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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
01kf | There is a heap of 330 stones. Nick and Mike play the following game. They, in turn (Nick is the first), remove the stones from the heap. Per move it is allowed to remove exactly 1 or exactly $m$ or exactly $n$ stones. The player wins if he removes the last stone. Before the start Nick fixes the value of $n$ ($1 < n < ... | [
"We separate the numbers from 2 to 9 into the pairs (2, 7), (3, 8), (5, 6), (4, 9). In order to win Mike can use the following rule: if Nick fixes one of the numbers from any of the pairs, then Mike fixes the other number from the same pair.\n\nNow we will solve the problem moving backward. We write all numbers fro... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Mike | |
014g | Problem:
Let the medians of the triangle $A B C$ intersect at the point $M$. A line $t$ through $M$ intersects the circumcircle of $A B C$ at $X$ and $Y$ so that $A$ and $C$ lie on the same side of $t$. Prove that $B X \cdot B Y = A X \cdot A Y + C X \cdot C Y$. | [
"Solution:\nLet us start with a lemma: If the diagonals of an inscribed quadrilateral $A B C D$ intersect at $O$, then\n$$\n\\frac{A B \\cdot B C}{A D \\cdot D C} = \\frac{B O}{O D}.\n$$\nIndeed,\n$$\n\\frac{A B \\cdot B C}{A D \\cdot D C} = \\frac{\\frac{1}{2} A B \\cdot B C \\cdot \\sin B}{\\frac{1}{2} A D \\cdot... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
000d | Se tienen 2 cuadriláteros convexos iguales de papel: $ABCD$ y $A'B'C'D'$ ($AB = A'B'$, $BC = B'C'$, $CD = C'D'$, $DA = D'A'$). Se corta el cuadrilátero $ABCD$ por la diagonal $AC$ y se corta el cuadrilátero $A'B'C'D'$ por la diagonal $B'D'$, obteniendo así cuatro trozos de papel.
a) Indica un procedimiento, que no dep... | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Translation"
] | español | proof only | null | |
0473 | Fix an irrational number $\alpha > 1$ and a positive integer $L$ such that $L > \frac{\alpha^2}{\alpha - 1}$.
Given an integer $x_1 > L$, define a sequence $\{x_n\}$ as follows: for every integer $n \ge 1$,
$$
x_{n+1} = \begin{cases} \lfloor \alpha x_n \rfloor, & \text{if } x_n \le L, \\ \lfloor \frac{x_n}{\alpha} \rfl... | [
"*Proof*. First, since $\\alpha$ is an irrational number, the integer part operation in the definition of $x_{n+1}$ always makes the corresponding number strictly smaller. For any integer $u$ satisfying $\\frac{1}{\\alpha-1} < u \\le L$, we have $\\lfloor \\alpha u \\rfloor > \\alpha u - 1 > u$ and $\\lfloor \\alph... | China | 2024 CMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0l3u | Amy, Bomani, Charlie, and Daria work in a chocolate factory. On Monday Amy, Bomani, and Charlie started working at 1:00 PM and were able to pack $4$, $3$, and $3$ packages, respectively, every $3$ minutes. At some later time, Daria joined the group, and Daria was able to pack $5$ packages every $4$ minutes. Together, t... | [
"Every $3$ minutes, Amy, Bomani, and Charlie together packed $10$ packages. From $1{:}00$ PM to $2{:}45$ PM, a span of $60 + 45 = 105$ minutes, these three packers packed $\\frac{105}{3} \\cdot 10 = 350$ packages. This means that Daria must have packed $450 - 350 = 100$ packages. The time needed for Daria to pack $... | United States | AMC 10 A | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | A | |
0ciw | Solve in $\mathbb{R}$ the equation $\cos[x] + \arccos{x} = x$. | [] | Romania | 75th NMO | [
"Precalculus > Trigonometric functions",
"Calculus > Differential Calculus > Applications"
] | English | proof and answer | 0.9396 | |
0imv | Problem:
Five people are crowding into a booth against a wall at a noisy restaurant. If at most three can fit on one side, how many seating arrangements accommodate them all? | [
"Solution:\n\nAnswer: $240$. Three people will sit on one side and two sit on the other, giving a factor of two. Then there are $5!$ ways to permute the people."
] | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 240 | |
0d5l | أثبت أنّه لا يمكن كتابة كثيرة الحدود $P(x)=\left(x^{2}-12 x+11\right)^{4}+23$ كاصل ضرب ثلاث كثيرات حدود غير ثابتة، معاملاتُها صحيحة. | [] | Saudi Arabia | SAMC 2015 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English, Arabic | proof only | null | |
0fao | Problem:
Half the cells of a $2m \times n$ board are colored black and the other half are colored white. The cells at the opposite ends of the main diagonal are different colors. The center of each black cell is connected to the center of every other black cell by a straight line segment, and similarly for the white c... | [
"Solution:\n\nSuppose we have an odd number of arbitrary points $A_1, A_2, \\ldots, A_{2k + 1}$ then we claim that if we take the vector $A_iA_j$ for $i < j$ and $j - i$ odd and the vector $A_jA_i$ for $i < j$ and $j - i$ even, then we get the sum of the vectors zero. We prove the claim by induction. It is true for... | Soviet Union | 1st CIS | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Transformations > Rotation",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
0ba5 | For every positive integer $n$ let $\tau(n)$ denote the number of its positive factors. Determine all $n \in \mathbb{N}$ that satisfy the equality $\tau(n) = \frac{n}{3}$. | [
"If $n \\in \\mathbb{N}$ satisfies the condition $\\tau(n) = \\frac{n}{3}$, then $3 \\mid n$. Put $n = 3k$, $k \\in \\mathbb{N}$. If $k$ is even then $\\frac{k}{2} = \\frac{n}{6}$ is a factor of $n$. Even if all the positive numbers smaller than $\\frac{n}{6}$ are factors of $n$ and the numbers $\\frac{n}{5}, \\fra... | Romania | 62nd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 9, 18, 24 | |
0ep8 | The largest number below is
(A) $2^3$ (B) $\frac{15}{2}$ (C) $\sqrt{81}$ (D) $4^2$ (E) $\frac{31}{4}$ | [
"Answer D.\nAfter simplification, the numbers are (A) $8$, (B) $7.5$, (C) $9$, (D) $16$, (E) $7.75$."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | MCQ | D | |
0gen | 一個城市是平面上的一個點。假設平面上有 $n \ge 2$ 個城市。假設對於每個城市 $X$，都存在另一個城市 $N(X)$，使得 $X$ 到 $N(X)$ 的距離嚴格小於 $X$ 到任何其他城市的距離。政府在所有的城市 $X$ 與其 $N(X)$ 之間建有道路，除此之外城市之間沒有其他道路。已知我們可以從任何一個城市，經過一系列的道路，抵達任何一個其他城市。我們稱一個城市 $Y$ 是一個近郊，若且唯若存在城市 $X$ 使得 $Y = N(X)$。試證明至少有 $(n-2)/4$ 個近郊。
A city is a point on the plane. Suppose there are $n \ge 2$ cities. Sup... | [
"讓我們以城市頂點，$(X, N(X))$ 為邊建立有向圖 $G$。基於 $G$ 連通且共有 $n$ 個邊，此圖恰有一個環。這表示我們有至多一對城市 $(A, B)$ 滿足 $A = N(B)$ 且 $B = N(A)$。\n\n讓我們考慮以下關鍵引理：\n\n**引理：** 如果 $B = N(A)$，則存在至多 4 個異於 $B$ 的城市滿足 $A = N(X)$。\n\n**證明：** 假設 $X_1, \\cdots, X_t$ 為所有異於 $B$ 且滿足 $A = N(X_i)$ 的城市。由定義知 $\\overline{X_iA} < \\overline{X_iB}$，意味著 $X_i$ 與 $A$ 必須在 $... | Taiwan | 2021 數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
0f44 | Problem:
An integer is initially written at each vertex of a cube. A move is to add $1$ to the numbers at two vertices connected by an edge. Is it possible to equalise the numbers by a series of moves in the following cases?
(1) The initial numbers are $0$, except for one vertex which is $1$.
(2) The initial numbers... | [] | Soviet Union | 15th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | (1) Not possible. (2) Not possible. (3) Possible. | |
08wm | Let $k$ be a positive integer. Players $A$ and $B$ play a game according to the following rule: Initially, a chess piece is placed at the origin $(0, 0)$ of the $xy$-plane. The player $A$ will start the game followed by $B$ and repeat, each choosing a strategy among those specified by the following rules:
* Possible st... | [
"We will show that for any positive integer $k$ $A$ has strategies to win the game no matter how $B$ chooses his strategies.\nIn the sequel, we restrict the possibilities for $A$ to mark only those lattice points in the set $J = \\{(x, y) : x + y = 2^{k+1}k\\}$. Also, we allow $A$ not to choose any lattice point to... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | All positive integers k | |
0k8n | Problem:
Can the sum of three fourth powers end with the four digits 2019? (A fourth power is an integer of the form $n^{4}$, where $n$ is an integer.) | [
"Solution:\n\nNo, in fact it cannot even end in the digit 9. The possible last digits of a fourth power are $0, 1, 5, 6$. No combination of three of these add up to 9."
] | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | No | |
0ge1 | 令 $O$ 為正三角形 $ABC$ 的中心。設 $P_1, P_2$ 為 $\odot(BOC)$ 的上異於 $B, O, C$ 的兩點, 並依 $B, P_1, P_2, O, C$ 此順序落在 $\odot(BOC)$ 上。延長 $BP_1, CP_1$ 分別交邊 $CA, AB$ 於 $R, S$, $AP_1$ 與 $RS$ 交於 $Q_1$, 並以類比方式定義 $Q_2$。設 $U$ 為 $\odot(OP_1Q_1)$ 與 $\odot(OP_2Q_2)$ 異於 $O$ 的交點。
證明：$2 \angle Q_2UQ_1 + \angle Q_2OQ_1 = 360^\circ$。
Let $O$ be the cen... | [
"由 $\\angle RP_1S = 120^\\circ = 180^\\circ - \\angle SAR$ 知 $A, S, P_1, R$ 共於一圓 $\\Gamma_1$。由 $\\angle RBA = \\angle SCA$ 知 $\\overline{AR} = \\overline{SB}$, 同理有 $\\overline{RC} = \\overline{AS}$, 所以易知 $\\overline{OR} = \\overline{OS}$。因此由 $AO$ 為 $\\angle RAS$ 的內角平分線可得 $O \\in \\Gamma_1$。\n\n令 $M$ 為 $\\overline{B... | Taiwan | 2020 Taiwan IMO 1J | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01fi | Determine if there is an integer $k \ge 2$ such that if we partition the set $\{2, 3, \dots, k\}$ in two parts, then at least one of the parts contains numbers $a, b$ and $c$ with $ab = c$? (We allow $a = b$.) If such a number $k$ exist, find the least $k$ with this property. | [
"We show first that $k = 32$ is such a number. Consider a partition $\\{U, V\\}$ of the set $\\{2, 3, \\dots, 32\\}$ where we may assume that $2 \\in U$. Towards contradiction, suppose that none of the parts contains numbers $a, b$ and $c$ with the desired property. As $2 \\in U$ and $2 \\cdot 2 = 4$, we have $4 \\... | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 32 | |
00ai | For an integer $m \ge 3$ set $S(m) = 1 + \frac{1}{3} + \dots + \frac{1}{m}$ (the fraction $1/m$ does not participate in the sum). Let $n \ge 3$ and $k \ge 3$. Compare the numbers $S(nk)$ and $S(n) + S(k)$. | [
"We show that $S(nk) < S(n) + S(k)$. Cancel the summand of $S(k)$ on both sides of this inequality, then add $\\frac{1}{2}$ to both sides and rearrange. This yields the equivalent inequality\n$$\n\\frac{1}{k+1} + \\frac{1}{k+2} + \\dots + \\frac{1}{nk} + \\frac{1}{2} < 1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\f... | Argentina | Argentine National Olympiad 2016 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | S(nk) < S(n) + S(k) | |
06pv | Let $S \subseteq \mathbb{R}$ be a set of real numbers. We say that a pair $(f, g)$ of functions from $S$ into $S$ is a Spanish Couple on $S$, if they satisfy the following conditions:
(i) Both functions are strictly increasing, i.e. $f(x) < f(y)$ and $g(x) < g(y)$ for all $x, y \in S$ with $x < y$;
(ii) The inequality ... | [
"We show that the answer is NO for part (a), and YES for part (b).\n\na.\nThroughout the solution, we will use the notation $g_{k}(x) = \\overbrace{g(g(\\ldots g}^{k}(x) \\ldots))$, including $g_{0}(x) = x$ as well.\nSuppose that there exists a Spanish Couple $(f, g)$ on the set $\\mathbb{N}$. From property (i) we ... | IMO | 49th International Mathematical Olympiad Spain | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a: no; b: yes | |
0a7z | Problem:
Let $A$ be a set of seven positive numbers. Determine the maximal number of triples $(x, y, z)$ of elements of $A$ satisfying $x<y$ and $x+y=z$. | [
"Solution:\n\nLet $0 < a_1 < a_2 < \\ldots < a_7$ be the elements of the set $A$. If $(a_i, a_j, a_k)$ is a triple of the kind required in the problem, then $a_i < a_j < a_i + a_j = a_k$. There are at most $k-1$ pairs $(a_i, a_j)$ such that $a_i + a_j = a_k$. The number of pairs satisfying $a_i < a_j$ is at most $\... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 11 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 9 | |
06ux | Consider 2018 pairwise crossing circles no three of which are concurrent. These circles subdivide the plane into regions bounded by circular edges that meet at vertices. Notice that there are an even number of vertices on each circle. Given the circle, alternately colour the vertices on that circle red and blue. In doi... | [
"Letting $n=2018$, we will show that, if every region has at least one non-yellow vertex, then every circle contains at most $n+\\lfloor\\sqrt{n-2}\\rfloor-2$ yellow points. In the case at hand, the latter equals $2018+44-2=2060$, contradicting the hypothesis.\n\nConsider the natural geometric graph $G$ associated ... | IMO | IMO Shortlisted Problems | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0jv4 | Problem:
A nonempty set $S$ is called well-filled if for every $m \in S$, there are fewer than $\frac{1}{2} m$ elements of $S$ which are less than $m$. Determine the number of well-filled subsets of $\{1,2, \ldots, 42\}$. | [
"Solution:\nLet $a_{n}$ be the number of well-filled subsets whose maximum element is $n$ (setting $a_{0}=1$). Then it's easy to see that\n$$\n\\begin{aligned}\n& a_{2k+1} = a_{2k} + a_{2k-1} + \\cdots + a_{0} \\\\\n& a_{2k+2} = \\left(a_{2k+1} - C_{k}\\right) + a_{2k} + \\cdots + a_{0}\n\\end{aligned}\n$$\nwhere $... | United States | HMMT February 2016 | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | binomial(43, 21) - 1 | |
0gwu | Find out how many solutions may equation
$$
[a^2, b^2] + [b^2, c^2] + [c^2, a^2] = (a^2, b^2)(b^2, c^2)(c^2, a^2)
$$
have in natural numbers, if $[m, n]$ and $(m, n)$ stands respectively for the LCM and GCD of the natural numbers $m$ and $n$. | [
"Let $a = px$, $b = py$, $c = pz$ where $p$, $x$, $y$, $z$ are pairwise coprime numbers. Then the equation will be as follows: $(pxy)^2 + (pxz)^2 + (pyz)^2 = (p^2)^3 \\Leftrightarrow x^2y^2 + x^2z^2 + y^2z^2 = p^4$\n$$ \\Leftrightarrow x^2(y^2 + z^2) = (p^2 - yz)(p^2 + yz). $$\nIf $p^2 = y^2 + yz + z^2$, $x^2(y^2 +... | Ukraine | Ukrajina 2008 | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | Infinitely many solutions | |
03oh | If the inequality
$$
sin^2 x + a \cos x + a^2 \ge 1 + \cos x
$$
holds for any $x \in \mathbb{R}$, the range of values for negative $a$ is
______. | [
"$a + a^2 \\ge 2$ when $x = 0$. So $a \\le -2$ (because $a < 0$). When $a \\le -2$, we have\n$$\n\\begin{aligned}\na^2 + a \\cos x &\\ge a^2 + a \\ge 2 \\ge \\cos^2 x + \\cos x \\\\\n&= 1 + \\cos x - \\sin^2 x,\n\\end{aligned}\n$$\nthat is,\n$$\n\\sin^2 x + a \\cos x + a^2 \\ge 1 + \\cos x.\n$$\nHence, the range of... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a ≤ -2 | |
0g7v | 令 $\mathbb{Z}$ 和 $\mathbb{Q}$ 分別為整數與有理數所形成的集合。對於 $\mathbb{Q}$ 的子集合 $X, Y$, 定義 $X + Y$ 為:
$$
X + Y := \{x + y \mid x \in X, y \in Y\}.
$$
(a) 試問能否將 $\mathbb{Z}$ 分割成三個非空子集 $A, B, C$ 使得 $A+B, B+C, C+A$ 互斥?
(b) 試問能否將 $\mathbb{Q}$ 分割成三個非空子集 $A, B, C$ 使得 $A+B, B+C, C+A$ 互斥? | [
"(a) 是。下為一例：\n$$\nA = \\{3k \\mid k \\in \\mathbb{Z}\\}, \\quad B = \\{3k + 1 \\mid k \\in \\mathbb{Z}\\}, \\quad C = \\{3k + 2 \\mid k \\in \\mathbb{Z}\\}.\n$$\n\n(b) 否。假設 $\\mathbb{Q}$ 可分割成三個非空子集 $A, B, C$ 使得 $A+B, B+C, C+A$ 互斥。注意到對於所有 $a \\in A, b \\in B, c \\in C$ 有\n$$\na + b - c \\in C, \\quad b + c - a \\in ... | Taiwan | 二〇一三數學奧林匹亞競賽第三階段選訓營, 模擬競賽(二) | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Other"
] | null | proof and answer | a) Yes. b) No. | |
01me | All cells of a $9 \times 10$ board are colored red, blue, and green. In any row of the board the number of red cells is no smaller than the number of blue cells and no smaller than the number of green ones. In any column of the board the number of blue cells is no smaller than the number of red cells and no smaller tha... | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | smallest = 10, greatest = 18 | |
02fv | 15 positive integers smaller than $1998$ are relatively prime (no pair has a common factor larger than $1$). Show that at least one of them must be prime. | [
"Suppose there are $15$ such integers, none prime. If we list the primes in order, the $15$th is $p_{15} = 47$. Now let $p(n)$ be the smallest prime dividing $n$. Take $N$ to be that integer $n$ among the $15$ which has the largest $p(n)$. Then since $N$ is not prime and $p(n)$ is its smallest prime factor, we must... | Brazil | XX OBM | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0kpn | Problem:
Points $A$, $B$, $C$, $M$ are such that $B$, $M$, $C$ lie on a line, $AB = AC = 353$, and $BM = MC$. Point $I$ is on segment $AM$ such that the circle centered at $I$ through $M$ has radius $106$ and is tangent to $AB$ and $AC$. Given that there are two possible areas of right triangle $AMB$, find the larger ... | [
"Solution:\n\nThe two configurations that work are the configuration when $B = C$ and angle $A$ is right and the configuration when $B = C$ and angle $B$ is right. In the second case, the height from $M$ is $2 \\cdot 106$, which is larger than in the other case. The answer is thus $106 \\cdot 353 = 37418$.\n\n![](a... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 37418 | |
03sp | Suppose $\log_a (2x^2 + x - 1) > \log_a 2 - 1$. Then the range of $x$ is ( ).
(A) $\frac{1}{2} < x < 1$
(B) $x > \frac{1}{2}$ and $x \ne 1$
(C) $x > 1$ | [
"From\n$$\n\\begin{cases}\nx > 0, \\\\\nx \\ne 1, \\\\\n2x^2 + x - 1 > 0\n\\end{cases}\n$$\nwe get $x > \\frac{1}{2}$, $x \\ne 1$.\n\nFurthermore,\n$\\log_a (2x^2 + x - 1) > \\log_a 2 - 1 \\Rightarrow \\log_a (2x^3 + x^2 - x) > \\log_a 2 \\Rightarrow \\begin{cases} 0 < x < 1, \\\\ 2x^3 + x^2 - x < 2, \\end{cases}$ ... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | MCQ | B | |
0hz5 | Problem:
Find the minimum possible value of the largest of $x y$, $1-x-y+x y$, and $x+y-2 x y$ if $0 \leq x \leq y \leq 1$. | [
"Solution:\n\nI claim the answer is $\\frac{4}{9}$. Let $s = x + y$, $p = x y$, so $x$ and $y$ are $\\frac{s \\pm \\sqrt{s^{2} - 4p}}{2}$. Since $x$ and $y$ are real, $s^{2} - 4p \\geq 0$.\n\nIf one of the three quantities is less than or equal to $\\frac{1}{9}$, then at least one of the others is at least $\\frac{... | United States | Harvard-MIT Math Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 4/9 | |
066y | Find all triads $(x, y, z)$ of positive integers satisfying the equation:
$$
\frac{1}{x} + \frac{2}{y} - \frac{4}{z} = 1
$$ | [
"If $x \\ge 3$ and $y \\ge 3$, then we have:\n$$\n\\frac{1}{x} + \\frac{2}{y} - \\frac{4}{z} \\le \\frac{1}{3} + \\frac{2}{3} - \\frac{4}{z} = 1 - \\frac{4}{z} < 1,\n$$\nand hence the equation is not satisfied. Hence we may have: $x \\le 2$ or $y \\le 2$.\n\n* For $x=1$ we have: $\\frac{2}{y} - \\frac{4}{z} = 0 \\L... | Greece | Hellenic Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All positive integer triples are: (1, k, 2k) with k a positive integer; (l, 2, 4l) with l a positive integer; (3, 1, 3); and (2, 3, 24). | |
00cx | En el triángulo isósceles $ABC$ sean $D$ y $E$ puntos en los lados $AB$ y $AC$, respectivamente, tales que las rectas $BE$ y $CD$ se cortan en $F$. Además, los triángulos $AEB$ y $ADC$ son iguales y tienen $AD=AE=10$ y $AB=AC=30$.
Calcular $\frac{\text{área}(ADFE)}{\text{área}(ABC)}$. | [
"Como $AD=AE$, el punto $F$ está a igual distancia de los lados $AC$ y $AB$; llamemos $h$ a esa distancia.\nEntonces $\\frac{\\text{área}(ADF)}{\\text{área}(AEF)} = \\frac{10h}{2}$, por lo tanto $\\frac{\\text{área}(ADFE)}{\\text{área}(ADF)} = 2 \\frac{\\text{área}(ADF)}{\\text{área}(ADF)} = 10h$.\n\nPor otra parte... | Argentina | Nacional OMA | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | Spanish | proof and answer | 1/6 | |
03fd | For a positive integer $n$, denote with $b(n)$ the smallest positive integer $k$, such that there exist integers $a_1, a_2, \dots, a_k$, satisfying $n = a_1^{33} + a_2^{33} + \dots + a_k^{33}$. Determine whether the set of positive integers $n$ is finite or infinite, which satisfy:
$$
\text{a) } b(n) = 12; \quad \text{... | [
"a) From Fermat's theorem and $y^2 \\equiv 1 \\pmod{67} \\Leftrightarrow y \\equiv \\pm 1 \\pmod{67}$ it follows that any student number gives a remainder of $0$, $1$ or $66$ when divided by $67$. Let us consider the numbers $12^{66k+1}$, where $k \\in \\mathbb{N}$. They are presented as the sum of $12$ student num... | Bulgaria | Bulgarian Spring Tournament | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | a) infinite; b) finite (empty) | |
0416 | Do there exist integers $a$, $b$ and $c$ such that $a^2bc + 2$, $ab^2c + 2$, $abc^2 + 2$ are perfect squares? | [
"No. Suppose the contrary that there are such integers $a$, $b$ and $c$.\nIf one of them is even, say $a$, then $a^2bc + 2 \\equiv 2 \\pmod 4$, which contradicts the assumption that $a^2bc + 2$ is a perfect square. We may then assume that $a$, $b$ and $c$ are odd, so they are either $1$ or $3 \\pmod 4$. It follows ... | China | China Western Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | No | |
0c2k | For any integer $k \in \mathbb{Z}$, we define the polynomial $F_k = X^4 + 2(1 - k)X^2 + (1 + k)^2$. Find all integers $k \in \mathbb{Z}$, such that $F_k$ is irreducible over $\mathbb{Z}$ but is reducible over $\mathbb{Z}_p$, for any prime $p$. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | All integers k such that k is neither a perfect square nor a negative perfect square. | |
0l1b | Pablo will decorate each of 6 identical white balls with either a striped or a dotted pattern, using either red or blue paint. He will decide on the color and pattern for each ball by flipping a fair coin for each of the 12 decisions he must make. After the paint dries, he will place the 6 balls in an urn. Frida will r... | [
"There are $4^6$ ways to paint the balls, each equally likely. It remains to count the number of paintings for which the two given events are independent.\n\n* If all the balls are red, then $P(\\text{red}) = 1$ and the events are independent regardless of $P(\\text{striped})$. The same reasoning applies if all the... | United States | 2024 AMC 12 B | [
"Statistics > Probability > Counting Methods > Combinations",
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | A | |
09ba | $ABC$ гурвалжны $AA_1$ нь медиан бөгөөд $AB < AC$ бол $\angle BAA_1 > \angle A_1AC$ гэж харуул. | [
"\n\n$A$-цэгийг $A_1$-цэгийн хувьд төвийн тэгш хэмтэй хувиргаж $M$-цэгийг байгуулан $M$-г $B$, $C$ цэгүүдтэй холбоё.\n\n$BA_1 = A_1C$ тул $\\triangle ABA_1 = \\triangle MCA_1$ ба $\\triangle AA_1C = \\triangle MA_1B$ болно.\n\nЭндээс $AC = BM$ ба $AB = CM$ байна. Мөн $\\angle A_1AC = \\angl... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Transformations > Rotation"
] | Mongolian | proof only | null | |
056w | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that, for every real number $a$, the function $g(x) = f(x + a)$ is either an even or odd function. | [
"Denote $f(0) = c$. We show that $f(x) = c$ for every real number $x$. Fix a real number $a$ arbitrarily; by assumption, the function $g(x) = f(x + \\frac{a}{2})$ is either an even or odd function. If this $g$ is even then\n$$\nf(a) = f\\left(\\frac{a}{2} + \\frac{a}{2}\\right) = g\\left(\\frac{a}{2}\\right) = g\\l... | Estonia | Final Round of National Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | All constant functions f(x) = c for some real constant c. | |
03ef | Let $p, q$ be coprime integers, such that $\frac{p}{q} \le 1$. For which $p, q$, there exist even integers $b_1, b_2, \dots, b_n$, such that
$$
\frac{p}{q} = \frac{1}{b_1 + \frac{1}{b_2 + \frac{1}{b_3 + \dots}}}?
$$ | [
"Set $b_i = 2\\ell_i$, all of them are even. At the first step we get the pairs\n$$\nA := \\{(p, q) : (p, q) = (1, 2\\ell_1), \\ell_1 \\in \\mathbb{Z}. \\quad (1)\\}.\n$$\nAt each subsequent step we expand the set $A$ by adding additional pairs $(p', q')$\ndefined as\n$$\n\\left\\{ (p', q') : \\frac{p'}{q'} = \\fra... | Bulgaria | Autumn tournament | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Exactly those coprime integers p, q with |p| < |q| and with one of p, q even and the other odd (excluding |p| = |q|, i.e., ±1). | |
0eyz | Problem:
A finite sequence of 0s and 1s has the following properties: (1) for any $i < j$, the sequences of length 5 beginning at position $i$ and position $j$ are different; (2) if you add an additional digit at either the start or end of the sequence, then (1) no longer holds. Prove that the first 4 digits of the seq... | [
"Solution:\nLet the last 4 digits be $a b c d$. Then the 5 digit sequences $a b c d 0$ and $a b c d 1$ must occur somewhere. If neither of them are at the beginning then there are three 5 digit sequences $x a b c d$, two of which must therefore be the same, contradicting (1). Hence $a b c d$ are the first 4 digits.... | Soviet Union | 3rd ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
07np | Prove that
$$
1 - \frac{1}{2012} \left( \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2013} \right) > \frac{1}{\sqrt[2012]{2013}}.
$$ | [
"Because\n$$\n1 - \\frac{1}{2012} \\left( \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{2013} \\right) = \\frac{1}{2012} \\left( 1 - \\frac{1}{2} + 1 - \\frac{1}{3} + \\dots + 1 - \\frac{1}{2013} \\right) \\\\\n= \\frac{1}{2012} \\left( \\frac{1}{2} + \\frac{2}{3} + \\dots + \\frac{2012}{2013} \\right)\n$$\nthe ... | Ireland | Ireland | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0ety | Let $\triangle ABC$ be a triangle with $\angle B \neq \angle C$. The incircle $I$ of a triangle $ABC$ touches the sides $BC, CA, AB$ at the points $D, E, F$, respectively. Let $P$ be the intersection of $AD$ and the incircle $I$, which is different from $D$.
Let $Q$ be the intersection of the line $EF$ and the line pas... | [
"Let $Q'$ be the intersection of the line passing $A$ and parallel to $BC$ and the line passing $P$ and perpendicular to $AD$. Let $U$ be the intersection of $DI$ and $AQ'$, and $V$ be the intersection of $PQ'$ and the circle $I$. ($V \\neq P$)\nSince $\\angle VPD = 90^\\circ$, four points $D, I, V, U$ lie on a lin... | South Korea | Korean Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point... | English | proof only | null | |
0aia | Two tangents are drawn from a point $M$ to circle $k$, that touch it at points $G$ and $H$. If $O$ is the center of $k$ and $K$ is the orthocenter of the triangle $MGH$, prove that $\angle GMH = \angle OGK$.
Од точка $M$ кон кружница $k$ се повлечени две тангенти со допирни точки $G$ и $H$. Ако $O$ е центарот на $k$ и... | [
"Let us notice that $K$ must lie on $OM$. From $HK \\perp GM$ and $OG \\perp GM$, it follows that $HK \\parallel OG$. Analogously $OH \\parallel GK$. From $\\overline{OG} = \\overline{OH}$, it follows that $OHKG$ is a rhombus. Let us notice that $O$, $H$, $M$ and $G$ lie on the circle with diameter $OM$. Hence $\\a... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05p8 | Problem:
Soit $a_{0}, a_{1}, \ldots, a_{99}, b_{0}, b_{1}, \ldots, b_{99}$ des réels strictement positifs.
Pour $k=0,1, \ldots, 198$, on pose $S_{k} = \sum_{i=0}^{198} a_{i} b_{k-i}$, avec $a_{j} = 0$ et $b_{j} = 0$ si $j < 0$ ou $j > 99$.
Est-il possible que les nombres $S_{0}, S_{1}, \ldots, S_{198}$ soient tous éga... | [
"Solution:\n\nNon. Raisonnons par l'absurde. En effet, comme $a_{0} b_{0} = a_{0} b_{99} + \\cdots + a_{99} b_{0}$ où les $\\cdots$ représentent des termes strictement positifs, on a $a_{0} b_{0} > a_{0} b_{99}$ donc $b_{0} > b_{99}$, et de même $a_{0} > a_{99}$. Par conséquent, $S_{0} = a_{0} b_{0} > a_{99} b_{99}... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | No | |
0dkm | Find all periodic sequences $a_1 a_2, \dots$ of real numbers such that the following conditions hold for all $n \ge 1$:
$$
a_{n+2} + a_n^2 = a_n + a_{n+1}^2 \quad \text{and} \quad |a_{n+1} - a_n| \le 1.
$$ | [
"Answer: The sequences satisfying the conditions of the problem are:\n$$\nc, -c, c, -c, \\dots \\\\\nd, d, d, d, \\dots\n$$\nwhere $c \\in \\left[-\\frac{1}{2}, \\frac{1}{2}\\right]$ and $d$ is any real number.\n\nWe rewrite the first condition as\n$$\na_{n+2} + a_{n+1} = (a_{n+1} + a_n)(a_{n+1} - a_n + 1)\n$$\nIf ... | Saudi Arabia | Saudi Booklet | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | All constant sequences d, d, d, … with any real d; and all alternating two-periodic sequences c, −c, c, −c, … with |c| ≤ 1/2. | |
092x | Problem:
Let $n \geqslant 3$ be an integer. A labelling of the $n$ vertices, the $n$ sides and the interior of a regular $n$-gon by $2n+1$ distinct integers is called memorable if the following conditions hold:
1. Each side has a label that is the arithmetic mean of the labels of its endpoints.
2. The interior of the... | [
"Solution:\n\nWe prove that the desired $n$'s are precisely those divisible by $4$.\n\nFix $n$ and assume such labelling exists. Without loss of generality, the labels form a set $\\{0,1, \\ldots, 2n\\}$. A maximum can't be obtained by averaging, so number $2n$ labels a vertex. In order for the side labels to be in... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All integers at least three that are divisible by four | |
0dal | Let $n > 2$ be a positive integer. Consider $n$ bags of candy, each of them has exactly 1 candy. Ali and Omar take turns playing the following game (Ali moves first): At each turn, the player takes two bags containing the numbers of candy as $x, y$ for some coprime integers $x, y$ and then merges them into one bag. Who... | [
"We shall prove that for all $n > 2$, Omar always has the strategy to win.\n\nFirst, Ali has to merge some two bags of 1 candy into one bag of 2 candies. We prove that Omar can turn the state of all bags into: one bag contains odd number of candies and the others just have one candy each.\n\nIndeed, in the second t... | Saudi Arabia | Team selection tests for JBMO 2018 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | Omar always has a winning strategy for all starting counts greater than two. | |
02ph | Problem:
Montar a tabela de um torneio em que todas as $n$ equipes se enfrentam ao longo de $n-1$ rodadas (como, por exemplo, em cada turno do Brasileirão) é um problema matemático bastante elaborado e que possui vários métodos de solução. Nesta questão, vamos conhecer uma dessas abordagens.
Vamos considerar um tornei... | [
"Solution:\n\n(a)\n\n$$\n\\begin{gathered}\n\\left(\\begin{array}{c}\n1 \\times \\infty \\\\\n2 \\times 5 \\\\\n3 \\times 4\n\\end{array}\\right) \\rightarrow \\left(\\begin{array}{c}\n2 \\times \\infty \\\\\n3 \\times 1 \\\\\n4 \\times 5\n\\end{array}\\right) \\rightarrow \\left(\\begin{array}{c}\n3 \\times \\inft... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Algorithms"
] | null | final answer only | a) For 6 teams (5 rounds):
1) 1 vs ∞, 2 vs 5, 3 vs 4
2) 2 vs ∞, 3 vs 1, 4 vs 5
3) 3 vs ∞, 4 vs 2, 5 vs 1
4) 4 vs ∞, 5 vs 3, 1 vs 2
5) 5 vs ∞, 1 vs 4, 2 vs 3
b) For 8 teams (7 rounds):
1) 1 vs ∞, 2 vs 7, 3 vs 6, 4 vs 5
2) 2 vs ∞, 3 vs 1, 4 vs 7, 5 vs 6
3) 3 vs ∞, 4 vs 2, 5 vs 1, 6 vs 7
4) 4 vs ∞, 5 vs 3, 6 vs 2, 7 vs 1... | |
07q1 | Suppose a doubly infinite sequence of real numbers
$$
\dots, a_{-2}, a_{-1}, a_0, a_1, a_2, \dots
$$
has the property that
$$
a_{n+3} = \frac{a_n + a_{n+1} + a_{n+2}}{3}, \quad \text{for all integers } n.
$$
Show that if this sequence is bounded (i.e., if there exists a number $R$ such that $|a_n| \le R$ for all $n$), ... | [
"Assume that the sequence is not constant. Define a function $D: \\mathbb{Z} \\to [0, \\infty)$ by the equation\n$$\nD(n) = |a_{n+1} - a_n| + |a_{n+2} - a_{n+1}| + |a_{n+2} - a_n|,\n$$\nor equivalently,\n$$\nD(n) = 2(x_n - y_n), \\quad (3)\n$$\nwhere $x_n$ and $y_n$ are the largest and the smallest, respectively, o... | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
090v | Ana y Benito juegan a un juego que consta de 2020 rondas. Inicialmente, en la mesa hay 2020 cartas, numeradas de 1 a 2020, y Ana tiene una carta adicional con el número 0. En la ronda $k$-ésima, el jugador que no tiene la carta $k-1$ decide si toma la carta $k$ o si se la entrega al otro jugador. El número de cada cart... | [
"Ambos jugadores pueden forzar el empate. Dividimos el juego en 505 etapas, cada una con cuatro rondas consecutivas de la forma\n$$\n\\{k, k+1, k+2, k+3\\}.\n$$\nVamos a demostrar que cada jugador puede conseguir al menos la mitad de los puntos de cada etapa, independientemente de qué jugador tenga la carta $k-1$. ... | Mexico | LVI Olimpiada Matemática Española (Concurso Final) | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Spanish | proof and answer | Both players can force a tie. | |
0eew | Problem:
Obseg pravokotnika je $4~\mathrm{cm}$, velikost kota med diagonalama pa $60^\circ$. Nariši skico pravokotnika z označenim kotom med diagonalama. Izračunaj dolžini obeh stranic tega pravokotnika. Rezultat naj bo natančen in zapisan v obliki okrajšanih ulomkov z racionaliziranimi imenovalci. Nalogo reši brez up... | [
"Solution:\n\nSkica pravokotnika z označenim kotom med diagonalama\n\n\n\nUpoštevanje $2a + 2b = 4$\n\nZapis ali upoštevanje $\\tan 30^\\circ = \\frac{b}{a}$\n\nZapis ali upoštevanje $\\tan 30^\\circ = \\frac{\\sqrt{3}}{3}$\n\nReševanje sistema enačb:\n\n$2a + 2b = 4$\n\n$\\tan 30^\\circ = ... | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | a = 3 - sqrt(3) cm, b = sqrt(3) - 1 cm | |
002d | Sean $m$, $n$ enteros positivos. En un tablero de $m+n$ cuadrículado en cuadrados de $1 \times 1$, consideramos todos los caminos que van del vértice superior derecho al inferior izquierdo, recorriendo líneas de la cuadrícula exclusivamente en las direcciones $\leftarrow$ y $\downarrow$.
Se define el área de un camino... | [] | Argentina | 15ª Olimpiada Matemática del Cono Sur | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | español | proof and answer | C(m+n, m) / p | |
0ky2 | Problem:
One hundred points labeled $1$ to $100$ are arranged in a $10 \times 10$ grid such that adjacent points are one unit apart. The labels are increasing left to right, top to bottom (so the first row has labels $1$ to $10$, the second row has labels $11$ to $20$, and so on).
Convex polygon $\mathcal{P}$ has the... | [
"Solution:\n\nThe vertices of the smallest $\\mathcal{P}$ are located at the points on the grid corresponding to the numbers $7, 21, 91, 98$, and $70$. The entire grid has area $81$, and the portion of the grid not in $\\mathcal{P}$ is composed of three triangles of areas $6, 9, 3$. Thus the area of $\\mathcal{P}$ ... | United States | HMMT February 2023 | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 63 | |
00kh | We are given a right-angled triangle $MNP$ with right angle in $P$. Let $k_M$ be the circle with center $M$ and radius $MP$, and let $k_N$ be the circle with center $N$ and radius $NP$.
Let $A$ and $B$ be the common points of $k_M$ and the line $MN$, and let $C$ and $D$ be the common points of $k_N$ and the line $MN$, ... | [
"Let $\\alpha = \\angle NMP$ and $\\beta = \\angle MNP$. Because the sum of angles in $MNP$ is $180^\\circ$, we obtain $\\alpha + \\beta = 90^\\circ$. Since $BP$ is a chord of the circle through $P$ and with mid-point $M$, we have\n$$\n\\angle BAP = \\frac{1}{2} \\angle BMP = \\frac{1}{2} \\alpha.\n$$\nSimilarly, f... | Austria | Austria 2014 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
015w | Assume that triangle $ABC$ is not equilateral and that both $\beta = \angle ABC$ and $\gamma = \angle ACB$ are larger than $30^\circ$. Let $O$ be the orthocentre of triangle $ABC$. Let the triangles $ACB'$ and $ABC'$ be equilateral with $B$ and $B'$ on opposite sides of $AC$ and $C$ and $C'$ on opposite sides of $AB$. ... | [
"Let the triangles $ACB'''$ and $ABC'''$ be equilateral with $B$ and $B'''$ on the same side of $AC$ and $C$ and $C'''$ on the same side of $AB$. Note that $O$ is an interior point of segment $B'''B'$ and $B'''O = \\frac{1}{2}\\left(1 - \\frac{\\tan(90^\\circ - \\beta)}{\\tan 60^\\circ}\\right)B'''B'$. Segment $OB'... | Baltic Way | Baltic Way SHL | [
"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",
"Geometry > Plane Geometry > Triangles > Triangle tri... | null | proof only | null | |
0glj | Let $m$ and $n$ be positive integers. Prove that if $m^{4n+1} - 1$ is a prime number then there exists an integer $t \ge 0$ such that $n = 2^t$. | [
"Let $p$ be an odd prime where $p \\mid n$. Then,\n$$\n\\begin{aligned}\n4^n + 1 &= (4^{\\frac{n}{p}})^p + 1 = (4^{\\frac{n}{p}} + 1)((4^{\\frac{n}{p}})^{p-1} - (4^{\\frac{n}{p}})^{p-2} + \\dots - 4^{\\frac{n}{p}} + 1) \\\\\n&= (4^{\\frac{n}{p}} + 1)x\n\\end{aligned}\n$$\n\nwhere $x = (4^{\\frac{n}{p}})^{p-1} - (4^... | Thailand | The 13th Thailand Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0l3g | Problem:
A positive integer $n$ is stacked if $2 n$ has the same number of digits as $n$ and the digits of $2 n$ are multiples of the corresponding digits of $n$. For example, $1203$ is stacked because $2 \times 1203 = 2406$, and $2, 4, 0, 6$ are multiples of $1, 2, 0, 3$, respectively. Compute the number of stacked in... | [
"Solution:\nWe do casework on the number of digits of $n$.\n\nOne digit. There are $4$ one-digit stacked integers: $1, 2, 3, 4$.\n\nTwo digits. Suppose $n = \\overline{ab}$ is a two-digit integer. If $a < 5$ and $b < 5$, then the digits of $2 n$ are double the respective digits of $n$, so $n$ is stacked; there are ... | United States | HMMT November | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 135 | |
05rz | Problem:
Soit $\mathcal{C}$ un cercle de rayon $1$, et soit $T$ un nombre réel. On dit qu'un ensemble de triangles est $T$-méraire s'il satisfait les trois conditions suivantes :
$\triangleright$ les sommets de chaque triangle appartiennent à $\mathcal{C}$ ;
$\triangleright$ les triangles sont d'intérieurs deux à deu... | [
"Solution:\n\nNous allons démontrer que les réels recherchés sont exactement les réels $\\mathbf{T} \\leqslant 4$. Pour ce faire, fixons un réel $\\mathbf{T} \\leqslant 4$. On commence par montrer qu'il existe des ensembles $T$-méraires de n'importe quelle taille, grâce à la construction suivante. Soit $[AB]$ un di... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | T ≤ 4 | |
0ceu | Fix an integer $n \ge 2$ and let $a_1, a_2, \dots, a_n$ be real numbers in the closed interval $[1, 2024]$. Prove that
$$
\sum_{i=1}^{n} \frac{1}{a_i} (a_1 + a_2 + \dots + a_i) > \frac{1}{44} n(n + 33).
$$ | [
"Let $A_j = \\{i: 2^{j-1} \\le a_i < 2^j\\}, j = 1, 2, \\dots, 11$. The $A_j$ form a partition of the index set $\\{1, 2, \\dots, n\\}$.\n\nNote that, for every $j$ in the range 1 through 11, if $A_j$ is non-empty, then\n$$\n\\begin{aligned}\n\\sum_{i \\in A_j} \\frac{1}{a_i} (a_1 + a_2 + \\dots + a_i) &= \\sum_{i ... | Romania | Eighteenth STARS OF MATHEMATICS Competition | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
03cv | A positive integer $n$ is called balanced if it is not a prime and for any integer $k$ in the interval $[1, \sqrt{n}-1]$ the number of ways to choose $k$ persons from a group of $n$ people is divisible by $n$. If $m$ and $n$ are balanced five digit positive integers find the smallest value of the difference $|m-n|$. | [
"Answer. 202. As in problem 9.3 we obtain that any 5-digit balanced number is of the forms: $p^2$ or $p(p+2)$ where $p$ and $p+2$ are primes. Since $p(p+2) = (p+1)^2 - 1 < (p+1)^2$ it follows that the smallest positive difference between two 5-digit numbers equals $2p$ where $p$ is the smallest prime number for whi... | Bulgaria | Bulgaria 2022 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 202 | |
017w | There is a pile of $1000$ matches. Two players each take turns and can take up to $5$ matches. It is also allowed at most $10$ times during the whole game to take $6$ matches. (There are no restrictions who uses this possibility, for example $1$ exceptional move can be done by the first player, and, say, $3$ moves by t... | [
"Let $r$ be the number of the remaining exceptional moves in the current position (at the beginning of the game $r=10$ and $r$ decreases during the game). The winning strategy of the second player is the following. After his move the number of matches in the pile must have the form $6n + r$, where $n > r$, or $7n$,... | Baltic Way | BALTIC WAY | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Second player wins | |
099b | The sequence $\{x_n\}_{n \geq 1}$ is defined as follows: $x_1 = x_2 = 1$ and $x_{n+2} = x_{n+1} + x_n + 2\sqrt{x_{n+1} x_n + 3}$ for all $n \geq 2$. Prove that $x_n$ is integer for all $n \geq 2$.
(proposed by Bat. Bayarjargal) | [
"We will show that by induction method.\nFor $n = 1$ is trivial. Now suppose that $x_1, x_2, ..., x_n, x_{n+1}$ are integers. From the given recurrence we get $x_n - x_{n+1} - x_{n+2} = 2\\sqrt{x_{n+1} \\cdot x_{n+2} + 3}$. Hence $2\\sqrt{x_{n+1} \\cdot x_{n+2} + 3}$ is integer. This is showing us $x_{n+3} = x_{n+2... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0kcz | Problem:
Let $ABCD$ be a cyclic quadrilateral, and let segments $AC$ and $BD$ intersect at $E$. Let $W$ and $Y$ be the feet of the altitudes from $E$ to sides $DA$ and $BC$, respectively, and let $X$ and $Z$ be the midpoints of sides $AB$ and $CD$, respectively. Given that the area of $AED$ is $9$, the area of $BEC$ is... | [
"Solution:\nReflect $E$ across $DA$ to $E_{W}$, and across $BC$ to $E_{Y}$. As $ABCD$ is cyclic, $\\triangle AED$ and $\\triangle BEC$ are similar. Thus $E_{W}AED$ and $EBE_{Y}C$ are similar too.\nNow since $W$ is the midpoint of $E_{W}E$, $X$ is the midpoint of $AB$, $Y$ is the midpoint of $EE_{Y}$, and $Z$ is the... | United States | HMMT February 2020 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 17 + (15/2)*sqrt(3) | |
0amd | Problem:
There are 100 people in a room. 60 of them claim to be good at math, but only 50 are actually good at math. If 30 of them correctly deny that they are good at math, how many people are good at math but refuse to admit it? | [] | Philippines | AREA STAGE | [
"Math Word Problems"
] | null | final answer only | 10 | |
0azn | Problem:
On the line containing diameter $AB$ of a circle, a point $P$ is chosen outside of this circle, with $P$ closer to $A$ than $B$. One of the two tangent lines through $P$ is drawn. Let $D$ and $E$ be two points on the tangent line such that $AD$ and $BE$ are perpendicular to it. If $DE = 6$, find the area of tr... | [
"Solution:\n$FB = \\sqrt{AB^{2} - AF^{2}} = \\sqrt{64 - 36} = 2\\sqrt{7}$\n\nThe radius $OC$ is the midsegment of trapezoid $ADED$. Hence $[ABED] = 4 \\times 6 = 24$. Moreover\n$$\n\\begin{aligned}\n[ABED] & = [ABF] + [ADEF] \\\\\n24 & = \\frac{2\\sqrt{7} \\times 6}{2} + AD \\times 6 \\\\\nAD & = 4 - \\sqrt{7} \\\\... | Philippines | 20th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 12 + (69/14) * sqrt(7) | |
0fwe | Problem:
Auf einem Kreis $k$ liegen fünf verschiedene Punkte $A, M, B, C$ und $D$ in dieser Reihenfolge und es gelte $M A = M B$. Die Geraden $A C$ und $M D$ schneiden sich in $P$, und die Geraden $B D$ und $M C$ schneiden sich in $Q$. Die Gerade $P Q$ schneide $k$ in $X$ und $Y$. Zeige $M X = M Y$. | [
"Solution:\n\n\nAbbildung 2: Skizze zur Lösung von Aufgabe 5\n\nWegen $M A = M B$ genügt es zu zeigen, dass $X Y$ parallel zu $A B$ ist (die Punkte $X$ und $Y$ sind dann wie $A$ und $B$ spiegelsymmetrisch bezüglich der Geraden durch $M$ und dem Mittelpunkt von $k$ angeordnet).\n\nDie Punkte... | Switzerland | Vorrundenprüfung | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fnm | Sean $a$, $b$ números positivos. Probar que
$$
a + b \geq \sqrt{ab} + \sqrt{\frac{a^2 + b^2}{2}}
$$ | [
"La desigualdad equivale a\n$$\n\\frac{\\sqrt{ab} + \\sqrt{\\frac{a^2+b^2}{2}}}{2} \\le \\frac{a+b}{2}.\n$$\nSi aplicamos la desigualdad entre las medias aritmética y geométrica al miembro de la izquierda obtenemos\n$$\n\\frac{\\sqrt{ab} + \\sqrt{\\frac{a^2+b^2}{2}}}{2} \\le \\sqrt{\\frac{ab + \\frac{a^2+b^2}{2}}{2... | Spain | L Olimpiada Matemática Española | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Spanish | proof only | null | |
07ee | Let $a, a_1, ..., a_n$ be positive integers. We know that for any positive integer $k$ where $ak + 1$ is a square number, at least one of the numbers $a_1k + 1, ..., a_nk + 1$ is also a square. Prove that $a \in \{a_1, ..., a_n\}$. | [
"**Lemma.** If $P$ is a polynomial with integer coefficients and a square leading coefficient which for infinitely many numbers $n$, $P(n)$ is a square number, then there exists a $Q \\in \\mathbb{Z}[x]$ which\n$$\nP(x) = Q(x)^2\n$$\nAssume $P(x)$ has the following representation\n$$\nP(x) = c_n x^n + c_{n-1} x^{n-... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Pigeonhole prin... | English | proof only | null | |
0l0o | Suppose that $a_1 = 2$ and the sequence $(a_n)$ satisfies the recurrence relation
$$
\frac{a_n - 1}{n - 1} = \frac{a_{n-1} + 1}{(n-1) + 1}
$$
for all $n \ge 2$. What is the greatest integer less than or equal to
$$
\sum_{n=1}^{100} a_n^2?
$$ | [
"Computing the first few terms of this sequence gives $a_1 = 2$, $a_2 = \\frac{5}{2}$, $a_3 = \\frac{10}{3}$, and $a_4 = \\frac{17}{4}$, so it appears that $a_n = n + \\frac{1}{n}$. Indeed, this is correct, because the recurrence relation is satisfied:\n$$\n\\frac{a_{n-1} + 1}{(n-1) + 1} = \\frac{n - 1 + \\frac{1}{... | United States | AMC 12 A | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 338551 | |
05mu | Problem:
On veut colorier les parties à trois éléments de $\{1,2,3,4,5,6,7\}$, de sorte que si deux de ces parties n'ont pas d'élément en commun alors elles soient de couleurs différentes. Quel est le nombre minimum de couleurs pour réaliser cet objectif? | [
"Solution:\n\nConsidérons la suite de parties $\\{1,2,3\\},\\{4,5,6\\},\\{1,2,7\\},\\{3,4,6\\},\\{1,5,7\\},\\{2,3,6\\},\\{4,5,7\\}$, $\\{1,2,3\\}$.\nChaque partie doit avoir une couleur différente de la suivante, donc déjà il y a au moins deux couleurs. S'il n'y avait qu'exactement deux couleurs, alors les couleurs... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | 3 | |
00v2 | Let $ABC$ be an acute triangle and $P$ be a point inside the triangle such that $\angle APB = \angle BPC = \angle CPA$. Denote with $S$ the area and with $\alpha, \beta, \gamma$ the angles of $\triangle ABC$. Prove that
$$
\frac{1}{\sin \alpha} + \frac{1}{\sin \beta} + \frac{1}{\sin \gamma} \ge \frac{PA^2 + PB^2 + PC^2... | [
"The inequality can be rewritten as\n$$\n2S \\left( \\frac{1}{\\sin \\alpha} + \\frac{1}{\\sin \\beta} + \\frac{1}{\\sin \\gamma} - \\frac{4}{\\sqrt{3}} \\right) \\ge PA^2 + PB^2 + PC^2.\n$$\nNote that $AB \\cdot AC \\cdot \\sin \\alpha = AB \\cdot BC \\cdot \\sin \\beta = AC \\cdot BC \\cdot \\sin \\gamma = 2S$ an... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Napoleon and Fermat points",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry >... | English | proof and answer | Equality occurs if and only if the triangle is equilateral. | |
0iz8 | Problem:
16 progamers are playing in a single elimination tournament. Each player has a different skill level and when two play against each other the one with the higher skill level will always win. Each round, each progamer plays a match against another and the loser is eliminated. This continues until only one rema... | [
"Solution:\n\nAnswer: 9\n\nEach finalist must be better than the person he beat in the semifinals, both of the people they beat in the second round, and all 4 of the people any of those people beat in the first round. So, none of the 7 worst players can possibly make it to the finals. Any of the 9 best players can ... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 9 | |
001m | En el cuadrado $ABCD$, sean $P$ y $Q$ puntos pertenecientes a los lados $BC$ y $CD$ respectivamente, distintos de los extremos, tales que $BP = CQ$. Se consideran puntos $X$ e $Y$, $X \neq Y$, pertenecientes a los segmentos $AP$ y $AQ$ respectivamente. Demuestre que, cualesquiera sean $X$ e $Y$, existe un triángulo cuy... | [] | Argentina | XVIII Olimpiada Iberoamericana de Matemática | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | español | proof only | null | |
0imf | Problem:
The sequence $\{a_n\}_{n \geq 1}$ is defined by $a_{n+2} = 7 a_{n+1} - a_n$ for positive integers $n$ with initial values $a_1 = 1$ and $a_2 = 8$. Another sequence, $\{b_n\}$, is defined by the rule $b_{n+2} = 3 b_{n+1} - b_n$ for positive integers $n$ together with the values $b_1 = 1$ and $b_2 = 2$. Find $\... | [
"Solution:\n\nAnswer: 89. We show by induction that $a_n = F_{4n-2}$ and $b_n = F_{2n-1}$, where $F_k$ is the $k$\\text{th}$ Fibonacci number. The base cases are clear. As for the inductive steps, note that\n$$\nF_{k+2} = F_{k+1} + F_k = 2 F_k + F_{k-1} = 3 F_k - F_{k-2}\n$$\nand\n$$\nF_{k+4} = 3 F_{k+2} - F_k = 8 ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | final answer only | 89 | |
07y6 | Problem:
Sia $ABC$ un triangolo, e siano $D$ ed $E$ le proiezioni di $A$ sulle bisettrici uscenti da $B$ e $C$. Dimostrare che $DE$ è parallelo a $BC$.
 | [
"Solution:\n\nSiano $F$ e $G$ i punti d'intersezione fra la retta $BC$ e le rette $AE$ e $AD$ rispettivamente. Il segmento $CE$ è bisettrice e altezza relativa al lato $AF$ del triangolo $ACF$, dunque è anche mediana (e il triangolo $ACF$ è isoscele sulla base $AF$): abbiamo $AE = EF$. Similmente, $BD$ è bisettrice... | Italy | null | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > ... | null | proof only | null | |
0gv8 | In the Cartesian plane $xOy$, draw the locus of all points $M(x;y)$ such that
$$
(x^2 - 1)(|y| - 1) \geq 0.
$$ | [
"$$\n(x^2 - 1)(|y| - 1) \\geq 0 \\iff \\left[ \\begin{cases} x^2 - 1 \\geq 0, \\\\ |y| - 1 \\geq 0; \\\\ x^2 - 1 \\leq 0, \\\\ |y| - 1 \\leq 0; \\end{cases} \\right] \\iff \\left[ \\begin{cases} |x| \\geq 1, \\\\ |y| \\geq 1; \\\\ |x| \\leq 1, \\\\ |y| \\leq 1. \\end{cases} \\right]\n$$"
] | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Locus = {(x, y): (|x| ≥ 1 and |y| ≥ 1) or (|x| ≤ 1 and |y| ≤ 1)}; equivalently, the union of the closed square |x| ≤ 1, |y| ≤ 1 and the four closed exterior corner regions |x| ≥ 1, |y| ≥ 1. | |
0ew1 | Problem:
Given an $m \times n$ array of real numbers. You may change the sign of all numbers in a row or of all numbers in a column. Prove that by repeated changes you can obtain an array with all row and column sums non-negative. | [
"Solution:\nThe array has $mn$ entries. Call an array that can be obtained by repeated changes a reachable array. A reachable array differs from the original only in that some or all of the signs of its $mn$ entries may be different. There are at most $2^{mn}$ different reachable arrays. For each reachable array ca... | Soviet Union | 1st ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof only | null | |
0g8t | 某國有數個城市, 其中若干個城市之間有航線相連; 航線都是雙向的。已知從該國中任選兩個城市, 都可以從其中一個城市, 透過一系列航線抵達另一個城市。定義兩個城市的距離為從一個城市抵達另一個城市所需的最少航線數量。已知對於任何一個城市, 至多都只有 100 個城市與其距離恰為 3。試證: 不存在一個城市, 有超過 2550 個其他城市與其距離恰為 4。
In some country several pairs of cities are connected by direct two-way flights. It is possible to go from any city to any other city by a seq... | [
"定義 $d(a, b)$ 為城市 $a$ 和 $b$ 的距離，並令\n$$\nS_i(a) = \\{c : d(a, c) = i\\}\n$$\n為與城市 $a$ 距離恰為 $i$ 的城市所成集合。\n\n以下歸謬：假設存在城市 $x$, 使得 $D = S_4(x)$ 滿足 $|D| \\geq 2551$。令 $A = S_1(x)$。我們稱集合 $A'$ 對於 $A$ 是必經的, 若且唯若每一個在 $D$ 中的城市, 都可以在 4 段航程後抵達 $x$, 且途中經過 $A'$ 的某個城市 (舉例來說, $A$ 對於自己當然是必經的。) 換言之, $D$ 的每個城市都跟某個 $A'$ 的城市距離恰為 3, 也就是 ... | Taiwan | 二〇一四數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
07vu | Let $S = \mathbb{N} \cup \{1/n \mid n \in \mathbb{N}\}$ be the set of all positive integers and their reciprocals. A function $f : S \to S$, defined on $S$ and with values in $S$, is called *semi-reciprocal* if $f(f(x)) = 1/x$ for all $x \in S$.
a. Find a semi-reciprocal function.
b. Show that for every semi-reciproc... | [
"We first note that if $f$ is a semi-reciprocal function and $f(1) = a \\in S$, then $f(a) = f(f(1)) = 1$. Hence, $a = f(1) = f(f(a)) = 1/a$, and so $a^2 = 1$. Since $S$ does not contain negative numbers, we must have $a = 1$, i.e. $f(1) = 1$ for each semi-reciprocal function $f$.\n\nIf $a \\in S$ and $b = f(a)$, t... | Ireland | IRL_ABooklet_2023 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | One example: f(1) = 1 and for all integers n ≥ 1,
- f(2n) = 2n + 1,
- f(2n + 1) = 1/(2n),
- f(1/(2n)) = 1/(2n + 1),
- f(1/(2n + 1)) = 2n.
For any semi-reciprocal function, the unique fixed point is 1. | |
06g5 | Let $n$ be a positive integer. When $\frac{n}{567}$ is expressed as a decimal, it is a recurring decimal with smallest period $k$. Find the sum of all possible values of $k$. (Note: We say that $k$ is a *period* of a recurring decimal if starting from some place the digits of the decimal repeat every $k$ digits. For in... | [
"The answer is $37$.\n\nWe first prove the following result.\n**Claim.** Consider a positive rational number $\\frac{a}{b}$ where $(10a, b) = 1$. The smallest period of this number when expressed as a decimal number is the order of $10$ modulo $b$.\n*Proof.* Note that $\\frac{a}{b}$ has period $k$ if and only if th... | Hong Kong | IMO HK TST | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | 37 | |
0d47 | The 2013 numbers
$$
\frac{1}{1 \times 2}, \frac{1}{2 \times 3}, \frac{1}{3 \times 4}, \ldots, \frac{1}{2013 \times 2014}
$$
are arranged randomly on a circle.
a. Prove that there exist ten consecutive numbers on the circle whose sum is less than $\frac{1}{4000}$.
b. Prove that there exist ten consecutive numbers on t... | [
"a.\nConsider 201 disjoint blocks $B_{1}, B_{2}, \\ldots, B_{201}$ consisting each of 10 numbers, consecutive on the circle. By the pigeonhole principle, there exists a block $B_{k}$, for some $1 \\leq k \\leq 201$, not containing any of the 200 numbers\n$$\n\\frac{1}{1 \\times 2}, \\frac{1}{2 \\times 3}, \\ldots, ... | Saudi Arabia | SAMC | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English, Arabic | proof only | null | |
07jb | Two circles $\omega_1$ and $\omega_2$ with the same radius will meet at $P$ and $Q$. Points $B$ and $C$ are inside the circles $\omega_1$ and $\omega_2$, respectively. Points $X$ and $Y$ different from $P$ are on $\omega_1$ and $\omega_2$, respectively such that $\angle CPQ = \angle CXQ$ and $\angle BPQ = \angle BYQ$. ... | [
"Since the circles are identical, we have\n$$\n\\angle PYQ = \\angle PXQ, \\angle PCQ = \\angle PBQ\n$$\nWe shall firstly prove that $X$ is the orthocenter of $PCQ$. If we denote above mentioned center by $X'$, it follows that $\\angle CX'Q = \\angle CPQ$ and $\\angle PX'Q = 180^\\circ - \\angle PCQ = 180^\\circ - ... | Iran | 41th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / C... | null | proof only | null | |
0lgj | Problem:
Prove that for some positive integer $n$ the remainder of $3^{n}$ when divided by $2^{n}$ is greater than $10^{2021}$. | [
"Solution:\n\nWe choose a positive integer $M$ such that $2^{M}>10^{2022}$, and consider the remainder of $3^{M}$ when divided by $2^{M}$:\n$$\n3^{M} \\equiv r \\quad\\left(\\bmod 2^{M}\\right),\\ 0<r<2^{M}\n$$\nIf $r>10^{2021}$, then $M$ is the desired number. Otherwise we choose the smallest integer $k$ for which... | Zhautykov Olympiad | Zhautykov Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0iw1 | Problem:
Let $C$ be the circle of radius $12$ centered at $(0,0)$. What is the length of the shortest path in the plane between $(8 \sqrt{3}, 0)$ and $(0,12 \sqrt{2})$ that does not pass through the interior of $C$? | [
"Solution:\n$12 + 4 \\sqrt{3} + \\pi$\n\nThe shortest path consists of a tangent to the circle, a circular arc, and then another tangent. The first tangent, from $(8 \\sqrt{3}, 0)$ to the circle, has length $4 \\sqrt{3}$, because it is a leg of a $30$-$60$-$90$ right triangle. The $15^{\\circ}$ arc has length $\\fr... | United States | 2nd Annual Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 12 + 4 \sqrt{3} + \pi | |
0fn0 | Calcula la suma de los inversos de los dos mil trece primeros términos de la sucesión de término general
$$
a_n = 1 - \frac{1}{4n^2}
$$ | [
"El término general se puede escribir como\n$$\na_n = \\frac{4n^2 - 1}{4n^2} = \\frac{(2n - 1)(2n + 1)}{4n^2}\n$$\ny su inverso es\n$$\n\\frac{1}{a_n} = \\frac{4n^2}{(2n - 1)(2n + 1)} = \\frac{n}{2n - 1} + \\frac{n}{2n + 1}\n$$\nHemos de calcular\n$$\n\\begin{aligned}\nS &= \\frac{1}{a_1} + \\frac{1}{a_2} + \\frac{... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Spanish | proof and answer | 8108364/4027 | |
04hr | Determine positive integer $n$ such that the sum of his two smallest divisors is $6$ and the sum of his two largest divisors is $1122$. | [] | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 935 | |
05uk | Problem:
Soit $n \geqslant 6$ un entier. On a disposé, dans le plan, $n$ disques $D_{1}, D_{2}, \ldots, D_{n}$ deux à deux disjoints, de rayons $r_{1} \geqslant r_{2} \geqslant \ldots \geqslant r_{n}$. Pour tout entier $i \leqslant n$, on considère un point $P_{i}$ à l'intérieur du disque $D_{i}$. Enfin, soit $A$ un p... | [
"Solution:\n\nAu vu d'un tel énoncé, et ne sachant pas nécessairement par où commencer, on souhaite utiliser une récurrence sur $n$. Dans ces conditions, il s'avère que la propriété clé de notre récurrence sera la suivante.\n\nLemme. Si $n=6$, il existe un entier $i$ tel que $A P_{i} \\geqslant r_{6}$.\n\nDémonstra... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Discrete Mathematics > Combinatorics > In... | null | proof only | null | |
0b8t | Let $f: [-1, 1] \to \mathbb{R}$ be a continuous function having finite derivative at $0$, and
$$
I(h) = \int_{-h}^{h} f(x) \, dx, \quad h \in [0, 1].
$$
Prove that
a) There exists $M > 0$, such that $|I(h) - 2f(0)h| \leq Mh^2$, for any $h \in [0, 1]$.
b) The sequence $(a_n)_{n \geq 1}$, defined by $a_n = \sum_{k=1}^{... | [
"a) The continuous function $\\varphi : (0, 1] \\to \\mathbb{R}$, $\\varphi(h) = \\frac{I(h) - 2f(0)h}{h^2}$, may be prolonged by continuity at $0$, since\n$$\n\\begin{align*} \n\\lim_{h \\to 0} \\varphi(h) &= \\lim_{h \\to 0} \\frac{(I(h) - 2f(0)h)'}{2h} = \\lim_{h \\to 0} \\frac{f(h) + f(-h) - 2f(0)}{2h} = \\\\ \... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null | |
022o | Problem:
Um jogo é composto das seguintes regras:
i) Em cada rodada, ocorre o lançamento de um dado comum não viciado.
ii) Se sair o número $3$, então o jogador $A$ ganha.
iii) Se sair um dos números do conjunto $\{4,5,6\}$, então o jogador $B$ ganha.
iv) Se sair um dos números do conjunto $\{1,2\}$, então o dado é la... | [
"Solution:\n\nSeja $P_{i}(B)$ a probabilidade do jogador $B$ vencer na rodada $i$, com $i$ inteiro positivo, e $P_{i}(\\overline{A+B})$ a probabilidade de $A$ e $B$ não vencerem na rodada $i$. Portanto, temos que\n$$\n\\begin{aligned}\n& P_{1}(B)=\\frac{3}{6}=\\frac{1}{2} \\\\\n& P_{2}(B)=P_{1}(\\overline{A+B}) \\c... | Brazil | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 3/4 | |
0jf6 | Problem:
Let $z_{0} + z_{1} + z_{2} + \cdots$ be an infinite complex geometric series such that $z_{0} = 1$ and $z_{2013} = \frac{1}{2013^{2013}}$. Find the sum of all possible sums of this series. | [
"Solution:\n\nAnswer: $\\frac{2013^{2014}}{2013^{2013}-1}$\n\nClearly, the possible common ratios are the $2013$ roots $r_{1}, r_{2}, \\ldots, r_{2013}$ of the equation $r^{2013} = \\frac{1}{2013^{2013}}$. We want the sum of the values of $x_{n} = \\frac{1}{1 - r_{n}}$, so we consider the polynomial whose roots are... | United States | HMMT 2013 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 2013^{2014}/(2013^{2013}-1) | |
08es | Problem:
Siano $a, b, c$ interi, ciascuno compreso fra $1$ e $2021$ (estremi inclusi), che soddisfano l'equazione
$$
\sqrt{a} + \sqrt{b} = \sqrt{a + c \sqrt{b}}.
$$
Quanti sono i possibili valori distinti di $c$?
(A) 130
(B) 132
(C) 133
(D) 1936
(E) 2025 | [
"Solution:\n\nLa risposta è (A). Elevando al quadrato l'equazione del testo otteniamo la condizione equivalente $a + 2 \\sqrt{a b} + b = a + c \\sqrt{b}$, da cui sottraendo $a$ e dividendo per $\\sqrt{b}$ si arriva a\n$$\n2 \\sqrt{a} + \\sqrt{b} = c.\n$$\nAffinché $c$ sia intero, entrambi $a$ e $b$ devono essere de... | Italy | Italian Mathematical Olympiad, February Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | A | |
095t | Problem:
Fie $A$ mulţimea tuturor produselor de trei numere naturale impare consecutive, iar $B$ mulţimea tuturor produselor de două numere naturale impare consecutive. Să se determine, dacă $A$ conţine vre-un număr, care ar fi cu 2018 mai mare decât un careva număr din $B$. | [
"Solution:\n\nDin trei numere impare consecutive unul se divide cu $3$, prin urmare, oricare număr $a \\in A$ are forma $3m$, pentru un careva număr întreg $m$. Oricare număr $b \\in B$ are forma $(2n-1)(2n+1)$. Conform condiţiei, $a = b + 2018$, adică $3m = (2n-1)(2n+1) + 2018 = 4n^2 + 2017$, de unde\n$$\n(2n)^2 =... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
03pu | (1) Prove that there exist five nonnegative real numbers $a$, $b$, $c$, $d$ and $e$ with their sum equal to $1$ such that for any arrangement of these numbers around a circle, there are always two neighboring numbers with their product not less than $\frac{1}{9}$.
(2) Prove that for any five nonnegative real numbers w... | [
"(1) Let $a = b = c = \\frac{1}{3}$, $d = e = 0$, it is easy to see that, when arranging them around a circle, we can always get two neighboring numbers of $\\frac{1}{3}$, and their product is $\\frac{1}{9}$.\n\n(2) For any five nonnegative real numbers $a$, $b$, $c$, $d$ and $e$ with their sum equal to $1$, withou... | China | China Girls' Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0eo7 | Let $n > 1$ be an integer. An $n \times n$-square is divided into $n^2$ unit squares. Of these smaller squares, $n$ are coloured green and $n$ are coloured blue. All remaining squares are coloured white. Are there more such colourings for which there are no two green squares in a row, and no two blue squares in a colum... | [
"Suppose that $n$ squares have been coloured green, no two of them in the same row. This leaves $n-1$ empty squares in each row, which means that there are $(n-1)^n$ ways to colour $n$ of the remaining $n^2-n$ squares blue such that there are no two blue squares in the same row.\n\nOn the other hand, let $x_i$ be t... | South Africa | The South African Mathematical Olympiad Third Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | There are more colorings with no two green squares in a row and no two blue squares in a row than with no two green squares in a row and no two blue squares in a column. | |
09dr | $ABCD$ гүдгэр дөрвөн өнцөгтөд багтсан $\gamma$ тойрог $AD$ ба $CD$ талуудыг харгалзан $P$ ба $Q$ цэгт шүргэнэ. $BD$ диагналь нь $\gamma$ тойрогтой $X$ ба $Y$ цэгт огтлолцдог бөгөөд $XY$ хэрчмийн дундаж цэг нь $M$ бол $\angle AMP = \angle CMQ$ гэж батал. | [
"ММО-48, Хангийн IV давааны ВЗ бодлогыг үзнэ үү."
] | Mongolia | ММО-48 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Spiral similarity"
] | Mongolian | proof only | null | |
09jz | A family has four children with pairwise distinct positive integer ages such that the sum of the ages of any set of the children is different from the sum of the ages of any other set of the children. What is the minimum possible age of the eldest child? | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 8 | |
0l9b | Find all polynomials $P(x)$ with real coefficients, satisfying the relation
$$
(x^3 + 3x^2 + 3x + 2)P(x - 1) = (x^3 - 3x^2 + 3x - 2)P(x)
$$
for every real number $x$. | [
"We have:\n$$\n(x^3 + 3x^2 + 3x + 2)P(x - 1) = (x^3 - 3x^2 + 3x - 2)P(x) \\quad \\forall x \\in \\mathbb{R} \\quad (1)\n$$\n$$\n\\Leftrightarrow (x + 2)(x^2 + x + 1)P(x - 1) = (x - 2)(x^2 - x + 1)P(x) \\quad \\forall x \\in \\mathbb{R} \\quad (2)\n$$\nBy substituting $x = -2$ into (2), we get $0 = -28 \\cdot P(-2)$... | Vietnam | 2003 Vietnamese Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | P(x) = c(x - 1)x(x + 1)(x + 2)(x^2 + x + 1), where c is any real constant | |
09g3 | Let $a$ and $b$ be relatively prime positive integers such that $(a, b) \ne (2, 1)$. Show that
$$
\text{rad}(a^n + b^n) \ne \text{rad}(a^m + b^m),
$$
for any distinct positive integers $m$ and $n$. Here $\text{rad}(c)$ denotes the distinct primes dividing the integer $c$. | [
"Assume that there is a quadruple $(a, b, n, m)$ such that\n$$\n\\mathrm{rad}(a^n + b^n) = \\mathrm{rad}(a^m + b^m).\n$$\nLet $d$ be the greatest common divisor of $a^n + b^n$ and $a^m + b^m$. We claim that\n$$\nd = \\begin{cases} a^{(n,m)} + b^{(n,m)}, & v_2(n) = v_2(m) \\\\ (a+b, 2), & v_2(n) \\neq v_2(m) \\end{c... | Mongolia | 2015 Mongolian IMO Team Selection Tests | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0iqv | Problem:
$ABCDE$ is a regular pentagon inscribed in a circle of radius $1$. What is the area of the set of points inside the circle that are farther from $A$ than they are from any other vertex? | [
"Solution:\n\nAnswer: $\\frac{\\pi}{5}$\n\nDraw the perpendicular bisectors of all the sides and diagonals of the pentagon with one endpoint at $A$. These lines all intersect in the center of the circle, because they are the set of points equidistant from two points on the circle. Now, a given point is farther from... | United States | 1st Annual Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | pi/5 | |
0a0v | For a three-digit positive integer, we can multiply the three digits together. We call the result the *digit product* of that number. For example, $123$ has digit product $1 \times 2 \times 3 = 6$ and $524$ has digit product $5 \times 2 \times 4 = 40$. A number cannot begin with the digit $0$.
Determine the three-digit... | [
"$175$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 175 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.