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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0cj1 | Determine all functions $f : (0, \infty) \to (0, \infty)$ such that $f(x^y) = (f(x))^{f(y)}$, for any positive real numbers $x$ and $y$. | [] | Romania | 75th NMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 1 for all x > 0, and f(x) = x for all x > 0 | |
09oa | Let $n \ge 3$ be a given integer, and $S \ge 1$ be a multiple of $10^n - 3$. Suppose that the decimal expansion of $S$ has $m$ digits and $2n - 1 > m > n$. Show that $S$ has at least 3 different digits.
(Bayarmagnai Gombodorj) | [
"Assume to the contrary that there exists an integer $A = \\overline{a_m \\dots a_1}$ such that $\\overline{a_m \\dots a_1}$ has at most 2 different digits and $d \\mid A$, where $d$ denotes $10^n - 3$. Consider $S = \\overline{a_n \\dots a_1} + 3 \\times \\overline{a_m \\dots a_{n+1}}$ which exists since $m > n$. ... | Mongolia | MMO2025 Round 4 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0g90 | 對任意自然數 $n$, 設 $a_n = \sum_{k=1}^{\infty} \left[ \frac{n+2^{k-1}}{2^k} \right]$, 其中 $[x]$ 表示不超過 $x$ 的最大整數。試求 $a_{2015}$ 之值。
For any positive integer $n$, let $a_n = \sum_{k=1}^{\infty} \left[ \frac{n+2^{k-1}}{2^k} \right]$, where $[x]$ is the largest integer that is equal or less than $x$. Determine the value of $a_{20... | [
"分類討論易知\n$$\n\\left[ x + \\frac{1}{2} \\right] = [2x] - [x]\n$$\n對任意實數 $x$ 皆成立，因此\n$$\n\\left[ \\frac{n + 2^{k-1}}{2^k} \\right] = \\left[ \\frac{n}{2^{k-1}} \\right] - \\left[ \\frac{n}{2^k} \\right],\n$$\n從而我們有\n$$\n\\sum_{k=1}^{L} \\left[ \\frac{n + 2^{k-1}}{2^k} \\right] = [n] - \\left[ \\frac{n}{2^L} \\right],... | Taiwan | 二〇一五數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 2015 | |
0h6x | Let us denote by $deg(n) = \alpha_1 + \alpha_2 + \ldots + \alpha_k$ the degree of the number $n = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_k^{\alpha_k}$, where $p_i$ are pairwise different prime numbers, and $\alpha_1, \alpha_2, \ldots, \alpha_k$ are positive integers. Prove that there exist $2016$ consecutive positive i... | [
"First let us prove the following lemma.\n\n**Lemma.** There are $l$ consecutive positive integers, among which there are no numbers with the degree less than $t$ ($t \\ge 2$).\n\n*Proof* is conducted by Mathematical Induction with respect to $t$.\n\nBase for $t=2$. The statement is equivalent to the fact that ther... | Ukraine | UkraineMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0jg2 | Problem:
For a positive integer $n > 2$, consider the $n-1$ fractions
$$
\frac{2}{1}, \frac{3}{2}, \cdots, \frac{n}{n-1}
$$
The product of these fractions equals $n$, but if you reciprocate (i.e. turn upside down) some of the fractions, the product will change. Can you make the product equal $1$? Find all values of $n$... | [
"Solution:\nWe will show that this is possible exactly when $n$ is a perfect square larger than $1$. Suppose that we can reciprocate some of the fractions so that the resulting product is $1$. Let $r$ represent the product of the fractions that we will reciprocate and $t$ represent the product of the fractions that... | United States | Bay Area Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n is a perfect square greater than 1 | |
0ghj | There is an equilateral triangle $ABC$ on the plane. Three straight lines pass through $A$, $B$ and $C$, respectively, such that the intersections of these lines form an equilateral triangle inside $ABC$. On each turn, Ming chooses a two-line intersection inside $ABC$, and draw the straight line determined by the inter... | [
"答案為 $7651$；一般性地，對於第 $6n$ 回合，最大可能值為 $3n^2 + 3n + 1$。\n\n讓我們對這些直線座標化：對於通過點 $A$ 的直線 $\\ell$，標記為 $(1, r)$，其中\n$$\nr = \\frac{\\sin \\angle(AB, \\ell)}{\\sin \\angle(\\ell, AC)}.\n$$\n類似地定義 $(2, r)$ 與 $(3, r)$。注意到這表示：\n- 存在 $t > 0$ 使得起始的三條直線為 $(1, t)$、$(2, t)$ 與 $(3, t)$；\n- 由角元希瓦定理，$(1, r_1)$、$(2, r_2)$ 與 $(3, r_3)$ 三... | Taiwan | 2023 數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Chinese (Traditional) | proof and answer | 7651 | |
0ad4 | $ABCDE$ is pentagon where $K$, $L$, $M$, $N$ are the midpoints of $AB$, $BC$, $CD$, $DE$ respectively. Let $P$, $Q$, $F$ be the midpoints of $KM$, $LN$, $AD$, respectively. Prove that $PQ$ and $AE$ are parallel and $\overline{AE} = 4\overline{PQ}$. | [
"The quadrilateral $KFML$ is parallelogram. The point $P$ is the midpoint for $KM$ and $P \\in LF$ and also the midpoint for $LF$. In the triangle $LFN$, we\n\n\nhave $PQ \\parallel FN$ and $\\overline{PQ} = \\frac{1}{2}\\overline{FN}$. In the triangle $ADE$, we have $FN \\parallel AE$ and ... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | PQ is parallel to AE and AE = 4·PQ | |
0dxm | Problem:
Dokaži, da je polinom $p(x) = (x-2)^{100} + (x-1)^{50} - 1$ deljiv s polinomom $g(x) = x^{2} - 3x + 2$. | [
"Solution:\n\nNajprej funkcijski predpis $g(x)$ razstavimo: $g(x) = (x-2)(x-1)$. Ničli polinoma $g(x)$ sta $x_{1} = 2$ in $x_{2} = 1$. Izračunamo vrednost $p(2)$ in $p(1)$. Ugotovimo, da sta to ničli polinoma $p(x)$. Ker sta $2$ in $1$ ničli obeh danih polinomov, velja trditev, da $g(x)$ deli $p(x)$.\n\nZapis $g(x)... | Slovenia | 7. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0ap4 | Problem:
Two friends, Marco and Ian, are talking about their ages.
Ian says, "My age is a zero of a polynomial with integer coefficients."
Having seen the polynomial $p(x)$ Ian was talking about, Marco exclaims, "You mean, you are seven years old? Oops, sorry I miscalculated! $p(7)=77$ and not zero."
"Yes, I am old... | [
"Solution:\n\nLet $a$ be Ian's age. Then\n$$\np(x) = (x - a) q(x)\n$$\nwhere $q(x)$ is a polynomial with integer coefficients.\n\nSince $p(7) = 77$, we have\n$$\np(7) = (7 - a) q(7) = 77 = 7 \\cdot 11\n$$\nSince $q(7)$ is an integer and $7 - a < 0$, we restrict\n$$\na - 7 \\in \\{1, 7, 11, 77\\}\n$$\nLet $b$ be the... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 14 | |
00np | The numbers $1, 2, \dots, 2020$ and $2021$ are written on a blackboard. The following operation is executed:
Two numbers are chosen, both are erased and replaced by the absolute value of their difference.
This operation is repeated until there is only one number left on the blackboard.
a) Show that $2021$ can be the f... | [
"(a) Let us first choose the following $1010$ pairs of numbers:\n$(1, 2); (3, 4); \\ldots; (2019, 2020)$.\nThe absolute value of the difference within each of these pairs is $1$. After applying the operation for each of these pairs, the number $2021$ and $1010$ times the number $1$ remain on the blackboard. Now we ... | Austria | Austrian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0iba | Problem:
Find all positive integer solutions $(m, n)$ to the following equation:
$$
m^{2} = 1! + 2! + \cdots + n!.
$$ | [
"Solution:\n$(1,1), (3,3)$\nA square must end in the digit $0, 1, 4, 5, 6$, or $9$. If $n \\geq 4$, then $1! + 2! + \\cdots + n!$ ends in the digit $3$, so cannot be a square. A simple check for the remaining cases reveals that the only solutions are $(1,1)$ and $(3,3)$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | [(1, 1), (3, 3)] | |
07el | Find the least possible value of $n$, such that one can place $1, 2, \ldots, n$ in each cell of an $18 \times 18$ table, such that each number is used at least once, and in each row or column, there exist neither two equal nor two consecutive numbers. | [
"First, we prove that $n = 36$ is not enough. Assume that numbers $1, 2, \\ldots, 36$ are assigned to the cells of the table with the described conditions. Since the number $2$ is used, so even numbers (i.e. $2, 4, 6, \\ldots, 36$) must be put in its row and its column. Similarly, the number $35$ is used, so we hav... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 37 | |
0hd1 | A polynomial $x^2 + 1$ is written on the blackboard. Every day Kate wipes off currently written polynomial $F(x)$ and writes one of the $F^2(x) + 1$ and $F(x^2 + 1)$ instead, choosing at her discretion. Prove that the constant (free) term will exceed $2^{2^{222}}$ in one year.
(Arseniy Nikolaev) | [
"We are going to prove a lemma prior to the problem itself: the polynomial which will be written after $k$ days does not depend on Kate's choices and equals to\n$$\nF_k = \\underbrace{(\\dots((x^2 + 1)^2 + 1)^2 + \\dots)^2}_{k} + 1.\n$$\n\nUsing induction one can directly verify this lemma. Alternatively, one can u... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0fxx | Problem:
Gegeben ist ein beliebiger Bodengrundriss, der aus $n$ Einheitsquadraten zusammengesetzt ist. Albert und Berta möchten diesen Boden mit Kacheln bedecken, wobei jede Kachel die Form eines $1 \times 2$-Dominos oder eines T-Tetrominos hat. Albert hat nur Kacheln von einer Farbe zur Verfügung, Berta hingegen hat ... | [
"Solution:\n\nWir nennen Alberts Überdeckungen farblos und jene von Berta farbig. Sei $A$ bzw. $B$ die Menge der farblosen bzw. farbigen Überdeckungen. Durch \"Vergessen der Farbe\" erhält man eine Abbildung $\\varphi: B \\rightarrow A$ und wir behaupten, dass diese Abbildung $2^{\\frac{n}{2}} : 1$ ist. Wir müssen ... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 2^{n/2} | |
0ihw | Problem:
Suppose $f$ and $g$ are differentiable functions such that
$$
x g(f(x)) f'(g(x)) g'(x) = f(g(x)) g'(f(x)) f'(x)
$$
for all real $x$. Moreover, $f$ is nonnegative and $g$ is positive. Furthermore,
$$
\int_{0}^{a} f(g(x)) dx = 1 - \frac{e^{-2a}}{2}
$$
for all reals $a$. Given that $g(f(0)) = 1$, compute the valu... | [
"Solution:\nDifferentiating the given integral with respect to $a$ gives $f(g(a)) = e^{-2a}$. Now\n$$\nx \\frac{d[\\ln (f(g(x)))]}{dx} = x \\frac{f'(g(x)) g'(x)}{f(g(x))} = \\frac{g'(f(x)) f'(x)}{g(f(x))} = \\frac{d[\\ln (g(f(x)))]}{dx}\n$$\nwhere the second equals sign follows from the given. Since $\\ln (f(g(x)))... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | e^{-16} | |
0gaw | 令 $n \ge 2$ 為一正整數。有一個 $n \times n$ 的棋盤狀地區, 每一小格都是一座公園。每座公園裡都有若干隻貓 (貓的數量是非負整數)。為了管理貓咪, 管理處會對公園進行操作。每次操作, 管理處會選擇一座公園:
(1) 所選的公園, 其貓咪數量必須大於或等於該公園的相鄰公園數。
(2) 選定公園 $A$ 後, 對於該公園的每座相鄰公園 $B$, 管理處都從 $A$ 赶一隻貓到 $B$ (我們稱兩座公園相鄰, 若且唯若它們有公共邊。)
令 $m$ 為所有公園內的貓咪數量。試求最小的 $m$, 使得存在一種起始的貓咪分布, 讓管理處可以藉由適當的選擇每次操作的公園來進行無限次操作。 | [
"$\\underline{2n(n-1)}$。\n\n- 首先給出最小值的構造法：\n\n* 最左下角的公園沒有貓；\n* 最下方一橫排及最左方一直列上的公園各有一隻；\n* 其餘公園各兩隻貓。\n\n則共有 $$(n-1) + (n-1) + 2(n-2)^2 = 2n(n-1)$$ 隻貓，考慮從最上方橫排開始，每排從最右邊的公園到最左邊的公園各做一次操作，接著換第二橫排……依序下去。易檢驗在操作完全部的格子後，會回到起始狀態，故可以操作無限次。\n\n- 接著證明至少要 $2n(n-1)$ 隻貓。\n\n1. 首先發現每一個公園都要被操作過無窮多次。否則，考慮所有操作無限次的公園所成集合 $I$，其內的總貓數必遞減（因... | Taiwan | 二〇一七數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 2n(n-1) | |
0ezf | Problem:
$n$ is a $17$ digit number. $m$ is derived from $n$ by taking its decimal digits in the reverse order. Show that at least one digit of $n + m$ is even. | [
"Solution:\n\nLet the number be $n$ with digits $d_1d_2\\ldots d_{17}$, so that the reversed number $m$ has digits $d_{17}d_{16}\\ldots d_1$. Let the digits of $n + m$ be $a_0a_1\\ldots a_{17}$, where $a_0$ may be zero. Let the carry forward when adding digits to get $a_i$ be $c_{i - 1}$, so that, in general, $c_i ... | Soviet Union | 4th ASU | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof only | null | |
04ev | Prove that the number whose decimal representation consists of 2187 digits 1 is divisible by 2187. | [
"Let $N$ be the number whose decimal representation consists of 2187 digits 1. Then\n$$\nN = \\underbrace{111\\ldots 1}_{2187\\ \\text{digits}} = \\frac{10^{2187} - 1}{9}.\n$$\n\nWe want to show that $2187$ divides $N$.\n\nNote that $2187 = 3^7$.\n\nIt suffices to show that $3^7$ divides $N$.\n\nSince $N = \\frac{1... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
09z1 | Petra, Quinten, Rakhi, Salome, and Teun organise a badminton tournament consisting of five rounds. In each round, two players play against each other and a third player is the referee. The other two players rest during the round. Everyone plays twice and is the referee once. Nobody plays two matches in a row and the re... | [] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Logic"
] | English | MCQ | B | |
0jyf | Problem:
Consider an equilateral triangular grid $G$ with 20 points on a side, where each row consists of points spaced 1 unit apart. More specifically, there is a single point in the first row, two points in the second row, ..., and 20 points in the last row, for a total of 210 points. Let $S$ be a closed non-selfint... | [
"Solution:\n\nImagine deforming the triangle lattice such that now it looks like a lattice of 45-45-90 right triangles with legs of length 1. Note that by doing this, the area has multiplied by $\\frac{2}{\\sqrt{3}}$, so we need to readjust our answer on the isosceles triangle lattice by a factor of $\\frac{\\sqrt{... | United States | February 2017 | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem"
] | null | proof and answer | 52√3 | |
0cws | Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function. Let us call a chord a segment of integer length parallel to the axis $Ox$ whose endpoints belong to the graph $y = f(x)$. It is known that the graph $y = f(x)$ has exactly $N$ chords, and moreover, among them there is a chord of length $2025$. Find the least ... | [
"**Ответ.** 4049.\n\nFor a natural number $n$, define $g_n(x) = f(x+n) - f(x)$. Then the number of chords of length $n$ equals the number of zeros of the function $g_n(x)$.\n\nAs an *example*, consider the following piecewise linear function $f$: $f(x) = x$ for $x \\le 2024\\frac{9}{10}$ and $f(x) = 20249 \\cdot |2... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | Russian | proof and answer | 4049 | |
0eno | In triangle $ABC$, $\angle A = 60^\circ$. Let $E$ and $F$ be points on the extensions of $AB$ and $AC$ such that $BE = CF = BC$. The circumcircle of $ACE$ intersects $EF$ in $K$ (different from $E$). Prove that $K$ lies on the bisector of $\angle BAC$. | [
"Let $T$ be the intersection of $BF$ and $CE$. We have that $\\angle BCE + \\angle BEC = \\angle ABC$, and since $BC = BE$, it follows that $\\angle BCE = \\frac{1}{2} \\angle ABC$. Similarly, $\\angle CBF = \\frac{1}{2} \\angle ACB$. Hence\n$$\n\\angle CTF = \\angle BCE + \\angle CBF = \\frac{1}{2}(\\angle ABC + \... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0ky7 | Consider pairs $(f, g)$ of functions from the set of nonnegative integers to itself such that
* $f(0) \ge f(1) \ge f(2) \ge \dots \ge f(300) \ge 0$;
* $f(0) + f(1) + f(2) + \dots + f(300) \le 300$;
* for any 20 nonnegative integers $n_1, n_2, \dots, n_{20}$, not necessarily distinct, we have
$$
g(n_1 + n_2 + \dots + n_... | [
"Replace $300 = \\frac{24 \\cdot 25}{2}$ with $\\frac{s(s+1)}{2}$ where $s = 24$, and 20 with $k$. The answer is $115440 = \\frac{ks(ks+1)}{2}$. Equality is achieved at $f(n) = \\max(s-n, 0)$ and $g(n) = \\max(ks-n, 0)$. To prove\n$$\ng(n_1 + \\dots + n_k) \\le f(n_1) + \\dots + f(n_k),\n$$\nwrite it as\n$$\n\\max(... | United States | USA TST | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 115440 | |
04rp | Find all functions $f: \mathbb{R} \setminus \{0\} \to \mathbb{R}$ such that for all non-zero numbers $x, y$,
$$
x \cdot f(xy) + f(-y) = x \cdot f(x).
$$ | [
"Substituting $x = 1$ gives\n$$\nf(y) + f(-y) = f(1).\n$$\nDenoting $f(1) = a$, we have $f(-y) = a - f(y)$. Substituting $y = -1$, we get\n$$\nx \\cdot f(-x) + f(1) = x \\cdot f(x),\n$$\ni. e.,\n$$\nx(a - f(x)) + a = x \\cdot f(x),\n$$\nhence\n$$\nf(x) = \\frac{a(x+1)}{2x} = \\frac{a}{2}\\left(1+\\frac{1}{x}\\right... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All solutions are f(x) = c(1 + 1/x) for real constants c and all nonzero real x. | |
03ob | Find all positive integers $n$ such that
$$
n^4 - 4n^3 + 22n^2 - 36n + 18
$$
is a perfect square. | [
"We write\n$$\n\\begin{aligned}\nA &= n^4 - 4n^3 + 22n^2 - 36n + 18 \\\\\n&= (n^2 - 2n)^2 + 18(n^2 - 2n) + 18.\n\\end{aligned}\n$$\nLet $n^2 - 2n = x$, $A = y^2$, where $y$ is a nonnegative integer. Then\n$$\n(x+9)^2 - 63 = y^2.\n$$\nThat is, $(x+9-y)(x+9+y) = 63$.\nIt can only be $(x+9-y, x+9+y) = (1, 63)$, $(3, 2... | China | China Western Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 1, 3 | |
0hcy | Let $a$, $b$, $c$ be natural numbers. Prove that there exists an integer nonnegative number $k$, such that $\text{GCD}(a^k + bc, b^k + ca, c^k + ab) > 1$. | [
"Let $p$ be some prime divisor of the number $abc + 1$. Let's prove that $k = p - 2$ satisfies the statement of the problem. Since $p$ doesn't divide any of the numbers $a$, $b$, $c$, by Fermat's little theorem: $a^{p-1} \\equiv b^{p-1} \\equiv c^{p-1} \\equiv 1 \\pmod{p}$. Hence $p$ divides each number $a^{p-1} + ... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
006f | Sea $ABC$ un triángulo con $AB < AC$. La circunferencia inscripta al triángulo es tangente a $BC$ en $X$, a $CA$ en $Y$ y a $AB$ en $Z$. Sea $U$ el punto medio del arco $\overarc{BC}$ que contiene a $A$ de la circunferencia circunscripta al triángulo $ABC$. La recta $UX$ corta nuevamente a la circunferencia circunscrip... | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Brocard ... | Spanish | proof only | null | |
03he | Problem:
A bus route consists of a circular road of circumference $10$ miles and a straight road of length $1$ mile which runs from a terminus to the point $Q$ on the circular road (see diagram). It is served by two buses, each of which requires $20$ minutes for the round trip. Bus No. $1$, upon leaving the terminus, ... | [] | Canada | Canadian Mathematical Olympiad | [
"Math Word Problems",
"Precalculus > Functions"
] | null | proof and answer | Let v = 12 miles per 20 minutes. For 0 ≤ x < 1 and 11 < x < 12, w(x) = 10 + (5/3)·min{x, 12 − x}. For 1 ≤ x ≤ 11, w(x) = max(30 − (5/3)x, 10 + (5/3)x). In particular: w(2) = 80/3 minutes; w(4) = 70/3 minutes. The maximum of w(x) occurs at x = 1 and x = 11, with value 85/3 minutes. Graph sketch: piecewise linear—on [0,1... | |
0c9g | Problem:
Aflaţi ultima cifră a numărului natural $n$, ştiind că penultima cifră a lui $n^{2}$ este 9. | [] | Romania | Olimpiada Națională GAZETA MATEMATICĂ | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 4 or 6 | |
089o | Problem:
Quante sono le coppie di interi ordinate $(x, y)$ tali che $x y = 4(y^{2} + x)$?
(A) 0
(B) 1
(C) 2
(D) 7
(E) 14 | [
"Solution:\n\nLa risposta è **(E)**. Riscriviamo l'espressione come $x y - 4 x = 4 y^{2}$, ovvero $x(y-4) = 4 y^{2}$. $y = 4$ non è soluzione, dunque possiamo dividere per $y-4$ e ottenere l'equazione\n$$\nx = \\frac{4 y^{2}}{y-4}.\n$$\nOsservando che $y^{2} = (y+4)(y-4) + 16$ possiamo ulteriormente riscrivere l'eq... | Italy | Progetto Olimpiadi della Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | MCQ | E | |
00wa | Problem:
In two piles there are $72$ and $30$ sweets respectively. Two students take, one after another, some sweets from one of the piles. Each time the number of sweets taken from a pile must be an integer multiple of the number of sweets in the other pile. Is it the beginner of the game or his adversary who can alw... | [
"Solution:\n\nNote that one of the players must have a winning strategy. Assume that it is the player making the second move who has it. Then his strategy will assure taking the last sweet also in the case when the beginner takes $2 \\cdot 30$ sweets as his first move. But now, if the beginner takes $1 \\cdot 30$ s... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | the beginner (first player) | |
0bnx | Determine all positive integers $n$ such that all positive integers less than or equal to $n$ and prime to $n$ are pairwise coprime. | [
"Notice that $5$ is the first index $k$ such that $p_1 \\cdots p_{k-1} > p_k^2$. Now, if $p_1 \\cdots p_{k-1} > p_k^2$ for some index $k \\ge 5$, then $p_1 \\cdots p_{k-1}p_k > p_k^3 > 4p_k^2 > p_{k+1}^2$, by the Bertrand-Tchebysheff theorem, so $p_1 \\cdots p_{k-1} > p_k^2$ for all indices $k \\ge 5$.\n\nConsequen... | Romania | 2015 Danube Mathematical Competition | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 1, 2, 3, 4, 6, 8, 12, 18, 24, 30 | |
0ffe | Problem:
¿Cuántos números de tres cifras (es decir, mayores que $99$ y menores que $1000$) hay que tengan su cifra central mayor que las otras dos? ¿Cuántos de ellos tienen además las tres cifras distintas? | [
"Solution:\n\na. Las cifras pueden repetirse. Sea $n > 1$ la cifra central. La de la izquierda puede ser una cifra cualquiera entre $1$ y $n-1$. La de la derecha puede ser una cualquiera entre $0$ y $n-1$. En total hay\n$$\n\\sum_{2}^{9} n(n-1) = \\sum_{1}^{8} n(n+1) = \\sum_{1}^{8} n^{2} + \\sum_{1}^{8} n = \\frac... | Spain | Olimpiadas Matemáticas Españolas | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 240; 204 | |
03vf | Find all functions $f: (0, +\infty) \to (0, +\infty)$ ($f$ is a function mapping positive real numbers to positive real numbers) such that
$$
\frac{(f(w))^2 + (f(x))^2}{f(y^2) + f(z^2)} = \frac{w^2 + x^2}{y^2 + z^2}
$$
for all positive real numbers $w, x, y, z$ satisfying
$$wx = yz.$$ | [
"Take\n$$\nw = x = y = z = 1,\n$$\nthen we get $(f(1))^2 = f(1)$, so $f(1) = 1$.\nFor any real number $t > 0$, let $w = t$, $x = 1$, $y = z = \\sqrt{t}$,\nwe get\n$$\n\\frac{(f(t))^2 + 1}{2f(t)} = \\frac{t^2 + 1}{2t},\n$$\nwhich implies $(tf(t) - 1)(f(t) - t) = 0$.\nSo, for any $t > 0$,\n$$\nf(t) = t \\quad \\text{... | China | International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x)=x for all x>0, or f(x)=1/x for all x>0 | |
0d0w | Prove that for every positive real numbers $x, y, z$ the following inequality holds
$$
\frac{9}{x+y+z} - \frac{1}{xyz} \le 2.
$$ | [
"From the AM-GM inequality we have\n$$\n\\frac{9}{x+y+z} - \\frac{1}{xyz} \\le \\frac{3}{\\sqrt[3]{xyz}} - \\frac{1}{xyz}. \\quad (1)\n$$\nLet $t = \\frac{1}{\\sqrt[3]{xyz}}$. According to inequality (1) it is sufficient to prove that $3t - t^3 \\le 2$ for $t > 0$. The last inequality is equivalent to $t^3 - 3t + 2... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
07wt | Let $a$, $b$ be two distinct positive integers. Prove that the number
$$
\frac{(a+b)^3}{a^3 - a^2b + ab^2 - b^3}
$$
is not an integer. | [
"Suppose the given fraction is equal to the integer $k$, then\n$$\n(a+b)^3 = k(a^2+b^2)(a-b). \\qquad (14)\n$$\nFirst note that replacing $a$ by $da$ and $b$ by $db$ does not change the value of $k$, because numerator and denominator are both homogeneous of degree three. Therefore, we may assume $\\gcd(a, b) = 1$. ... | Ireland | IRL_ABooklet_2024 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order... | null | proof only | null | |
0bds | Let $ABC$ be a triangle and let points $D, E \in (BC)$, $F, G \in (CA)$, $H, I \in (AB)$ such that $BD = CE$, $CF = AG$ and $AH = BI$. Consider $M, N, P$ the midpoint of the segments $GH$, $DI$, $EF$ respectively and let $M'$ be the intersection point of the lines $AM$ and $BC$.
a. Show that $\frac{BM'}{CM'} = \frac{A... | [
"a. Set $m = \\frac{BM'}{CM'}$. We have $\\overrightarrow{AM} = \\frac{1}{2}\\overrightarrow{AH} + \\frac{1}{2}\\overrightarrow{AG} = \\frac{1}{2}\\frac{AH}{AB} \\cdot \\overrightarrow{AB} + \\frac{1}{2}\\frac{AG}{AC} \\cdot \\overrightarrow{AC}$ and $\\overrightarrow{AM'} = \\frac{1}{m+1}\\overrightarrow{AB} + \\f... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | BM'/CM' = (AG/AH) * (AB/AC), and the lines AM, BN, and CP are concurrent. | |
09la | Prove that, when a crooked dice is thrown twice, the probability of obtaining the same parity faces on both rolls is not less than the probability of obtaining different parity faces on the two rolls. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0e62 | Find all pairs of natural numbers $m$ and $n$ such that $2m^2 + n^2$ divides $3mn + 3m$. | [
"Suppose $3mn + 3m = k(2m^2 + n^2)$ where $k$ is a natural number. The inequality between the arithmetic and geometric means gives\n$$\n3mn + 3m = 3m(n + 1) = k(2m^2 + n^2) \\geq 2\\sqrt{2}kmn,\n$$\nhence\n$$\nk \\leq \\frac{3(n + 1)}{2\\sqrt{2}n} = \\frac{3}{2\\sqrt{2}}\\left(1 + \\frac{1}{n}\\right) \\leq \\frac{... | Slovenia | Selection Examinations for the IMO 2012 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (1, 1), (4, 2), (4, 10) | |
0311 | Problem:
Spunem că o mulțime $A \subset \mathbb{R}$ cu cel puțin trei elemente este liberă de progresii aritmetice dacă pentru orice $a, b, c \in A$ distincte, avem $a+b \neq 2c$.
Arătați că mulțimea $\{0,1,2, \ldots, 3^{8}-1\}$ conține o submulțime $A$ cu cel puțin 256 elemente, liberă de progresii aritmetice. | [
"Solution:\n\nAvem $6560 = 3^{8} - 1$.\nOrice număr din mulțimea $\\{0,1,2, \\ldots, 6560\\}$ se scrie în baza 3 cu 8 cifre (0, 1 sau 2). Definim mulțimea $A$ ca mulțimea numerelor care au doar cifrele 0 și 1, numărul lor fiind $2^{8} = 256$ și arătăm că ea este liberă de progresii aritmetice.\n\nDacă prin absurd a... | Brazil | Primul baraj de selecție pentru OBMJ | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
01ga | Find all functions $f : \mathbb{R} \setminus \{0\} \to \mathbb{R}$ so that
$$
f\left(\frac{1}{x}\right) \ge 1 - f(x) \ge x^2 f(x)
$$
for all $x \in \mathbb{R} \setminus \{0\}$. | [
"Answer: $f(x) = \\frac{1}{x^2 + 1}$.\nWe have\n$$\n1 - f(x) \\ge x^2 f(x) \\iff \\frac{1}{x^2 + 1} \\ge f(x).\n$$\n\nIf we substitute $z = \\frac{1}{x}$ into $f(\\frac{1}{x}) \\ge 1 - f(x)$ we get this:\n$$\nf(z) \\ge 1 - f\\left(\\frac{1}{z}\\right) \\quad \\forall z \\in \\mathbb{R} \\setminus \\{0\\}.\n$$\nWe c... | Baltic Way | Baltic Way 2020 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1/(x^2 + 1) | |
00gh | In a small town, there are $n \times n$ houses indexed by $(i, j)$ for $1 \leq i, j \leq n$ with $(1,1)$ being the house at the top left corner, where $i$ and $j$ are the row and column indices, respectively. At time $0$, a fire breaks out at the house indexed by $(1, c)$, where $c \leq \frac{n}{2}$. During each subseq... | [
"At most $n^{2}+c^{2}-n c-c$ houses can be saved. This can be achieved under the following order of defending:\n$$\n\\begin{gather*}\n(2, c), (2, c+1); (3, c-1), (3, c+2); (4, c-2), (4, c+3); \\ldots \\\\\n(c+1, 1), (c+1, 2c); (c+1, 2c+1), \\ldots, (c+1, n)\n\\end{gather*}\n$$\n\nUnder this strategy, there are\n\n2... | Asia Pacific Mathematics Olympiad (APMO) | XVII APMO - March, 2005 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Other... | null | proof and answer | n^2 + c^2 - n c - c | |
098z | Problem:
Determinați toate funcțiile $f: \mathbb{R} \rightarrow \mathbb{R}$, care satisfac proprietatea $f(x+a) \cdot f(x+b) = x$, $\forall x \in \mathbb{R}$, unde $a, b \in \mathbb{R}$ sunt numere arbitrare fixate. | [
"Solution:\n\nDin $f(x+a) \\cdot f(x+b) = x$ pentru $x = 0$ obținem $f(a) \\cdot f(b) = 0 \\Rightarrow \\left[\\begin{array}{l}f(a) = 0 \\\\ f(b) = 0\\end{array}\\right.$.\n\nPentru $x = a-b$ obținem că $f(2a-b) \\cdot f(a) = a-b \\Leftrightarrow a-b = 0 \\Leftrightarrow a = b$.\n\nPentru $x = b-a$ obținem că $f(b)... | Moldova | OLIMPIADA REPUBLICANĂ LA MATEMATICĂ | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0cjc | Let $N$ be a positive integer. On a board, we initially have two numbers: a red $0$ and a blue $0$. We define the following procedure:
At each step, a natural number $k$ is chosen (not necessarily distinct from previous choices). Let $x$ be the blue number and $y$ the red number. Replace them with:
$$
x \rightarrow x ... | [
"Let $M_k$ denote the move that increases the blue number by $k+1$ and the red number by $k^2+2$. We show that for any $k \\ge 2$, the move $M_k$ can be replaced by moves involving only $M_0$ and $M_1$ in such a way that the total increase in the blue number is the same, but the red number increases by less than $k... | Romania | 75th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | N + floor((N+1)/2) | |
0cns | Real numbers $a$, $b$ and $c$ satisfy the conditions
$$
(a + b)(b + c)(c + a) = abc, \\
(a^3 + b^3)(b^3 + c^3)(c^3 + a^3) = a^3b^3c^3.
$$
Prove that $abc = 0$. | [
"Заметим сначала, что для любых различных чисел $x$ и $y$ выполняется неравенство\n$$\nx^2 - xy + y^2 > |xy| \\quad (1)\n$$\nВ самом деле, если $|xy| = xy$, то неравенство (1) равносильно неравенству $(x - y)^2 > 0$, которое верно; если же $|xy| = -xy$, то (1) равносильно неравенству $x^2 + y^2 > 0$.\n\nПредположим... | Russia | Russian mathematical olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English; Russian | proof only | null | |
0gmv | For a sequence of integers, determine the maximal number of distinct values that can be attained infinitely many times by this sequence if there exists a positive integer $N$ such that, for any $n \ge N$,
$$
a_n = |\{i \mid 1 \le i < n \text{ and } a_i + i \ge n\}|$$ | [] | Turkey | XIII. National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 2 | |
01iq | The Euclidean plane is dissected into several disjoint *bounded* regions (not further subdivided) by $n$ intersecting equilateral triangles of side length $1$. Determine the smallest value of $n$ such that the number of parts thus arising can be at least $2023$.
The picture shows four intersecting triangles.
![](atta... | [
"We will prove the following claim:\n**Claim:** For $n \\in \\mathbb{Z}^+$, the maximal number of bounded regions of the plane generated by $n$ triangles (of whatever size and shape) is $1 + 3n(n-1)$.\n\nOnce the claim is established, we can easily see that the inequality $1 + 3n(n-1) \\ge 2023$ is equivalent to $n... | Baltic Way | Baltic Way 2023 Shortlist | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | 27 | |
0h6z | A circle, inscribed into a triangle $ABC$, touches its sides $AB$, $BC$ and $CA$ in the points $N$, $P$, $K$ respectively. A segment $BK$ intersects the inscribed circle the second time in a point $L$.
Let us define points $T = AL \cap NK$, $Q = CL \cap KP$. Prove that straight lines $BK$, $NQ$ and $PT$ intersect in on... | [
"Let us denote the angles as shown in fig. 26. For triangles $AKT$ and $ATN$ let us apply Law of sines and get:\n$$\n\\frac{NT}{TK} = \\frac{AN \\cdot \\sin \\alpha_1}{AK \\cdot \\sin \\alpha_2} = \\frac{\\sin \\alpha_1}{\\sin \\alpha_2}\n$$\nSimilarly we can get:\n$$\n\\frac{NF}{FP} = \\frac{NB \\cdot \\sin \\beta... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jtg | Problem:
For positive integers $a, b$, $a \uparrow \uparrow b$ is defined as follows: $a \uparrow \uparrow 1 = a$, and $a \uparrow \uparrow b = a^{a \uparrow \uparrow (b-1)}$ if $b > 1$.
Find the smallest positive integer $n$ for which there exists a positive integer $a$ such that $a \uparrow \uparrow 6 \not\equiv a \... | [
"Solution:\n\nWe see that the smallest such $n$ must be a prime power, because if two numbers are distinct mod $n$, they must be distinct mod at least one of the prime powers that divide $n$. For $k \\geq 2$, if $a \\uparrow \\uparrow k$ and $a \\uparrow \\uparrow (k+1)$ are distinct $\\bmod p^{r}$, then $a \\uparr... | United States | HMMT February | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"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"
] | null | proof and answer | 283 | |
09k3 | Let $a$, $b$, $c$ be positive integers and let $P_N = (a^2 + N)(b^2 + N)(c^2 + N)$ for any integer $N$. Show that
(1) there is a positive integer $N$ so that $P_N$ is not a perfect square,
(2) there is a positive integer $N$ so that $P_N$ is a perfect square. | [
"(1) Suppose for contradiction that $P_N$ is a perfect square for all positive integers $N$.\n\nLet $N = 1$. Then $P_1 = (a^2 + 1)(b^2 + 1)(c^2 + 1)$. Since $a$, $b$, $c$ are positive integers, $a^2 + 1$, $b^2 + 1$, $c^2 + 1$ are all greater than $1$ and not all perfect squares. In particular, for $a = 1$, $a^2 + 1... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0g24 | Problem:
Soit $n$ un nombre naturel pair. On partitionne les nombres $1,2, \ldots, n^{2}$ en deux ensembles $A$ et $B$ de taille égale, de telle manière que chacun des $n^{2}$ nombres appartient à exactement un des deux ensembles. Soient $S_{A}$ et $S_{B}$ la somme de tous les éléments dans $A$ et $B$ respectivement. ... | [
"Solution:\n\nSoit $k$ le nombre entier tel que $S_{A}=39 k$ et $S_{B}=64 k$. On a alors que la somme de tous les nombres de 1 à $n^{2}$ vaut\n$$\n\\frac{n^{2}\\left(n^{2}+1\\right)}{2}=S_{A}+S_{B}=39 k+64 k=103 k\n$$\nAinsi, $103 \\left\\lvert\\, \\frac{n^{2}\\left(n^{2}+1\\right)}{2}\\right.$, et comme 103 est un... | Switzerland | SMO-Selektion | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n is divisible by 206 | |
0gsx | In triangle $ABC$, the incenter is $I$ and the incircle touches the sides $BC, AC, AB$ at $D, E, F$, respectively. For a line $\ell$ passing through $I$, the feet of the perpendiculars dropped from $A, B, C$ onto $\ell$ are $X, Y, Z$ respectively. Show that the lines $DX, EY, FZ$ are concurrent. | [
"In the solution, we use directed lengths. Let $\\ell$ intersect $EF$, $DF$, $DE$ at $K$, $L$, $M$ respectively and let $IX = x$, $IY = y$ and $IZ = z$; hence $IK = r^2/x$, $IL = r^2/y$ and $IM = r^2/z$ where $r$ is the radius of the incircle. Consequently we have\n$$\n(1) \\quad \\frac{MX}{LX} \\cdot \\frac{KY}{MY... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, ince... | null | proof only | null | |
07ri | Two players, Abigail and Zoe, play a game. Abigail chooses an odd number less than $800$ and writes it on the blackboard. Zoe now chooses a different odd number less than $1200$ and writes it on the blackboard. Each in turn replaces their number by the difference between their number and the smallest power of $2$ bigge... | [
"Let $n$ be the current number of one of the players. Because both players start with an odd number, $n$ will be odd as it remains positive throughout the game. If $n = \\sum_{i=0}^{k} b_i 2^i$ with $b_i \\in \\{0, 1\\}$ is the binary representation of $n$, then $b_0 = 1$. If we assume $b_k = 1$, then the smallest ... | Ireland | Irish | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Zoe | |
0979 | Problem:
Fie $ABC$ un triunghi echilateral fixat. Pentru fiecare dreaptă arbitrară $l$ ce trece prin vârful $B$ se consideră punctele $D_{l}$ și $E_{l}$, ce reprezintă piciorul perpendicularei duse din punctul $A$, respectiv $C$, la dreapta $l$. Să se determine locul geometric al punctelor $P_{l}$, ce formează un triun... | [
"Solution:\n\nSunt posibile diferite localizări ale dreptei $l$. Considerăm cazul de localizare a dreptei $l$, reprezentat în desen; alte cazuri se consideră în mod analogic. Fie $O$ este piciorul perpendicularei duse din $B$ la $AC$ (și mijlocul laturii $AC$).\n\nDeoarece $\\angle BDA = \\... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | The locus is the circle centered at B with radius equal to BO, the distance from B to the midpoint of AC. | |
07it | Find all functions $f : \mathbb{C} \to \mathbb{C}$ such that for all complex numbers $x, y$:
$$
f(f(x) + y f(y)) = x + |y|^2
$$ | [
"Plugging $y = 0$ it follows that $f(f(x)) = x$, hence the function is bijective. Plugging $(x, y) = (0, 1)$ it follows that $f(f(0) + f(1)) = 1 = f(f(1))$ using injectivity to obtain $f(0) = 0$. Now, taking $f$ from both sides and using $f(f(x)) = x$ to obtain $f(x) + y f(y) = f(x + |y|^2)$. Plugging $x = 0$ yield... | Iran | 41th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | f(z) = C·\overline{z} for all z in \mathbb{C}, where |C| = 1 | |
06rs | Let $n \geq 1$ be an integer. What is the maximum number of disjoint pairs of elements of the set $\{1,2, \ldots, n\}$ such that the sums of the different pairs are different integers not exceeding $n$? | [
"Consider $x$ such pairs in $\\{1,2, \\ldots, n\\}$. The sum $S$ of the $2x$ numbers in them is at least $1+2+\\cdots+2x$ since the pairs are disjoint. On the other hand $S \\leq n+(n-1)+\\cdots+(n-x+1)$ because the sums of the pairs are different and do not exceed $n$. This gives the inequality\n$$\n\\frac{2x(2x+1... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | floor((2n−1)/5) | |
0bah | Let $n$ be a positive integer. Prove that every integer between $1$ and $n!$ can be written as the sum of at most $n$ distinct positive divisors of $n!$. | [
"We use induction on $n$; the base case $n=1$ is obvious.\n\nSuppose now that the conclusion holds for some positive integer $n$ and consider $m \\le (n+1)!$. Then $m = (n+1)q + r$, $q, r \\in \\mathbb{N}$, $0 \\le r \\le n$. It is easily noticed that $q \\le n!$, therefore $q$ can be written as a sum of (at most) ... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
08f2 | Problem:
Viale Marconi è lungo $800~\mathrm{m}$; a $200~\mathrm{m}$ da un estremo c'è un parchimetro, a $100~\mathrm{m}$ dall'estremo opposto c'è un negozio di vestiti. Astolfo vuole parcheggiare la sua macchina, prendere un biglietto al parchimetro, tornare alla macchina per esporlo sul parabrezza, visitare il negozi... | [
"Solution:\n\nLa risposta è (E). Notiamo innanzitutto che la distanza fra parchimetro e negozio è di $500~\\mathrm{m}$. Se Astolfo parcheggia in un punto compreso fra il parchimetro ed il negozio, diciamo a distanza $x$ dal parchimetro (e quindi distanza $500~\\mathrm{m}-x$ dal negozio), allora dovrà percorrere a p... | Italy | Gara di Febbraio | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | E | |
0ir8 | Problem:
What is the largest $x$ such that $x^{2}$ divides $24 \cdot 35 \cdot 46 \cdot 57$? | [
"Solution:\nWe factor the product as $2^{4} \\cdot 3^{2} \\cdot 5 \\cdot 7 \\cdot 19 \\cdot 23$. If $x^{2}$ divides this product, $x$ can have at most 2 factors of $2$, 1 factor of $3$, and no factors of any other prime. So $2^{2} \\cdot 3 = 12$ is the largest value of $x$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 12 | |
0d2j | A Saudi company has two offices. One office is located in Riyadh and the other in Jeddah. To insure the connection between the two offices, the company has designated from each office a number of correspondents so that
(a) each pair of correspondents from the same office share exactly one common correspondent from the ... | [
"Assume that we have two correspondents, $R$ from Riyadh and $J$ from Jeddah, who are not connected to each other. Let $\\mathcal{J}_R$ be the set of correspondents from Jeddah connected to $R$ and $\\mathcal{R}_J$ the set of correspondents from Riyadh connected to $J$. Define the function\n$$\nf: \\mathcal{J}_R \\... | Saudi Arabia | Selection tests for the International Mathematical Olympiad 2013 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 8 | |
01xl | The sequence $a_1, a_2, a_3, \dots$ of positive integers is defined in the following way: $a_1 = 63$ and for each $n \ge 2$ the number $a_n$ is the smallest positive integer divisible by $n$, which is not less than $a_{n-1}$. (For example, $a_2 = 64$, $a_3 = 66, a_4 = 68, a_5 = 70$.)
Prove that each positive integer ca... | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0akv | Let $(a_n)_{n=1}^\infty$ be a sequence of positive numbers defined with: $a_1 = 1$, $a_2 = 2$ and $\frac{a_{n+1}^4}{a_n^3} = 2a_{n+2} - a_{n+1}$. Prove that for all positive integers $N > 1$ it holds that
$$
\sum_{k=1}^{N} \frac{a_k^2}{a_{k+1}} < 3.
$$ | [] | North Macedonia | Team Selection Test for BMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0gxl | For which values of parameter $a$ the system of equations
$$
\begin{cases} x + y + z = 0 \\ (xy + yz) + a xz = 0 \end{cases} \text{ has a unique solution?}
$$ | [
"It is evident that for every value of parameter $a$ there exists solution $x = y = z = 0$, and so all we need is to find out when this solution is unique.\n\nFrom the first equation we have that $y = -x - z$. Substitute $(-x - z)$ instead of $y$ into the second equation:\n\n$-x^2 - 2xz - z^2 + a xz = 0$\n\nor\n\n$... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (0,4) | |
06fv | In a chess tournament there were $n$ ($n > 10$) participants. Each participant plays against another exactly once. If a game ends in a draw, each participant gets 1 point, otherwise the winner gets 2 points and the loser gets none. At the end of the tournament, every participant found that half of his/her score came fr... | [
"$n$ can only be 25.\n\nWe first provide an example for the case $n = 25$. Consider 10 participants $A_1, A_2, \\dots, A_{10}$ in group $\\mathcal{A}$, and 15 participants $B_1, B_2, \\dots, B_{15}$ in group $\\mathcal{B}$. Suppose the games played between 2 participants in the same group end in a draw. For the gam... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 25 | |
03fm | $S_n = \{a \mid a < n, \exists k \in \mathbb{N} : 2^k \equiv a \pmod{n}\}$.
Are there different odd numbers $m$ and $r$ such that $S_m = S_r$? | [
"No! There exists a natural number $s$ such that $2^s \\equiv 1 \\pmod{n}$ (for example $s = \\varphi(n)$ from Euler's theorem or because the power series of $2$ modulo $n$ is periodic). We have $2^{s-1} \\equiv \\frac{n+1}{2} \\pmod{n}$ and so $x = \\frac{n+1}{2} \\in S_n$, but $2x = n+1 > n$ is not in $S_n$. Also... | Bulgaria | Bulgarian Spring Tournament | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | No | |
09ku | Let $n, m$ be positive integers. A sequence of $mn$ integers written on a circle is called nice if the sum of any $m$ consecutive integers is a power of $m$. Show that
(1) for any nice sequence of $mn$ ($m \ge 2$) integers, one can delete $m$ consecutive integers so that the remaining sequence of $m(n-1)$ integers is ... | [
"Suppose that $a_0, a_1, \\dots, a_{mn-1}$ is a nice sequence, where we take indices modulo $mn$. First of all, we claim that if $a_t = \\max\\{a_0, a_1, \\dots, a_{mn-1}\\}$ then $a_{t+1-m} = a_{t+1}, a_{t+2-m} = a_{t+2}, \\dots, a_{t-1} = a_{t+(m-1)}$ (call these $2(m-1)$ terms $(m-1)$-pairs). Let $S_i = a_i + \\... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0gxf | For every positive integer $m$ denote by $\sigma(m)$ the sum of all its divisors, and by $\phi(m)$ the number of positive integers less than $m$ and coprime with $m$. Prove that there exists infinitely many such positive integer numbers $n$ for which $\phi(\sigma(n)) > n$. | [
"Consider number $n = 2^{p-1}$, where $p$ is a prime number. Then $\\sigma(n) = 2^p - 1$. Let's prove that $\\phi(2^p - 1) > 2^{p-1}$ for infinitely many primes $p$.\n\nLet $m = 2^p - 1 = p_1^{\\alpha_1} \\dots p_k^{\\alpha_k}$. Then\n$$\n\\phi(m) = (2^p - 1) \\frac{p_1 - 1}{p_1} \\dots \\frac{p_k - 1}{p_k}.\n$$\nN... | Ukraine | The Problems of Ukrainian Authors | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibilit... | English | proof only | null | |
0d26 | For positive real numbers $a$, $b$ and $c$, prove that
$$
\frac{a^{3}}{a^{2}+a b+b^{2}}+\frac{b^{3}}{b^{2}+b c+c^{2}}+\frac{c^{3}}{c^{2}+c a+a^{2}} \geq \frac{a+b+c}{3}
$$ | [
"We have\n$$\n\\begin{aligned}\n& \\frac{a^{3}}{a^{2}+a b+b^{2}}+\\frac{b^{3}}{b^{2}+b c+c^{2}}+\\frac{c^{3}}{c^{2}+c a+a^{2}} \\\\\n& \\quad=\\frac{a^{4}}{a^{3}+a^{2} b+a b^{2}}+\\frac{b^{4}}{b^{3}+b^{2} c+b c^{2}}+\\frac{c^{4}}{c^{3}+c^{2} a+c a^{2}} \\\\\n& \\quad \\geq \\frac{\\left(a^{2}+b^{2}+c^{2}\\right)^{2... | Saudi Arabia | Selection tests for the Gulf Mathematical Olympiad 2013 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0e1r | Let $p$ and $q$ be two polynomials of degree $3$ with integer coefficients, such that the leading coefficients are relatively prime. Let $a$ be a rational number such that $p(a)$ and $q(a)$ are integers. Prove that $a$ is also an integer. | [
"Let $p(x) = b_1 x^3 + c_1 x^2 + d_1 x + e_1$ and $q(x) = b_2 x^3 + c_2 x^2 + d_2 x + e_2$. Let us write $a = \\frac{r}{s}$ where $r$ and $s$ are relatively prime integers and $s > 0$. Denote $p(a) = m$ and $q(a) = n$. Then\n$$\nb_1 \\cdot \\frac{r^3}{s^3} + c_1 \\cdot \\frac{r^2}{s^2} + d_1 \\cdot \\frac{r}{s} + e... | Slovenia | National Math Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
02ki | Problem:
Esmeralda escreveu em ordem crescente todos os números de $1$ a $999$, sem separá-los, formando o número mostrado a seguir: $12345678910111213\ldots 997998999$. Nesse número, quantas vezes aparece o agrupamento "21", nesta ordem? | [
"Solution:\n\nVamos primeiro listar os números que têm o agrupamento $21$ no meio de sua representação decimal:\n$21, 121, 221, \\ldots, 921 \\rightarrow 10$ números\n$210, 211, \\ldots, 219 \\rightarrow 10$ números\n\nTemos também que contar os agrupamentos $21$ obtidos a partir de um par de números consecutivos t... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Other"
] | null | final answer only | 31 | |
0h6p | The twins Pete and Ostap had had an argue and started to go to school different ways. Pete goes $210$ meters to the South, then $70$ meters to the East and reaches the school. Ostap goes to the North for a while and then he head to the school directly. How many meters Ostap has to go to the North if both twins have equ... | [
"Let us denote the points: home - $D$, school - $S$, the point where Pete turns to the East - $B$, the point where Ostap begins to walk straight to the school - $N$ (fig. 12). Then $DB = 210$, $BS = 70$, $DN = x$, $NS = y$. Also we have $x + y = 280$.\n\n\n\nBy Pythagorean theorem: $BN^2 + ... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 30 | |
08yf | Suppose a regular pentagon $ABCDE$ is given. Let $F$ be the point of intersection of the lines $BE$ and $AC$. Suppose that a straight line going through $F$ intersects sides $AB$ and $CD$ at the points $G$, $H$, respectively, and $BG = 4$ and $CH = 5$ hold. Find the length of the line segment $AG$. Here by $XY$ we deno... | [
"$2\\sqrt{6} - 2$\n\nLet $I$ be the point of intersection of the line segments $CE$ and $FH$. Let also $AG = a$. Five vertices of the regular pentagon $ABCDE$ lie on the circumference of the same circle, and these points divide the circumference of this circle into five equal parts. Consequently, we have\n$$\n\\ang... | Japan | 2019 Japan Mathematical Olympiad First Stage | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2√6 - 2 | |
06dk | Let $p$ be an odd prime such that $p \equiv 1 \pmod 4$. Evaluate with reasons, $\sum_{k=1}^{\frac{p-1}{2}} \left\{ \frac{k^2}{p} \right\}$, where $\{x\} = x - [x]$, $[x]$ being the greatest integer not exceeding $x$. | [
"The answer is $\\frac{p-1}{4}$.\nNote that $1^2, 2^2, \\dots, \\left(\\frac{p-1}{2}\\right)^2$ are pairwise incongruent modulo $p$ since\n$$\nx^2 \\equiv y^2 \\pmod{p} \\Rightarrow x \\equiv \\pm y \\pmod{p}.\n$$\nAs there are exactly $\\frac{p-1}{2}$ nonzero quadratic residues, $1^2, 2^2, \\dots, \\left(\\frac{p-... | Hong Kong | CHKMO | [
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | (p-1)/4 | |
0gp8 | Between any two cities of country $A$ consisting of $2011$ cities and country $B$ consisting of $2011$ cities there is a unique direct two way flight organized by some airway company. For each given city there are at most $19$ different airway companies operating flights related to this city. Determine the maximum poss... | [
"**The answer is $212$.**\nLet $K_{2011,2011}$ be a complete bipartite graph in which all vertices of a set $A$ with $|A| = 2011$ are connected to all vertices of $B$ with $|B| = 2011$. We prove that there exists a monochromatic connected subgraph with $212$ vertices if edges of the graph $K_{2011,2011}$ are colore... | Turkey | 19th Turkish Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | 212 | |
0ed7 | Assume that $n \ge 3$ people of different ages sit around a round table. We call any pair consisting of two people $A$ and $B$ a *newbie pair*, if
(i) $A$ and $B$ do not sit next to one another and
(ii) on at least one of the arcs connecting $A$ and $B$ along the table edge all people are older than $A$ and $B$.
Determ... | [
"Denote the people in the order of ascending age by $A_1, A_2, \\dots, A_n$. Using induction on $n$ we show that regardless of the seating arrangement there are exactly $n-3$ newbie pairs.\n\nIn the base case of $n=3$ every person is adjacent to every other person so there are no newbie pairs.\n\nFor the induction ... | Slovenia | Slovenija 2016 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | n - 3 | |
0bk0 | Consider real numbers $a_1, a_2, \dots, a_{2n}$ whose sum is $0$. Prove that among the pairs $(a_i, a_j)$, $i < j$, with $i, j \in \{1, 2, \dots, 2n\}$, there exist at least $2n - 1$ pairs such that $a_i + a_j \ge 0$. | [
"* If $a_n + a_{2n-1} \\ge 0$, then all the sums $a_i + a_{2n-1}$ with $i = \\overline{n}, 2n-2$ as well as all the sums $a_i + a_{2n}$ with $i = \\overline{n}, 2n-1$ are non-negative. In total, there are at least $(n-1)+n = 2n-1$ non-negative sums.\n\n* If $a_n + a_{2n-1} < 0$, then $a_1 + \\dots + a_{n-1} + a_{n+... | Romania | THE 2014 DANUBE MATHEMATICAL COMPETITION | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2n - 1 | |
0691 | Let $ABC$ be an equilateral triangle, and let $P$ be some point in its circumcircle. Determine, with reasons, all the positive integers $n$ such that the sum
$$
S_n P = |PA|^n + |PB|^n + |PC|^n
$$
is independent of the choice of the point $P$. | [] | Greece | 20th Mediterranean Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | 2 and 4 | |
02n3 | Problem:
Calcule o valor da soma
$$
S = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \cdots + \frac{1}{2006 \cdot 2007} + \frac{1}{2007 \cdot 2008}
$$ | [
"Solution:\n\nInicialmente, observe que $\\frac{1}{k \\cdot (k+1)} = \\frac{1}{k} - \\frac{1}{k+1}$. Logo,\n$$\n\\frac{1}{1 \\cdot 2} = 1 - \\frac{1}{2}, \\quad \\frac{1}{2 \\cdot 3} = \\frac{1}{2} - \\frac{1}{3}, \\quad \\ldots, \\quad \\frac{1}{2007 \\cdot 2008} = \\frac{1}{2007} - \\frac{1}{2008}\n$$\nAssim, tem... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2007/2008 | |
04gh | Let $ABC$ be an equilateral triangle with sides of length $1$. The point $X$ on the ray $AB$ and the point $Y$ on the ray $AC$ are chosen, so that $|AX|$ and $|AY|$ are positive integers.
Can the radius of the circumcircle of the triangle $AXY$ be $\sqrt{2014}$? | [
"Let us assume that the radius of the circumcircle of the triangle $AXY$ is equal to $\\sqrt{2014}$.\nIn the triangle $AXY$ the angle opposite to the side $XY$ is $60^\\circ$ because the triangle $ABC$ is equilateral.\nIf $R$ is the radius of the circumcircle of the triangle $AXY$, then\n$$\n|XY| = 2R \\sin 60^\\ci... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | No | |
000b | Sea $ABC$ un triángulo tal que el ángulo $BAC = 45^\circ$. Sean $P$ y $Q$ puntos interiores del triángulo $ABC$ tales que $ABQ = BQP = PBC$ y $ACQ = CQP = PCB$. Sean $D$ y $E$ los pies de las perpendiculares trazadas desde $P$ a los lados $CA$ y $AB$, respectivamente. Demostrar que $Q$ es el ortocentro del triángulo $A... | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | español | proof only | null | |
0i1p | Problem:
Evaluate the definite integral
$$\int_{-1}^{+1} \frac{2 u^{332}+u^{998}+4 u^{1664} \sin u^{691}}{1+u^{666}} \, \mathrm{d}u.$$ | [
"Solution:\n\nThe term $\\frac{4 u^{1664} \\sin u^{691}}{1+u^{666}}$ is odd in $u$, so its integral is $0$.\n\nNow make the substitution $u = v^{1/333} \\Rightarrow \\mathrm{d}u = \\frac{1}{333} v^{-332/333} \\mathrm{d}v$ to find that\n$$\n\\int_{-1}^{+1} \\frac{2 u^{332} + u^{998}}{1 + u^{666}} \\, \\mathrm{d}u = ... | United States | Harvard-MIT Math Tournament | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Trigonometric functions"
] | null | proof and answer | 2/333 * (1 + π/4) | |
0dec | In the popular game of Minesweeper, some fields of an $a \times b$ board are marked with a mine and on all the remaining fields the number of adjacent fields that contain a mine is recorded. Two fields are considered adjacent if they share a common vertex. For which $k \in \{0, 1, 2, 3, 4, 5, 6, 7, 8\}$ is it possible ... | [
"For $k = 0$ this is impossible, since at least one non-mine field will be adjacent to the bombs unless the entire field is covered with bombs.\n\n* For $k = 1$ we take $a = 1, b = 3 \\cdot 1011 - 1$, and 1011 mines spaced two-spaces apart, with one on one of the endpoints, so the total number of fields marked 1 is... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Other"
] | null | proof and answer | {1, 2, 3, 4, 5, 6, 8} | |
0lc9 | Let $n$ be a positive integer. Prove that there is only one tuple of $n+2$ real numbers $(\alpha_1, \alpha_2, \alpha_3, \dots, \alpha_n, \alpha_{n+1})$ such that $\alpha_0 = \alpha_{n+1} = 0$, $|\alpha_i| \le \frac{\pi}{6}$ and
$$
1 + \sin \alpha_{i+1} + 3 \sin \alpha_{i-1} = 10 \sin \alpha_i - 2 \sin 3\alpha_i
$$
for ... | [] | Vietnam | Vietnamese Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0hfb | Find all triples of positive integers $a$, $b$, $c$, such that
$$
a + (a, b) = b + (b, c) = c + (c, a),
$$
where by $(x, y)$ we denote the largest common divisor of integers $x$, $y$. | [
"If numbers $a$, $b$, $c$ have a common divisor, we can divide by it and get a triple of integers $a_1$, $b_1$, $c_1$, whose largest common divisor is $1$. As $a_1 + (a_1, b_1) = b_1 + (b_1, c_1)$, $(b_1, c_1)$ is divisible by $(a_1, b_1)$. But $((a_1, b_1), (b_1, c_1)) = 1$, as explained above, so $(a_1, b_1) = 1$... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | (n, n, n) for any positive integer n | |
04tb | We have $n^2$ empty boxes; each of them having square base. The height and the width of each box belongs to $\{1, 2, \dots, n\}$ and every two boxes differ in at least one of these two dimensions. One box fits into another one if both its dimensions are smaller and at least one is smaller by at least 2. In this way, we... | [
"We will show that the required minimal number of shelves is $3n-2$. This answer is clearly correct for $n=1$ and $n=2$ since for such $n$ we have $n^2 = 3n - 2$ and each box has to be stored on a different shelf. From now on, let $n \\ge 3$.\nWe identify boxes with points in an $n \\times n$ grid: A box of width $... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Algorithms"
] | English | proof and answer | 3n - 2 | |
0fey | Problem:
Los vértices del cuadrilátero convexo $ABCD$ están situados en una circunferencia. Sus diagonales $AC$ y $BD$ se cortan en el punto $E$. Sea $O_{1}$ el centro del círculo inscrito en el triángulo $ABC$, y $O_{2}$ el centro del círculo inscrito en el triángulo $ABD$. La recta $O_{1}O_{2}$ corta a $EB$ en $M$ y... | [
"Solution:\n\nEl centro del círculo inscrito es el punto de intersección de las bisectrices, así que\n$$\n\\angle O_{1}BA = \\angle O_{1}BC = \\beta, \\quad \\angle O_{1}AB = \\angle O_{1}AC = \\tau, \\quad \\angle O_{2}BA = \\angle O_{2}BD = \\alpha, \\quad \\angle O_{2}AB = \\angle O_{2}AB = \\gamma\n$$\nEntonces... | Spain | TANDA II | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cvo | Is it true that for any nonzero integer numbers $a$ and $b$ the system of equations
$$
\begin{cases}
\tan(13x) \tan(ay) = 1, \\
\tan(21x) \tan(by) = 1
\end{cases}
$$
has at least one solution? | [
"Покажем, что система не будет иметь решений при $a = 8$, $b = 13$. Действительно, из уравнений системы вытекает, что\n$$\n13x + ay = \\frac{\\pi}{2} + \\pi k, \\quad 21x + by = \\frac{\\pi}{2} + \\pi \\ell\n$$\nпри целых $k$ и $\\ell$. Отсюда следует\n$$\n(21a - 13b)y = 21(13x + ay) - 13(21x + by) = \\pi(4 + 21k -... | Russia | Final round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English; Russian | proof and answer | No; for a = 8 and b = 13 the system has no solution. | |
083j | Problem:
Un villaggio è costituito da abitazioni isolate, collegate da strade. Ognuna di queste strade è un sentiero che collega due abitazioni (e tra due abitazioni vi è al più un sentiero che le collega). Le abitazioni sono di due tipi: centrali e periferiche. Ogni abitazione centrale è collegata esattamente ad altr... | [] | Italy | Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 24 | |
0cp7 | Given an acute-angled triangle $ABC$ such that its median $AM$ is larger than the side $AB$. Prove that one can cut triangle $ABC$ into three parts and rearrange them so that they will form a rhombus.
В остроугольном треугольнике $ABC$ медиана $AM$ длиннее стороны $AB$. Докажите, что треугольник $ABC$ можно разрезать ... | [
"Пусть $N$ — середина стороны $AC$, а $K$ — точка на прямой $MN$ такая, что $MK = MN$. Тогда треугольники $MNC$ и $MKB$ симметричны относительно $M$ и потому равны. Проведём разрез по средней линии $MN$; переложив треугольник $MNC$ так, чтобы он совпал с $\\triangle MKB$, получаем параллелограмм $ANKB$ (см. рис. 16... | Russia | Final round | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English; Russian | proof only | null | |
07ug | For any positive integer $N$ we denote by $\overline\{N\}$ the number obtained by writing the digits of $N$ from right to left. For instance, if $N = 2395$ then $\overline\{N\} = 5932$ and if $N = 3780$ then $\overline\{N\} = 873$.
Find all positive integers $N$ such that $N - \overline\{N\} = 9$. | [
"If $N$ has one digit this is not possible. If $N$ has two digits, say $N = ab$, then $\\overline\\{N\\} = ba$ so that $N - \\overline\\{N\\} = 9(a - b)$ which yields $a - b = 1$ so the numbers are\n$$\nN = 10, 21, 32, 43, 54, 65, 76, 87, 98.\n$$\nAssume there are numbers $N$ with the above property that have $m \\... | Ireland | IRL_ABooklet | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 10, 21, 32, 43, 54, 65, 76, 87, 98 | |
0kov | Let $S$ be the set of all rational numbers that can be expressed as a repeating decimal in the form $0.abcd$, where at least one of the digits $a, b, c$, or $d$ is nonzero. Let $N$ be the number of distinct numerators obtained when numbers in $S$ are written as fractions in lowest terms. For example, both $4$ and $410$... | [] | United States | 2022 AIME I | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Discrete Mathematics > Combinatorics > Inclusion-exclus... | null | proof and answer | 392 | |
0hz8 | Problem:
You are given 16 pieces of paper numbered $16,15, \ldots, 2,1$ in that order. You want to put them in the order $1,2, \ldots, 15,16$ switching only two adjacent pieces of paper at a time. What is the minimum number of switches necessary? | [
"Solution:\n\nPiece $16$ has to move to the back $15$ times, piece $15$ has to move to the back $14$ times, $\\ldots$. Piece $2$ has to move to the back $1$ time, piece $1$ has to move to the back $0$ times. Since only one piece can move back in each switch, we must have at least $15+14+\\ldots+1=\\mathbf{120}$ swi... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 120 | |
08xq | In a certain school, there is a committee consisting of 4 members. There are 4 sub-committees for this committee, and each committee member will be asked to chair a different sub-committee. Suppose each committee member has 2 sub-committees he prefers to chair. It is known that there are exactly 2 ways to assign sub-co... | [
"Consider a graph which consists of 4 vertices, representing 4 sub-committees, and edges, representing 4 members, each of which connects a pair of vertices, corresponding to two sub-committees a committee member prefers to serve in. Let us call a method of choosing one of the end points for each of the edges to be ... | Japan | Japan Mathematical Olympiad Initial Round | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 936 | |
05p6 | Problem:
Déterminer tous les entiers $n \geqslant 3$ pour lesquels il existe $n$ polynômes $P_{1}, \ldots, P_{n}$ à coefficients réels et tels que, pour tous $i, j, k$ distincts : $P_{i}+P_{j}$ n'a aucune racine réelle et $P_{i}+P_{j}+P_{k}$ admet au moins une racine réelle. | [
"Solution:\n\nProuvons que seul $n=3$ a la propriété désirée.\n\nTout d'abord, en posant $P_{1}(x)=3x^{2}+3$ et $P_{2}(x)=P_{3}(x)=-x^{2}-2$, on vérifie aisément que les polynômes $P_{1}(x)+P_{2}(x)=P_{1}(x)+P_{3}(x)=2x^{2}+1$ et $P_{2}(x)+P_{3}(x)=-2x^{2}-4$ n'ont aucune racine réelle, tandis que $P_{1}(x)+P_{2}(x... | France | Olympiades Françaises de Mathématiques - Envoi 2 (Algèbre) | [
"Algebra > Algebraic Expressions > Polynomials",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3 | |
0498 | Mario has written a 30-digit number whose sum of digits is 123. Then he wrote all the digits again in some other order following the original number. Prove that the 60-digit number he obtained is not a perfect square. | [] | Croatia | CroatianCompetitions2011 | [
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
097r | Problem:
Numerele reale $x$, $y$, $z$ și $A$ verifică relația
$$
A = \sqrt{\frac{1}{(2x - y - z)^2} + \frac{1}{(2y - z - x)^2} + \frac{1}{(2z - x - y)^2}}
$$
Arătați că dacă $x$, $y$ și $z$ sunt numere raționale, atunci și $A$ este un număr rațional. | [
"Solution:\n\nNotăm $2x - y - z = a$, $2y - z - x = b$, $2z - x - y = c$.\nAtunci $a + b + c = 0$ și deoarece $x, y, z \\in \\mathbb{Q}$, avem $a, b, c \\in \\mathbb{Q}$.\nMai mult,\n$$\n\\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} = \\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right)^2 - 2\\left(\\frac{1... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
03u5 | Point $D$ lies inside triangle $ABC$ such that $\angle DAC = \angle DCA = 30^\circ$ and $\angle DBA = 60^\circ$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF = 2FC$. Prove that $DE \perp EF$. | [
"Let $G$ and $M$ be the midpoints of segments $AF$ and $AC$ respectively.\n\nIn right triangle $ADM$, $\\angle ADM = 60^\\circ$ and $AM = \\sqrt{3}DM$.\n\nNote that $AM = 3GM$. Hence $DM = \\sqrt{3}GM$ and $DG = 2GM = AG = GF$. By symmetry, $DF = GF$ and $DFG$ is an isosceles triangle.\n\n\... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
08gg | Problem:
Un quadrilatero convesso $ABCD$ è tale che $AB = 14$, $BC = 24$, $CD = 26$ e $DA = 16$. La sua area è $360$. Trovare l'area della regione di piano interna al quadrilatero costituita dai punti distanti al massimo $3$ dal suo perimetro.
(A) $100$
(B) $120$
(C) $120 + 9\pi$
(D) $156$
(E) $200$ | [
"Solution:\n\nLa risposta è $(\\mathbf{E})$. La soluzione del problema si basa su due considerazioni chiave. Innanzitutto, notando che $AB + CD = BC + DA = 40$, si deduce che il quadrilatero $ABCD$ è circoscritto a una circonferenza. In particolare, detto $r$ il raggio di tale circonferenza, vale $r = \\frac{360}{4... | Italy | Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | MCQ | E | |
06wh | A hunter and an invisible rabbit play a game on an infinite square grid. First the hunter fixes a colouring of the cells with finitely many colours. The rabbit then secretly chooses a cell to start in. Every minute, the rabbit reports the colour of its current cell to the hunter, and then secretly moves to an adjacent ... | [
"Answer: Yes, there exists a colouring that yields a winning strategy for the hunter.\n\nA central idea is that several colourings $C_{1}, C_{2}, \\ldots, C_{k}$ can be merged together into a single product colouring $C_{1} \\times C_{2} \\times \\cdots \\times C_{k}$ as follows: the colours in the product colourin... | IMO | IMO 2021 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Yes, there exists a colouring that yields a winning strategy for the hunter. | |
09wp | Problem:
Voor een geheel getal $n \geq 3$ bekijken we een cirkel met $n$ punten erop. We plaatsen een positief geheel getal bij elk punt, waarbij de getallen niet noodzakelijk verschillend hoeven te zijn. Zo'n plaatsing van getallen heet stabiel als drie getallen naast elkaar altijd product $n$ hebben. Voor hoeveel wa... | [
"Solution:\n\nStel dat $n$ geen veelvoud van 3 is en dat we een stabiele plaatsing hebben van de getallen $a_{1}, a_{2}, \\ldots, a_{n}$, die in die volgorde op de cirkel liggen. Er geldt dan $a_{i} a_{i+1} a_{i+2}=n$ voor alle $i$, waarbij we de indices modulo $n$ rekenen. Dus\n$$\na_{i+1} a_{i+2} a_{i+3}=n=a_{i} ... | Netherlands | Selectietoets | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 680 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.