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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0bb0 | A positive integer $N$ has the digits $1, 2, 3, 4, 5, 6$ and $7$, so that each digit $i$, $i \in \{1, 2, 3, 4, 5, 6, 7\}$ occurs $4i$ times in the decimal representation of $N$. Prove that $N$ is not a perfect square. | [
"$N$ has $1 \\cdot 4 = 4$ digits equal to $1$, $2 \\cdot 4 = 8$ digits equal to $2$, $\\dots$, $7 \\cdot 4 = 28$ digits equal to $7$, so the sum of its digits equals $S = 4(1^2 + 2^2 + \\dots + 7^2) = 560$. Since $560 = 3 \\cdot 186 + 2$, the number $N$ is not a square, and the remainder left by a perfect square up... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
06wu | Let $k \geqslant 2$ be an integer. Find the smallest integer $n \geqslant k+1$ with the property that there exists a set of $n$ distinct real numbers such that each of its elements can be written as a sum of $k$ other distinct elements of the set. | [
"First we show that $n \\geqslant k+4$. Suppose that there exists such a set with $n$ numbers and denote them by $a_{1} < a_{2} < \\cdots < a_{n}$.\nNote that in order to express $a_{1}$ as a sum of $k$ distinct elements of the set, we must have $a_{1} \\geqslant a_{2} + \\cdots + a_{k+1}$ and, similarly for $a_{n}... | IMO | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | k+4 | |
0f16 | Problem:
$ABCD$ is a square. $P$ is on the segment $AB$ and $Q$ is on the segment $BC$ such that $BP = BQ$. $H$ lies on $PC$ such that $BHC$ is a right angle. Show that $DHQ$ is a right angle. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0bk4 | Suppose $(G, \cdot)$ is a finite group with unity $e$, $a$ is an element in $G \setminus \{e\}$ and $p$ is a prime number such that $x^{p+1} = a^{-1} x a$, for all $x \in G$.
a) Show that there is $k \in \mathbb{N}^*$ such that $\text{ord}(G) = p^k$.
b) Prove that $H = \{x \in G \mid x^p = e\}$ is a subgroup of $G$ a... | [
"a) If $x, y \\in G$, then $(xy)^{p+1} = a^{-1} x y a = a^{-1} x x a^{-1} y a = x^{p+1} y^{p+1}$. We can write $x(yx)^p y = x^{p+1} y^{p+1}$, then $(yx)^p = x^p y^p$. For $x = a$ we get $a^p = e$ so by the preceding equality $(y a)^p = y^p$. Multiplying at left by $y a$ we obtain $y a y^p = (y a)^{p+1} = y^{p+1} a$... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
06qb | Let $n$ be a positive integer. Show that the numbers
$$
\binom{2^{n}-1}{0}, \quad\binom{2^{n}-1}{1}, \quad\binom{2^{n}-1}{2}, \quad \ldots, \quad\binom{2^{n}-1}{2^{n-1}-1}
$$
are congruent modulo $2^{n}$ to $1,3,5, \ldots, 2^{n}-1$ in some order. | [
"It is well-known that all these numbers are odd. So the assertion that their remainders $(\\bmod\\ 2^{n})$ make up a permutation of $\\{1,3, \\ldots, 2^{n}-1\\}$ is equivalent just to saying that these remainders are all distinct. We begin by showing that\n$$\n\\begin{equation*}\n\\binom{2^{n}-1}{2 k}+\\binom{2^{n... | IMO | 49th International Mathematical Olympiad Spain | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
08bz | Problem:
Sia $ABCD$ un quadrilatero convesso, $F$ un punto sul segmento $CD$, $E$ il punto di intersezione di $AC$ con $BF$. È noto che $AB = FC$, $AE = 14$, $BE = 10\sqrt{2}$, $\widehat{BAC} = \widehat{BFD}$, $\widehat{BEA} = 45^\circ$. Quanto misura il segmento $EF$? | [
"Solution:\n\nLa risposta è $6$. Si prolunghi il segmento $BF$ dalla parte di $F$, e su tale prolungamento si consideri il punto $K$, tale che $FK = 14$. L'angolo $\\widehat{CFK}$ è uguale all'angolo $\\widehat{DFB}$, in quanto opposti al vertice per costruzione, per cui i triangoli $ABE$ e $FCK$ sono simili e, in ... | Italy | Gara di Febbraio | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 6 | |
0ajd | Let $A$ and $B$ be two identical convex polygons, each having area $2015$. The polygon $A$ is divided into polygons $A_1, A_2, \dots, A_{2015}$ with positive area, and the polygon $B$ into polygons $B_1, B_2, \dots, B_{2015}$ with positive area. The polygons $A_1, A_2, \dots, A_{2015}, B_1, B_2, \dots, B_{2015}$ are co... | [
"After the overlapping of the polygons we get $C_{ij} = A_i \\cap B_j$, $i, j \\in \\{1, 2, \\dots, 2015\\}$. The polygons $B_1, B_2, \\dots, B_{2015}$ can be colored in $2015!$ ways.\n\nLet a coloring of $A_1, A_2, \\dots, A_{2015}$ be given. For an arbitrary coloring of $B_1, B_2, \\dots, B_{2015}$ which we denot... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0jjf | Problem:
Find the number of ordered quadruples of positive integers $(a, b, c, d)$ such that $a$, $b$, $c$, and $d$ are all (not necessarily distinct) factors of $30$ and $a b c d > 900$. | [
"Solution:\nAnswer: $1940$\n\nSince $a b c d > 900 \\Longleftrightarrow \\frac{30}{a} \\frac{30}{b} \\frac{30}{c} \\frac{30}{d} < 900$, and there are $\\binom{4}{2}^{3}$ solutions to $a b c d = 2^{2} 3^{2} 5^{2}$, the answer is $\\frac{1}{2}\\left(8^{4} - \\binom{4}{2}^{3}\\right) = 1940$ by symmetry."
] | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1940 | |
0jxv | Problem:
Find the smallest possible value of $x+y$ where $x, y \geq 1$ and $x$ and $y$ are integers that satisfy $x^{2}-29 y^{2}=1$ | [
"Solution:\nContinued fraction convergents to $\\sqrt{29}$ are $5, \\frac{11}{2}, \\frac{16}{3}, \\frac{27}{5}, \\frac{70}{13}$ and you get $70^{2}-29 \\cdot 13^{2}=-1$ so since $(70+13 \\sqrt{29})^{2}=9801+1820 \\sqrt{29}$ the answer is $9801+1820=11621$"
] | United States | February 2017 | [
"Number Theory > Diophantine Equations > Pell's equations"
] | null | final answer only | 11621 | |
0j4n | Problem:
A sequence of integers $\{a_{i}\}$ is defined as follows: $a_{i} = i$ for all $1 \leq i \leq 5$, and $a_{i} = a_{1} a_{2} \cdots a_{i-1} - 1$ for all $i > 5$. Evaluate $a_{1} a_{2} \cdots a_{2011} - \sum_{i=1}^{2011} a_{i}^{2}$. | [
"Solution:\nAnswer: $-1941$\n\nFor all $i \\geq 6$, we have $a_{i} = a_{1} a_{2} \\cdots a_{i-1} - 1$. So\n$$\n\\begin{aligned}\na_{i+1} & = a_{1} a_{2} \\cdots a_{i} - 1 \\\\\n& = \\left(a_{1} a_{2} \\cdots a_{i-1}\\right) a_{i} - 1 \\\\\n& = \\left(a_{i} + 1\\right) a_{i} - 1 \\\\\n& = a_{i}^{2} + a_{i} - 1.\n\\e... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | -1941 | |
0dae | Let $I, O$ be the incenter, circumcenter of triangle $ABC$ and $A_{1}, B_{1}, C_{1}$ be arbitrary points on the segments $AI, BI, CI$ respectively. The perpendicular bisectors of $AA_{1}, BB_{1}, CC_{1}$ intersect each other at $X, Y$ and $Z$. Prove that the circumcenter of triangle $XYZ$ coincides with $O$ if and only... | [
"Denote $\\angle B = \\beta$ then we have\n$$\n\\angle XYZ = 180^\\circ - \\angle AIC = 180^\\circ - \\left(90^\\circ + \\frac{\\beta}{2}\\right) = 90^\\circ - \\frac{\\beta}{2}.\n$$\nSo if $O$ is the circumcenter of $XYZ$ then\n$$\n\\angle OXZ = 90^\\circ - \\left(90^\\circ - \\frac{\\beta}{2}\\right) = \\frac{\\b... | Saudi Arabia | Team selection tests for GMO 2018 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
00m2 | Man bestimme alle Polynome $P(x)$, die reelle Koeffizienten haben und die folgenden zwei Bedingungen erfüllen:
a. $P(2017) = 2016$ und
b. $(P(x) + 1)^2 = P(x^2 + 1)$ für alle reellen Zahlen $x$.
(Walther Janous) | [
"Mit $Q(x) := P(x) + 1$ erhalten wir $Q(2017) = 2017$ und $(Q(x))^2 = Q(x^2 + 1) - 1$ für alle $x \\in \\mathbb{R}$, was sich zu $Q(x^2 + 1) = Q(x)^2 + 1$ für alle $x \\in \\mathbb{R}$ umformen lässt.\nWir definieren die Folge $(x_n)_{n \\ge 0}$ rekursiv durch $x_0 = 2017$ und $x_{n+1} = x_n^2 + 1$ für alle $n \\ge... | Austria | 48. Österreichische Mathematik-Olympiade Bundeswettbewerb für Fortgeschrittene, Teil 1 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials"
] | German | proof and answer | P(x) = x - 1 | |
0b1v | Problem:
Determine all positive integers $k$ for which there exist positive integers $r$ and $s$ that satisfy the equation
$$
\left(k^{2}-6 k+11\right)^{r-1}=(2 k-7)^{s}.
$$ | [
"Solution:\nClearly, if $r=1$, then $2 k-7=1$ or $2 k-7=-1$. Thus, two solutions are $k=4$ and $k=3$.\n\nFurthermore, notice that if $k=2$, then $3^{r-1}=(-3)^{s}$, which has a solution for $r$ and $s$. Thus, another solution is $k=2$.\n\nFor $r \\geq 2$, notice that $k^{2}-6 k+11=(k-3)^{2}+2 \\geq 2$ and $k^{2}-6 ... | Philippines | $22^{ ext {nd }}$ Philippine Mathematical Olympiad National Stage, Written Phase | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 2, 3, 4, 8 | |
01ho | Given an equilateral triangle, find all positive integers $n$ such that it can be partitioned into $n$ equilateral triangles (not necessarily of the same size). | [
"An equilateral triangle can be partitioned into one piece, that is a partition with the triangle itself as the only piece.\nWe also note that the cases $n = 6$ and $n = 8$ are possible as figure 5 shows\n\n\nFigure 5: The cases $n = 6$ and $n = 8$\n\nAssume that the original triangle can b... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | All positive integers except 2, 3, and 5 | |
0ktm | Problem:
Let $ABCD$ be a trapezoid such that $AB \parallel CD$, $\angle BAC = 25^{\circ}$, $\angle ABC = 125^{\circ}$, and $AB + AD = CD$. Compute $\angle ADC$. | [
"Solution:\n\nConstruct the parallelogram $ABED$. From the condition $AB + AD = CD$, we get that $EC = AD = EB$. Thus,\n$$\n\\angle ADC = \\angle BEC = 180^{\\circ} - 2 \\angle BCE = 180^{\\circ} - 2 \\cdot 55^{\\circ} = 70^{\\circ}.\n$$"
] | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 70 | |
0bu3 | Problem:
Fie $a, b, c \in \mathbb{C}^*$, distincte şi având acelaşi modul, astfel încât
$$
a^{2}+b^{2}+c^{2}+a b+a c+b c=0
$$
Demonstraţi că $a, b, c$ reprezintă afixele vârfurilor unui triunghi dreptunghic sau echilateral. | [
"Solution:\nPutem presupune $|a|=|b|=|c|=1$. Ipoteza devine $(a+b+c)^{2}=a b+b c+c a$, de unde $(a+b+c)^{2}=a b c\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)$ sau $(a+b+c)^{2}=a b c \\overline{(a+b+c)}$. Prin trecere la modul avem $|a+b+c| \\in\\{0,1\\}$.\n\nDacă $|a+b+c|=0$, deducem că ortocentrul triungh... | Romania | Olimpiada Naţională de Matematică | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
028e | Problem:
Um queijo triangular - Osvaldo comprou um queijo em forma de um triângulo equilátero. Ele quer dividir o queijo igualmente entre ele e seus quatro primos. Faça um desenho indicando como ele deve fazer essa divisão. | [
"Solution:\n\nUm queijo triangular - Para dividir o queijo em 5 partes iguais, é suficiente dividi-lo em $5k$ partes iguais e dar $k$ partes a cada um. Uma forma de fazer essa partição, é mostrada na figura, onde o queijo foi partido em $25 = 5 \\times 5$ triângulos.\n\n"
] | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0enl | Sixty points, of which thirty are coloured red, twenty are coloured blue, and ten are coloured green, are marked on a circle. These points divide the circle into sixty arcs. Each of these arcs is assigned a number according to the colours of its endpoints: an arc between a red and a green point is assigned a number 1, ... | [
"Let the score of a red point be $0$, the score of a green point be $1$, and the score of a blue point be $2$. Note that the number assigned to an arc is at most the sum of the scores of the endpoints. This means that the sum of all the numbers assigned to the arcs is at most twice the sum of all the sixty scores, ... | South Africa | South African Mathematical Olympiad Third Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 100 | |
0id1 | Problem:
Find (with proof) all monic polynomials $f(x)$ with integer coefficients that satisfy the following two conditions.
1. $f(0)=2004$.
2. If $x$ is irrational, then $f(x)$ is also irrational.
(Notes: A polynomial is monic if its highest degree term has coefficient 1. Thus, $f(x)=x^{4}-5 x^{3}-4 x+7$ is an exampl... | [
"Solution:\nThe polynomial $x+2004$ certainly meets the two conditions. In fact, this is the only one. We will prove this using three ingredients: the infinitude of primes, the Rational Roots Theorem for polynomials, and the approximation principle that $x^{n}$ dominates any polynomial of lower degree, for large en... | United States | Bay Area Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | f(x) = x + 2004 | |
0f70 | Problem:
A $1$ or a $-1$ is put into each cell of an $n \times n$ array as follows. A $-1$ is put into each of the cells around the perimeter. An unoccupied cell is then chosen arbitrarily. It is given the product of the four cells which are closest to it in each of the four directions. For example, if the cells below ... | [] | Soviet Union | 20th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Minimum = ceil((n−2)^2 / 2); Maximum = (n−2)^2 | |
07s4 | Define a sequence $x_n : n = 1, 2, 3, \dots$ by $x_1 = 1$ and $x_n = x_{n-1} + \sqrt{n}$ for $n \ge 2$. Show that:
$$
\sum_{n=1}^{2018} \frac{1}{x_n} < 3.
$$ | [
"$x_n \\ge \\frac{(n+1)(n+2)}{6}$\n\nThis is true for $n=1$. Now suppose $n \\ge 2$ and suppose that, inductively:\n$$\nx_{n-1} \\ge \\frac{n(n+1)}{6}\n$$\nNow as $n \\ge 2$, we have:\n$$\n\\frac{n+1}{n} = 1 + \\frac{1}{n} \\le \\frac{3}{2}\n$$\nMultiplying each side by $\\frac{n(n+1)}{9}$:\n$$\n\\frac{(n+1)^2}{9} ... | Ireland | Irish | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
086y | Problem:
Determinare il massimo intero positivo $k$ tale che $k^{2}$ divide $\frac{n !}{(n-6) !}$ per ogni $n>6$. | [
"Solution:\n\nLa risposta è 12. Sia $N=\\frac{n !}{(n-6) !}=(n-5)(n-4)(n-3)(n-2)(n-1) n$.\n\nCerchiamo innanzitutto qual è il massimo $q$ tale che $q$ divide $N, \\forall n$. Tra sei interi consecutivi tre sono divisibili per 2, di questi tre, uno è sicuramente divisibile per 4. Dunque $2^{4}$ divide $N$. Tra sei i... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 12 | |
0da0 | Two positive integers $m$ and $n$ are called similar if one of them can be obtained from the other one by swapping two digits (note that a 0-digit cannot be swapped with the leading digit). Find the greatest integer $N$ such that $N$ is divisible by 13 and any number similar to $N$ is not divisible by 13. | [
"Let $k$ be the number of digits of $N$ and let $N = \\sum_{i=0}^{k-1} 10^{i} \\cdot d_{i}$ where $d_{i}$ are digits. If $M$ is obtained by swapping the $i$-th and the $j$-th digits, then\n$$\nM - N = (10^{j} - 10^{i}) \\cdot (d_{i} - d_{j}) .\n$$\nTherefore $13 \\mid M - N$ implies that $d_{i} = d_{j}$ or $6 \\mid... | Saudi Arabia | Team selection tests for GMO 2018 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | 9876230 | |
05rv | Problem:
Soit $n$ un entier naturel. Un escalier de taille $n$ est constitué de petits carrés $1 \times 1$, avec 1 carré pour la première marche, 2 carrés pour la deuxième marche, et ainsi de suite, jusqu'à $n$ carrés pour la $n^{\text{ème}}$ marche.
On dispose de pierres carrées (de côté entier) de toutes les tailles... | [
"Solution:\n\nCommençons par quelques définitions et observations générales. Dans la suite, on note $(i, j)$ le petit carré $1 \\times 1$ situé au $j$ème étage de la $i^{\\text{ème}}$ marche. On appellera carré supérieur chaque petit carré $(k, k)$, c'est-à-dire chaque carré situé tout en haut d'une marche.\n\nOn c... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | 1) f(n) = n holds exactly for n = 2^k − 1 with k ≥ 0.
2) f(n) = n + 1 holds exactly for n = 2^{k+1} − 2^{k'} − 1 with k ≥ k' ≥ 0. | |
01vn | For all positive integers $n$ and $m$ prove the inequality
$$
|n\sqrt{n^2+1} - m| \ge \sqrt{2} - 1.
$$ | [
"Let $M(n, m) = |n\\sqrt{n^2+1} - m|$. Using an obvious chain of inequalities\n$$\nn^2 < n\\sqrt{n^2+1} < n^2 + \\frac{1}{2},\n$$\nwe obtain\n$$\nn^2 - m < n\\sqrt{n^2+1} - m < n^2 - m + \\frac{1}{2}.\n$$\nTherefore, if $m \\neq n^2$ then $M(n, m) > 1/2$. Since $1/2 > \\sqrt{2}-1$, the required inequality is proved... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
09sp | Problem:
Bepaal alle drietallen $(x, y, z)$ van niet-negatieve reële getallen die voldoen aan het stelsel vergelijkingen
$$
\begin{aligned}
& x^{2}-y=(z-1)^{2} \\
& y^{2}-z=(x-1)^{2} \\
& z^{2}-x=(y-1)^{2}
\end{aligned}
$$ | [
"Solution:\nHaakjes uitwerken en alles bij elkaar optellen geeft\n$$\nx^{2}+y^{2}+z^{2}-(x+y+z)=x^{2}+y^{2}+z^{2}-2(x+y+z)+3,\n$$\ndus $x+y+z=3$. Zonder verlies van algemeenheid nemen we aan dat $x \\leq y, z$. Dan geldt $0 \\leq x \\leq 1$. Dus $x^{2} \\leq x$, dus $x^{2}-y \\leq x-y \\leq 0$. Anderzijds is $x^{2}... | Netherlands | Selectietoets | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (1, 1, 1) | |
0gqp | Find all pairs of positive odd integers $(m, n)$ satisfying $n \mid 3m + 1$ and $m \mid n^2 + 3$. | [
"Conditions $n \\mid 3m + 1$ and $m \\mid n^2 + 3$ we label by (1) and (2).\nBy (1) $3$ and $n$ are coprime: $(3, n) = 1$ (3).\nLet $n \\le 9$. Due to (3) $n$ can take $1, 5, 7$.\nIf $n = 1$ from (2) $m \\mid 4$ and since $m$ is odd we get $m = 1$. $(m, n) = (1, 1)$ satisfies the conditions.\nIf $n = 5$ (1) and (2)... | Turkey | Team Selection Test | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (1,1), (49,37), (43,13) | |
0h1s | In volleyball tournament there are 8 teams, that play one-round tournament (each team plays exactly one game with another). Each win worth 1 point, each lose worth 0 points, there are no draws in volleyball. After tournament is finished, if the difference between the first and the second place, does not exceed 1 point,... | [
"If we assume, that there were no extra games then the difference between 1-st and 2-nd, 3-d and 4-th, 5-th and 6-th, and 7-th and 8-th is at least 2 points and therefore the difference between 1-st and 8-th places is at least 8 points, while the first place can not have more than 7 points. The following example sh... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1 | |
0gls | Let $ABCD$ be a cyclic quadrilateral. Let $O$ be the circumcenter of the quadrilateral $ABCD$. The diagonals $AC$ and $BD$ intersect at $G$. Let $P, Q, R$ and $S$ be the circumcenters of triangles $AGB, BGC, CGD$ and $DGA$ respectively. The lines $PR$ and $QS$ intersect at $M$. Show that $M$ is the midpoint of $G$ and ... | [
"First we show that $PORG$ is a parallelogram.\n\nSince $R$ and $O$ lie on the perpendicular bisector of chord $CD$, we have $OR$ is perpendicular to $CD$.\nLet $L$ be the intersection of $PG$ and $CD$, and $X$ be the midpoint of $BG$.\nSince $\\angle(LP, PX) = \\angle(GP, PX) = \\angle(GA,... | Thailand | The 14th Thailand Mathematical Olympiad | [
"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 | |
085h | Problem:
Ogni anno un gran numero di studenti partecipa alle Olimpiadi Internazionali di Matematica. Un dodicesimo di essi vince una medaglia d'oro, un altro sesto vince una medaglia d'argento, un ulteriore quarto vince una medaglia di bronzo e la restante metà vince una stretta di mano. Se incontriamo un gruppo di se... | [
"Solution:\n\nLa risposta è $\\mathbf{( C )}$. Contiamo innanzitutto il numero di modi in cui si possono suddividere 6 studenti in modo che due di essi abbiano vinto la medaglia d'oro, due la medaglia d'argento e due non abbiano vinto alcuna medaglia. I due che hanno vinto la medaglia d'oro possono essere scelti in... | Italy | Olimpiadi di Matematica | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | MCQ | (C) | |
0dww | Problem:
Naj bo $T$ poljubna točka znotraj kvadrata $A B C D$, točke $A^{\prime}, B^{\prime}, C^{\prime}$ in $D^{\prime}$ pa druga presečišča premic $T A, T B, T C$ in $T D$ z očrtano krožnico kvadrata $A B C D$. Dokaži, da je $|A^{\prime} B^{\prime}| \cdot |C^{\prime} D^{\prime}| = |A^{\prime} D^{\prime}| \cdot |B^{\... | [
"Solution:\n\nZaradi tetivnosti imamo mnogo enakih obodnih kotov in zato tudi mnogo parov podobnih trikotnikov: $\\triangle A B T \\sim \\triangle B^{\\prime} A^{\\prime} T$, $\\triangle B C T \\sim \\triangle C^{\\prime} B^{\\prime} T$, $\\triangle C D T \\sim \\triangle D^{\\prime} C^{\\prime} T$ in $\\triangle D... | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a0n | A baker baked a big square cake. He wants to cut the cake. He cuts the cake only lengthwise or widthwise, all the way from one edge to the opposite edge.
At least how many cuts does the baker need in order to cut the cake into exactly $180$ pieces?
A) $25$ B) $26$ C) $27$ D) $28$ E) $29$ | [
"A) $25$"
] | Netherlands | Junior Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | MCQ | A | |
0dmb | Problem:
Нека је $k$ природан број. За сваку функцију $f: \mathbb{N} \rightarrow \mathbb{N}$, нека је низ функција $\left(f_{m}\right)_{m \geqslant 1}$ дефинисан са $f_{1}=f$ и $f_{m+1}=f \circ f_{m}$ за $m \geqslant 1$. Функција $f$ је $k$-фина уколико за све $n \in \mathbb{N}$ важи
$$
f_{k}(n)=f(n)^{k}
$$
a) За кој... | [
"Solution:\n\nСвака функција је $1$-фина, па је одговор на оба дела задатка потврдан. Нека је надаље $k \\geqslant 2$. Свака $k$-фина функција је $1$-$1$ јер из $f(m)=f(n)$ следи $m^{k}=f_{k}(m)=f_{k}(n)=n^{k}$, тј. $m=n$.\n\na) Одговор: ДА. Конструишимо функцију $f$ индуктивно на следећи начин. Нека је $n$ најмањи... | Serbia | СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | a) For all k. b) Only for k equal to 1. | |
01t7 | Given a graph with $n$ ($n \ge 4$) vertices. It is known that for any two of the vertices there is a vertex connected with none of these two vertices.
Find the greatest possible number of the edges in the graph.
(E. Barabanov) | [
"(Solution by Y. Dubovik, L. Manzhulina, I. Pchalintsau, B. Serankou.)\nFirst, for the sake of convenience, we reformulate the problem as follows. Any\ntwo of $n$ vertices of the graph ($n \\ge 4$) are connected with an edge either of\nred or of blue color. It is known that for any two vertices there exists a verte... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | ((n-2)(n-1))/2 - floor(n/2) | |
068z | Let $AB\Gamma$ an equilateral triangle of side $\alpha$. Let $\Delta$, $E$ and $Z$ the midpoints of the sides $AB$, $B\Gamma$ and $\Gamma A$, respectively. Let $H$ the symmetric point of $\Delta$ with respect to the line $B\Gamma$. We color the points $A$, $B$, $\Gamma$, $\Delta$, $E$, $Z$, $H$ using one of the two col... | [
"a.\nThere are defined totally seven equilateral triangles from the given points. Since $\\Delta E = EZ = Z\\Delta = \\frac{\\alpha}{2}$ the seven equilateral triangles are\n\nFigure 7\n\nFigure 8\n$AB\\Gamma$, $A\\Delta Z$, $B\\Delta E$, $E\\Gamma Z$, $\\Delta EZ$,... | Greece | SELECTION EXAMINATION | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) 7; b) Yes; c) No | |
0lbb | Calculate the number of the arrangements of 5 girls $G_1$, $G_2$, $G_3$, $G_4$ and $G_5$ and 12 boys in a row satisfying the following conditions:
1. The order of the girls from left to right is $G_1$, $G_2$, $G_3$, $G_4$ and $G_5$.
2. There are at least 3 boys between $G_1$ and $G_2$.
3. There are at least 1 boy and a... | [
"Recall that the number of natural solutions of the equation\n$$\n\\sum_{i=1}^{n} x_i = m\n$$\nis $\\binom{m+n-1}{n-1}$. We will use this fact to calculate the number of the arrangements of boys and girls satisfying the given conditions.\n\nLet $x_i$ be the number of boys standing between $G_{i-1}$ and $G_i$ for $i... | Vietnam | Vietnam Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | Vietnamese | proof and answer | 556120857600 | |
0l64 | Problem:
Let $\omega_{1}$ and $\omega_{2}$ be two circles intersecting at distinct points $A$ and $B$. Point $X$ varies along $\omega_{1}$, and point $Y$ on $\omega_{2}$ is chosen such that $AB$ bisects the angle $\angle XAY$. Prove that as $X$ varies along $\omega_{1}$, the circumcenter of $\triangle AXY$ (if it exis... | [
"Solution:\n\n\n\nLet $O_{1}$, $O_{2}$, and $O$ be the centers of $\\omega_{1}$, $\\omega_{2}$, and the circumcircle of $\\triangle AXY$, respectively.\nWe claim that triangle $O O_{1}O_{2}$ is isosceles with $O O_{1} = O O_{2}$, and thus in particular $O$ always lies on the perpendicular b... | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configuration... | null | proof only | null | |
04y5 | Find all positive real numbers $c$ such that there are infinitely many pairs of positive integers $(n, m)$ satisfying the following conditions: $n \ge m + c\sqrt{m-1} + 1$ and among numbers $n, n+1, \dots, 2n-m$ there is no square of an integer. | [
"We prove that $c$ satisfies the condition in the statement if and only if $c \\le 2$.\n\nLet us first consider any $c \\le 2$. For any positive integer $k$, define\n$$\nn = k^2 + 1 \\quad \\text{and} \\quad m = (k-1)^2 + 1\n$$\nObserve that\n$$\nm + c\\sqrt{m-1} + 1 \\le k^2 - 2k + 2 + 2(k-1) + 1 = k^2 + 1 = n\n$$... | Czech-Polish-Slovak Mathematical Match | null | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | all positive real c with c ≤ 2 | |
0jej | Problem:
Gary plays the following game with a fair $n$-sided die whose faces are labeled with the positive integers between $1$ and $n$, inclusive: if $n=1$, he stops; otherwise he rolls the die, and starts over with a $k$-sided die, where $k$ is the number his $n$-sided die lands on. (In particular, if he gets $k=1$,... | [
"Solution:\n\nLet $a_n$ be the expected number of rolls starting with an $n$-sided die. We see immediately that $a_1 = 0$, and $a_n = 1 + \\frac{1}{n} \\sum_{i=1}^{n} a_i$ for $n > 1$. Thus $a_2 = 2$, and for $n \\geq 3$, $a_n = 1 + \\frac{1}{n} a_n + \\frac{n-1}{n}(a_{n-1} - 1)$, or $a_n = a_{n-1} + \\frac{1}{n-1}... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 197/60 | |
034p | Problem:
Find all values of the real parameter $a$ such that the equation
$$
a(\sin 2x + 1) + 1 = (a-3)(\sin x + \cos x)
$$
has a solution. | [
"Solution:\nWe write the given equation as\n$$\n2a y^2 - \\sqrt{2}(a-3) y + 1 = 0\n$$\nwhere\n$$\ny = \\frac{\\sqrt{2}}{2}(\\sin x + \\cos x) = \\sin\\left(x + 45^\\circ\\right) \\in [-1,1]\n$$\nIt has a solution if and only if (1) has a solution in the interval $[-1,1]$.\n\nFor $a=0$ the equation (1) is linear and... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (-∞, 1] ∪ [9, ∞) | |
0ftv | Problem:
Auf einem gewöhnlichen Schachbrett stehen 17 Türme. Zeige, dass man stets drei Türme auswählen kann, die sich gegenseitig nicht bedrohen. (Ein Turm kann in einem Zug beliebig viele Felder nach links, rechts, oben oder unten ziehen. Ein Turm bedroht einen anderen, falls er in einem Zug auf das Feld des anderen... | [
"Solution:\n\nZwei Türme bedrohen sich genau dann, wenn sie in derselben Zeile oder Spalte stehen. Unterteile die 64 Felder wie folgt in 8 Schubfächer:\n\n| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 2 | 3 | 4 | 5 | 6 | 7 | 8 | 1 |\n| 3 | 4 | 5 | 6 | 7 | 8 | 1 | 2 ... | Switzerland | Vorrundenprüfung | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0fyj | Problem:
Finde alle Funktionen $f: \mathbb{R} \rightarrow \mathbb{R}$, sodass für alle reellen $x, y$ die folgende Gleichung erfüllt ist:
$$
f(f(x)) + f(f(y)) = 2y + f(x-y)
$$ | [
"Solution:\nWir setzen $a = f(0)$. Mit $x = y$ folgt\n$$\nf(f(x)) = x + \\frac{a}{2}\n$$\nSetzt man dies auf der linken Seite der Gleichung ein und vereinfacht, dann folgt\n$$\nf(x-y) = (x-y) + a\n$$\nalso $f(z) = z + a$ für alle reellen $z$. Einsetzen in die ursprüngliche Gleichung zeigt, dass $a = 0$ sein muss, d... | Switzerland | SMO Finalrunde | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x | |
0bpf | Problem:
Fie $a \geq 0$ şi $\left(x_{n}\right)_{n \geq 1}$ un şir de numere reale. Să se arate că dacă şirul $\left(\frac{x_{1}+\cdots+x_{n}}{n^{a}}\right)_{n \geq 1}$ este mărginit, atunci şirul $\left(y_{n}\right)_{n \geq 1}$, definit prin $y_{n}=\frac{x_{1}}{1^{b}}+\frac{x_{2}}{2^{b}}+\cdots+\frac{x_{n}}{n^{b}}$, e... | [
"Solution:\n\nNotăm $S_{n}=\\sum_{k=1}^{n} x_{k},\\ n \\in \\mathbb{N}^{*}$. Conform ipotezei, există $c>0$ astfel ca $\\left|S_{n}\\right| \\leq c n^{a},\\ \\forall n \\in \\mathbb{N}^{*}$. Pentru $n, p \\in \\mathbb{N}^{*}$, avem:\n$$\n\\begin{gathered}\n\\left|y_{n+p}-y_{n}\\right|=\\left|\\sum_{k=n+1}^{n+p} \\f... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
0dcl | Let $ABC$ be an acute nonisosceles triangle with incenter $I$ and $(d)$ is an arbitrary line tangent to $(I)$ at $K$. The lines passes through $I$, perpendicular to $IA$, $IB$, $IC$ cut $(d)$ at $A_{1}$, $B_{1}$, $C_{1}$ respectively. Suppose that $(d)$ cuts $BC$, $CA$, $AB$ at $M$, $N$, $P$ respectively. The lines thr... | [
"Suppose that $(d)$ is tangent to $(I)$ at $K$. Denote $D$, $E$, $F$ as the tangent points of $(I)$ with the sides $BC$, $CA$, $AB$. Let $V$ be the projection of $K$ onto $A_{1}I$ and $U = AI \\cap EF$. It is easy to see that\n$$\nIU \\cdot IA = IV \\cdot IA_{1} = r^{2}\n$$\nwith $r$ is the radius of $(I)$.\n\nCons... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral ... | English | proof only | null | |
0561 | There are $n$ candies on the table. On every turn, a player eats a number of candies that is greater than $1$ and divides the number of candies on the table at the start of the turn, but must leave at least $1$ candy on the table. Two players take alternate turns and the player who is unable to make a move loses. Find ... | [
"Define all even numbers which are not odd powers of $2$ as *good* and the rest of the positive integers as *bad*. We show that the player before whose turn the number of candies is good has a move which yields in a bad number of candies, whereas the player before whose turn the number of candies is bad either has ... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | All even n that are not of the form 2^(2k+1). | |
0c3e | For $n \ge 2$, consider $n$ boxes aligned from left to right. In each box, one puts a ball that can be red, blue or white such that the following condition is fulfilled: Each box is neighboring at least one box containing a ball of the same color. We denote by $I_n$ the number of such configurations.
a) Determine $I_{... | [
"Obviously $a_2 = 1$, $a_3 = 1$. For $a \\ge 4$, if the first box contains a red ball, then so does the second one. Now we distinguish two types of configurations: those containing a red ball in the third box, and those with a blue or a white ball in the third box. In order to count the configurations of the first ... | Romania | 69th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | I_11 = 1023; I_n = 2^{n-1} + (-1)^n | |
0jje | Problem:
Let $ABC$ be a triangle with $AB = AC = 5$ and $BC = 6$. Denote by $\omega$ the circumcircle of $ABC$. We draw a circle $\Omega$ which is externally tangent to $\omega$ as well as to the lines $AB$ and $AC$ (such a circle is called an $A$-mixtilinear excircle). Find the radius of $\Omega$. | [
"Solution:\n\nLet $M$ be the midpoint of $BC$. Let $D$ be the point diametrically opposite $A$ on the circumcircle, and let the $A$-mixtilinear excircle be tangent to lines $AB$ and $AC$ at $X$ and $Y$. Let $O$ be the center of the $A$-mixtilinear excircle.\n\nNotice that $\\triangle AOX \\sim \\triangle ABM$. If w... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 75/8 | |
0i53 | Problem:
A game is played with two players and an initial stack of $n$ pennies ($n \geq 3$). The players take turns choosing one of the stacks of pennies on the table and splitting it into two stacks. The winner is the player who makes a move that causes all stacks to be of height 1 or 2. For which starting values of ... | [
"Solution:\n\nPlayer 1 wins if and only if $n=3$ or $n$ is even; player 2 wins for all odd $n>3$. We can easily check this for the first few cases, say up to $n=6$, and then we can proceed by induction.\n\n- If $n>6$ is even, player 1 creates a pile of size 1 and a pile of size $n-1$. Since $n-1$ is odd, player 1 w... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | The first player wins exactly when the starting size is even or equals three; for all other odd starting sizes greater than three, the second player wins. | |
0aqe | Problem:
For what real numbers $r$ does the system of equations
$$
\left\{\begin{aligned}
x^{2} &= y^{2} \\
(x - r)^{2} + y^{2} &= 1
\end{aligned}\right.
$$
have no solutions? | [] | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | |r| > sqrt(2) | |
0hrw | Problem:
Certain cities are connected by roads connecting pairs of them. The roads intersect only at the cities. A subset of the roads is called important if destroying those roads would make it so that there are two cities such that it is impossible to go from the first to the second. A subset $S$ of the roads is cal... | [
"Solution:\n\nLet $G$ be the graph having the cities as its vertices and the roads as its edges. (This just restates the problem in graph-theoretic language, which we use from now on.) We will look at the components of the graph $G$ and its subgraphs, where a component of a graph is defined to be a maximal connecte... | United States | Berkeley Math Circle Take-Home Contest | [
"Discrete Mathematics > Graph Theory > Menger's theorem / max-flow, min-cut"
] | null | proof only | null | |
02s0 | Problem:
Daniel inventou uma brincadeira na qual é permitido apenas realizar as seguintes operações:
- somar quatro unidades;
- multiplicar por quatro;
- elevar ao quadrado.
Começando de um certo número, Daniel desafia um amigo a obter um outro número realizando sucessivamente qualquer uma das operações permitidas.
Po... | [
"Solution:\n\na) As operações realizadas em cada etapa aparecem acima de cada seta abaixo:\n\n\nb) Pode-se obter o $9$ a partir do $5$ somando-se $4$. A partir daí pode-se usar as três últimas etapas do item anterior como ilustrado abaixo:\n\n\nc) Observe que somar ... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0k9n | Problem:
Let $a$, $b$, $c$ be positive real numbers. Prove that
$$
\frac{1}{a(1+b)} + \frac{1}{b(1+c)} + \frac{1}{c(1+a)} \geq \frac{3}{1+abc},
$$
and that equality occurs if and only if $a = b = c = 1$. | [
"Solution:\n\nAdding $\\frac{3}{1+abc}$ to both sides, the desired inequality is equivalent to\n$$\n\\frac{1}{a(1+b)} + \\frac{1}{b(1+c)} + \\frac{1}{c(1+a)} + \\frac{3}{1+abc} \\geq \\frac{6}{1+abc}\n$$\nWe note that\n$$\n\\begin{aligned}\n& \\frac{1}{a(1+b)} + \\frac{1}{1+abc} = \\frac{1}{1+abc} \\left( \\frac{1+... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | a = b = c = 1 | |
0ko3 | Problem:
A regular $2022$-gon has perimeter $6.28$. To the nearest positive integer, compute the area of the $2022$-gon. | [
"Solution:\nNote that the area of a regular $2022$-gon is approximately equal to the area of its circumcircle, and the perimeter of a regular $2022$-gon approximately equals the perimeter of its circumcircle. Since the perimeter is $6.28 \\approx 2\\pi$, the circumradius $R \\approx 1$, so the area of the $2022$-go... | United States | HMMT February | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 3 | |
08om | Problem:
Let $x, y, z$ be real numbers, satisfying the relations
$$
\left\{\begin{array}{l}
x \geq 20 \\
y \geq 40 \\
z \geq 1675 \\
x+y+z=2015
\end{array}\right.
$$
Find the greatest value of the product $P = x \cdot y \cdot z$. | [
"Solution:\nBy virtue of $z \\geq 1675$ we have\n$$\ny+z<2015 \\Leftrightarrow y<2015-z \\leq 2015-1675<1675\n$$\nIt follows that $(1675-y) \\cdot (1675-z) \\leq 0 \\Leftrightarrow y \\cdot z \\leq 1675 \\cdot (y+z-1675)$.\nBy using the inequality $u \\cdot v \\leq \\left(\\frac{u+v}{2}\\right)^2$ for all real numb... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 48407500 | |
0jnj | Problem:
Find
$$
\sum_{k=0}^{\infty}\left\lfloor\frac{1+\sqrt{\frac{2000000}{4^{k}}}}{2}\right\rfloor
$$
where $\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$. | [
"Solution:\nThe $k$th floor (for $k \\geq 0$) counts the number of positive integer solutions to $4^{k}(2x-1)^{2} \\leq 2 \\cdot 10^{6}$. So summing over all $k$, we want the number of integer solutions to $4^{k}(2x-1)^{2} \\leq 2 \\cdot 10^{6}$ with $k \\geq 0$ and $x \\geq 1$. But each positive integer can be uni... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 1414 | |
0357 | Problem:
Let $ABCD$ be a convex quadrilateral. The orthogonal projections of $D$ on the lines $BC$ and $BA$ are denoted by $A_1$ and $C_1$, respectively.
The segment $A_1C_1$ meets the diagonal $AC$ at an interior point $B_1$ such that $DB_1 \geq DA_1$. Prove that the quadrilateral $ABCD$ is cyclic if and only if
$$
\... | [
"Solution:\n\nLet $ABCD$ be a cyclic quadrilateral. Then the Simson theorem for $\\triangle ABC$ gives $DB_1 \\perp AC$. Hence $\\angle B_1C_1D = \\angle B_1AD = \\angle CBD$, $\\angle B_1DC_1 = \\angle B_1AC_1 = \\angle CDB$ and therefore $\\triangle B_1C_1D \\sim \\triangle CBD$.\n\nAnalogously $\\triangle B_1A_1... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02gi | Show that there is a set of $2002$ distinct positive integers such that the sum of one or more elements of the set is never a square, cube, or higher power. | [
"Let $p$ be a prime and $A = \\{p, 2p, 3p, \\dots, 2002p\\}$. The sum of any quantity of numbers from $A$ is at most $p + 2p + \\dots + 2002p = 1001 \\cdot 2003p$. Choose any $p > 1001 \\cdot 2003$ and we are done, because every sum of numbers from $A$ is a multiple of $p$ but not of $p^2$, and cannot be a perfect ... | Brazil | XXIV OBM | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0k4e | Problem:
An equilateral hexagon with side length $1$ has interior angles $90^{\circ}, 120^{\circ}, 150^{\circ}, 90^{\circ}, 120^{\circ}, 150^{\circ}$ in that order. Find its area. | [
"Solution:\n\nAnswer: $\\frac{3+\\sqrt{3}}{2}$\n\nThe area of this hexagon is the area of a $\\frac{3}{2} \\times \\left(1+\\frac{\\sqrt{3}}{2}\\right)$ rectangle (with the $90^{\\circ}$ angles of the hexagon at opposite vertices) minus the area of an equilateral triangle with side length $1$. Then this is\n$$\n\\f... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | (3+sqrt(3))/2 | |
0gkf | Let $x$, $y$, $z$ be positive real numbers. Prove that
$$
\frac{x}{\sqrt{2(x^2 + y^2)}} + \frac{y}{\sqrt{2(y^2 + z^2)}} + \frac{z}{\sqrt{2(z^2 + x^2)}} < \frac{4x^2 + y^2}{x^2 + 4y^2} + \frac{4y^2 + z^2}{y^2 + 4z^2} + \frac{4z^2 + x^2}{z^2 + 4x^2} < 9.
$$ | [
"**Solution.** To prove the second inequality, we may assume $x = \\max\\{x, y, z\\}$. Then we get\n$$\n\\frac{4z^2 + x^2}{z^2 + 4x^2} \\le 1, \\quad \\frac{4x^2 + y^2}{x^2 + 4y^2} < 4, \\quad \\text{and} \\quad \\frac{4y^2 + z^2}{y^2 + 4z^2} < 4.\n$$\nSo the desired inequality follows.\n\nNext we prove the first i... | Thailand | Thai Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0dow | Prove the inequality
$$
4 \left( \sqrt[3]{\frac{a}{b}} + \sqrt[3]{\frac{b}{c}} + \sqrt[3]{\frac{c}{a}} \right) \le 3 \left( 2 + a + b + c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)^{2/3}
$$
for positive real numbers $a$, $b$ and $c$ with $abc = 1$. | [
"Let $x$, $y$, $z$ be positive real numbers such that $a = x/y$, $b = y/z$, $c = z/x$. After substitution and simplifying the given inequality is transformed to the following one:\n$$\n4(x + y + z)^{\\sqrt{3}xyz} \\le 3(x + y)^{2/3}(y + z)^{2/3}(z + x)^{2/3}\n$$\nFor triangle with the sides lengths $u = x + y$, $v ... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle i... | English | proof only | null | |
07ps | Find all pairs $(m, n)$ of integers that satisfy $m^2 + n^2 = 65$. Use this to prove the inequalities
$$
\frac{4 + 4\sqrt{10} + 6\sqrt{2}}{\sqrt{65}} < \pi < \frac{4}{11}(3 + \sqrt{2} + 2\sqrt{5}).
$$ | [
"The 16 solutions of $x^2 + y^2 = 65$ in integers are\n$$\n(\\pm 8, \\pm 1), (\\pm 1, \\pm 8), (\\pm 4, \\pm 7), (\\pm 7, \\pm 4).\n$$\n\nThese form the vertices of a hexadecagon. Calculating the perimeter of this hexadecagon gives rise to the lower bound for $\\pi$ as follows. Let $P_1 = (8, -1)$, $P_2 = (8, 1)$, ... | Ireland | Ireland | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization i... | null | proof and answer | All integer solutions: (±8, ±1), (±1, ±8), (±4, ±7), (±7, ±4). The bounds: (4 + 4√10 + 6√2)/√65 < π < (4/11)(3 + √2 + 2√5). | |
0gv7 | Let $\alpha, \beta, \gamma \in [0; \frac{\pi}{2}]$ satisfy the conditions
$$
\sin \alpha + \sin \beta + \sin \gamma = 1, \sin \alpha \cos 2\alpha + \sin \beta \cos 2\beta + \sin \gamma \cos 2\gamma = -1.
$$
Find all the possible values of the expression $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$. | [
"З умови задачі випливають рівності\n$$\n\\sin \\alpha - 2\\sin^3 \\alpha + \\sin \\beta - 2\\sin^3 \\beta + \\sin \\gamma - 2\\sin^3 \\gamma = -1, \\\\ \\sin^3 \\alpha + \\sin^3 \\beta + \\sin^3 \\gamma = 1.\n$$\nОскільки для $\\varphi \\in [0; \\frac{\\pi}{2}]$ справджуються нерівності $\\sin^3 \\varphi \\le \\si... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | 1 | |
0756 | Let $ABCDE$ be a convex pentagon in which $BC \parallel AE$, $AB = BC + AE$ and $\angle ABC = \angle CDE$. Let $M$ be the mid-point of $CE$ and let $O$ be the circumcentre of triangle $BCD$. Suppose $\angle DMO = 90^\circ$. Prove that $2\angle BDA = \angle CDE$. | [
"Let the circum-circle of $BCD$ be $\\Gamma$ and let the circle with diameter $OD$ be $\\Gamma'$. Let the mid-point of $OD$ be $O'$ (which is also the centre of $\\Gamma'$). Now $OO'$ passes through $D$ and hence $\\Gamma'$ and $\\Gamma$ are tangent to each other at $D$. Hence there is a homothety with centre $D$ t... | India | Indija TS | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Ge... | English | proof only | null | |
01qx | Consider a fixed circle $\Gamma$ with three fixed points $A$, $B$, and $C$ on it. Also let us fix a real number $\lambda \in (0,1)$. For a variable point $P \notin \{A, B, C\}$ on $\Gamma$, let $M$ be the point on the segment $CP$ such that $CM = \lambda \cdot CP$. Let $Q$ be the second point of intersection of the cir... | [
"3. See IMO-2014 Shortlist, Problem G4."
] | Belarus | SELECTION and TRAINING SESSION | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous > Constructions an... | English | proof only | null | |
00at | A circle of length $999$ is divided into unit arcs by $999$ black points. Then $d$ arcs with lengths $1$, $2$, ..., $d$ are placed on the circle, with their end points black, so that none of these arcs contains another (otherwise the arcs may overlap). Find all $d$ for which such a configuration exists. | [
"Consider the problem for a circle of length $n$ divided into unit arcs by $n$ black points, and $d$ arcs of lengths $1$, $2$, ..., $d$ with black endpoints such that none of them contains another. We show that the maximal admissible $d$ equals $\\lfloor \\frac{n+1}{2} \\rfloor$; here $\\lfloor \\cdot \\rfloor$ den... | Argentina | Argentine National Olympiad 2016 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | d = 1, 2, ..., 500 | |
05bt | Let $D$ be a point on the side $BC$ of an acute triangle $ABC$ and let $E$ be a point on the line segment $AD$. Let $F$ and $G$ be the feet of the altitudes drawn from the vertex $D$ in triangles $ABD$ and $ACD$, respectively. The line $BE$ intersects the circumcircle of the triangle $DEG$ at point $H \neq E$. Prove th... | [
"Since $\\angle AFD = \\angle AGD = 90^\\circ$, the quadrilateral $AFDG$ is cyclic (Fig. 28). Using inscribed angles in the circumcircle of the quadrilateral $AFDG$, we get\n$$\n\\angle BFG = 180^\\circ - \\angle AFG = 180^\\circ - \\angle ADG = 180^\\circ - \\angle EDG.\n$$\n\n\nFig. 28\n\... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05ks | Problem:
Soit $n \in \mathbb{N}^{*}$. On appelle $S$ l'ensemble des suites de $2 n$ chiffres comportant $n$ zéros et autant de uns. Deux suites de $S$ sont voisines lorsqu'il suffit de changer la position d'un chiffre de l'une pour obtenir l'autre. Par exemple, 11100010 et 10110010 sont voisines puisqu'en décalant le ... | [
"Solution:\n\nOn traite séparément les inégalités.\n\n- Montrons que $\\frac{1}{2 n^{2}+1}|S| \\leq|T|$. Soit $s$ un élément de $S$. Soit $s_{i, j}$ la suite que l'on peut obtenir à partir de $s$, en déplaçant le $i$-ème chiffre de $s$ pour le mettre en position $j$. Si $s_{i, j} \\neq s$, alors $j \\neq i$. En out... | France | Envoi de combinatoire | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0gxu | Eight teams are playing single round-robin tournament, that is each team plays every other team once. And at that the schedule is arranged by rounds, in every round all teams play. What is the least possible number of rounds should finish to determine the best and the worst teams (that is, for any results of the games ... | [
"a) Lets show that at least 6 rounds are required. It is easy to construct an example for 6 rounds. We need to show that 5 rounds are not enough. Indeed, even if the best team won all games, to ensure first place, it should have at least 3 point advantage. Similarly, the worst team should have at least 3 point disa... | Ukraine | The Problems of Ukrainian Authors | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) 6 rounds; b) 5 rounds | |
0esk | Let one dollar cost $R12$, and let one British pound cost $R21$. What does one pound cost in dollars?
(A) 1.25 (B) 1.30 (C) 1.50 (D) 1.60 (E) 1.75 | [
"One rand is $\\frac{1}{12}$ of a dollar, so one pound is $\\frac{21}{12} = \\frac{7}{4} = 1.75$ dollars."
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | English | MCQ | E | |
03xt | Determine all possible values of positive integer $n$, such that there are $n$ different 3-element subsets $A_1, A_2, \dots, A_n$ of the set $\{1, 2, \dots, n\}$, with $|A_i \cap A_j| \neq 1$ for all $i \neq j$. | [
"The set of positive integers satisfying the given condition consists of all positive multiples of $4$. We first prove that $n = 4k$ ($k \\in \\mathbb{Z}_+$) satisfies the condition. Define $A_1, A_2, \\dots, A_{4k}$ as follows: $A_{4i-j} = \\{4i-3, 4i-2, 4i-1, 4i\\} \\setminus \\{4i-j\\}$, for all $1 \\le i \\le k... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All positive multiples of 4 | |
0ive | Problem:
Consider an equilateral triangle and a square both inscribed in a unit circle such that one side of the square is parallel to one side of the triangle. Compute the area of the convex heptagon formed by the vertices of both the triangle and the square. | [
"Solution:\n\nAnswer: $\\frac{3+\\sqrt{3}}{2}$\n\n\n\nConsider the diagram above. We see that the shape is a square plus 3 triangles. The top and bottom triangles have base $\\sqrt{2}$ and height $\\frac{1}{2}(\\sqrt{3}-\\sqrt{2})$, and the triangle on the side has the same base and height ... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | (3 + sqrt(3)) / 2 | |
0cez | A *diameter* of a finite planar set is any line segment of maximal Euclidean length having both end points in that set. A *lattice point* in the Cartesian plane is one whose coordinates are both integral. Given an integer $n \ge 2$, prove that a set of $n$ lattice points in the plane has at most $n-1$ diameters. | [
"Consider the *diameter graph* $\\Gamma$ on $n$ pairwise distinct lattice points in the plane, i.e., the geometric graph on those points, whose edges are the diameters of the configuration. We will prove that $\\Gamma$ has at least one vertex of degree 1. Removal of one such and the edge it is incident to, allows t... | Romania | Twentieth IMAR Mathematical Competition | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Algebrai... | English | proof only | null | |
0jac | Problem:
Franklin has four bags, numbered 1 through 4. Initially, the first bag contains fifteen balls, numbered 1 through 15, and the other bags are empty. Franklin randomly pulls a pair of balls out of the first bag, throws away the ball with the lower number, and moves the ball with the higher number into the secon... | [
"Solution:\n\nAnswer: $\\frac{2}{3}$\n\nPretend there is a 16th ball numbered 16. This process is equivalent to randomly drawing a tournament bracket for the 16 balls, and playing a tournament where the higher ranked ball always wins. The probability that a ball is left in a bag at the end is the probability that i... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 2/3 | |
0ese | The square of the difference between the largest and smallest of three consecutive integers is
(A) 1 (B) 9 (C) 4 (D) 100 (E) 2 | [
"Any three consecutive integers are of the form $n$, $n+1$, $n+2$. The difference between the largest and the smallest is $(n+2) - n = 2$, and the square of the difference is $4$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | C | |
07m5 | Suppose $a, b, c$ are the side lengths of a triangle $ABC$. Let
$$
x = \frac{b+c}{2}, \quad y = \frac{c+a}{2}, \quad z = \frac{a+b}{2}.
$$
Show that $x, y, z$ are the side lengths of a triangle $XYZ$, with the same perimeter as $ABC$, but with a bigger area, unless $ABC$ is equilateral. | [
"First of all, for instance, $x < y + z$, because\n$$\n2x = b + c < 2a + b + c = (c + a) + (a + b) = 2y + 2z.\n$$\nThus, $x, y, z$ are the side lengths of a triangle. Next, $x + y + z = a + b + c = 2s$, in the usual notation. Hence, $\\triangle ABC$ and $\\triangle XYZ$ have the same perimeter. Now, by Heron's form... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0g4b | Problem:
Determine all pairs $(m, p)$ of a positive integer $m$ and a prime number $p$ satisfying the equation
$$
p^{2} + p m = m^{3}
$$ | [
"Solution:\nRewriting the equation as $p^{2} = m (m^{2} - p)$, we see that $m$ must divide $p^{2}$. However, since $p$ is prime, the only positive factors of $p^{2}$ are $1$, $p$ and $p^{2}$. We now check each case separately:\n\nCase $m = 1$:\nThe equation becomes $p^{2} + p = 1$. Since this would imply that $p$ d... | Switzerland | Second round 2022 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2, 2) | |
07u0 | A sequence whose first term is positive is constructed so that any given term is the area of a square whose perimeter is the preceding term. If the first three terms form an arithmetic progression, determine all possible values of the first term. | [
"A square of side length $x$ has perimeter $4x$ and area $x^2$. If the preceding term of the sequence is $T_{n-1}$, then its successor is $T_n = (T_{n-1}/4)^2$. Thus the first three terms are\n$$\na, \\quad \\frac{a^2}{16} = \\frac{a^2}{2^4} \\quad \\text{and} \\quad \\frac{a^4}{4096} = \\frac{a^4}{2^{12}}.\n$$\nSi... | Ireland | IRL_ABooklet | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 16 and 8(sqrt(5) - 1) | |
0jlq | Problem:
Let $\mathcal{C}$ be a circle in the $xy$ plane with radius $1$ and center $(0,0,0)$, and let $P$ be a point in space with coordinates $(3,4,8)$. Find the largest possible radius of a sphere that is contained entirely in the slanted cone with base $\mathcal{C}$ and vertex $P$. | [
"Solution:\n$3-\\sqrt{5}$\n\nConsider the plane passing through $P$ that is perpendicular to the plane of the circle. The intersection of the plane with the cone and sphere is a cross section consisting of a circle inscribed in a triangle with a vertex $P$. By symmetry, this circle is a great circle of the sphere, ... | United States | HMMT 2014 | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 3 - sqrt(5) | |
01de | Let $ABC$ be a triangle and $D$ and $E$ points on the lines $CA$ and $BA$ such that $CD = AB$, $BE = AC$ and $A$, $D$ and $E$ lie on the same side of $BC$. Let $I$ be the incenter of $ABC$ and let $H$ be a point such that $I$ is the orthocenter of $BCI$. Show that $D$, $E$ and $H$ are collinear. | [
"Let the point $A'$ be such that $ABA'C$ is a parallelogram with $AB \\parallel A'C$ and $AC \\parallel A'B$. Denote $\\alpha = \\angle BAC = \\angle CA'B$.\nSince $CD = AB = CA'$, we find that $CDA'$ is an isosceles triangle. As $\\angle DCA' = 180^\\circ - \\alpha$, we deduce that $\\angle A'DC = \\angle CA'B = \... | Baltic Way | Baltic Way 2016 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
049w | Every cell of a $1000 \times 1000$ table is colored black or white. The difference between the number of black and white cells is $2012$. Prove that there exists a $2 \times 2$ square that contains an odd number of white cells. | [
"Assume the contrary, that every $2 \\times 2$ square contains an even number of black cells. Let us compare two adjacent rows.\nIf the first cell in the lower row is the same color as the first cell in the upper row (e.g. black), then the second cells in these rows are also the same color (either black or white). ... | Croatia | CroatianCompetitions2011 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0l89 | Find all polynomials $f$ with integer coefficients such that for all positive integers $n$, $n$ divides $\underbrace{f(f(\dots(f(0))\dots))}_{n+1 \text{ f's}} -1$. | [] | United States | USA TST Selection Test for 67th IMO and 15th EGMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All integer-coefficient polynomials f such that either f(x) = x + 1 for all x, or f satisfies f(1) = 1 and f(f(0)) = 1. | |
0crf | In a plane, $n$ convex $k$-gons are given. Every two of them share a common point, and every two of them are homothetic with a positive ratio. Prove that there exists a point belonging to at least $1 + \frac{n-1}{2k}$ of these $k$-gons. | [
"Lemma.\nLet $P$ and $P'$ be intersecting convex polygons, homothetic with a positive ratio. Then one of the vertices of one of them lies in the other.\n\nProof. If one of the polygons is entirely contained in the other, the statement is obvious. Otherwise, there exists a side $AB$ of polygon $P$ that intersects th... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0acf | In the first barrel there is $5$ hl and $25$ l wine. In the second barrel there is $3$ times more wine than in the first one and in the third there is $1$ hl and $75$ l less wine than in the first one. How much wine is there in each of the barrels? | [
"In the first barrel there is $5$ hl $25$ l $= 525$ l of wine. So in the second barrel there is $3 \\cdot 525$ l $= 1575$ l of wine and in the third $5$ hl $25$ l $- 1$ hl $75$ l $= 525$ l $- 175$ l $= 350$ l of wine. In the three barrels there is $525$ l $+ 1575$ l $+ 350$ l $= 2450$ l $= 24$ hl $50$ l of wine in ... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | First barrel: 525 liters; Second barrel: 1575 liters; Third barrel: 350 liters | |
06fy | Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Show that
$$
\frac{1}{a^2(b+c)} + \frac{1}{b^2(c+a)} + \frac{1}{c^2(a+b)} \ge \frac{3}{2}.
$$ | [
"Let $x = bc$, $y = ca$ and $z = ab$. The condition becomes $xyz = 1$. Note that\n$$\n\\sum_{\\text{cyc}} \\frac{1}{a^2(b+c)} = \\sum_{\\text{cyc}} \\frac{bc}{a(b+c)} = \\sum_{\\text{cyc}} \\frac{x}{y+z}.\n$$\n\nIt remains to prove\n$$\n\\frac{x}{y+z} + \\frac{y}{z+x} + \\frac{z}{x+y} \\ge \\frac{3}{2}.\n$$\n\nThis... | Hong Kong | IMO HK TST | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
00hg | Let $m$ be a fixed positive integer. The infinite sequence $\{a_{n}\}_{n \geq 1}$ is defined in the following way: $a_{1}$ is a positive integer, and for every integer $n \geq 1$ we have
$$
a_{n+1}= \begin{cases}a_{n}^{2}+2^{m} & \text{ if } a_{n}<2^{m} \\ a_{n} / 2 & \text{ if } a_{n} \geq 2^{m}\end{cases}
$$
For each... | [
"Suppose that for integers $m$ and $a_{1}$ all the terms of the sequence are integers. For each $i \\geq 1$, write the $i$th term of the sequence as $a_{i}=b_{i} 2^{c_{i}}$ where $b_{i}$ is the largest odd divisor of $a_{i}$ (the \"odd part\" of $a_{i}$) and $c_{i}$ is a nonnegative integer.\n\nLemma 1. The sequenc... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | Only when m = 2, and then exactly for starting values a1 = 2^ℓ with ℓ ≥ 1; for all other m there is no valid starting value. | |
0e10 | Let $ABC$ be an acute triangle such that $|AB| > |BC| > |AC|$. Let $D$ be a point different from $C$ on the segment $BC$, such that $|AC| = |AD|$. Let $H$ denote the orthocentre of the triangle $ABC$, and let $A_1, B_1$ be the feet of the altitudes from $A$ and $B$, respectively. The line $DH$ intersects the line $AC$ ... | [
"The triangle $CAD$ is isosceles since $|AC| = |AD|$. The line $AA_1$ is the altitude in this isosceles triangle, so $\\angle HDA = \\angle ACH$.\n\n\n\nIn the quadrilateral $HA_1CB_1$ we have $\\angle CA_1H = \\frac{\\pi}{2} = \\angle CB_1H$, so this quadrilateral is cyclic and $\\angle B_... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01n5 | Find all pairs $(n; m)$ of positive integers $n$ and $m$ satisfying the equality $n^2 + n + 1 = (m^2 + m - 3)(m^2 - m + 5)$. | [
"Answer: $(n; m) = (4; 2)$.\nBy condition,\n$$\nn^2 + n + 1 = (m^2 + n - 3)(m^2 - n + 5) = m^4 + m^2 + 8m - 15.\n$$\nConsider the obtained equation\n$$\nn^2 + n - (m^4 + m^2 + 8m - 16) = 0 \\quad (1)\n$$\nas a quadratic equation with respect to $n$. It has positive integer roots only if the determinant $D = 4m^4+4m... | Belarus | Belorusija 2012 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (4; 2) | |
05eu | Problem:
Trouver tous les triplets $\{p, q, r\}$ de nombres premiers tels que les 3 différences
$$
|p-q|,\ |q-r|,\ |r-p|
$$
soient également des nombres premiers. | [
"Solution:\n\nNotons que les trois nombres doivent être deux à deux distincts puisque $0$ n'est pas un nombre premier. On peut donc supposer, quitte à échanger l'ordre des variables, que $p > q > r$. Un nombre premier est impair ou égal à $2$.\n\nOn suppose que $p$, $q$ et $r$ sont tous impairs. Alors $p-q$, $q-r$ ... | France | ENVOi 3 : ARITHMÉTIQUE | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | {2, 5, 7} | |
06iw | Let $\{a_n\}$ be a sequence of positive integers. It is given that $a_1 = 1$, and for $n \ge 1$, $a_{n+1}$ is the smallest positive integer greater than $a_n$ which satisfies the following condition: for any integers $i, j, k$, with $1 \le i, j, k \le n+1$, $a_i + a_j \ne 3a_k$. Find $a_{2015}$. | [
"We have $a_{2015} = 4531$.\nWe claim that\n$$\na_{4k+1} = 9k + 1,\\quad a_{4k+2} = 9k + 3,\\quad a_{4k+3} = 9k + 4,\\quad a_{4k+4} = 9k + 7\n$$\nfor any integer $k \\ge 0$. The base cases $k = 0, 1$ can be verified directly. Indeed, the first 8 terms are\n1, 3, 4, 7, 10, 12, 13, 16.\nNow, assume the claim holds fo... | Hong Kong | 1997-2023 IMO HK TST | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 4531 | |
06h7 | Given a line $\ell$ and four points $P$, $Q$, $R$ and $S$ (in that order) on the line, describe a straight edge (unmarked ruler) and compass construction producing a square $ABCD$ such that $P$ lies on the line $AD$, $Q$ on the line $BC$, $R$ on the line $AB$ and $S$ on the line $CD$. | [
"Note that we can use compass and ruler to find the perpendicular bisector of two given points. Thus, we can use this to locate the midpoint of two points, and draw the circle having a given segment as a diameter.\n\nConstruct the circles with diameter $PS$ and $QR$ respectively. Construct the midpoints $M$ and $N$... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | null | |
0jpa | Problem:
Starting with the number $0$, Casey performs an infinite sequence of moves as follows: he chooses a number from $\{1, 2\}$ at random (each with probability $\frac{1}{2}$) and adds it to the current number. Let $p_{m}$ be the probability that Casey ever reaches the number $m$. Find $p_{20} - p_{15}$. | [
"Solution:\n\nWe note that the only way $n$ does not appear in the sequence is if $n-1$ and then $n+1$ appears. Hence, we have $p_{0} = 1$, and $p_{n} = 1 - \\frac{1}{2} p_{n-1}$ for $n > 0$. This gives $p_{n} - \\frac{2}{3} = -\\frac{1}{2} (p_{n-1} - \\frac{2}{3})$, so that\n$$\np_{n} = \\frac{2}{3} + \\frac{1}{3}... | United States | HMMT February 2015 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 11/2^20 | |
0hdx | Show that one can choose 8 pairwise distinct numbers among $1, 2, \ldots, 10000$, such that none of them is a perfect square and that no sum of the several of them is a perfect square. | [
"Consider the following 7 numbers: $2^1, 2^3, \\ldots, 2^{13} = 8192 < 10000$. Clearly, the sum of any subset of them is such that the highest power of $2$ that divides the sum is odd, thus, it is not a perfect square.\n\nAdd number $3$ to the chosen numbers. Suppose it is possible to choose several numbers such th... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
02ix | Problem:
Na lista de frações, no quadro ao lado, temos:
- 2 frações cuja soma é $\frac{5}{2}$
- 2 frações cuja diferença é $\frac{5}{2}$
- 2 frações cujo produto é $\frac{5}{2}$
- 2 frações cujo quociente é $\frac{5}{2}$
Encontre a fração que está sobrando.
| $\frac{5}{4}$ | $\frac{17}{6}$ | $\frac{-5}{4}$ | $\frac{1... | [
"Solution:\n\na. 2 frações cuja diferença é $\\frac{5}{2}$:\n$\\frac{5}{4} - \\left(-\\frac{5}{4}\\right) = \\frac{5}{4} + \\frac{5}{4} = \\frac{10}{4} = \\frac{5}{2}$\n\nb. 2 frações cujo produto é $\\frac{5}{2}$:\n$\\frac{10}{7} \\times \\frac{14}{8} = \\frac{10}{7} \\times \\frac{7}{4} = \\frac{10}{4} = \\frac{5... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | -3/2 | |
061k | Problem:
Man ermittle alle Lösungen der Gleichung $x^{2y} + (x+1)^{2y} = (x+2)^{2y}$ mit $x, y \in \mathbb{N}$. | [
"Solution:\n\nMan erkennt leicht, dass weder $x$, noch $y$, Null sein können.\nFür $y=1$ erhält man aus $x^{2} + (x+1)^{2} = (x+2)^{2}$ die Gleichung $x^{2} - 2x - 3 = 0$, von der nur die Lösung $x=3$ in Frage kommt.\n\nSei nun $y > 1$.\nDa $x$ und $x+2$ dieselbe Parität haben, ist $x+1$ eine gerade und demnach $x$... | Germany | Auswahlwettbewerb zur IMO 2002 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | x=3, y=1 | |
0dqj | Determine all couples of coprime numbers $(p, q)$ satisfying:
$$
p^3 - q^5 = (p+q)^2.
$$ | [
"Consider the equation modulo $q$, which reduces to $p^3 \\equiv p^2 \\pmod{q}$. Since $p$ and $q$ are coprime ($p \\neq q$ clearly), we must have $p \\equiv 1 \\pmod{q}$.\n\nConsider now the equation modulo $q^2$. Then the equation becomes $p^3 \\equiv p^2 + 2pq \\pmod{q^2}$, and since $\\gcd(p, q) = 1$, $p^2 \\eq... | Singapore | Singapore International Mathematical Olympiad Committee National Team Selection Test | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (7, 3) | |
0gvz | A circle is divided into $2006$ equal arcs by $2006$ points. Baron Munchausen claims that he can construct a closed polygonal curve with the set of vertices consisting of these $2006$ points such that amongst its $2006$ edges there cannot be found any two, which are parallel to each other. Is his claim true or false? | [
"Відповідь: ні, барон помиляється. Позначимо $2006$ точок поділу числами $0, 1, 2, \\ldots, 2005$, записаними послідовно. Ланки ламаної будемо позначати номерами її кінців. Нескладно перевірити, що якщо $i + j = k + l$ (mod $2006$), то ланки $i j$ та $k l$ будуть паралельними. Припустимо, що барон Мюнхгаузен правий... | Ukraine | Ukrainian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | false | |
0cu4 | Each cell of a $2016 \times 2016$ table is painted either in white or in black. We say that a positive integer $k$ is lucky if $k \le 2016$, and each of cellular $k \times k$ squares within the table contains exactly $k$ black cells. (For example, if all the cells are black, then only the number 1 is lucky.) Determine ... | [
"Let $a$ and $b$ be lucky. If $b \\ge 2a$, then a $b \\times b$ square contains $[b/a]^2$ squares of size $a \\times a$, so at least $[b/a]^2 \\cdot a > b$ black cells. This is impossible. Thus, the ratio of any two lucky numbers is less than $2$, which yields that there are at most $1008$ of them.\n\nAn example is... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 1008 | |
07f2 | We have a rectangle with its sides being a mirror. A light ray enters from one of the corners of the rectangle and after being reflected several times, gets to the opposite corner of its starting point. Prove that the light ray has passed the center (Intersection of diagonals) of the rectangle. | [
"First note that if the line $l$ has slope $a$, then the reflection of $l$ with respect to any line which is parallel to one of the axes has the slope $-a$.\n\nNow assume that at the start the ray has slope $a$. Then the ray always has the slope $\\pm a$. Now at the starting point a line with slope $-a$ lies outsid... | Iran | 37th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations"
] | English | proof only | null | |
04fd | Let $ABC$ be an acute triangle with the orthocentre $H$. Let $D$ be such a point that the quadrilateral $AHCD$ is a parallelogram. Let $p$ be a line perpendicular to the line $AB$ passing through the midpoint $A_1$ of the segment $BC$. Let us denote the intersection of $p$ and $AB$ by $E$ and the midpoint of the segmen... | [
"Since $AHCD$ is a parallelogram we have $\\angle ADC = \\angle CHA = 180^\\circ - \\beta$, which implies that $D$ lies on the circumference of the triangle $ABC$. Also, $\\angle ACD = \\angle HAC = 90^\\circ - \\gamma$.\nSince $ABCD$ is cyclic we have $\\angle ABD = \\angle ACD$, and since lines $AG$ and $BD$ are ... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
07cw | An arbitrary point $P$ lies on side $BC$ of triangle $ABC$. Angle bisectors of $\widehat{APB}$ and $\widehat{APC}$ intersect the external angle bisector of $\widehat{A}$ at $X$ and $Y$, respectively. Circumcircle of triangle $PXY$ meets $BC$ for the second time at $Q$. Prove that $\widehat{BAP} = \widehat{CAQ}$. | [
"Let $Q'$ be a point on side $BC$ such that $\\overline{BAP} = \\overline{CAQ'} = \\alpha$.\n\n\n\nNote that $X, Y$ also lie on the exterior angle bisector of $\\overline{PAQ'}$, that's because\n$$\n\\overline{PAX} = \\overline{Q'AY} = \\left(90^\\circ - \\frac{\\hat{A}}{2}\\right) + \\alph... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null |
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