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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
06gh | Let $A_1A_2\cdots A_n$ be a cyclic polygon ($n \ge 3$). Find the maximum number of distinct acute-angled triangles whose vertices are chosen from $A_1, A_2, \dots, A_n$. | [
"The maximum number of acute-angled triangles is $\\frac{n(n-1)(n+1)}{24}$ if $n$ is odd and is $\\frac{n(n-2)(n+2)}{24}$ if $n$ is even.\n\nWe first count the number $N_1$ of non-acute triangles with $A_1$ as one vertex such that $\\angle A_1 < 90^\\circ$. Let $\\ell$ be the diameter of the circle which passes thr... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof and answer | Maximum number of acute triangles equals n(n-1)(n+1)/24 for odd n and n(n-2)(n+2)/24 for even n. | |
02gx | Consider all the ways of writing exactly ten times each one of the numbers $0, 1, 2, 3, \ldots, 9$ in the squares of a $10 \times 10$ board.
Find the greatest integer $n$ with the property that there is always a row or a column with $n$ different numbers. | [
"Let's count in two ways the number of ordered pairs $(d, l)$, where $d$ is a digit and $l$ is a row or column containing $d$. For simplicity, let a *line* be a row or a column. Since there are $10$ occurrences of $d$, they are present in at least $7$ lines (the intersections of the rows and columns must cover all ... | Brazil | XXVI OBM | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 4 | |
04l4 | Does there exist a positive integer $m$ such that $2^{m^2} - 4$ is divisible by $7$? (Ukraine 2013) | [] | Croatia | Mathematical competitions in Croatia | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | No; there is no positive integer m such that 2^{m^2} − 4 is divisible by 7. | |
078r | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that for all real numbers $a, b, c$, we have
$$
f(a + b + c)f(ab + bc + ca) - f(a)f(b)f(c) = f(a + b)f(b + c)f(c + a).
$$ | [
"Let $\\mathcal{P}(a, b, c)$ denote the expression\n$$\nf(a + b + c)f(ab + bc + ca) = f(a)f(b)f(c) + f(a + b)f(b + c)f(c + a).\n$$\n\n$\\mathcal{P}(0, 0, 0) \\Rightarrow f(0)^2 = 2f(0)^3.$ So, $f(0) = \\frac{1}{2}$ or $f(0) = 0$. Exclude the solution $f \\equiv 0$.\n\n**Case 1** ($f(0) = \\frac{1}{2}$):\n$\\mathcal... | India | IMO TST | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | All solutions are: (1) f(x) = 1/2 for all real x; (2) f(0) = 1/2 and f(x) = 0 for all nonzero x; (3) f(x) = x for all real x. | |
026m | Problem:
Fração e porcentagem - Se na fração $\frac{x}{y}$ diminuirmos o numerador de $40\%$ e o denominador $y$ de $60\%$, então a fração $\frac{x}{y}$ :
(A) diminui $20\%$
(B) aumenta $20\%$
(C) diminui $50\%$
(D) aumenta $50\%$ | [
"Solution:\n\nA opção correta é (D).\nSe um número $x$ é diminuído de $40\\%$, ele passa a valer $60\\%$ de $x$, ou seja: $0,6x$. Do mesmo modo, quando um número $y$ é diminuído de $60\\%$, ele passa a valer $0,4y$. Portanto, a fração $\\frac{x}{y}$ passa a ter o valor $\\frac{0,6x}{0,4y} = \\frac{6}{4} \\frac{x}{y... | Brazil | Nível 3 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | D | |
0hef | Positive integer $n$ has no more than $2020$ digits. Prove that there exist $m$ such palindromes that their sum equals $n$, where $m \le 13$.
A palindrome is a number which is read the same both left to right and right to left, i.e. $1001$, $9$ and $767$ are palindromes, while $1212$ and $110$ are not. | [
"First, we prove a lemma:\n\n**Lemma 1.** Suppose there is a $k$-digit positive integer $X$, $k > 2$. Then such a palindrome can be subtracted from it, that the remaining number consists of no more than $\\lfloor \\frac{1}{2}k+1 \\rfloor$ digits.\n\n*Proof.* Suppose the first $\\lfloor \\frac{k+1}{2} \\rfloor$ digi... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
01jk | Let $\mathbb{Z}^+$ be the set of positive integers and $(F_n)_{n \in \mathbb{Z}^+}$ be the Fibonacci sequence defined by $F_1 = F_2 = 1$ and $F_{n+1} = F_n + F_{n-1}$ for $n \ge 2$. Consider the number
$$
N = 2^{2022}F_{2023} - 2^{2021}F_{2022} + 2^{2020}F_{2021} + \dots + 2F_2 + F_1.
$$
Prove that the binary expansion... | [
"Since $F_n$ is monotonically increasing, it is clear that\n$$\n0 < N < 2^{2022}F_{2023}.\n$$\nMoreover, since $F_{n+1} \\le 2F_n$, it is clear that $F_{2023} < 2^{2023}$ and hence $N$ has at most 4045 binary digits. It will thus suffice to prove that the last 2023 binary digits of $N$ are 1's, in other words, that... | Baltic Way | Baltic Way 2023 Shortlist | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0b15 | Problem:
For each positive integer $n$, let $\varphi(n)$ be the number of positive integers from $1$ to $n$ that are relatively prime to $n$. Evaluate
$$
\sum_{n=1}^{\infty} \frac{\varphi(n) 4^{n}}{7^{n}-4^{n}}
$$ | [
"Solution:\nWe compute\n$$\n\\sum_{n=1}^{\\infty} \\frac{\\varphi(n) 4^{n}}{7^{n}-4^{n}} = \\sum_{n=1}^{\\infty} \\varphi(n) \\frac{\\left(\\frac{4}{7}\\right)^{n}}{1-\\left(\\frac{4}{7}\\right)^{n}} = \\sum_{n=1}^{\\infty} \\varphi(n) \\sum_{k=1}^{\\infty}\\left(\\frac{4}{7}\\right)^{n k}\n$$\nInterchanging the or... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 28/9 | |
02to | Problem:
Na figura abaixo, os trapézios retângulos $ABCD$ e $AEFG$, com $BC \parallel EF$ e $CD \parallel FG$, possuem a mesma área. Sabendo que $BC=4$, $AD=7$, $CT=1$ e $TD=2$, determine a medida do segmento $DG$.
 | [
"Solution:\n\nTrace uma perpendicular por $B$ ao lado $DA$, intersectando $EF$ em $Y$ e $AD$ em $X$. Chamemos $DG$ de $x$ e $EY$ de $y$. Veja a figura abaixo:\n\nOs triângulos $\\triangle EYB$ e $\\triangle AXB$ possuem os mesmos ângulos e, consequentemente, são semelhantes. Desse modo, $\\... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 9/4 | |
0bmj | Consider $n$ positive, not necessarily distinct, integers $a_1, a_2, \dots, a_n$ whose sum is $2S$. A positive integer $k$ is called a separator if one can choose $k$ indexes from $\{1, 2, \dots, n\}$ such that the sum of the corresponding numbers is $S$. What is the maximum possible number of separators? | [
"If 1 is a separator, then we can not have any other separators than 1 and $n-1$.\nFor $n=1$ we can not have any separators.\nFor $n=2$, only 1 can be a separator (if the two numbers are equal).\nFor $n=3$ only 1 and 2 can be separators (for example, in the case of the numbers 1, 2, 3; $1+2+3=3$).\nFor $n=4$ we can... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | For n = 1: 0; n = 2: 1; n = 3: 2; n = 4: 2; and for n ≥ 5: n − 3. | |
0jzh | Problem:
Lunasa, Merlin, and Lyrica each have a distinct hat. Every day, two of these three people, selected randomly, switch their hats. What is the probability that, after 2017 days, every person has their own hat back? | [
"Solution:\n\nImagine that the three hats are the vertices of an equilateral triangle. Then each day the exchange is equivalent to reflecting the triangle along one of its three symmetry axes, which changes the orientation of the triangle (from clockwise to counterclockwise or vice versa). Thus, an even number of s... | United States | HMMT November | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 0 | |
069n | Triangle $ABC$ is such that $AB < AC$. The perpendicular bisector of side $BC$ intersects lines $AB$ and $AC$ at points $P$ and $Q$, respectively. Let $H$ be the orthocentre of triangle $ABC$, and let $M$ and $N$ be the midpoints of segments $BC$ and $PQ$, respectively. Prove that lines $HM$ and $AN$ meet on the circum... | [] | Greece | 23rd Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
012z | Problem:
All the positive divisors of a positive integer $n$ are stored into an array in increasing order. Mary has to write a program which decides for an arbitrarily chosen divisor $d>1$ whether it is a prime. Let $n$ have $k$ divisors not greater than $d$. Mary claims that it suffices to check divisibility of $d$ b... | [
"Solution:\n\nLet $d>1$ be a divisor of $n$. Suppose Mary's program outputs \"composite\" for $d$. That means it has found a divisor of $d$ greater than 1. Since $d>1$, the array contains at least 2 divisors of $d$, namely 1 and $d$. Thus Mary's program does not check divisibility of $d$ by $d$ (the first half gets... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
07s8 | Suppose $a$, $b$, $c$ are the side lengths of a triangle. Prove that
$$
(a+b+c)(a^2+b^2+c^2) + a(b-c)^2 + b(c-a)^2 + c(a-b)^2 \le 3(a^3+b^3+c^3),
$$
with equality iff the triangle is equilateral. | [
"$$\n\\begin{align*}\n& 3(a^3 + b^3 + c^3) - (a + b + c)(a^2 + b^2 + c^2) \\\\\n&= 2(a^3 + b^3 + c^3) + a^2(b + c) + b^2(c + a) + c^2(a + b) \\\\\n&= (a^3 + b^3 - a^2b - ab^2) + (b^3 + c^3 - b^2c - c^2b) + (c^3 + a^3 - c^2a - a^2c) \\\\\n&= (a + b)(a - b)^2 + (b + c)(b - c)^2 + (c + a)(c - a)^2 \\\\\n&\\geq (a + b ... | Ireland | Irish | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
05kn | Problem:
Combien existe-t-il de couples d'entiers strictement positifs $(a, b)$ tels que
$$
\frac{1}{a} + \frac{1}{b} = \frac{1}{2014} ?
$$ | [
"Solution:\n\nSoient $a, b$ deux entiers strictement positifs tels que $\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{2014}$. On a en particulier $a, b > 2014$. On peut donc multiplier l'équation par $ab$ : on cherche en fait les entiers $a, b > 2014$ tels que $ab - 2014a - 2014b = 0$ ou encore, de manière équivalente, t... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 27 | |
0ijm | Let $n$ be a positive integer. Find, with proof, the least positive integer $d_n$ which cannot be expressed in the form
$$
\sum_{i=1}^{n} (-1)^{a_i} 2^{b_i},
$$
where $a_i$ and $b_i$ are nonnegative integers for each $i$. | [
"The answer is $d_n = \\dfrac{2^{2n+1} + 1}{3}$. We first show that $d_n$ cannot be obtained. For any $p$ let $t(p)$ be the minimum $n$ required to express $p$ in the desired form and call any realization of this minimum a minimal representation. If $p$ is even, any sequence of $b_i$ that can produce $p$ must conta... | United States | Team Selection Test | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (2^{2n+1} + 1) / 3 | |
0j3d | Problem:
Let $f(n) = \sum_{k=2}^{\infty} \frac{1}{k^{n} \cdot k!}$. Calculate $\sum_{n=2}^{\infty} f(n)$. | [
"Solution:\nAnswer: $3-e$\n$$\n\\begin{aligned}\n\\sum_{n=2}^{\\infty} f(n) & = \\sum_{k=2}^{\\infty} \\sum_{n=2}^{\\infty} \\frac{1}{k^{n} \\cdot k!} \\\\\n& = \\sum_{k=2}^{\\infty} \\frac{1}{k!} \\sum_{n=2}^{\\infty} \\frac{1}{k^{n}} \\\\\n& = \\sum_{k=2}^{\\infty} \\frac{1}{k!} \\cdot \\frac{1}{k(k-1)} \\\\\n& =... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 3 - e | |
0keb | Problem:
A triangle and a circle are in the same plane. Show that the area of the intersection of the triangle and the circle is at most one third of the area of the triangle plus one half of the area of the circle. | [
"Solution:\n\nLet $\\triangle$ denote the triangle, $\\circ$ denote the circle, and $\\triangle'$ denote the reflection of the triangle in the center of the circle. Letting overline denote complement,\n$$\n\\begin{aligned}\n[\\triangle \\cap \\circ] & \\leq \\frac{1}{3}[\\triangle]+\\frac{1}{2}[\\mathrm{o}] \\\\\n\... | United States | HMIC | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
08yn | For all quadruples $(x, y, z, w)$ consisting of integers $1 \le x, y, z, w \le 1000$, we consider adding up the maximum value of $xy + zw$, $xz + yw$, $xw + yz$ and denote the sum by $M$. Similarly, for all quadruples $(x, y, z, w)$ consisting of integers $1 \le x, y, z, w \le 1000$, we consider adding up the minimum v... | [
"$$\n\\boxed{20412}\n$$\nFor 3 real numbers $a, b, c$, the difference of the maximum value and the minimum value of them is $\\frac{|a-b|+|b-c|+|c-a|}{2}$. Using $|(xy + zw) - (xz + yw)| = |x - w||y - z|$, we have\n$$\nM - m = \\frac{1}{2} \\sum_{x,y,z,w=1}^{1000} \\left( |x - w||y - z| + |x - y||z - w| + |x - z||y... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 20412 | |
0ghn | 正整數 $n$ 與 $k$ 滿足 $n > 2023k^3$。貓貓國有 $n$ 座城市, 其中每對城市之間有至多一條道路。已知該國的道路總數不少於 $2n^{3/2}$。證明: 我們可以選出 $3k+1$ 座城市, 使得以這些城市為兩端點的道路數量不少於 $4k$。
Integers $n$ and $k$ satisfy $n > 2023k^3$. Kingdom Kitty has $n$ cities, with at most one road between each pair of cities. It is known that the total number of roads in the kingdom is... | [
"讓我們先將 degree 最小的頂點依次從圖中移除，直到剩下的圖 $G' = (V', E')$ 中所有點的 degree 都大於 $n^{1/2}$。注意到以上動作至多移除 $n \\times n^{1/2}$，因此 $|E'| \\ge n^{3/2}$，從而 $m = |V'| \\ge \\sqrt{2|E'|} \\ge n^{3/4}$。\n任取 $V'$ 中的一個點 $v$，並令 $V_1$ 和 $V_2$ 分別為 $V$ 中距離 $v$ 單位 1 與 2 的點所形成的集合。將 $v$ 到 $V_1$ 的所有邊塗成藍色。此外，對於每個 $V_2$ 裡面的點 $y$，選定 $V_1$ 中與 $y$ 有連邊的一... | Taiwan | 2023 數學奧林匹亞競賽第二階段選訓營 | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Chinese (Traditional) | proof only | null | |
0891 | Problem:
In un triangolo acutangolo $ABC$ con $AB < AC$, la bisettrice che parte da $A$ interseca il lato $BC$ nel punto $P$. La parallela al lato $AB$ passante per $P$ interseca il lato $AC$ nel punto $Q$; su questa retta sia $R$ il punto che giace sulla semiretta uscente da $Q$ che non contiene $P$ e tale che $QR = ... | [
"Solution:\n\na.\nDimostriamo che non solo il triangolo $QAR$ è isoscele di base $AR$ per costruzione, ma anche il triangolo $PAQ$ lo è (con base $AP$).\n\n\n\nInfatti\n$$\nB\\widehat{A}P = P\\widehat{A}Q\n$$\nperché $AP$ è bisettrice, mentre\n$$\nB\\widehat{A}P = A\\widehat{P}Q\n$$\nperché... | Italy | UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA GARA di SECONDO LIVELLO | [
"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",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
0apz | Problem:
Find the number of real roots of the equation
$$
4 \cos (2007 a) = 2007 a
$$ | [
"Solution:\nLet $x = 2007 a$. Then the given equation becomes $4 \\cos x = x$. The graphs of the equations $y = 4 \\cos x$ and $y = x$ intersect at three points. Thus, the equation $4 \\cos x = x$ has three roots. Consequently, the equation $4 \\cos (2007 a) = 2007 a$ also has three roots."
] | Philippines | Tenth Philippine Mathematical Olympiad | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | null | proof and answer | 3 | |
0iy3 | Problem:
Let $n$ be a positive integer. Let $V_{n}$ be the set of all sequences of 0's and 1's of length $n$. Define $G_{n}$ to be the graph having vertex set $V_{n}$, such that two sequences are adjacent in $G_{n}$ if and only if they differ in either 1 or 2 places. For instance, if $n=3$, the sequences $(1,0,0)$, $(... | [
"Solution:\n\nWe will assume that there is a coloring with $n+1$ colors and derive a contradiction. For each string $s$, let $T_{s}$ be the set consisting of all strings that differ from $s$ in at most 1 place. Thus $T_{s}$ has size $n+1$ and all vertices in $T_{s}$ are adjacent. In particular, if there is an $(n+1... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
03jc | Problem:
For positive integers $n$ and $k$, define $F(n, k) = \sum_{r=1}^{n} r^{2k-1}$. Prove that $F(n, 1)$ divides $F(n, k)$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0clq | Determine the sets $S$ of positive integers satisfying the following two conditions:
a) For any positive integers $a, b, c$, if $ab + bc + ca$ is in $S$, then so are $a + b + c$ and $abc$;
b) The set $S$ contains an integer $N \ge 160$ such that $N - 2$ is not divisible by 4. | [
"We will prove that $S$ is the set of all positive integers. The argument hinges on the three facts below:\n\n(1) The set $S$ contains an integer $M \\ge 40$ divisible by 4.\n\n(2) If $4k$ belongs to $S$ for some integer $k \\ge 2$, then so does $4m$ for all positive integers $m < k$.\n\n(3) The set $S$ contains $4... | Romania | 75th NMO Selection Tests | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | S is the set of all positive integers. | |
04z2 | Let $n \ge 2$. Positive integers $a_1, a_2, \dots, a_n$ whose sum is even and which satisfy $a_i \le i$ for every $i = 1, 2, \dots, n$, are given. Prove that it is possible to choose signs in the expression $a_1 \pm a_2 \pm \dots \pm a_n$ in such a way that its value becomes $0$. (Seniors.) | [
"Prove the claim by induction on $n$. If $n=2$, then the only way to choose integers that satisfy the conditions of the problem is $a_1 = 1$ and $a_2 = 1$. In this case, $a_1 - a_2 = 0$.\n\nAssume now that the claim holds whenever $2 \\le n \\le k$ and show that it holds also for $n = k + 1$. Consider two cases.\n\... | Estonia | Estonija 2010 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms",
"Number Theory > Other"
] | null | proof only | null | |
013j | Problem:
$D$ is the midpoint of the side $BC$ of the given triangle $ABC$. $M$ is a point on the side $BC$ such that $\angle BAM = \angle DAC$. $L$ is the second intersection point of the circumcircle of the triangle $CAM$ with the side $AB$. $K$ is the second intersection point of the circumcircle of the triangle $BA... | [
"Solution:\n\nIt is sufficient to prove that $CK : LB = AC : AB$.\n\nThe triangles $ABC$ and $MKC$ are similar because they have common angle $C$ and $\\angle CMK = 180^{\\circ} - \\angle BMK = \\angle KAB$ (the latter equality is due to the observation that $\\angle BMK$ and $\\angle KAB$ are the opposite angles i... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0iz7 | Problem:
Suppose that $p(x)$ is a polynomial and that $p(x) - p'(x) = x^{2} + 2x + 1$. Compute $p(5)$. | [
"Solution:\nObserve that $p(x)$ must be quadratic. Let $p(x) = a x^{2} + b x + c$.\n\nComparing coefficients gives $a = 1$, $b - 2a = 2$, and $c - b = 1$.\n\nSo $b = 4$, $c = 5$, $p(x) = x^{2} + 4x + 5$ and $p(5) = 25 + 20 + 5 = 50$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 50 | |
05hw | Problem:
Soit $ABC$ un triangle équilatéral. Soit $X$ un point de la droite $(BC)$ différent de $B$ et $C$. Soient $Y$ et $Z$ deux points sur les droites $(AB)$ et $(AC)$ de telle sorte que les deux droites $(BZ)$ et $(CY)$ sont parallèles à la droite $(AX)$. La droite $(XY)$ intersecte la droite $(AC)$ en $M$ et la d... | [
"Solution:\n\nOn commence par énoncer deux lemmes qui vont être utiles dans la suite. Si l'utilité du premier lemme sautera aux yeux du lecteur, l'intérêt du deuxième lemme paraîtra sans doute moins clair.\n\nLemme 1. Soit $ABC$ un triangle et $\\omega$ le cercle $A$-exinscrit de ce triangle, on note $J$ le centre ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"... | null | proof only | null | |
08fg | Problem:
Alberto sceglie 2022 numeri interi $a_{1}, a_{2}, \ldots, a_{2022}$ (non necessariamente positivi e non necessariamente distinti) e li dispone su una tabella $2022 \times 2022$ in modo che nella casella $(i, j)$ ci sia il numero $a_{k}$, con $k$ uguale al massimo tra $i$ e $j$, come nella figura seguente (in ... | [
"Solution:\n\nPer ogni $k$ ammissibile, il minimo numero di percorsi da utilizzare per determinare con certezza $a_{k}$ è due.\n\nUn percorso non basta\nConsideriamo un percorso, ed indichiamo con $c_{i}$ il numero di caselle che contengono il valore $a_{i}$ attraversate dal percorso. Osserviamo che $c_{i} \\geq 1$... | Italy | XXXVIII Olimpiade Italiana di Matematica, Cesenatico | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | 2 | |
0l8r | Let be given an integer $n > 1$ and an orthonormal system of coordinates $Oxyz$ in space. Let $T$ be the set of all points $P(x, y, z)$, the coordinates $x, y, z$ of which are integers satisfying the conditions $1 \le x, y, z \le n$. Colour some points in $T$ so that if the point $A(x_0, y_0, z_0)$ had been coloured th... | [] | Vietnam | VIETNAMESE MATHEMATICAL COMPETITION FOR TEAM SELECTION | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | floor((3 n^2 + 1)/4) | |
03i4 | Problem:
Let $n$ be an integer. If the tens digit of $n^2$ is $7$, what is the units digit of $n^2$? | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 6 | |
0i7p | Problem:
Suppose $P(x)$ is a polynomial such that $P(1) = 1$ and
$$
\frac{P(2x)}{P(x+1)} = 8 - \frac{56}{x+7}
$$
for all real $x$ for which both sides are defined. Find $P(-1)$. | [
"Solution:\nCross-multiplying gives $(x+7) P(2x) = 8(x+7) P(x+1) - 56 P(x+1) = 8(x+7) P(x+1) - 56 P(x+1)$. But more simply, rearrange:\n\n$\\frac{P(2x)}{P(x+1)} = 8 - \\frac{56}{x+7}$\n\n$\\Rightarrow \\frac{P(2x)}{P(x+1)} = \\frac{8(x+7) - 56}{x+7} = \\frac{8x + 56 - 56}{x+7} = \\frac{8x}{x+7}$\n\nSo:\n\n$\\frac{P... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | -5/21 | |
03gv | Problem:
A set of balls is given. Each ball is coloured red or blue, and there is at least one of each colour. Each ball weighs either $1$ pound or $2$ pounds, and there is at least one of each weight. Prove that there are $2$ balls having different weights and different colours. | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0f5l | Problem:
A and B are acute angles such that $\sin^2 A + \sin^2 B = \sin (A + B)$. Show that $A + B = \pi / 2$. | [
"Solution:\nGiven $\\sin^2 A + \\sin^2 B = \\sin (A + B)$.\n\nWe know that $\\sin^2 A = 1 - \\cos^2 A$ and $\\sin^2 B = 1 - \\cos^2 B$, but let's try to use the sum-to-product identities.\n\nRecall that:\n$$\n\\sin^2 A + \\sin^2 B = \\frac{1 - \\cos 2A}{2} + \\frac{1 - \\cos 2B}{2} = 1 - \\frac{\\cos 2A + \\cos 2B}... | Soviet Union | 17th ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
03fr | Let $A_0B_0C_0$ be a triangle. For a positive integer $n \ge 1$, we define $A_n$ on the segment $B_{n-1}C_{n-1}$ such that $B_{n-1}A_n : C_{n-1}A_n = 2 : 1$ and $B_n, C_n$ are defined cyclically in a similar manner. Show that there exists an unique point $P$ that lies in the interior of all triangles $A_nB_nC_n$. | [
"(Dragomir Grozev) We have nested compact sets (closed triangles), so they have non empty intersection. We prove that they intersect in only one point. It's enough to prove that the three points $A_n, B_n, C_n$ converge to a common point $P$. Assume it's false. Then there exists some subsequences of $A_n, B_n, C_n$... | Bulgaria | Bulgarian National Olympiad - Regional Round | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09n6 | Let $\omega$ be the circumcircle of $ABC$ where $AB \neq AC$, and let $M$ be the midpoint of side $BC$. Tangent lines drawn at points $B$ and $C$ of circle $\omega$ intersect at point $T$. The circumcircle of triangle $AMT$ intersects line $BC$ again at point $N$. Let $S$ be the midpoint of $NT$. Prove that $SA$ is tan... | [
"Since *BTC* is isosceles, *TM* is altitude.\n$$\n\\angle CMT = \\angle NMT = 90^\\{\\circ\\}.\n$$\nThus, *S* is the circumcenter of triangle *AMT*, making *SA* = *ST* and $\\angle SAT = \\angle STA$. Considering *AT* as the *A*-symmedian of $ABC$, we know $\\angle BAM = \\angle TAC$.\n$$\n\\begin{align*}\n\\angle ... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0dn7 | Problem:
Нека је
$$
S_{n} = \left\{ \binom{n}{n}, \binom{2n}{n}, \binom{3n}{n}, \ldots, \binom{n^{2}}{n} \right\}, \quad \text{за } n \in \mathbb{N}
$$
a) Доказати да постоји бесконачно много сложених природних бројева $n$ таквих да $S_{n}$ није потпун систем остатака по модулу $n$.
б) Доказати да постоји бесконачно... | [
"Solution:\n\nа. Доказаћемо да $n = 2p$ задовољава услове, где је $p > 2$ прост број. Имамо\n$$\n\\binom{2kp}{2p} = k \\prod_{i=1}^{p-1} \\frac{2kp - i}{2p - i} \\cdot (2k-1) \\prod_{i=1}^{p-1} \\frac{2kp - p - i}{p - i} \\equiv k(2k-1) \\pmod{p}\n$$\nКонкретно, одавде је $\\binom{2kp}{2p}$ дељиво са $p$ за $k \\in... | Serbia | Serbian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0hkh | Problem:
Arrange the following number from smallest to largest: $2^{1000}, 3^{750}, 5^{500}$. | [
"Solution:\n\nSince all the numbers are positive, taking the $250$th root of each number will not change their ordering. The resulting numbers are $2^{4} = 16$, $3^{3} = 27$, and $5^{2} = 25$. These have the ordering $2^{4} < 5^{2} < 3^{3}$, so the ordering of the original numbers is $2^{1000} < 5^{500} < 3^{750}$.... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 2^1000 < 5^500 < 3^750 | |
0fuk | Problem:
Finde alle Funktionen $f: \mathbb{N} \rightarrow \mathbb{N}$, sodass $f(m)^2+f(n)$ ein Teiler ist von $(m^2+n)^2$ für alle $m, n \in \mathbb{N}$. | [
"Solution:\nMit $m=n=1$ folgt, dass $f(1)^2+f(1)$ ein Teiler von $4$ ist, folglich ist $f(1)=1$. Für $m=1$ erhalten wir\n$$\nf(n)+1 \\mid (n+1)^2\n$$\nFür $n=1$ erhalten wir\n$$\nf(m)^2+1 \\mid (m^2+1)^2\n$$\nSei $p$ eine Primzahl. Mit $n=p-1$ folgt aus (1), dass $f(p-1)+1$ ein Teiler von $p^2$ ist. Daher gilt $f(p... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analy... | null | proof and answer | f(n) = n for all n in N | |
00b4 | There are 13 weights, all of different colors, and a balance. Ana and Beto know that the weights are of $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $10$, $11$, $12$ and $13$ grams, but only Ana knows which color corresponds to each weight.
An *operation* consists in putting weights on each side of the balance so that... | [
"Let us see that the minimum number of operations that Ana has to make is $2$.\n\nIn the first operation, Ana balance eight weights in one side with three in the other. The weight of eight weights is at least $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36$, and the weight of three weights is at most $11 + 12 + 13 = 36$. Then,... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 2 | |
0j07 | Problem:
16 progamers are playing in another single elimination tournament. Each round, each of the remaining progamers plays against another and the loser is eliminated. Additionally, each time a progamer wins, he will have a ceremony to celebrate. A player's first ceremony is ten seconds long, and afterward each cer... | [
"Solution:\n\nAt the end of the first round, each of the $8$ winners has a $10$ second ceremony. After the second round, the $4$ winners have a $20$ second ceremony. The two remaining players have $30$ second ceremonies after the third round, and the winner has a $40$ second ceremony after the finals. So, all of th... | United States | Harvard-MIT November Tournament | [
"Math Word Problems"
] | null | final answer only | 260 | |
0b0w | Problem:
Find all triples $(a, b, c)$ of positive integers such that
$$
\begin{aligned}
a^{2}+b^{2} &= n \operatorname{lcm}(a, b)+n^{2} \\
b^{2}+c^{2} &= n \operatorname{lcm}(b, c)+n^{2} \\
c^{2}+a^{2} &= n \operatorname{lcm}(c, a)+n^{2}
\end{aligned}
$$
for some positive integer $n$. | [
"Solution:\nWe claim that the only triples that satisfy the system are those of the form $(k, k, k)$. It can be easily checked that all such triples are solutions, where $n=k$.\n\nConversely, suppose that $(a, b, c)$ is a solution. We then need to show that $a=b=c$.\n\nSuppose that there exists some integer $d>1$ s... | Philippines | 21st Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All triples with a = b = c = k for some positive integer k, with n = k. | |
03dd | Let $ABC$ be an acute triangle and $M$ is the midpoint of $AB$. A circle through the vertices $B$ and $C$ intersects the segments $CM$ and $BM$ in points $P$ and $Q$ respectively. Let $K$ be symmetrical point to $P$ regarding $M$. Let circumcircles of $\triangle AKM$ and $\triangle CQM$ meet again in point $X$, and cir... | [
"Obviously $AKBP$ is paralellogram and from quadrilateral $BQPC$ is inscribed follows $\\triangle AKC = \\triangle KPB = \\triangle AQC = \\varphi$. Hence $AKQC$ is inscribed quadrilateral with center $O$. Then $\\triangle AYC = (180^\\circ - \\triangle AYM) + (180^\\circ - \\triangle CYM) = \\triangle AKM + \\tria... | Bulgaria | Bulgaria 2022 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0i7m | Problem:
A point is chosen randomly with uniform distribution in the interior of a circle of radius $1$. What is its expected distance from the center of the circle? | [
"Solution:\nThe probability of the point falling between a distance $r$ and $r + d r$ from the center is the ratio of the area of the corresponding annulus to the area of the whole circle: $$\\frac{\\pi\\left[(r + d r)^2 - r^2\\right]}{\\pi} \\rightarrow \\frac{2 \\pi r d r}{\\pi} = 2 r d r$$ for small values of $d... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | 2/3 | |
02qj | Problem:
Em uma caixa há $10$ bolas idênticas, numeradas de $1$ a $10$. O número de cada bola corresponde a um dos pontos da figura, os quais dividem a circunferência em $10$ partes iguais. Nos itens a seguir, considere que as bolas são retiradas ao acaso, uma a uma e sem reposição.
a) Se f... | [
"Solution:\n\na) 1ª solução: O princípio multiplicativo mostra que o número de maneiras de retirar duas bolas, uma a uma, é $10 \\times 9 = 90$. Dessas retiradas, há dez para as quais o segmento determinado pelos pontos retirados é um diâmetro, a saber, $(1,6), (2,7), (3,8), (4,9), (5,10), (6,1), (7,2), (8,3), (9,4... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | a) 1/9, b) 1/3, c) 1/21 | |
09ll | Let $n, m \ge 3$ be odd numbers. A sequence of $mn-1$ integers is written on a circle in such a way that the sum of any $m$ consecutive integers is a power of $m$. Show that the sequence contains a term which is repeated at least $m+1$ times.
(Bayarmagnai Gombodorj) | [
"Assume that $a_0, a_1, \\dots, a_{k-1}$ is a sequence of integers satisfying the given condition, where we take indices modulo $k$. If each term of the sequence is divisible by $m$ then the sequence $a_0/m, a_1/m, \\dots a_{k-1}/m$ satisfies the given condition. Hence we can assume that $a_0$ is not divisible by $... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
02bj | Problem:
Numa família cada menino tem o mesmo número de irmãos que de irmãs, e cada menina tem o dobro de irmãos que de irmãs. Qual é a composição dessa família? | [
"Solution:\n\n3 meninas e 4 meninos"
] | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 3 girls and 4 boys | |
02ee | $P$ is a fixed point in the plane. $A$, $B$, $C$ are points such that $PA = 3$, $PB = 5$, $PC = 7$ and the area $ABC$ is as large as possible. Show that $P$ must be the orthocenter of $ABC$. | [
"Consider all points $A'$ such that $PA' = 3$. They lie on a circle with center $P$. The area of $A'BC$ is $\\frac{BC}{2}$ times the distance of $A'$ from $BC$. That distance is maximal for $A'P$ perpendicular to $BC$ (because the distance is the distance of $P$ from $BC$ is $PA' \\sin \\theta$, where $\\theta$ is ... | Brazil | X OBM | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geo... | English | proof only | null | |
00hn | Let $\mathbb{Z}$ denote the set of all integers. Find all polynomials $P(x)$ with integer coefficients that satisfy the following property:
For any infinite sequence $a_{1}, a_{2}, \ldots$ of integers in which each integer in $\mathbb{Z}$ appears exactly once, there exist indices $i<j$ and an integer $k$ such that $a_{... | [
"Part 1: All polynomials with $\\operatorname{deg} P=1$ satisfy the given property.\nSuppose $P(x)=c x+d$, and assume without loss of generality that $c>d \\geq 0$. Denote $s_{i}=a_{1}+a_{2}+ \\cdots+a_{i} (\\bmod c)$. It suffices to show that there exist indices $i$ and $j$ such that $j-i \\geq 2$ and $s_{j}-s_{i}... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Algebra > Algebraic Expressions > Polynomials",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Exactly the polynomials of degree one with integer coefficients. | |
0a68 | Problem:
Consider the sequence $a_{1}, a_{2}, a_{3}, \ldots$ defined by $a_{1} = 2024^{2024}$ and for each positive integer $n$,
\[
a_{n + 1} = \left|a_{n} - \sqrt{2}\right|.
\]
Prove that there exists an integer $k$ such that $a_{k + 2} = a_{k}$.
Here $|x|$ denotes the absolute value of $x$. | [
"Solution:\nAs long as $a_{n} \\geqslant \\sqrt{2}$ we have $a_{n + 1} = a_{n} - \\sqrt{2}$ and so the sequence begins with an arithmetic progression with common difference $d = - \\sqrt{2}$. Let $k$ be the first index such that $a_{k} < \\sqrt{2}$. So $a_{k - 1} \\geqslant \\sqrt{2}$. Therefore\n\\[\na_{k} = a_{k ... | New Zealand | NZMO Round Two | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
05k5 | Problem:
On considère un ensemble fini $E$ de garçons et de filles.
Une partie $G$ de garçons de $E$ est dite populaire si chaque fille de $E$ connaît au moins un des garçons de $G$. De même, une partie $F$ de filles est dite populaire si chaque garçon de $E$ connaît au moins une des filles de $F$. On suppose que si ... | [
"Solution:\n\nPremière solution. On dira qu'un ensemble $X$ de garçons est étranger à un groupe $Y$ de filles si aucun garçon de $X$ n'est connu d'une des filles de $Y$. De même, un ensemble $Y$ de filles sera dit étranger à un groupe $X$ de garçons si aucune fille de $Y$ n'est connue d'un des garçons de $X$. Puisq... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof only | null | |
0jki | Problem:
Victoria wants to order at least $550$ donuts from Dunkin' Donuts for the HMMT 2014 November contest. However, donuts only come in multiples of twelve. Assuming every twelve donuts cost $\$ 7.49$, what is the minimum amount Victoria needs to pay, in dollars? (Because HMMT is affiliated with MIT, the purchase ... | [
"Solution:\n\nThe smallest multiple of $12$ larger than $550$ is $552 = 12 \\cdot 46$. So the answer is $46 \\cdot \\$ 7.49$. To make the multiplication easier, we can write this as $46 \\cdot (\\$ 7.5 - \\$ 0.01) = \\$ 345 - \\$ 0.46 = \\$ 344.54$."
] | United States | HMMT November 2014 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 344.54 | |
0852 | Problem:
Determinare tutti i valori di $m, n, p$ tali che $p^{n}+144=m^{2}$, dove $m$ ed $n$ sono interi positivi e $p$ è un numero primo. | [
"Solution:\n\nI valori possibili di $(m, n, p)$ sono $(13,2,5)$, $(20,8,2)$ e $(15,4,3)$.\n\nRiscriviamo l'equazione nella forma $p^{n}=m^{2}-144=(m+12)(m-12)$. Poiché gli unici divisori di $p^{n}$ sono potenze di $p$, l'equazione è equivalente al sistema\n$$\n\\left\\{\n\\begin{array}{l}\nm+12=p^{a} \\\\\nm-12=p^{... | Italy | XXII OLIMPIADE ITALIANA DI MATEMATICA | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (m, n, p) = (13, 2, 5), (20, 8, 2), (15, 4, 3) | |
0kt4 | Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\lfloor \frac{n}{4} \rfloor$, $\lfloor \frac{n}{5} \rfloor$, and $\lfloor \frac{n}{6} \rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$. | [] | United States | AIME II | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 80 | |
0a61 | Problem:
Let $x$, $y$ and $z$ be real numbers such that: $x^{2} = y + 2$, and $y^{2} = z + 2$, and $z^{2} = x + 2$. Prove that $x + y + z$ is an integer. | [
"Solution:\nFirst we exclude $-1$ and $2$:\n$x = 2$ implies $y = 2$ implies $z = 2$ implies $x = 2$\n$x = -1$ implies $y = -1$ implies $z = -1$ implies $x = -1$\nIn both these cases we have $x + y + z$ being an integer. So henceforth we assume none of $x,y,z$ are $2$ nor $-1$. Now let $x,y,z$ be the roots of the fo... | New Zealand | NZMO Round Two | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0fnt | Consider increasing integer sequences with elements from $1, \dots, 10^9$. Such a sequence is Adriatic if its first element equals $1$ and if every element is at least twice the preceding element. A sequence is Tyrrhenian if its final element equals $10^6$ and if every element is strictly greater than the sum of all pr... | [
"Consider the Adriatic sequence $\\langle a_1, \\dots, a_n \\rangle$ starting with $a_1 = 1$. Construct a new sequence\n$$\n\\langle a_2 - 1, a_3 - a_2, \\dots, a_n - a_{n-1}, 10^6 \\rangle\n$$\nfrom it. Note that the new sequence is Tyrrhenian, as\n$$\n(a_2 - 1) + (a_3 - a_2) + \\dots + (a_n - a_{n-1}) = a_{k-1} -... | Spain | Mediterranean Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | Spanish | proof and answer | equal | |
0fj6 | Problem:
Un cristalero dispone de una pieza de vidrio de forma triangular. Usando sus conocimientos de geometría, sabe que podría cortar de ella un círculo de radio $r$. Demuestra que, para cualquier número natural $n$, de la pieza triangular puede obtener $n^{2}$ círculos de radio $\frac{r}{n}$ (suponiendo que se pued... | [
"Solution:\nEl círculo de mayor radio que se puede cortar de un triángulo viene determinado por la circunferencia inscrita. Del enunciado se deduce que el radio de la circunferencia inscrita del triángulo dado es mayor o igual que $r$.\n\nDividimos cada lado del triángulo en $n$ partes iguales. Por cada uno de esos... | Spain | Viernes 19 de enero de 2001 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0k3u | Problem:
Is it possible for the projection of the set of points $(x, y, z)$ with $0 \leq x, y, z \leq 1$ onto some two-dimensional plane to be a simple convex pentagon? | [
"Solution:\n\nIt is not possible. Consider $P$, the projection of $\\left(\\frac{1}{2}, \\frac{1}{2}, \\frac{1}{2}\\right)$ onto the plane. Since for any point $(x, y, z)$ in the cube, $(1-x, 1-y, 1-z)$ is also in the cube, and the midpoint of their projections will be the projection of their midpoint, which is $P$... | United States | HMMT February | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | No | |
06j9 | $ABCD$ is a trapezium with $AB \parallel CD$, $AB = 42$, $BC = 20$ and $DA = 15$. $P$ is a point on $AB$, and a circle with centre $P$ is tangent to both $BC$ and $AD$. Find $PA \times PB$. | [
"Let $r$ be the radius of the circle, and $E$ and $F$ be the points where the circle touches $AD$ and $BC$ respectively. Then $PE = PF = r$ and they are the heights of $\\triangle PAD$ and $\\triangle PBC$ from $P$.\nLet $h$ be the height of the trapezium. By considering the areas of $\\triangle PAD$ and $\\triangl... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 432 | |
0iuj | Problem:
8 students are practicing for a math contest, and they divide into pairs to take a practice test. In how many ways can they be split up? | [
"Solution:\n\nAnswer: $105$\n\nWe create the pairs one at a time. The first person has $7$ possible partners. Set this pair aside. Of the remaining six people, pick a person. He or she has $5$ possible partners. Set this pair aside. Of the remaining four people, pick a person. He or she has $3$ possible partners. S... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 105 | |
059b | Let $n \ge 2$ be a positive integer and let $S = \{1, 2, \dots, n\}$.
For $k = 1, 2, \dots, n-1$, we call two $k$-element subsets of $S$ neighbours, if they have $k-1$ elements in common (i.e. differ by exactly one element). Let $f(n,k)$ be the size of the largest possible collection of $k$-element subsets of $S$, in ... | [
"For any two subsets belonging to such a collection, the sets of the $k-1$ smallest elements must be different (or else they would be neighbours). There are $\\binom{n-1}{k-1}$ ways to choose the $k-1$ smallest elements, since the number $n$ cannot be one of them. Therefore $f(n,k) \\le \\binom{n-1}{k-1}$, as desir... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0j90 | Problem:
Let $P$ and $Q$ be points on line $l$ with $PQ = 12$. Two circles, $\omega$ and $\Omega$, are both tangent to $l$ at $P$ and are externally tangent to each other. A line through $Q$ intersects $\omega$ at $A$ and $B$, with $A$ closer to $Q$ than $B$, such that $AB = 10$. Similarly, another line through $Q$ in... | [
"Solution:\n\nAnswer: $\\frac{8}{9}$\n\nWe first apply the Power of a Point theorem repeatedly. Note that $QA \\cdot QB = QP^{2} = QC \\cdot QD$. Substituting in our known values, we obtain $QA(QA + 10) = 12^{2} = QC(QC + 7)$. Solving these quadratics, we get that $QA = 8$ and $QC = 9$.\n\nWe can see that $\\frac{A... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 8/9 | |
0blz | Problem:
a. Oldd meg az $x^{2}-x+\hat{2}=\hat{0}$ egyenletet, ha $x \in \mathbb{Z}_{7}$.
b. Határozd meg azokat az $n \geq 2$ természetes számokat, amelyekre az $x^{2}-x+\hat{2}=\hat{0}$ egyenletnek egyetlen megoldása van $x \in \mathbb{Z}_{n}$-ben!
Problem:
a. Rezolvaţi ecuaţia $x^{2}-x+\hat{2}=\hat{0}$, $x \in \m... | [
"Solution:\n\na. Cum $4$ şi $7$ sunt coprime, iar $(\\mathbb{Z}_{7},+, \\cdot)$ este corp, ecuaţia dată este echivalentă cu $\\hat{4} x^{2}-\\hat{4} x+\\hat{1}=\\hat{0}$, adică, $(\\hat{2} x-\\hat{1})^{2}=\\hat{0}$, deci $\\hat{2} x=\\hat{1}$, de unde $x=\\hat{4}$.\n\nb. Fie $n \\geq 2$ un număr natural pentru care... | Romania | Olimpiada Naţională de Matematică | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Abstract Algebra > Ring Theory"
] | null | proof and answer | Part a: x = 4 in integers modulo 7. Part b: n = 7. | |
05pj | Problem:
1) Pierre répartit les entiers $1,2, \ldots, 2012$ en deux groupes disjoints dont les sommes respectives des éléments sont égales.
Sans même regarder la répartition choisie par Pierre, Clara affirme alors que l'on peut éliminer deux nombres de chaque groupe de sorte que, dans chaque groupe, les sommes respect... | [
"Solution:\n\nOn colorie les éléments du groupe $G_{1}$ en rouge, et ceux de $G_{2}$ en bleu. Sans perte de généralité, on peut supposer que $1$ est rouge. Lorsque deux entiers consécutifs ne sont pas dans le même groupe, on dit que l'on a une alternance.\n\n- Si l'on a au moins quatre alternances, on peut trouver ... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0baq | Prove that the set $S = \{\lfloor n\pi \rfloor : n = 0, 1, 2, 3, \dots \}$ contains arithmetic progressions of any finite length, but no infinite arithmetic progressions. | [
"If $x$ is a real number, let $\\{x\\} = x - \\lfloor x \\rfloor$ denote the fractional part of $x$. Given an integer number $m \\ge 3$, there exists a positive integer number $n$ such that $\\{n\\pi\\} < 1/m$, for the set $\\{k\\pi : k = 0, 1, 2, 3, \\dots\\}$ is dense in the closed unit interval $[0, 1]$. Consequ... | Romania | 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Other"
] | null | proof only | null | |
00i5 | Let $P(x)$ be a non-constant polynomial with integer coefficients such that $P(0) \neq 0$. Let $a_{1}, a_{2}, a_{3}, \ldots$ be an infinite sequence of integers such that $P(i-j)$ divides $a_{i}-a_{j}$ for all distinct positive integers $i, j$. Prove that the sequence $a_{1}, a_{2}, a_{3}, \ldots$ must be constant, tha... | [
"Let $a_{0}=P(0) \\neq 0$ be the independent coefficient, i.e., the constant term of $P(x)$. Then there are infinitely many primes $p$ such that $p$ divides $P(k)$ but $p$ does not divide $k$. In fact, since $P(k)-a_{0}$ is a multiple of $k$, $\\operatorname{gcd}(P(k), k)=\\operatorname{gcd}\\left(k, a_{0}\\right) ... | Asia Pacific Mathematics Olympiad (APMO) | APMO 2025 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
0lgl | Problem:
Let $n \geq 2$ be an integer. Elwyn is given an $n \times n$ table filled with real numbers (each cell of the table contains exactly one number). We define a rook set as a set of $n$ cells of the table situated in $n$ distinct rows as well as in $n$ distinct columns. Assume that, for every rook set, the sum o... | [
"Solution:\n\nWe start with the following known consequence of Hall's lemma.\n\nLemma. Let $G=(U \\sqcup V, E)$ be a bipartite multigraph with parts $U$ and $V$, both of size $n$. Assume that each vertex has degree $k$; then the edges can be partitioned into $k$ perfect matchings.\n\nProof. Induction on $k$; the ba... | Zhautykov Olympiad | Zhautykov Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Vectors",
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
03gm | Problem:
Show that, for all integers $n$, $n^{2} + 2n + 12$ is not a multiple of $121$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
07sp | Determine the last (rightmost) three decimal digits of $n$ where:
$$
n = 1 \times 3 \times 5 \times 7 \times \dots \times 2019.
$$ | [
"Recall that the Chinese Remainder Theorem (CRT) states that, for any given integers $r$, $s$ and any two positive co-prime integers $a$, $b$ there exists exactly one integer $x$ that satisfies $0 \\le x < ab$, $x \\equiv r \\pmod a$ and $x \\equiv s \\pmod b$. We can apply CRT here with $a = 125$ and $b = 8$, beca... | Ireland | IRL_ABooklet_2020 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 875 | |
07vp | Find the largest solution of the equation $\{x\}^2 = \{x^2\}$ which is smaller than $2023$. Here $\{x\}$ is the fractional part of the number $x$, e.g. $\{22/7\} = 1/7$. | [
"By definition, $n = x - \\{x\\}$ is an integer. Then $x = n + \\{x\\}$ and $x^2 = n^2 + 2n\\{x\\} + \\{x\\}^2$. Hence $\\{x^2\\} = \\{2n\\{x\\} + \\{x\\}^2\\}$. If $\\{x\\}^2 = \\{x^2\\}$ then $2n\\{x\\}$ is an integer.\n\nAs $x < 2023$, the largest possible $n$ is $n = 2022$. We then look for the largest $\\{x\\}... | Ireland | IRL_ABooklet_2023 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 2022 + 4043/4044 | |
0ips | Problem:
A cat is going up a stairwell with ten stairs. However, instead of walking up the stairs one at a time, the cat jumps, going either two or three stairs up at each step (though if necessary, it will just walk the last step). How many different ways can the cat go from the bottom to the top? | [
"Solution:\nThe number of ways for the cat to get to the $i$th step is the number of ways for the cat to get to step $i-2$ plus the number of ways to get to step $i-3$, because for each way to get to step $i$, we can undo the last move the cat made to go back to one of these two steps. The cat can get to step $1$ i... | United States | 1st Annual Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 12 | |
0g0x | Problem:
Auf einem Kreis liegen $2n$ verschiedene Punkte. Die Zahlen $1$ bis $2n$ werden zufällig auf diese Punkte verteilt. Jeder Punkt wird mit genau einem anderen Punkt verbunden, sodass sich keine der entstehenden Verbindungsstrecken schneiden. Verbindet eine Strecke die Zahlen $a$ und $b$, so weisen wir der Strec... | [
"Solution:\n\nMan sieht, dass $$(n + (n+1) + \\ldots + 2n) - (1 + 2 + \\ldots + n) = n^{2}$$ ist. Also falls wir es schaffen, jede Zahl aus der zweiten Klammer mit einer aus der ersten Klammer (natürlich ohne Überkreuzungen) zu verbinden, sind wir fertig. Es gibt immer eine Zahl aus der ersten Klammer, die neben ei... | Switzerland | SMO - Finalrunde | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
04nw | Find the locus of the centres of all circles that are externally tangent to the circle that satisfies $x^2 + y^2 - 4y + 3 = 0$, and that are also tangent to the $x$-axis.
*(Anastazija Pažanin)* | [
"Let the given circle be $x^2 + y^2 - 4y + 3 = 0$. Rewrite it in standard form:\n\n$x^2 + (y^2 - 4y) + 3 = 0$\n$x^2 + (y - 2)^2 - 4 + 3 = 0$\n$x^2 + (y - 2)^2 - 1 = 0$\n$x^2 + (y - 2)^2 = 1$\n\nSo, the circle has centre $(0, 2)$ and radius $1$.\n\nLet the centre of the required circle be $(h, k)$ and its radius be ... | Croatia | Croatia_2018 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | x^2 - 6y + 3 = 0 | |
0av3 | Problem:
Let $n = 2^{23} 3^{17}$. How many factors of $n^{2}$ are less than $n$, but do not divide $n$? | [
"Solution:\nLet $n = 2^{23} 3^{17}$.\n\nThe number of factors of $n$ is $(23+1)(17+1) = 24 \\times 18 = 432$.\n\nThe number of factors of $n^2$ is $(2 \\times 23 + 1)(2 \\times 17 + 1) = 47 \\times 35 = 1645$.\n\nThe factors of $n^2$ come in pairs $(d, n^2/d)$, and exactly one of each pair is less than $n$ (unless ... | Philippines | 19th Philippine Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 391 | |
086w | Problem:
Nell'isola Chenonc'è ci sono 2009 abitanti, divisi in tre clan: i furfanti che mentono sempre, i cavalieri che non mentono mai, i paggi che mentono un giorno sì e uno no, in modo indipendente l'uno dall'altro. Un giorno chiedo a ciascuno degli abitanti quanti furfanti sono sull'isola. Il primo dice: "c'è alme... | [
"Solution:\n\nLa risposta è (D). Sia $f$ il numero di furfanti e $p$ il numero di paggi. In ognuna delle due liste risulteranno esattamente $f$ affermazioni vere (le prime $f$) ed esattamente $2009-f$ affermazioni false (le rimanenti). Quindi tra le due liste ci sono $4018-2f$ frasi false. Ora ogni furfante ha forn... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | D | |
03cb | In a mathematical olympiad students received marks for any of the four areas: algebra, geometry, number theory and combinatorics. Any two of the students have distinct marks for all four areas. A group of students is called *nice* if all students in the group can be ordered in increasing order simultaneously of at leas... | [
"Answer: 730.\n\n**Lemma.** A sequence $A = a_1, a_2, \\dots, a_k$ consisting of distinct numbers does not possess 10 term increasing subsequence if and only if its terms can be colored in 9 colors such that the members of one and the same color from a decreasing sequence.\n\n*Proof.* First, assume that such a colo... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 730 | |
074q | Problem:
Find all non-zero real numbers $x, y, z$ which satisfy the system of equations:
$$
\begin{aligned}
\left(x^{2}+x y+y^{2}\right)\left(y^{2}+y z+z^{2}\right)\left(z^{2}+z x+x^{2}\right) & =x y z \\
\left(x^{4}+x^{2} y^{2}+y^{4}\right)\left(y^{4}+y^{2} z^{2}+z^{4}\right)\left(z^{4}+z^{2} x^{2}+x^{4}\right) & =x^{... | [
"Solution:\nSince $x y z \\neq 0$, we can divide the second relation by the first. Observe that\n$$\nx^{4}+x^{2} y^{2}+y^{4}=\\left(x^{2}+x y+y^{2}\\right)\\left(x^{2}-x y+y^{2}\\right)\n$$\nholds for any $x, y$. Thus we get\n$$\n\\left(x^{2}-x y+y^{2}\\right)\\left(y^{2}-y z+z^{2}\\right)\\left(z^{2}-z x+x^{2}\\ri... | India | INMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x = y = z = 1/3 | |
0bqe | Four positive integers $a, b, c, d$ are not divisible by $5$ and the sum of their squares is divisible by $5$. Prove that
$$
N = (a^2 + b^2)(a^2 + c^2)(a^2 + d^2)(b^2 + c^2)(b^2 + d^2)(c^2 + d^2)
$$
is divisible by $625$. | [] | Romania | 67th NMO Shortlisted Problems | [
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0f31 | Problem:
$m$ and $n$ are relatively prime positive integers. The interval $[0, 1]$ is divided into $m + n$ equal subintervals. Show that each part except those at each end contains just one of the numbers $1/m$, $2/m$, $3/m$, $\ldots$, $(m-1)/m$, $1/n$, $2/n$, $\ldots$, $(n-1)/n$. | [] | Soviet Union | ASU | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Other"
] | null | proof only | null | |
0b4i | Problem:
Let $p = p_{1} p_{2} \ldots p_{6}$ be a permutation of the integers from $1$ to $6$. For any such permutation $p$, we count how many integers there are which have nothing bigger on its left. We let $f(p)$ be the number of these integers for the permutation $p$. For example, $f(612345) = 1$ because only $6$ ha... | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 18 | |
0dx5 | Problem:
Dan je trikotnik $ABC$ z oglišči $A(-1,2)$, $B(-2,-3)$ in $C(2,-1)$. Iz točke $A$ konstruiramo pravokotnico na stranico $BC$. Pravokotnica seka $BC$ v točki $E$. Natančno izračunaj dolžino daljice $AE$. | [
"Solution:\n\nUgotovimo, da je v danem trikotniku $v_{a} = d(A, E)$. Izračunamo ploščino trikotnika $ABC$: $S = 9$.\n\nDolžina stranice $a$ je enaka $d(B, C) = 2 \\sqrt{5}$.\n\nNato izrazimo $v_{a}$ iz ploščine:\n$$\nv_{a} = \\frac{2 \\cdot S}{a} = \\frac{9 \\sqrt{5}}{5}\n$$\n\nDolžina $v_{a} = \\frac{9 \\sqrt{5}}{... | Slovenia | 6. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 9√5/5 | |
0128 | Problem:
Solve the system of equations
$$
\left\{\begin{array}{l}
a^{3}+3 a b^{2}+3 a c^{2}-6 a b c=1 \\
b^{3}+3 b a^{2}+3 b c^{2}-6 a b c=1 \\
c^{3}+3 c a^{2}+3 c b^{2}-6 a b c=1
\end{array}\right.
$$
in real numbers. | [
"Solution:\nDenoting the left hand sides of the given equations as $A$, $B$ and $C$, the following equalities can easily be seen to hold:\n$$\n\\begin{aligned}\n-A+B+C & =(-a+b+c)^{3} \\\\\nA-B+C & =(a-b+c)^{3} \\\\\nA+B-C & =(a+b-c)^{3} .\n\\end{aligned}\n$$\nHence, the system of equations given in the problem is ... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | (1, 1, 1) | |
0d6o | Let $(O_{1})$, $(O_{2})$ be given two circles intersecting at $A$ and $B$. The tangent lines of $(O_{1})$ at $A$, $B$ intersect at $O$. Let $I$ be a point on the circle $(O_{1})$ but outside the circle $(O_{2})$. The lines $IA$, $IB$ intersect circle $(O_{2})$ at $C$, $D$. Denote by $M$ the midpoint of $CD$. Prove that... | [
"Denote $N$ as the midpoint of $AB$. Because $A$, $B$, $C$, $D$ belong to the same circle, then we have\n$$\n\\triangle IAB \\sim \\triangle IDC.\n$$\nSince $M$ is the midpoint of $CD$ and $N$ is the midpoint of $AB$ then $IM$, $IN$ are isogonal conjugate with respect to the angle $\\angle CID$.\n\nIn the other han... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | English | proof only | null | |
09u0 | At a conference, there were participants from four countries: the Netherlands, Belgium, Germany, and France. There were three times as many participants from the Netherlands as there were Belgians, and three times as many Germans as French. Five of the participants counted the total number of participants (including th... | [
"C) $368$"
] | Netherlands | Junior Mathematical Olympiad, September 2019 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | C | |
014k | Problem:
Does there exist a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers such that the sum of every $n$ consecutive elements is divisible by $n^{2}$ for every positive integer $n$? | [
"Solution:\n\nWe will show that whenever we have positive integers $a_{1}, \\ldots, a_{k}$ such that $n^{2} \\mid a_{i+1}+\\cdots+a_{i+n}$ for every $n \\leq k$ and $i \\leq k-n$, then it is possible to choose $a_{k+1}$ such that $n^{2} \\mid a_{i+1}+\\cdots+a_{i+n}$ for every $n \\leq k+1$ and $i \\leq k+1-n$. Thi... | Baltic Way | Baltic Way | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0fi8 | Problem:
Se consideran las parábolas $y = x^{2} + p x + q$ que cortan a los ejes de coordenadas en tres puntos diferentes, por los que se traza una circunferencia. Demostrar que todas las circunferencias trazadas al variar $p$ y $q$ en $\mathbb{R}$ pasan por un punto fijo, que se determinará. | [
"Solution:\n\nSean $\\alpha$ y $\\beta$ las raíces. Los tres puntos que definen la circunferencia son $A(\\alpha, 0)$, $B(\\beta, 0)$ y $C(0, q)$, cumpliéndose $\\alpha + \\beta - p = 0$ y $\\alpha \\beta = q$.\n\nLa mediatriz de $AB$ es la recta paralela al eje $OY$ de ecuación $x = -\\frac{p}{2}$.\n\nHallando la ... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Al... | null | proof and answer | (0, 1) | |
05kl | Problem:
Soient $D$ et $E$ des points appartenant respectivement aux intérieurs des côtés $[A B]$ et $[A C]$ d'un triangle $A B C$, tels que $D B = B C = C E$. Soient $F$ le point d'intersection des droites $C D$ et $B E$, $I$ le centre du cercle inscrit au triangle $A B C$, $H$ l'orthocentre du triangle $D E F$ et $M... | [
"Solution:\n\nPuisque $D B = B C = C E$, nous avons $(B I) \\perp (C D)$ et $(C I) \\perp (B E)$. Ainsi, $I$ est l'orthocentre du triangle $B F C$. Soit $K$ le point d'intersection des droites $(B I)$ et $(C D)$, et $L$ le point d'intersection des droites $(C I)$ et $(B E)$. Les triangles $I L B$ et $I K C$ sont se... | France | EGMO 2014, 1ère journée | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformat... | null | proof only | null | |
0hxv | Problem:
Find the area of the region bounded by the graphs of $y = x^{2}$, $y = x$, and $x = 2$. | [
"Solution:\nThere are two regions to consider. First, there is the region bounded by $y = x^{2}$ and $y = x$, in the interval $[0, 1]$. In this interval, the values of $y = x$ are greater than the values of $y = x^{2}$, thus the area is calculated by\n$$\n\\int_{0}^{1} (x - x^{2}) \\, dx\n$$\nSecond, there is the r... | United States | HMMT | [
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | final answer only | 1 | |
0imj | Problem:
$ABCD$ is a cyclic quadrilateral in which $AB = 4$, $BC = 3$, $CD = 2$, and $AD = 5$. Diagonals $AC$ and $BD$ intersect at $X$. A circle $\omega$ passes through $A$ and is tangent to $BD$ at $X$. $\omega$ intersects $AB$ and $AD$ at $Y$ and $Z$ respectively. Compute $YZ / BD$.
 | [
"Solution:\n\nAnswer: $\\frac{115}{143}$. Denote the lengths $AB, BC, CD$, and $DA$ by $a, b, c$, and $d$ respectively. Because $ABCD$ is cyclic, $\\triangle ABX \\sim \\triangle DCX$ and $\\triangle ADX \\sim \\triangle BCX$. It follows that $\\frac{AX}{DX} = \\frac{BX}{CX} = \\frac{a}{c}$ and $\\frac{AX}{BX} = \\... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 115/143 | |
08o4 | Problem:
Find all ordered triples $(x, y, z)$ of positive integers satisfying the equation
$$
\frac{1}{x^{2}}+\frac{y}{x z}+\frac{1}{z^{2}}=\frac{1}{2013}
$$ | [
"Solution:\nWe have $x^{2} z^{2}=2013\\left(x^{2}+x y z+z^{2}\\right)$. Let $d=\\operatorname{gcd}(x, z)$ and $x=d a, z=d b$. Then $a^{2} b^{2} d^{2}=2013\\left(a^{2}+a b y+b^{2}\\right)$.\nAs $\\operatorname{gcd}(a, b)=1$, we also have $\\operatorname{gcd}\\left(a^{2}, a^{2}+a b y+b^{2}\\right)=1$ and $\\operatorn... | JBMO | 17th Junior Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2013 n, 2013 n^2 - 2, 2013 n) for positive integer n | |
0hag | Andriy and Olesya write a natural number each on a chalkboard. It turns out, that number, written by Olesya, has sum of digits $2018$ and has precisely $1$ digit less than Andriy's number. It is also known, that difference of numbers, written by him, equals to one-digit number. What can be the number, written by Andriy... | [
"It is not hard to see, that Andriy's number can only be $\\overline{100...0a}$, and Olesya's – only: $\\overline{99...9b}$. Otherwise, the difference will not be a one-digit number. Really, if not all Olesya's digits, except for the last, are $9$, then after adding a one-digit number, the number of digits will not... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 10^225 or 10^225 + 1 | |
07so | Let $\mathbb{N}$ denote the strictly positive integers. A function $f : \mathbb{N} \to \mathbb{N}$ satisfies the following for all $n \in \mathbb{N}$:
$$
\begin{aligned}
f(1) &= 1 \\
f(f(n)) &= n \\
f(2n) &= 2f(n) + 1.
\end{aligned}
$$
Find the value of $f(2020)$. | [
"From $f(f(n)) = n$ we obtain for all integers $k, n > 0$ that $n = f(k)$ if and only if $f(n) = k$.\n\nWe first show that $f(2n+1) = 2f(n)$ for all $n > 0$. To see this, let $k = f(n)$. Then $n = f(k)$ and $f(2k) = 2f(k)+1 = 2n+1$ and so $2f(n) = 2k = f(2n+1)$.\n\nNext we show that $f(2^k n) = 2^k f(n) + 2^k - 1$ ... | Ireland | IRL_ABooklet_2020 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | 1051 | |
0gwt | Find all the functions $f: \mathbb{R} \to \mathbb{R}$ so that
$$ f(f(y) + 2 + x) + f(f(y) - x) = y f(y)(x + 1) $$
is true for any real numbers $x$ and $y$. | [
"Let's make substitution $x = -2 - t$ where $t$ is an arbitrary real number. We find that $f(f(y) - t) + f(f(y) + 2 + t) = -y f(y)(t + 1)$. We can see that the left side of the equation has not changed while a minus sign appeared on its right side. Thus for all $y$ and $t$ the following equation should be true: $y ... | Ukraine | Ukrajina 2008 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0 for all real x | |
0dyw | Problem:
Ploščina trikotnika $ABC$ z oglišči $A(-1,-6)$, $B(1,0)$, $C(3,-3)$ je $9$. Izračunaj dolžino težiščnice na stranico $b$ in razdaljo med nožiščem višine na stranico $b$ in razpoloviščem stranice $b$. | [
"Solution:\n\nIzračunamo koordinati razpolovišča stranice $AC$, ki je $R\\left(\\frac{-1+3}{2}, \\frac{-6+(-3)}{2}\\right)$. Izračunamo $t_{b} = d(B, R) = \\frac{9}{2}$. Izračunamo dolžino stranice $b = d(A, C) = 5$. Uporabimo obrazec $S = \\frac{b \\cdot v_{b}}{2}$ in dobimo $v_{b} = \\frac{18}{5}$. Upoštevamo zve... | Slovenia | Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 9/2, 27/10 | |
0ekx | Problem:
a) Izračunaj presečišči in velikost kota med krivuljama $y = -x^{-2}$ in $y = -\frac{x^{2}}{2} + \frac{1}{2}$.
b) Dana je funkcija $f(x) = \frac{x}{\ln x}$. Določi definicijsko območje funkcije $f$ in intervale, na katerih funkcija $f$ narašča in pada. | [
"Solution:\na)\nEnačimo krivulji: $-x^{-2} = -\\frac{x^{2}}{2} + \\frac{1}{2}$, nato množimo s skupnim imenovalcem in uredimo do oblike $x^{4} - x^{2} - 2 = 0$. Zapišemo v obliki produkta $(x^{2} - 2)(x^{2} + 1) = 0$, nato razstavimo še prvi oklepaj na $(x - \\sqrt{2})(x + \\sqrt{2})(x^{2} + 1) = 0$. Dobimo rešitvi... | Slovenia | 22. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a) Intersections: (√2, −1/2) and (−√2, −1/2). The angle between the curves at each intersection is 90 degrees.
b) Domain: (0, 1) ∪ (1, ∞). Increasing on (e, ∞). Decreasing on (0, 1) and (1, e). | |
02p4 | If $u_1, \dots, u_k \in \mathbb{R}^3$, denote by $C(u_1, \dots, u_k)$ the cone generated by $u_1, \dots, u_k$:
$$
C(u_1, \dots, u_k) = \{a_1 u_1 + \dots + a_k u_k; a_1, \dots, a_k \in [0, +\infty)\}.
$$
Let $v_1, v_2, v_3, v_4$ points randomly and independently chosen from the unit sphere $x^2 + y^2 + z^2 = 1$.
a. Wha... | [
"a. The probability that the cone of the four vectors is proper is $\\frac{7}{8}$ so the probability that the cone is all of $\\mathbb{R}^3$ is $\\frac{1}{8}$.\nConstruct a vector $u_{12}$ that is normal to the plane spanned by $v_1$ and $v_2$ oriented so that $v_3 \\cdot u_{12} > 0$. Then the half-space $\\{w \\mi... | Brazil | Brazilian Math Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Linear Algebra > Vectors"
] | English | proof and answer | a: 1/8; b: 1/2 | |
0dbs | Let $ABCDEF$ be a convex hexagon satisfying $AC = DF$, $CE = FB$ and $EA = BD$. Prove that the lines connecting the midpoints of opposite sides of the hexagon $ABCDEF$ intersect in one point. | [
"Let $M, N, P, Q, R, S$ be the midpoints of sides $AB$, $BC$, $CD$, $DE$, $EF$, $FA$, respectively, and $X, Y, Z$ be the midpoints of $AD$, $BE$, $CF$.\n\n\n\nSince $AE = BD$ and the midsegments in some triangles, we get\n$$\nXQ = YM = \\frac{1}{2} \\cdot AE = \\frac{1}{2} \\cdot BD = XM = ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0e7f | Problem:
Naj bo $a < 0$. Poenostavi izraz
$$
\frac{\sqrt{27 a^{2}} - \sqrt{12 a^{2}} + a \sqrt{108} + 5 \sqrt{3}}{5 \sqrt{3} a - 5 \sqrt{3}}
$$
Izračunaj vrednost izraza za $a = -2^{-1}$. | [] | Slovenia | Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | (a+1)/(a-1); at a = -2^{-1}: -1/3 | |
0bby | Given a continuous function $f: \mathbb{R} \to \mathbb{R}$, denote, for each interval $[a, b]$, $m_{ab} = \min_{x \in [a, b]} f(x)$ and $M_{ab} = \max_{x \in [a, b]} f(x)$. Find all the continuous functions $f: \mathbb{R} \to \mathbb{R}$ such that, for every $a < b$, $f(\frac{a+b}{2}) = \frac{m_{ab} + M_{ab}}{2}$. | [
"Let $f: \\mathbb{R} \\to \\mathbb{R}$ be continuous and satisfy, for every $a < b$,\n$$\nf\\left(\\frac{a+b}{2}\\right) = \\frac{m_{ab} + M_{ab}}{2}\n$$\nwhere $m_{ab} = \\min_{x \\in [a, b]} f(x)$ and $M_{ab} = \\max_{x \\in [a, b]} f(x)$.\n\nLet us fix $a < b$ and consider the function $f$ on $[a, b]$.\n\nSince ... | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All affine functions f(x) = c x + d with real constants c and d. |
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