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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0exv | Problem:
A group of children are arranged into two equal rows. Every child in the back row is taller than the child standing in front of him in the other row. Prove that this remains true if each row is rearranged so that the children increase in height from left to right. | [
"Solution:\n\nRearrange the children in the back row into order, and rearrange the front row in the same way, so that each child stays in front of the same child in the back row. Denote heights in the back row by $a_i$ and heights in the front row by $b_i$. So we have $a_1 \\leq a_2 \\leq \\ldots \\leq a_n$, and $a... | Soviet Union | 6th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
00y8 | Problem:
Let $\alpha, \beta, \gamma$ be the angles of a triangle opposite to its sides with lengths $a, b$ and $c$, respectively. Prove the inequality
$$
a \cdot\left(\frac{1}{\beta}+\frac{1}{\gamma}\right)+b \cdot\left(\frac{1}{\gamma}+\frac{1}{\alpha}\right)+c \cdot\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \geq 2... | [
"Solution:\nClearly, the inequality $a > b$ implies $\\alpha > \\beta$ and similarly $a < b$ implies $\\alpha < \\beta$, hence $(a-b)(\\alpha-\\beta) \\geq 0$ and $a \\alpha + b \\beta \\geq a \\beta + b \\alpha$. Dividing the last equality by $\\alpha \\beta$ we get\n$$\n\\frac{a}{\\beta} + \\frac{b}{\\alpha} \\ge... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
0d02 | Let $a$, $b$, $c$, $d$ be integers with $a$, $c$ both nonzero, and define $x_n = \gcd(an + b, cn + d)$ for all positive integers $n$. Show that the sequence $x_1, x_2, x_3, \dots$ is unbounded if and only if $ad = bc$. | [
"Note that\n$$\n\\begin{aligned}\n\\frac{\\gcd(acn + bc, acn + ad)}{ac} &\\le \\gcd(an + b, cn + d) \\\\\n&\\le \\gcd(acn + bc, acn + ad).\n\\end{aligned}\n$$\nSuppose $ad = bc$. We have\n$$\n\\begin{aligned}\n\\gcd(an + b, cn + d) &\\ge \\frac{\\gcd(acn + bc, acn + ad)}{ac} \\\\\n&= \\frac{\\gcd(acn + bc, acn + bc... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
08k6 | Problem:
Find all positive integers $x, y$ satisfying the equation
$$
9\left(x^{2}+y^{2}+1\right)+2(3 x y+2)=2005
$$ | [
"Solution:\nThe given equation can be written into the form\n$$\n2(x+y)^{2}+(x-y)^{2}=664\n$$\nTherefore, both numbers $x+y$ and $x-y$ are even.\nLet $x+y=2 m$ and $x-y=2 t$, $t \\in \\mathbb{Z}$.\nNow from (1) we have that $t$ and $t^{2}$ are even and $m$ is odd.\nSo, if $t=2 k$, $k \\in \\mathbb{Z}$ and $m=2 n+1$... | JBMO | OJBM | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (11, 7) and (7, 11) | |
0isa | Problem:
Say that $\frac{a}{b}$ is a positive rational number in simplest form, with $a \neq 1$. Further, say that $n$ is an integer such that:
$$
\frac{1}{n}>\frac{a}{b}>\frac{1}{n+1}
$$
Show that when $\frac{a}{b}-\frac{1}{n+1}$ is written in simplest form, its numerator is smaller than $a$. | [
"Solution:\n\n$\\frac{a}{b}-\\frac{1}{n+1}=\\frac{a(n+1)-b}{b(n+1)}$. Therefore, when we write it in simplest form, its numerator will be at most $a(n+1)-b$. We claim that $a(n+1)-b<a$. Indeed, this is the same as $a n-b<0 \\Longleftrightarrow a n<b \\Longleftrightarrow \\frac{b}{a}>n$, which is given."
] | United States | Harvard-MIT November Tournament | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
05q4 | Problem:
On note $P(n)$ le plus grand diviseur premier de $n$. Montrer qu'il existe une infinité d'entiers $n$ tels que $P(n-1)<P(n)<P(n+1)$. | [
"Solution:\n\nSoit $p$ un nombre premier impair. On cherche un entier $n$ de la forme $p^{2^{k}}$, avec $k$ entier, tel que $P(n-1)<P(n)<P(n+1)$.\n\nSoit $k>l \\geq 0$, on a\n$$\n\\frac{p^{2^{k}}+1}{2}=\\frac{p^{2^{l}}+1}{2}\\left(p^{2^{k}-2^{l}}-p^{2^{k}-2 \\cdot 2^{l}}+\\cdots-p^{2 \\cdot 2^{l}}+p^{2^{l}}-1\\righ... | France | Envoi 1: Arithmétique | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0hwb | Problem:
Let $x$, $y$, and $z$ be positive integers satisfying $x y = z^{2} + 1$. Prove that there are integers $a$, $b$, $c$, and $d$ such that $x = a^{2} + b^{2}$, $y = c^{2} + d^{2}$, and $z = a c + b d$. | [
"Solution:\nWe will use strong induction on $z$. Assume first that $z = 1$. Then $x y = 1^{2} + 1 = 2$, so $x$ and $y$ are $1$ and $2$ in some order. If $x = 2$ and $y = 1$, then there is a solution $a = b = c = 1$, $d = 0$; if $x = 1$ and $y = 2$, we may take $a = c = d = 1$, $b = 0$.\n\nNow assume that $z > 1$ an... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
00tp | Can every positive rational number $q$ be written as
$$
\frac{a^{2021} + b^{2023}}{c^{2022} + d^{2024}},
$$
where $a, b, c, d$ are all positive integers? | [
"The answer is yes. Set $a = x^{2023}$, $b = x^{2021}$ and $c = y^{2024}$, $d = y^{2022}$ for some integers $x, y$ and let $q = \\frac{m}{n}$ in lowest terms. Then we could try to solve\n$$\n\\frac{a^{2021} + b^{2023}}{c^{2022} + d^{2024}} = \\frac{2x^{2021 \\times 2023}}{2y^{2022 \\times 2024}} = \\frac{x^{2021 \\... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | Yes | |
0i3i | Problem:
A real number $x$ is randomly chosen in the interval $\left[-15 \frac{1}{2}, 15 \frac{1}{2}\right]$. Find the probability that the closest integer to $x$ is odd. | [
"Solution:\nBy using a graphical method, we can see that, for real $x$ on $\\left[-n-\\frac{1}{2}, n+\\frac{1}{2}\\right]$, $n$ an even integer, the probability that the closest integer to $x$ is odd is $\\frac{n}{2 n+1}$. The desired probability is $\\left(\\frac{15}{31}\\right)$."
] | United States | Harvard-MIT Math Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 16/31 | |
0dpp | An infinite strictly increasing sequence of positive integers $\{a_n\}_{n\ge 1}$ is given. It is also given that $a_n \le n + 2020$ and $n^3 a_n - 1$ is divisible by $a_{n+1}$ for any positive integer $n$. Prove that $a_n = n$ for any positive integer $n$. (Kanat Satylkhanov) | [
"By induction on $n$ it is easy to show that $a_n \\ge n$ for any $n$. Suppose that there exists a positive integer $k$ such that $a_k > k$. Let's choose such positive integer $m$ that $m : 2021!$ and $m > k$. Then for any $i = 2, 3, \\dots, 2021$, $\\text{GCD}(m, m + i) > 1$. It follows from the problem statement ... | Silk Road Mathematics Competition | XIX Silk Road Mathematical Competition | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
05oa | Problem:
Déterminer la valeur minimale de
$$
\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{b+c+d}{a}\right\rfloor+\left\lfloor\frac{c+d+a}{b}\right\rfloor+\left\lfloor\frac{d+a+b}{c}\right\rfloor
$$
lorsque $a, b, c, d$ décrivent $\mathbb{N}^*$. | [
"Solution:\n\nLe minimum cherché est $9$.\nEn effet, si $a, b, c, d$ sont des entiers naturels non nuls, on pose\n$$\nf(a, b, c, d)=\\left\\lfloor\\frac{a+b+c}{d}\\right\\rfloor+\\left\\lfloor\\frac{b+c+d}{a}\\right\\rfloor+\\left\\lfloor\\frac{c+d+a}{b}\\right\\rfloor+\\left\\lfloor\\frac{d+a+b}{c}\\right\\rfloor.... | France | Olympiades Françaises de Mathématiques - Envoi 2 (Algèbre) | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 9 | |
0l71 | Problem:
Compute the number of ways to pick two rectangles in a $5 \times 5$ grid of squares such that the edges of the rectangles lie on the lines of the grid and the rectangles do not overlap at their interiors, edges, or vertices. The order in which the rectangles are chosen does not matter. | [
"Solution:\n\nA rectangle can be specified by two intervals, one specifying its horizontal extent ($x$-coordinates of left and right sides) and one specifying its vertical extent ($y$-coordinates of bottom and top sides). For the rectangles to not overlap, we need either the horizontal intervals or the vertical int... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 6300 | |
0c9t | Problem:
Fie $a, b, c, d \in \mathbb{N}$, $d \neq 0$ şi fie funcţia $f: \mathbb{N} \rightarrow \mathbb{N}$, definită prin
$$
f(n) = \left[ \frac{a n + b}{c n + d} \right], \text{ oricare ar fi } n \in \mathbb{N}
$$
unde $[x]$ reprezintă partea întreagă a numărului real $x$. Demonstraţi că următoarele afirmaţii sunt ech... | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ Etapa a III-a | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
08u0 | $OA = 2$, $OB = 3$, $OC = 4$, $\angle AOB = \angle BOC = \angle COA$
are satisfied. Furthermore, there is a point $X$ on the base $ABC$ for which
$\angle AOX = \angle BOX = \angle COX = 30^\circ$.
Determine the volume of this triangular pyramid $OABC$. Here, by $PQ$ we denote the length of the line segment $PQ$. | [
"Choose points $P$, $Q$ on the line segments $OB$, $OC$, respectively in such a way that $OP = OQ = 2$. Let us compare the volumes of the triangular pyramids $OABC$ and $OAPQ$. Considering the triangles $\\triangle OBC$ and $\\triangle OPQ$ as their respective bases and $A$ as their top vertex, we have that\n$$\n\\... | Japan | Japan Junior Mathematical Olympiad First Round | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 9/4 | |
0c5k | Determine the positive integers $n$ that satisfy the following property: for every positive divisor $d$ of $n$, $d + 1$ is a divisor of $n + 1$. | [
"We prove that the numbers that have the given property are $1$ and the odd prime numbers. It is clear that all these numbers do indeed have the desired property and also that $2$ does not have it.\n\nConversely, let us consider a composite number $n$ and prove that it does not have the given property. If $n$ is co... | Romania | Stars of Mathematics Competition | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | n = 1 or n is an odd prime | |
04rg | We are given a sheet of paper in the form of a rectangle $x \times y$, where $x$ and $y$ are integer numbers larger than $1$. Let us draw a lattice of $x \times y$ unit squares on the sheet. Rolling up the rectangle and gluing it along its opposite sides we shape a lateral surface of a circular cylinder. Join each two ... | [
"We will compute the requested number $P$ of all internal segments for the cylinder formed by gluing the rectangle $x \\times y$ along the opposite sides of length $y$.\n\nThis cylinder has two bases of perimeter $x$ and its lateral sides are of length $y$. We will use an obvious formula $P = P_0 - P_1 - P_2$, wher... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Other"
] | English | proof and answer | For the cylinder formed by gluing along the sides of length y (bases have perimeter x): P = x(x−1)(y^2 + 2y − 1)/2. For the cylinder formed by gluing along the sides of length x (bases have perimeter y): Q = y(y−1)(x^2 + 2x − 1)/2. If x > y, then P > Q; the cylinder whose bases have perimeter x has more internal segmen... | |
00tq | A natural number $n$ is given. Determine all $(n-1)$-tuples of nonnegative integers $a_1, a_2, \dots, a_{n-1}$ such that
$$
\left[ \frac{m}{2^n - 1} \right] + \left[ \frac{2m + a_1}{2^n - 1} \right] + \left[ \frac{2^2m + a_2}{2^n - 1} \right] + \left[ \frac{2^3m + a_3}{2^n - 1} \right] + \dots + \left[ \frac{2^{n-1}m +... | [
"**Solution 1.** We will show that there is a unique such $n$-tuple: $a_k = 2^{n-1} + 2^{k-1} - 1$ for $k = 1, \\dots, n-1$.\nWrite $N = 2^n - 1$ and $f_k(x) = \\left[ \\frac{2^k x + a_k}{N} \\right]$ for $k = 0, 1, \\dots, n-1$, where $a_0 = 0$. Since\n$$\n\\sum_{k=0}^{n-1} f_k(m) - \\sum_{k=0}^{n-1} f_k(m-1) = 1,... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | a_k = 2^{n-1} + 2^{k-1} - 1 for k = 1, 2, ..., n-1 | |
01dn | Do there exist positive integers $a, b, c$, such they have no common divisor and
$$
ab + bc + ca = (a + b - c)(b + c - a)(c + a - b)?
$$ | [
"We show that all of $a, b$ and $c$ have the number $3$ as a common factor. First suppose that one of $a, b, c$ is divisible by $3$. By symmetry, we may assume that $a \\equiv 0 \\pmod{3}$. Then the equation implies $bc \\equiv (b-c)(b+c)(c-b) \\pmod{3}$. If neither of $b$ and $c$ is divisible by $3$, this gives $b... | Baltic Way | Baltic Way 2016 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | No; any solution would force all three integers to be divisible by three, so none exist without a common divisor. | |
0gmo | Let $A$ be a point on a circle with center $O$ and $B$ be the midpoint of $[OA]$. Let $C$ and $D$ be points on the circle and on the same side of $OA$ satisfying the condition $\overline\{CBO\} = \overline\{DBA\}$. Show that the symmetric point of the midpoint of $[CD]$ with respect to the point $B$ is on the circle. | [] | Turkey | IMO TEAM SELECTION EXAM | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ipb | Problem:
Evaluate the infinite sum
$$
\sum_{n=0}^{\infty} \binom{2 n}{n} \frac{1}{5^{n}}
$$ | [
"Solution:\nAnswer: $\\sqrt{5}$\n\nFirst Solution: Note that\n$$\n\\begin{aligned}\n\\binom{2 n}{n} & =\\frac{(2 n)!}{n!\\cdot n!}=\\frac{(2 n)(2 n-2)(2 n-4) \\cdots(2)}{n!} \\cdot \\frac{(2 n-1)(2 n-3)(2 n-5) \\cdots(1)}{n!} \\\\\n& =2^{n} \\cdot \\frac{(-2)^{n}}{n!}\\left(-\\frac{1}{2}\\right)\\left(-\\frac{1}{2}... | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | proof and answer | sqrt(5) | |
0hgr | In the left-lower corner $1 \times 1$ cell of the $m \times n$ board ($m, n \geq 3$) there is a black chip, and in the left-upper and right-lower cells, there are white chips. In his move, Petrik can move the black chip to the adjacent by side cell twice in a row, and Vasyl can either move one of the white chips to the... | [
"**Answer:** if $m$ and $n$ have the same parity, Vasyl wins, else Petrik wins.\n\nWlog the number of rows doesn't exceed the number of columns, so $m \\leq n$. Enumerate the rows from bottom to top as 1, 2, ..., $m$, and columns from left to right as 1, 2, ..., $n$, then each cell has two coordinates – numbers of ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Vasyl wins if and only if m and n have the same parity; otherwise Petrik wins. | |
0jb6 | Problem:
Let $ABC$ be a triangle with $AB < AC$. Let $M$ be the midpoint of $BC$. Line $l$ is drawn through $M$ so that it is perpendicular to $AM$, and intersects line $AB$ at point $X$ and line $AC$ at point $Y$. Prove that $\angle BAC = 90^{\circ}$ if and only if quadrilateral $XBYC$ is cyclic. | [
"Solution:\n\n\n\nFirst, note that $XBYC$ cyclic is equivalent to $\\measuredangle BXM = \\measuredangle ACB$. However, note that $\\measuredangle BXM = 90^{\\circ} - \\measuredangle BAM$, so $XBYC$ cyclic is in turn equivalent to $\\measuredangle BAM + \\measuredangle ACB = 90^{\\circ}$.\n... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0hck | A $3 \times 3$ table has positive integers in its cells so that the sum of numbers in any two cells that share common side is a factorial of a positive integer. Show that there are at least $3$ equal numbers in the table.
The factorial of a positive integer $n$ is a product $1 \cdot 2 \cdot 3 \cdots n$.
![](attached_i... | [
"We will start with the following lemma.\n**Lemma 1.** There exists a diagonal that has two equal numbers for any $2 \\times 2$ square.\n**Proof.** Let $X$ be the greatest number in $2 \\times 2$ square. Let his neighbors in this $2 \\times 2$ square be $A$ and $B$. Clearly, they are located on a diagonal, so if $A... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof only | null | |
03h3 | Problem:
Given three distinct unit circles, each of which is tangent to the other two, find the radii of the circles which are tangent to all three circles. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | radii = 2/√3 − 1 and 1 + 2/√3 | |
082f | Problem:
Sia data una scacchiera di 100 righe e 100 colonne, con tutte le caselle bianche.
a. È possibile colorare un numero dispari di caselle in modo tale che ogni casella colorata abbia un numero dispari di caselle colorate adiacenti?
b. È possibile colorare alcune caselle in modo tale che un numero dispari di es... | [
"Solution:\n\na.\nNon è possibile. Infatti, siano $C_{1}, \\ldots, C_{m}$ le caselle colorate, sia $a_{i}$ il numero delle caselle colorate adiacenti a $C_{i}$ per $i=1, \\ldots, m$ e sia $l$ il numero dei lati in comune a due caselle adiacenti colorate. Abbiamo $a_{1}+\\cdots+a_{m}=2l$, poiché ogni lato è in comun... | Italy | Cesenatico Gara Individuale | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a: no; b: yes; c: no | |
0dlo | Define the function $S: \mathbb{Z}^+ \to \mathbb{Z}^+$ to be the sum of prime divisors (not necessarily distinct) of $n$ (if $v_p(n) = a$ then add $p \cdot a$ to the sum). Prove that
$$
\frac{\tau(n!)}{\tau((n-1)!)} \le \frac{4S(n)}{n}.
$$ | [] | Saudi Arabia | Saudi Booklet | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0h0q | Solve the equation $[x^2] - 2x + 1 = 0$, where $[x^2]$ denotes the greatest integer that does not exceed $x^2$. | [
"We have $x = \\frac{[x^2] + 1}{2}$. Hence, $x = t$, or $x = t + \\frac{1}{2}$, where $t$ is an integer.\n\nIf $x = t$, we have\n$$\n[x^2] - 2x + 1 = [t^2] - 2t + 1 = t^2 - 2t + 1 = (t - 1)^2 = 0 \\Leftrightarrow t = 1 \\Leftrightarrow x = 1.\n$$\n\nIf $x = t + \\frac{1}{2}$, then\n$$\n[x^2] - 2x + 1 = [(t + \\frac... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 1/2, 1, 3/2 | |
07ms | 2011 is prime and the sum of its digits equals its number of digits. Find all smaller primes with this property; a leading zero is not allowed. | [
"*Case 1:* All digits equal to $1$. One solution is $11$. We rule out $1$, $111$ as it is divisible by $3$, and $1111$ which is divisible by $11$.\n\nIn all other cases there is at least one zero. Zeros are not allowed in the leading position, or in the final position (multiples of $10$ are non-prime). Thus there a... | Ireland | Ireland | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 11, 1021, 1201 | |
08wu | Players $A$ and $B$ wrote down two positive integers each on a black board. The product of the two numbers $A$ wrote is twice the sum of the two numbers $B$ wrote, and the product of the two numbers $B$ wrote is twice the sum of the two numbers $A$ wrote, and furthermore, the sum of the two numbers $A$ wrote is greater... | [
"$8, 9, 10, 13, 17, 19, 27$\n\nLet $a$ be the sum of two numbers the player $A$ wrote, and $b$ be the sum of the two numbers $B$ wrote. If we let $t, s$ be the two numbers $A$ wrote, then $t+s = a$ and $ts = 2b$, so $t, s$ must be the solutions of the quadratic equation $x^2 - a x + 2b = 0$. Similarly, the two numb... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | [8, 9, 10, 13, 17, 19, 27] | |
0end | Let $p$ be an odd prime number and let $(a_1, a_2, \dots, a_p)$ and $(b_1, b_2, \dots, b_p)$ be arbitrary arrangements of the $p$-tuple $(0, 1, \dots, p-1)$. For each $i$, let $c_i$ be the non-negative remainder when the product $a_i b_i$ is divided by $p$. Show that $(c_1, c_2, \dots, c_p)$ cannot be a rearrangement o... | [
"If $a_i = b_j = 0$, then $c_i = c_j = 0$. If $(c_1, c_2, \\dots, c_p)$ is a rearrangement of $(0, 1, \\dots, p-1)$, then no entry can appear more than once. Hence $i = j$.\n\nWithout loss of generality we may suppose that $a_p = b_p = 0$ so that $c_p = 0$. Then we have\n$$\n(p-1)! \\equiv_p a_1 a_2 \\dots a_{p-1} ... | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0j9p | Problem:
Niffy's favorite number is a positive integer, and Stebbysaurus is trying to guess what it is. Niffy tells her that when expressed in decimal without any leading zeros, her favorite number satisfies the following:
- Adding $1$ to the number results in an integer divisible by $210$.
- The sum of the digits of ... | [
"Solution:\n\nAnswer: $1010309$\n\nNote that Niffy's favorite number must end in $9$, since adding $1$ makes it divisible by $10$. Also, the sum of the digits of Niffy's favorite number must be even (because it is equal to twice the number of digits) and congruent to $2$ modulo $3$ (because adding $1$ gives a multi... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 1010309 | |
0f0d | Problem:
9 lines each divide a square into two quadrilaterals with areas $\frac{2}{5}$ and $\frac{3}{5}$ that of the square. Show that 3 of the lines meet in a point. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
021z | Problem:
Se $\frac{x}{y}=2$, então $\frac{x-y}{x}$ é igual a:
(a) -1
(b) $-\frac{1}{2}$
(c) $\frac{1}{2}$
(d) 1
(e) 2 | [
"Solution:\nTemos:\n$$\n\\frac{x-y}{x}=\\frac{x}{x}-\\frac{y}{x}=1-\\frac{y}{x}\n$$\nComo $\\frac{x}{y}=2$ temos que $\\frac{y}{x}=\\frac{1}{2}$, assim\n$$\n\\frac{x-y}{x}=\\frac{x}{x}-\\frac{y}{x}=1-\\frac{1}{2}=\\frac{1}{2}\n$$\nA opção correta é (c).\n\nSolução 2: Se $\\frac{x}{y}=2$, então $x=2 y$. Logo\n$$\n\\... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | c | |
06cf | The diagonals of a quadrilateral $ABCD$ intersect at $K$. $M, N$ are the midpoints of $AB$ and $CD$ respectively. Let $k, m, n$ be the lines through $K, M, N$ perpendicular to $AD, BD$ and $AC$ respectively. Show that if $k, m, n$ are concurrent, then $A, B, C, D$ are concyclic. | [] | Hong Kong | HKG TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0feq | Problem:
En el triángulo $ABC$ se traza la bisectriz interior $CD$. Se sabe que el centro del círculo inscrito en el triángulo $BCD$ coincide con el centro del círculo circunscrito del triángulo $ABC$. Calcular los ángulos del triángulo $ABC$. | [
"Solution:\n\nSea $O$ el centro común de los círculos mencionados en el enunciado. Por la hipótesis, $BO$ y $CO$ son bisectrices de los ángulos $\\angle ABC$ y $\\angle BCD$ ($O$ es el centro del círculo inscrito en $BCD$).\n\nAdemás $AO = BO = CO$ ($O$ es el centro del círculo circunscrito a $ABC$).\n\nDe aquí, si... | Spain | TANDA I | [
"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 | A = 72°, B = 36°, C = 72° | |
0cq3 | On a $8 \times 8$ chessboard, a closed broken line is drawn; its vertices are the centers of the cells, and each segment connects the centers of two cells sharing a common point (in particular, they may share a common side). This broken line bounds a polygon $P$. Prove that the total area of white parts of $P$ equals t... | [
"Проведём пунктиром вертикальные и горизонтальные линии через центры клеток доски. На получившейся пунктирной сетке каждое звено нашей ломаной соединяет узлы, соседние по вертикали, горизонтали или диагонали. Поэтому пунктирные прямые разбивают область, ограниченную ломаной, на единичные квадратики и половинки квад... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English, Russian | proof only | null | |
066w | In a mathematical competition some competitors are friends; friendship is always mutual, that is to say that is when $A$ is a friend of $B$, then also $B$ is a friend of $A$. We say that $n \ge 3$ different competitors $A_1, A_2, \dots, A_n$ form a weekly-friendly circle if $A_i$ is not a friend of $A_{i+1}$, for $1 \l... | [] | Greece | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0a15 | A chess board consists of $8 \times 8$ squares. A knight is placed on a corner square. The knight makes jumps of either two squares horizontally and one square vertically, or two squares vertically and one square horizontally. On each square of the chess board, we write down how many jumps the knight needs to reach it.... | [
"C) 4"
] | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | English | MCQ | C | |
0dme | Problem:
У скупу целих бројева решити једначину
$$
12^{x} + y^{4} = 2008^{z}.
$$ | [
"Solution:\n\nЗа $x < 0$ или $z \\leq 0$ једино решење је тривијално $(0, 0, 0)$. Надаље је $z > 0$. Како је $2008 = 2^{3} \\cdot 251$, обе стране једначине су дељиве са $251$. Претпоставимо да је $x = 2x_{1}$ парно. Тада $\\left(2^{x_{1}}\\right)^{2} \\equiv -\\left(y^{2}\\right)^{2} \\pmod{251}$, што дизанјем на ... | Serbia | СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | (0, 0, 0) | |
08t8 | Compute the value of
$$
\frac{\sum_{n=1}^{99} (\sqrt{10 + \sqrt{n}})}{\sum_{n=1}^{99} (\sqrt{10 - \sqrt{n}})}
$$ | [
"$[1 + \\sqrt{2}]$\nLet us denote by *S* and *T* the numerator and the denominator, respectively of the fraction to be computed. Namely,\n$$\nS = \\sum_{n=1}^{99} (\\sqrt{10 + \\sqrt{n}}), \\quad T = \\sum_{n=1}^{99} (\\sqrt{10 - \\sqrt{n}}).\n$$\nFirst, we note that for a pair of positive real numbers *a*, *b*, we... | Japan | Japan Mathematical Olympiad First Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | final answer only | 1 + sqrt(2) | |
0kkk | Problem:
Let $n$ be the answer to this problem. Hexagon $A B C D E F$ is inscribed in a circle of radius $90$. The area of $A B C D E F$ is $8 n$, $A B = B C = D E = E F$, and $C D = F A$. Find the area of triangle $A B C$. | [
"Solution:\n\n\n\nLet $O$ be the center of the circle, and let $O B$ intersect $A C$ at point $M$; note $O B$ is the perpendicular bisector of $A C$. Since triangles $A B C$ and $D E F$ are congruent, $A C D F$ has area $6 n$, meaning that $A O C$ has area $\\frac{3 n}{2}$. It follows that ... | United States | HMMT November 2021 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2592 | |
0ccc | Find all the quadruples $(a, b, c, d)$ of integers, so that $a + b + c = 2d$ and $\sqrt{ab} + \sqrt{bc} + \sqrt{ca} = d^2$. | [
"If $a \\ge b \\ge c \\ge 0$, then $d \\ge 0$. From $d^2 = \\sqrt{ab} + \\sqrt{ac} + \\sqrt{bc} \\le a + b + c \\le 2a + 2b + 2c \\le 4d$, at least one of the equalities being strict, follows $d \\in \\{0, 1, 2, 3\\}$.\n\nCase I: $d = 0$. Then $a = b = c = 0$.\n\nCase II: $d = 1$. Then $a + b + c = 2$ and $\\sqrt{a... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All quadruples are (0, 0, 0, 0), (0, 1, 1, 1), (1, 0, 1, 1), (1, 1, 0, 1), (0, -1, -1, -1), (-1, 0, -1, -1), (-1, -1, 0, -1). | |
0fpf | En Rectilandia hay $n$ ciudades situadas a lo largo de una carretera recta. Cada ciudad tiene asignados dos robots: un d-robot, situado a su derecha y que solo se desplaza en ese sentido, y un i-robot, situado a su izquierda, y que solo se desplaza en ese sentido. Cada uno de los $2n$ robots tiene asignado un número, y... | [
"Designamos a las $n$ ciudades, situadas de izquierda a derecha, como $C_1, C_2, \\dots, C_n$. Observemos en primer lugar que si $C_i$ vence a $C_j$, también vence a cualquier ciudad situada entre ambas. Demostraremos el resultado por inducción sobre el número de ciudades. El caso base $n=1$ es trivial. Con $n$ ciu... | Spain | PREPARACIÓN BARCELONA CONTEST | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Spanish | proof only | null | |
0ku6 | Problem:
Let $ABCD$ be a convex trapezoid such that $\angle ABC = \angle BCD = 90^{\circ}$, $AB = 3$, $BC = 6$, and $CD = 12$. Among all points $X$ inside the trapezoid satisfying $\angle XBC = \angle XDA$, compute the minimum possible value of $CX$. | [
"Solution:\n\n\n\nLet $P = AD \\cap BC$. Then, the given angle condition $\\angle XBC = \\angle XAD$ implies that $\\angle XBD + \\angle XPD = 180^{\\circ}$, so $X$ always lies on circle $\\odot(PBD)$, which is fixed. Thus, we see that the locus of $X$ is the arc $\\widehat{BD}$ of $\\odot(... | United States | HMMT November | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | sqrt(113) - sqrt(65) | |
04lv | Let $z_1$ and $z_2$ be complex numbers such that $|z_1| = |z_2| = 1$, and let $a$ and $b$ be real numbers such that $a + b = 1$. Prove
$$
|a z_1 + b z_2| \ge \frac{1}{2} |z_1 + z_2|.
$$ | [] | Croatia | Mathematical competitions in Croatia | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof only | null | |
01jy | Given real non-zero numbers $a$, $b$, $c$ ($b > 0$) so that two distinct roots of the equation $ax^2 + bx + c = 0$ are also the roots of the equation $x^3 + bx^2 + ax + c = 0$.
Prove that a) $abc > 16$; b) $abc \ge \frac{3125}{108}$. (V. Karamzin) | [
"Let $x_1$, $x_2$ be distinct roots of the equation $ax^2 + bx + c = 0$, i.e. be zeroes of the functions $f(x) = ax^2 + bx + c$ and $g(x) = x^3 + bx^2 + ax + c$. By condition, $f(0) = g(0) = -c \\neq 0$. Let $F(x) = g(x) - f(x)$. Then $0$, $x_1$, $x_2$ are the distinct zeroes of the polynomial $F(x)$. So $F(x) = x(... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof only | null | |
070r | Problem:
Given two circles $C$ and $C'$, we say that $C$ bisects $C'$ if their common chord is a diameter of $C'$. Show that for any two circles which are not concentric, there are infinitely many circles which bisect them both. Find the locus of the centers of the bisecting circles. | [
"Solution:\n\nLet $C$, $C'$ have centers $O$, $O'$ respectively and radii $r$, $r'$ respectively. Let a circle center $P$ bisect $C$. Suppose it meets $C$ at $A$ and $B$. Then $AB$ is perpendicular to $OP$ and is a diameter of $C$. Hence $PA^{2} = OP^{2} + r^{2}$. Conversely, the circle center $P$, radius $\\sqrt{O... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0f38 | Problem:
Given a finite set $X$ of points in the plane. $S$ is a set of vectors $AB$ where $(A, B)$ are some pairs of points in $X$. For every point $A$ the number of vectors $AB$ (starting at $A$) in $S$ equals the number of vectors $CA$ (ending at $A$) in $S$. Show that the sum of the vectors in $S$ is zero. | [
"Solution:\n\nLet $X = \\{P_1, P_2, \\ldots, P_n\\}$. For each vector $AB$ in $S$, let $\\vec{v}_{AB}$ denote the vector from $A$ to $B$.\n\nThe sum of all vectors in $S$ is:\n$$\n\\sum_{AB \\in S} \\vec{v}_{AB} = \\sum_{AB \\in S} (\\vec{OB} - \\vec{OA})\n$$\nwhere $O$ is the origin and $\\vec{OA}$ is the position... | Soviet Union | ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
0cdk | Consider a tetrahedron $ABCD$ with $\widehat{BAC} + \widehat{CAD} + \widehat{DAB} = 180^\circ$ and $\widehat{ABC} \equiv \widehat{DAB}$. If the projection of the vertex $D$ on the plane $(ABC)$ is the orthocenter of the triangle $ABC$, prove that $AB = AC$ and $DB = DC$. | [
"Let $BE$ and $CF$ be the altitudes from $B$ and $C$, respectively, of the triangle $ABC$, and denote by $H$ their meeting point. The three perpendiculars theorem implies that $DE \\perp AC$ and $DF \\perp AB$.\nUnfold the tetrahedron on the plane $(ABC)$ and denote by $D_1$ the image of the vertex $D$ of the trian... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0f3e | Problem:
Let $f(n)$ be the sum of $n$ and its digits. For example, $f(34) = 41$. Is there an integer such that $f(n) = 1980$? Show that given any positive integer $m$ we can find $n$ such that $f(n) = m$ or $m + 1$. | [
"Solution:\nIf the last digit of $n$ is not $9$, then $f(n + 1) = f(n) + 2$. If the last digit of $n$ is $9$, then $f(n + 1) < f(n)$. On the other hand, $f$ clearly achieves arbitrarily large values. Also $f(1) = 1$.\n\nNow consider any $m > 1$. Let $M$ be the smallest integer such that $f(M) > m$. Then $f(M - 1) \... | Soviet Union | ASU | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | Yes | |
0kml | Problem:
Define the function $f: \mathbb{R} \rightarrow \mathbb{R}$ by
$$
f(x)= \begin{cases}\frac{1}{x^{2}+\sqrt{x^{4}+2 x}} & \text{ if } x \notin(-\sqrt[3]{2}, 0] \\ 0 & \text{ otherwise }\end{cases}
$$
The sum of all real numbers $x$ for which $f^{10}(x)=1$ can be written as $\frac{a+b \sqrt{c}}{d}$, where $a, b, ... | [
"Solution:\n\nIf $x \\in(-\\sqrt[3]{2}, 0]$, it is evidently not a solution, so let us assume otherwise. Then, we find\n$$\nf(x)=\\frac{\\sqrt{x^{4}+2 x}-x^{2}}{2 x}\n$$\nwhich implies that $x f(x)^{2}+x^{2} f(x)-1 / 2=0$, by reverse engineering the quadratic formula. Therefore, if $x>0, f(x)$ is the unique positiv... | United States | HMMT November 2021 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 932 | |
07l1 | Let $ABC$ be an isosceles triangle with $|AB| = |AC|$. If $D$ is a point on the circumcircle of $\triangle ABC$ on the same side of $BC$ as $A$, prove
$$
|AB| + |AC| \ge |DB| + |DC|
$$
with equality if and only if $D = A$. Hence prove, of all triangles described in a circle, the equilateral triangle has the greatest pe... | [
"Let $O$ be the centre of the circle and $r$ be its radius. Draw $OE \\perp AB$, $OF \\perp BD$ and $OG \\perp DC$. Then $|AE| = r \\cos \\angle BAO$. Noting that $\\angle BAO = \\angle CAO$ and $E$ is the midpoint of $AB$, it follows that\n$$\n|AB| + |AC| = 4r \\cos \\angle BAO.\n$$\n\nUsing $|DF| = r \\cos \\angl... | Ireland | Irska | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07ce | Vertices of a regular $n$-gon have been colored blue and red such that for each rotation of the $n$-gon, the number of vertices that have different colors before and after the rotation is less than 32 percent of the number of vertices. Prove that either the number of blue vertices or the number of red vertices is less ... | [
"Obviously, there are exactly $n-1$ different nontrivial rotations of a regular $n$-gon (rotations of $\\frac{2\\pi i}{n}$ angles, for $1 \\le i \\le n-1$). Denote the number of blue and red vertices of the polygon by $b$ and $r$, respectively $(a+b=n)$, and the number of vertices that have different colors before ... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Transformations > Rotation",
"Algebra > Equations and Inequalities > Lin... | null | proof only | null | |
0aey | Ана замислила еден број. Тој број го помножила со $7$, потоа му додала $6$, добиениот резултат го поделила со $5$ и го добила бројот $53$. Откриј кој број го змислила Ана! | [
"Прв начин: Нека бројот што го замислила Ана се означи со $x$. Тогаш, од условот во задачата се добива равенката $(7x+6):5=53$, од каде $x=37$.\n\nВтор начин: Реализирајќи ги условите од задачата од назад кон напред се добива $(53 \\times 5-6): 7 = 37$."
] | North Macedonia | Регионален натпревар по математика за основно образование | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | Macedonian, English | proof and answer | 37 | |
0ko7 | Problem:
Two linear functions $f(x)$ and $g(x)$ satisfy the properties that for all $x$,
- $f(x) + g(x) = 2$
- $f(f(x)) = g(g(x))$
and $f(0) = 2022$. Compute $f(1)$. | [
"Solution:\n\nFirstly, $f(x)$ and $g(x)$ must intersect - otherwise, $f(x) = g(x) = 1$, which can't be true.\n\nSecondly, suppose they intersect at $a$, so that $f(a) = g(a) = c$. Then $f(c) = g(c) \\Longrightarrow f(c) = g(c) = 1$. But then, $a = c$, and $c = 1$. So $f(1) = 1$, and we're done.",
"Solution:\n\nWe... | United States | HMMT November 2022 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 1 | |
0kh1 | Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts 210 equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts 42 steps of the same size from the fro... | [
"Let $L$ denote the required length of the ship. Assume Emily's speed is $1$ and let $v < 1$ denote the speed of the ship relative to the riverbank. When Emily walks from the back to the front, it takes $210$ steps, so $210(1 - v) = L$. Similarly, when Emily walks from the front to the back of the ship, it takes $4... | United States | AMC 10 A | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | A | |
07y3 | Suppose $a$, $b$, $c$ are real numbers such that $a + b + c = 1$ and $a^2 + b^2 + c^2 = 1$.
Prove that $a^3 + b^3 + c^3 \le 1$. | [
"First note that\n$$\n0 = (a + b + c)^2 - (a^2 + b^2 + c^2) = 2(ab + bc + ca).\n$$\nUsing $a + b = 1 - c$, we then obtain\n$$\n0 = ab + bc + ca = ab + c(a + b) = ab + c(1 - c), \\quad \\text{ i.e. } \\ ab = c(c - 1)\n$$\nand so $abc = c^2(c - 1) \\le 0$ because $a^2 + b^2 + c^2 = 1$ implies $c \\le 1$. Finally,\n$$... | Ireland | IRL_ABooklet_2025 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
07vg | Consider $\triangle ABC$ with $\angle A > 90^\circ$ and two points $D$ and $E$ on segment $BC$ so that $\angle BAD = \angle CAE$. If the circumcircles of $\triangle BAD$ and $\triangle CAE$ are tangent to each other prove that $\angle BAE = \angle CAD = 90^\circ$. | [
"Extend $AD$ and $AE$ to meet the circumcircle of $\\triangle ABC$ at $F$ and $G$, respectively. Let $O$ be the intersection point of $BG$ and $CF$.\n\n\n\nSince $\\angle BAD = \\angle CAE$, we have\n$$\n\\angle BCO = \\angle OGF = \\angle BAD = \\angle CAE = \\angle CBO = \\angle OFG,\n$$\... | Ireland | IRL_ABooklet_2023 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00lm | Let $(a_n)_{n \ge 0}$ be the sequence of rational numbers with $a_0 = 2016$ and
$$
a_{n+1} = a_n + \frac{2}{a_n}
$$
Show that the sequence does not contain a square of a rational number. | [
"We look at this sequence modulo $5$. This is possible as long as $a_n \\ne 0 \\pmod 5$ so that the next element is defined modulo $5$. If we start to compute the elements modulo $5$ we obtain\n$$\na_0 \\equiv 1 \\pmod 5,\n$$\n$$\na_1 \\equiv 1+2 \\equiv 3 \\pmod 5,\n$$\n$$\na_2 \\equiv 3+2 \\cdot 3^{-1} \\equiv 3+... | Austria | 48th Austrian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
09jk | Are there sequences of nonnegative integers $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$ satisfying the following property:
$$
a_0 = b_0 = 0 \text{ and for any } k \ge 1
$$
(i) $a_k$ is the smallest positive integer that cannot be written as $a_i + b_j$ for some $0 \le i \le k-1$ and $j \ge 0$, and
(ii) $b_k$ is... | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0ixy | Problem:
Let $\triangle ABC$ be an acute triangle with angles $\alpha, \beta$, and $\gamma$. Prove that
$$
\frac{\cos \alpha}{\cos (\beta-\gamma)}+\frac{\cos \beta}{\cos (\gamma-\alpha)}+\frac{\cos \gamma}{\cos (\alpha-\beta)} \geq \frac{3}{2}
$$ | [
"Solution:\nSince $\\cos \\gamma = -\\cos (\\alpha+\\beta) = -\\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta$, and $\\cos (\\alpha-\\beta) = \\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta$ the given inequality transforms to\n$$\n\\begin{aligned}\n& \\frac{\\sin \\beta \\sin \\gamma - \\cos \\beta ... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Mi... | null | proof only | null | |
0292 | Problem:
Na figura ao lado, o triângulo $\triangle ABC$ é retângulo e os semicírculos desenhados têm diâmetros $AB$, $BC$ e $AC$.
Mostre que a área sombreada é igual à área do triângulo $\triangle ABC$.
 | [
"Solution:\n\nSejam $T$ a área do triângulo $\\triangle ABC$, $a$ e $c$ as áreas sombreadas na figura dada e $b$ e $d$ as áreas compreendidas entre os catetos do triângulo e o semicírculo de diâmetro $AB$.\n\nA área $a+b$ é a área do semicírculo de diâmetro $AB$:\n$$\na+b=\\frac{1}{2} \\pi\\left(\\frac{AB}{2}\\righ... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles"
] | null | proof only | null | |
0bob | In a regular tetrahedron $ABCD$ consider planes that are parallel to its faces such that each edge is divided into 6 equal segments. These planes determine, on the edges of the tetrahedron, on its faces and in its interior a set consisting of 80 points of intersection. Denote this set by $V$.
Find the maximum number of... | [
"Suppose, WLOG, that the height of $ABCD$ equals $6$. Denote by $k$ the maximal number of elements of the set $W$ and by $M_1, \\dots, M_k$ its elements. For each $i \\in \\{1, 2, \\dots, k\\}$ denote by $a_i, b_i, c_i$, and $d_i$ the distances from the point $M_i$ to the planes $(BCD), (ACD), (ABD)$ and $(ABC)$ re... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 8 | |
0fe6 | Problem:
Se considera un tetraedro regular como el de la figura. Si el punto $E$ recorre la arista $AB$. ¿Cuándo el ángulo $\widehat{CED}$ es máximo?
 | [
"Solution:\nSupongamos que el tetraedro tiene arista de longitud $1$, sea $\\alpha$ el ángulo $CED$ y $x$ la longitud del segmento $AE$.\n\nSi aplicamos el Teorema del coseno al triángulo $AEC$ tenemos:\n$$\nEC^{2} = x^{2} + 1 - 2x \\cos 60^{\\circ} = x^{2} + 1 - x\n$$\nPor simetría se tiene que la longitud $EC$ es... | Spain | null | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | When the point is the midpoint of the edge AB. | |
0a5v | Problem:
Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = abc$. Prove that at least one of $a$, $b$ or $c$ is greater than $\frac{17}{10}$. | [
"Solution:\nWlog assume $a \\geq b \\geq c$. Therefore $a + b + c \\geq 3c$. Now for a proof by contradiction, assume $a \\leq \\frac{17}{10}$ and $b \\leq \\frac{17}{10}$. Since $\\left(\\frac{17}{10}\\right)^2 = \\frac{289}{100} < 3$, it follows that $\\frac{17}{10} < \\sqrt{3}$ and therefore $ab \\leq 3$. Howeve... | New Zealand | NZMO Round Two | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0i9j | Problem:
A tightrope walker stands in the center of a rope of length $32$ meters. Every minute she walks forward one meter with probability $3/4$ and backward one meter with probability $1/4$. What is the probability that she reaches the end in front of her before the end behind her? | [
"Solution:\nAfter one minute, she is three times as likely to be one meter forward as one meter back. After two minutes she is either in the same place, two meters forward, or two meters back. The chance of being two meters forward is clearly $(3/4)^2$ and thus $3^2 = 9$ times greater than the chance of being two b... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 3^16 / (3^16 + 1) | |
03wx | Let $m$ be an integer greater than $1$, and let $n$ be an odd number with $3 \le n < 2m$. Numbers $a_{i,j}$ ($i, j \in \mathbb{N}$, $1 \le i \le m$, $1 \le j \le n$) satisfy:
(1) For every $1 \le j \le n$, $a_{1,j}$, $a_{2,j}$, ..., $a_{m,j}$ is a permutation of $1, 2, ..., m$;
(2) $|a_{i,j} - a_{i,j+1}| \le 1$ for eve... | [
"Let $n = 2l + 1$. Since $3 \\le n < 2m$, we have $1 \\le l \\le m - 1$. We first estimate the lower bound of $M$.\nBy the condition (1), there exists a unique $1 \\le i_0 \\le m$, such that $a_{i_0, l+1} = m$. Consider $a_{i_0, l}$ and $a_{i_0, l+2}$.\n\nCase 1: At least one of $a_{i_0, l}$ and $a_{i_0, l+2}$ is $... | China | China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | (2l+1)m - l^2 where n = 2l + 1 | |
0fu1 | Problem:
Bestimme alle natürlichen Zahlen $n$ mit genau 100 verschiedenen positiven Teilern, sodass mindestens 10 dieser Teiler aufeinanderfolgende Zahlen sind. | [
"Solution:\n\nVon 10 aufeinanderfolgenden ganzen Zahlen ist sicher eine durch $9$, eine durch $8$, eine durch $7$ und eine durch $5$ teilbar. Daher ist $n$ durch $2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7$ teilbar. Sei nun\n$$\nn = 2^{a} \\cdot 3^{b} \\cdot 5^{c} \\cdot 7^{d} \\cdot p_{1}^{e_{1}} \\cdots p_{r}^{e_{r}}\n... | Switzerland | Vorrundenprüfung | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 45360 | |
0j0o | Problem:
How many functions $f:\{1,2,3,4,5\} \rightarrow \{1,2,3,4,5\}$ have the property that $f(\{1,2,3\})$ and $f(f(\{1,2,3\}))$ are disjoint? | [
"Solution:\nLet $f(\\{1,2,3\\})$ be $A$. Then $A \\cap f(A) = \\emptyset$, so $A$ must be a subset of $\\{4,5\\}$.\n\nIf $A = \\{4,5\\}$, there are $2^3 - 2$ ways to assign each element in $\\{1,2,3\\}$ to a value in $\\{4,5\\}$, and $9$ ways to assign each element of $\\{4,5\\}$ to a value in $\\{1,2,3\\}$, for a ... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics"
] | null | final answer only | 94 | |
02or | Consider the polynomial $f(x) = x^3 + x^2 - 4x + 1$.
a. Prove that if $r$ is a root of $f(x)$ then $r^2 + r - 3$ is also a root of $f(x)$.
b. Let $\alpha$, $\beta$, $\gamma$ be the three roots of $f(x)$, in some order. Determine all possible values of
$$
\frac{\alpha}{\beta} + \frac{\beta}{\gamma} + \frac{\gamma}{\al... | [
"a. First notice that, since $r$ is a root, then $r^3 + r^2 - 4r + 1 = 0 \\iff r^2 + r - 3 = 1 - \\frac{1}{r}$. So we need to prove that\n$$\n\\begin{aligned}\n& \\left(1 - \\frac{1}{r}\\right)^3 + \\left(1 - \\frac{1}{r}\\right)^2 - 4\\left(1 - \\frac{1}{r}\\right) + 1 = 0 \\\\\n\\iff & 1 - \\frac{3}{r} + \\frac{3... | Brazil | Brazilian Math Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | 3 and -10 | |
0hhv | There are 100 cards, each with one of the numbers $1$, $2$, $\ldots$, $100$ written on it, such that each number appears on exactly one card. The cards are stacked in such a way that the numbers $1$, $2$, $\ldots$, $100$ are written from top to bottom in order. Petrik rearranges the cards according to the following rul... | [
"Let us consider two moves: $2k-1$ and $2k$. Suppose that before the first of them there was a layout of cards $a_1, a_2, \\ldots, a_{2k-1}, a_{2k}, a_{2k+1}, \\ldots, a_{100}$, then after two moves we will have the following changes:\n$$\na_{2k-1}, a_{2k-2}, \\dots, a_1, a_{2k}, a_{2k+1}, \\dots, a_{100} \\to a_{2... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof only | null | |
0jyk | Problem:
Let $a$, $b$, $c$ be non-negative real numbers such that $ab + bc + ca = 3$. Suppose that
$$
a^3 b + b^3 c + c^3 a + 2abc(a + b + c) = \frac{9}{2}
$$
What is the minimum possible value of $ab^3 + bc^3 + ca^3$? | [
"Solution:\n\nExpanding the inequality $\\sum_{\\text{cyc}} ab(b + c - 2a)^2 \\geq 0$ gives\n$$\n\\left(\\sum_{\\text{cyc}} ab^3\\right) + 4\\left(\\sum_{\\text{cyc}} a^3 b\\right) - 4\\left(\\sum_{\\text{cyc}} a^2 b^2\\right) - abc(a + b + c) \\geq 0\n$$\nUsing $\\left(\\sum_{\\text{cyc}} a^3 b\\right) + 2abc(a + ... | United States | February 2017 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 18 | |
0j3t | Problem:
Indecisive Andy starts out at the midpoint of the 1-unit-long segment $\overline{H T}$. He flips 2010 coins. On each flip, if the coin is heads, he moves halfway towards endpoint $H$, and if the coin is tails, he moves halfway towards endpoint $T$. After his 2010 moves, what is the expected distance between A... | [
"Solution:\n\nLet Andy's position be $x$ units from the $H$ end after 2009 flips. If Andy moves towards the $H$ end, he ends up at $\\frac{x}{2}$, a distance of $\\frac{1-x}{2}$ from the midpoint. If Andy moves towards the $T$ end, he ends up at $\\frac{1+x}{2}$, a distance of $\\frac{x}{2}$ from the midpoint. His ... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 1/4 | |
0jjp | Problem:
Let $ABC$ be a triangle with $\angle B = 90^{\circ}$. Given that there exists a point $D$ on $AC$ such that $AD = DC$ and $BD = BC$, compute the value of the ratio $\frac{AB}{BC}$. | [
"Solution:\n\n$\\sqrt{3}$\n\n$D$ is the circumcenter of $ABC$ because it is the midpoint of the hypotenuse. Therefore, $DB = DA = DC$ because they are all radii of the circumcircle, so $DBC$ is an equilateral triangle, and $\\angle C = 60^{\\circ}$. This means that $ABC$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ t... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | sqrt(3) | |
0azt | Problem:
Suppose that $\{a_{n}\}_{n \geq 1}$ is an arithmetic sequence of real numbers such that
$$
\begin{array}{r}
a_{1}+a_{2}+a_{3}+a_{4}+\cdots+a_{10}=20 \\
a_{1}+a_{4}+a_{9}+a_{16}+\cdots+a_{100}=18
\end{array}
$$
Compute $a_{1}+a_{8}+a_{27}+a_{64}+\cdots+a_{1000}$. | [
"Solution:\n\nLet $a_n = a_1 + (n-1)d$ for some $a_1$ and common difference $d$.\n\nFirst, $a_1 + a_2 + \\cdots + a_{10} = 20$.\n\nThe sum of the first $10$ terms of an arithmetic sequence is:\n$$\nS_{10} = \\frac{10}{2}(a_1 + a_{10}) = 5(a_1 + a_{10})\n$$\nBut $a_{10} = a_1 + 9d$, so:\n$$\nS_{10} = 5(a_1 + a_1 + 9... | Philippines | 21st PMO Area Stage | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2 | |
0iux | Let $n$ be a positive integer. Determine the size of the largest subset of $\{-n, -n+1, \dots, n-1, n\}$ which does not contain three elements $a, b, c$ (not necessarily distinct) satisfying $a+b+c=0$. | [
"The maximum size is $n$ if $n$ is even, and $n+1$ if $n$ is odd, achieved by the subset\n$$\n\\{-n, \\dots, -\\lfloor \\frac{n}{2} \\rfloor - 1, \\lfloor \\frac{n}{2} \\rfloor + 1, \\dots, n\\}.\n$$\n**Lemma 1.** Let $A, B$ be finite nonempty subsets of $\\mathbb{Z}$. Then the set $A + B = \\{a + b : a \\in A, b \... | United States | USAMO 2009 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof and answer | Maximum size: n when n is even, and n+1 when n is odd. One extremal subset is all integers in {−n, …, n} except those with absolute value at most floor(n/2), i.e., {−n, …, −floor(n/2)−1, floor(n/2)+1, …, n}. | |
04dn | Legs of a right triangle are of lengths $6$ cm and $8$ cm. What is the radius of its circumcircle? | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | final answer only | 5 cm | |
0akn | Problem:
Let $ABC$ be an equilateral triangle. From the vertex $A$ we draw a ray towards the interior of the triangle such that the ray reaches one of the sides of the triangle. When the ray reaches a side, it then bounces off following the law of reflection, that is, if it arrives with a directed angle $\alpha$, it l... | [] | North Macedonia | Asian-Pacific Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | All odd positive integers | |
018a | Let $x$, $y$, $z$, $t$ be positive real numbers such that $xyzt = 1$ and
$$
\frac{x}{y} + \frac{y}{z} + \frac{z}{t} + \frac{t}{x} \leq x + y + z + t.
$$
Prove that
$$
\frac{y}{x} + \frac{z}{y} + \frac{t}{z} + \frac{x}{t} \geq x + y + z + t.
$$ | [
"By the arithmetic mean-geometric mean inequality we have\n$$\nx = \\sqrt[4]{x^4} = \\sqrt[4]{\\frac{x^4}{xyzt}} = \\sqrt[4]{\\frac{x^3}{yzt}} = \\sqrt[4]{\\frac{x}{y} \\cdot \\frac{x}{t} \\cdot \\frac{t}{z} \\cdot \\frac{x}{t}} \\leq \\frac{1}{4} \\left( \\frac{x}{y} + \\frac{x}{t} + \\frac{t}{z} + \\frac{x}{t} \\... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0kiw | Problem:
Suppose you have an equilateral triangle divided into 9 smaller equilateral triangles as shown, with the bottom side horizontal. Starting from the top corner labeled $A$, you must walk to the bottom right corner labeled $B$, and are only allowed to take steps along the edges down to the left, down to the righ... | [
"Solution:\n\nThe number of paths from the top vertex to itself or to any vertex on the left side of the triangle is $1$. For each other vertex, the number of paths can be found by adding the number of paths to the vertex directly to its left and to the two vertices above it. Thus, the following triangle of numbers... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | 22 | |
023s | Problem:
Ana, Beto e Carolina vão participar do programa de televisão "Descubra a cor do seu chapéu". No programa, eles se posicionam em roda e sobre a cabeça de cada um será colocado um chapéu azul ou verde. Cada um pode ver os chapéus dos outros, mas não a cor do seu próprio chapéu. Em seguida, cada um deles escreve... | [
"Solution:\n\n(a) Seguindo a ideia de Carolina, cada pessoa tem duas cores possíveis de chapéu, no total há $2 \\cdot 2 \\cdot 2=8$ possibilidades de chutes. Entre eles, há apenas um caso favorável: acertarem seus chutes. Portanto, a probabilidade de ganhar o carro é $\\frac{1}{8}$.\n\n(b) Com a ideia de Beto, há a... | Brazil | NÍVEL 3 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | (a) 1/8; (b) 1/2; (c) Example strategy: each player passes when the two hats they see are of different colors; if the two hats they see are the same color, they guess their own hat is the opposite color. This wins in 6 of the 8 cases, giving probability 3/4. | |
075j | Problem:
Let $ABC$ be a triangle. An interior point $P$ of $ABC$ is said to be good if we can find exactly $27$ rays emanating from $P$ intersecting the sides of the triangle $ABC$ such that the triangle is divided by these rays into $27$ smaller triangles of equal area. Determine the number of good points for a given... | [
"Solution:\n\nLet $P$ be a good point. Let $l, m, n$ be respectively the number of parts the sides $BC, CA, AB$ are divided by the rays starting from $P$. Note that a ray must pass through each of the vertices of the triangle $ABC$; otherwise we get some quadrilaterals.\n\nLet $h_1$ be the distance of $P$ from $BC$... | India | INMO | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 325 | |
0c46 | Problem:
Fie $ABC$ un triunghi ascuţitunghic în care $AB < AC$, $G$ centrul său de greutate şi $D$ piciorul înălţimii din $A$. Dreapta $DG$ intersectează arcul mic $BC$ al cercului circumscris triunghiului $ABC$ în punctul $E$. Arătaţi că dreapta $AB$ este tangentă cercului circumscris triunghiului $BDE$. | [
"Solution:\n\nFie $F$ celălalt punct de intersecţie a dreptei $DG$ cu cercul circumscris triunghiului $ABC$. Concluzia este echivalentă cu $\\angle BEF \\equiv \\angle ABC$, deci cu congruenţa arcelor $AC$ şi $BF$, ceea ce revine la congruenţa arcelor $AB$ şi $FC$, deci la $AF \\parallel BC$ (cu alte cuvinte la $AB... | Romania | Al cincilea test de selecţie pentru OBMJ | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circums... | null | proof only | null | |
03qp | Let $ABC$ be an obtuse triangle inscribed in a circle of radius $1$. Prove that triangle $ABC$ can be covered by an isosceles right-angled triangle with hypotenuse $\sqrt{2} + 1$. (posed by Leng Gangsong) | [
"Without loss of generality, we may assume that $\\angle C > 90^\\circ$. Since $\\angle A + \\angle B < 90^\\circ$, we may assume without loss of generality that $\\angle A < 45^\\circ$.\n\nWe can then construct a semicircle $\\omega$ with $AB$ as its diameter such that point $C$ lies inside $\\omega$. Let $O$ be t... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0hfy | In the triangle *ABC* the point *N* is the midpoint of the median *CM*. The point *X* satisfies the following conditions: $\angle XMC = \angle MBC$, $\angle XCM = \angle MCB$ and points *X* and *B* lie from different sides with respect to the line *CM*. Let $\omega$ be the circumscribed circle of $\triangle AMX$. Prove... | [
"a) as from the statement $\\triangle XMC \\sim \\triangle MBC$, we get\n$$\n\\begin{aligned} \n\\angle AMX &= 180^\\circ - \\angle XMC - \\angle BMC = \\\\ \n&= 180^\\circ - \\angle XMC - \\angle MXC = \\angle MCX \n\\end{aligned}\n$$\n\nand $\\frac{AM}{MX} = \\frac{BM}{MX} = \\frac{MC}{CX} \\Rightarrow \\triangle... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
0h8q | Does there exist such value of $x \in (0, \frac{\pi}{2})$, for which numbers $\sin x$, $\cos x$ and $\operatorname{tg} x$ form a geometric progression? | [
"Let us write the equation that is necessary and sufficient for these three numbers to form geometric progression:\n$$\n\\sin x \\cdot \\operatorname{tg} x = \\cos^2 x.\n$$\nThen, $\\sin^2 x = \\cos^3 x$ and if we denote by $t = \\cos x \\in (0; 1)$, then we obtain the equation:\n$$\n1 - t^2 = t^3 \\Rightarrow f(t)... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof and answer | Yes | |
0e5t | Let $a$ and $b$ be different real numbers such that the equations $x^2 + a x + b = 0$ and $x^2 + b x + a = 0$ have some common solution. How much is $a + b$? | [
"If we equalize the left sides of the quadratic equations, we get $x^2 + a x + b = x^2 + b x + a$. Hence $a x + b = b x + a$ and $(x - 1)(a - b) = 0$. Because $a \\neq b$, it follows that $x = 1$. If we now substitute $x = 1$ in one of the equations, we get $a + b = -1$."
] | Slovenia | National Math Olympiad 2012 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | -1 | |
0d8a | In the garden of Wonderland, there are $2016$ apples, $2017$ bananas and $2018$ oranges. Two monkeys Adu and Bakar play the following game: alternatively each of them takes and eats one fruit of any kind except for the one that he took in previous turn (in the first turn, each of them can take a fruit of any kind). Who... | [
"Adu has the strategy to win the game.\nFirst, he takes one banana then we have $2016$ apples, $2016$ bananas and $2018$ oranges. In the next turns, we have some cases:\n1. If Bakar takes apple or orange, then Adu takes the same. We can see that the number of fruits in these type are always even, then whenever Baka... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Adu | |
0bgg | Problem:
O configurație de 4027 de puncte se numește columbiană dacă are 2013 puncte colorate cu roșu și 2014 puncte colorate cu albastru, și nu conține trei puncte coliniare. O mulțime finită de drepte din plan împarte planul în regiuni. O mulțime de drepte se numește bună pentru o configurație columbiană dacă următo... | [] | Romania | Romania Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 63 | |
00h6 | Let $ABC$ be a triangle, and let $D$ be a point on side $BC$. A line through $D$ intersects side $AB$ at $X$ and ray $AC$ at $Y$. The circumcircle of triangle $BXD$ intersects the circumcircle $\omega$ of triangle $ABC$ again at point $Z \neq B$. The lines $ZD$ and $ZY$ intersect $\omega$ again at $V$ and $W$, respecti... | [
"Suppose $XY$ intersects $\\omega$ at points $P$ and $Q$, where $Q$ lies between $X$ and $Y$. We will show that $V$ and $W$ are the reflections of $A$ and $B$ with respect to the perpendicular bisector of $PQ$. From this, it follows that $AVWB$ is an isosceles trapezoid and hence $AB = VW$.\n\nFirst, note that\n$$\... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0emr | Two evenly matched teams are engaged in a series of games, which ends as soon as a team wins four games. Is it more likely for the series to end in exactly six games or in exactly seven games? | [
"Let the two teams be $A$ and $B$, and represent each possible series as a bit string, where a $1$ represents a win by team $A$ and a $0$ a win by team $B$. Let $S$ be the set of all possible bit strings. We need to determine whether there are more bit strings of length $6$ or length $7$ in $S$. For each bit string... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | exactly seven games | |
08wl | Let $n$ be a positive integer. Suppose there are infinitely many cards and on each of these cards one integer greater than or equal to $0$ is written. Suppose also for each integer $l$ greater than or equal to $0$, there are exactly $n$ cards which have the number $l$ written on them. Let us consider repeating the proc... | [
"We will show that the smallest $n$ satisfying the condition of the problem is $2012^2$.\n\nFirst, let us show that if $n = 2012^2$, then there exists a method of selecting cards at each stage so as to satisfy the condition of the problem. For this purpose, let for each $k \\ge 1$, select $2012$ cards to be drawn i... | Japan | Japan Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 2012^2 | |
04qv | Show that for each natural number $k$ there exist at most finitely many triples of mutually distinct primes $p, q, r$ for which the number $qr - k$ is a multiple of $p$, the number $pr - k$ is a multiple of $q$, and the number $pq - k$ is a multiple of $r$. | [
"(Mutually distinct) primes $p, q, r$ satisfy the desired conditions if and only if the number $pq + pr + qr - k$ is divisible by each of the primes $p, q, r$; that is, by the product $pqr$. The equality $pq + pr + qr - k = n \\cdot pqr$, for a suitable integer $n$, can be rewritten as $k = pq + pr + qr - n \\cdot ... | Czech Republic | Czech-Slovak-Polish Match | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0hka | Problem:
A scalene triangle has side lengths which are all prime numbers. What is the smallest possible perimeter it could have? | [
"Solution:\nThe answer is $3 + 5 + 7 = 15$. This is achievable since the three sides $\\{3, 5, 7\\}$ form the sides of a triangle.\n\nTo show it is best possible, note that $\\{3, 5, 7\\}$ are the three smallest odd primes; so any smaller triangle would need to have a side of length $2$. However, the two longer len... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 15 | |
029v | Problem:
O perímetro de um retângulo $ABCD$ é $20~\mathrm{m}$. O menor comprimento, em metros, que a diagonal $AC$ pode ter é:
(a) $0$
(b) $\sqrt{50}$
(c) $10$
(d) $\sqrt{200}$
(e) $20 \sqrt{5}$ | [
"Solution:\n\nDenotemos por $a$ e $b$ os comprimentos dos lados do retângulo, assim $2a + 2b = 20$, logo $a + b = 10$. Por outro lado, o quadrado do comprimento da diagonal pode ser calculado usando o teorema de Pitágoras, assim $d^{2} = a^{2} + b^{2}$. Como\n\n$$\n\\begin{aligned}\n2d^{2} &= 2a^{2} + 2b^{2} = (a^{... | Brazil | null | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | b | |
0iov | Problem:
Positive real numbers $x$, $y$ satisfy the equations $x^{2} + y^{2} = 1$ and $x^{4} + y^{4} = \frac{17}{18}$. Find $x y$. | [
"Solution:\n\nWe have\n$$\n2 x^{2} y^{2} = (x^{2} + y^{2})^{2} - (x^{4} + y^{4}) = 1^{2} - \\frac{17}{18} = 1 - \\frac{17}{18} = \\frac{1}{18}.\n$$\nSo\n$$\nx^{2} y^{2} = \\frac{1}{36} \\implies x y = \\frac{1}{6}.\n$$"
] | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 1/6 | |
0el4 | Problem:
V trikotniku $ABC$ je $\angle ACB = 32^\circ$. Na nosilki stranice $AB$ ležita točki $D$ in $E$, za kateri velja $|AD| = |AC|$ in $|BE| = |BC|$ (glej sliko). Koliko stopinj je velikost kota $\angle DCE$?
(A) 90
(B) 96
(C) 100
(D) 106
(E) 116
 | [
"Solution:\n\nOznačimo $\\angle BAC = \\alpha$. Tedaj je $\\angle CBA = 180^\\circ - 32^\\circ - \\alpha = 148^\\circ - \\alpha$. Torej je $\\angle CAD = 180^\\circ - \\angle BAC = 180^\\circ - \\alpha$ in $\\angle EBC = 180^\\circ - \\angle CBA = \\alpha + 32^\\circ$. Trikotnika $CAD$ in $EBC$ sta enakokraka z vrh... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | D | |
0d6x | Let $a, b$ be given two real numbers with $a \neq 0$. Find all polynomials $P$ with real coefficients such that
$$
x P(x-a) = (x-b) P(x) \text{ for all } x \in \mathbb{R}.
$$ | [
"Let $P \\in \\mathbb{R}[x]$ satisfy the given condition.\n\nFirst, we notice that if $b=0$ then $x P(x-a) = x P(x) \\Rightarrow P(x-a) = P(x)$ (for all $x \\neq 0$), hence $P(x) \\equiv$ a constant (since $a \\neq 0$), and we can recheck that any constant polynomial satisfies the given condition.\n\nNow, let $b \\... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All real-coefficient polynomials P are:
- If b = 0: P(x) = c for any real constant c.
- If k = b/a is a natural number: P(x) = c · x(x − a)(x − 2a) ··· (x − (k − 1)a) for any real constant c.
- If k = b/a is not a natural number: P(x) ≡ 0. | |
0e4v | Problem:
Naj bo $P$ taka točka znotraj trikotnika $ABC$, da je $\angle CBP = \angle PAC$. Presečišče premice $AP$ s stranico $BC$ označimo z $D$, presečišče premice $BP$ s stranico $AC$ pa z $E$. Trikotnikoma $ADC$ in $BEC$ očrtani krožnici se sekata v točkah $C$ in $F$. Pokaži, da je premica $CP$ simetrala kota $DFE$... | [
"Solution:\n\nS pomočjo obodnih kotov v štirikotnikih $AFDC$ in $BCEF$ ter dane enakosti $\\angle CBP = \\angle PAC$ izpeljemo\n$$\n\\begin{aligned}\n\\angle CFE &= \\angle CBE = \\angle CBP = \\angle PAC \\\\\n&= \\angle DAC = \\angle DFC\n\\end{aligned}\n$$\nzato je premica $CF$ simetrala kota $\\angle DEF$. Poka... | Slovenia | 55. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07r0 | A function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ is called *loggy* if it satisfies the following two conditions:
(i) $f(xy) \equiv f(x) + f(y) \pmod{8}$ for all $x, y \in \mathbb{Z}$ that are not divisible by $17$;
(ii) $f(x + 17) \equiv f(x) \pmod{8}$ for all $x \in \mathbb{Z}$.
Determine, with proof,
a. if there ex... | [
"a. Because $6^2 \\equiv 2 \\pmod{17}$, conditions (i) and (ii) imply\n$$\nf(2) \\equiv f(6^2) \\equiv 2f(6) \\pmod{8}\n$$\nIf a loggy function $f$ satisfies $f(2) \\equiv 1 \\pmod{8}$, we get $2f(6) \\equiv 1 \\pmod{8}$. But the congruence $2x \\equiv 1 \\pmod{8}$ has no solution $x \\in \\mathbb{Z}$. Hence, there... | Ireland | Ireland_2017 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | a. No such function exists. b. Yes, such a function exists. |
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