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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
08vd | Let $N$ be a positive integer. $N$ squares are lined up contiguously from left to right. Students A and B play a game according to the following rules:
(1) To start off, A will write 1 non-negative integer into each of the $N$ squares.
(2) The game ends when the following condition is achieved: For every $i$ satisfying... | [
"We will show that the player B can finish the game no matter how A plays.\nSuppose we represent by $(x_1, x_2, \\dots, x_N)$ the list of numbers written into the $N$ boxes from left to right. Let also $x_0 = 0$. For this list we call the index $i$ $(1 \\le i \\le N-1)$ the **index of completion** if it is the smal... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Yes; B can always force the game to end in finitely many moves. | |
0kd2 | Problem:
Let $\{a_{i}\}_{i \geq 0}$ be a sequence of real numbers defined by
$$
a_{n+1} = a_{n}^2 - \frac{1}{2^{2020 \cdot 2^{n}} - 1}
$$
for $n \geq 0$. Determine the largest value for $a_{0}$ such that $\{a_{i}\}_{i \geq 0}$ is bounded. | [
"Solution:\nLet $a_{0} = \\frac{1}{\\sqrt{2}^{2020}}\\left(t + \\frac{1}{t}\\right)$, with $t \\geq 1$. (If $a_{0} < \\frac{1}{\\sqrt{2}^{2018}}$ then no real $t$ exists, but we ignore these values because $a_{0}$ is smaller.) Then, we can prove by induction that\n$$\na_{n} = \\frac{1}{\\sqrt{2}^{2020 \\cdot 2^{n}}... | United States | HMMT February 2020 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1 + 1/2^{2020} | |
06gp | Let $ABC$ be a scalene triangle (with no two sides equal). A circle touching sides $AB$ and $BC$ intersects the median $AM$ at $P$ and $Q$. Another circle touching sides $AC$ and $BC$ also intersects $AM$ at $P$ and $Q$. Prove that the two circles are the same circle. | [
"Let $\\Gamma_1$ be the circle touching $AB$ and $BC$, and let $\\Gamma_2$ be the circle touching $AC$ and $BC$. Let $\\Gamma_1$ touch $AB$ and $BC$ at $X$ and $D$ respectively. Let $\\Gamma_2$ touch $AC$ and $BC$ at $Y$ and $E$ respectively.\n\n\n\nSuppose on the contrary that $\\Gamma_1 \... | Hong Kong | Year 2012 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0djp | One has $2022$ yellow, $2022$ red and $2022$ blue sticks. It is known that if we pick three sticks of different color, then it is possible to construct a triangle using these sticks. Prove that there exists a color such that: if we pick $3$ sticks of that color, then it is possible to construct a triangle using these s... | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0b1b | Problem:
In $\triangle XYZ$, let $A$ be a point on (segment) $YZ$ such that $XA$ is perpendicular to $YZ$. Let $M$ and $N$ be the incenters of triangles $XYA$ and $XZA$, respectively. If $YZ = 28$, $XA = 24$, and $YA = 10$, what is the length of $MN$? | [] | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2*sqrt(26) | |
0ep5 | How many of the integers between $97$ and $199$ are multiples of $2$ or $3$?
(A) $33$ (B) $40$ (C) $55$ (D) $60$ (E) $68$ | [
"**E** Between (and including) $98$ and $200$ there are $51$ multiples of $2$; between $98$ and $199$ there are $34$ multiples of $3$. Between $102$ and $198$ there are $17$ multiples of $6$. The number we seek is $51 + 34 - 17 = 68$"
] | South Africa | South African Mathematics Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Other"
] | English | MCQ | E | |
02w2 | Problem:
Neste problema, iremos estudar quantos fatores 2 aparecem na fatoração de números da forma $5^{2^{n}}-1$.
a. Sejam $x$ e $y$ dois números inteiros ímpares. Prove que $x^{2}+y^{2}$ possui exatamente um fator 2 em sua fatoração em primos.
b. Usando a fatoração $a^{2}-b^{2}=(a-b)(a+b)$, determine quantos fator... | [
"Solution:\n\na. Como $x$ e $y$ são ímpares, então existem inteiros $x_{0}$ e $y_{0}$ tais que $x=2 x_{0}+1$ e $y=2 y_{0}+1$. Daí,\n$$\n\\begin{aligned}\nx^{2}+y^{2} & =\\left(2 x_{0}+1\\right)^{2}+\\left(2 y_{0}+1\\right)^{2} \\\\\n& =4 x_{0}^{2}+4 x_{0}+1+4 y_{0}^{2}+4 y_{0}+1 \\\\\n& =4 x_{0}^{2}+4 x_{0}+4 y_{0}... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) Exactly one factor of two. b) Four factors of two. c) Two thousand nineteen factors of two. d) The decimal representation has six consecutive zeros. | |
039a | Find all integers $n \ge 3$ such that: for any two positive integers $m < n-1$, $r < n-1$ there exist $m$ distinct elements of the set $\{1, 2, \dots, n-1\}$ whose sum is congruent to $r$ modulo $n$. | [] | Bulgaria | Second selection test for IMO 2007, Vietnam | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | All odd integers n ≥ 3 | |
0ah6 | Find all prime numbers of the form $\overbrace{11...1}^{2n}/11$, where $n$ is a natural number. | [
"We use the equality: $\\overbrace{11...1}^{k} = \\frac{10^k - 1}{9}$. We get\n$$\n\\frac{\\overbrace{11...1}^{2n}}{11} = \\frac{10^{2n} - 1}{11 \\cdot 9} = \\frac{(10^n - 1)(10^n + 1)}{11 \\cdot 9}.\n$$\nFor $n=1$, $\\overbrace{11...1}^{2n}/11 = \\frac{11}{11} = 1$ which is not a prime number. For $n=2$,\n$$\n\\fr... | North Macedonia | XVI Junior Macedonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English | proof and answer | 101 | |
0hbf | Suppose there are $2019$ points $A_1, A_2, \dots, A_{2019}$ in a circle, that form the regular $2019$-gon $A_1A_2\dots A_{2019}$. Olesya and Andrew take turns, Olesya is the first to go. According to certain rules, Olesya makes a move to form an obtuse triangle, and Andrew – an acute triangle. The rules for forming tri... | [
"Let's place the circle so that the diameter through the vertex $A_1$ is vertical and this vertex itself is located below. Let's renumber all the vertices as shown on Fig. 44:\n\n$A_i = B_{i-1},\\ i = 1, 1010$ and\n$A_i = B_{1010-i},\\ i = 1011, 2019$.\n\nThe strategy of Olesya is as follows: she always marks a ver... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof and answer | Olesya | |
02ei | $n$ is a positive integer such that $\frac{n(n+1)}{3}$ is a square. Show that $n$ is a multiple of $3$, and $n+1$ and $\frac{n}{3}$ are squares. | [
"Suppose $\\frac{n(n+1)}{3} = m^2$, so $n(n+1) = 3m^2$. $n$ and $n+1$ are coprime so either\n\n(1) $n = 3a^2$ and $n+1 = b^2$, or\n\n(2) $n = a^2$ and $n+1 = 3b^2$.\n\nBut (2) implies that $3b^2 - 1$ is a square, whereas squares must be $0$ or $1$ mod $3$. So we must have (1)."
] | Brazil | XI OBM | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0f1u | Problem:
$S$ is a sphere with unit radius. $P$ is a plane through the center. For any point $x$ on the sphere, $f(x)$ is the perpendicular distance from $x$ to $P$. Show that if $x$, $y$, $z$ are the ends of three mutually perpendicular radii, then $f(x)^2 + f(y)^2 + f(z)^2 = 1$ (*).
Now let $g(x)$ be any function on... | [] | Soviet Union | ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
09gi | Let $P$ be point inside of the triangle $ABC$. Let $S$ be the perpendicular foot from $P$ to $AB$ and $T$ be the perpendicular foot from $P$ to $BC$. Let line $AC$ intersected to the circumcircle of $STA$ and circumcircle of $STC$ at point $X$ and $Y$, respectively. If $R$ is the intersection of the lines $SY$ and $TX$... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0eba | Problem:
Izračunaj tak $x$, da bo zaporedje $x+2, 3x, x^{2}+5$ geometrijsko in bodo členi zaporedja cela števila. Izračunaj količnik in zapiši zaporedje. | [
"Solution:\n\nNaj bo zaporedje $a_1 = x+2$, $a_2 = 3x$, $a_3 = x^2 + 5$.\n\nKer je zaporedje geometrijsko, velja:\n$$\n\\frac{a_2}{a_1} = \\frac{a_3}{a_2}\n$$\n\nVstavimo izraze:\n$$\n\\frac{3x}{x+2} = \\frac{x^2 + 5}{3x}\n$$\n\nKrižno pomnožimo:\n$$\n(3x)^2 = (x+2)(x^2 + 5)\n$$\n$$\n9x^2 = (x+2)(x^2 + 5)\n$$\n$$\n... | Slovenia | 15. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | x = 2; common ratio = 3/2; sequence: 4, 6, 9 | |
09we | Sabine has a very large collection of shells. She decides to give part of her collection to her sister.
On the first day, she lines up all her shells. She takes the shells that are in a position that is a perfect square (the first, fourth, ninth, sixteenth, etc. shell), and gives them to her sister. On the second day,... | [
"As $n > 1$. The next day, she therefore gives $n - 1$ shells to her sister and is left with $n^2 - n - (n - 1) = (n - 1)^2$ shells, again a perfect square. We see that the numbers of shells that Sabine is left with are alternately a perfect square and a number that is not a perfect square.\n\nLet $d$ be the first ... | Netherlands | Final Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 2020 | |
0fl3 | Problem:
Una sucesión pucelana es una sucesión creciente de dieciséis números impares positivos consecutivos, cuya suma es un cubo perfecto. ¿Cuántas sucesiones pucelanas tienen solamente números de tres cifras? | [
"Solution:\n\nSea la sucesión $n, n+2, \\ldots, n+30$. Entonces la suma es $\\frac{1}{2} 16(2 n+30) = 8(2 n+30)$. Por tanto, es necesario que $2 n+30$ sea un cubo perfecto. Ahora hay que contar el número de tales $n$ que son impares y verifican $101 \\leq n \\leq 969$. Los cubos pares entre 232 y 1968 son 512, 1000... | Spain | XLVI Olimpiada Matemática Española Fase nacional | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3 | |
0j0j | Problem:
Let $g_{1}(x) = \frac{1}{3}(1 + x + x^{2} + \cdots)$ for all values of $x$ for which the right hand side converges. Let $g_{n}(x) = g_{1}(g_{n-1}(x))$ for all integers $n \geq 2$. What is the largest integer $r$ such that $g_{r}(x)$ is defined for some real number $x$? | [
"Solution:\nNotice that the series is geometric with ratio $x$, so it converges if $-1 < x < 1$. Also notice that where $g_{1}(x)$ is defined, it is equal to $\\frac{1}{3(1-x)}$. The image of $g_{1}(x)$ is then the interval $\\left(\\frac{1}{6}, \\infty\\right)$. The image of $g_{2}(x)$ is simply the values of $g_{... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 5 | |
0184 | Let $P$ be a polynomial of degree $2011$. Show that there exists an arithmetic sequence $x_1, x_2, \dots, x_{2011}$ such that
$$
\sum_{k=1}^{2011} P(x_k) = 2011.
$$ | [
"If $P_1(x) = P(x) - 1$ and if for any $2011$ numbers $x_k$ we have $\\sum P_1(x_k) = 0$, then for the same numbers $\\sum P(x_k) = 2011$. So it is sufficient to show that for any polynomial $P(x)$ of degree $2011$ there is an arithmetic sequence $(x_k)$ of $2011$ terms such that $\\sum P(x_k) = 0$.\n\nNow, being a... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof only | null | |
0le2 | Let $ABC$ be a triangle with circumcircle $(O)$, incircle $(I)$, and excircle $(J)$ with respect to vertex $A$. Consider $D, E, F$ to be the tangent points of $(J)$ with $BC, CA, AB$, respectively.
a. Let $L$ be the midpoint of $BC$. The circle with diameter $LJ$ intersects $DE, DF$ again at $K, H$, respectively. Prov... | [
"(a) Denote $(I)$ as the incircle of triangle $ABC$. Let $D', F'$ be the tangent points of $(I)$ with $BC, BA$. $DW$ is the diameter of $(J)$. We have\n$$\n\\frac{AI}{AJ} = \\frac{IF'}{JF} = \\frac{ID'}{JW}\n$$\nhence $A, D', W$ are collinear. Denote $D_1$ as the intersection of $ID'$ and $AD$. We have $JD = JW$, s... | Vietnam | VN IMO Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0b06 | Problem:
A sequence $\{a_n\}_{n \geq 1}$ of positive integers satisfies the recurrence relation $a_{n+1} = n \left\lfloor \frac{a_n}{n} \right\rfloor + 1$ for all integers $n \geq 1$. If $a_4 = 34$, find the sum of all the possible values of $a_1$. | [] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 130 | |
067i | Determine all possible pair of positive integers $x$, $y$ satisfying the equation: $xy(x+y-10)-3x^2-2y^2+21x+16y=60$. | [
"1. The equation is written:\n$$\nxy(x+y-10)-3x^2-2y^2+21x+16y=60 \\\\\n\\Leftrightarrow (y-3)x^2+(y^2-10y+21)x=2y^2-16y+60 \\\\\n\\Leftrightarrow (y-3)x^2+(y-7)(y-3)x=2(y-3)(y-5)+30 \\\\\n\\Leftrightarrow (y-3)x^2+(y-7)(y-3)x-2(y-3)(y-5)=30 \\\\\n\\Leftrightarrow (y-3)[x^2+(y-7)x-2(y-5)]=30 \\\\\n\\Leftrightarrow ... | Greece | SELECTION EXAMINATION | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (4, 6), (5, 5), (3, 8), (7, 4) | |
0i4q | Problem:
Find $\lim_{h \rightarrow 0} \frac{x^{2}-(x+h)^{2}}{h}$. | [
"Solution:\nThis equals $\\lim_{h \\rightarrow 0} \\frac{x^{2}-x^{2}-h^{2}-2 h x}{h} = \\lim_{h \\rightarrow 0} -h - 2x = -2x$.\n\nAlternate Solution: This is the definition of the derivative of $-x^{2}$ with respect to $x$, which is $-2x$."
] | United States | Harvard-MIT Math Tournament | [
"Calculus > Differential Calculus > Derivatives"
] | null | final answer only | -2x | |
0if4 | Problem:
Suppose $0 < a \leq b$ and $4 \nmid m n$. Prove that the number of ways in which an $m \times n$ rectangle can be partitioned into dominoes of type $(a, b)$ is even. | [
"Solution:\n\nIf the rectangle is tileable, it can be partitioned into an odd number of dominoes. Consider the reflection of the partitioned rectangle over one axis. This gives another partition of the rectangle. In fact, it cannot be the same partition, for suppose it were. Then we can pair each domino with its re... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof only | null | |
0hnq | Problem:
The integers from $1$ to $13$ are arranged around several rings such that every number appears once and every ring contains at least one two-digit number. Prove that there exist three one-digit numbers adjacent to one another on one ring. | [
"Solution:\n\nFor any $n \\in S = 1, 2, \\ldots, 13$, define $f(n)$ to be the number immediately clockwise of $n$ on the same ring, where $f(n) = n$ if $n$ lies on a one-element ring. Notice that $f$ is a bijective function, since every number is immediately clockwise of exactly one number. Notice that there are fo... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0h7m | a) Does there exist a set $A$ that has $2016$ pairwise different positive integers and such a property: for every non-empty subset $B \subset A$ that is not equal to $A$ and every non-empty subset $C \subset (A \setminus B)$ the sum of the elements of subset $B$ is not divisible by the sum of the elements of subset $C$... | [
"a)\nLet us firstly choose numbers in such a way: $2, 3, 5, \\ldots, p_{2016}$, where $p_i$ is a prime number in increasing order. Then we will write down all the possible combinations of subsets $B$ and $C$. Its amount is finite. Let there are $N$ such pairs. Successively we will change the elements in a particula... | Ukraine | UkraineMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) Yes; b) Yes | |
0dzg | Let $ABCD$ be a convex quadrilateral such that the triangle $BCD$ is acute and $|AB| = |AD|$. Denote the intersection of the bisector of the angle $\angle CAD$ with the side $CD$ by $K$ and the intersection of the bisector of the angle $\angle BAC$ with the side $BC$ by $L$. Let $K'$ and $L'$ be the orthogonal projecti... | [
"We use the Law of sines for the triangle $BAL$,\n$$\n\\frac{\\sin(\\angle BAL)}{|BL|} = \\frac{\\sin(\\angle ALB)}{|AB|},\n$$\nand for the triangle $CAL$,\n$$\n\\frac{\\sin(\\angle LAC)}{|CL|} = \\frac{\\sin(\\angle CLA)}{|AC|}.\n$$\nSince $\\angle ALB = \\pi - \\angle CLA$, we have $\\sin \\angle ALB = \\sin \\an... | Slovenia | Slovenija 2008 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0b7r | Let $\ell$ be a line, and let $\gamma$ and $\gamma'$ be two circles. The line $\ell$ meets $\gamma$ at points $A$ and $B$, and $\gamma'$ at points $A'$ and $B'$. The tangents to $\gamma$ at $A$ and $B$ meet at point $C$, and the tangents to $\gamma'$ at $A'$ and $B'$ meet at point $C'$. The lines $\ell$ and $CC'$ meet ... | [
"Let the lines $CX$ and $C'X'$ meet at point $Q$. The line $CX$ meets $\\ell$ at $D$, and $\\gamma$ a second time at $Y$; similarly, the line $C'X'$ meets $\\ell$ at $D'$, and $\\gamma'$ a second time at $Y'$. Notice that the cross-ratios $(CDXY)$ and $(C'D'X'Y')$ are both harmonic, for $C$ and $C'$ are the poles o... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry >... | English | proof only | null | |
0hmx | Problem:
Let $ABC$ be an isosceles triangle with $AB = AC$, and let $P$ be a point moving along the side $BC$. Drop the heights $PX$, $PY$ from $P$ to the sides $AB$ and $AC$. Prove that the sum $PX + PY$ remains constant as $P$ moves.
 | [
"Solution:\n\nWe can cut triangle $ABC$ into triangles $ABP$ and $APC$, so\n$$\n\\text{Area } ABC = \\text{Area } ABP + \\text{Area } APC = \\frac{AB \\cdot PX}{2} + \\frac{AC \\cdot PY}{2}.\n$$\nBut sides $AB$ and $AC$ have the same length, so this can be rewritten as\n$$\n\\frac{AB \\cdot PX}{2} + \\frac{AB \\cdo... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
01g6 | A positive integer $N$ is called a *lighthouse number* if it has the following property. Any positive integer not exceeding $N$ which is relatively prime to it has at most two prime factors (possibly equal). Determine whether the lighthouse numbers are infinitely or finitely many and, in the latter case, find the great... | [
"**Answer:** The greatest lighthouse number is 1260.\n\nDenote by $p_n$ the $n$-th prime number. We first set out to prove that, for $n \\ge 5$, the inequality\n$$\np_{n+1}^3 \\le p_1 \\cdots p_n, \\quad n \\ge 5, \\qquad (\\text{Eq-8})\n$$\nholds. As is readily verified by computation, the inequality holds for $n ... | Baltic Way | Baltic Way 2019 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | Finitely many; the greatest lighthouse number is 1260. | |
0a6q | Problem:
Let $a, b, c$ be positive real numbers satisfying $a b c = 1$. Determine the smallest possible value of
$$
\frac{a^{2} + 2025}{a^{3}(b + c)} + \frac{b^{2} + 2025}{b^{3}(c + a)} + \frac{c^{2} + 2025}{c^{3}(a + b)}
$$ | [
"Solution:\nNote,\n$$\n\\begin{array}{r l r}\n\\frac{1}{a^{3}(b+c)}+\\frac{1}{b^{3}(a+c)}+\\frac{1}{c^{3}(a+b)} = \\frac{\\left(\\frac{1}{a}\\right)^{2}}{a(b+c)} + \\frac{\\left(\\frac{1}{b}\\right)^{2}}{b(a+c)} + \\frac{\\left(\\frac{1}{c}\\right)^{2}}{c(a+b)} \\\\\n& & \\geq \\frac{\\left(\\frac{1}{a} + \\frac{1}... | New Zealand | NZMO Round Two | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 3039 | |
01ai | Let $ABC$ be a triangle, and let $X$, $Y$, $Z$ be points on $BC$, $CA$, $AB$, respectively. Suppose that $AX$, $BY$ and $CZ$ intersect in a point $P$. Prove that
$$
\frac{AP}{AX} + \frac{BP}{BY} + \frac{CP}{CZ} = 2.
$$ | [
"Assign weights $p, q, r$ to the vertices $A, B, C$, respectively, in such a way that the centre of mass is $P$. Draw the line $L$ through $P$, parallel to $BC$. Let $d$ denote the distance of $A$ to $L$, and let $h$ denote the distance of $A$ to $BC$. Since the triangle will be in static equilibrium when balanced ... | Baltic Way | Baltic Way 2013 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | null | proof only | null | |
0bt3 | Let $K$ be a finite field of cardinality $q \ge 3$, and let $S$ be the set of degree $q-2$ polynomials in $K[X]$ whose coefficients are all non-zero and pairwise distinct. Determine the number of polynomials in $S$ having $q-2$ distinct roots in $K$.
Mariean Andronache | [
"The required number is $(q-1)\\varphi(q-1)$, where $\\varphi$ is Euler's totient function. Since every polynomial in $S$ is associated in divisibility with a unique polynomial $f$ in $S$ such that $f(0) = 1$, it is sufficient to count the polynomials of the form $f = 1 + a_1X + \\dots + a_{q-2}X^{q-2}$ satisfying ... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Number-Theoretic Functi... | English | proof and answer | (q-1)φ(q-1) | |
0g0w | Problem:
Sei $a_{n}$ eine Folge natürlicher Zahlen definiert durch $a_{1}=m$ und $a_{n}=a_{n-1}^{2}+1$ für $n>1$. Ein Paar $(a_{k}, a_{l})$ nennen wir interessant, falls
(i) $0<l-k<2016$,
(ii) $a_{k}$ teilt $a_{l}$.
Zeige, dass ein $m$ existiert, sodass die Folge $a_{n}$ kein interessantes Paar enthält. | [
"Solution:\n\nDie zweite Bedingung ist genau dann erfüllt, wenn $a_{l}$ kongruent zu $0$ modulo $a_{k}$ ist. Schauen wir also die möglichen Resultate von $a_{l}$ modulo $a_{k}$ an. Für $l=k+1, k+2$ berechnen wir:\n$$\n\\begin{array}{rr}\na_{k+1}=a_{k}^{2}+1 \\equiv 1 & (a_{k}) \\\\\na_{k+2}=a_{k+1}^{2}+1 \\equiv 1+... | Switzerland | SMO - Finalrunde | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | null | proof only | null | |
016x | Let $1 < r < 2$ be a rational number. Prove that there exist three integers $k, m, n$ such that
$$
r = \frac{k^3 + m^3}{k^3 + n^3}
$$ | [
"Let $r = \\frac{p}{q}$ with $1 \\leq q \\leq p \\leq 2q$. First notice that for any $k, m$ and $n$,\n$$\n\\frac{k^3 + m^3}{k^3 + n^3} = \\frac{k+m}{k+n} \\cdot \\frac{k^2 - km + m^2}{k^2 - kn + n^2}\n$$\nWe have that $k^2 - km + m^2 = k^2 - kn + n^2$ if and only if $m^2 - n^2 = k(m - n)$. Now assume $k = m + n$ an... | Baltic Way | BALTIC WAY | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof only | null | |
01mi | Do there exist integers $x$ and $y$ satisfying the equality $2x^3 + y^3 = 2011$? | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | No | |
0fiq | Problem:
Prueba que la longitud de los catetos de un triángulo rectángulo isósceles es siempre igual a la suma de los radios de sus circunferencias inscrita y circunscrita. | [
"Solution:\n\nSea $ABC$ un triángulo rectángulo e isósceles con ángulo recto en el vértice $C$. Sea $O$ el centro de la circunferencia inscrita, la cual consideramos tangente a la hipotenusa $AB$ en su punto medio $M$ y al lado $AC$ en el punto $P$. El radio de esta circunferencia es $r=\\o... | Spain | XXXV Olimpiada Matemática Española | [
"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"
] | null | proof only | null | |
05em | Problem:
Trouver tous les entiers $p$ tels que $p$, $p+2$ et $p+4$ soient tous les 3 premiers.
Un nombre premier est un entier $\geqslant 2$ qui n'est divisible que par 1 et lui-même. | [
"Solution:\n\nTout d'abord, on peut s'attendre à ce qu'il n'y en ait que très peu. On souhaite obtenir des informations sur ces nombres premiers.\n\nNotons qu'un nombre premier divisible par $3$ est a fortiori égal à $3$.\n\nOn considère 3 cas en fonction du reste de la division de $p$ par $3$ (modulo $3$) :\n\n- $... | France | ENVOi 3 : ARITHMÉTIQUE | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 3 | |
0ebp | Problem:
Poišči vse polinome $p$ lihe stopnje $z$ realnimi koeficienti, za katere velja
$$
p(p(x)) \leq (p(x))^{3}
$$
za vse $x \in \mathbb{R}$ in ki imajo koeficient pri $x^{2}$ enak 0. | [
"Solution:\n\nPišimo $p(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + \\cdots + a_{0}$, kjer je $a_{n} \\neq 0$. Vodilni člen polinoma $p(p(x))$ je enak $a_{n} (a_{n} x^{n})^{n} = a_{n}^{n+1} x^{n^{2}}$, vodilni člen polinoma $p(x)^{3}$ pa $(a_{n} x^{n})^{3} = a_{n}^{3} x^{3n}$, oba sta lihe stopnje. Ker je polinom $p(x)^{3... | Slovenia | 59. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | p(x) = x^3 + b with b ≤ 0 | |
0382 | Problem:
After a volleyball tournament (every two teams played exactly once) with $n$ teams it turned out that for any two teams $A$ and $B$, such that $B$ wins over $A$, there exist positive integer $t$ and teams $C_{1}, C_{2}, \ldots, C_{t}$, such that $A$ wins over $C_{1}, C_{1}$ wins over $C_{2}, \ldots, C_{t}$ wi... | [
"Solution:\n\nWe first show that there exist teams $A, B$ and $C$, such that $A$ wins over $B, B$ wins over $C$ and $C$ wins over $A$. Suppose the contrary and take the shortest cycle of $m \\geq 4$ teams $A, B, C_{1}, \\ldots, C_{t}$ (i.e. $t \\geq 2$), such that $A$ wins over $C_{1}, C_{1}$ wins over $C_{2}, \\ld... | Bulgaria | Team selection test for 23. BMO | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0iqr | Problem:
Evaluate the sum
$$
\cos \left(\frac{2 \pi}{18}\right)+\cos \left(\frac{4 \pi}{18}\right)+\cdots+\cos \left(\frac{34 \pi}{18}\right)
$$ | [
"Solution:\nIf $k < 18$, then we can pair $\\cos \\left(\\frac{k \\pi}{18}\\right)$ with $\\cos \\left(\\frac{(18-k) \\pi}{18}\\right)$, and these two terms sum to $0$. If $k > 18$, then the pair $\\cos \\left(\\frac{k \\pi}{18}\\right)$ and $\\cos \\left(\\frac{(36-k) \\pi}{18}\\right)$ also sums to $0$. So, the o... | United States | 1st Annual Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | -1 | |
08ip | Problem:
Let $a, b, c$ be positive numbers such that $a b + b c + c a = 3$. Prove that
$$
a + b + c \geq a b c + 2
$$
Problem:
Fie $a$, $b$, $c$ numere pozitive astfel că $a b + b c + c a = 3$. Demonstrați că $a + b + c \geq a b c + 2$. | [
"Solution:\nEliminating $c$ gives\n$$\na + b + c - a b c = a + b + (1 - a b) c = a + b + \\frac{(1 - a b)(3 - a b)}{a + b}\n$$\nPut $x = \\sqrt{a b}$. Then $a + b \\geq 2x$, and since $1 < x^{2} < 3$, $\\frac{(1 - a b)(3 - a b)}{a + b} \\geq \\frac{(1 - x^{2})(3 - x^{2})}{2x}$.\nIt then suffices to prove that\n$$\n... | JBMO | 7th JBMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0bqf | Let $f : \mathbb{R} \to \mathbb{R}$ be a function so that for every $x, y \in \mathbb{R}$, $x < y$, there exists $z \in (x, y)$ so that
$$
(y - x)f(z) \le (y - z)f(x) + (z - x)f(y).
$$
a) Give an example of a non-convex function $f$ with the given property.
b) Prove that a continuous function with the given property i... | [] | Romania | 67th NMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | Example for part a: f equals 1 on rational numbers and 0 on irrational numbers. For part b: any continuous function with the stated property is convex. | |
0cri | A convex heptagon is given. One calculates the cosines of four arbitrary its angles, the sines of three remaining angles, and then finds the sum of these 7 numbers. It happens that this sum is independent of the choice of the four angles. Prove that this heptagon has four equal angles. | [
"Consider one of the sums described in the problem. Now, swap the arguments of one sine and one cosine (call these arguments $\\alpha$ and $\\beta$, respectively); the sum changes by\n\n$$(\\sin \\beta + \\cos \\alpha) - (\\sin \\alpha + \\cos \\beta) = \\sqrt{2}(\\sin(\\beta - \\pi/4) - \\sin(\\alpha - \\pi/4)).$$... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
080e | Problem:
In un tetraedro regolare di vertici $A, B, C, D$, indichiamo con $P$ e $Q$ i centri delle due facce che hanno in comune lo spigolo $A B$. Qual è il rapporto tra il volume del tetraedro iniziale e quello del tetraedro che ha per vertici i punti $A, B, P$ e $Q$ ?
Problem:
Sia $n$ il più piccolo intero positi... | [] | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinat... | null | final answer only | ratio = 3*sqrt(3); n = 315 | |
02g8 | Let $\sigma(n)$ be the sum of all positive divisors of $n$, where $n$ is a positive integer (for instance, $\sigma(6) = 12$ and $\sigma(11) = 12$). We say that $n$ is *almost perfect* if $\sigma(n) = 2n - 1$ (for instance, 4 is almost perfect since $\sigma(4) = 7$). Let $n \mod k$ be the remainder of the division of $n... | [
"$$\n\\begin{align*}\ns(n) = s(n-1) &\\iff \\sum_{1 \\le k \\le n} n \\bmod k = \\sum_{1 \\le k \\le n-1} (n-1) \\bmod k \\\\\n&\\iff \\sum_{1 \\le k \\le n} n \\bmod k = \\sum_{1 \\le k \\le n-1} ((n \\bmod k) - 1) + \\sum_{\\substack{1 \\le k \\le n-1 \\\\ k \\nmid n}} k \\\\\n&\\iff \\sum_{1 \\le k \\le n-1} 1 =... | Brazil | XXII OBM | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0bgu | Problem:
Fie $\mathbb{Q}_{>0}$ mulțimea numerelor raționale strict pozitive. Fie $f: \mathbb{Q}_{>0} \rightarrow \mathbb{R}$ o funcție ce îndeplinește următoarele trei condiții:
(i) Pentru orice $x, y \in \mathbb{Q}_{>0}$ avem $f(x) f(y) \geq f(x y)$,
(ii) Pentru orice $x, y \in \mathbb{Q}_{>0}$ avem $f(x+y) \geq f(... | [] | Romania | Romania Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | f(x) = x for all positive rational x | |
05zr | Problem:
Déterminer tous les entiers $n$ ayant la propriété suivante : si l'on pose $a_{k} = \operatorname{ppcm}(k, k+1, \ldots, k+n-1)$, alors la suite $(a_{k})$ est croissante. | [
"Solution:\n\nMontrons que les naturels ayant cette propriété sont exactement $1$ et $2$.\n\nDéjà, si $n=1$, $a_{k}=k$ est une suite croissante.\n\nEt si $n=2$, puisque deux entiers consécutifs sont premiers entre eux, $a_{k}=k(k+1)$ est une suite croissante également.\n\nSi $n=3$, on remarque que $a_{5}=210 > a_{6... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 1 and 2 | |
00id | Determine all triples of real numbers $(x, y, z)$, such that the equation
$$
4x^4 - x^2 (4y^4 + 4z^4 - 1) - 2xyz + y^8 + 2y^4z^4 + y^2z^2 + z^8 = 0
$$
holds. | [
"We first note that\n$$\n\\begin{aligned}\n& 4x^4 - x^2(4y^4 + 4z^4 - 1) - 2xyz + y^8 + 2y^4z^4 + y^2z^2 + z^8 \\\\\n& = (4x^4 + y^8 + z^8 - 4x^2y^4 - 4x^2z^4 + 2y^4z^4) + (x^2 - 2xyz + y^2z^2) \\\\\n& = (2x^2 - y^4 - z^4)^2 + (x - yz)^2.\n\\end{aligned}\n$$\nThe given equation is therefore equivalent to\n$$\n(2x^2... | Austria | Austria 2010 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | { (t^2, t, t) | t in R } ∪ { (-t^2, t, -t) | t in R } | |
0dl6 | Let $a > b$ be odd positive integers and $k$ be a positive integer such that $a^{1001} + k$ and $b^{1001} + k$ are powers of $2$. Prove that $a > b^{1001} + k$. | [] | Saudi Arabia | Saudi Booklet | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
01lb | Let $I$ be the incenter of the acute-angled non-isosceles triangle $ABC$. Let the incircle touch the side $AB$ at point $Q$. Point $T$ is marked on the side $AB$ so that $IT \parallel CQ$. The line through $T$ touches the incircle at point $K$ (different from $Q$) and meets the lines $CA$ and $CB$ at points $L$ and $N$... | [
"Let $x = \\angle CQI$, $y = \\angle QCI$, $2\\alpha = \\angle BAC$, $2\\beta = \\angle ABC$, $2\\gamma = \\angle ACB$. Let $r$ be the inradius of the triangle $ABC$. It is easy to see that\n\n\n\n$$\n\\angle CIQ = 360^\\circ - 90^\\circ - 2\\alpha - \\gamma =\n$$\n$$\n= 270^\\circ - (\\alp... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Method... | English | proof only | null | |
07bs | We call a spatial polygon (a polygon in the three dimensional space) **Latticelike** if its edges are parallel to the coordinate axes.
a) For an arbitrary Latticelike polygon, any two consecutive edges form a right angle that lies in either $xy$, $yz$ or $zx$ plane. Prove that the number of ... | [
"First, we set some notations.\n\n| | Description |\n|---|---------------------------------------------|\n| A | Number of sides parallel to the $x$ axis |\n| B | Number of sides parallel to the $y$ axis |\n| C | Number of sides parallel to the $z$ axis |\n| X | Number of a... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry >... | null | proof only | null | |
0jjy | Problem:
Compute the number of ordered quintuples of nonnegative integers $\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right)$ such that $0 \leq a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \leq 7$ and $5$ divides $2^{a_{1}}+2^{a_{2}}+2^{a_{3}}+2^{a_{4}}+2^{a_{5}}$. | [
"Solution:\n\nLet $f(n)$ denote the number of $n$-tuples $\\left(a_{1}, \\ldots, a_{n}\\right)$ such that $0 \\leq a_{1}, \\ldots, a_{n} \\leq 7$ and $5 \\mid 2^{a_{1}}+\\ldots+2^{a_{n}}$. To compute $f(n+1)$ from $f(n)$, we note that given any $n$-tuple $\\left(a_{1}, \\ldots, a_{n}\\right)$ such that $0 \\leq a_{... | United States | HMMT 2014 | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 6528 | |
0001 | Sean $a$, $b$ y $c$ números reales positivos. Demostrar que
$$
\left(\frac{a}{b+c} + \frac{1}{2}\right) \left(\frac{b}{c+a} + \frac{1}{2}\right) \left(\frac{c}{a+b} + \frac{1}{2}\right) \ge 1
$$ | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | español | proof only | null | |
08dg | Problem:
Sia $n$ un intero maggiore di $2$. Si vogliono colorare di rosso esattamente $n+1$ dei numeri $1, 2, \cdots, 2n-1, 2n$ in modo tale che non ci siano tre numeri distinti $x, y, z$ colorati di rosso che soddisfano l'uguaglianza $x+y=z$. Dimostrare che esiste uno e un solo modo di scegliere i numeri da colorare ... | [
"Solution:\n\nSia $A$ l'insieme dei numeri da colorare di rosso. Se $A = \\{n, n+1, \\ldots, 2n\\}$ non ce ne sono tre per i quali $x+y=z$, in quanto per ogni $x, y \\in A$ con $x \\neq y$ si ha $x+y > 2n$.\n\nRimane dunque da dimostrare che questa è l'unica scelta possibile per l'insieme $A$.\n\nDimostriamo l'enun... | Italy | XXXV Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | {n, n+1, ..., 2n} | |
0i52 | Problem:
A sequence $s_{0}, s_{1}, s_{2}, s_{3}, \ldots$ is defined by $s_{0}=s_{1}=1$ and, for every positive integer $n$, $s_{2 n}=s_{n}$, $s_{4 n+1}=s_{2 n+1}$, $s_{4 n-1}=s_{2 n-1}+s_{2 n-1}^{2} / s_{n-1}$. What is the value of $s_{1000}$? | [
"Solution:\nSome experimentation with small values may suggest that $s_{n}=k!$, where $k$ is the number of ones in the binary representation of $n$, and this formula is in fact provable by a straightforward induction. Since $1000_{10}=1111101000_{2}$, with six ones, $s_{1000}=6!=720$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | final answer only | 720 | |
0dmw | Problem:
Нека је $n$ непаран природан број такав да су бројеви $\varphi(n)$ и $\varphi(n+1)$ степени броја два ($\varphi(n)$ је број природних бројева не већих од $n$ и узајамно простих са $n$). Доказати да је $n+1$ степен броја два или је $n=5$. | [
"Solution:\n\nАко је $n=\\prod_{i=1}^{k} p_{i}^{r_{k}}$ канонска факторизација $n$, важи $\\varphi(n)=\\prod_{i=1}^{k} p_{i}^{r_{k}-1}\\left(p_{i}-1\\right)$, па пошто $n$ нема других простих чинилаца осим двојке, мора бити $a_{i}=1$ и $p_{i}-1=2^{b_{i}}$ за свако $i$ и неке $b_{i}$. Како $2^{b_{i}}+1$ може бити пр... | Serbia | СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0i2s | Problem:
There are $3,999,999$ cities in Antarctica, and some pairs of them are connected by roads. It is known that, given any two cities, there is a sequence of roads leading from one to the other. Prove that the cities can be divided into $1999$ groups (of $2001$ cities each) such that, given any two cities in the ... | [
"Solution:\n\nFirst, we provide some relevant graph-theoretic background. Any finite, connected graph can be turned into a tree (a connected graph without cycles) by removing some edges. Proof: If our graph has a cycle, any edge of that cycle can be removed without disconnecting the graph. So remove this edge, leav... | United States | Berkeley Math Circle Monthly Contest #7 | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0j2k | Problem:
Acute triangle $A B C$ has $\angle B A C < 45^{\circ}$. Point $D$ lies in the interior of triangle $A B C$ so that $B D = C D$ and $\angle B D C = 4 \angle B A C$. Point $E$ is the reflection of $C$ across line $A B$, and point $F$ is the reflection of $B$ across line $A C$. Prove that lines $A D$ and $E F$ a... | [
"Solution:\n\nBegin by reflecting $C$ over $A F$ to point $G$, as shown in the diagram.\n\n\n\nAs usual we denote $\\alpha = \\angle B A C$, $\\beta = \\angle A B C$, and $\\gamma = \\angle A C B$.\n\nIn order to prove that $A D \\perp E F$, we will show that $\\triangle A D C \\sim \\trian... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
07zh | Problem:
Alberto vuole organizzare per questa sera una partita di poker.
Egli sa che Bruno e Barbara si recano insieme in palestra una sera su tre, e che Carla, Corrado, Dario e Davide sono impegnati una sera su due (ma non necessariamente negli stessi giorni).
Inoltre, sa che Dario non vuole giocare con Davide poiché... | [
"Solution:\n\nVi sono $\\frac{2}{3}$ di probabilità che Bruno e Barbara possano giocare; in tal caso basta che sia disponibile anche uno solo degli altri amici. Che uno fissato degli amici non possa giocare è un evento di probabilità $\\frac{1}{2}$, e dunque che nessuno dei quattro possa giocare è un evento di prob... | Italy | GARA NAZIONALE di MATEMATICA | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 11/16 | |
0eq7 | It is said that you can only fold a piece of paper in half 7 times. Harold folds a sheet of paper in half 5 times and then makes a hole in the folded paper. How many holes does the sheet of paper have after it is unfolded again?
(A) 32
(B) 36
(C) 81
(D) 50
(E) 64 | [
"Each fold doubles the number of layers that will be pierced. There will be $2^5$ layers and therefore $2^5 = 32$ holes."
] | South Africa | South African Mathematics Olympiad | [
"Math Word Problems"
] | English | MCQ | A | |
0iar | Let $ABCD$ be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least $60^{\circ}$. Prove that
$$
\frac{1}{3} |AB^3 - AD^3| \leq |BC^3 - CD^3| \leq 3|AB^3 - AD^3|.
$$
When does equality hold? | [
"By symmetry, we only need to prove the first inequality.\nBecause quadrilateral $ABCD$ has an incircle, we have $AB + CD = BC + AD$, or $AB - AD = BC - CD$. It suffices to prove that\n$$\n\\frac{1}{3}(AB^2 + AB \\cdot AD + AD^2) \\leq BC^2 + BC \\cdot CD + CD^2.\n$$\nBy the given condition, $60^{\\circ} \\leq \\an... | United States | USA IMO | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | Equality holds if and only if AB = AD (equivalently BC = CD), i.e., ABCD is a tangential kite. | |
0jq0 | Problem:
Chords $AB$ and $CD$ of a circle are perpendicular and intersect at a point $P$. If $AP = 6$, $BP = 12$, and $CD = 22$, find the area of the circle. | [
"Solution:\n\nLet $O$ be the center of the circle and let $M$ be the midpoint of segment $AB$ and let $N$ be the midpoint of segment $CD$. Since quadrilateral $OMPN$ is a rectangle we have that $ON = MP = AM - AP = 3$ so\n$$\nOC = \\sqrt{ON^2 + NC^2} = \\sqrt{9 + 121} = \\sqrt{130}\n$$\nHence the desired area is $1... | United States | HMMT November 2015 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 130π | |
03ry | The circles $O_1$ and $O_2$ meet at points $A$ and $B$. The line $DC$ passes through $O_1$, intersects the circle $O_1$ at $D$ and is a tangent to the circle $O_2$ at $C$. Also, $CA$ is a tangent to the circle $O_1$ at $A$. The secant $AE$ of the circle $O_1$ is perpendicular to $DC$. $AF$ is perpendicular to $DE$ and ... | [
"Let $AE$ intersect $DC$ at point $H$,\n\n\nand $AF$ intersect $BD$ at point $G$. Join $AB$, $BC$, $BH$, $BE$, $CE$ and $GH$. By symmetry, $CE$ is also a tangent line of the circle $O_1$ and $H$ is the midpoint of $AE$.\n\nSince $AF \\perp DE$, we have\n$$\n\\angle AGB = \\frac{\\pi}{2} - \... | China | China Western Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ax2 | Problem:
Mr. Curry is playing basketball. Suppose that, on the average, he makes two successful three-point shots out of every five attempts. What is the probability that he will be able to make at least two successful 3-point shots in a row out of four attempts?
Answer: $\frac{44}{125}$ | [
"Solution:\n\nThere are only a few possibilities to the contrary. Note that if Mr. Curry shoots at least three, for certain, two of them will be consecutive. Hence, we need consider only the cases wherein he makes two shots or less.\n\nNo shots made: The probability of this happening is simply $\\left(\\frac{3}{5}\... | Philippines | Philippine Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Combinations",
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 44/125 | |
06o1 | Let $x$, $y$, $z$ be real numbers such that $x + y + z \neq 0$. Find the minimum value of
$$
\frac{|x| + |x + 4y| + |y + 7z| + 2|z|}{|x + y + z|}
$$ | [
"By the triangle inequality, we have\n$$\n\\begin{aligned}\n\\frac{|x| + |x + 4y| + |y + 7z| + 2|z|}{|x + y + z|} &\\ge \\frac{|x| + \\frac{4}{11}|x + 4y| + \\frac{1}{11}|-y - 7z| + 2|z|}{|x + y + z|} \\\\\n&\\ge \\frac{|x + \\frac{4}{11}(x + 4y) + \\frac{1}{11}(-y - 7z) + 2z|}{|x + y + z|} \\\\\n&= \\frac{15}{11}.... | Hong Kong | IMO HK TST | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof and answer | 15/11 | |
0auf | Problem:
Trapezoid $ABCD$ has right angles at $C$ and $D$, and $AD > BC$. Let $E$ and $F$ be the points on $AD$ and $AB$, respectively, such that $\angle BED$ and $\angle DFA$ are right angles. Let $G$ be the point of intersection of segments $BE$ and $DF$. If $\angle CED = 58^\circ$ and $\angle FDE = 41^\circ$, what ... | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 17° | |
000x | Sea la sucesión \{$a_n$\} definida de la siguiente manera:
$$
a_1 = 1
$$
$$
a_2 = 3
$$
$$
a_{n+2} = 2a_{n+1}a_n + 1 \text{ ; para todo } n \ge 1
$$
Probar que la máxima potencia de 2 que divide a $a_{4006} - a_{4005}$ es $2^{2003}$. | [] | Argentina | XIV Olimpiada Matemática de Países del Cono Sur | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | español | proof and answer | 2^{2003} | |
0huy | Problem:
The Fibonacci sequence $F_{1}, F_{2}, F_{3}, \ldots$ is defined by $F_{1}=F_{2}=1$ and $F_{n+2}=F_{n+1}+F_{n}$. Find the least positive integer $t$ such that for all $n>0$, $F_{n}=F_{n+t}$. | [
"Solution:\n\n60 ."
] | United States | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | final answer only | 60 | |
0cwd | Let $ABC$ be a triangle. A point $D$ is chosen on the external angle bisector of angle $ABC$ lying inside the angle $ABC$. Assume that $\angle BCD = 60^\circ$ and $CD = 2AB$. Let $M$ be the midpoint of $BD$. Prove that the triangle $AMC$ is isosceles.
(A. Kuznetsov) | [
"**Первое решение.** Обозначим $\\angle CBM = \\alpha$. Поскольку $BM$ — внешняя биссектриса угла $ABC$, то $\\angle ABM = 180^\\circ - \\alpha$ и $\\alpha < 90^\\circ$ (см. рис. 6).\n\nОпустим из точки $D$ перпендикуляр $DH$ на прямую $BC$. Так как в треугольнике $BCD$ углы при вершинах $B$ и $C$ острые, точка $H$... | Russia | Regional round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English; Russian | proof only | null | |
05ub | Problem:
Soit $ABCD$ un quadrilatère convexe dont les angles en $B$ et en $D$ sont obtus, et dont les angles en $A$ et en $C$ sont égaux l'un à l'autre. Soit $E$ et $F$ les symétriques de $A$ par rapport à $(BC)$ et $(CD)$. Soit $K$ et $L$ les points d'intersection de $(BD)$ avec $(AE)$ et $(AF)$. Démontrer que les ce... | [
"Solution:\n\nLa manière la plus pratique de construire le point $A$ est d'en construire tout d'abord son symétrique $A'$ par rapport à $(BD)$, puisque $A'$ est cocyclique avec $B$, $C$ et $D$. Ce faisant, on constate que $A'$ semble appartenir aux deux cercles circonscrits aux triangles $BEK$ et $DFL$, et une chas... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0l3b | Problem:
Let $\mathbb{Q}$ be the set of rational numbers. Given a rational number $a \neq 0$, find, with proof, all functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$ satisfying the equation
$$
f(f(x)+a y)=a f(y)+x
$$
for all $x, y \in \mathbb{Q}$. | [
"Solution:\nLet $P(x, y)$ denote the functional equation. From $P(x, 0)$, we have $f(f(x))=x+a f(0)$. Thus, the tripling trick gives $f(x+a f(0))=f(f(f(x)))=f(x)+a f(0)$.\nNow, here is the main idea: $P(f(x), y)$ gives\n$$\n\\begin{aligned}\nf(f(f(x))+a y) & =a f(y)+f(x) \\\\\nf(x+a f(0)+a y) & =f(x)+a f(y) \\\\\nf... | United States | HMMT February 2024 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | For any nonzero rational a: the solutions are f(x) = x and f(x) = −x. Additionally, if a = 2, all functions f(x) = x + c for any rational c also satisfy the equation. | |
05eh | Problem:
Soit $n \geqslant 1$ un entier tel que le quotient de $2^{n}$ par $n$ est une puissance de $2$. Montrer que $n$ est une puissance de $2$. | [
"Solution:\n\nPar hypothèse, $2^{n} / n$ est une puissance de $2$ (ici, on parle bien de quotient, et non pas de quotient dans la division euclidienne). Il existe donc un entier $k \\geqslant 0$ tel que $2^{n} = 2^{k} n$. Ainsi, $n = 2^{n - k}$, de sorte que $n$ est bien une puissance de $2$."
] | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES, ENVOI No. 3 | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
03m5 | Let $P(x)$ and $Q(x)$ be polynomials with integer coefficients. Let $a_n = n! + n$. Show that if $P(a_n)/Q(a_n)$ is an integer for every $n$, then $P(n)/Q(n)$ is an integer for every integer $n$ such that $Q(n) \neq 0$. | [
"Imagine dividing $P(x)$ by $Q(x)$. We find that\n$$\n\\frac{P(x)}{Q(x)} = A(x) + \\frac{R(x)}{Q(x)},\n$$\nwhere $A(x)$ and $R(x)$ are polynomials with rational coefficients, and $R(x)$ is either identically $0$ or has degree less than the degree of $Q(x)$.\nBy bringing the coefficients of $A(x)$ to their least com... | Canada | Kanada 2010 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0l9p | Find all triples of natural numbers $(x, y, n)$ satisfying the relation
$$
\frac{x! + y!}{n!} = 3^n
$$
(with the convention $0! = 1$). | [
"The relation in the problem can be written in the form\n$$\nx! + y! = 3^n n! \\quad (1)\n$$\nSuppose that $(x, y, n)$ is a triple of natural numbers satisfying (1).\nIt is easily seen that $n \\ge 1$ and w.l.g. we can suppose that $x \\le y$. We must now consider the following cases.\n\n1) 1st case: $x \\le n$\n$$... | Vietnam | Vietnam Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | [(0, 2, 1), (2, 0, 1), (1, 2, 1), (2, 1, 1)] | |
0461 | Let $x_1, x_2, \dots, x_{11}$ be nonnegative real numbers which add up to one. For $i = 1, 2, \dots, 11$, let
$$
y_i = \begin{cases} x_i + x_{i+1}, & \text{if } i \text{ is odd,} \\ x_i + x_{i+1} + x_{i+2}, & \text{if } i \text{ is even,} \end{cases}
$$
where $x_{12} = x_1$. Set $F(x_1, x_2, \dots, x_{11}) = y_1y_2\cdo... | [
"*Proof.* The function $F$, viewed as a continuous function with respect to $x_1, x_2, \\cdots, x_{11}$, has a maximum value on the bounded closed set\n$$\n\\Omega = \\{(x_1, x_2, \\cdots, x_{11}) \\in (\\mathbb{R}_{\\ge 0})^{11} \\mid x_1 + x_2 + \\cdots + x_{11} = 1\\}.\n$$\nAssume that $F$ has a maximum value at... | China | 2022 CGMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof only | null | |
0a8r | Problem:
27 balls, labelled by numbers from 1 to 27, are in a red, blue or yellow bowl. Find the possible numbers of balls in the red bowl, if the averages of the labels in the red, blue, and yellow bowl are 15, 3 and 18, respectively. | [
"Solution:\n\nLet $R$, $B$, and $Y$, respectively, be the numbers of balls in the red, blue, and yellow bowl. The mean value condition implies $B \\leq 5$ (there are at most two balls with a number $<3$, so there can be at most two balls with a number $>3$).\n\n$R$, $B$ and $Y$ satisfy the equations\n$$\n\\begin{al... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 18 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 11, 16, 21 | |
09k6 | Let $M$ be the midpoint of side $AC$ of triangle $ABC$. Let $P$ be a point inside triangle $ABC$ such that $\angle BAP = \angle BCP$. The line $CP$ intersects side $AB$ at point $R$. Let $Q$ be the foot of the perpendicular drawn from $B$ to line $CP$. If $Q$ lies inside triangle $ABC$ and $QM = \frac{PC}{2}$ then prov... | [
"\nLet $K$, $N$, and $L$ be the midpoints of segments $BP$, $BC$, and $PC$, respectively. Since $AB$ is parallel to $MN$ and $AP$ is parallel to $ML$, we have $\\angle BAP = \\angle NML$.\n\nConsidering the parallelogram $KNCL$, $\\angle BCP = \\angle BCL = \\angle NKL$. Moreover, since $\\... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0jfj | Problem:
Let $S$ be the set of integers of the form $2^{x} + 2^{y} + 2^{z}$, where $x, y, z$ are pairwise distinct non-negative integers. Determine the 100th smallest element of $S$. | [
"Solution:\n$S$ is the set of positive integers with exactly three ones in its binary representation. The number of such integers with at most $d$ total bits is $\\binom{d}{3}$, and noting that $\\binom{9}{3} = 84$ and $\\binom{10}{3} = 120$, we want the 16th smallest integer of the form $2^{9} + 2^{x} + 2^{y}$, wh... | United States | HMMT | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 577 | |
0ip7 | Problem:
Determine all pairs $(a, b)$ of real numbers such that $10, a, b, a b$ is an arithmetic progression. | [
"Solution:\n\nThe answer is $(4, -2)$ and $\\left(\\frac{5}{2}, -5\\right)$.\n\nSince $10, a, b$ is an arithmetic progression, we have\n$$\na = \\frac{1}{2}(10 + b).\n$$\nAlso, we have $a + a b = 2 b$, and so\n$$\na(1 + b) = 2b.\n$$\nSubstituting the expression for $a$ gives\n$$\n(10 + b)(1 + b) = 4b.\n$$\nSolving ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (4, -2) and (5/2, -5) | |
01tl | An array of $N$ ($N \ge 3$) integer numbers is written on a blackboard. Per move one change the array using the following rule: let $s$ be the arithmetic mean of all numbers of the current array, then each number which is less than or equal to $s-1$ is increased by 1 and each number which is greater than or equal to $s... | [
"Let $S_i$ denote the arithmetic mean of all $n$ numbers written on the blackboard after $i$ moves, $M_i$ and $m_i$ be the greatest and the smallest of the written numbers.\nShow that if $M_i - m_i \\ge 2$ then\n$$\nM_{i+1} - m_{i+1} < M_i - m_i. \\quad (1)\n$$\nIndeed, since $M_i - m_i \\ge 2$, we see that the seg... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0gmy | For a positive integer $n$, find in how many different ways a $2 \times n$ rectangle can be partitioned into rectangles with integer side lengths. | [] | Turkey | Team Selection Examination for the International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2·3^{n−1} | |
09ob | Consider the equation $(x^2 - 3)^2 = x + 3$.
(1) Find the integer solutions of the equation.
(2) Find the product of the non-integer solutions of the equation. | [] | Mongolia | MMO2025 Round 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | Integer solutions: 1, -2; product of non-integer solutions: -3 | |
0376 | Problem:
Solve the system
$$
\left\lvert\,
\begin{aligned}
& \left(4^{\sqrt{x^{2}+x}}+7 \cdot 2^{\sqrt{x^{2}+x}}-1\right) \sin (\pi y)=7|\sin (\pi y)| \\
& x^{2}+4 x+y^{2}=0
\end{aligned}
\right.
$$ | [
"Solution:\nThe system is defined for $x \\in (-\\infty, -1] \\cup [0, +\\infty)$ and every $y$. We consider three cases.\n\nCase 1. Let $\\sin (\\pi y) > 0$. The first equation gives\n$$\n4^{\\sqrt{x^{2}+x}} + 7 \\cdot 2^{\\sqrt{x^{2}+x}} - 1 = 7\n$$\nand setting $t = 2^{\\sqrt{x^{2}+x}} > 0$ we obtain the equatio... | Bulgaria | Spring Mathematical Competition | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (-1, -√3), (0, 0), (-4, 0), (-2 - √3, 1), (-2 - √3, -1), (-2, 2), (-2, -2) | |
019e | Nonnegative integers $a$ and $b$ have the following property: $d(na) \ge d(nb)$ for each positive integer $n$ (where $d(k)$ is the number of divisors of $k$). Prove that $a$ is divisible by $b$. | [
"Let $a = p_1^{\\alpha_1} \\dots p_m^{\\alpha_m}$, $b = p_1^{\\beta_1} \\dots p_m^{\\beta_m}$ be the prime decompositions of these numbers (we assume that some $\\alpha_k, \\beta_k$ can be equal to $0$). Let us check that for each $k$ $\\alpha_k \\ge \\beta_k$. Indeed, if the inequality does not hold for some $k$, ... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
00sb | Prove that for all positive integer $n$, there is a positive integer $m$, that $7^n \mid 3^m + 5^m - 1$. | [
"We prove this by induction on $n$. The case $n=1$ is indeed trivial for $m=1$.\n\nAssume that the statement of the problem holds true for $n$, and we have $3^m + 5^m - 1 = 7^n$ for some positive integer $l$ which is not divisible by $7$ (if not we are done). Since $3^6 \\equiv 1 \\pmod{7}$ and $5^6 \\equiv 1 \\pmo... | Balkan Mathematical Olympiad | BMO 2017 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest co... | English | proof only | null | |
0hpw | Problem:
How many roots does $\arctan x = x^{2} - 1.6$ have, where the arctan function is defined in the range $-\frac{\pi}{2} < \arctan x < \frac{\pi}{2}$? | [
"Solution:\n\n2 ."
] | United States | null | [
"Precalculus > Functions",
"Precalculus > Trigonometric functions"
] | null | final answer only | 2 | |
0cg3 | $$
3^{\log_5(5x-10)} - 2 = 5^{-1+\log_3 x}.
$$ | [
"We notice that the equation has solutions $x_1 = 3$ and $x_2 = 27$. We show that the equation does not have any other solutions.\n\nContinuation A.\nUsing the properties of logarithms, the equation becomes $15 \\cdot 3^{\\log_5(x-2)} = 10 + 5^{\\log_3 x}$, or $15(x-2)^{\\log_5 3} = 10 + x^{\\log_3 5}$, with $x > 2... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | 3, 27 | |
09m0 | Find all ordered triples of positive integers $x \le y \le z$ such that $x^2 + y^2 + z^2 = 99\ldots 9$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | The only possibilities are when the right-hand side is 9 or 99. For 9: (1, 2, 2). For 99: (1, 7, 7), (5, 5, 7), (3, 3, 9). There are no solutions when the number of nines is at least three. | |
0761 | Problem:
Let $n$ be a positive integer. Call a nonempty subset $S$ of $\{1,2, \ldots, n\}$ good if the arithmetic mean of the elements of $S$ is also an integer. Further let $t_{n}$ denote the number of good subsets of $\{1,2, \ldots, n\}$. Prove that $t_{n}$ and $n$ are both odd or both even. | [
"Solution:\n\nWe show that $T_{n}-n$ is even. Note that the subsets $\\{1\\},\\{2\\}, \\cdots,\\{n\\}$ are good. Among the other good subsets, let $A$ be the collection of subsets with an integer average which belongs to the subset, and let $B$ be the collection of subsets with an integer average which is not a mem... | India | INMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof only | null | |
0a09 | Problem:
Zij $\triangle ABC$ een driehoek met een rechte hoek in $C$ en $|AC| > |BC|$, zij $I$ het middelpunt van de ingeschreven cirkel en zij $H$ de projectie van $C$ op het lijnstuk $AB$. De ingeschreven cirkel $\omega$ van $\triangle ABC$ raakt de zijden $BC$, $CA$ en $AB$ in respectievelijk de punten $A_{1}$, $B_... | [
"Solution:\n\n\n\nMerk op dat de lijnstukken $BC_{1}$ en $BA_{1}$ even lang zijn omdat het raaklijnen zijn vanuit $B$ aan de zelfde cirkel $\\omega$. Dat betekent dat $\\triangle A_{1}BC_{1}$ een gelijkbenige driehoek is met tophoek $B$. Omdat de bissectrice en hoogtelijn in een gelijkbenig... | Netherlands | IMO-selectietoets III | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0jk8 | Problem:
Consider a permutation $(a_{1}, a_{2}, a_{3}, a_{4}, a_{5})$ of $\{1,2,3,4,5\}$. We say the tuple $(a_{1}, a_{2}, a_{3}, a_{4}, a_{5})$ is flawless if for all $1 \leq i<j<k \leq 5$, the sequence $(a_{i}, a_{j}, a_{k})$ is not an arithmetic progression (in that order). Find the number of flawless 5-tuples. | [
"Solution:\n\nWe do casework on the position of $3$.\n\n- If $a_{1}=3$, then the condition is that $4$ must appear after $5$ and $2$ must appear after $1$. It is easy to check there are six ways to do this.\n\n- If $a_{2}=3$, then there are no solutions; since there must be an index $i \\geq 3$ with $a_{i}=6-a_{1}$... | United States | HMMT November 2014 | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | 20 | |
00t1 | Let $\triangle ABC$ be an isosceles triangle with $AB = AC$ and $\angle A = 45^\circ$. Its circumcircle $(c)$ has center $O$, $M$ is the midpoint of $BC$ and $D$ is the foot of the perpendicular from $C$ to $AB$. With center $C$ and radius $CD$ we draw a circle which internally intersects $AC$ at the point $F$ and the ... | [
"\n\nLet $K$ be the intersection point of $ZE$ and $AC$. Then $K$ lies on the radical axis of the two circles, so it has equal powers to both circles. The power of the point $K$ with respect to the one circle is $KA \\cdot KC$, while the power to the other circle is $R^2 - KC^2 = CD^2 - KC^... | Balkan Mathematical Olympiad | BMO Short List | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line,... | English | proof only | null | |
0inl | Problem:
Determine the number of ways to select a positive number of squares on an $8 \times 8$ chessboard such that no two lie in the same row or the same column and no chosen square lies to the left of and below another chosen square. | [
"Solution:\n\nAnswer: $12869 = \\binom{16}{8} - 1$. If $k$ is the number of squares chosen, then there are $\\binom{8}{k}$ ways to choose $k$ columns, and $\\binom{8}{k}$ ways to choose $k$ rows, and this would uniquely determine the set of squares selected. Thus the answer is\n$$\n\\sum_{k=1}^{8} \\binom{8}{k} \\b... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 12869 | |
087t | Problem:
Qual è la seconda cifra (partendo da sinistra) del numero $\left(10^{16}+1\right)\left(10^{8}+1\right)\left(10^{4}+1\right)\left(10^{2}+1\right)(10+1)$ ?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 . | [
"Solution:\nLa risposta è $\\mathbf{( B )}$. Si possono calcolare direttamente tutte le cifre del numero. Si ha $10^{16}+1=\\frac{10^{32}-1}{10^{16}-1}$ (è il prodotto notevole $(a-1)(a+1)=a^{2}-1$ ), e scomposizioni analoghe per gli altri termini. Quindi\n$$\n\\begin{aligned}\n& \\left(10^{16}+1\\right)\\left(10^{... | Italy | UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA Progetto Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | B | |
0csh | The teacher gave to Pete four distinct positive integers. Pete has calculated the greatest common divisor of every two of these numbers. He obtained six numbers: $1$, $2$, $3$, $4$, $5$, and $N$, where $N > 5$. Find the least possible value of $N$. | [
"Число $N$ может равняться $14$, как показывает, например, четвёрка чисел $4$, $15$, $70$, $84$. Осталось показать, что $N \\ge 14$.\n\n**Лемма.** Среди попарных НОД четырёх чисел не может быть ровно двух чисел, делящихся на некоторое натуральное $k$.\n**Доказательство.** Если среди исходных четырёх чисел есть не б... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 14 | |
0b0u | Problem:
How many positive perfect cubes are divisors of the product $1! \cdot 2! \cdot 3! \cdots 10!$? | [
"Solution:\n\nWe have $N := 1! \\cdot 2! \\cdot 3! \\cdots 10! = 2^{38} 3^{17} 5^{7} 7^{4}$. Thus, a positive divisor of $N$ that is a perfect cube must be of the form $2^{3a} 3^{3b} 5^{3c} 7^{3d}$ for some nonnegative integers $a, b, c, d$.\n\nWe see that $3a \\leq 38$, $3b \\leq 17$, $3c \\leq 7$ and $3d \\leq 4$... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 468 | |
09h3 | Let $a_1, a_2, \dots, a_{2n+1}$ be given integers with sum zero. If any partial sum of these numbers is non zero and $1 \le |a_i| \le n+1$ for each $i$, then find these integers. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0b5q | Prove that $|a - b\sqrt{c}| < \frac{1}{2b}$ if and only if $|a^2 - b^2c| < \sqrt{c}$. | [
"The first relation also writes as $-\\frac{1}{2b} < a - b\\sqrt{c} < \\frac{1}{2b}$, i.e. $\\frac{2ab - 1}{2b^2} < \\sqrt{c} < \\frac{2ab + 1}{2b^2}$.\n\nThe second relation also writes as $-\\sqrt{c} < a^2 - b^2c < \\sqrt{c}$, or $b^2c + \\sqrt{c} - a^2 > 0$ and $b^2c - \\sqrt{c} - a^2 < 0$, i.e. $\\frac{\\sqrt{1... | Romania | Local Mathematical Competitions | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
06kd | Consider the sequences with $2016$ terms formed by the digits $1$, $2$, $3$ and $4$. Find the number of those sequences containing an even number of the digit $1$. | [
"There are $2^{2015} + 2^{4031}$ such sequences.\n\nLet $a_n$ be the number of such sequences with $n$ terms instead of $2016$. Since there are $4^n$ sequences formed by $1$, $2$, $3$, $4$, there are $4^n - a_n$ sequences with an odd number of $1$'s among these.\n\nConsider a sequence with $n$ terms formed by $1$, ... | Hong Kong | HKG TST | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2^{2015} + 2^{4031} | |
0ggj | 令 $n \ge 3$ 為一正整數。一個正整數 $m \ge n+1$ 被稱為 $n$ 色數, 若且唯若我們可以將正 $m$ 邊形的每一個頂點塗成 $n$ 色中的其中一個顏色, 使得這個正 $m$ 邊形的任何連續 $n+1$ 個頂點都包含全部 $n$ 種顏色。
證明對於任何 $n \ge 3$, 非 $n$ 色數只有有限多個, 並求最大的非 $n$ 色數。 | [
"最大的非 $n$ 色數為 $m_{\\text{max}} = n(n-1) - 1$。\n\n假設 $m = m_{\\text{max}}$，則必然有某個顏色 $C$ 在至多 $n-2$ 個頂點上，而這些頂點將所有非 $C$ 色的頂點切割為至多 $n-2$ 個連續片段。由於非 $C$ 色頂點至少有 $n(n-1) - 1 - (n-2) = (n-1)^2 > n(n-2)$ 個，因此其中必然有一個非 $C$ 色的連續片段的長度至少為 $n$，從而 $m_{\\text{max}}$ 是非 $n$ 色數。\n\n現在假設 $m \\ge m_{\\text{max}} + 1 = n(n-1)$。依據除法原理，有 $m... | Taiwan | 2022 數學奧林匹亞競賽第一階段選訓營, 獨立研究(一) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Chinese; English | proof and answer | n(n-1) - 1 |
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