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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0j3k | Problem:
Two circles with radius one are drawn in the coordinate plane, one with center $(0,1)$ and the other with center $(2, y)$, for some real number $y$ between $0$ and $1$. A third circle is drawn so as to be tangent to both of the other two circles as well as the $x$-axis. What is the smallest possible radius fo... | [
"Solution:\n\nAnswer: $3-2 \\sqrt{2}$\n\nSuppose that the smaller circle has radius $r$. Call the three circles (in order from left to right) $O_{1}$, $O_{2}$, and $O_{3}$. The distance between the centers of $O_{1}$ and $O_{2}$ is $1+r$, and the distance in their $y$-coordinates is $1-r$. Therefore, by the Pythago... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3-2*sqrt(2) | |
04bw | Determine all positive integers smaller than $1000$ that are equal to the sum of squares of their digits. | [
"Let $n$ be a positive integer less than $1000$ such that $n$ is equal to the sum of the squares of its digits.\n\nLet $n$ have digits $a$, $b$, $c$ (possibly with leading zeros), so $n = 100a + 10b + c$, where $0 \\leq a \\leq 9$, $0 \\leq b \\leq 9$, $0 \\leq c \\leq 9$, and $n < 1000$.\n\nWe want $n = a^2 + b^2 ... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 1, 4, 9 | |
04hs | Let $x$ and $y$ be distinct real numbers such that $2xy + 1 \neq 0$, and let
$$
A = \frac{6x^2y^2 + xy - 1}{2xy + 1} \quad \text{and} \quad B = \frac{x(x^2 - 1) - y(y^2 - 1)}{x - y}.
$$
Determine which number is larger, $A$ or $B$. | [] | Croatia | Croatia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | B > A | |
0dvi | Problem:
Dano je praštevilo $p$. Poišči vsa naravna števila $x$ in $y$, ki zadoščajo enačbi $p \cdot (x-5) = x \cdot y$. | [
"Solution:\n\nKer je $p$ praštevilo, ločimo 2 možnosti: $p$ deli $x$ ali pa $y$.\n\nOglejmo si najprej možnost, ko $p$ deli $x$. Tedaj lahko zapišemo $x = p x'$, kjer je $x'$ pozitivno število, in tako dobimo $p x' - 5 = x' y$, kar zapišemo kot $x'(p - y) = 5$.\n\nPotem je $x' = 1$ in $y = p - 5$ ali $x' = 5$ in $y... | Slovenia | 47. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All solutions in positive integers are: (x, y) = (p, p − 5) for primes p greater than five, and (x, y) = (5p, p − 1) for every prime p. | |
0hj1 | Problem:
Define a "word" to be a string of at most ten letters taken from the English alphabet. (The letters do not have to be distinct.) Prove that the number of "words" is divisible by $27$. | [
"Solution:\n\nThe total number of $9$- and $10$-letter words is divisible by $27$, because they can be partitioned into groups of $27$ with each group containing a $9$-letter word and the twenty-six $10$-letter words formed by adding a letter at the end of it.\n\nFor the same reason, the number of $7$- and $8$-lett... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
09db | $4 \times n$ тэгш өнцөгтийг $1 \times 2$ тэгш өнцөгтүүдээр зүйж хүчх нийт боломжийн тоог ол. | [
"Нийт боломжийн тоог $a_n$ гэе. $4 \\times n$ тэгш өнцөгтийн зүүлийн баганд дээд талд нь эсвэл доод талд $1 \\times 2$ тэгш өнцөгт босоогоор байрлах боломжийн тоог $b_n$ гээ. Мөн $4 \\times n$ тэгш өнцөгтийн сүүлийн баганд $1 \\times 2$ тэгш өнцөг яг голд нь байрлах нийт боломжийн тоог $c_n$ гээ. Сүүлийн баганд нь ... | Mongolia | ММО-48 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | Mongolian | proof and answer | Let a_n be the number of tilings of the 4 by n rectangle with 1 by 2 dominoes. Then a_n satisfies the recurrence a_n = a_{n-1} + 5 a_{n-2} + a_{n-3} - a_{n-4} with initial values a_0 = 1, a_1 = 1, a_2 = 5, a_3 = 11. An explicit closed form is given by
a_n = (1/√29)(-x_1^{n+1} - x_2^{n+1} + x_3^{n+1} + x_4^{n+1}),
whe... | |
0f8c | Problem:
Let $m$, $n$, $k$ be positive integers with $m \geq n$ and $1 + 2 + \ldots + n = mk$. Prove that the numbers $1, 2, \ldots, n$ can be divided into $k$ groups in such a way that the sum of the numbers in each group equals $m$. | [
"Solution:\n\nInduction on $n$, then $m$. For $n = 1$, $2$ there is nothing to prove. Assume the result is proved for $< n$ and consider the case $n$.\n\nIf $n$ is odd, we have $n = n - 1 + 1 = n - 2 + 2 = \\ldots = (n + 1)/2 + (n - 1)/2$, so the result is true for $m = n$, $k = (n + 1)/2$.\n\nIf $n$ is even, we ha... | Soviet Union | 22nd ASU | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
08g9 | Problem:
Si consideri il numero $N=1000\ldots 0001$ che consiste nella cifra uno seguita da 2023 zeri, a loro volta seguiti dalla cifra uno. Quanti sono i divisori propri di $N$ (ovvero, i divisori strettamente compresi fra 1 ed $N$ ) che si scrivono anch'essi come una cifra 1 seguita da un qualche numero positivo di ... | [
"Solution:\n\nLa risposta è (A). Osserviamo che $N=10^{2024}+1$ e che un suo divisore del tipo richiesto si scrive come $10^{n}+1$ per qualche $1<n<2024$. Assumiamo quindi che $10^{n}+1$ divida $N$ e scriviamo\n$$\nN=10^{2024}+1=\\left(10^{n}+1\\right)\\left(10^{2024-n}\\right)+r_{1},\n$$\nda cui $r_{1}=-\\left(10^... | Italy | Italian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | A | |
0kl1 | 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) $4^{3032}$ | [
"Solution:\n**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$$\n\\frac{n}{4} = \\frac{4^{3033}}{4} = 4^{3032}.\n$$\nAll the other choices are less than $4^{3032}$. Choices (A) and (D) have the same base but a lesser exponent. Choice (B) is in... | United States | AMC 12 B | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | E | |
0bo8 | Prove that any set of 20 consecutive positive integers contains an integer $k$ such that $n\sqrt{k} \cdot \{n\sqrt{k}\} > 5/2$ for every positive integer $n$. Here $\{x\}$ denotes the fractional part of the real number $x$, that is, the difference of $x$ and the largest integer not exceeding $x$. | [
"Fix a set of 20 consecutive positive integers, and choose a member $k \\equiv 15 \\pmod{20}$. We shall prove that $k$ satisfies the required condition.\nFix a positive integer $n$, and notice that $k$ is not a square, since $k \\equiv 3 \\pmod{4}$, to write $m < n\\sqrt{k} < m + 1$ for some positive integer $m$, s... | Romania | 2015 BALKAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0ke8 | Problem:
Suppose we have a convex polygon in which all interior angles are integers when measured in degrees, and the interior angles at every two consecutive vertices differ by exactly $1^{\circ}$. If the greatest and least interior angles in the polygon are $M^{\circ}$ and $m^{\circ}$, what is the maximum possible v... | [
"Solution:\n\nThe answer is $18$.\n\nTo justify this answer, we will find it helpful to discuss the exterior angles rather than the interior angles. Consecutive exterior angles must still be integers and must still differ by $1^{\\circ}$, and the value we seek is equal to the difference between the greatest and lea... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 18 | |
0950 | Problem:
Să se afle toate funcțiile continue $f: i \rightarrow i$, ce verifică relația $f(x)=5 f(5 x)-5 x, \forall x \in i$. | [
"Solution:\n\nConsiderăm funcția $g: i \\rightarrow i$, $g(x)=f(x)-\\frac{5}{24} x$, $\\forall x \\in i$. Avem $g(5 x)=f(5 x)-\\frac{25}{24} x$, $\\forall x \\in i$. Deoarece $f(5 x)=\\frac{1}{5} f(x)+x$, $\\forall x \\in i$, rezultă că $g(5 x)=\\frac{1}{5} f(x)-\\frac{1}{24} x=\\frac{1}{5} g(x)$, $\\forall x \\in ... | Moldova | A 61-a OLIMPIAD DE MATEMATICA A REPUBLICII MOLDOVA | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = (5/24) x | |
04dv | Let $a$ be a complex number such that
$a^5 + a + 1 = 0.$
What values can the expression $a^2 (a - 1)$ take on? | [
"We have\n$$\n\\begin{aligned}\na^5 + a + 1 &= (a^5 - a^4) + (a^4 - a^3) + (a^3 - a^2) + a^2 + a + 1 \\\\\n&= (a^2 + a + 1)(a^3 - a^2 + 1).\n\\end{aligned}\n$$\nIf $a^2 + a + 1 = 0$, then $a^2(a - 1) = -(a + 1)(a - 1) = 1 - a^2 = a + 2$.\nSince $a = \\frac{-1 \\pm i\\sqrt{3}}{2}$, it follows that $a^2(a - 1) = \\fr... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | -1, 3/2 + (√3/2)i, 3/2 - (√3/2)i | |
0e70 | Find all natural numbers $n$ and prime numbers $p$ such that $\sqrt{n} + \frac{p}{\sqrt{n}}$ is the square of a natural number. | [
"Denote $\\sqrt{n} + \\frac{p}{\\sqrt{n}} = k^2$ where $k$ is a natural number.\n\nSquaring both sides of the equation gives us\n$$\nn + 2p + \\frac{p^2}{n} = k^4.\n$$\nHence $n$ must divide $p^2$. Since $p$ is prime, we conclude $n = 1$, $n = p$ or $n = p^2$.\n\nIf $n = p$, we get the equation $p + 2p + p = k^4$ o... | Slovenia | National Math Olympiad 2012 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (n, p) = (1, 3) and (9, 3) | |
01xr | The point $M$ is the midpoint of the side $BC$ of the triangle $ABC$. A circle is passing through $B$, is tangent to the line $AM$ at $M$ and intersects the segment $AB$ secondary at the point $P$.
Prove that the circle, passing through $A$, $P$ and the midpoint of the segment $AM$, is tangent to the line $AC$. | [
"Let $N$ be the midpoint of $AM$ and $D$ be the reflection of $A$ in $M$. We will prove that the quadrilateral $BPND$ is cyclic by showing the equality\n\n$$\nAP \\cdot AB = AN \\cdot AD.\n$$\nSince $AM$ is tangent to the circumcircle of the triangle $BPM$,\n$$\nAP \\cdot AB = AM^2 = 2AN \\... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0l2o | Problem:
Sally the snail sits on the $3 \times 24$ lattice of points $(i, j)$ for all $1 \leq i \leq 3$ and $1 \leq j \leq 24$. She wants to visit every point in the lattice exactly once. In a move, Sally can move to a point in the lattice exactly one unit away. Given that Sally starts at $(2,1)$, compute the number o... | [
"Solution:\n\nOn her first turn, Sally cannot continue moving down the middle row. She must turn either to the bottom row or the top row. WLOG, she turns to the top row, and enters the cell $(3,1)$ and we will multiply by 2 later. Then, we can see that the path must finish in $(1,1)$. So, we will follow these two b... | United States | HMMT February 2024 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Graph Theory"
] | null | final answer only | 4096 | |
022v | Problem:
As circunferências $\mathcal{C}_1$ e $\mathcal{C}_2$ são tangentes à reta $\ell$ nos pontos $A$ e $B$ e tangentes entre si no ponto $C$. Prove que o triângulo $ABC$ é retângulo. | [
"Solution:\n\nComo as circunferências são tangentes, então o ponto de tangência $C$ e os centros $O_1$ e $O_2$ pertencem a uma mesma reta. Além disso, como as circunferências são tangentes a $\\ell$, então $O_1A$ e $O_2B$ são perpendiculares a $\\ell$ e, portanto, paralelas.\n\nSeja $\\alpha$ a medida do ângulo $O_... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00xo | Problem:
An equilateral triangle $ABC$ is divided into $100$ congruent equilateral triangles. What is the greatest number of vertices of small triangles that can be chosen so that no two of them lie on a line that is parallel to any of the sides of the triangle $ABC$? | [
"Solution:\n\n\nFigure 2\n\nAn example for $7$ vertices is shown in Figure 2. Now assume we have chosen $8$ vertices satisfying the conditions of the problem. Let the height of each small triangle be equal to $1$ and denote by $a_{i}, b_{i}, c_{i}$ the distance of the $i$th point from the t... | Baltic Way | Baltic Way 1993 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Triangles",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 7 | |
0len | A student divides $30$ marbles into $5$ boxes labelled $1, 2, 3, 4, 5$ (there may be a box without marble).
a. How many ways are there to divide marbles into boxes (two ways are different if there is a box with different number of marbles)?
b. After dividing, this student paints those marbles by a number of colors (e... | [
"a.\nIt is well known that there are $\\binom{n+k-1}{k-1}$ ways to divide $n$ marbles into $k$ boxes. In this case, the answer is $\\binom{34}{4}$.\n\nb.\nLet $m$ be the number of colors, $x_1, x_2, \\dots, x_m$ be the number of boxes containing marble with color $1, 2, \\dots, m$ respectively. We now count the num... | Vietnam | VMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | a) C(34, 4).
b) At least 10 colors are necessary.
c) One valid configuration using exactly 10 colors: Box 1 uses colors 1,2,3,4,5,6; Box 2 uses colors 1,2,3,7,9,10; Box 3 uses colors 1,4,5,8,9,10; Box 4 uses colors 2,4,6,7,8,10; Box 5 uses colors 3,5,6,7,8,9. | |
0gzb | For positive numbers $a$, $b$ with $a + b = ab$ prove inequality:
$$
\frac{a}{b^2 + 4} + \frac{b}{a^2 + 4} \ge \frac{1}{2}.
$$ | [
"$ab = a + b \\ge 2\\sqrt{ab} \\Rightarrow ab \\ge 4$, then\n$$\n\\begin{aligned}\n\\frac{a}{b^2+4} + \\frac{b}{a^2+4} &\\ge \\frac{a}{b^2+ab} + \\frac{b}{a^2+ab} = \\\\\n&= \\frac{a}{b(a+b)} + \\frac{b}{a(a+b)} = \\frac{a^2+b^2}{(a+b)^2} \\ge \\frac{1}{2}.\n\\end{aligned}\n$$"
] | Ukraine | The Problems of Ukrainian Authors | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0h25 | For nonnegative real numbers $a, b, c$, with the sum that does not exceed $2$, prove
$$
ab(a^2 + b^2) + bc(b^2 + c^2) + ca(c^2 + a^2) \leq 2.
$$ | [
"It is enough to prove the inequality for $a + b + c = 2$.\nLet us denote $x = ab + bc + ac$, then\n$$\na^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ac) = 4 - 2(ab + bc + ac) = 2(2 - x).\n$$\nFrom the obvious inequality $x(2 - x) \\leq 1$, we get\n$$\n\\begin{aligned}\n2 + 2abc &\\geq 2 \\geq 2x(2 - x) = (ab + bc ... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0d0p | Find all positive integers $n$ and primes $p \ge 5$ such that
$$
(2p)^n + 1
$$
is a perfect cube. | [
"If $(2p)^n + 1 = a^3$, then we have\n$$\n(2p)^n = a^3 - 1 = (a-1)(a^2+a+1) = (a-1)[(a-1)^2+3(a-1)+3].\n$$\nSince $a$ is odd, it follows that $a^2+a+1$ is also odd, hence $2^n \\nmid a-1$. This means that we have $a = 2^n p^k + 1$ for some integer $k \\ge 0$.\n\nWe obtain\n$$\n2^n p^n = 2^n p^k (2^{2n} p^{2k} + 3 \... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial oper... | English | proof and answer | n = 1, p = 13 | |
03kx | Problem:
Let $ABC$ be a triangle with $\angle BAC = 40^\circ$ and $\angle ABC = 60^\circ$. Let $D$ and $E$ be the points lying on the sides $AC$ and $AB$, respectively, such that $\angle CBD = 40^\circ$ and $\angle BCE = 70^\circ$. Let $F$ be the point of intersection of the lines $BD$ and $CE$. Show that the line $AF... | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
09xh | Let $ABC$ be an acute non-isosceles triangle with orthocentre $H$. Let $O$ be the circumcentre of triangle $ABC$, and let $K$ be the circumcentre of triangle $AHO$. Prove that the reflection of $K$ in $OH$ lies on $BC$. | [
"We consider the configuration as in the figure. Other configurations are treated analogously. Denote by $D$ the second intersection of $AH$ with the circumcircle of $\\triangle ABC$. Denote by $S$ the second intersection of the circumcircles of $ABC$ and $AHO$. (Because $\\triangle ABC$ is acute, both $O$ and $H$ ... | Netherlands | IMO Team Selection Test 1 | [
"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 | |
04rx | Find all positive integers $n$ which are not powers of $2$ and which satisfy the equation $n = 3D + 5d$, where $D$ (and $d$) denote the greatest (and the least) numbers among all odd divisors of $n$ which are larger than $1$.
(Tomáš Jurík) | [
"Let $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$ be the prime factorization of a satisfactory number $n$. Here $p_1 < p_2 < \\dots < p_k$ are all the prime divisors of $n$ and the exponents $\\alpha_i$ are positive integers. The given equation implies that $p_1 = 2$ (otherwise $D = n$ which contrad... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 60, 100, and 8p for any odd prime p | |
0a27 | Mila stands on an infinitely large board divided into squares and starts moving. An $n$-jump is a movement in which Mila moves one square left, right, up or down and then $n$ squares in a direction perpendicular to that. Below is an example where Mila starts in the middle box on the left and first does a $1$-jump, foll... | [
"Colour the squares on the board alternately white and black, like on a chess board. If $n$ is odd, then an $n$-jump always goes to a square of the same colour as the starting square, and if $n$ is even precisely to a square of the other colour. Suppose Mila starts on a white square. If Mila makes a total of $m$ ju... | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | All positive integers congruent to zero or one modulo four with at least four steps; that is, m in {4, 5, 8, 9, 12, 13, …}. | |
0d3l | Find the minimum of $\sum_{k=0}^{40}\left(x+\frac{k}{2}\right)^{2}$ where $x$ is a real number. | [
"We have\n$$\n\\begin{aligned}\n\\sum_{k=0}^{40}\\left(x+\\frac{k}{2}\\right)^{2} & =(x+10)^{2}+\\sum_{k=1}^{20}\\left(((x+10)+\\frac{k}{2})^{2}+((x+10)-\\frac{k}{2})^{2}\\right) \\\\\n& =(x+10)^{2}+2 \\sum_{k=1}^{20}\\left((x+10)^{2}+\\left(\\frac{k}{2}\\right)^{2}\\right) \\\\\n& =41(x+10)^{2}+\\frac{1}{2} \\sum_... | Saudi Arabia | SAMC | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English, Arabic | proof and answer | 1435 | |
0a5z | Problem:
For any positive integer $n$, let $f(n)$ be the number of subsets of $\{1,2,\ldots ,n\}$ whose sum is equal to $n$. Does there exist infinitely many positive integers $m$ such that $f(m) = f(m + 1)$? (Note that each element in a subset must be distinct.) | [
"Solution:\n\nLet $S(n)$ be the set of such subsets. Consider the map from $S(n)$ to $S(n + 1)$ that adds one to the largest element of each $A \\in S(n)$. This map is an injection (needs proof but easy) and not a surjection provided that $S(n + 1)$ contains a set whose largest and second largest elements differ by... | New Zealand | NZMO Round Two | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | No | |
0hzn | Problem:
A semicircle is inscribed in a semicircle of radius $2$ as shown. Find the radius of the smaller semicircle.
 | [
"Solution:\nDraw a line from the center of the smaller semicircle to the center of the larger one, and a line from the center of the larger semicircle to one of the other points of intersection of the two semicircles. We now have a right triangle whose legs are both the radius of the smaller semicircle and whose hy... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(2) | |
0a8c | Problem:
The persons $P_{1}, P_{2}, \ldots, P_{n-1}, P_{n}$ sit around a table, in this order, and each one of them has a number of coins. In the start, $P_{1}$ has one coin more than $P_{2}$, $P_{2}$ has one coin more than $P_{3}$, etc., up to $P_{n-1}$ who has one coin more than $P_{n}$. Now $P_{1}$ gives one coin t... | [
"Solution:\n\nAssume that $P_{n}$ has $m$ coins in the start. Then $P_{n-1}$ has $m+1$ coins, ... and $P_{1}$ has $m+n-1$ coins. In every move a player receives $k$ coins and gives $k+1$ coins away, so her net loss is one coin. After the first round, when $P_{n}$ has given $n$ coins to $P_{1}$, $P_{n}$ has $m-1$ co... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 14 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | Either 3 persons with 6 coins in total, or 9 persons with 63 coins in total. | |
079m | $S$ and $T$ are two trees without having a vertex of degree $2$. Each edge of them has a positive number named the *length* of the edge. The *distance* between two vertices is the sum of the *length* of the edges of the path between them. We call the vertices with degree $1$, *leaf*. $f$ is an injective and surjective ... | [
"It is obvious that the total number of the leaves of $S$ and $T$ are equal. We solve the problem using induction. The base case for $2$ leaves is trivial because there is no vertex of degree $2$. Now suppose that $S, T$ have $k$ leaves. Let\n\n$$\nS' = S - \\{u\\}, \\quad T' = T - \\{f(u)\\}\n$$\n\nin which $u$ is... | Iran | 27th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ea0 | Find the smallest positive integer of the form $3a^2 - ab^2 - 2b - 4$, where $a$ and $b$ are some positive integers. | [
"The answer is $2$. If $a = 4$ and $b = 3$, then $3a^2 - ab^2 - 2b - 4 = 2$.\n\nIt therefore suffices to show that the equation $3a^2 - ab^2 - 2b - 4 = 1$ has no solutions in positive integers. We can rewrite the equation as $3a^2 - ab^2 = 2b + 5$. The right-hand side is odd, so the left-hand side must be odd as we... | Slovenia | National Math Olympiad in Slovenia | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2 | |
09ix | Let $A_n$ denote the number of ways to color the vertices of an $n$-gon in four colors – red, green, blue and yellow – so that each three consecutive vertices are colored in three different colors. Prove that $A_7 = A_8$. | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
04z4 | Call *pure* any positive integer $n$ that does not occur in any integer sequence $c_0, c_1, c_2, \dots$, where $0 < c_0 < n$ and
$$
c_i = \begin{cases} \frac{1}{2}c_{i-1} & \text{if } c_{i-1} \text{ is even,} \\ 3c_{i-1} - 1 & \text{if } c_{i-1} \text{ is odd,} \end{cases}
$$
for every $i \ge 1$. (For instance, $10$ is... | [
"a) Note that $3c_{i-1} - 1$ is never divisible by $3$ and if $\\frac{1}{2}c_{i-1}$ is divisible by $3$, then also $c_{i-1}$ is divisible by $3$. Thus, if some term $c_k = n$ is divisible by $3$, then, up to it, only dividing by $2$ is used to build the terms (i.e., $c_i = \\frac{1}{2}c_{i-1}$ for every $i$ such th... | Estonia | Estonija 2010 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof only | null | |
0asv | Problem:
Triangle $ABC$ is right-angled at $C$, and point $D$ on $AC$ is the foot of the bisector of $\angle B$. If $AB = 6\ \mathrm{cm}$ and the area of $\triangle ABD$ is $4.5\ \mathrm{cm}^2$, what is the length, in $\mathrm{cm}$, of $CD$? | [
"1.5"
] | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 1.5 | |
09xy | An integer $n$ is a *combi number* if each pair of distinct digits from the set of all possible digits $0$ to $9$ appear at least once in the number as neighbouring digits. For example, in a combi number the digits $3$ and $5$ have to appear somewhere next to each other. It does not matter whether they appear in the or... | [
"$50$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Algorithms"
] | English | proof and answer | 50 | |
09yw | Let $n$ be a positive integer. For a real $x \ge 1$, assume that $\lfloor x^{n+1} \rfloor, \lfloor x^{n+2} \rfloor, \dots, \lfloor x^{4n} \rfloor$ are all squares of positive integers. Prove that $\lfloor x \rfloor$ is also the square of a positive integer.
Here $\lfloor z \rfloor$ is the greatest integer smaller than... | [
"We first prove the statement for $n = 1$. Write $x = a + r$, with $a \\ge 1$ an integer and $0 \\le r < 1$. Suppose $\\lfloor x^2 \\rfloor, \\lfloor x^3 \\rfloor$ and $\\lfloor x^4 \\rfloor$ are squares. Then we have $a \\le x < a + 1$, from which it follows that $a^2 \\le x^2 < (a+1)^2$. Hence, $x^2$ is squeezed ... | Netherlands | IMO Team Selection Test 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
04uv | Circles $\Omega_1$ and $\Omega_2$ with different radii intersect at two points, denote one of them by $P$. A variable line $\ell$ passing through $P$ intersects the arc of $\Omega_1$ which is outside of $\Omega_2$ at $X_1$, and the arc of $\Omega_2$ which is outside of $\Omega_1$ at $X_2$. Let $R$ be the point on segme... | [] | Czech Republic | Czech-Polish-Slovak Match | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
013o | Problem:
Find all sets $X$ consisting of at least two positive integers such that for every pair $m, n \in X$, where $n>m$, there exists $k \in X$ such that $n = m k^{2}$. | [
"Solution:\nThe sets $\\{m, m^{3}\\}$, where $m>1$.\n\nLet $X$ be a set satisfying the condition of the problem and let $n>m$ be the two smallest elements in the set $X$. There has to exist a $k \\in X$ so that $n = m k^{2}$, but as $m \\leq k \\leq n$, either $k = n$ or $k = m$. The first case gives $m = n = 1$, a... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All sets of the form {m, m^3} with m > 1. | |
0ao4 | Problem:
Find the range of the function $f(x) = \frac{6}{5 \sqrt{x^{2} - 10x + 29} - 2}$.
(a) $[-1/2, 3/4]$
(b) $(0, 3/4]$
(c) $(1/2, 4/3]$
(d) $[-1/2, 0) \cup (0, 4/3]$ | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | (b) | |
0dfo | Let polynomial
$$
P(x) = \underbrace{((\dots((x-2)^2 - 2)^2 - \dots)^2 - 2)^2}_{k \text{ times}}
$$
is given. Find coefficient at $x^2$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | final answer only | (16^k - 4^k)/12 | |
0c38 | $$\int_{0}^{x} \left( 1 + \frac{t}{1!} + \frac{t^2}{2!} + \dots + \frac{t^{2n+1}}{(2n+1)!} \right) \cdot \frac{1}{1+t^2} \, dt < \frac{\arctan x}{x} \int_{0}^{x} e^t \, dt,$$
for every $x > 0$. | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof only | null | |
04or | A square board whose squares are coloured in either black or white is called *beautiful* if a rotation by $90^\circ$ does not change its appearance.
How many different $5 \times 5$ beautiful boards are there? (Azra Tafro) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 128 | |
02nr | Problem:
Um triângulo tem vértices $A=(3,0)$, $B=(0,3)$ e $C$, onde $C$ está na reta de equação $x+y=7$. Qual é a área desse triângulo? | [
"Solution:\n\nObserve que a altura $h$ relativa ao lado $AB$ de todos os triângulos $\\triangle ABC$ que têm o vértice $C$ na reta $x+y=7$ é sempre a mesma, pois a reta $x+y=7$ é paralela à reta $x+y=3$ que passa por $A$ e $B$. Logo, todos esses triângulos têm a mesma área, a saber,\n$$\n\\frac{1}{2}(AB \\times h)\... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 6 | |
07my | $ABCD$ is a trapezium, in which a circle can be inscribed, with $AB$ parallel to $DC$ and $AD$ equal to but not parallel to $BC$. The inscribed circle touches $AB$, $BC$, $CD$ and $DA$ at $X$, $Y$, $Z$ and $W$ respectively. $XZ$ and $WY$ intersect at $P$. Prove that the diagonals of $ABCD$ pass through $P$. | [
"Let $E$ be the intersection point of the lines $AD$ and $BC$. As $|AD| = |BC|$, the triangle $ABE$ is isosceles with axis of symmetry the line through $X$ and $E$. In particular, $A$, $P$, $C$ are collinear iff $B$, $P$, $D$ are collinear.\n\n\n\nAccording to the converse of Menelaus' Theo... | Ireland | Ireland | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Pla... | English | proof only | null | |
02g1 | Let $ABCDE$ be a regular pentagon such that the star $ACEBD$ has area $1$. Let $P$ be the point of intersection of $AC$ and $BE$ and $Q$ be the point of intersection of $BD$ and $CE$. Find the area of $APQD$. | [
"Let $AD$ meet $EC$ and $EB$ at $R$ and $S$ respectively. $PQDR$ is a parallelogram, so the area of $PQD$ equals the area of $RQD$, which by its turn equals the area of $ERS$. Therefore the area of $APQD$ equals the area of $APDRES$, which is half of the area of the whole star."
] | Brazil | XXI OBM | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 1/2 | |
04f9 | We are placing red and blue balls on the circle. In the beginning there are exactly two red balls on the circle. We are allowed to perform the following operations:
i) add one red ball and change the colour of its neighbouring two balls (red to blue and vice versa);
ii) remove one red ball and change the colour of its ... | [
"a. We consider three different cases of adding a red ball depending on the colours of its neighbours.\nIf we add the ball between two blue balls, then we have the case $PP \\rightarrow CCC$ in which the number of blue balls decreases by two. If we add the ball between two red balls, we have $CC \\rightarrow PCP$ a... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) No; b) No | |
0hqi | Problem:
Two villages, $A$ and $B$, lie on opposite sides of a straight river in the positions shown. There will be a market in village $B$, and residents of village $A$ wish to attend. The people of village $A$ would like to build a bridge across and perpendicular to the river so that the total route walked by reside... | [
"Solution:\n\nFor convenience, we label the point $X$, the foot of the perpendicular from $A$ to the river. We also label $Y$, the point where the perpendicular $B Y$ to the river through $B$ meets the parallel $A Y$ to the river through $A$. The key construction is to translate $B$ southward $1 \\mathrm{~km}$ to p... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | Place the bridge where the line from village A to the point obtained by translating village B perpendicular to the river by the river’s width meets the bank; the minimal total route length is 6 km. | |
0a5f | Problem:
Find all pairs of integers $x, y$ such that
$$y^{5} + 2x y = x^{2} + 2y^{4}.$$ | [
"Solution:\nRearrange and factorize to get\n$$y^{2}(y - 1)(y^{2} - y - 1) = (x - y)^{2}.$$\nNote that $y$ and $(y - 1)$ are coprime (their greatest common divisor is 1) because they are consecutive integers. Note since $y(y - 1)$ and $(y^{2} - y - 1)$ are consecutive integers, we see that $(y^{2} - y - 1)$ is copri... | New Zealand | NZMO Round Two | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | [(0, 0), (1, 1), (0, 2), (4, 2)] | |
04xj | On the blackboard $n$ nonnegative integers have been written, such that their greatest common divisor is equal to $1$. In one step we can erase two numbers $x$, $y$ such that $x \ge y$, and replace them with numbers $x - y$, $2y$. Determine, for which sequences of original $n$ integers one can lead to a situation, in w... | [
"The answer: the sum of the numbers has to be a power of $2$, or they are all zeroes. From this moment we assume that the numbers are not all zeroes, otherwise we are done without making any step.\n\nLet $S$ be the sum of the numbers on the blackboard and $D$ be their greatest common divisor at some moment. At the ... | Czech-Polish-Slovak Mathematical Match | 11-th Czech-Slovak-Polish Match, 2011 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | Exactly those initial sequences whose sum is a power of two (or the trivial case where all entries are zero). | |
0i14 | Problem:
Let $f, g$ be two functions from the set $\mathbf{R}$ of real numbers to itself, such that $f(x)<g(x)$ for all $x \in \mathbf{R}$. Prove that there exists an infinite subset $S \subseteq \mathbf{R}$ such that $f(x)<g(y)$ for all $x, y \in S$. | [
"Solution:\n\nNote that, for every $x$, we can choose a rational number $h(x)$ such that $f(x)<h(x)<g(x)$. Proof: since $g(x)-f(x)>0$, we can choose an integer $n$ larger than $1/(g(x)-f(x))$. Then the open interval $(n f(x), n g(x))$ has width $n(g(x)-f(x))>1$, so it contains some integer $m$. So we have $f(x)<m/n... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Logic"
] | null | proof only | null | |
0flk | Problem:
Un cuadrado $C$ se recubre completamente con un número entero de cuadrados de lado unidad, sin solapamientos. Si uno coloca dentro de $C$ y sin solapamientos tantos cuadrados como sea posible de área $2$, con los lados paralelos a los lados de $C$, se puede cubrir las ocho novenas partes del área del cuadrado... | [
"Solution:\n\nSea $l$ el lado del cuadrado y $n$ el número máximo de cuadrados de área $2$ que caben en cada lado del cuadrado. $l$ y $n$ son enteros. Las condiciones del problema son que\n$$\n\\left\\{\n\\begin{array}{c}\n2 n^{2} = \\frac{8}{9} l^{2} \\\\\nl^{2} < 2(n+1)^{2}\n\\end{array}\n\\right.\n$$\nPoniendo $... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | Side length l ∈ {3, 6, 9, 12, 15, 18, 21, 24} | |
0jez | Problem:
Given a point $p$ and a line segment $l$, let $d(p, l)$ be the distance between them. Let $A$, $B$, and $C$ be points in the plane such that $AB = 6$, $BC = 8$, $AC = 10$. What is the area of the region in the $(x, y)$-plane formed by the ordered pairs $(x, y)$ such that there exists a point $P$ inside triang... | [
"Solution:\n\nAnswer: $\\frac{288}{5}$.\n\nPlace $ABC$ in the coordinate plane so that $A = (0,6)$, $B = (0,0)$, $C = (8,0)$. Consider a point $P = (a, b)$ inside triangle $ABC$. Clearly, $d(P, AB) = a$, $d(P, BC) = b$.\n\nNow, we see that the area of triangle $ABC$ is $\\frac{6 \\cdot 8}{2} = 24$, but may also be ... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 288/5 | |
06jj | Let $O$ be the circumcentre of a triangle $ABC$, and let $\ell$ be the line going through the midpoint of the side $BC$ and is perpendicular to the bisector of $\angle BAC$. Determine the value of $\angle BAC$ if the line $\ell$ goes through the midpoint of the line segment $AO$. | [
"We have $\\angle BAC = 120^\\circ$.\nLet $M$ be the midpoint of $BC$. Suppose $l$ meets the altitude from $A$, the angle bisector of $\\angle BAC$, and the line $AO$ at $E$, $P$, $N$ respectively.\nNote that $AE // MO$. In order that $EM$ bisects $AO$, the quadrilateral $AEOM$ must be convex. Therefore, $\\angle B... | Hong Kong | Year 2016 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 120° | |
020w | Problem:
A divisor $d$ of a positive integer $n$ is said to be a close divisor of $n$ if $\sqrt{n} < d < 2 \sqrt{n}$. Does there exist a positive integer with exactly 2020 close divisors? | [
"Solution:\nLet $m$ be an odd integer with exactly 2020 positive divisors and which is (automatically) not a square. For example, $m = 3^{2019}$ suffices, but of course there are many alternatives. Now consider $n = 2^{k} m$, for some integer $k$ such that $2^{k} > m$. Any divisor of $n$ is then of the form $2^{\\e... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | Yes | |
0hwx | Problem:
Let $\sigma(n)$ denote the sum of the positive divisors of $n$. We say $n$ is perfect if $\sigma(n)=2 n$. If $n$ is a positive integer such that
$$
\frac{\sigma(n)}{n}=\frac{5}{3}
$$
show that $5 n$ is an odd perfect number. | [
"Solution:\nNote that $\\sigma(n) / n$ is the sum of the reciprocals of the divisors of $n$. Since $3 \\sigma(n)=5 n$, $n$ must be divisible by $3$. If $n$ were even, then $n$ would be divisible by $6$, so we would get\n$$\n\\frac{\\sigma(n)}{n} \\geq 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}=2>\\frac{5}{3}\n$$\na c... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0e6m | Peter breeds horses and cows on his farm. One day, the number of horses was the same as the number of cows and greater than $0$. Then Peter bought some more cows so that the number of cows increased by $50\%$. Now only $30\%$ of the animals are horses. How many horses does Peter have on his farm?
(A) $8$ (B) $9$ (C) $1... | [
"Say at the beginning there were $x$ cows and $x$ horses on the farm. When the number of cows increased by $50\\%$, there were $x + \\frac{1}{2}x = \\frac{3}{2}x$ cows. Hence the ratio of horses was equal to $x/(x + \\frac{3}{2}x) = \\frac{2}{5}$, which is $40\\%$ and not $30\\%$. The correct answer is thus (E)."
] | Slovenia | National Math Olympiad 2012 | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | E | |
0396 | Let $m > 1$ and $n > 1$ be odd integers. Distinct real numbers are written in the cells of a table of $m$ rows and $n$ columns. A number is called *good* if:
1. it is the largest in its row (column);
2. it is the middle number of its column (row).
What is the maximal number of good numbers? | [
"Let $N_c$ be the set of good numbers that are largest in their column and middle in their row. Since no two of these numbers belong to one and the same row we have $|N_c| \\le m$. Denote by $a$ the largest element of $N_c$ and let $a_1, a_2, \\dots, a_{\\frac{n-1}{2}}$ be the numbers in the row of $a$ that are gre... | Bulgaria | Spring Mathematical Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | min{m, (n+1)/2} + min{n, (m+1)/2} | |
0dx3 | Problem:
1. Dani sta enačbi $2 \log_{3}(x+2) - \log_{3} x = 2$ in $25^{x} = 0{,}008$.
a) Reši enačbi.
b) Rešitve enačb so ničle polinoma $p$ tretje stopnje. Graf tega polinoma poteka tudi skozi točko $(2,7)$. Zapiši predpis polinoma $p$ v ničelni obliki. | [
"Solution:\n\na)\nLogaritemsko enačbo preoblikujemo:\n\n$$\n2 \\log_{3}(x+2) - \\log_{3} x = 2\n$$\n\nUporabimo lastnosti logaritmov:\n$$\n\\log_{3}(x+2)^2 - \\log_{3} x = 2\n$$\n$$\n\\log_{3} \\frac{(x+2)^2}{x} = 2\n$$\n$$\n\\frac{(x+2)^2}{x} = 3^2\n$$\n$$\n\\frac{x^2 + 4x + 4}{x} = 9\n$$\n$$\nx^2 + 4x + 4 = 9x\n$... | Slovenia | 6. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | Equation 1 solutions: x = 1 and x = 4. Equation 2 solution: x = -3/2. The polynomial is p(x) = -(x - 4)(x - 1)(x + 3/2). | |
03fi | The nonnegative real numbers $x, y, z$ are such that $(x + y)(y+z)(z+x) = 1$. We denote by $m$ and $M$ respectively the smallest and largest possible values of the expression $A = (xy + yz + zx)(x + y + z)$.
a) Find $m$ and $M$.
b) Is there a triple of nonnegative rational numbers $(x, y, z)$ satisfying the given equ... | [
"a) The inequality $(xy+yz+zx)(x+y+z) \\ge 1 = (x+y)(y+z)(z+x)$ is equivalent to $xyz \\ge 0$. Equality is reached only when one of the variables, say $x$, is $0$ and the other two (in this case $y$ and $z$) are whatever with $yz(y+z) = 1$; one possibility is $y=1$ and $z = \\frac{\\sqrt{5}-1}{2}$. The inequality $... | Bulgaria | Bulgarian Spring Tournament | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Algebraic Number Theory > Unique factorization"
] | English | proof and answer | m = 1, M = 9/8; No, the minimum cannot be attained by any triple of nonnegative rational numbers. | |
056q | Prove that, for any natural number $n$, either $3^{2n} - 3^{n+1} + 3^n - 3$ or $3^{2n} - 3^{n+1} + 3^n + 1$ is divisible by 32. | [
"Note that $3^{2n} - 3^{n+1} + 3^n - 3 = (3^n - 3)(3^n + 1)$.\n\nIf $n$ is odd then $3^n - 3 = 3(3^{n-1} - 1) = 3\\left(3^{\\frac{n-1}{2}} - 1\\right)\\left(3^{\\frac{n-1}{2}} + 1\\right)$.\n\nAs all powers of $3$ are odd, $3^{\\frac{n-1}{2}} - 1$ and $3^{\\frac{n-1}{2}} + 1$ are consecutive even numbers. One of th... | Estonia | Final Round of National Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0jaf | Problem:
Let $S$ be the set $\{1,2, \ldots, 2012\}$. A perfectutation is a bijective function $h$ from $S$ to itself such that there exists an $a \in S$ such that $h(a) \neq a$, and that for any pair of integers $a \in S$ and $b \in S$ such that $h(a) \neq a, h(b) \neq b$, there exists a positive integer $k$ such that ... | [
"Solution:\nAnswer: 2\n\nNote that both $f$ and $g$, when written in cycle notation, must contain exactly one cycle that contains more than 1 element. Assume $f$ has $k$ fixed points, and that the other $2012-k$ elements form a cycle, (of which there are $(2011-k)!$ ways).\n\nThen note that if $f$ fixes $a$ then $f... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | 2 | |
0btn | Let $\triangle ABC$ be an acute triangle, and let $M$ be the midpoint of the side $AC$. A circle through $B$ and $M$ meets the sides $AB$ and $BC$ again at $P$ and $Q$, respectively. The reflection $T$ of $B$ across the midpoint of the segment $PQ$ lies on the circle $ABC$. Evaluate the ratio $BT/BM$. | [
"The required ratio equals $\\sqrt{2}$. To prove this, let $S$ be the midpoint of the segment $PQ$, and let $B'$ be the reflection of $B$ across $M$. Clearly, $ABCB'$ is a parallelogram, $\\angle ABB' = \\angle PQM$, and $\\angle BB'A = \\angle B'BC = \\angle MPQ$, so the triangles $ABB'$ and $MQP$ are similar. Sin... | Romania | 67th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | sqrt(2) | |
0b1g | Problem:
A wooden rectangular brick with dimensions $3$ units by $a$ units by $b$ units is painted blue on all six faces and then cut into $3ab$ unit cubes. Exactly $1/8$ of these unit cubes have all their faces unpainted. Given that $a$ and $b$ are positive integers, what is the volume of the brick? | [] | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Other 3D problems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 96 | |
064m | Problem:
Für eine feste positive ganze Zahl $k$ sei $K$ die Menge aller Gitterpunkte $(x, y)$ in der Ebene, deren beide Koordinaten $x$ und $y$ nichtnegative ganze Zahlen kleiner als $2k$ sind. Es gilt also $|K| = 4k^{2}$.
Eine Menge $V$ bestehe nun aus $k^{2}$ nicht-ausgearteten Vierecken mit folgenden Eigenschaften... | [
"Solution:\n\nJeder Punkt aus $K$ ist Eckpunkt eines eindeutig definierten zentralen Quadrats. Daher ist die Menge $Q$ aller zentralen Quadrate akzeptabel. Wir zeigen $S(V) \\leq S(Q) = S(k)$, woraus die Antwort folgt.\n\nWir benutzen das folgende\n\nLemma 1. Für jedes Viereck $V = A_{1}A_{2}A_{3}A_{4}$ und jeden b... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade 2022 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
... | null | proof and answer | k^2(2k-1)(2k+1)/3 | |
0laz | Given an integer $n \ge 3$. Consider $n$ real numbers $x_1, x_2, \dots, x_n$ satisfying at the same time the following conditions:
i/ $x_1 \ge x_2 \ge \dots \ge x_n$;
ii/ $\sum_{i=1}^{n} x_i = 0$;
iii/ $\sum_{i=1}^{n} x_i^2 = n(n-1)$.
Find the maximum and the minimum values of the sum $S = x_1 + x_2$. | [] | Vietnam | IMO2011 Selection | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Minimum S = n − 2; Maximum S = sqrt(2(n − 1)(n − 2)) | |
0d1f | Triangle $ABC$ is inscribed in circle $\omega$. Point $D$ is midpoint of side $AC$, and point $M$ lies on segment $BD$ with $DM = 2BM$. Ray $AM$ meets side $BC$ at $E$, and ray $CM$ meets side $BA$ at $F$. Ray $FE$ intersects $\omega$ at $N$. Suppose that $AM \perp CM$. Prove that $ADEF$ is cyclic if and only if line $... | [
"Denote by $a$, $b$, $c$ the side lengths, and by $m_a$, $m_b$, $m_c$ the lengths of the medians of the triangle $ABC$. Since $MD$ is median in the right-angled triangle $AMC$, it follows that\n$$\n2m_b/3 = MD = AD = CD = b/2,\n$$\nso $m_b = 3b/4$, which means that\n$$\n(3b/4)^2 = m_b^2 = (a^2 + c^2)/2 - b^2/4.\n$$... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0ktt | Problem:
Four people are playing rock-paper-scissors. They each play one of the three options (rock, paper, or scissors) independently at random, with equal probability of each choice. Compute the probability that someone beats everyone else.
(In rock-paper-scissors, a player that plays rock beats a player that plays... | [
"Solution:\n\nAs the four players and three events are symmetric, the probability a particular player makes a particular move and beats everyone else is the same regardless of the choice of player or move. So, focusing on one such scenario, the desired probability is $12$ times the probability that player $1$ plays... | United States | HMMT November | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 4/27 | |
0ke1 | Problem:
Consider a sheet of paper in the shape of an equilateral triangle creased along the dashed lines as in the figure below on the left. Folding over each of the three corners along the dashed lines creates a new object which is uniformly four layers thick, as in the figure below on the right. The number in each ... | [
"Solution:\n\na. Divide the sides of the triangle into three equal parts and connect them with lines as in the figure on the left below.\n\n\n\nNow crease and fold along the dotted lines to form the figure in the center. Finally, fold along the dashed lines in the center figure to form the ... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
09i9 | How many positive integers not exceeding $11111100111$ can be written using only digits $0$ and $1$ in base $10$? | [] | Mongolia | Mongolian Mathematical Olympiad Round 1 | [
"Discrete Mathematics > Other"
] | null | proof and answer | 2023 | |
07tz | For which positive integers $n$ is it possible to split the set of integers $t$ that satisfy $1 \le t \le n^{2022}$ into $n$ sets of equal size such that the sum of the 2021-th powers of the elements is the same for each set? | [
"**Solution 1.** When $n = 1$ this is obviously possible. We will show that this is possible for all integers $n \\ge 2$ by proving a more general version using induction. The number 2021 will be replaced by an integer $k \\ge 0$ on which we will carry out the induction. For this approach to work we will consider p... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other"
] | null | proof and answer | all positive integers | |
07eo | Let $P(x)$ be a non-constant polynomial with real coefficients such that $P(0) = 0$. For all positive real numbers $M$, prove that there is a positive integer $d$ such that for any monic polynomial $Q(x)$ of degree greater than or equal to $d$, total number of integers $k$ where
$$
|P(Q(k))| \leq M,
$$
is at most $\deg... | [
"It is clear that the solutions of the inequality, $|P(x)| \\le M$, is a subset of an interval of the form $(-a, a)$ for some positive real number $a$. Now, assume $\\deg Q(x) = d \\ge m$. Consider, integers $x_0 < \\dots < x_d$. Then by Lagrange's interpolation formula, one can find that\n$$\nQ(x) = \\sum_{i=0}^{d... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0fmi | Hallar todas las soluciones enteras $(x, y)$ de la ecuación
$$
y^k = x^2 + x
$$
donde $k$ es un número entero dado mayor que 1. | [
"Puesto que $y^k = x^2 + x = x(x+1)$ y $\\mathrm{mcd}(x, x+1) = 1$ resulta que tanto $x+1$ como $x$ deben ser potencias $k$-ésimas de un entero. Pero los dos únicos números enteros consecutivos que son potencias $k$-ésimas, con $k > 1$ son $0$ y $1$ o bien $-1$ y $0$. Las dos únicas soluciones son, pues, $x = 0$, $... | Spain | Olimpiada Matemática Española | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Spanish | proof and answer | (x, y) = (0, 0) and (−1, 0) | |
0cxi | Quadrilateral $ABCD$ with perpendicular diagonals $AC$ and $BD$ is inscribed in a circle. Altitude $DE$ in triangle $ABD$ intersects diagonal $AC$ in $F$. Prove that $FB = BC$. | [
"\n\n$F$ is the orthocenter of $\\triangle ABD$, that is $BF \\perp AD$. We have $\\widehat{BCA} \\equiv \\widehat{BDA}$ ($ABCD$ is cyclic) and $\\widehat{BDA} = \\frac{\\pi}{2} - \\widehat{PAD}$. Also, $\\widehat{BFP} = \\widehat{AFB'} = \\frac{\\pi}{2} - \\widehat{FAB'} = \\frac{\\pi}{2} ... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry ... | English | proof only | null | |
081t | Problem:
Sia dato un triangolo $ABC$. Si indichino con $M$ ed $N$ i punti medi rispettivamente dei lati $AC$ e $BC$. Siano inoltre $S$ e $T$ rispettivamente punti sui lati $AC$ e $BC$ tali che:
$$
AS = \frac{1}{3} AC \quad BT = \frac{1}{3} BC.
$$
Dimostrare che le bisettrici degli angoli $\angle AST$ e $\angle BTS$ si... | [
"Solution:\n\nSi supponga che le bisettrici degli angoli $\\angle AST$ e $\\angle BTS$ si incontrino in un punto $P$ del lato $AB$.\nDal momento che $\\frac{AS}{AC} = \\frac{BT}{BC}$, per il teorema di Talete, $ST$ e $AB$ sono segmenti paralleli, pertanto gli angoli $P\\widehat{S}T$ e $S\\widehat{P}A$ sono uguali. ... | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02xa | Problem:
O frasco de um perfume tem formato cilíndrico e uma embalagem em forma de flor deve ser construída para acondicioná-lo. Para a confecção desta embalagem será utilizada uma folha quadrada de lado $n$, quatro arcos de circunferência com centros nos vértices do quadrado e raio medindo $n$ e uma circunferência ta... | [
"Solution:\n\na) O raio do arco mede $n$ e a diagonal do quadrado mede $n \\sqrt{2}$, então o raio da circunferência é $\\frac{n-(n \\sqrt{2}-n)}{2}=\\frac{2 n-n \\sqrt{2}}{2}$.\n\nb) Dividindo o quadrado como na figura, temos que a soma das áreas $A, B$ e $C$ é $\\frac{n^{2}}{2}$. Como $B=\\frac{\\pi \\cdot n^{2}}... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | a) r = n*(2 - sqrt(2))/2. b) Area = n^2*(4 - 4*pi/3 - sqrt(3)). | |
09dl | a sequence $a_0, a_1, \dots$ is defined by $a_0 = 1$ and for $k \ge 1$, $a_{2k} = (-1)^k a_k$; for $k \ge 0$, $a_{2k+1} = -a_k$. Prove that for every $n \ge 0$, $a_0 + a_1 + \dots + a_n \ge 0$. | [] | Mongolia | ММО-48 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Mongolian | proof only | null | |
05tg | Problem:
Martin cherche à remplir chaque case d'une grille rectangulaire ayant 8 lignes et $n$ colonnes avec l'une des quatre lettres $\{P\}, \{O\}, \{F\}$ et $M$ de sorte que pour toute paire de lignes distinctes, il existe au plus une colonne telle que ses intersections avec les deux lignes sont des cases ayant la m... | [
"Solution:\n\nDans ce problème, on cherche le plus grand entier satisfaisant une certaine propriété. Supposons que l'on veuille montrer que le plus grand entier recherché est l'entier $c$. Pour montrer que $c$ est bien le plus grand entier, on va d'une part montrer que si un entier $n$ satisfait la propriété, alors... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 7 | |
0fl1 | Problem:
En el interior de una circunferencia de centro $O$ y radio $r$, se toman dos puntos $A$ y $B$, simétricos respecto de $O$. Se considera un punto variable $P$ sobre esta circunferencia y se traza la cuerda $P P^{\prime}$, perpendicular a $A P$. Sea $C$ el punto simétrico de $B$ respecto de $P P^{\prime}$. Hall... | [
"Solution:\n\nEstablezcamos primero que $A C$ es constante.\n\nMétodo 1.\nSe obtiene $C$ a partir de $A$ aplicando un giro de $180^{\\circ}$ con centro en $O$ seguido de la simetría de eje $P P^{\\prime}$.\n\n\n\nDescomponiendo el giro en producto de dos simetrías de ejes perpendiculares $e... | Spain | Fase Nacional de la XLV Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0clj | Let $ABCD A'B'C'D'$ be a cube. On the segments $BC$ and $DD'$ we take the points $M$ and $N$ respectively, such that $BM = DN$. Prove that line $A'M$ is perpendicular to plane $(AB'N)$.
Cătălin Barbu
 | [
"$BC \\perp (ABB')$ and $AB' \\subset (ABB')$ yields $AB' \\perp BC$. Since $AB' \\perp A'B$ (diagonals of the square $ABB'A'$), we get $AB' \\perp (A'BC)$. Because $A'M \\subset (A'BC)$, it follows that $A'M \\perp AB'$. (1)\n\nLet $E \\in (AD)$ be such that $AE = BM$. Then $ABME$ is a rectangle, so $AB \\parallel... | Romania | 75th Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0d4g | Prove that there exists a positive integer $n$ such that the last digits of $n^{3}$ are $\square 201320132013$. | [
"Notice that $a = 201320132013$ and $10^{12}$ are coprime. By Euler's theorem\n$$\na^{\\phi\\left(10^{12}\\right)} \\equiv 1 \\quad \\bmod \\left(10^{12}\\right)\n$$\nLet $b = a^{\\phi\\left(10^{12}\\right) - 1}$. We have $a b \\equiv 1 \\bmod \\left(10^{12}\\right)$. Because $b$ and $10^{12}$ are coprime, again by... | Saudi Arabia | SAMC | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | English, Arabic | proof only | null | |
0l24 | Problem:
Over all pairs of complex numbers $(x, y)$ satisfying the equations
$$
x + 2y^{2} = x^{4} \quad \text{and} \quad y + 2x^{2} = y^{4}
$$
compute the minimum possible real part of $x$. | [
"Solution:\nNote the following observations:\n(a) if $(x, y)$ is a solution then $(\\omega x, \\omega^{2} y)$ is also a solution if $\\omega^{3}=1$ and $\\omega \\neq 1$.\n(b) we have some solutions $(x, x)$ where $x$ is a solution of $x^{4}-2x^{2}-x=0$.\n\nThese are really the only necessary observations and the f... | United States | HMMT February 2024 Guts Round | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | ((1 - sqrt(33))/2)^(1/3) | |
026j | Problem:
Determine all integer numbers $x$ and $y$ such that:
$$
\frac{1}{x} + \frac{1}{y} = \frac{1}{19}
$$ | [
"Solution:\n\nA equação é equivalente a $x y = 19(x + y)$. Uma vez que estamos procurando soluções inteiras e $19$ é um número primo, esta igualdade implica que $x$ ou $y$ devem ser divisíveis por $19$. Como a equação é simétrica em relação às variáveis $x$ e $y$, podemos supor que $x$ é divisível por $19$. Isto é,... | Brazil | Nível 2 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (38, 38), (380, 20), (20, 380), (-342, 18), (18, -342) | |
0cxu | Consider the sequence $a_{1} = 1$ and $a_{n+1} = \frac{3 a_{n}^{2} + 1}{2} - a_{n}$ for $n = 1, 2, \ldots$. Prove that if $n$ is a power of $3$ then $n$ divides $a_{n}$. | [
"The recursive relation is equivalent to\n$$\n3 a_{n+1} - 1 = \\frac{1}{2} \\left(9 a_{n}^{2} - 6 a_{n} + 1\\right), \\quad n = 1, 2, \\ldots\n$$\nLet $x_{n} = 3 a_{n} - 1$, $n = 1, 2, \\ldots$, and get $x_{1} = 8 = 2^{3}$,\n$$\nx_{n+1} = \\frac{1}{2} x_{n}^{2}, \\quad n = 1, 2, \\ldots\n$$\nLet $x_{n} = 2^{y_{n}}$... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
06zk | Problem:
Let $p(m, n)$ be the polynomial $2 m^{2} - 6 m n + 5 n^{2}$. The range of $p$ is the set of all integers $k$ such that $k = p(m, n)$ for some integers $m$, $n$. Find which members of $\{1,2, \ldots, 100\}$ are in the range of $p$. Show that if $h$ and $k$ are in the range of $p$, then so is $h k$. | [
"Solution:\n\nWe have $p(m, n) = (m - 2 n)^{2} + (m - n)^{2}$, so $p(2 a - b, a - b) = a^{2} + b^{2}$. Hence the range of $p$ is just the sums of two squares.\n\n$\\left(a^{2} + b^{2}\\right)\\left(c^{2} + d^{2}\\right) = (a c - b d)^{2} + (a d + b c)^{2}$, which establishes that if $h$ and $k$ are in the range, th... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Algebraic Number Theory > Quadratic forms",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | The integers in 1 to 100 that are in the range are: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37, 40, 41, 45, 49, 50, 52, 53, 58, 61, 64, 65, 68, 72, 73, 74, 80, 81, 82, 85, 89, 90, 97, 98, 100. | |
0dzq | Problem:
Marko je postavil kola navpično na vodoravna tla tako, da je eden segal $1~\mathrm{m}$ v višino, drugi pa $2~\mathrm{m}$ v višino. Od vrha vsakega kola do točke, kjer je bil drugi kol zabit v vodoravna tla, je napel vrv. Na kateri višini od tal sta se vrvi križali? | [
"Solution:\n\nOb narisani skici ugotovimo podobnost trikotnikov. Zapišemo razmerji, npr: $h : 2 = y : x$ in $h : 1 = (x - y) : x$. Izračunamo $h = \\frac{2}{3}~\\mathrm{m}$, kar je višina, na kateri se sekata vrvici.\n\n$h =$ iskana višina,\n$y =$ vodoravna razdalja med nižjim kolom in nožiščem $h$-ja,\n$x =$ vodor... | Slovenia | Državno tekmovanje | [
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 2/3 m | |
07ls | Prove that $\frac{1}{\pi} \tan^{-1}(\sqrt{2})$ is not a rational number. | [
"Let $A = \\tan^{-1}(\\sqrt{2})$ and assume $\\frac{A}{\\pi} = \\frac{m}{n}$ with integers $m, n$. This gives $nA = m\\pi$ and $\\tan(A) = \\sqrt{2}$. The strategy is to study the sequence $\\tan(kA)$ for $k = 1, 2, 3, \\dots$. We first observe that $nA = m\\pi$ implies that this sequence contains only finitely man... | Ireland | Irska | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0fz1 | Problem:
In einem Dreieck $A B C$ mit $A B \neq A C$ sei $D$ die Projektion von $A$ auf $B C$. Ferner seien $E, F$ die Mittelpunkte der Strecken $A D$ bzw. $B C$ und $G$ die Projektion von $B$ auf $A F$. Zeige, dass die Gerade $E F$ die Tangente im Punkt $F$ an den Umkreis des Dreiecks $G F C$ ist. | [
"Solution:\n\n1. Lösung Sei $\\alpha=\\angle A F E$, dann genügt es nach dem Tangentenwinkelsatz zu zeigen, dass $\\angle G C F=\\alpha$. Konstruiere $D'$ als die Spiegelung von $D$ an $F$. Es gilt nun $F D=F D'$ und nach Konstruktion von $E$ folgt $E F \\parallel A D'$. Also ist $\\angle D A F=\\alpha$ als Wechsel... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kil | Menkara has a $4 \times 6$ index card. If she shortens the length of one side of this card by $1$ inch, the card would have area $18$ square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by $1$ inch?
(A) $16$ (B) $17$ (C) $18$ (D) $19$ (E) $20$ | [] | United States | AMC 12 A | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | E | |
0bln | The diagonals of the parallelogram $ABCD$ meet at $O$. The bisectors of the angles $DAC$ and $DBC$ meet at $T$. It is known that $TD + TC = TO$. Find the measures of the angles of triangle $ABT$.
Claudiu Militaru | [
"The hypothesis shows that $DOCT$ is a parallelogram. From $AO \\parallel DT$ follows $\\angle DTA = \\angle OAT = \\angle DAT$, hence $DA = DT$.\n\nThis leads to $DA = DT = OC$; in the same way $BC = CT = OD$, therefore $BD = AC$. So $ABCD$ is a rectangle, $AOTD$ is a rhombus, triangle $AOD$ is equilateral and tri... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | Each angle is 60 degrees | |
075t | A marker is placed at the origin of an integer lattice. Calvin and Hobbes play the following game. Calvin starts the game and each of them takes turns alternatively. At each turn, one can choose two (not necessarily distinct) integers $a$, $b$, neither of which was chosen earlier by any player and move the marker by $a... | [
"Let $A_n$ denote the set of chosen integers after $n$ turns. We claim (by induction) that after Calvin's move he can ensure that if $a \\in A_n$ then $-a \\in A_n$, and that the marker is at $(r, -r)$ for some non-zero integer $r$ in $A_n$.\n\nLet Calvin move the marker to $(-1, 1)$ in his first turn. Suppose that... | India | Indija TS 2013 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
00jm | Let $ABC$ be a non-isosceles triangle with circumcenter $U$ and incenter $I$. Assume that the bisector of the segment $UI$ passes through the common point of the angle bisector of $\gamma = \angle ACB$ with the circumcircle of $ABC$. Prove that $\gamma$ is the second largest angle in the triangle $ABC$.
G. Baron, Vienn... | [
"Let $w_\\gamma$ be the angle bisector of $\\gamma$, $k$ the circumcircle of $ABC$ and $D = k \\cap w_\\gamma$. Since $\\angle DCB = \\angle DCA$ we certainly have $|DA| = |DB|$. Considering the triangle $DBI$, we note that $\\angle CDB = \\angle CAB = \\alpha$. Also, $\\angle DBI = \\angle DBA + \\angle ABI = \\an... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jl0 | Problem:
Let $M$ denote the number of positive integers which divide $2014!$, and let $N$ be the integer closest to $\ln (M)$. Estimate the value of $N$. If your answer is a positive integer $A$, your score on this problem will be the larger of $0$ and $\left\lfloor 20-\frac{1}{8}|A-N|\right\rfloor$. Otherwise, your s... | [
"Solution:\n\nAnswer: 439 Combining Legendre's Formula and the standard prime approximations, the answer is\n$$\n\\prod_{p}\\left(1+\\frac{2014-s_{p}(2014)}{p-1}\\right)\n$$\nwhere $s_{p}(n)$ denotes the sum of the base $p$-digits of $n$.\n\nEstimate $\\ln 1000 \\approx 8$, and $\\ln 2014 \\approx 9$. Using the Pri... | United States | HMMT November 2014 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 439 | |
0leb | Given a system of equations on $\mathbb{R}$
$$
\begin{cases}
x - ay = yz \\
y - az = zx \\
z - ax = xy
\end{cases}
$$
a) Solve that system when $a = 0$.
b) Prove that the system has 5 different roots when $a > 1$. | [
"a. For $a = 0$, we have\n$$\n\\begin{cases}\nx = yz, \\\\\ny = zx, \\\\\nz = xy.\n\\end{cases}\n$$\nIf one of three numbers is equal to $0$ then the other numbers are equal to $0$ too. We consider the case $xyz \\neq 0$ and multiply the equations, side by side, we get $xyz = 1$ then\n$$\nx^2 = y^2 = z^2 = 1.\n$$\n... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof and answer | a) Solutions: (0, 0, 0); (1, 1, 1); and the three permutations of (−1, −1, 1). b) For parameter greater than one, the system has exactly five distinct real solutions. | |
04qs | Let $ABCD$ be an equilateral trapezoid with sides $AB$ and $CD$. The incircle of the triangle $BCD$ touches $CD$ at $E$. Point $F$ is chosen on the bisector of the angle $\angle DAC$ such that the lines $EF$ and $CD$ are perpendicular. The circumcircle of the triangle $ACF$ intersects the line $CD$ again at $G$. Prove ... | [
"\nLet us show that $|FA| = |FG|$. We will proceed: from behind. On the extension of $CD$ we take the point $P$ such that $|DP| = |DA|$ and similarly on extension of $DC$ we take the point $Q$ such that $|CQ| = |CA|$. Then, using well known properties of the incircle, we have\n$$\n\\begin{a... | Czech Republic | 6-th Czech-Slovak Match | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0j6n | Problem:
Let $ABC$ be a triangle, and let points $P$ and $Q$ lie on $BC$ such that $P$ is closer to $B$ than $Q$ is. Suppose that the radii of the incircles of triangles $ABP$, $APQ$, and $AQC$ are all equal to $1$, and that the radii of the corresponding excircles opposite $A$ are $3$, $6$, and $5$, respectively. If t... | [
"Solution:\nLet $t$ denote the radius of the excircle of triangle $ABC$ opposite $A$.\n\nLemma: Let $ABC$ be a triangle, and let $r$ and $r_A$ be the inradius and exradius opposite $A$. Then\n$$\n\\frac{r}{r_A} = \\tan \\frac{B}{2} \\tan \\frac{C}{2}\n$$\nProof. Let $I$ and $J$ denote the incenter and the excenter ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 135 | |
0cmn | An increasing sequence of positive integers $\{a_n\}$ and a positive integer $k$ are given. For each positive integer $n$ the following conditions hold: the number $a_n$ is divisible either by $1005$ or by $1006$, $a_n$ is not divisible by $97$, and $a_{n+1} - a_n \le k$. Find the least possible value of $k$. (A. Golov... | [
"**Ответ.** $k = 2010$.\n\nLet us denote our sequence by $(a_n)$. Clearly, $a_1 < 1005 \\cdot 1006 \\cdot 97 \\cdot N = D$ for some natural $N$. Then there exists such $n$ that $a_n \\le D$, but $a_{n+1} > D$ (at the same time, $a_n \\ne D$ by the condition). But the largest numbers less than $D$ and divisible by $... | Russia | Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization"
] | English; Russian | proof and answer | 2010 | |
072b | Problem:
Prove that the number of 5-tuples of positive integers $(a, b, c, d, e)$ satisfying the equation
$$
a b c d e = 5(b c d e + a c d e + a b d e + a b c e + a b c d)
$$
is an odd integer. | [
"Solution:\nWe write the equation in the form:\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} + \\frac{1}{e} = \\frac{1}{5}\n$$\nThe number of five tuples $(a, b, c, d, e)$ which satisfy the given relation and for which $a \\neq b$ is even, because if $(a, b, c, d, e)$ is a solution, then so is $(b,... | India | INMO 2004 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0exm | Problem:
A tangent to the inscribed circle of a triangle drawn parallel to one of the sides meets the other two sides at $X$ and $Y$. What is the maximum length $XY$, if the triangle has perimeter $p$? | [
"Solution:\nLet $BC$ be the side parallel to $XY$, $h$ the length of the altitude from $A$, and $r$ the radius of the incircle. Then $XY/BC = (h - 2r)/h$. But $r p = h BC$. So $XY = \\frac{(p - 2BC) BC}{p} = \\frac{p^2}{8} - \\frac{2(BC - p/4)^2}{p}$. So the maximum occurs when $BC = p/4$ and has value $p/8$."
] | Soviet Union | 5th ASU | [
"Geometry > Plane Geometry > Circles > Tangents",
"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 > Transformati... | null | proof and answer | p/8 | |
05go | Problem:
Soit $ABC$ un triangle isocèle et obtus en $A$. Soit $\Gamma$ le cercle de centre $B$ passant par $A$, et $\Omega$ le cercle de centre $C$ passant par $A$. Soit $D$ le point d'intersection du cercle $\Gamma$ avec le segment $[BC]$, $E$ le deuxième point d'intersection de la droite $(AD)$ avec le cercle $\Omeg... | [
"Solution:\n\nSoit $X$ le point d'intersection du cercle $\\Omega$ avec le segment $[BC]$. Les points $B$ et $C$ sont symétriques par rapport à la médiatrice du segment $[BC]$ donc les cercles $\\Gamma$ et $\\Omega$ le sont aussi. Il vient que $D$ et $X$ sont symétriques par rapport à la médiatrice du segment $[BC]... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null |
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