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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
064c | Problem:
Wir betrachten ein Dreieck $ABC$ und einen Punkt $P$ in dessen Inneren. Die Spiegelpunkte von $P$ an den Seiten $\overline{BC}$, $\overline{CA}$ und $\overline{AB}$ seien mit $A_1$, $B_1$ und $C_1$ bezeichnet. Ferner sei $\Omega$ der Umkreis des Dreiecks $A_1B_1C_1$ und schließlich seien $A_2$, $B_2$ und $C_2... | [
"Solution:\n\nWir arbeiten mit gerichteten Winkeln modulo $180^{\\circ}$.\nIn dieser Skizze entspricht $T$ dem Punkt $P$.\n\n\nSchritt 1: Zunächst stellen wir fest, dass es sich bei $CA$ und $CB$ um die Mittelsenkrechten der Strecken $\\overline{PB_1}$ und $\\overline{PA_1}$ handelt, also i... | Germany | 2. Auswahlklausur | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
0ebe | Problem:
Poišči vse pare realnih števil $x$ in $y$, ki zadoščajo enačbama
$$
\begin{aligned}
& x+\frac{1}{y-x}=1 \\
& y+\frac{1}{x-y}=2
\end{aligned}
$$ | [
"Solution:\n\n1. način. Enačbi seštejemo in dobimo $x+y=3$. Od tod izrazimo $y=3-x$ in vstavimo v prvo enačbo, da dobimo $x+\\frac{1}{3-2x}=1$. Odpravimo ulomke in preoblikujemo do $2x^{2}-5x+2=0$. Levo stran te enačbe lahko razstavimo $2(x-2)\\left(x-\\frac{1}{2}\\right)=0$. Sledi $x=2$ ali $x=\\frac{1}{2}$. V prv... | Slovenia | 59. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (2, 1) and (1/2, 5/2) | |
0exk | Problem:
The beam of a lighthouse on a small rock penetrates to a fixed distance $d$. As the beam rotates the extremity of the beam moves with velocity $v$. Prove that a ship with speed at most $v/8$ cannot reach the rock without being illuminated. | [
"Solution:\n\nLet the lighthouse be at $L$. Take time $t = 0$ at the moment the boat starts its run, so that at $t = 0$ it is at $S$ a distance $d$ from $L$, and thereafter it is at a distance less than $d$. Take $A$ and $B$ a distance $d$ from $L$ so that $ALBS$ is a semicircle with diameter $AB$ and $S$ the midpo... | Soviet Union | 5th ASU | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
03lv | Problem:
Find all functions $f$ defined on the natural numbers that take values among the natural numbers for which
$$
(f(n))^{p} \equiv n \pmod{f(p)}
$$
for all $n \in \mathbf{N}$ and all prime numbers $p$. | [
"Solution:\nThe substitution $n = p$, a prime, yields $p \\equiv (f(p))^{p} \\equiv 0 \\pmod{f(p)}$, so that $p$ is divisible by $f(p)$. Hence, for each prime $p$, $f(p) = 1$ or $f(p) = p$.\n\nLet $S = \\{p : p \\text{ is prime and } f(p) = p\\}$. If $S$ is infinite, then $f(n)^{p} \\equiv n \\pmod{p}$ for infinite... | Canada | 40th Canadian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All functions f: N -> N satisfying exactly one of the following:
1) f(n) = n for all n.
2) f(p) = 1 for every prime p (with f arbitrary on composite inputs).
3) f(2) = 2, f(p) = 1 for every odd prime p, and f(n) has the same parity as n for all n. | |
0hzd | Problem:
Find all twice differentiable functions $f(x)$ such that $f^{\prime \prime}(x)=0$, $f(0)=19$, and $f(1)=99$. | [
"Solution:\nSince $f^{\\prime \\prime}(x)=0$ we must have $f(x)=a x + b$ for some real numbers $a, b$.\n\nThus $f(0) = b = 19$ and $f(1) = a + 19 = 99$, so $a = 80$.\n\nTherefore $f(x) = 80 x + 19$."
] | United States | Harvard-MIT Math Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Equations > ODEs"
] | null | proof and answer | f(x) = 80x + 19 | |
03e4 | Find the smallest possible number of divisors a positive integer $n$ may have, which satisfies the following conditions:
1.
$24 \mid n+1;$
2. The sum of the squares of all divisors of $n$ is divisible by $48$ (1 and $n$ are included). | [] | Bulgaria | Autumn tournament | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 48 | |
07m1 | The numbers $1, 2, 3, \dots, (2n)^2$ are written in the unit squares of a $2n \times 2n$ array, $n \ge 3$. Prove that there exist $n+1$ columns in the array such that in each of them any number is less than the sum of the remaining $2n-1$ numbers in that column. | [
"Assume that there exist $n$ columns such that in each of them one can find a number that is greater than or equal to the sum of the remaining $2n - 1$.\nTherefore, in these columns, say $i = 1, 2, 3, \\dots, n$, we must have $a_i \\ge c_i$, where $a_i$ denotes the maximum value in column $i$, and $c_i$ denotes the... | Ireland | Irish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0hqj | Problem:
The numbers $1, 2, \ldots, 50$ are written on a blackboard. We may erase two numbers $a$ and $b$, and replace both with $a + b + 2 a b$; we repeat this operation until only one number remains. Prove that the value of this last number does not depend on how the operations were performed. | [
"Solution:\n\nSuppose at some point the numbers on the board are $x_{1}, \\ldots, x_{k}$. We claim that the quantity $\\left(2 x_{1}+1\\right)\\left(2 x_{2}+1\\right) \\ldots\\left(2 x_{k}+1\\right)$ does not change. Indeed, this follows from the identity $1+2(a+b+2 a b)=(1+2 a)(1+2 b)$.\n\nThus, if there is exactl... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0ki5 | An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$, which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of the pillars at $A$, $B$, and $C$ are $12$, $9$, and $10$ m... | [] | United States | AMC 10 A | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | MCQ | D | |
0gad | 對於一個由有限多個正整數所成的集合 $A$, 我們將它分割成兩個非空的子集 $A_1$ 和 $A_2$。我們稱 $(A_1, A_2)$ 是個好分割, 若且唯若 $A_1$ 所有元素的最小公倍數等於 $A_2$ 所有元素的最大公因數。試求最小的 $n$, 使得存在一個由 $n$ 個正整數所成的集合, 其恰好有 2015 個好分割。
For a finite set $A$ of positive integers, we call a partition of $A$ into two disjoint nonempty subsets $A_1$ and $A_2$ are good if the least common mu... | [
"答案：3024\n\n假設 $A = \\{a_1, a_2, \\cdots, a_n\\}$, 其中 $a_1 < a_2 < \\cdots < a_n$。對任何非空的正整數集 $B$, 令 $l(B)$ 和 $g(B)$ 分別代表其所有元素的最小公倍數和最大公因數。\n\n考慮 $A$ 的所有好分割 $(A_1, A_2)$。由定義，對於所有 $a_i \\in A_1$ 和 $a_j \\in A_2$，必有 $a_i \\le l(A_1) = g(A_2) \\le a_j$，因此必存在 $1 \\le k < n$ 使得 $A_1 = \\{a_1, a_2, \\cdots, a_k\\}$ 且 $A_2... | Taiwan | 二〇一六數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3024 | |
0kux | Problem:
A prime number $p$ is mundane if there exist positive integers $a$ and $b$ less than $\frac{p}{2}$ such that $\frac{a b-1}{p}$ is a positive integer. Find, with proof, all prime numbers that are not mundane. | [
"Solution:\nThe cases $p=11$, $p=17$, $p=19$ fail by $3 \\cdot 4$, $3 \\cdot 6$, and $4 \\cdot 5$, respectively, so assume that $p \\geq 21$. The key idea is the following identity:\n$$\n\\frac{1}{10}-\\left(-\\frac{2}{5}\\right)=\\frac{1}{2}\n$$\nTo see how to utilize this, notice that $10<\\frac{p}{2}$ and $-\\fr... | United States | HMIC 2023 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 3, 5, 7, 13 | |
0bbb | Let $g : \mathbb{R} \to \mathbb{R}$ be a continuous, decreasing function, such that $g(\mathbb{R}) = (-\infty, 0)$. Prove that there are no continuous functions $f : \mathbb{R} \to \mathbb{R}$ such that the equality $f \circ f \circ \dots \circ f = g$ is true for some integer $k \ge 2$. | [
"Suppose that such a function $f$ exists. Injectivity of $g$ implies the injectivity of the continuous function $f$, which in turn is strictly monotone. As $g$ is increasing we conclude that $f$ is increasing and $k$ is an odd number. Moreover, $f$ is not surjective.\n\nDenote by $f^{[k]} = f \\circ f \\circ \\dots... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0hkl | Problem:
Find the units digit of $17^{2021}$. | [
"Solution:\nThe units digits of powers of $17$ cycle: $7, 9, 3, 1, 7, 9, 3, 1, \\ldots$, so the units digit of $17^{n}$ is $1$ whenever $n$ is a multiple of $4$. Since $2020$ is a multiple of $4$, $17^{2020}$ has units digit $1$, so $17^{2021}$ has units digit $7$."
] | United States | Berkeley Math Circle | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 7 | |
0d3k | Find all functions $f: \mathbb{N} \rightarrow (0, \infty)$ such that $f(4) = 4$ and
$$
\frac{1}{f(1) f(2)} + \frac{1}{f(2) f(3)} + \cdots + \frac{1}{f(n) f(n+1)} = \frac{f(n)}{f(n+1)}, \quad \forall n \in \mathbb{N},
$$
where $\mathbb{N} = \{1, 2, \ldots\}$ is the set of positive integers. | [
"Taking $n = 1$, we obtain $\\frac{1}{f(1)} = f(1)$. Because $f(1) > 0$, we deduce that $f(1) = 1$.\n\nTaking $n = 2$, we obtain $f(3) + 1 = f(2)^2$, and taking $n = 3$ we obtain $4 f(3) + 4 + f(2) = f(2) f(3)^2$. Because $f(2) \\neq 0$, this is equivalent to\n$$\nf(3) + 1 = f(2)^2, \\quad \\text{ and } \\quad 4 f(... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English, Arabic | proof and answer | f(n) = n for all positive integers n | |
0kar | Problem:
Let $S, P, A, C, E$ be (not necessarily distinct) decimal digits where $E \neq 0$. Given that $N=\sqrt{\overline{E S C A P E}}$ is a positive integer, find the minimum possible value of $N$. | [
"Solution:\n\nSince $E \\neq 0$, the 6-digit number $\\overline{E S C A P E}$ is at least $10^{5}$, so $N \\geq 317$. If $N$ were 317 or 318, the last digit of $N^{2}$ would not match the first digit of $N^{2}$, which contradicts the condition. However, $N=319$ will work, since the first and last digit of $N^{2}$ a... | United States | HMMT February 2019 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 319 | |
0658 | If the number $K = \frac{9n^2 + 31}{n^2 + 7}$ is integer, find the possible values of $n \in \mathbb{Z}$. | [
"We have\n$$\nK = \\frac{9n^2 + 31}{n^2 + 7} = \\frac{9(n^2 + 7) - 32}{n^2 + 7} = 9 - \\frac{32}{n^2 + 7}.\n$$\nSince $K$ is integer, it follows that $n^2 + 7$ is a divisor of $32$ and taking in mind that $n^2 + 7 \\ge 8$, we conclude:\n$$\nn^2 + 7 \\in \\{8, 16, 32\\} \\Leftrightarrow n^2 \\in \\{1, 9, 25\\} \\Lef... | Greece | 26th Hellenic Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | {-5, -3, -1, 1, 3, 5} | |
0hxg | Problem:
Do there exist positive integers $a_{1}, \ldots, a_{100}$ such that for each $k=1, \ldots, 100$, the number $a_{1}+\cdots+a_{k}$ has exactly $a_{k}$ divisors? | [
"Solution:\n\nAnswer: yes.\n\nThe idea is to define the partial sums instead as follows. Let $d(n)$ denote the divisor function. Let $s_{N}$ be suitably large, then define by downwards recursion\n$$\ns_{n}=s_{n+1}-d\\left(s_{n+1}\\right)\n$$\nwith $N$ set such that $s_{0}=1$. Then $s_{k}=a_{1}+\\cdots+a_{k}$ works ... | United States | Berkeley Math Circle | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | yes | |
061o | Problem:
Gegeben ist ein Dreieck $ABC$ und ein Punkt $M$ so, dass die Geraden $MA$, $MB$, $MC$ die Geraden $BC$, $CA$, $AB$ (in dieser Reihenfolge) in $D$, $E$ beziehungsweise $F$ schneiden.
Man beweise, dass es dann stets die Zahlen $\varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3}$ aus $\{-1,1\}$ gibt, so dass gilt... | [
"Solution:\n\nPunkt $M$ ($M \\notin \\{A, B, C\\}$) kann entweder auf einer der vorgegebenen Geraden, oder in einer der sieben Gebiete liegen, in denen die Ebene des Dreiecks $ABC$ durch die Geraden $AB$, $BC$, $CA$ geteilt wird.\n\n\n\nEs ist stets\n$$\n\\begin{aligned}\n& \\frac{MD}{AD} =... | Germany | Auswahlwettbewerb zur IMO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0c9q | Problem:
Pe o tablă sunt scrise numerele de forma $n(n+1)$, cu $n=1,2,3, \ldots, 2020$. Un copil alege trei numere $a, b$ și $c$ de pe tablă, le șterge și scrie pe tablă numărul $\frac{a b c}{a b+a c+b c}$. După 1009 astfel de operații, unul dintre numerele rămase pe tablă este 47.
a) Calculați suma inverselor numerel... | [] | Romania | Olimpiada Națională GAZETA MATEMATICĂ | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | a) 2020/2021; b) 2021/1977 | |
0h1a | Find all integer $n$, that satisfy the following equality:
$$
(n-1)(n-3)(n-5)\dots(n-2011) = n(n+2)(n+4)\dots(n+2010).
$$ | [
"If $n$ is even, then LHS is odd, and RHS is even, and vice versa (for odd $n$)."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | no integer solutions | |
0cd2 | A domino is a rectangle formed by two unit squares that share a common side. A number of $18$ dominoes fit together to tile a $6 \times 6$ square. Show that some line crossing the interior of the square crosses the interior of no domino. Is it possible that such a line be unique? | [
"Let the square be $[0, 6] \\times [0, 6]$. We first show that either some grid-vertical $x = i$, $i = 1, 2, 3, 4, 5$, or some grid-horizontal $y = j$, $j = 1, 2, 3, 4, 5$, crosses no tile. Let $m_i$ and $n_j$ be the number of tiles crossed by the grid-vertical $x = i$ and the grid-horizontal $y = j$, respectively.... | Romania | THE Sixteenth STARS OF MATHEMATICS Competition | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Yes | |
0hyp | Problem:
On the bases $AB$ and $CD$ of a trapezoid $ABCD$ draw two squares externally to $ABCD$. Let $O$ be the intersection point of the diagonals $AC$ and $BD$, and let $O_{1}$ and $O_{2}$ be the centers of the two squares. Prove that $O_{1}$, $O$ and $O_{2}$ lie on a line (i.e. they are collinear; see Figure.) | [
"Solution:\n\nThe idea is to show that $\\triangle OCO_{2} \\sim \\triangle OAO_{1}$. Indeed, first notice that $\\triangle AO_{1}B \\sim \\triangle CO_{2}D$ — both are right isosceles triangles. Therefore, $AO_{1} : CO_{2} = AB : CD$. But $\\triangle AOB \\sim \\triangle COD$ ($AB \\parallel CD \\Rightarrow$ all t... | United States | BAMO | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0hfs | Prove that for any integer $n$ there is a monic quadratic polynomial $x^2 + bx + c$ with integer coefficients which attains values $n, n^2, n^3$ at some three integer points. | [
"$$\nf(x_1) = n,\\ f(x_2) = n^2,\\ f(x_3) = n^3.\n$$\nThen for a polynomial $g(x) = f(x) - n$ we have $g(x_1) = 0$, $g(x_2) = n^2 - n$, $g(x_3) = n^3 - n$, so $g(x) = (x - x_1)(x - t)$ for some integer $t$. Then we need to find $x_1, x_2, x_3, t$, such that $g(x_2) = (x_2 - x_1)(x_2 - t) = n^2 - n$, $g(x_3) = (x_3 ... | Ukraine | Problems from Ukrainian Authors | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
03h0 | Problem:
Prove that the equation $x^{3} + 11^{3} = y^{3}$ has no solution in positive integers $x$ and $y$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Algebraic Number Theory > Unique factorization",
"Number Theory > Algebraic Number Theory > Quadratic fields"
] | null | proof only | null | |
03hl | Problem:
For each real number $r$, $[r]$ denotes the largest integer less than or equal to $r$, e.g., $[6] = 6$, $[\pi] = 3$, $[-1.5] = -2$. Indicate on the $(x, y)$-plane the set of all points $(x, y)$ for which $[x]^2 + [y]^2 = 4$. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | {(x,y): x in [2,3), y in [0,1)} ∪ {(x,y): x in [-2,-1), y in [0,1)} ∪ {(x,y): x in [0,1), y in [2,3)} ∪ {(x,y): x in [0,1), y in [-2,-1)} | |
07wa | Suppose $f : \{0,1\}^{10} \to \mathbb{R}$, i.e. $f(x_1, x_2, x_3, \dots, x_{10})$ is defined whenever $x_i \in \{0,1\}$ for each $1 \le i \le 10$. We are not given these values $f(x_1, \dots, x_{10})$, but, for every choice of $x_1, \dots, x_{10} \in \{0,1\}$, we know each of the following ten sums of two values of $f$... | [
"We consider the natural generalisation with $10$ replaced everywhere by $n$ for some $n > 1$, $n \\in \\mathbb{N}$. It is convenient to use vector notation, writing $f(\\underline{x})$ in place of $f(x_1, \\dots, x_n)$. To show that the given sums are insufficient to compute all $f$-values, it suffices to note tha... | Ireland | IRL_ABooklet_2023 | [
"Algebra > Linear Algebra > Vectors",
"Algebra > Linear Algebra > Linear transformations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | One additional sum suffices; for example, f(0,0,0,0,0,0,0,0,0,0) + f(1,1,0,0,0,0,0,0,0,0). More generally, any fixed extra sum f(a) + f(b) where a and b differ in an even number of positions (equivalently have the same parity of coordinate sum) suffices to determine all values of f when combined with the given sums. | |
0it5 | Problem:
Let $P_{1}, P_{2}, \ldots, P_{8}$ be 8 distinct points on a circle. Determine the number of possible configurations made by drawing a set of line segments connecting pairs of these 8 points, such that: (1) each $P_{i}$ is the endpoint of at most one segment and (2) no two segments intersect. (The configuration... | [
"Solution:\nAnswer: 323\nLet $f(n)$ denote the number of valid configurations when there are $n$ points on the circle. Let $P$ be one of the points. If $P$ is not the end point of an edge, then there are $f(n-1)$ ways to connect the remaining $n-1$ points. If $P$ belongs to an edge that separates the circle so that... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | proof and answer | 323 | |
054c | Prove that for any positive integer $n$, $2 \cdot \sqrt{3} \cdot \sqrt[3]{4} \cdot \dots \cdot \sqrt[n-1]{n} > n$. | [
"For $2 \\le k \\le n$, the GM-HM inequality for the numbers $k, \\dots, k, 1$, with $k$ repeated $k-2$ times, gives\n$$\nk - 1 = \\frac{(k - 1)^2}{k - 1} = \\frac{k(k - 2) + 1}{k - 1} \\ge \\sqrt[k-1]{k^{k-2}} = \\sqrt[k-1]{\\frac{k^{k-1}}{k}} = \\frac{k}{\\sqrt[k-1]{k}}\n$$\nHence $\\sqrt[k-1]{k} \\ge \\frac{k}{k... | Estonia | Estonian Math Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
02cw | Problem:
Uma das diagonais de um quadrilátero inscritível é um diâmetro de seu círculo circunscrito.
Verifique que as interseções $E$ e $F$ das retas perpendiculares por $A$ e $C$, respectivamente, à reta $BD$ satisfazem
$$
DE = BF
$$ | [
"Solution:\nEstenda as perpendiculares $CF$ e $AE$ para a segunda interseção com o círculo nos pontos $H$ e $G$, respectivamente. Como $AG \\parallel CH$ e $AC$ é um diâmetro, segue que $AGCH$ é um retângulo. Seja $M$ o ponto médio de $EF$. Por ser uma corda da circunferência, temos $OM$ perpendicular a $BD$. Usand... | Brazil | null | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0agt | Let $ABCD$ be an inscribed quadrangle, and let $BC$ and $AD$ intersect at point $P$. The point $Q$ belongs to the line $BP$ in such a way that $\overline{PQ} = \overline{BP}$, and $CAQR$ and $DBCS$ are parallelograms. Prove that the points $C, Q, R$ and $S$ are concyclic. | [
"Obviously, it is enough to show that\n$$\n\\angle RQC = \\angle RSC\n$$\n(*)\nFrom the conditions of the problem we have\n$$\n\\angle RQC = \\angle ACQ = \\angle ACB = \\angle ADB\n$$\n(1)\nWe choose a point $T$, such that $QABT$ is a parallelogram. Then $\\overline{BT} = \\overline{AQ} = \\overline{CR}$ and $\\ov... | North Macedonia | IMO Team Selection Test | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08tt | Determine the smallest possible positive integer $n$ for which the 4 numbers $14n$, $16n$, $18n$, $20n$ have exactly the same number of positive factors. | [
"Let us first prove the following Theorem:\n\n**Theorem**\nSuppose that a positive integer $n$ has the prime factorization $n = p_1^{e_1} p_2^{e_2} \\cdots p_m^{e_m}$. (Here, $p_1, p_2, \\cdots, p_m$ are distinct prime numbers and $e_1, e_2, \\cdots, e_m$ are positive integers.) Then the total number of factors of ... | Japan | Japan Junior Mathematical Olympiad First Round | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 30 | |
093k | Problem:
Prove that every integer from $1$ to $2019$ can be represented as an arithmetic expression consisting of up to $17$ symbols $2$ and an arbitrary number of additions, subtractions, multiplications, divisions and brackets. The $2$'s may not be used for any other operation, for example to form multi-digit number... | [
"Solution:\n\nWe will first prove by induction that every even number less than $2^{n}$ can be written with at most $\\frac{3}{2} n-1$ $2$'s. This is certainly true for $n=2$ and $n=3$ with $2=2$, $4=2+2$ and $6=2+2+2$.\n\nLet $k \\geq 8$ be an even number $<2^{n}$. If it is divisible by $4$, it can be written as $... | Middle European Mathematical Olympiad (MEMO) | MEMO Team Competition | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Other"
] | null | proof only | null | |
0lcl | The teacher has $2013$ candies of $11$ different types. She distributes the candies to her students such that no student obtains more than one candy of each type. She then asks each pair of students to write down on the board the number of candy types that they have in common. Let $M$ be the sum of written numbers.
a)... | [
"a) Suppose that there are $m$ students labeled by $1,2,3,\\ldots,m$ and $11$ types of candy denoted by $a_1,a_2,a_3,\\ldots,a_{11}$. Let $X=\\{a_1,a_2,a_3,\\ldots,a_{11}\\}$ and $A_1,A_2,\\ldots,A_m$ be the set of candy types that the students $1,2,3,\\ldots,m$ received, respectively. We have $A_1,A_2,\\ldots,A_m ... | Vietnam | Vietnamese Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | a) 183183; b) 224115 (achieved when three types have 223 candies and six types have 224 candies). | |
01ig | Let $D := \mathbb{R} \setminus \{0,1\}$. Find all functions $f:D \rightarrow D$ which satisfy for any $x, y \in \mathbb{R}$ with $x, xy \in D$ the equation
$$
f(f(xy)) = 1 - \frac{1}{y f(f(f(x)))}
$$ | [
"Plugging in $y = \\frac{a}{x}$ for $a, x \\in D$ gives\n$$\nf(f(a)) = 1 - \\frac{x}{a f(f(f(x)))}.\n$$\nOn the other hand, $y = 1$ and $x = a$ gives for $a \\in D$\n$$\nf(f(a)) = 1 - \\frac{1}{f(f(f(a)))}.\n$$\nfrom which we conclude $\\frac{x}{f(f(f(x)))}$ is constant for all $x \\in D$. Thus it follows $f(f(f(x)... | Baltic Way | Baltic Way 2023 Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 1/(1 - x) for all x in D | |
06cj | Let $I$ and $O$ be the incentre and circumcentre of $\triangle ABC$, respectively. Assume $\triangle ABC$ is not equilateral (so $I \neq O$). Prove that $\angle AIO \leq 90^\circ$ if and only if $2BC \leq AB + CA$. | [
"Let $AI$ meet $(ABC)$ again at $D$. Recall that $DB = DI = DC$. Applying Ptolemy's theorem to $ABCD$, we obtain\n$$\nAD \\times BC = AB \\times CD + AC \\times BD = DI(AB + AC).\n$$\nNote that $\\angle AIO \\le 90^\\circ$ if and only if $AI \\ge ID$. This is equivalent to\n$$\n2 \\le \\frac{AD}{DI} = \\frac{AB + A... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
06m8 | If $x^2 - y^2 = \frac{1}{22}$ and $x \neq 0$, find the greatest possible value of $\frac{1 - 22xy}{x^2}$. | [
"**Answer:** $\\frac{55}{2}$\nLet $x = \\frac{1}{\\sqrt{22}} \\sec \\theta$ and $y = \\frac{1}{\\sqrt{22}} \\tan \\theta$. Then $x^2 - y^2 = \\frac{1}{22}$ and\n$$\n\\begin{aligned}\n\\frac{1 - 22xy}{x^2} &= \\frac{1 - \\sec \\theta \\tan \\theta}{\\frac{1}{22} \\sec^2 \\theta} \\\\\n&= 22(\\cos^2 \\theta - \\sin \... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 55/2 | |
0dtq | Let $a_1, a_2, \dots$ be a sequence of positive numbers satisfying, for any positive integers $k, l, m, n$ such that $k + n = m + l$,
$$
\frac{a_k + a_n}{1 + a_k a_n} = \frac{a_m + a_l}{1 + a_m a_l}.
$$
Show that there exist positive numbers $b, c$ so that $b \le a_n \le c$ for any positive integer $n$. | [
"Let $A_{k+n} = \\frac{a_k+a_n}{1+a_k a_n}$. So for any $n$, $A_n = \\frac{a_1+a_{n-1}}{1+a_1 a_{n-1}}$. Consider the function\n$$\nf(x) = \\frac{a_1 + x}{1 + a_1 x}\n$$\nwhere $x > 0$. Since $f(x) = \\frac{1}{a_1} \\frac{a_1^2+a_1x}{1+a_1x} = \\frac{a_1(a_1+a_1x)}{a_1+a_1^2x}$, it follows that\n$$\nf(x) \\geq \\be... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
05zz | Problem:
Déterminer toutes les fonctions $f: \mathbb{R} \rightarrow \mathbb{R}$ telles que pour tous $x, y$ réels, on ait
$$
f\left(x^{2}+x y+f\left(y^{2}\right)\right)=x f(y)+f\left(x^{2}\right)+y^{2}
$$ | [
"Solution:\nEn posant $y=0$, on trouve $f\\left(x^{2}+f(0)\\right)=x f(0)+f\\left(x^{2}\\right)$. En remplaçant $x$ par $-x$ dans cette équation, on trouve $f\\left(x^{2}+f(0)\\right)=-x f(0)+f\\left(x^{2}\\right)$, et en combinant on obtient donc $2 x f(0)=0$ pour tout $x$, soit $f(0)=0$.\n\nEn posant $x=0$, on tr... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x and f(x) = -x | |
0g78 | 是否能夠將所有的自然數分成6個兩兩互斥的子集合 $A_1, A_2, \dots, A_6$, 使得滿足 $x+2y=5z$ 的任意正整數 $x, y, z$, 都不同會時落在某一個 $A_i$ 之中? | [
"可以。\n令 $A_i$ 為所有形如 $7^m(7k + i)$ 的正整數所成的集合，其中 $m,k$ 為非負整數，$i = 1,2,3,4,5,6$。下證 $A_1,A_2,A_3,A_4,A_5,A_6$ 滿足題設。\n\n現設 $x,y,z \\in A_i$ 且滿足 $x+2y=5z$；將 $x,y,z$ 分別寫成\n$$\nx = 7^{m_1}(7k_1 + i), \\quad y = 7^{m_2}(7k_2 + i), \\quad z = 7^{m_3}(7k_3 + i), \\quad m_j, k_j \\in \\mathbb{N} \\cup \\{0\\},\n$$\n且令 $\\alpha... | Taiwan | 二0一三數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 可以 | |
0kbg | Problem:
Two hexagons are attached to form a new polygon $P$. Compute the minimum number of sides that $P$ can have. | [
"Solution:\n\nA triangle can be split into two hexagons; in other words, two concave hexagons can be attached to form a triangle, like the following:\n\n"
] | United States | HMMO 2020 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 3 | |
0dvz | Problem:
Poenostavi:
$$
\frac{a^{3}-1}{1+\frac{1}{a-\frac{a}{a+1}}}
$$ | [
"Solution:\n\nIzraz poenostavimo:\n$$\n\\frac{a^{3}-1}{1+\\frac{1}{a-\\frac{a}{a+1}}} = \\frac{a^{3}-1}{1+\\frac{1}{\\frac{a^{2}}{a+1}}} = \\frac{a^{3}-1}{\\frac{a^{2}+a+1}{a^{2}}} = \\frac{(a-1)\\left(a^{2}+a+1\\right) a^{2}}{a^{2}+a+1} = a^{2}(a-1)\n$$\n\nPoenostavljeno do oblike: $\\frac{a^{3}-1}{1+\\frac{1}{\\f... | Slovenia | 4. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | a^2(a-1) | |
04fr | A box contains $k$ balls marked by $\binom{k}{1}$ for all $k = 1, 2, \dots, 50$. The balls are drawn from the box without looking. What is the minimal number of balls that need to be drawn to be sure that at least 10 balls with the same mark have been drawn? (AHSME 1994) | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 415 | |
06n2 | Find the largest real root of the equation $2x^2 + 6x + 9 = 7x\sqrt{2x + 3}$. | [
"Let $u = \\sqrt{2x+3}$. Then the equation becomes $2x^2 + 3u^2 = 7xu$, or $(2x-u)(x-3u) = 0$. Hence either $u = 2x$ or $x = 3u$.\n\n* If $u = 2x$, we get $2x = \\sqrt{2x+3}$. Squaring both sides gives $4x^2 = 2x + 3$, which leads to $x = \\frac{1 \\pm \\sqrt{13}}{4}$.\n\n* If $x = 3u$, we get $\\frac{x}{3} = \\sqr... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Intermediate Algebra > Other"
] | English | final answer only | 9 + 6√3 | |
0e7m | Positive real numbers $x$ and $y$ satisfy
$$
2013^{\log_3 x} = y^{\log_5 2013} \quad \text{and} \quad \log_{\frac{1}{2}} x + \log_{\frac{1}{2}} y > 0.
$$
Which of the numbers $x$ and $y$ is greater? | [
"Taking logarithms on both sides of the equation and taking into account that $\\log a^b = b \\log a$ and $\\log_a b = \\frac{\\log b}{\\log a}$, we get\n$$\n\\frac{\\log x \\log 2013}{\\log 3} = \\frac{\\log y \\log 2013}{\\log 5},\n$$\nor\n$$\n\\log y = \\frac{\\log 5}{\\log 3} \\log x = \\log_3 5 \\log x.\n$$\nF... | Slovenia | National Math Olympiad 2013 - Final Round | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | x > y | |
0huu | Problem:
Aerith and Bob take turns picking a nonnegative integer, each time subtracting a (positive) divisor from the other's last number. The first person to pick $0$ loses. For example, if Aerith reached $2020$ on some turn, Bob could pick $2020-20=2000$, as $20$ is a divisor of $2020$.
Continuing this example (with... | [
"Solution:\n\nIf $1998 = 2000 - 2$ were a losing position, Aerith would pick it. Otherwise, she can choose $1999 = 2000 - 1$. And since this is prime, Bob must pick either $1999 - 1999 = 0$ or $1999 - 1 = 1998$. In any case, Aerith wins."
] | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | Aerith | |
0146 | Problem:
Let $K$ and $N$ be positive integers with $1 \leq K \leq N$. A deck of $N$ different playing cards is shuffled by repeating the operation of reversing the order of the $K$ topmost cards and moving these to the bottom of the deck. Prove that the deck will be back in its initial order after a number of operatio... | [
"Solution:\n\nLet $N = q \\cdot K + r$, $0 \\leq r < K$, and let us number the cards $1, 2, \\ldots, N$, starting from the one at the bottom of the deck. First we find out how the cards $1, 2, \\ldots, K$ are moving in the deck.\n\nIf $i \\leq r$ then the card $i$ is moving along the cycle\n$$\n\\begin{aligned}\n& ... | Baltic Way | Baltic Way 2005 | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Number Theory > Other"
] | null | proof only | null | |
0jtb | Problem:
Let $t=2016$ and $p=\ln 2$. Evaluate in closed form the sum
$$
\sum_{k=1}^{\infty}\left(1-\sum_{n=0}^{k-1} \frac{e^{-t} t^{n}}{n!}\right)(1-p)^{k-1} p
$$ | [
"Solution:\nLet $q=1-p$. Then\n$$\n\\begin{aligned}\n\\sum_{k=1}^{\\infty}\\left(1-\\sum_{n=0}^{k-1} \\frac{e^{-t} t^{n}}{n!}\\right) q^{k-1} p & =\\sum_{k=1}^{\\infty} q^{k-1} p-\\sum_{k=1}^{\\infty} \\sum_{n=0}^{k-1} \\frac{e^{-t} t^{n}}{n!} q^{k-1} p \\\\\n& =1-\\sum_{k=1}^{\\infty} \\sum_{n=0}^{k-1} \\frac{e^{-... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 1 - (1/2)^{2016} | |
0cfv | a) Give an example of a finite nonabelian group $G$, with unit element $e$, having a proper subgroup $H$ with the property that $x^2 = e$ for any $x \in G \setminus H$.
b) Let $G$ be a finite group, and $H$ a proper subgroup such that $x^2 = e$ for any $x \in G \setminus H$. If $|G| > 2 \cdot |H|$, show that the group... | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Abstract Algebra > Group Theory"
] | English | proof and answer | a) For example, take G = S3 and H = A3; all elements outside H are transpositions and square to the identity. (Alternatively, G = the dihedral group of order 8 with H the rotation subgroup.) b) Under the given conditions, G is commutative. | |
038u | The edge $DA$ of a pyramid $ABCD$ is perpendicular to its base $ABC$, $(ABD) \perp (BCD)$, $\angle BDC = 45^\circ$ and $DB = 2$. Find $\angle ADB$ if the sum of the squares of lateral faces of the pyramid equals $8$. | [
"Since $(ABD) \\perp (ABC), (BCD)$, then $(ABC) \\perp BC$ and hence $\\angle ABC = \\angle DBC = 90^\\circ$. Since $\\angle BDC = 45^\\circ$, then $BC = BD = 2$. Let $\\angle ADB = \\alpha$. Then $AB = 2 \\sin \\alpha$, $AD = 2 \\cos \\alpha$ and $AC = 2\\sqrt{\\sin^2 \\alpha + 1}$. So\n$$\nS_{ABD} = \\frac{AB \\c... | Bulgaria | Spring Mathematical Tournament | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | 45° | |
0kfi | Problem:
Let $P(x)$ be the unique polynomial of degree at most $2020$ satisfying $P\left(k^{2}\right)=k$ for $k=0,1,2, \ldots, 2020$. Compute $P\left(2021^{2}\right)$. | [
"Solution:\nSince $P(0)=0$, we see that $P$ has no constant term. Let $Q(x)=\\frac{P\\left(x^{2}\\right)-x}{x}$ be a polynomial with degree at most $4039$. From the given values of $P$, we see that $Q(k)=0$ and $Q(-k)=-2$ for $k=1,2,3, \\ldots, 2020$.\nNow, consider the polynomial $R(x)=Q(x+1)-Q(x)$, which has degr... | United States | HMMT February 2020 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2021 - binom(4040, 2020) | |
0h3t | Inscribed circle of triangle *ABC* touches its sides *AB*, *BC*, *CA* at the points *K*, *N*, *M* respectively. Given that $\angle MKC = \angle MNA$, prove that triangle *ABC* is isosceles. | [
"Нехай прямі $AN$ і $CK$ вдруге перетинають вписане коло трикутника $ABC$ в точках $P$ і $Q$ відповідно. За теоремами про вписаний кут та кут між дотичною й хордою одержуємо:\n$$\n\\angle PNM = \\angle PQM = \\angle PMA,\n$$\n$$\n\\angle QKM = \\angle QPM = \\angle QMC.\n$$\nЗа умовою задачі, $\\angle PNM = \\angle... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ... | English | proof only | null | |
0aw4 | Problem:
Let $f(x)$ be a function such that $f(1)=1$, $f(2)=2$ and $f(x+2)=f(x+1)-f(x)$. Find $f(2016)$. | [
"Solution:\n$$\n\\begin{aligned}\n& f(1)=1 \\\\\n& f(2)=2 \\\\\n& f(3)=f(2)-f(1)=2-1=1 \\\\\n& f(4)=f(3)-f(2)=1-2=-1 \\\\\n& f(5)=f(4)-f(3)=-1-1=-2 \\\\\n& f(6)=f(5)-f(4)=-2-(-1)=-1 \\\\\n& f(7)=f(6)-f(5)=-1-(-2)=1 \\\\\n& f(8)=f(7)-f(6)=1-(-1)=2\n\\end{aligned}\n$$\nObserve that this pattern will repeat itself eve... | Philippines | 18th PMO National Stage Oral Phase | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | -1 | |
058j | The lines tangent to the circumcircle of triangle $ABC$ at points $B$ and $C$ intersect at point $D$. The circumcircle of triangle $BCD$ intersects the lines $AB$ and $AC$ the second time at points $K$ and $L$, respectively. Prove that the line $AD$ bisects the line segment $KL$. | [
"We prove that $AKDL$ is a parallelogram; this implies the desired claim since $AD$ and $KL$ are diagonals of this quadrilateral. We prove at first that $KD \\parallel AL$.\n\nIf $K$ lies between $A$ and $B$ (Fig. 19) then by inscribed angles $\\angle BAC = \\angle BCD$ and $\\angle BCD = \\angle BKD$. Consequently... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0jpe | Problem:
Two random points are chosen on a segment and the segment is divided at each of these two points. Of the three segments obtained, find the probability that the largest segment is more than three times longer than the smallest segment. | [
"Solution:\n\nAnswer: $\\frac{27}{35}$\n\nWe interpret the problem with geometric probability. Let the three segments have lengths $x$, $y$, $1-x-y$ and assume WLOG that $x \\geq y \\geq 1-x-y$. Every possible $(x, y)$ can be found in the triangle determined by the points $\\left(\\frac{1}{3}, \\frac{1}{3}\\right)$... | United States | HMMT November 2015 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 27/35 | |
0h70 | The country Plato has the shape of a convex polygon in vertices of which there are border towers. Trucker needs to drive through all towers, and for every kilometre he is ought to pay one puylyk (state currency) to swindlers from the government (route does not have to be closed, the only condition is to visit each towe... | [
"If the state has the shape of an equilateral triangle, then obviously, trucker can do the job and pay only 2000 puylyks (in this case the route of the trucker lies along the two sides of the triangle).\n\nNow let us prove that in any case the trucker will have to pay at least 2000 puylyks. Suppose that there is a ... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | null | proof and answer | 2000 | |
01du | Consider $m \ge 3$ positive real numbers $g_1, \dots, g_m$, each number being less than the sum of the others. For any subset $M \subseteq \{1, \dots, m\}$, denote
$$
S_M = \sum_{k \in M} g_k.
$$
Find all $m$ for which it is always possible to partition the indices $1, \dots, m$ into three sets $A, B, C$, with the prop... | [
"Answer: The partition is always possible precisely when $m \\ne 4$.\nFor $m = 3$ it is trivially possible, and for $m = 4$ the four equal numbers $g, g, g, g$ provide a counter-example. Henceforth, we assume $m \\ge 5$.\nAmong all possible partitions $A \\sqcup B \\sqcup C = \\{1, \\dots, m\\}$ such that\n$$\nS_A ... | Baltic Way | Baltic Way shortlist | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All m except 4 (i.e., m = 3 and all m ≥ 5). | |
0cxe | Find all triples $(x, y, z)$ of real numbers such that
$$
x^{2}+y^{2}+z^{2}+1=xy+yz+zx+|x-2y+z| .
$$ | [
"We can write\n$$\nx^{2}+y^{2}+z^{2}+1=xy+yz+zx+|x-y+z-y|\n$$\nhence\n$$\n(x-y)^{2}+(y-z)^{2}+(z-x)^{2}+2=2|x-y+z-y| .\n$$\nIt follows\n$$\n(x-y)^{2}+(y-z)^{2}+(z-x)^{2}+2 \\leq 2|x-y|+2|y-z| .\n$$\nThe last relation is equivalent to\n$$\n(|x-y|-1)^{2}+(|y-z|-1)^{2}+(z-x)^{2} \\leq 0 .\n$$\nWe get $|x-y|=1$, $|y-z|... | Saudi Arabia | SAMC | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | All triples of the form (a, a − 1, a) or (a, a + 1, a) for any real a. | |
0bid | For every positive integer $n$, let $\sigma(n)$ denote the sum of all positive divisors of $n$ (1 and $n$, inclusive). Show that a positive integer $n$, which has at most two distinct prime factors, satisfies the condition $\sigma(n) = 2n - 2$ if and only if $n = 2^k(2^{k+1} + 1)$, where $k$ is a non-negative integer a... | [
"$$\n1 + \\frac{1}{p} \\le \\frac{\\sigma(p^l)}{p^l} = \\frac{p - \\frac{1}{p^l}}{p-1} < \\frac{p}{p-1},\n$$\n\n$$\n(2^{k+1} - 1) \\left(1 + \\frac{1}{p}\\right) \\le \\frac{\\sigma(n)}{p^l} = 2^{k+1} - \\frac{2}{p^l} < (2^{k+1} - 1) \\frac{p}{p-1}.\n$$\nBy the first inequality, $(2^{k+1}-1)(1+1/p) < 2^{k+1}$, so $... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | n = 2^k(2^{k+1} + 1) with k a non-negative integer and 2^{k+1} + 1 prime | |
0k19 | Problem:
Let a positive integer $n$ be called a cubic square if there exist positive integers $a, b$ with $n = \operatorname{gcd}\left(a^{2}, b^{3}\right)$. Count the number of cubic squares between 1 and 100 inclusive. | [
"Solution:\nThis is easily equivalent to $v_{p}(n) \\not \\equiv 1,5 \\pmod{6}$ for all primes $p$. We just count: $p \\geq 11 \\Longrightarrow v_{p}(n)=1$ is clear, so we only look at the prime factorizations with primes from $\\{2,3,5,7\\}$. This is easy to compute: we obtain 13."
] | United States | HMMT February 2018 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 13 | |
0fyn | Problem:
Finde alle Funktionen $f: \mathbb{R} \rightarrow \mathbb{R}$, die für alle reellen $x, y$ die folgende Gleichung erfüllen:
$$
f\left(x^{4}+y^{4}\right)=x f\left(x^{3}\right)+y^{2} f\left(y^{2}\right)
$$ | [
"Solution:\nMit $x=y=0$ erhält man $f(0)=0$. Mit $y=0$ folgt weiter\n$$\nx f\\left(x^{3}\\right)=f\\left(x^{4}\\right)=f\\left((-x)^{4}\\right)=-x f\\left(-x^{3}\\right)\n$$\nalso ist $f$ eine ungerade Funktion. Mit $x=0$ erhält man $f\\left(y^{4}\\right)=y^{2} f\\left(y^{2}\\right)$, also\n$$\nf\\left(z^{2}\\right... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = a x for all real x, where a is any real constant | |
0hhr | Find all pairs of natural numbers $(m, n)$ for which $2^n - 13^m$ is a cube of a natural number. | [
"Let $2^n - 13^m = k^3$. Then $2^n \\equiv k^3 \\pmod{13}$. If $n$ is not divisible by $3$, then $2^n$ can be written as $4l^3$ or $2l^3$. Then either $4$ or $2$ is a cube of some integer modulo $13$. But by simple checking, we can see that only remainders $0$, $\\pm1$, $\\pm5$ can be cubes of integers modulo $13$.... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomi... | English | proof and answer | (m, n) = (2, 9) | |
0csb | 99 positive integers are arranged in a circle. For every two neighboring numbers, either they differ by 1, or they differ by 2, or one of them is twice the other. Prove that one of these numbers is divisible by 3. | [
"Suppose that none of the numbers is divisible by $3$. Then each number gives a remainder $1$ or $2$ when divided by $3$.\n\nBut numbers giving the same nonzero remainder modulo $3$ cannot differ by $1$ or $2$; nor can one be twice the other. Therefore, neighboring numbers must give different remainders modulo $3$,... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0hkp | Problem:
There are 25 people at a party and every pair of them is either friends or strangers. Prove that there are two people at the party who have the same number of friends. | [
"Solution:\n\nAssume that no two people have the same number of friends. Because the number of friends a person can have ranges from $0$ to $24$, we can label the people with their numbers of friends and every label will be used once. Now consider the people labeled $0$ and $24$. They cannot be friends, because the... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
08ng | Problem:
Let $M N P Q$ be a square of side length $1$, and $A, B, C, D$ points on the sides $M N$, $N P$, $P Q$, and $Q M$ respectively such that $A C \cdot B D = \frac{5}{4}$. Can the set $\{A B, B C, C D, D A\}$ be partitioned into two subsets $S_{1}$ and $S_{2}$ of two elements each, so that each one of the sums of ... | [
"Solution:\nThe answer is negative.\nSuppose such a partitioning was possible (Figure 7). Then $A B + B C + C D + D A \\in \\mathbb{N}$. But $(A B + B C) + (C D + D A) > A C + A C \\geq 2$, hence $A B + B C + C D + D A > 2$.\n\nOn the other hand, $A B + B C + C D + D A < (A N + N B) + (B P + P C) + (C Q + Q D) + (D... | JBMO | JBMO Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | No | |
0b92 | Let $a, b, c, d$ be positive integers, and let $p = a + b + c + d$. Prove that if $p$ is a prime, then $p$ is not a divisor of $ab - cd$. | [
"Consider the relation $(a+c)(b+c) = ab + ac + bc + c^2 = (a+b+c+d)c + ab - cd = pc + ab - cd$.\nIf $p$ divides $ab - cd$, then $p$ divides $a+c$ or $b+c$.\nOn the other hand, $0 < a+c < p$, and $0 < b+c < p$, a contradiction."
] | Romania | Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0as3 | Problem:
Let $x + \frac{1}{x} = \sqrt{2}$. Find the value of $x^{8} + \frac{1}{x^{8}}$. | [
"Solution:\nLet $x + \\frac{1}{x} = 2 \\cos \\theta$. Then\n$$\nx^{2} - 2 \\cos \\theta \\cdot x + 1 = 0\n$$\nwhich gives $x = \\cos \\theta \\pm i \\sin \\theta$.\n\nTherefore,\n$$\nx^{n} = \\cos n\\theta \\pm i \\sin n\\theta\n$$\nand\n$$\nx^{-n} = \\cos n\\theta \\mp i \\sin n\\theta.\n$$\nThus,\n$$\nx^{n} + \\f... | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | 2 | |
0il3 | Problem:
Show a way to construct an equiangular hexagon with side lengths $1,2,3,4,5$, and $6$ (not necessarily in that order). | [
"Solution:\n\nThe trick is to view an equiangular hexagon as an equilateral triangle with its corners cut off. Consider an equilateral triangle with side length $9$, and cut off equilateral triangles of side length $1$, $2$, and $3$ from its corners. This yields an equiangular hexagon with sides of length $1,6,2,4,... | United States | Harvard-MIT Mathematics Tournament, Team Round B | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
02ai | Problem:
Três cilindros têm alturas e raios das bases iguais a $10~\mathrm{cm} \times 10~\mathrm{cm}$, $10~\mathrm{cm} \times 5~\mathrm{cm}$ e $20~\mathrm{cm} \times 5~\mathrm{cm}$, e volumes $V_{1}, V_{2}$ e $V_{3}$, respectivamente.
a) Escreva em ordem crescente os volumes $V_{1}, V_{2}$ ... | [
"Solution:\n\n(a) Dado que o volume de um cilindro de raio $R$ e altura $h$ é $\\pi R^{2} h$, temos que os volumes $V_{1}, V_{2}$ e $V_{3}$ são:\n\n$V_{1} = \\pi \\times 10^{2} \\times 10 = 1000\\pi$\n\n$V_{2} = \\pi \\times 5^{2} \\times 10 = 250\\pi$\n\n$V_{3} = \\pi \\times 5^{2} \\times 20 = 500\\pi$\n\nAssim, ... | Brazil | Lista 2 | [
"Geometry > Solid Geometry > Volume"
] | null | final answer only | (a) V2 < V3 < V1. (b) One example: radius 5 cm and height 15 cm. (c) One example: radius 8 cm and height 10 cm. | |
0deg | Let $A$ be a point lies outside circle $(O)$ and tangent lines $AB$, $AC$ of $(O)$. Consider points $D$, $E$, $M$ on $(O)$ such that $MD = ME$. The line $DE$ cuts $MB$, $MC$ at $R$, $S$. Take $X \in OB$, $Y \in OC$ such that $RX$, $RY \perp DE$. Prove that $XY \perp AM$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof only | null | |
075e | Let $a \ge b$ and $c \ge d$ be real numbers. Prove that the equation
$$
(x + a)(x + d) + (x + b)(x + c) = 0
$$
has real roots. | [
"Let $f(x) = (x + a)(x + d) + (x + b)(x + c)$. The leading coefficient of $f(x)$ is $2$. Hence $f(x)$ is positive for large values of $x$. We have\n$$\n\\begin{aligned}\nf(-a) + f(-b) &= (b-a)(c-a) + (a-b)(d-b) = (a-b)(-c+a+d-b), \\\\\nf(-c) + f(-d) &= (a-c)(d-c) + (b-d)(c-d) = (c-d)(-a+c+b-d).\n\\end{aligned}\n$$\... | India | Indija TS 2012 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
05m3 | Problem:
Déterminer tous les nombres réels $x, y, z$ satisfaisant le système d'équations suivant :
$$
\begin{cases}
x = \sqrt{2y + 3} \\
y = \sqrt{2z + 3} \\
z = \sqrt{2x + 3}
\end{cases}
$$ | [
"Solution:\n\nIl est évident que les nombres $x, y, z$ doivent être strictement positifs. Les deux premières équations donnent $x^{2} = 2y + 3$ et $y^{2} = 2z + 3$. En les soustrayant, on obtient $x^{2} - y^{2} = 2(y - z)$. On en déduit que si $x \\leqslant y$ alors $y \\leqslant z$, et de même si $y \\leqslant z$ ... | France | Olympiades FRANçaises DE Mathématiques | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x = y = z = 3 | |
0fmz | Demuestra que el producto de los dos mil trece primeros términos de la sucesión
$$
a_n = 1 + \frac{1}{n^3}
$$
no llega a valer 3. | [
"Veamos por inducción que $p_n = a_1 \\cdot a_2 \\cdot a_3 \\dots a_n \\le 3 - \\frac{1}{n}$ y, así, quedará probado para el caso particular $n = 2013$ que se pide en el enunciado.\n\nPara $n = 1$ es $p_1 = a_1 = 1 + \\frac{1}{1^3} = 2 \\le 3 - \\frac{1}{1}$.\n\nSupongamos que es cierto para $n = k$, $p_k = a_1 \\c... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Spanish | proof only | null | |
0a0b | Problem:
Op een vismarkt staan 10 kraampjes die allemaal dezelfde 10 vissoorten verkopen. Alle vissen zijn gevangen in de Noordzee of de Middellandse Zee, en elk kraampje heeft per vissoort maar één zee van afkomst. Een aantal, $k$, klanten koopt van elk kraampje één vis zo dat ze één vis van elke soort hebben. Verder... | [
"Solution:\n\nHet antwoord is $2^{10}-10$. Ten eerste merken we op dat er $2^{10}$ mogelijke combinaties zijn voor de afkomsten per vissoort. We gaan laten zien dat er altijd minstens 10 uitzonderingen zijn (mogelijkheden die afvallen) en dat er een marktopzet is waarbij er precies 10 uitzonderingen zijn.\n\nWe ord... | Netherlands | Selectietoets | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof and answer | 1014 | |
0hnh | Problem:
A man is standing on a platform and sees his train move such that after $t$ seconds it is $2 t^{2} + d_{0}$ feet from his original position, where $d_{0}$ is some number. Call the smallest (constant) speed at which the man have to run so that he catches the train $v$. In terms of $n$, find the $n$th smallest v... | [
"Solution:\nThe train's distance from the man's original position is $t^{2} + d_{0}$, and the man's distance from his original position if he runs at speed $v$ is $v t$ at time $t$. We need to find where $t^{2} + d_{0} = v t$ has a solution. Note that this is a quadratic equation with discriminant $D = \\sqrt{v^{2}... | United States | null | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 4^{n-1} | |
0aly | Problem:
Six boy-girl pairs are to be formed from a group of six boys and six girls. In how many ways can this be done? | [] | Philippines | AREA STAGE | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | final answer only | 720 | |
0272 | Problem:
Um número inteiro positivo é chamado "equilibrado" se ele tem quatro algarismos, e um desses algarismos é igual à média dos outros três. Por exemplo: o número $2631$ é equilibrado porque $3$ é a média de $2, 6$ e $1$; $4444$ também é equilibrado porque $4$ é a média de $4, 4$ e $4$.
a) Encontre os três menore... | [
"Solution:\nUm número de quatro algarismos é equilibrado quando um de seus algarismos, digamos $a$, é a média dos outros três algarismos, digamos $b, c$ e $d$, ou seja, quando $(b+c+d) / 3 = a$ ou, equivalentemente, quando $(a+b+c+d) / 4 = a$. Vemos assim que um número $N$ de quatro algarismos é equilibrado quando ... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | a) 1003, 1012, 1021; b) 90 | |
08sc | How many ways are there to cut a cube $S$ into tetrahedron $\{T_1, \dots, T_k\}$ with following properties?
(1) Every vertex of $T_1, \dots, T_k$ is one of the vertices of $S$.
(2) For every $i \neq j$, the intersection of $T_i$ and $T_j$ is a common face of them, a common edge of them, a common vertex of them or empty... | [
"Consider a division of cube $S = ABCD - EFGH$ into $T_1, \\cdots, T_k$ with properties in the problem. Then one of the below holds.\n* $\\triangle ABC$ and $\\triangle ACD$ is a face of a tetrahedron.\n* $\\triangle ABD$ and $\\triangle BCD$ is a face of a tetrahedron.\nBy symmetry we can get the answer by countin... | Japan | Japan 2007 | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 74 | |
05cb | The diagonals of the quadrilateral *ABDE* meet at *C*. The segments *AB* and *CE* are of equal length 8 cm, and the segments *AE* and *CD* are also of equal length. The perimeter of the triangle *CDE* is 35 cm. Given that $\angle BAC = \angle AEC$, find the perimeter of the pentagon *ABCDE*. | [
"Using the condition of the problem, we get\n$$\n\\angle BAE = \\angle BAC + \\angle CAE = \\angle AEC + \\angle CAE\n$$\n(Fig. 22). From the triangle $ACE$ we get $\\angle AEC + \\angle CAE = \\angle DCE$. Thus $\\angle BAE = \\angle DCE$. At the same time, $AE = CD$ and $AB = CE$. Consequently, the triangles $AEB... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | English | proof and answer | 54 cm | |
083l | Problem:
Quante sono le coppie di interi positivi $ (x, y) $ che verificano l'equazione
$$
x^{2}+y^{2}-2004 x-2004 y+2 x y-2005=0 ?
$$
Nota: se $x \neq y$, le coppie $(x, y)$ e $(y, x)$ sono da considerarsi diverse. | [] | Italy | UNIONE MATEMATICA ITALIANA Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO BIENNIO | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | final answer only | 2004 | |
00y5 | Problem:
Let $NS$ and $EW$ be two perpendicular diameters of a circle $\mathcal{C}$. A line $l$ touches $\mathcal{C}$ at point $S$. Let $A$ and $B$ be two points on $\mathcal{C}$, symmetric with respect to the diameter $EW$. Denote the intersection points of $l$ with the lines $NA$ and $NB$ by $A'$ and $B'$, respectiv... | [
"Solution:\n\nWe have $\\angle NAS = \\angle NBS = 90^{\\circ}$ (see Figure 1). Thus, the triangles $NA'S$ and $NSA$ are similar. Also, the triangles $B'NS$ and $SNB$ are similar and the triangles $NSA$ and $SNB$ are congruent. Hence, the triangles $NA'S$ and $B'NS$ are similar which implies $\\frac{SA'}{SN} = \\fr... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
02tv | Problem:
Umberto e Doisberto jogam em um tabuleiro $3 \times n$ colocando dominós sempre cobrindo duas casas adjacentes (com lado em comum) do tabuleiro. Umberto faz a primeira jogada, Doisberto faz a segunda e eles seguem jogando alternadamente. Perde o jogador que não conseguir jogar. Para cada um dos casos abaixo, ... | [
"Solution:\n\na) Doisberto pode sempre garantir a vitória. Basta ele realizar um movimento que complete um quadrado $2 \\times 2$ a partir do primeiro dominó de Umberto.\n\n\n\nVeja na figura que sobram 5 casas. Independente da jogada de Umberto, na jogada seguinte de Doisberto, o jogo acab... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) Doisberto
b) Umberto | |
0hb3 | From natural numbers $2, 3, 4, \ldots, 2019$ one constructs $1009$ fractions, and chooses the maximum of these. What is the minimum possible value of this maximum fraction?
(Bogdan Rublyov) | [
"First of all, note that this value can be achieved with the following fractions choice:\n$$\n\\frac{1}{1011}, \\frac{2}{1012}, \\frac{3}{1013}, \\dots, \\frac{1010}{2019}.\n$$\n\nFor the sake of contradiction, let's suppose that the smaller value of the maximum fraction can be obtained in some other way. Clearly, ... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1010/2019 | |
0kx6 | Problem:
A function $g$ is ever more than a function $h$ if, for all real numbers $x$, we have $g(x) \geq h(x)$. Consider all quadratic functions $f(x)$ such that $f(1)=16$ and $f(x)$ is ever more than both $(x+3)^2$ and $x^2+9$. Across all such quadratic functions $f$, compute the minimum value of $f(0)$.
Proposed b... | [
"Solution:\n\nLet $g(x) = (x+3)^2$ and $h(x) = x^2 + 9$. Then $f(1) = g(1) = 16$. Thus, $f(x) - g(x)$ has a root at $x = 1$. Since $f$ is ever more than $g$, this means that in fact\n$$\nf(x) - g(x) = c(x-1)^2\n$$\nfor some constant $c$.\n\nNow\n$$\nf(x) - h(x) = (f(x) - g(x)) + (g(x) - h(x)) = c(x-1)^2 + 6x = c x^... | United States | HMMT November | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 21/2 | |
017i | Bob tries to create a game for two players. He has decided that the game is to be played on a board with $n \times m$ squares. The first player marks not more than $x$ squares on the board. Then the second player tries to find $p$ squares in a row, horizontally, vertically or diagonally that have not been marked by the... | [
"We will first show that there are no $p$ squares in a row that are not marked. Consider any fixed square $(x_0, y_0)$. The column containing this square has coordinates $(x_0, y)$, and solving the system\n$$\n\\begin{aligned}\ni + ap &= x_0 \\\\\n2i + bp &= y\n\\end{aligned}\n$$\ngives $y = bp + 2(x_0 - ap) = 2x_0... | Baltic Way | BALTIC WAY | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0hoo | Problem:
Two congruent line segments $AB$ and $CD$ intersect at a point $E$. The perpendicular bisectors of $AC$ and $BD$ intersect at a point $F$ in the interior of $\angle AEC$. Prove that $EF$ bisects $\angle AEC$. | [
"Solution:\n\nSince $F$ is on the perpendicular bisectors of $AC$ and $BD$, $FA = FC$ and $FB = FD$. Also $AB = CD$ is given. Thus $\\triangle FAB \\cong \\triangle FCD$ by SSS. Since corresponding heights of congruent triangles are equal, $F$ is equidistant from $AB$ and $CD$, implying that $F$ is on the bisector ... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ffo | Problem:
Demostrar que
$$
0 \leq y z + z x + x y - 2 x y z \leq \frac{7}{27}
$$
donde $x$, $y$, $z$ son números reales no negativos que cumplen $x + y + z = 1$. | [] | Spain | International Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0evd | Let $I$ be the incenter of a triangle $ABC$ whose incircle is tangent to the sides $BC$, $AC$, $AB$ at $D$, $E$, $F$, respectively. Suppose the circumcircle of the triangle $ABC$ intersects the line $EF$ at $P$ and $Q$. If $O_1$ and $O_2$ are the circumcenters of the triangles $IAB$ and $IAC$, respectively, show that t... | [
"Let $M$ be the midpoint of the side $BC$. First, we want to show that $M$ is on the circumcircle of the triangle $DPQ$. For the intersection point $X$ of $PQ$ and $BC$, Menelaus theorem guarantees\n$$\n\\frac{BX}{XC} = \\frac{CE}{EA} = \\frac{AF}{FB} = 1.\n$$\nSince $AF = AE$, $BD = BF$, $CD = CE$ and $BX \\cdot C... | South Korea | Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Rad... | English | proof only | null | |
013i | Problem:
A ray emanating from the vertex $A$ of the triangle $ABC$ intersects the side $BC$ at $X$ and the circumcircle of $ABC$ at $Y$. Prove that
$$
\frac{1}{AX} + \frac{1}{XY} \geq \frac{4}{BC}.
$$ | [
"Solution:\nFrom the GM-HM inequality we have\n$$\n\\frac{1}{AX} + \\frac{1}{XY} \\geq \\frac{2}{\\sqrt{AX \\cdot XY}}\n$$\nAs $BC$ and $AY$ are chords intersecting at $X$ we have $AX \\cdot XY = BX \\cdot XC$. Therefore (1) transforms into\n$$\n\\frac{1}{AX} + \\frac{1}{XY} \\geq \\frac{2}{\\sqrt{BX \\cdot XC}}\n$... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
07a4 | Let $\alpha$ be a real number and $a_1, a_2, a_3, \dots$ a strictly increasing sequence of positive integers such that for every $n \in \mathbb{N}$, $a_n \le n^\alpha$. A prime number $q$ is called *golden* if there is a positive integer $m$ such that $q \mid a_m$. Suppose that $q_1 < q_2 < q_3 < \dots$ are all *golden... | [
"a) Denote by $t$ the number of *golden* prime numbers less than or equal to $1390^n$. We want to show that $t \\ge n$. Suppose that $S$ is collection of all natural numbers less than or equal to $1390^n$ with prime factors from the set $\\{q_1, q_2, \\dots, q_t\\}$. Obviously each element of $S$ can be written in ... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0gnn | Let $ABC$ be a given triangle. Let $D$ be a point on $[BC]$ such that $|AD| = \frac{|BD|^2}{|AB| + |AD|} = \frac{|CD|^2}{|AC| + |AD|}$ and $E$ be a point such that $D \in [AE]$ and $|CD| = \frac{|DE|^2}{|CD| + |CE|}$. Prove that $|AE| = |AB| + |AC|$. (Ali Doğanaksoy). | [
"**Lemma.** Let $ABC$ be a triangle with $|AB|^2 + |AB| \\cdot |AC| = |BC|^2$. Then $m(\\widehat{A}) = 2m(\\widehat{C})$.\n**Proof:** Let $D$ be an intersection of interior angle bisector of $\\widehat{A}$ with $BC$. Then $|BD| = |AB| \\cdot k$, $|CD| = |AC| \\cdot k$. $|AB| \\cdot (|AB| + |BC|) \\cdot k = |BC|^2 \... | Turkey | Team Selection Test for IMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
02uh | Problem:
A bissetriz de um ângulo é uma semirreta com origem no vértice de um ângulo que o divide em dois outros ângulos congruentes. Por exemplo, no desenho abaixo, a semirreta $OC$ é bissetriz do ângulo $\angle AOB$.
a) A diferença entre dois ângulos consecutivos mas não adjacentes é $100^\... | [
"Solution:\n\na) Sejam $\\angle BAD=2x$ e $\\angle BAC=2y$ os ângulos adjacentes.\n\nO ângulo entre as bissetrizes é\n$$\n\\begin{aligned}\n\\angle EAF &= \\angle EAB - \\angle FAB \\\\\n&= x - y \\\\\n&= \\frac{2x}{2} - \\frac{2y}{2} \\\\\n&= \\frac{\\angle CAB}{2} - \\frac{\\angle DAB}{2}... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | a) 50 degrees; b) 30 degrees | |
000v | Sean $C$ y $D$ dos puntos de la semicircunferencia de diámetro $AB$ tales que $B$ y $C$ están en semiplanos distintos respecto de la recta $AD$. Denotemos $M$, $N$ y $P$ los puntos medios de $AC$, $DB$ y $CD$, respectivamente. Sean $O_A$ y $O_B$ los circuncentros de los triángulos $ACP$ y $BDP$. Demuestre que las recta... | [] | Argentina | XVIII Olimpíada Iberoamericana de Matemática | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, or... | español | proof only | null | |
08vg | Let $p$, $q$ be positive integers. To start off, we write the integer $1$ on a blackboard. Then, we repeat the following operation:
**Operation:** Choose $p$ or $q$ and replace the number written on the blackboard by the number obtained by adding the chosen number to it.
Find a condition on $p$, $q$ which guarantees th... | [
"We will show that the condition we seek is that $p$ and $q$ are not relatively prime.\n\nSo, first let us show that if $p$ and $q$ are not relatively prime, then the operation can be continued indefinitely without writing on the blackboard any multiples of $p$ or any multiples of $q$. If $p$ and $q$ are not relati... | Japan | Japan Junior Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | The operation is possible indefinitely if and only if gcd(p, q) ≥ 2 (i.e., p and q are not relatively prime). | |
0emb | If $x^2 + y^2 = 1$ and $x, y > 0$, prove that $x^3 + y^3 \ge \sqrt{2}xy$. | [
"First notice that by the AM-GM inequality we have\n$$\n\\frac{x^5 y + x^3 y^3}{2} \\geqslant x^4 y^2 \\quad \\text{and} \\quad \\frac{y^5 x + y^3 x^3}{2} \\geqslant y^4 x^2.\n$$\nAlso note that\n$$\n\\begin{align*}\nx^6 + y^6 &\\geqslant x^5 y + x y^5 \\\\\n\\Leftrightarrow x^6 - x^5 y + y^6 - x y^5 &\\geqslant 0 ... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof only | null | |
0he2 | Let quasi-square be a figure that consists of an $n \times n$-square with one more $1 \times 1$ square attached along one of its sides, so that the unit square shares a side with one of the unit squares of an $n \times n$-square as well as shares a vertex with one of the corner squares of the $n \times n$-square. Thus,... | [
"Fig. 4 shows an example of filling the plane with quasi-squares in which an extra $1 \\times 1$-square shares a side with an edge square. Fig. 5 shows an example for quasi-squares which have unit square sharing the vertex but not the side with the edge square of an $n \\times n$-square.\n\n![](attached_image_2.png... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Other"
] | null | proof and answer | All integers n ≥ 3 | |
0h13 | Solve the equation:
$$
\cos \pi x = \left[ \frac{x}{2} - \left[ \frac{x}{2} \right] - \frac{1}{2} \right].
$$
Here $[a]$ stands for the greatest integer number that does not exceed $a$. | [
"**Answer:** $x = \\frac{3}{2} + 2n, n \\in \\mathbb{Z}$.\n\nSince $a - [a] = \\{a\\}$, where $\\{a\\}$ is the fractional part of $a$, we can rewrite our equation in the following way:\n$$\n\\cos \\pi x = \\left[ \\frac{\\{x\\}}{2} - \\frac{1}{2} \\right].\n$$\nObviously, $0 \\leq \\frac{\\{x\\}}{2} < 1$ for every ... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | x = 3/2 + 2n, where n is any integer | |
0ked | Problem:
For each positive integer $n$, let $a_{n}$ be the smallest nonnegative integer such that there is only one positive integer at most $n$ that is relatively prime to all of $n, n+1, \ldots, n+a_{n}$. If $n<100$, compute the largest possible value of $n-a_{n}$. | [
"Solution:\n\nNote that $1$ is relatively prime to all positive integers. Therefore, the definition of $a_{n}$ can equivalently be stated as: \"$a_{n}$ is the smallest nonnegative integer such that for all integers $x$, $2 \\leq x \\leq n$, $x$ shares a prime factor with at least one of $n, n+1, \\ldots, n+a_{n}$.\... | United States | HMMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 16 | |
06h4 | In a chess tournament there are $n$ players (where $n > 1$ is odd), and every two players play against each other exactly once. It is known that exactly $n$ games end in a tie. For any set $S$ of players including $A$ and $B$, we say that $A$ *admires* $B$ in $S$ if
(a) $A$ does not beat $B$; or
(b) there exists a sequ... | [
"The answer is $\\frac{n(n-3)(n^2+6n-31)}{48}$.\n\nFor each $k$, let $d_k$ be the number of games that player $k$ wins. Note that\n$$\nS := d_1 + d_2 + \\dots + d_n = \\binom{n}{2} - n = \\frac{n(n-3)}{2}.\n$$\nObserve that for a set of four players, if one of them beats everybody else, then the set is not harmonic... | Hong Kong | CHKMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof and answer | n(n-3)(n^2+6n-31)/48 | |
0d8h | Denote by $\{x\}$ the fractional part of a real number $x$, that is $\{x\} = x - \lfloor x \rfloor$ where $\lfloor x \rfloor$ is the maximum integer not greater than $x$. Prove that
1. For every integer $n$, we have $\{n \sqrt{17}\} > \frac{1}{2 \sqrt{17} \cdot n}$.
2. The value $\frac{1}{2 \sqrt{17}}$ is the largest c... | [
"1) For all $n \\in \\mathbb{Z}^{+}$, we have $n \\sqrt{17} \\notin \\mathbb{Z}$ then $[n \\sqrt{17}] < n \\sqrt{17}$ or\n$$\n[n \\sqrt{17}]^{2} < \\left(n \\sqrt{17}\\right)^{2} \\forall n.\n$$\nThis implies that\n$$\n\\begin{aligned}\n& 17 n^{2} - [n \\sqrt{17}]^{2} \\geq 1 \\\\\n& \\Leftrightarrow 17 n^{2} - (n ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 1/(2√17) | |
0kgp | Problem:
Three faces $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ of a unit cube share a common vertex. Suppose the projections of $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ onto a fixed plane $\mathcal{P}$ have areas $x, y, z$, respectively. If $x: y: z=6: 10: 15$, then $x+y+z$ can be written as $\frac{m}{n}$, where $m, n$ ar... | [
"Solution:\nIntroduce coordinates so that $\\mathcal{X}, \\mathcal{Y}, \\mathcal{Z}$ are normal to $(1,0,0), (0,1,0)$, and $(0,0,1)$, respectively. Also, suppose that $\\mathcal{P}$ is normal to unit vector $(\\alpha, \\beta, \\gamma)$ with $\\alpha, \\beta, \\gamma \\geq 0$.\n\nSince the area of $\\mathcal{X}$ is ... | United States | HMMT November 2021 Team Round | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 3119 |
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