brazil | XXIII OBM | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
00ht | Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved. | [
"Since the expression is cyclic, we could WLOG $a=\\max \\{a, b, c, d\\}$. Let\n$$\nS(a, b, c, d)=(a-b)(b-c)(c-d)(d-a)\n$$\nNote that we have given $(a, b, c, d)$ such that $S(a, b, c, d)=-\\frac{1}{8}$. Therefore, to prove that $S(a, b, c, d) \\geq -\\frac{1}{8}$, we just need to consider the case where $S(a, b, c... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | Minimum value: -1/8. Equality is attained by the eight quadruples obtained from the base quadruple (1/4 + sqrt(3)/4, -1/4 - sqrt(3)/4, 1/4 - sqrt(3)/4, -1/4 + sqrt(3)/4) by cyclic permutation of entries and by simultaneously negating all entries. | |
00rn | Given that $ABC$ is a triangle where $AB < AC$. On the half-lines $BA$ and $CA$ we take points $F$ and $E$ respectively such that $BF = CE = BC$. Let $M, N$ and $H$ be the mid-points of the segments $BF, CE$ and $BC$ respectively and $K$ and $O$ be the circumcircles of the triangles $ABC$ and $MNH$ respectively. We ass... | [
"The circumcenter of the triangle $\\triangle MNH$ coincides with the incenter of the triangle $\\triangle ABC$ because the triangles $\\triangle BMH$ and $\\triangle NHC$ are isosceles and therefore the perpendiculars of the $MH, HN$ are also the bisectors of the angles $\\angle ABC, \\angle ACB$, respectively.\n\... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"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 > Distance chasin... | null | proof only | null | |
0e47 | Let $n > 1$ be a positive integer. The first term of the infinite progression $(a_k)$ of positive integers is $a_1 = n$. For all $k > 1$ we have either $a_k = 2a_{k-1} + 1$ or $a_k = 2a_{k-1} - 1$. Prove that not all terms of this progression are prime numbers. | [
"Suppose for contradiction that all terms of the progression $(a_k)$ are prime numbers.\n\nLet us consider the two possible recursions:\n\n- $a_k = 2a_{k-1} + 1$\n- $a_k = 2a_{k-1} - 1$\n\nLet $a_1 = n > 1$.\n\nLet us compute $a_2$:\n- $a_2 = 2n + 1$ or $a_2 = 2n - 1$\n\nLet us consider the sequence where we always... | Slovenia | Selection Examinations for the IMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | null | proof only | null | |
06d9 | For each integer $k \ge 4$, prove that if $F(x)$ is a polynomial with integer coefficients satisfying the condition $0 \le F(c) \le k$ for every $c = 0, 1, \dots, k+1$, then
$$
F(0) = F(1) = \dots = F(k+1).
$$ | [
"Note that $(k+1)-0 = F(k+1)-F(0)$. Since $|F(k+1)-F(0)| \\le k$, we must have $F(k+1) = F(0) = d$ for some constant $d$. Let $F(x) = d + x(x-k-1)G(x)$ for some polynomial $G$ with integer coefficients.\n\nFor $2 \\le n \\le k-1$, we have $F(n) = d+n(n-k-1)G(n)$, and so $n(k+1-n) \\mid F(n)-d$.\nAgain, $|F(n)-d| \\... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
07xi | A positive integer $n$ is an *anchor* if every digit of $n$ (when written in base 10) is an odd number. Show that there exists an anchor $n$ such that the product $m \cdot n$ is not an anchor for any anchor $m > 1$. | [
"**Solution 1.** One such integer is $n = 91$. We show there is no anchor $m > 1$ such that $m \\times n$ is also an anchor for this value of $n$. Let $m > 1$ be an anchor. We consider separately the cases $3 \\le m \\le 9$ and $m \\ge 11$.\nFor small anchors $m$, we have $3 \\times 91 = 273$, $5 \\times 91 = 455$,... | Ireland | IRL_ABooklet_2025 | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
02m4 | Let $n > 3$ be a fixed integer and $x_1, x_2, \dots, x_n$ be positive real numbers. Find, in terms of $n$, all possible real values of
$$
\frac{x_1}{x_n + x_1 + x_2} + \frac{x_2}{x_1 + x_2 + x_3} + \dots + \frac{x_{n-1}}{x_{n-2} + x_{n-1} + x_n} + \frac{x_n}{x_{n-1} + x_n + x_1}
$$ | [
"The answer is all real numbers in the interval $]1, \\lfloor n/2 \\rfloor[$. For simplicity, let\n$$\nE = \\frac{x_1}{x_n + x_1 + x_2} + \\frac{x_2}{x_1 + x_2 + x_3} + \\frac{x_3}{x_2 + x_3 + x_4} + \\dots + \\frac{x_n}{x_{n-1} + x_n + x_1}.\n$$\nLet's prove first the lower bound. Let $S = x_1 + x_2 + \\dots + x_n... | Brazil | XXXI Brazilian Math Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | (1, floor(n/2)) | |
049o | Let $ABC$ be a triangle such that $|AB| = |AC|$. Let $D$ be a point on the side $AC$ such that $|AD| < |CD|$, and let $P$ be a point on the segment $BD$ such that $\angle APC = 90^\circ$. If $\angle ABP = \angle BCP$, determine $|AD| : |CD|$.
(Stipe Vidak) | [
"Let us denote $\\angle ABP = \\angle BCP = \\varphi$ and $\\angle PBC = \\angle PCA = \\psi$.\nLet $T$ be the midpoint of the segment $AC$. Since $AC$ is the hypotenuse of the right triangle $APC$, it follows that $|AT| = |PT| = |CT|$. Hence $\\angle CPT = \\psi$.\n\n\n\nSince $\\angle DPC... | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 1:2 | |
04mw | Determine all triples $(p, q, r)$ of prime numbers such that
$$
p^q = r - 1.
$$ | [] | Croatia | Croatia_2018 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (2, 2, 5) | |
0fqd | A set $T$ of integers is called *orensano* if there are integers $a < b < c$ such that $a, c \in T$ and $b \notin T$.
Find the number of orensanos subsets $T$ of $\{1, 2, \dots, 2019\}$. | [] | Spain | SPANISH MATHEMATICAL OLYMPIAD (FINAL ROUND) | [
"Discrete Mathematics > Other"
] | English | proof and answer | 2^2019 - 2039191 | |
0l1z | Problem:
Rectangle $R$ with area $20$ and diagonal of length $7$ is translated $2$ units in some direction to form a new rectangle $R'$. The vertices of $R$ and $R'$ that are not contained in the other rectangle form a convex hexagon. Compute the maximum possible area of this hexagon. | [
"Solution:\n\n\n\nDissect the hexagon as shown above, so that it consists of a parallelogram and two triangles which are each half the original rectangle. The parallelogram has side lengths $7$ and $2$, so its maximum possible area is $14$. As the two triangles combined always have the same... | United States | HMMT November | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 34 | |
0l06 | Problem:
An equilateral triangle is inscribed in a circle $\omega$. A chord of $\omega$ is cut by the perimeter of the triangle into three segments of lengths $55$, $121$, and $55$ in that order. Compute the sum of all possible side lengths of the triangle. | [
"Solution:\n\n\n\nNote that the chord splits two of the sides into segments of lengths $a, b$ and $c, d$, where segments of length $a$ and $c$ are incident to the same vertex of the equilateral triangle. Moreover, $a+b = c+d$ (as the triangle is equilateral) and $ab = cd = 55 \\cdot 176$ by... | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 410 | |
066x | A triangle $AB\Gamma$ is given, $c(O,R)$ is its circumcircle and let $\Delta$ a point on the side $B\Gamma$ different than the midpoint of $B\Gamma$. The circumcircle of the triangle $BO\Delta$, say $c_1$, intersects the circle $c(O,R)$ at $K$ and the line $AB$ at $Z$. The circumcircle of the triangle $GO\Delta$, say $... | [
"We will prove that the circle $c_3$ passes from the center $O$ of $c(O,R)$.\nFrom the inscribed quadrilateral $O\\Delta\\Gamma E$ (in the circle $c_3$) we get: $\\hat{O}_1 = \\hat{\\Gamma}$.\nFrom the inscribed quadrilateral $O\\Delta BZ$ (in the circle $c_1$) we get: $\\hat{O}_2 = \\hat{B}$.\nSumming up the above... | Greece | Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nin... | English | proof only | null | |
00iz | We are given a non-isosceles triangle $ABC$ with incenter $I$. Show that the circumcircle $k$ of $AIB$ is not tangent to the lines $CA$ or $CB$.
The second common point of $k$ with $CA$ is named $P$ and the second with $CB$ is named $Q$. Prove that the points $A, B, P$ and $Q$ are (not necessarily in this order) verti... | [
"It is well known that the angle bisector in $C$ and the bisector of the side $AB$ intersect in a point on the circumcircle of a triangle $ABC$. Let this point be $M$. It is certainly equidistant from $A$ and $B$. We now consider the triangle $AIM$. Naming the angles in $A$, $B$ and $C$, $\\alpha$, $\\beta$ and $\\... | Austria | AustriaMO2011 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
03f1 | Find all pairs $(x, y)$ of real numbers for which
$$
4y^4 + x^4 + 12y^3 + 5x^2(y^2 + 1) + y^2 + 4 = 12y.
$$ | [
"We have the inequalities $x^4 \\geq 0$, $5x^2(y^2 + 1) \\geq 0$ and $4y^4 + 12y^3 + y^2 - 12y + 4 = (2y - 1)^2(y + 2)^2 \\geq 0$. The sum of the left sides is $0$ if and only if each of them is equal to $0$. The first two lead to $x = 0$, and the third to $y = -2$ or $y = \\frac{1}{2}$."
] | Bulgaria | 3 Bulgarian Spring Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | proof and answer | (0, -2) and (0, 1/2) | |
0dnl | Problem:
За цео број $a, a \neq 0$, означимо са $v_{2}(a)$ највећи ненегативан цео број $k$ такав да $2^{k} \mid a$. За дато $n \in \mathbb{N}$ одредити највећу могућу кардиналност подскупа $A$ скупа $\{1,2,3, \ldots, 2^{n}\}$ са следећим својством:
$$
\text{за све } x, y \in A, x \neq y, \text{ број } v_{2}(x-y) \tex... | [
"Solution:\n\nДоказаћемо индукцијом по $k$ да скуп $A$ садржи највише $2^{k}$ различитих елемената по модулу $2^{2k}$. То тривијално важи за $k=0$. Нека је $k>0$. По индуктивној претпоставци, елементи скупа $A$ дају највише $2^{k-1}$ остатака по модулу $2^{2k-2}$. Претпоставимо да елементи $A$ дају више од $2^{k}$ ... | Serbia | 9. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2^{floor((n+1)/2)} | |
0dhc | Let $\phi_n(m) = \phi(\phi_{n-1}(m))$, where $\phi_1(m) = \phi(m)$ is the Euler totient function, and set $\omega(m)$ the smallest number $n$ such that $\phi_n(m) = 1$. If $m < 2^\alpha$, then prove that $\omega(m) \le \alpha$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | English | proof only | null | |
00nk | Suppose that $p$ is an odd prime number and $M$ a set of $\frac{p^2+1}{2}$ integer squares.
Investigate if one can choose $p$ elements of this set so that the arithmetic mean of these $p$ elements is an integer.
(Walther Janous) | [
"The idea is to choose from the $\\frac{p^2+1}{2}$ square numbers $p$ numbers that are in the same residue class modulo $p$. Obviously, the sum of these $p$ numbers is then divisible by $p$ and thus the arithmetic mean is an integer.\nIt is known that the square numbers do not run through all residue classes modulo... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0j9q | Problem:
Let triangle $ABC$ have $AB = 5$, $BC = 6$, and $AC = 7$, with circumcenter $O$. Extend ray $AB$ to point $D$ such that $BD = 5$, and extend ray $BC$ to point $E$ such that $OD = OE$. Find $CE$. | [
"Solution:\n\nBecause $OD = OE$, $D$ and $E$ have equal power with respect to the circle, so $(EC)(EB) = (DB)(DA) = 50$. Letting $EC = x$, we have $x(x + 6) = 50$, and taking the positive root gives $x = \\sqrt{59} - 3$."
] | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof and answer | sqrt(59) - 3 | |
0bcl | Problem:
O mulţime $A$ de numere întregi se zice sumă-plină dacă $A \subseteq A+A$, adică orice element $a \in A$ este suma unei perechi (nu neapărat unice) de elemente (nu neapărat distincte) $b, c \in A$. O mulţime $A$ de numere întregi se zice liberă-de-sume-zero dacă $0$ este singurul număr întreg care nu poate fi... | [] | Romania | Olimpiada europeana de matematica a fetelor | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | No, such a set does not exist. | |
0cko | Consider the squares $ABCD$ and $BEFG$, such that $B$ lies on the segment $AE$ and $G$ lies on the segment $BC$. Let $H$ be the intersection of the lines $DF$ and $EG$. The perpendicular from $H$ to the line $DF$ intersects the lines $AE$ and $BC$ at points $I$ and $J$, respectively. Prove that the quadrilateral $DIFJ$... | [
"$\\angle GBF = \\angle DBC = 45^\\circ$, thus the triangle $BDF$ is right-angled at $B$. Since $EG$ is the perpendicular bisector of $BF$ in triangle $BDF$, it follows that $H$ is the midpoint of the hypotenuse $DF$.\n\nLet $K$ be the projection of $H$ onto $AE$. Since $H$ is the midpoint of $DF$ and $HK \\paralle... | Romania | 75th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry ... | English | proof only | null | |
0iki | Problem:
A triangle with vertices at $(1003,0)$, $(1004,3)$, and $(1005,1)$ in the $xy$-plane is revolved all the way around the $y$-axis. Find the volume of the solid thus obtained. | [
"Solution:\nLet $T \\subset \\mathbb{R}^2$ denote the triangle, including its interior. Then $T$'s area is $5/2$, and its centroid is $(1004, 4/3)$, so\n$$\n\\int_{(x, y) \\in T} x \\, dx \\, dy = \\frac{5}{2} \\cdot 1004 = 2510\n$$\nWe are interested in the volume\n$$\n\\int_{(x, y) \\in T} 2\\pi x \\, dx \\, dy\n... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 5020π | |
02qo | Problem:
A Figura I mostra um quadrado de $40~\mathrm{cm}^2$ cortado em cinco triângulos retângulos isósceles, um quadrado e um paralelogramo, formando as sete peças do jogo Tangran. Com elas é possível formar a Figura II, que tem um buraco sombreado. Qual é a área do buraco?
Figura I
![](atta... | [
"Solution:\nAbaixo vemos as figuras do enunciado da questão. A descrição das peças da Figura I implica que os pontos $M$ e $N$ são pontos médios dos lados $AB$ e $AC$. A Figura III, onde $P$ é o ponto médio de $BC$, mostra que a área do triângulo $AMN$ é igual à quarta parte da área do triângulo $ABC$, que por sua ... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | C | |
07ex | Numbers $m$ and $n$ are given positive integers. There are $mn$ people in a party, standing in the shape of an $m \times n$ grid. Some of these people are police officers and the rest are the guests. Some of the guests may be criminals. The goal is to determine whether there is a criminal between the guests or not.
Tw... | [
"Answer. $\\left\\lfloor \\frac{mn}{2} \\right\\rfloor + 1$.\n\nThere are two steps for the proof, first showing that this number is enough, and for any smaller number of officers, show a way for some of the guests to be criminals and to threaten the officers such that the existence of the criminals remains unknown... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | floor(mn/2) + 1 | |
01k4 | Find $\left\{ \frac{2009!}{2011!} \right\}$. (Here $\{x\}$ means the fractional part of $x$.) | [
"Answer: $\\frac{1}{2011}$.\nSince $2011$ is a prime number, we see that $2010! \\equiv -1 \\pmod{2011}$ (Wilson's theorem). Hence $2010! \\equiv 2010 \\pmod{2011}$, which gives"
] | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1/2011 | |
0j32 | Problem:
$ABC$ is a right triangle with $\angle A = 30^{\circ}$ and circumcircle $O$. Circles $\omega_1$, $\omega_2$, and $\omega_3$ lie outside $ABC$ and are tangent to $O$ at $T_1$, $T_2$, and $T_3$ respectively and to $AB$, $BC$, and $CA$ at $S_1$, $S_2$, and $S_3$, respectively. Lines $T_1 S_1$, $T_2 S_2$, and $T_... | [
"Solution:\n\nAnswer: $\\frac{\\sqrt{3}+1}{2}$\n\nLet $[PQR]$ denote the area of $\\triangle PQR$. The key to this problem is the following fact: $[PQR] = \\frac{1}{2} PQ \\cdot PR \\sin \\angle QPR$.\n\nAssume that the radius of $O$ is $1$. Since $\\angle A = 30^{\\circ}$, we have $BC = 1$ and $AB = \\sqrt{3}$. So... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | null | proof and answer | (sqrt(3)+1)/2 | |
0en9 | Let circles $\Gamma_1$ and $\Gamma_2$ intersect at $D$ and $P$. The common tangent of the two circles closest to the point $D$ touches $\Gamma_1$ at $A$ and $\Gamma_2$ at $B$. The line $AD$ intersects $\Gamma_2$ for the second time in $C$. Let $M$ be the middle of the line segment $BC$. Prove that $\angle DPM = \angle ... | [
"\nLet $S$ be the intersection of $PD$ and $AB$. Then $S$ lies on the radical axis of the two circles, so $AS = SB$, that is, $PS$ is a median in triangle $PAB$.\nNow, by the tan-chord theorem,\n$$\n\\angle BAP = 180^\\circ - \\angle ADP = \\angle CDP = \\angle CBP.\n$$\nAlso, $\\angle ABP ... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09p4 | Find all triples of positive integers $(m, n, p)$ with $m, n$ positive integers and $p$ a prime number such that $m^{2025} + n^{2024} = pmn$. | [
"Answer: $(m, n, p) = (1, 1, 2)$.\n\nSetting $k = 2024$, $m_1 = m/d$ and $n_1 = n/d$, the given identity becomes $d^{k-2}(dm_1^{k+1} + n_1^k) = pm_1n_1$, where $d$ is the greatest common divisor $(m, n)$ of $n$ and $m$. Put $a = n_1/b$ and $c = d/b$, where $b$ denotes $(d, n_1)$. Then we get $(cb)^{k-2}(cm_1^{k+1} ... | Mongolia | MMO2025 Round 4 | [
"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"
] | English | proof and answer | (1, 1, 2) | |
0kv5 | Problem:
Five people take a true-or-false test with five questions. Each person randomly guesses on every question. Given that, for each question, a majority of test-takers answered it correctly, let $p$ be the probability that every person answers exactly three questions correctly. Suppose that $p=\frac{a}{2^{b}}$ wh... | [
"Solution:\n\nThere are a total of $16^{5}$ ways for the people to collectively ace the test. Consider groups of people who share the same problems that they got incorrect. We either have a group of 2 and a group of 3, or a group 5.\n\nIn the first case, we can pick the group of two in $\\binom{5}{2}$ ways, the pro... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 25517 | |
085t | Problem:
Sia $P(x) = x^{3} + a x^{2} + b x + c$. Sapendo che la somma di due delle radici del polinomio vale zero, quale fra le seguenti relazioni tra i coefficienti di $P(x)$ è sempre vera?
(A) $a b c = 0$
(B) $c = a b$
(C) $c = a + b$
(D) $b^{2} = a c$
(E) nessuna delle risposte precedenti è corretta. | [
"Solution:\n\nLa risposta è (B). Dette $x_{0}$, $x_{1}$ e $-x_{1}$ le radici del polinomio, deve aversi, per ogni $x$, che $(x - x_{0})(x - x_{1})(x + x_{1}) = x^{3} + a x^{2} + b x + c$. Svolgendo i prodotti e uguagliando i coefficienti dei termini dello stesso grado (principio di identità dei polinomi), si ha $-x... | Italy | Olimpiadi di Matematica | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | B | |
0680 | Find all injective functions $f : \mathbb{R} \to \mathbb{R}$ such that for every real number $x$ and every positive integer $n$,
$$
\left| \sum_{i=1}^{n} i \left( f(x+i+1) - f(x+i) \right) \right| < 2016
$$ | [] | Greece | 33rd Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | No such injective function exists. | |
0i5m | Problem:
Find the area of the region in the first quadrant $x>0, y>0$ bounded above the graph of $y=\arcsin (x)$ and below the graph of $y=\arccos (x)$. | [
"Solution:\nWe can integrate over $y$ rather than $x$. In particular, the solution is\n$$\n\\int_{0}^{\\pi / 4} \\sin y \\, dy + \\int_{\\pi / 4}^{\\pi / 2} \\cos y \\, dy = \\left(1 - \\frac{1}{\\sqrt{2}}\\right) 2 = 2 - \\sqrt{2}.\n$$"
] | United States | Harvard-MIT Math Tournament | [
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Trigonometric functions"
] | null | proof and answer | 2 - sqrt(2) | |
092i | Problem:
Find all pairs of positive integers $(a, b)$ such that
$$
a! + b! = a^{b} + b^{a}
$$ | [
"Solution:\nIf $a = b$, the equation reduces to $a! = a^{a}$. Since $a^{a} > a!$ for $a \\geqslant 2$, the only solution in this case is $a = b = 1$.\n\nIf $a = 1$, the equation reduces to $b! = b$, which gives an additional solution $a = 1, b = 2$.\n\nWe prove $a = b = 1$; $a = 1, b = 2$ and $a = 2, b = 1$ are the... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (a, b) = (1, 1), (1, 2), (2, 1) | |
032l | Problem:
Let $O$ and $G$ be respectively the circumcenter and the centroid of $\triangle ABC$ and let $M$ be the midpoint of the side $AB$. If $OG \perp CM$, prove that $\triangle ABC$ is isosceles. | [
"Solution:\nSet $\\vec{a}=\\overrightarrow{OA}$, $\\vec{b}=\\overrightarrow{OB}$, $\\vec{c}=\\overrightarrow{OC}$. We have that $\\overrightarrow{OM}=\\frac{1}{2}(\\vec{a}+\\vec{b})$ and hence\n$$\n\\overrightarrow{OG}=\\frac{1}{6}(3\\vec{a}+\\vec{b}+2\\vec{c})\n$$\nOn the other hand, $\\overrightarrow{CM}=\\frac{1... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
01td | Let $\mathcal{A}$ be a nonempty set of positive integers. We say that a positive integer $n$ is *special* if there exists a unique subset $\mathcal{B}$ of the set $\mathcal{A}$ such that
(i) the number of the elements in $\mathcal{B}$ is odd;
(ii) the sum of all elements of $\mathcal{B}$ is equal to $n$.
Prove that th... | [
"3. See IMO-2015 Shortlist, Problem C6."
] | Belarus | 66th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Generating functions"
] | English | proof only | null | |
0d3s | Let $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$ be nonzero real numbers. Prove that the polynomial
$$
P(X) = \prod_{k=0}^{4} \left( a_{k+1} X^{4} + a_{k+2} X^{3} + a_{k+3} X^{2} + a_{k+4} X + a_{k+5} \right),
$$
where $a_{5+i} = a_{i}$ for $i = 1, 2, 3, 4$, has a root with negative real part. | [
"Assume, to the contrary, that all roots of the polynomial $P(X)$ have nonegative real parts. We deduce that the real parts of the sums of the roots of its factors\n$$\na_{k} X^{4} + a_{k+1} X^{3} + a_{k+2} X^{2} + a_{k+3} X + a_{k+4}\n$$\nfor $k = 1, 2, 3, 4$, are nonegative. Therefore, by Vieta's relations, we ha... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Complex numbers"
] | English, Arabic | proof only | null | |
0bbz | The cells of a square $2011 \times 2011$ array are labelled with the integers $1, 2, \dots, 2011^2$, in such a way that every label is used exactly once. We then identify the left-hand and right-hand edges, and then the top and bottom, in the normal way to form a torus (the surface of a doughnut). Determine the largest... | [
"For the toroidal case, it is clear the statement of the problem is referring to the cells of a $\\mathbb{Z}_N \\times \\mathbb{Z}_N$ lattice on the surface of the torus, labeled with the numbers $1, 2, \\dots, N^2$, where one has to determine the least possible maximal absolute value $M$ of the difference of label... | Romania | 2011 Fourth ROMANIAN MASTER OF MATHEMATICS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 4021 | |
0d7m | Show that there are infinitely many positive integers $n$ such that $n$ has at least two prime divisors and $20^{n}+16^{n}$ is divisible by $n^{2}$. | [
"We will construct (by induction on $k$) the infinite increasing sequence $\\left(n_{k}\\right)_{k \\geq 1}$ of odd positive integers such that any $n_{k}$ satisfies $4^{n_{k}}+5^{n_{k}}$ is divisible by $n_{k}^{2}$.\n\nWe take $n_{1}=1$ and $n_{2}=3$.\n\nAssume we already have $n_{k}$, that is $4^{n_{k}}+5^{n_{k}}... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof only | null | |
0920 | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(x f(x)+2 y)=f\left(x^{2}\right)+f(y)+x+y-1
$$
for all $x, y \in \mathbb{R}$. | [
"Solution:\nPutting $x=y=0$, we get $f(0)=1$.\n\nPutting $x=0, y=z$, we get\n$$\nf(2 z)=f(z)+z\n$$\n\nPutting $x=z, y=-z f(z)$, we get\n$$\nf\\left(z^{2}\\right)=z f(z)-z+1\n$$\n\nReplacing $z$ by $2 z$ in (2) and using (1), we obtain\n$$\nf\\left(4 z^{2}\\right)=2 z f(2 z)-2 z+1=2 z(f(z)+z)-2 z+1=2 z f(z)+2 z^{2}-... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x + 1 | |
0048 | Consideramos todas las sucesiones infinitas $(x_n)_{n \ge 1}$ de enteros positivos que satisfacen la recurrencia
$$
x_{n+2} = \text{mcd}(x_{n+1}, x_n) + 2006,
$$
para todo $n = 1, 2, \dots$. Aquí $\text{mcd}(u, v)$ denota el máximo común divisor de $u$ y $v$.
¿Puede ocurrir que una sucesión de este tipo contenga exacta... | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | Español | proof and answer | No | |
0f4a | Problem:
$ABCDEF$ is a prism. Its base $ABC$ and its top $DEF$ are congruent equilateral triangles. The side edges are $AD$, $BE$ and $CF$. Find all points on the base which are equidistant from the three lines $AE$, $BF$ and $CD$. | [] | Soviet Union | 15th ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0elz | Solve the system of equations:
$xy = x + y;$
$x^2 + y^2 = 1$ | [
"We are trying to solve the equations\n$$\nxy = x + y\n$$\n$$\n1 = x^2 + y^2.\n$$\n\nSquaring both sides of $xy = x + y$ and using $x^2 + y^2 = 1$ to simplify gives\n$$\nx^2 y^2 = x^2 + 2xy + y^2 = 1 + 2xy \\Rightarrow (xy)^2 - 2(xy) - 1 = 0.\n$$\nThis gives $xy = 1 \\pm \\sqrt{2}$ or $y = \\frac{1\\pm\\sqrt{2}}{x}... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | (x, y) = ((1 - √2 + √(2√2 - 1))/2, (1 - √2 - √(2√2 - 1))/2) and (x, y) = ((1 - √2 - √(2√2 - 1))/2, (1 - √2 + √(2√2 - 1))/2) | |
09ft | For a positive integer $n$, denote by $c_n$ the number of triples $(x, y, z)$ of integers such that $0 \le x \le y \le z \le x+y$ and $x+y+z=n$. Prove that, for $n \ge 2$,
$$
n \cdot c_n \le 9 \cdot (c_0 + c_1 + \dots + c_{n-2}).
$$ | [
"Let $X_n := \\{(x, y, z) \\in \\mathbb{Z}^3 \\mid 0 \\le x \\le y \\le z \\le x+y \\text{ and } x+y+z=n\\}$. Then $c_n = |X_n|$. Let $Y_n := \\{(p, q, r) \\in \\mathbb{Z}^3 \\mid 0 \\le p, q, r \\text{ and } 2p+3q+4r=n\\}$ and define the maps $f: X_n \\to Y_n$ by\n$$\nf(x, y, z) := (y - x, x + y - z, z - y)\n$$\na... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | proof only | null | |
0hue | Problem:
Find the smallest prime $p > 100$ for which there exists an integer $a > 1$ such that $p$ divides $\frac{a^{89} - 1}{a - 1}$. | [
"Solution:\nThe answer is $p = 179$. To see this works, take $a = 4$; by Fermat's little theorem, $4^{89} - 1 = 2^{178} - 1$ is divisible by $179$.\n\nNow suppose $a^{89} \\equiv 1 \\pmod{p}$. We consider two cases:\n- If $a \\equiv 1 \\pmod{p}$, then\n$$\n0 \\equiv 1 + a + \\cdots + a^{88} \\equiv 89 \\pmod{p}\n$$... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 179 | |
045n | Find all positive integers $k$ such that there are finitely many triangles on the Descartes coordinate plane such that
(1) the center of mass of each triangle is an integral point;
(2) the intersection of any two triangles is either the empty set, a common vertex, or an edge joining two common vertices; and
(3) the uni... | [
"The desired positive integers $k$ are those divisible by $3$.\n\nFirst assume that $k = 3t$ for $t \\in \\mathbb{N}$. Consider the square with vertices $(0,0)$, $(3t, 3t)$, $(3t, 0)$, and $(0, 3t)$. Divide it into $t^2$ different smaller squares with the same side length $3$ along the lines $x = 3i$ ($i = 1, \\dot... | China | 2022 China Team Selection Test for IMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | all positive integers divisible by 3 | |
00xq | Problem:
On each face of two dice some positive integer is written. The two dice are thrown and the numbers on the top faces are added. Determine whether one can select the integers on the faces so that the possible sums are $2,3,4,5,6,7,8,9,10,11,12,13$, all equally likely? | [
"Solution:\n\nWe can write $1, 2, 3, 4, 5, 6$ on the sides of one die and $1, 1, 1, 7, 7, 7$ on the sides of the other. Then each of the 12 possible sums appears in exactly 3 cases."
] | Baltic Way | Baltic Way 1993 | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | Yes. Label one die with 1, 2, 3, 4, 5, 6 and the other with 1, 1, 1, 7, 7, 7; then each sum from 2 to 13 occurs exactly 3 times out of 36. | |
09c3 | Хэрэв $a, b, c > 0$ тоонуудын хувьд $ab+bc+ca = 1$ бол
$$
\sqrt[3]{\frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a} \le \frac{1}{abc}
$$
гэж батал. | [
"Эхлээд\n$$\n3(ab \\cdot ac + ab \\cdot bc + ac \\cdot bc) \\leq (ab + ac + bc)^2\n$$\nтэнцэтгэл бишийг баталъя.\n$$\nab \\cdot ac + ab \\cdot bc + ac \\cdot bc \\leq a^2b^2 + a^2c^2 + b^2c^2\n$$\n$$\n0 \\leq (ab + ac)^2 + (ab + bc)^2 + (ac + bc)^2\n$$\nболж батлагдлаа.\n$$\n3abc(a + b + c) = 3(ab \\cdot ac + ab \\... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Mongolian | proof only | null | |
0eb9 | Find all pairs of positive integers $a$ and $b$ which satisfy $2a^b = ab + 3$. | [
"Since $b$ is a positive integer, $a$ divides $2a^b$ and $ab$, and hence also $3$. Since $3$ is a prime number, we have $a = 1$ or $a = 3$.\n\nIf $a = 1$ we get equation $2 = b + 3$, which does not have solutions in positive integers.\n\nTherefore $a = 3$ and we get equation $2 \\cdot 3^b = 3b + 3$ which gives $2 \... | Slovenia | National Math Olympiad 2015 – Final Round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | a = 3, b = 1 | |
04q8 | Cryptogram of a positive integer $n$ is an $n$-tuple $a = (a_1, a_2, \dots, a_n)$ of non-negative integers such that
$$
a_1 + 2a_2 + \dots + n a_n = n.
$$
Let $\mathcal{K}_n$ be the set of all cryptograms of the number $n$. For $a \in \mathcal{K}_n$ let $J(a)$ denote the number of occurrences of the number 1 in the cry... | [
"Let $k_n = |\\mathcal{K}_n|$, $j_n = \\sum_{a \\in \\mathcal{K}_n} J(a)$ and $d_{n+1} = \\sum_{a \\in \\mathcal{K}_{n+1}} a_2$. We need to prove that $j_n = d_{n+1}$ for every positive integer $n$. We prove the statement by complete mathematical induction.\n\nIt is clear that $j_1 = 1 = d_2$ and $j_2 = 1 = d_3$.\n... | Croatia | Croatian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
06bb | Prove that there is an infinite number of triads of positive integers $(x, y, z)$ such that
$$
x^2 + y^2 + z^2 + xy + yz + zx = 6xyz.
$$ | [
"We write the given relation in the form\n$$\nx^2 + y^2 + xy = (6xy - x - y - z)z^2.\n$$\n\nWe observe that substituting $z$ by $6xy - x - y - z$, leaves invariant the right part of the equality. Thus, if $(x, y, z)$ is a solution of the given equation with $x > y > z$, then $(6xy - x - y - z, x, y)$, with $6xy - x... | Greece | Hellenic Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof only | null | |
06rp | Let $k$ be a positive integer and set $n=2^{k}+1$. Prove that $n$ is a prime number if and only if the following holds: there is a permutation $a_{1}, \ldots, a_{n-1}$ of the numbers $1,2, \ldots, n-1$ and a sequence of integers $g_{1}, g_{2}, \ldots, g_{n-1}$ such that $n$ divides $g_{i}^{a_{i}}-a_{i+1}$ for every $i ... | [
"Let $N=\\{1,2, \\ldots, n-1\\}$. For $a, b \\in N$, we say that $b$ follows $a$ if there exists an integer $g$ such that $b \\equiv g^{a} (\\bmod n)$ and denote this property as $a \\rightarrow b$. This way we have a directed graph with $N$ as set of vertices. If $a_{1}, \\ldots, a_{n-1}$ is a permutation of $1,2,... | IMO | 52nd International Mathematical Olympiad 2011 Shortlist | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primit... | null | proof only | null | |
0klc | Problem:
The sum of the digits of the time 19 minutes ago is two less than the sum of the digits of the time right now. Find the sum of the digits of the time in 19 minutes. (Here, we use a standard 12-hour clock of the form hh:mm.) | [
"Solution:\n\nLet's say the time 19 minutes ago is hours and $m$ minutes, so the sum of the digits is equivalent to $h + m \\bmod 9$. If $m \\leq 40$, then the time right now is hours and $m + 19$ minutes, so the sum of digits is equivalent $\\bmod 9$ to $h + m + 19 \\equiv h + m + 1 \\pmod{9}$, which is impossible... | United States | HMMT November 2021 | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 11 | |
03e0 | Find all positive integers $k$, so that there exists a polynomial $f(x)$ with rational coefficients, such that for all sufficiently large $n$,
$$
f(n) = \operatorname{lcm}(n + 1, n + 2, \dots, n + k).
$$ | [
"For $k=1$ and $k=2$, the required polynomials are $f(x) = x+1$ and $f(x) = (x+1)(x+2)$, respectively. Let $k \\ge 3$ and assume that such a polynomial $f(x)$ exists. For any prime number $p$, its degree in $\\operatorname{lcm}(n+1, n+2, \\dots, n+k)$ is $\\max\\{\\alpha_1, \\alpha_2, \\dots, \\alpha_{k-1}\\}$, whe... | Bulgaria | 1 Autumn tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | k = 1 or k = 2 | |
0c44 | Problem:
Arătaţi că o funcţie continuă $f: \mathbb{R} \rightarrow \mathbb{R}$ este crescătoare dacă şi numai dacă
$$
(c-b) \int_{a}^{b} f(x) \, \mathrm{d} x \leq (b-a) \int_{b}^{c} f(x) \, \mathrm{d} x
$$
oricare ar fi numerele reale $a < b < c$. | [
"Solution:\nDacă $f$ este crescătoare şi $a < b < c$, atunci\n$$\n(c-b) \\int_{a}^{b} f(x) \\, \\mathrm{d} x \\leq (c-b)(b-a) f(b) = (b-a)(c-b) f(b) \\leq (b-a) \\int_{b}^{c} f(x) \\, \\mathrm{d} x\n$$\n\nReciproc, fie $a$ şi $b$ două numere reale, astfel încât $a < b$, şi fie $F: \\mathbb{R} \\rightarrow \\mathbb{... | Romania | Olimpiada Naţională de Matematică Etapa Judeţeană şi a Municipiului Bucureşti | [
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | proof only | null |
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