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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0hha | Find all such functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for any real $x, y$ the following equality holds:
$$
f(x + y f(x + y)) = f(y^2) + x f(y) + f(x).
$$ | [
"**Answer:** $f(x) = x$, $f(x) = 0$.\n\nLet us denote the given equation by (1):\n$$\nf(x + y f(x + y)) = f(y^2) + x f(y) + f(x). \\quad (1)\n$$\n\nSubstitute $x = y = 0$ into (1):\n$$\nf(0) = 2f(0) \\implies f(0) = 0.\n$$\n\nSubstitute $x = 0$ into (1):\n$$\nf(y f(y)) = f(y^2). \\quad (2)\n$$\n\nSubstitute $y = -x... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x or f(x) = 0 | |
01g3 | Let odd positive integer $a$ be not a perfect square, $b$, and $c$ be odd primes and
$$
a^2 + a + 1 = 3(b^2 + b + 1)(c^2 + c + 1).
$$
Prove that at least one of $b^2 + b + 1$ and $c^2 + c + 1$ is composite. | [
"Assume that $b^2 + b + 1$ and $c^2 + c + 1$ are primes. Then\n$$\na^2 + a + 1 \\equiv 0 \\pmod{b^2 + b + 1}.\n$$\nThis quadratic equation modulo prime has at most 2 roots. We must have\neither $a \\equiv b \\pmod{b^2 + b + 1}$ or $a \\equiv b^2 \\pmod{b^2 + b + 1}$\n(in the latter case $a^2 + a + 1 \\equiv b^4 + b... | Baltic Way | Baltic Way 2019 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
04dh | By erasing the first two digits of a positive integer $n$ we obtain a number that is 73 times smaller. Determine the smallest two such numbers $n$. | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 365 and 1825 | |
0kug | Problem:
Acute triangle $ABC$ has circumcenter $O$. The bisector of $\angle ABC$ and the altitude from $C$ to side $AB$ intersect at $X$. Suppose that there is a circle passing through $B$, $O$, $X$, and $C$. If $\angle BAC = n^{\circ}$, where $n$ is a positive integer, compute the largest possible value of $n$. | [
"Solution:\n\nWe have $\\angle XBC = B/2$ and $\\angle XCB = 90^{\\circ} - B$. Thus, $\\angle BXC = 90^{\\circ} + B/2$. We have $\\angle BOC = 2A$, so\n$$\n90^{\\circ} + B/2 = 2A\n$$\nThis gives $B = 4A - 180^{\\circ}$, which gives $C = 360^{\\circ} - 5A$.\n\nIn order for $0^{\\circ} < B < 90^{\\circ}$, we need $45... | United States | HMMT February 2023 | [
"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 and answer | 67 | |
00fr | Find the smallest positive integer $n$ with the following property: There does not exist an arithmetic progression of $1999$ terms of real numbers containing exactly $n$ integers. | [
"We first note that the integer terms of any arithmetic progression are \"equally spaced\", because if the $i$th term $a_{i}$ and the $(i+j)$th term $a_{i+j}$ of an arithmetic progression are both integers, then so is the $(i+2j)$th term $a_{i+2j} = a_{i+j} + (a_{i+j} - a_{i})$.\n\nThus, by scaling and translation,... | Asia Pacific Mathematics Olympiad (APMO) | XI APMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 70 | |
0bt6 | a) Prove that there exist non-periodical functions $f : \mathbb{R} \to \mathbb{R}$ such that
$$
f(x + 1) + f(x - 1) = \sqrt{5}f(x),
$$
for all $x \in \mathbb{R}$.
b) Prove that any function $g : \mathbb{R} \to \mathbb{R}$ such that
$$
g(x + 1) + g(x - 1) = \sqrt{3}g(x),
$$
for all $x \in \mathbb{R}$, is periodical. | [
"a) It is natural to look for solutions of the exponential type $f(x) = a^x$, where $a > 0$. If this is a solution, we get $a + a^{-1} = \\sqrt{5}$, and thus $a = \\frac{\\sqrt{5}+1}{2}$. It is easy to check that the functions $f_1: \\mathbb{R} \\to \\mathbb{R}$, $f_1(x) = (\\frac{\\sqrt{5}-1}{2})^x$ and $f_2: \\ma... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
074x | Let $\triangle ABC$ be a triangle. Let $AD$, $BE$, $CF$ be cevians such that $\angle BAD = \angle CBE = \angle ACF$. Suppose these cevians concur at a point $\Omega$. (Such a point exists for each triangle and it is called a Brocard point.) Prove that
$$
\frac{A\Omega^2}{BC^2} + \frac{B\Omega^2}{CA^2} + \frac{C\Omega^2... | [
"Note that $\\angle B\\Omega D = \\angle AB\\Omega + BA\\Omega = B - \\alpha + \\alpha = B$, where $\\alpha$ is the Brocard angle. Thus triangles $B\\Omega D$ and $ABD$ are similar. It follows that $BD^2 = AD \\cdot \\Omega D$. Let $BD : DC = x : y$, $CE : EA = z : x$, $AF : FB = y : z$. Then $BD = a x/(x+y)$. We a... | India | Indija TS 2010 | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
064v | Let $ABCD$ be a convex quadrilateral with $\angle DAC = \angle BDC = 36^\circ$, $\angle CBD = 18^\circ$ and $\angle BAC = 72^\circ$. The diagonals $AC$ and $BD$ intersect at point $P$. Determine the measure of $\angle APD$. | [
"In the rays $DA$ and $BA$ we take a point $E$ and $Z$, respectively, such that $AC = AE = AZ$.\n\nSince $\\widehat{DEC} = \\frac{\\widehat{DAC}}{2} = 18^\\circ = \\widehat{CBD}$, the quadrilateral $DEBC$ is cyclic.\n\nSimilarly, the quadrilateral $CBZD$ is cyclic, because\n$$\n\\widehat{AZC} = \\frac{\\widehat{BAC... | Greece | 11th Junior Balkan Mathematical Olympiad | [
"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",
"Geometry > Plane Geometry > Miscellaneou... | English | proof and answer | 108 | |
02fy | Two players play a game as follows. The first player chooses two non-zero integers $A$ and $B$. The second player forms a quadratic with $A$, $B$ and $1998$ as coefficients (in any order). The first player wins iff the equation has two distinct rational roots. Show that the first player can always win. | [
"Choose $A$ and $B$ such that $A + B + 1998 = 0$. Then $1$ is a root of the quadratic equation no matter how the second player arranges the coefficients.\n\nThe other root is also rational, because the product of the roots, the quotient of two of the rational coefficients $A$, $B$, $1998$, is rational."
] | Brazil | XX OBM | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0en8 | Let $ABCD$ be a square and $X$ a point such that $A$ and $X$ are on opposite sides of $CD$. The lines $AX$ and $BX$ intersect $CD$ in $Y$ and $Z$ respectively. If the area of $ABCD$ is $1$ and the area of $XYZ$ is $\frac{2}{3}$, determine the length of $YZ$.
 | [
"Let the length of $YZ$ be $x$, and let $y$ be the associated height of triangle $XYZ$. Note that triangles $XYZ$ and $XAB$ are similar. Since corresponding sides $YZ$ and $AB$ have lengths $x$ and $1$ respectively, the heights $y$ and $1+y$ have to satisfy\n$$\n\\frac{y}{1+y} = \\frac{x}{1} = x.\n$$\nHence the are... | South Africa | South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 2/3 | |
0cd9 | Given a triangle $ABC$, let $H$ and $O$ be its orthocentre and circumcentre, respectively. Let $K$ be the midpoint of the line segment $AH$. Let further $\ell$ be a line through $O$, and let $P$ and $Q$ be the orthogonal projections of $B$ and $C$ onto $\ell$, respectively. Prove that $KP + KQ \ge BC$. | [
"Fix the origin at $O$ and the real axis along $\\ell$. As usual lower case letters denote the complex coordinate of the corresponding point in the configuration. For convenience, let $|a| = |b| = |c| = 1$. Clearly,\n\n$$\nk = a + \\frac{1}{2}(b + c), \\quad p = a + \\frac{1}{2}\\left(b + \\frac{1}{b}\\right) \\qua... | Romania | THE Fifteenth ROMANIAN MASTER OF MATHEMATICS | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
0dgm | Twenty children are queueing for ice cream that is sold at SR5 per cone. Ten of the children have a SR5 coin, the others want to pay with a R10 bill. At the beginning, the ice cream man does not have any change. How many possible arrangements of the twenty kids in a queue are there so that the ice cream man will never ... | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 16796 | |
0hep | Given positive numbers $a, b, c$ that satisfy the condition:
$$
a^2 + b^2 + c^2 + abc = 4,
$$
prove the following inequality:
$$
(4 - a^2)(4 - b^2)(4 - c^2)a^2 b^2 c^2 \leq (2a + bc)(2b + ca)(2c + ab).
$$ | [
"First, we will prove the following lemmas:\n\n**Lemma 1.** If $\\alpha, \\beta, \\gamma \\in (0^\\circ, 90^\\circ)$ and $\\cos^2 \\alpha + \\cos^2 \\beta + \\cos^2 \\gamma + 2\\cos\\alpha\\cos\\beta\\cos\\gamma = 1$, then $\\alpha, \\beta, \\gamma$ are the angles of an acute triangle.\n\n*Proof.* Consider the equa... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0jkr | Problem:
Let
$$
A=\frac{1}{6}\left(\left(\log_{2}(3)\right)^{3}-\left(\log_{2}(6)\right)^{3}-\left(\log_{2}(12)\right)^{3}+\left(\log_{2}(24)\right)^{3}\right)
$$
Compute $2^{A}$. | [
"Solution:\nAnswer: $72$\nLet $a=\\log_{2}(3)$, so $2^{a}=3$ and \n$$\nA=\\frac{1}{6}\\left[a^{3}-(a+1)^{3}-(a+2)^{3}+(a+3)^{3}\\right].\n$$\nBut $(x+1)^{3}-x^{3}=3x^{2}+3x+1$, so\n$$\nA=\\frac{1}{6}\\left[3(a+2)^{2}+3(a+2)-3a^{2}-3a\\right]=\\frac{1}{2}[4a+4+2]=2a+3.\n$$\nThus $2^{A}=(2^{a})^{2}(2^{3})=9 \\cdot 8=... | United States | HMMT 2014 | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 72 | |
0euh | In a table, there are $2009$ coins each with one side white and the other side black. At the beginning all coins are aligned in a row with all their white sides up except one with its black side up. In each step we choose a coin with black side up and flip the two adjacent coins. If we choose the outermost coin, then w... | [
"For each coin in the table we give a positive integer one by one from the leftmost one. We show that if the black coin is in the $s = 1005$th place, then we can make each coin's black side up after some steps. If we choose $s, s-1, s+1$ one by one, then all five coins having numbers from $s-2$ to $s+2$ become blac... | South Korea | 22nd Korean Mathematical Olympiad Final Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 1005 | |
01gf | Find all triples $x, y, z$ of real numbers such that the following system holds:
$$
\begin{cases} x^3 + y = z^2 \\ y^3 + z = x^2 \\ z^3 + x = y^2 \end{cases}
$$ | [
"Answer. The only triple satisfying given system of equations is $(x, y, z) = (0, 0, 0)$.\n\nFirst, consider the case $x, y, z \\ge 0$. Note that for all $t \\ge 0$ we have $t^3 + t \\ge 2t^2$. Summing up the three equations we obtain $x^3 + x + y^3 + y + z^3 + z = x^2 + y^2 + z^2$. By our observation, the lefthand... | Baltic Way | Baltic Way 2020 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | (0, 0, 0) | |
0klg | Let $n = 8^{2022}$. Which of the following is equal to $\frac{n}{4}$?
(A) $4^{1010}$ (B) $2^{2022}$ (C) $8^{2018}$ (D) $4^{3031}$ (E) $4^{3032}$ | [
"**Answer (E):** Note that $8 = 2^3$, so\n$$\nn = 8^{2022} = (2^3)^{2022} = 2^{6066} = (2^2)^{3033} = 4^{3033}.\n$$\nThus\n$$\n\\frac{n}{4} = \\frac{4^{3033}}{4} = 4^{3032}.\n$$\n\nAll the other choices are less than $4^{3032}$. Choices (A) and (D) have the same base but a lesser exponent. Choice (B) is incorrect b... | United States | Fall 2021 AMC 10 B | [
"Algebra > Intermediate Algebra > Exponential functions"
] | English | MCQ | E | |
005p | En un país hay $100$ distritos electorales, todos con la misma cantidad de votantes. Cada distrito elige un diputado para el parlamento nacional entre $3$ candidatos que representan a los $3$ partidos políticos $A$, $B$ y $C$. A nivel nacional, los partidos $A$, $B$ y $C$ tienen la adhesión de exactamente $60\%$, $30\%... | [] | Argentina | Argentina 2008 | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | Minimum districts A wins: 20. Maximum districts C could win: 39. | |
02de | Given a sheet of paper and the use of a rule, compass and pencil, show how to draw a straight line that passes through two given points, if the length of the ruler and the maximum opening of the compass are both less than half the distance between the two points. You may not fold the paper. | [
"Note that we can draw an arbitrarily long line through a given point by repeatedly extending a short line. We can also find the midpoint of an arbitrary line segment. For suppose the segment is $PQ$. Take a distance $k$ which is less than the maximum opening of the compass and less than the length of the ruler. St... | Brazil | III OBM | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0krc | Problem:
Call a simple graph $G$ quasi-colorable if we can color each edge blue, red, green, and white such that
- for each vertex $v$ of degree $3$ in $G$, the three edges containing $v$ as an endpoint are either colored blue, red, and green, or all three edges are white,
- not all edges are white.
A connected graph ... | [
"Solution:\n\nConsider a graph $G$ such that $\\frac{a}{b} > \\frac{1}{4}$. Note that the number of edges is $\\frac{4a + 3b}{2}$. Additionally, if any two vertices of degree $4$ are adjacent, we can simply color that edge red and every other edge in $G$ white to get a valid quasi-coloring. Thus, suppose no two ver... | United States | HMIC | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | 1/4 | |
0k1r | Problem:
Compute the number of integers $n \in \{1,2, \ldots, 300\}$ such that $n$ is the product of two distinct primes, and is also the length of the longest leg of some nondegenerate right triangle with integer side lengths. | [
"Solution:\nLet $n = p \\cdot q$ for primes $p < q$. If $n$ is the second largest side of a right triangle there exist integers $c, a$ such that $a < p q$ and $(p q)^2 = c^2 - a^2 = (c-a)(c+a)$. Since $c-a < c+a$ there are three cases for the values of $c-a, c+a$, and in each case we determine when $a < p q$.\n\na.... | United States | HMMT February 2018 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 13 | |
0h5x | Let $ABC$ be an acute-angled triangle and altitudes $AA_1$ and $BB_1$ intersect at $H$. Consider circles $w_1$ and $w_2$ with centers $H$ and $B$ and with radii $HB_1$ and $BB_1$ respectively. Let $CN$ and $CK$ be the tangent lines from $C$ to circles $w_1$ and $w_2$ respectively $(N \neq B_1, K \neq B_1)$. Prove that ... | [
"Let $CC_1$ be the altitude (Fig. 42). Since quadrilateral $AB_1HC_1$ is cyclic, we have $\\angle A = 180^\\circ - \\angle C_1HB_1 = \\angle B_1HC_1$. Since $\\angle B_1HC_1 = \\angle NHC_1$, we obtain $\\angle A = \\angle CHN_1$.\n\nTaking into account $\\angle HA_1C = 90^\\circ$, $\\angle HNC = 90^\\circ$, $\\ang... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"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 > Concurrency and Colline... | English | proof only | null | |
0bfy | Let $a$, $b$, $c$, $n$ be four integers, where $n \ge 2$, and let $p$ be a prime dividing both $a^2 + ab + b^2$ and $a^n + b^n + c^n$, but not $a + b + c$; for instance, $a \equiv b \equiv -1 \pmod{3}$, $c \equiv 1 \pmod{3}$, $n$ a positive even integer, and $p = 3$ or $a = 4$, $b = 7$, $c = -13$, $n = 5$, and $p = 31$... | [
"Throughout the proof congruences are taken modulo $p$. Begin by ruling out the case $p = 2$. If $p = 2$, then $a^2 + ab + b^2$ and $a^n + b^n + c^n$ are both even, and $a + b + c$ is odd. The first condition forces both $a$ and $b$ even, so $c$ is also even by the second, contradicting the third. Consequently, $p$... | Romania | The Danube Mathematical Competition | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0b0l | Problem:
Gari took a 6-item multiple choice test with 3 choices per item, labelled $A$, $B$, and $C$. After the test, he tried to recall his answers to the items. He only remembered that he never answered three consecutive $A$'s, he never answered three consecutive $B$'s, and he did not leave any item blank. How many ... | [
"Solution:\n\nThe problem is equivalent to that of finding the number of ternary strings of length $6$ that do not contain any $3$ consecutive $0$'s or $3$ consecutive $1$'s. Using the principle of inclusion and exclusion, it suffices to count the number of ternary strings of length $6$ that contain at least $3$ co... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 569 | |
01xu | The circles $\omega_1$ and $\omega_2$ centered at $O_1$ and $O_2$, respectively, intersect at the point $X$. The line $O_1X$ intersects $\omega_2$ at the points $X$ and $B$, and the line $XO_2$ intersects $\omega_1$ at the points $X$ and $A$. The line $O_1A$ intersects $\omega_1$ at the points $A$ and $D$, and the line... | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
026p | Problem:
Uma urna tem 9 bolas, numeradas com os números de 1 a 9. José e Maria retiram simultaneamente uma bola da urna. Com as bolas retiradas eles formam um número de 2 algarismos, sendo que o número que está escrito na bola de José é o algarismo das dezenas e o número que está escrito na bola de Maria é o algarismo... | [
"Solution:\n\nSejam $a$ e $b$ os números escritos nas bolas retiradas por José e Maria, respectivamente. Existem então 9 possibilidades para $a$ e 8 possibilidades para $b$. Deste modo, existem $9 \\times 8 = 72$ possibilidades para o número $ab$.\n\nPor outro lado, para contar quantos destes números são pares, pre... | Brazil | Nível 3 | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | 4/9 | |
0hsd | Problem:
A $2 \times n$ grid has a light bulb in each square. Each bulb has a switch that flips the state of its corresponding bulb as well as all (horizontally or vertically) adjacent bulbs. Determine whether it is always possible, regardless of the initial state of the bulbs, to turn all the bulbs off if
a) $n=2011... | [
"Solution:\n\nThe answer to both questions is no. Call a lamp \"hot\" if it is in one of the squares marked $\\times$ in the following two patterns:\n\nFor $2 \\times 2011$ :\n\n| $\\times$ | | $\\times$ | | $\\times$ | $\\cdots$ | $\\times$ | | $\\times$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) No; b) No | |
0ebm | Let $a_1, a_2, a_3, \dots$ be a sequence of non-zero real numbers which satisfies $a_n^2 = -a_{n+1}a_{n-1}$ for all positive integers $n$, $n \ge 2$. Prove that the sequence $a_2, a_4, a_6, \dots$ is geometric. | [
"For each $n \\ge 2$ we can rearrange the given recursive formula to $\\frac{a_{n+1}}{a_n} = -\\frac{a_n}{a_{n-1}}$. Using this we calculate\n$$\n\\frac{a_{2n+2}}{a_{2n}} = \\frac{a_{2n+2}}{a_{2n+1}} \\cdot \\frac{a_{2n+1}}{a_{2n}} = \\left(-\\frac{a_{2n+1}}{a_{2n}}\\right) \\cdot \\left(-\\frac{a_{2n}}{a_{2n-1}}\\... | Slovenia | National Math Olympiad 2015 – Final Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0hnn | Problem:
How many different combinations of 4 marbles can be made from 5 indistinguishable red marbles, 4 indistinguishable blue marbles, and 2 indistinguishable black marbles? | [
"Solution:\n\n$5 + 4 + 3 = 12$."
] | United States | null | [
"Discrete Mathematics > Combinatorics"
] | null | final answer only | 12 | |
0l0g | Problem:
Compute the smallest integer $n > 72$ that has the same set of prime divisors as $72$. | [
"Solution:\n\nThe prime divisors of $72$ are $2$ and $3$, and we note that $72 = 2^{3} \\cdot 3^{2}$. Since we need at least one factor of $2$ and one factor of $3$, we just need to check the multiples of $6$ from $72$ onwards.\n\n$13 \\cdot 6$, $14 \\cdot 6$, and $15 \\cdot 6$ have prime factors of $13$, $7$, and ... | United States | HMMT November 2024 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 96 | |
0j3p | Problem:
How many functions $f:\{1,2, \ldots, 10\} \rightarrow \{1,2, \ldots, 10\}$ satisfy the property that $f(i)+f(j)=11$ for all values of $i$ and $j$ such that $i+j=11$. | [
"Solution:\nTo construct such a function $f$, we just need to choose a value for $f(x)$ from $\\{1,2, \\ldots, 10\\}$ for each $x \\in \\{1,2, \\ldots, 10\\}$. But the condition that $f(i)+f(j)=11$ whenever $i+j=11$ means that\n$$\n\\begin{aligned}\nf(10) & = 11 - f(1) \\\\\nf(9) & = 11 - f(2) \\\\\n& \\vdots \\\\\... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 100000 | |
02p6 | Let $(x_n)_{n \ge 0}$ be a sequence of integer numbers that fulfills a linear recursion of order $k$ for a fixed positive integer $k$, i.e., there exists real constant numbers $c_1, c_2, \dots, c_k$ such that $x_{n+k} = \sum_{r=1}^k c_r x_{n+k-r}$, $\forall n \ge 0$. Suppose $k$ is the minimum positive integer with thi... | [
"Let $(a_n^{(i)})$ be the sequence of integers obtained by shifting $a_n$ by $i$ positions, i.e., $a_n^{(i)} = a_{n+i}$. Then\n$$\n(a_n^{(0)}), (a_n^{(1)}), \\dots, (a_n^{(k)})\n$$\nis a basis for the $k$-dimensional $C$-vector space of sequences $(b_n)$ satisfying\n$$\nb_{n+k} = c_{k-1}b_{n+k-1} + \\dots + c_0b_n ... | Brazil | Brazilian Math Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Number Theory > Algebraic Number Theory > Algebraic numbers",
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Vectors"
] | English | proof only | null | |
021h | Problem:
Uma lista de seis inteiros positivos $p, q, r, s, t, u$ satisfaz $p<q<r<s<t<u$. Existem exatamente 15 pares de números que podem ser formados escolhendo dois números diferentes desta lista. As somas destes 15 pares de números são:
$$
25,30,38,41,49,52,54,63,68,76,79,90,95,103,117
$$
Determine o valor da soma $... | [] | Brazil | Desafios | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 54 | |
0hlx | Problem:
Let $ABCD$ be a convex quadrilateral. Let $I_{A}$ be the incenter of $BCD$ and define $I_{B}, I_{C}, I_{D}$ similarly. Show that $AC \perp I_{B}I_{D}$ if and only if $I_{A}I_{C} \perp BD$. | [
"Solution:\n\nWe show that both are true if and only if $AB + CD = BC + DA$.\n\nLet's first consider the triangle $ABD$. We know the incircle of $ABD$ is centered at $I_{C}$. Let $X, Y, T_{A}$ denote the points where the incircle is tangent to $AB$, $AD$, and $BD$, respectively. We have that $AX = AY$, $BX = BT_{A}... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0fkl | Problem:
Se consideran un cubo de $1~\mathrm{cm}$ de arista y dos vértices $A$ y $B$ diagonalmente opuestos de una cara del cubo. Se denomina camino de longitud $n$ a una sucesión de $n+1$ vértices de forma que dos consecutivos están a $1~\mathrm{cm}$ de distancia. Entonces: ¿Cuál de los siguientes números es mayor: e... | [
"Solution:\n\nLlamemos $a_{n}$ al número de caminos de longitud $n~\\mathrm{cm}$ que empiezan y acaban en $A$, y llamemos $b_{n}$ al número de caminos de longitud $n~\\mathrm{cm}$ que empiezan en $A$ y acaban en $B$.\n\nEl $(n-1)$-ésimo vértice de un camino que termina en $A$ sólo puede ser el propio $A$, o cualqui... | Spain | XLV Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | The number of length one thousand paths from A back to A is larger by exactly one than the number from A to B. | |
0g5d | 在 $\triangle ABC$ 的邊 $BC$ 的延長線取一點 $D$ 使得 $CD = AC$。$\triangle ACD$ 的外接圓與以 $BC$ 為直徑的圓交於 $C$, $P$ 兩點，直線 $BP$ 與 $AC$ 交於 $E$，直線 $CP$ 與 $AB$ 交於 $F$。求證：$D$, $E$, $F$ 三點共線。 | [
"如圖所示。因點 $P$ 在 $\\triangle ACD$ 的外接圓上，且 $AC = CD$，所以\n$$\n\\angle APF = \\angle ADC = \\angle CAD = \\angle CPD,\n$$\n即 $PC$ 是 $\\angle APD$ 的外角平分線。於是，設直線 $AP$ 交 $BC$ 於 $Q$，則 $PC$ 平分 $\\angle QPD$。又點 $P$ 在以 $BC$ 為直徑的圓上，所以 $BP \\perp PC$，從而 $BP$ 為 $\\angle QPD$ 的外角平分線，如此可得\n$$\n\\frac{BD}{DC} = -\\frac{BQ}{QC}.\n$$\... | Taiwan | 二〇一一數學奧林匹亞競賽第一階段選訓營,獨立研究(一) | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0i2w | In a triangle $ABC$, let segment $AP$ bisect $\angle BAC$, with $P$ on side $BC$, and let segment $BQ$ bisect $\angle ABC$, with $Q$ on side $CA$. It is known that $\angle BAC = 60^\circ$ and that $AB + BP = AQ + QB$. What are the possible angles of triangle $ABC$? | [
"**First Solution.** (by Reid Barton and Gabriel Carroll) Extend segment $AB$ through $B$ to $R$ so that $BR = BP$, and construct $S$ on ray $AQ$ so that $AS = AR$.\n\n\n\n**Lemma 1.** Points $B$, $P$, $S$ are collinear, and consequently, $S$ coincides with $C$.\n\n*Proof.* Because $BR = BP... | United States | USA IMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analyti... | English | proof and answer | ∠ABC = 80°, ∠BCA = 40°, ∠CAB = 60° | |
0gyp | a) Prove that for every natural number $n$ there exist natural $m$, $k$, that satisfy the equation
$$
k + m^k + n^{m^k} = 2009^n.
$$
b) Prove that there exist infinitely many natural $n$, for which such pair $m, k$ is unique. | [
"a.\nIf we take $m=1$, then we will get that $k+1+n^1=2009^n$ or that $k=2009^n-1-n$. It is clear that this will be a solution: $(n, m, k) = (n, 1, 2009^n - n - 1)$.\n\nb.\nIf $m=1$, then our equation comes to $k+1+n=2009^n$, which holds for the only value of $k$. Thus, new solutions can exist only when $m \\ge 2$.... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof only | null | |
0kxv | Problem:
There are $n \geq 2$ coins, each with a different positive integer value. Call an integer $m$ sticky if some subset of these $n$ coins have total value $m$. We call the entire set of coins a stick if all the sticky numbers form a consecutive range of integers. Compute the minimum total value of a stick across... | [
"Solution:\n\nSort a stick by increasing value. Note that all sticks must contain $1$ by necessity, or the largest and second largest sticky values would not be consecutive. So, let's say a stick's highest coin value is $a$, and all the other terms have a value of $S$. If $a \\geq S+2$, we cannot build $S+1$, but w... | United States | HMMT November 2023 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 199 | |
0br1 | In triangle $ABC$, the measures of the angles $B$ and $A$ are directly proportional to $2$ and $3$. Take $M$ and $N$ on the sides $[AB]$ and $[BC]$ respectively, so that $BN = NM = MC = MA$. Prove that:
a) one of the obtained triangles is equilateral;
b) $CM \perp AB$. | [] | Romania | 67th NMO Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof only | null | |
0dbn | Positive integer $n$ is called 'nice' if the sum of squares of its divisors is equal to $(n+3)^2$ (for example $287=7 \cdot 41$ is 'nice', since $1^2+7^2+41^2+287^2= 84100=(287+3)^2$). Prove that if for two primes $p$ and $q$ (not necessarily distinct) the number $n=pq$ is 'nice', then both $n+2$ and $2(n+1)$ are perfe... | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
00t8 | Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that
$$
f(x + f(x) + f(y)) = 2f(x) + y
$$
holds for all $x, y \in \mathbb{R}^+$. | [
"We will show that $f(x) = x$ for every $x \\in \\mathbb{R}^+$. It is easy to check that this function satisfies the equation.\nWe write $P(x, y)$ for the assertion that $f(x + f(x) + f(y)) = 2f(x) + y$.\nWe first show that $f$ is injective. So assume $f(a) = f(b)$. Now $P(1, a)$ and $P(1, b)$ show that\n$$\n2f(1) ... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = x for all x > 0 | |
0l91 | Let be given an integer $n \ge 2$ and a rectangular board of size $n \times 2n$ ($n$ rows, $2n$ columns). One labels randomly $n^2$ little squares of the board. Prove that for every integer $k$ such that $1 < k \le \lfloor n/2 \rfloor + 1$, there always exist $k$ rows of the board so that the rectangular board of size ... | [] | Vietnam | CONTEST FOR THE SELECTION OF VIETNAMESE INTERNATIONAL MATHEMATICAL OLYMPIAD TEAM | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0htz | Problem:
Pentagon $A B C D E$ is cyclic, i.e., inscribed in a circle. Diagonals $A C$ and $B D$ meet at $P$, and diagonals $A D$ and $C E$ meet at $Q$. Triangles $A B P$, $A E Q$, $C D P$, $C D Q$, and $A P Q$ have equal areas. Prove that the pentagon is regular. | [] | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06z2 | Problem:
$P$ is a fixed point in the plane. Show that amongst triangles $ABC$ such that $PA = 3$, $PB = 5$, $PC = 7$, those with the largest perimeter have $P$ as incenter. | [
"Solution:\n\n\nGiven points $P$, $B$, $C$ and a fixed circle center $P$, we show that the point $A$ on the circle which maximises $AB + AC$ is such that $PA$ bisects angle $BAC$. Consider a point $A'$ close to $A$. Then the change in $AB + AC$ as we move $A$ to $A'$ is $AA'(\\sin PAC - \\s... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Pla... | null | proof only | null | |
09ck | Тойрогт $A$, $B$, $C$ цэгүүдэд огтлөцөх гурван хөвч татав. $[AB]$, $[AC]$ царагуудыг болон ба өгсөн тойргийг $A_1$ цэгт гадаад байдлаар шүргэсэн тойрог авч үзье. Аналогоор $B_1$, $C_1$ цэгүүдийг тодорхойлъё. $AA_1$, $BB_1$, $CC_1$ шулуунууд нэг цэгт огтлөцөн гэж батал. | [
"Өгсөн тойргийг $\\omega$, $[AB]$, $[AC]$ царагууд болон $\\omega$-г $A_1$ цэгт шүргэх тойргийг $\\omega_A$; $ABC$ гурвалжинд багтсан тойргийг $\\gamma$ гэе. $X$ төвтэй $t$ коэффициенттэй гомотетийг $H_X^m$ гэе. $\\gamma$-г $\\omega$-д буулгах сөрөг коэффициенттэй гомотетийн төвийг $A_0$ гэе.\n$$\n\\begin{aligned}\... | Mongolia | ОУМО-53 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and ... | Mongolian | proof only | null | |
0278 | Problem:
Tabuleiro de xadrez - De quantas maneiras podemos colocar dois bispos de mesma cor num tabuleiro de xadrez em filas, colunas e casas de cores distintas? | [
"Solution:\n\nUm tabuleiro de xadrez é um quadrado reticulado de 64 quadradinhos, denominados casas, sendo 32 casas pretas e 32 brancas, posicionados alternadamente. Uma das peças do xadrez recebe o nome de bispo, havendo um par deles para cada jogador. Um dos dois bispos de um jogador só se movimenta pelas casas p... | Brazil | Nível 2 | [
"Discrete Mathematics > Other"
] | null | proof and answer | 768 | |
07mo | Determine the least positive integer $k$ such that $2010$ can be written as a sum of $k$ palindromic primes.
[A palindromic number is a positive integer—like $16361$—whose decimal representation remains the same when its digits are reversed.] | [
"Since $2010$ is neither a prime nor a palindromic number, $k > 1$. Suppose $k = 2$. Then there are two palindromic primes $a, b$ such that $2010 = a + b$. Since $1005$ isn't prime, $a \\neq b$. Say $a < b$. Then $2 \\le a < 1005 < b \\le 2009$. But palindromic numbers $b > 2$ ending in $2$ are even, and so can't b... | Ireland | Irish Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other"
] | null | proof and answer | 4 | |
02cb | Problem:
A estrada que liga dois vilarejos em uma montanha é formada somente por trechos de subida ou descida. Um ônibus sempre viaja a $15~\mathrm{km}/\mathrm{h}$ em trechos de subida e a $30~\mathrm{km}/\mathrm{h}$ em trechos de descida. Encontre a distância entre os vilarejos se o ônibus leva exatamente 4 horas par... | [
"Solution:\n\nObserve que os trechos de subida no percurso de ida são exatamente os trechos de descida para a volta e vice-versa. Assim, em uma viagem de ida e volta a distância percorrida nas subidas é igual à distância percorrida nas descidas.\n\nChamemos de $d$ a distância entre os dois vilarejos. Como a distânc... | Brazil | Desafios | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 40 km | |
02sy | Problem:
Abaixo, veem-se círculos grandes e pequenos. Os círculos grandes têm raio $2$, e os círculos pequenos têm raio $1$. Qual a área da região pintada de cinza?
Observação: A área de um círculo de raio $r$ é igual a $\pi r^{2}$. | [
"Solution:\n\nVamos numerar as regiões:\n\n\n\nAs regiões $2$ e $4$ têm mesma área, pois uma pode ser obtida refletindo a outra. Além disso, as regiões $1$ e $6$ também possuem mesma área. Logo, a área da região em cinza acima é a mesma da região pintada em cinza abaixo:\n\n![](attached_ima... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Geometry > Plane Geometry > Circles"
] | null | final answer only | 4π | |
0ix7 | Determine all functions $f$ from the set of positive integers to the set of positive integers such that, for all positive integers $a$ and $b$, there exists a non-degenerate triangle with sides of lengths $a$, $f(b)$ and $f(b+f(a)-1)$. | [
"If $f(b+v_2) = f(b)-1$, then setting $a=2$ and $b = b+v_2$ in the given condition leads to the triangular triple $\\{2, f(b+v_2) = f(b)-1, f(b+2v_2)\\}$. By the triangle inequality and (c), the only possible value of $f(b+2v_2)$ is $f(b)-2$. Likewise, we can deduce that $f(b+3v_2) = f(b)-3$, and so on. But this im... | United States | IMO 2009 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, ext... | null | proof and answer | f(n) = n for all positive integers n | |
052t | a) Does there exist an integer $c$ and a polynomial $P(x)$ with integer coefficients for which $P(c) \neq c$, but $P(P(c)) = c$?
b) Does there exist an integer $c$ and a polynomial $P(x)$ with integer coefficients for which $P(c) \neq c$ and $P(P(c)) \neq c$, but $P(P(P(c))) = c$? | [
"a) Take $P(x) = -x$ and $c = 1$, then $P(c) = -1 \\neq c$ and $P(P(c)) = P(-1) = 1 = c$.\n\nb) Suppose that there exist a polynomial with integer coefficients $P(x)$ and an integer $c$, for which $P(c) \\neq c$, $P(P(c)) \\neq c$ and $P(P(P(c))) = c$. Notice that $P(c) \\neq P(P(c))$, because otherwise $P(P(c)) = ... | Estonia | Open Contests | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) Yes: for example, P(x) = -x with c = 1. b) No: such a polynomial and integer do not exist. | |
082z | Problem:
In Italia le targhe automobilistiche sono composte da 2 lettere, seguite da 3 cifre e da altre 2 lettere. Nel paese di Ailati le cose vanno alla rovescia e le targhe sono composte da 2 cifre, seguite da 3 lettere e da altre 2 cifre. Supponendo che in entrambi i paesi si usino 10 cifre e 22 lettere (I, O, U, Q... | [
"Solution:\n\nLa risposta è (B). Ogni lettera può essere scelta in 22 modi, ogni cifra in 10 modi, e ogni scelta di una lettera o di una cifra può essere combinata con qualsiasi altra scelta di cifre e di lettere in altre posizioni. Pertanto le targhe italiane sono $10^{3} \\cdot 22^{4}$ e quelle di Ailati $10^{4} ... | Italy | Progetto Olimpiadi di Matematica 2003 GARA di SECONDO LIVELLO BIENNIO | [
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | B | |
0etp | Let $A$ be a set of real numbers satisfying the following:
(a) $\sqrt{n^2+1} \in A$ for all positive integers $n$;
(b) if $x \in A$ and $y \in A$, then $x - y \in A$.
Prove that every integer can be written as a product of two different elements in $A$. | [
"We first observe some natural consequences of property (b):\n(i) $0 \\in A$, since if $x \\in A$ is arbitrary, then $0 = x - x \\in A$.\n(ii) If $x \\in A$, then $-x = 0 - x \\in A$.\n(iii) If $x, y \\in A$, then $x + y = x - (-y) \\in A$.\n(iv) If $x \\in A$, then $kx \\in A$ for any integer $k$. This follows eas... | South Africa | The South African Mathematical Olympiad Third Round | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Number Theory > Other"
] | null | proof only | null | |
0kqu | Problem:
Find the number of ordered pairs $(A, B)$ such that the following conditions hold:
- $A$ and $B$ are disjoint subsets of $\{1,2, \ldots, 50\}$.
- $|A|=|B|=25$
- The median of $B$ is 1 more than the median of $A$. | [
"Solution:\nThe median of both sets, which we will call $a$ and $b$ respectively, are more than exactly 12 of the members in their own set. Since $a$ and $b$ are consecutive, they must also be higher than the lower half of the other set and lower than the higher half of the other set, meaning that they are both hig... | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 7312459672336 | |
02od | Problem:
A calculadora do Dodó tem uma tecla especial com o símbolo $\mathrm{R}$. Se o visor mostra um número $x$ diferente de $2$, ao apertar $\mathrm{R}$ aparece o valor de $\frac{2x-3}{x-2}$.
a. Se o Dodó colocar $4$ no visor e apertar $\mathrm{R}$, qual número vai aparecer?
b. Dodó colocou um número no visor e, ... | [] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | a) 5/2; b) 1 and 3; c) Yes. For any starting number other than two, pressing the key twice returns the original number (since f(f(x)) = x and f(x) is never two). | |
0e5l | Problem:
Tilen je metal igralno kocko. V tabeli je predstavil število metov za posmezne pike.
| število pik | 1 | 2 | 3 | 4 | 5 | 6 |
| :--- | :---: | :--- | :---: | :---: | :---: | :---: |
| število metov | $3(x-2)$ | $2x$ | $2(x-1)$ | $x+2$ | $2x+1$ | $x$ |
a) Kolikokrat je vrgel posamezno število pik, če število ... | [
"Solution:\n\na)\nIzračunamo skupno število vseh metov:\n$$\n3(x-2) + 2x + 2(x-1) + x+2 + 2x+1 + x = 11x - 5.\n$$\nZapišemo enačbo:\n$$\n10\\% (11x - 5) = x.\n$$\nRešimo linearno enačbo:\n$$\n0.1 (11x - 5) = x \\\\\n11x - 5 = 10x \\\\\nx = 5.\n$$\nZ izračunanim $x$ lahko nato določimo število metov za posamezno šte... | Slovenia | Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | a) Counts: 1 appears 9 times, 2 appears 10 times, 3 appears 8 times, 4 appears 7 times, 5 appears 11 times, 6 appears 5 times. b) Average number of pips: 3.32. c) Pie chart sector angles: 1 → 64.8°, 2 → 72°, 3 → 57.6°, 4 → 50.4°, 5 → 79.2°, 6 → 36°. | |
0bnz | Let $n$ be a positive integer and let $a_1, \dots, a_n$ be $n$ positive integers. Show that
$$
\sum_{k=1}^{n} \frac{\sqrt{a_k}}{1 + a_1 + \dots + a_k} < \sum_{k=1}^{n^2} \frac{1}{k}.
$$ | [
"Set $b_0 = 1$ and $b_k = 1 + a_1 + \\dots + a_k$, $k = 1, \\dots, n$, to obtain a strictly increasing string of positive integers $1 = b_0 < b_1 < \\dots < b_n$, and write the sum in the left-hand member in the form $\\sum_{k=1}^{n} (b_k - b_{k-1})^{1/2}/b_k$.\n\nNext, let $m = \\min\\{k: b_k > n^2\\} \\ge 1$; if ... | Romania | 2015 Ninth STARS OF MATHEMATICS Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
00g2 | Let $x$, $y$, $z$ be positive numbers such that
$$
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1.
$$
Show that
$$
\sqrt{x + y z} + \sqrt{y + z x} + \sqrt{z + x y} \geq \sqrt{x y z} + \sqrt{x} + \sqrt{y} + \sqrt{z}.
$$ | [
"$$\n\\begin{aligned}\n\\sum_{\\text{cyclic}} \\sqrt{x + y z} &= \\sqrt{x y z} \\sum_{\\text{cyclic}} \\sqrt{\\frac{1}{x} + \\frac{1}{y z}} \\\\\n&= \\sqrt{x y z} \\sum_{\\text{cyclic}} \\sqrt{\\frac{1}{x}\\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) + \\frac{1}{y z}} \\quad [1\\ \\text{mark.}] \\\\\n&=... | Asia Pacific Mathematics Olympiad (APMO) | XIV APMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0kc1 | Problem:
You have six blocks in a row, labeled $1$ through $6$, each with weight $1$. Call two blocks $x \leq y$ connected when, for all $x \leq z \leq y$, block $z$ has not been removed. While there is still at least one block remaining, you choose a remaining block uniformly at random and remove it. The cost of this ... | [
"Solution:\nNote that the total cost is the total number of ordered pairs $(x, y)$ with $1 \\leq x, y \\leq 6$ such that $x$ and $y$ are connected right before $x$ gets removed.\nThe probability that blocks $x$ and $y$ are connected just before block $x$ is removed is simply $\\frac{1}{|x-y|+1}$, since all of the $... | United States | HMMT February 2020 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 163/10 | |
040m | Let $n$ be a square-free positive even number, $k$ be an integer, $p$ be a prime number, satisfying $p < 2\sqrt{n}$, $p \nmid n$, $p \mid n + k^2$. Prove that $n$ can be written as $n = ab + bc + ca$, where $a, b, c$ are distinctive positive integers. | [
"Since $n$ is even, we have $p \\neq 2$. As $p \\nmid n$, we have $p \\nmid k$. We may assume without loss of generality $0 < k < p$. Set $a = k$, $b = p - k$, then $c = \\frac{n - k(p - k)}{p} = \\frac{n + k^2}{p} - k$.\n\nBy assumption, $c$ is an integer, and $a, b$ are distinct positive integers. It remains to b... | China | China Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0ad3 | Let $p$ be a given line. $A$ and $B$ are two points such that $A \notin p$ and $B \notin p$. Construct a point $C$ on the line $p$ such that $\overline{AC} = \overline{BC}$. | [
"The points that lie on the bisector of the segment $AB$ are on equal distance of its end points $A$ and $B$. Hence the desired point $C$ is the intersection of the bisector of $AB$ and the line $p$. The problem has a unique solution if $AB$ and $p$ are not perpendicular, if $AB$ and $p$ are perpendicular and $p$ i... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0gi5 | 試求所有同時滿足以下兩條件的正整數 $n$:
(a) $n$ 的正因數個數不是 8 的倍數；
(b) 對於所有整數 $x$, 有
$$
x^n \equiv x \mod n.
$$ | [
"答案為 $n=1$ 或任意質數。我們拆解為以下四步驟。\n\n1. $n$ 沒有平方因子\n\n證明：若 $p^2|n$, 取 $x=p$, 則 $p^2 \\nmid x$ 但 $p^2|p^{p^2} = x$, 矛盾。\n\n2. $n=1, p$ 或 $pq$, 其中 $p$ 和 $q$ 為質數\n\n證明：由於 $n$ 沒有平方因子，$n$ 必型如 $p_1p_2\\cdots p_m$，從而其質因數個數為 $2^m$。但因為其質因數個數不被 8 整除，故 $m<3$，得證。\n\n3. $n$ 不可能型如 $pq$\n\n證明：不失一般性假設 $p < q$，此時取 $x$ 為 $q$ 的原根 (既讓 $a^d... | Taiwan | 2023 年台灣數學奧林匹亞考試 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Numbe... | Chinese (Traditional) | proof and answer | n = 1 or n is any prime | |
0glp | A point $(x, y)$ in the plane is a *lattice* point if $x$ and $y$ are both integers. Let $n$ be a positive integer. Prove that there exists a disk in the plane containing exactly $n$ lattice points in its interior. | [] | Thailand | The 14th Thailand Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
0cir | Consider a triangle $ABC$ with $AC > AB$ and $\angle ACB = 60^\circ$. Let $D$ be a point on side $BC$ such that $AD$ is the angle bisector of $\angle BAC$ and $AB = AD$. Point $E$ lies on line $AD$ such that $D$ is between $A$ and $E$, and $AE = AC$. Point $F$ lies on line $AB$ such that $B$ is between $A$ and $F$. Poi... | [] | Romania | 75th NMO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0au9 | Problem:
Find the largest possible value of
$$
\left(\sin \theta_{1}\right)\left(\cos \theta_{2}\right)+\left(\sin \theta_{2}\right)\left(\cos \theta_{3}\right)+\cdots+\left(\sin \theta_{2013}\right)\left(\cos \theta_{2014}\right)+\left(\sin \theta_{2014}\right)\left(\cos \theta_{1}\right)
$$ | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | 1007 | |
0ccm | A kingdom has $10$ towns. Some pairs of different towns have bridges built between them, so that between any two different towns there is at most one bridge. Two towns between which there is a bridge will be called *ends* of that bridge.
a) Prove that there are at least two towns which are *ends* for the same number o... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | No | |
031y | Problem:
A set $C$ of positive integers is called good if for any integer $k$ there exist $a, b \in C, a \neq b$, such that the numbers $a+k$ and $b+k$ are not coprime. Prove that if the sum of the elements of a good set $C$ equals $2003$, then there exists $c \in C$ for which the set $C \setminus \{c\}$ is good. | [
"Solution:\nLet $p_1, p_2, \\ldots, p_n$ be all prime divisors of all possible differences of two distinct numbers of $C$.\n\nSuppose that for any $p_i$ there exists an integer $\\alpha_i$ such that $c \\equiv \\alpha_i \\pmod{p_i}$ for at most one $c \\in C$. It follows by the Chinese remainder theorem that there ... | Bulgaria | 52. Bulgarian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0kgy | Consider two concentric circles of radius $17$ and radius $19$. The larger circle has a chord, half of which lies inside the smaller circle. What is the length of the chord in the larger circle?
(A) $12\sqrt{2}$ (B) $10\sqrt{3}$ (C) $\sqrt{17} \cdot 19$ (D) $18$ (E) $8\sqrt{6}$ | [] | United States | AMC 12 A | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | E | |
0517 | Find the minimum number of colours required to paint all points with integer coordinates in the plane in such a way that no two points which are exactly five units apart have the same color. | [
"Obviously at least 2 colors are necessary. Color all points $(x, y)$ with even sum of coordinates with one color and all other points with another color. All points that are at distance 5 from $(x, y)$ are $(x \\pm 4, y \\pm 3)$, $(x \\pm 3, y \\pm 4)$, $(x \\pm 5, y)$, $(x, y \\pm 5)$. In each case, the sum of co... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2 | |
04th | Find the least real $m$, for which there exists real $a$, $b$ such that
$$
|x^2 + a x + b| \leq m(x^2 + 1)
$$
holds for any $x \in (-1, 1]$. | [
"Let us assume that $a$, $b$, $m$ satisfy the condition:\n$$\n\\forall x \\in (-1, 1): |f(x)| \\leq m(x^2 + 1), \\quad \\text{where} \\quad f(x) = x^2 + a x + b.\n$$\nFirstly we prove that at least one of the $f(1) - f(0) \\geq 0$ and $f(-1) - f(0) \\geq 0$ holds:\nfor arbitrary $f(x) = x^2 + a x + b$ there is\n$$\... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 1/3 | |
09bg | $1 \le a < b$ байх бодит тоонууд өгөгдөв. $[a^n, b^n]$ хэрчимд 2010-аас багагүй тооны бүхэл тоо ордог байх $n$ натурал тоо олдохыг батал. | [
"$b^n - a^n > 2010$ байх $n$ натурал тоо олдохыг батлая. $b - a = c$ гэе. Тэгвэл\n$$\nb^n - a^n = c \\cdot (b^{n-1} + b^{n-2}a + \\dots + ba^{n-2} + a^{n-1}) > c \\cdot a^{n-1} \\cdot n \\geq c \\cdot n\n$$\nболно.\n$$\nc \\cdot n > 2010 \\Leftrightarrow n < \\frac{2010}{c} = \\frac{2010}{b-a}\n$$\nбайхаар $n$-ийг ... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | Mongolian | proof only | null | |
0kyu | Problem:
A peacock is a ten-digit positive integer that uses each digit exactly once. Compute the number of peacocks that are exactly twice another peacock. | [
"Solution:\nWe begin with the following observation:\n\nClaim 1. Let $x$ be a peacock. Then, $2x$ is a peacock if and only if:\n- the multiplication $x \\cdot 2$ uses five carries,\n- each of the pairs of digits $(0,5), (1,6), (2,7), (3,8), (4,9)$ receives exactly one carry,\n- the leading digit is not $5,6,7,8,9$.... | United States | HMMT February 2024 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 184320 | |
0cwv | Find all positive integers $n$ for which there exists an *even* positive integer $a$ such that $(a-1)(a^2-1)\dots(a^n-1)$ is a perfect square. | [
"**Ответ.** For $n=1$ and $n=2$.\n\nFor $n=1$, any even $a$ of the form $m^2+1$ works, for example, $a=2$.\n\nFor $n=2$, any even $a$ of the form $m^2-1$ works, for example, $a=8$.\n\nAssume that for $n=3$, such a number $a$ exists. Then the number $(a-1)(a^2-1)(a^3-1) = (a-1)^3(a+1)(a^2+a+1)$ must be a perfect squ... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Russian | proof and answer | n = 1 and n = 2 | |
0axi | Problem:
Find the area of the region $\{(x, y) \in \mathbb{R}^{2} : |x| - |y| \leq 2$ and $|y| \leq 2\}$. | [
"Solution:\nIn the first quadrant, the region is a trapezoid with vertices at $(4,2)$, $(2,0)$, $(0,0)$, $(0,2)$; this trapezoid has area $6$. By symmetry, the region has area $4 \\times 6 = 24$ square units."
] | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 24 | |
0h75 | The bisector of an angle $BAC$ of acute angled triangle $ABC$ ($AC \neq AB$) intersects its circumcircle second time at the point $W$. Let $O$ be the circumcenter of $\triangle ABC$. Line $AW$ intersects second time the circumcircles of triangles $OWB$ and $OWC$ at points $N$ and $M$ respectively. Prove that $BN + MC =... | [
"Without loss of generality we can consider that $AC < AB$. From the conditions (fig. 18) $2\\angle CAW = \\angle COW = \\angle CMW$, hence $\\angle ACM = \\angle CAM$. Thus $AM = CM$. Similarly $AN = NB$. Then it is enough to prove that $AM + AN = AW$, thus $AM = NW$. Obviously, $CO = OB$, $CW = WB$. Thus $\\angle... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0l2b | Problem:
Barry picks infinitely many points inside a unit circle, each independently and uniformly at random, $P_{1}, P_{2}, \ldots$ Compute the expected value of $N$, where $N$ is the smallest integer such that $P_{N+1}$ is inside the convex hull formed by the points $P_{1}, P_{2}, \ldots, P_{N}$.
Submit a positive ... | [
"Solution:\n\nClearly, $N \\geq 3$, and let's scale the circle to have area $1$. We can see that the probability to not reach $N=4$ is equal to the probability that the fourth point is inside the convex hull of the past three points. That is, the probability is just one minus the expected area of those $N$ points. ... | United States | HMMT February 2024 Guts Round | [
"Discrete Mathematics > Combinatorics > Expected values",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | null | final answer only | null | |
03zw | Find the least positive integer $m$, such that for every prime number $p>3$, we have $105 < 9^p - 29^p + m$. | [] | China | China Western Invitational Mathematical Competition | [
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | No such positive integer m exists | |
06nf | There is a $3 \times 3$ table and $9$ cards printed with the numbers $-4$, $-3$, $-2$, $-1$, $0$, $1$, $2$, $3$, $4$. How many ways are there for these cards to be distributed to the cells of the table so that there is one card in each cell, and that the sum of the numbers of the three cards on each row, each column an... | [
"Answer: $32$\n\nWe first make a few observations.\n\n* The sum of the numbers on all cards is $0$. Hence each row sum being nonnegative is equivalent to each row sum being $0$.\n* For the same reason as above, each column sum must be $0$.\n* The card $-4$ cannot be placed in a corner cell. Indeed, if $-4$ is place... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Discrete Mathematics > Combinatorics",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 32 | |
02mj | Problem:
Sequência de algarismos - Todos os números naturais de $1$ em diante foram escritos consecutivamente, formando uma sequência de algarismos, como segue.
$$
1234567891011121314151617181920212223 \ldots
$$
Qual é o algarismo que aparece na posição de número $206788$ ? | [] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | final answer only | 7 | |
0100 | Problem:
Let $x_{1}=1$ and $x_{n+1}=x_{n}+\left\lfloor\frac{x_{n}}{n}\right\rfloor+2$ for $n=1,2,3, \ldots$, where $\lfloor x\rfloor$ denotes the largest integer not greater than $x$. Determine $x_{1997}$. | [
"Solution:\n$x_{1997}=23913$.\n\nNote that if $x_{n}=a n+b$ with $0 \\leqslant b<n$, then\n$$\nx_{n+1}=x_{n}+a+2=a(n+1)+b+2.\n$$\nHence if $x_{N}=A N$ for some positive integers $A$ and $N$, then for $i=0,1, \\ldots, N$ we have $x_{N+i}=A(N+i)+2 i$, and $x_{2 N}=(A+1) \\cdot 2 N$.\n\nSince for $N=1$ the condition $... | Baltic Way | Baltic Way 1997 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 23913 | |
0byb | Let $n \in \mathbb{N}$, $n \ge 2$, and $a_1, a_2, \dots, a_n, b_1, b_2, \dots, b_n$ be real positive numbers such that
$$
\frac{a_1}{b_1} \le \frac{a_2}{b_2} \le \dots \le \frac{a_n}{b_n}.
$$
Find the largest real $c$ so that $(a_1 - b_1c)x_1 + (a_2 - b_2c)x_2 + \dots + (a_n - b_nc)x_n \ge 0$, for every $x_1, x_2, \dot... | [
"Denoting $y_1 = x_1$, $y_{k+1} = x_{k+1} - x_k$, for $k \\in \\{1, 2, \\dots, n-1\\}$, we get $x_k = y_1 + y_2 + \\dots + y_k$, $\\text{cu } y_1 > 0, y_k \\ge 0, \\forall k = \\overline{2, n}$. With these notations, the inequality becomes $(a_1 + a_2 + \\dots + a_n - c(b_1 + b_2 + \\dots + b_n))y_1 + (a_2 + a_3 + ... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (a1 + a2 + ... + an) / (b1 + b2 + ... + bn) | |
048m | What is the minimal number of elements that should be removed from the set $\{2, 4, 6, 8, 10, 12, 14, 16\}$ so that the product of the remaining elements would be a perfect square? | [] | Croatia | CroatianCompetitions2011 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 3 | |
04yq | The lengths of the sides of a quadrilateral are $a$, $b$, $c$, $d$ and its area is $S$. Prove that $a^2 + b^2 + c^2 + d^2 \ge 4S$. For which quadrilaterals does the equality hold? | [
"Without loss of generality we can assume that $a$, $b$, $c$ and $d$ are the lengths of consecutive sides of the quadrilateral. A diagonal divides the quadrilateral into two triangles. From one partition we get the inequality $\\frac{ab}{2} + \\frac{cd}{2} \\ge S$, whence $ab + cd \\ge 2S$, and from the other parti... | Estonia | Estonija 2010 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > ... | null | proof and answer | squares | |
0hcc | Show that for positive numbers $x, y, z, a, b, c$ so that $x + y + z = a + b + c$ the following inequality holds:
$$
\frac{x}{a+b} + \frac{y}{b+c} + \frac{z}{c+a} + \frac{a}{x+z} + \frac{b}{x+y} + \frac{c}{y+z} > 2.
$$ | [
"The following completes the proof:\n$$\n\\frac{x}{a+b} + \\frac{y}{b+c} + \\frac{z}{c+a} + \\frac{a}{x+z} + \\frac{b}{x+y} + \\frac{c}{y+z} > \\\\\n\\frac{x}{a+b+c} + \\frac{y}{a+b+c} + \\frac{z}{a+b+c} + \\frac{a}{x+y+z} + \\frac{b}{x+y+z} + \\frac{c}{x+y+z} = \\\\\n= \\frac{x+y+z}{a+b+c} + \\frac{a+b+c}{x+y+z} =... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0fh4 | Problem:
Demostrar que dados siete números reales cualesquiera, se pueden elegir dos, digamos $a$ y $b$, de manera que
$$
\sqrt{3}|a-b|<|1+a b|
$$
Dar un ejemplo de seis números reales que no cumplan esta propiedad. | [
"Solution:\nPondremos $a=\\tan \\alpha$ y $b=\\tan \\beta$, y de este modo tendremos\n$$\n\\tan (\\alpha-\\beta)=\\frac{\\tan \\alpha-\\tan \\beta}{1+\\tan \\alpha \\tan \\beta}=\\frac{a-b}{1+a b}, \\quad |\\tan (\\alpha-\\beta)|<\\frac{1}{\\sqrt{3}}=\\tan \\frac{\\pi}{6}\n$$\nEsto nos dice que dados siete números ... | Spain | OME 25 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | An example is the six numbers: -2 - sqrt(3), -1, sqrt(3) - 2, 2 - sqrt(3), 1, 2 + sqrt(3). | |
02k9 | Problem:
Paulo quer comprar um sorvete com 4 bolas em uma sorveteria que dispõe de três sabores: açaí, baunilha e cajá. De quantos modos diferentes ele pode fazer a compra?
A) 6
B) 9
C) 12
D) 15
E) 18 | [
"Solution:\n\nVamos denotar cada sabor de sorvete pela sua letra inicial:\n$a \\rightarrow$ açaí, $b \\rightarrow$ baunilha, $c \\rightarrow$ cajá\n\nPara enumerar todas as possibilidades de compra do sorvete com quatro bolas, devemos considerar os seguintes casos:\n\n- 4 bolas do mesmo sabor\n- 3 bolas do mesmo sa... | Brazil | Brazilian Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | MCQ | D | |
09lk | Suppose there is a class with some students, and out of all the pairs formed from any four students, an even number are friend pairs. Show that it is possible for the students to enroll in clubs such that the following conditions are satisfied:
(i) Any two members of the same club are friends.
(ii) For any pair of stu... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
00fw | Compute the sum $S=\sum_{i=0}^{101} \frac{x_{i}^{3}}{1-3 x_{i}+3 x_{i}^{2}}$ for $x_{i}=\frac{i}{101}$.
Answer: $S=51$. | [
"Since $x_{101-i}=\\frac{101-i}{101}=1-\\frac{i}{101}=1-x_{i}$ and\n$$\n1-3 x_{i}+3 x_{i}^{2}=\\left(1-3 x_{i}+3 x_{i}^{2}-x_{i}^{3}\\right)+x_{i}^{3}=\\left(1-x_{i}\\right)^{3}+x_{i}^{3}=x_{101-i}^{3}+x_{i}^{3},\n$$\nwe have, by replacing $i$ by $101-i$ in the second sum,\n$$\n2 S=S+S=\\sum_{i=0}^{101} \\frac{x_{i... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 51 | |
0frx | Tenemos una fila de 203 casillas. Inicialmente la casilla de más a la izquierda contiene 203 fichas, y las demás están vacías. En cada movimiento podemos hacer una de estas dos operaciones:
* Tomar una ficha, y desplazarla a una casilla adyacente (a izquierda o derecha).
* Tomar exactamente 20 fichas de una misma casil... | [
"Consideramos la frontera entre la $n$-sima y la $n + 1$-sima casilla por la derecha. La cantidad neta de fichas que tiene que cruzar esa frontera es $n$. Vamos a contar cuántos movimientos han tenido que desplazar fichas a través de ella. Si $n = 20k + r$, donde $r$ es el residuo al dividir $n$ entre $20$, el núme... | Spain | LIX Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Spanish | proof only | null | |
0b9x | It is said that a positive integer $n > 1$ has the property ($p$) if in its prime factorization
$$
n = p_1^{\alpha_1} \cdots p_j^{\alpha_j}
$$
at least one of the prime factors $p_1, \dots, p_j$ has the exponent equal to 2.
a) Find the largest number $k$ for which there exist $k$ consecutive positive integers that do ... | [
"a) Among any 8 consecutive integers there exists one of the form $8j + 4$. This number has the property ($p$) because the factor 2 from its prime factorization has the exponent 2. Therefore there can be at most 7 consecutive positive integers that do not have the property ($p$). Since none of the numbers 29, 30, 3... | Romania | 62nd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 7 | |
054q | Find all positive integers $n$ such that a square can be cut into $n$ square pieces. | [
"A partition of a square into 1 square is trivial. If $n \\ge 2$ then a partition into $2n$ squares can be obtained by cutting $n$ squares of side length $\\frac{1}{n}$ from one side of the square and $n - 1$ more squares of the same size from a neighbouring side. One square of side length $\\frac{n-1}{n}$ of the s... | Estonia | National Olympiad Final Round | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n = 1, 4, or any integer n ≥ 6; no dissections exist for n = 2, 3, or 5 | |
0d41 | Prove that among any 16 perfect cubes we can always find two cubes whose difference is divisible by 91. | [
"Notice first that any perfect cube is congruent to either $0$, $1$ or $6$ modulo $7$ and it is congruent to either $0$, $1$, $5$, $8$, or $12$ modulo $13$. Because $7$ and $13$ are relatively prime numbers, by the Chinese Remainder theorem, a perfect cube is congruent to precisely one of $15 = 3 \\times 5$ differe... | Saudi Arabia | SAMC | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English, Arabic | proof only | null | |
0kud | Problem:
The number $770$ is written on a blackboard. Melody repeatedly performs moves, where a move consists of subtracting either $40$ or $41$ from the number on the board. She performs moves until the number is not positive, and then she stops. Let $N$ be the number of sequences of moves that Melody could perform. ... | [
"Solution:\n\nNotice that if we use the $41$ move nine times or less, we will have to make a total of $\\left\\lceil\\frac{770}{40}\\right\\rceil = 20$ moves, and if we use it ten times or more, we will have to make a total of $\\left\\lfloor\\frac{770}{40}\\right\\rfloor = 19$ moves. So, doing casework on the numb... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | final answer only | 318 | |
03cj | The quadrilateral $ABCD$ is inscribed. The point $H_1$ is the orthocenter of $\triangle ABC$ and the points $A_1$ and $B_1$ are symmetric to the points $A$ and $B$ with respect to the lines $BH_1$ and $AH_1$, respectively. The point $O_1$ is a center of the circumscribed circle of $\triangle A_1B_1H_1$. The point $H_2$... | [
"We will need the following lemma.\n\n**Lemma.** The point $O_1$ lies on the line $H_1O$, where $O$ is the center of the circumcircle of $\\triangle ABC$. The ratio $H_1O_1 : O_1O$ depends on $\\angle ACB$ only.\n\n**Proof.** Let $K$ and $L$ be the intersection points of $AH_1$ and $BH_1$ with the circumcircle of $... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Tr... | English | proof only | null | |
0019 | En una suma hay $1669$ fracciones y los denominadores reordenan los enteros desde $1$ hasta $1669$. Si el denominador es múltiplo de $5$, entonces el numerador es $4$ y la fracción figura con signo $-$; si el denominador no es múltiplo de $5$, entonces el numerador es $1$ y la fracción figura con signo $+$:
$$
1 + \fra... | [] | Argentina | XII Olimpiada Matemática Rioplatense | [
"Number Theory > Modular Arithmetic > Inverses mod n"
] | español | proof only | null | |
0bg1 | Show that, for every integer $r \ge 2$, there exists an $r$-chromatic simple graph (no loops, nor multiple edges) which has no cycle of less than 6 edges. | [
"The case $r = 2$ is clear: Any cycle of even length works. In the other cases, define a sequence of graphs $G_r$, $r \\ge 3$, as follows. The graph $G_3$ is a cycle of just 7 edges. (Any larger odd number would do.) When $G_r$ is defined, with $n_r$ vertices say, construct $G_{r+1}$ as follows. Consider\n\n$$\n\\b... | Romania | The Danube Mathematical Competition | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, ... | null | proof only | null | |
056c | Let $n \ge 2$ be a natural number. Laura's mathematics teacher likes group exercises. In Laura's class the teacher composed new groups in every lesson of statistics. When all of the statistics curriculum had been covered, it appeared that every two different students had belonged together into exactly one group and eve... | [
"By assumptions, one group contained $n + 1$ students while the whole class contains more than $n + 2$ students. As every two students belonged together into one group, more than one groups with more than one members must have occurred. If some group contained only one student $C$, this student must have belonged t... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n^2 + n + 1 | |
07ap | There are $n$ line segments in the plane such that no two of them intersect with each other and among the $2n$ endpoints of these line segments no three are collinear. Is it always possible to find a simple $2n$-gon such that its vertices are the endpoints of these line segments and also each of these line segments is ... | [] | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | Yes | |
0jdi | Problem:
Let $N$ be a positive integer whose decimal representation contains $11235$ as a contiguous substring, and let $k$ be a positive integer such that $10^{k} > N$. Find the minimum possible value of
$$
\frac{10^{k}-1}{\operatorname{gcd}\left(N, 10^{k}-1\right)}
$$ | [
"Solution:\nAnswer: $89$\n\nSet $m = \\frac{10^{k}-1}{\\operatorname{gcd}\\left(N, 10^{k}-1\\right)}$. Then, in lowest terms, $\\frac{N}{10^{k}-1} = \\frac{a}{m}$ for some integer $a$. On the other hand, the decimal expansion of $\\frac{N}{10^{k}-1}$ simply consists of the decimal expansion of $N$, possibly with so... | United States | HMMT | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Discrete Mathematics > Combinatorics > Generating functions",
"Number Theory > Oth... | null | proof and answer | 89 |
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