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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
01kg | The bisectors of the angles $BAD$ and $CDA$ of the trapezium $ABCD$ ($BC \parallel AD$) meet at a point on the perpendicular bisector of the side $BC$.
Prove that either $AB = CD$ or $AB + CD = AD$. | [
"Let $P$ and $M$ be marked on $AD$ and $BC$ so that $MP \\perp AD$ and $BM = MC$. By condition, $K$ belongs to $MP$. Let $L$ and $N$ be respectively the feet of the perpendiculars from $K$ to the lines $AB$ and $DC$. Two cases are possible:\n\n1) Both the points lie on the sides $AB$ and $DC$ (or on their extension... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0kzf | Problem:
Compute the smallest positive integer such that, no matter how you rearrange its digits (in base ten), the resulting number is a multiple of $63$. | [
"Solution:\n\nFirst, the number must be a multiple of $9$ and $7$. The first is easy to check and holds for all permutations. Note that when two adjacent digits $a$ and $b$ are swapped, the number changes by $9(a-b) \\cdot 10^{k}$ (we disregard sign), so $9(a-b)$ must also be a multiple of $63$ for all digits $a$ a... | United States | HMMT February 2024 Guts Round | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 111888 | |
0kqn | Problem:
Let $ABCD$ be a rectangle with $AB = 8$ and $AD = 20$. Two circles of radius $5$ are drawn with centers in the interior of the rectangle: one tangent to $AB$ and $AD$, and the other passing through both $C$ and $D$. What is the area inside the rectangle and outside of both circles? | [
"Solution:\nLet $O_{1}$ and $O_{2}$ be the centers of the circles, and let $M$ be the midpoint of $\\overline{CD}$. We can see that $\\triangle O_{2}MC$ and $\\triangle O_{2}MD$ are both $3$-$4$-$5$ right triangles. Now let $C'$ be the intersection of circle $O_{2}$ and $\\overline{BC}$ (that isn't $C$), and let $D... | United States | HMMT November | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 112 - 25π | |
01yn | The sequence of positive integers $a_1, a_2, a_3, \dots$, for each integer $i \ge 3$ satisfy the equality
$$
a_{i+1} = a_i + \gcd(a_{i-1}, a_{i-2}).
$$
Prove that there exist positive integers $N$ and $M$ such that for each $n \ge N$ holds $a_{n+1} - a_n = M$. (Palina Chernikava) | [
"Since the term $a_1$ is involved only in the summand $\\gcd(a_1, a_2)$ when calculating $a_4$, without loss of generality we can assume that $a_1 = \\gcd(a_1, a_2)$. If the numbers $a_1, a_2$ and $a_3$ are divisible by the same number, then all the numbers in the sequence $(a_n)$ are divided by this number and we ... | Belarus | Belarus2022 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
03yf | There are $n$ ($n \ge 3$) players in a table tennis tournament, in which any two players have a match. Player $A$ is called not out-performed by player $B$, if at least one of player $A$'s losers is not a $B$'s loser.
Determine, with proof, all possible values of $n$, such that the following case could happen: after fi... | [
"The answer is $n = 3$ or $n \\ge 5$.\n\n(1) For $n = 3$, suppose $A$, $B$ and $C$ are three players, and the result of three matches are as follows: $A$ wins $B$, $B$ wins $C$, and $C$ wins $A$. These results obviously satisfy the condition.\n\n(2) If $n = 4$, suppose that the condition holds, i.e., in view of the... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | n = 3 or n >= 5 | |
0azf | Problem:
How many pairs of positive integers $(a, b)$ are there, both not exceeding $10$, such that $a \leq \operatorname{gcd}(a, b) \leq b$ and $a \leq \operatorname{lcm}(a, b) \leq b$? | [
"Solution:\n\n$\\operatorname{gcd}(a, b) \\geq a \\Longrightarrow \\operatorname{gcd}(a, b) = a$. Likewise, $\\operatorname{lcm}(a, b) = b$. This can only happen if $b$ is a multiple of $a$. Hence the answer is $10 + 5 + 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 = 27$."
] | Philippines | 20th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 27 | |
0grw | A line $d$ is said to *focus* a triangle $T$ if there exists a point on the plane whose projections on the edges of the triangle $T$ all lie on the line $d$. Triangles $T_1$ and $T_2$ are *equivalent* if the set of focusing lines of $T_1$ is identical to the set of focusing lines of $T_2$. For any given triangle $T$ on... | [
"First, note that the focusing lines of a triangle are its Simson lines. We will make use of the following well-known lemmas about Simson lines and we state them without proof.\n\n**Lemma 1.** Fix a triangle. Let $P$ be a point on the circumcircle and let $H$ be the orthocenter. The Simson line associated to $P$ pa... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
0hv1 | Problem:
Find the sum of the infinite series
$$
\frac{1}{3^{2}-1^{2}}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right)+\frac{1}{5^{2}-3^{2}}\left(\frac{1}{3^{2}}-\frac{1}{5^{2}}\right)+\frac{1}{7^{2}-5^{2}}\left(\frac{1}{5^{2}}-\frac{1}{7^{2}}\right)+\cdots
$$ | [
"Solution:\n$1-\\frac{1}{3}+\\frac{1}{3}-\\frac{1}{5}+\\cdots=1$."
] | United States | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 1 | |
0am5 | Problem:
There are two values of $r$ such that $x^{4} - x^{3} - 18 x^{2} + 52 x + k$ has $x - r$ as a factor. If one of them is $r = 2$, what is the other value of $r$? | [] | Philippines | Area Stage | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | -5 | |
0fi4 | Problem:
En Port Aventura hay 16 agentes secretos. Cada uno de ellos vigila a alguno de sus colegas. Se sabe que si el agente $A$ vigila al agente $B$, entonces $B$ no vigila a $A$. Además, 10 agentes cualesquiera pueden ser numerados de forma que el primero vigila al segundo, éste vigila al tercero, ..., el décimo vi... | [
"Solution:\n\nDiremos que los agentes $A$ y $B$ son neutrales si $A$ no vigila a $B$ ni $B$ vigila a $A$.\nSean $A_{1}, A_{2}, \\ldots, A_{n}$ los agentes. Sea $a_{i}$ el número de agentes que vigilan a $A_{i}$, $b_{i}$ el número de agentes que son vigilados por $A_{i}$, y $c_{i}$ el número de agentes que son neutr... | Spain | Olimpiada Matemática Española | [
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
051s | Malle drew a rhombus $ABCD$ and chose points $E$ and $F$ on sides $AB$ and $BC$, respectively, such that the triangle $DEF$ is equilateral. Malle was very surprised when she discovered that there is another possibility to choose points $E$ and $F$ on sides $AB$ and $BC$, respectively, such that $DEF$ is equilateral. Wh... | [
"There is clearly only one way to choose points $E$ and $F$ on sides $AB$ and $BC$, respectively, such that $E$ and $F$ would be symmetrical with respect to the diagonal $BD$ and $\\angle EDF = 60^\\circ$. Thus it is possible in Malle's rhombus to choose $E$ and $F$ asymmetrically with respect to diagonal $BD$ such... | Estonia | Final Round of National Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 60° and 120° | |
0arv | Problem:
Find the sum of all even factors of $1152$. | [
"Solution:\n$1152 = 2^{7} 3^{2}$. The even factors of $1152$ are each of the form $2^{k} 3^{l}$, $1 \\leq k \\leq 7$, $0 \\leq l \\leq 2$. Hence the sum can be written as\n$$(2 + 2^{2} + \\cdots + 2^{7})(1 + 3 + 3^{2}) = 2 \\frac{1 - 2^{7}}{1 - 2} \\cdot 13 = 3302$$"
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 3302 | |
0hpx | Problem:
If two fair dice are tossed, what is the probability that their sum is divisible by $5$? | [
"Solution:\n\n$\\frac{1}{4}$."
] | United States | null | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 1/4 | |
0gmk | Set $M = \{(a, b, c, d) | a, b, c, d \in \{1, 2, 3, 4\} \text{ and } abcd > 1\}$. Find all values $(a_1, b_1)$ can assume if the sequence $(a_1, b_1, c_1, d_1)$, $(a_2, b_2, c_2, d_2)$, ..., $(a_{255}, b_{255}, c_{255}, d_{255})$ with $c_1 = d_1 = 1$ contains every element of $M$ exactly once and satisfies the conditio... | [] | Turkey | IMO TEAM SELECTION EXAM | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | All pairs with one odd and the other even: (1,2), (1,4), (2,1), (2,3), (3,2), (3,4), (4,1), (4,3). | |
08rd | A $3 \times 4$ grid is given. Answer the number of possible cases to write a number among $1$, $2$, $3$ or $4$ in each square so that
* no number appears twice (or more) in the same row, and
* no number appears twice (or more) in the same column. | [
"Denote by $a_{ij}$ the entry that lies in the $i$-th column and the $j$-th row. The given conditions yields that each $1$, $2$, $3$ and $4$ appear exactly once in each column.\nWe may assume without loss of generality that $a_{1j} = j$ for $j = 1, 2, 3, 4$. We would multiply the answer by $4!$ afterwards. We may a... | Japan | The 16th Japanese Mathematical Olympiad - The First Round | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 576 | |
0dic | Find all real $\alpha > 0$ with the following property: there exists a sequence $(x_1, x_2, x_3, ...)$ of positive real numbers such that $x_{n+2} = \sqrt{\alpha x_{n+1} - x_n}$ for all $n \ge 1$. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | α > 1 | |
0ddt | Given an infinite sequence of numbers $a_1, a_2, a_3, \ldots$ such that for each positive integer $k$, there exists positive integer $t$ for which $a_k = a_{k+t} = a_{k+2t} = \ldots$. Does this sequence must be periodic? | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | No | |
06l9 | Is it possible to choose $24$ distinct points in the space such that no three of them lie on the same line, and choose $2019$ distinct planes in a way that each plane passes through at least $3$ of the chosen points and each triple of points belongs to one of the chosen planes? | [
"Solution:\n\nNo. Suppose such a choice exists. Let $n_1, n_2, \\dots, n_{2019}$ be the number of points on the planes respectively. Note that $n_j \\ge 3$ for $1 \\le j \\le 2019$. It is clear that\n$$\n2024 = \\binom{24}{3} = \\binom{n_1}{3} + \\binom{n_2}{3} + \\dots + \\binom{n_{2019}}{3}.\n$$\n\nWe first show ... | Hong Kong | HKG TST | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | No | |
00ia | Let $a, b$ be real numbers with $0 \le a, b \le 1$. Prove the inequality
$$
\sqrt{a^3 b^3} + \sqrt{(1-a^2)(1-ab)(1-b^2)} \le 1.
$$
G. Baron, Vienna | [
"Since we are given that $0 \\le a, b \\le 1$ holds, it follows from the AM-GM inequality, that\n$$\n\\begin{aligned}\n& \\sqrt{a^3 b^3} + \\sqrt{(1-a^2)(1-ab)(1-b^2)} \\\\\n& \\le \\sqrt[3]{a^3 b^3} + \\sqrt[3]{(1-a^2)(1-ab)(1-b^2)} \\\\\n& = \\sqrt[3]{a^2 \\cdot ab \\cdot b^2} + \\sqrt[3]{(1-a^2)(1-ab)(1-b^2)} \\... | Austria | Austria 2010 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0f7y | Problem:
Let $\{A_1, A_2, \ldots, A_{2m+1}\}$ and $\{B_1, B_2, \ldots, B_{2n+1}\}$ be points in the plane such that the $2m + 2n + 2$ lines $A_1A_2, A_2A_3, \ldots, A_{2m}A_{2m+1}, A_{2m+1}A_1, B_1B_2, B_2B_3, \ldots, B_{2n}B_{2n+1}, B_{2n+1}B_1$ are all different and no three of them are concurrent. Show that we can ... | [] | Soviet Union | 21st ASU | [
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
0gc1 | 設三角形 $ABC$ 的外接圓為 $\Omega$, 外心為 $O$, 垂心為 $H$. 令 $S$ 為 $\Omega$ 上一點, 點 $P$ 在 $BC$ 上使得 $\angle ASP = 90^\circ$, 直線 $SH$ 與 $\triangle APS$ 的外接圓交於 $X \ne S$. 設 $OP$ 分別與 $CA$, $AB$ 交於 $Q$, $R$, $QY$, $RZ$ 為 $\triangle AQR$ 的高。
證明: $X$, $Y$, $Z$ 共線。 | [
"\n\n令 $AD$, $BE$, $CF$ 為 $\\triangle ABC$ 的高, $\\Gamma_A$, $\\Gamma_B$, $\\Gamma_C$ 分別為以 $\\overline{AP}$, $\\overline{BQ}$, $\\overline{CR}$ 為直徑的圓，則 $H$ 關於 $\\Gamma_A$, $\\Gamma_B$, $\\Gamma_C$ 的幂分別為 $HA \\cdot HD$, $HB \\cdot HE$, $HC \\cdot HF$，因此 $H$ 關於三個圓的幂相等。令 $A'$, $B'$ 分別為 $A$, $B$... | Taiwan | 2018 數學奧林匹亞競賽第二階段選訓營, 獨立研究(一) | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Ge... | null | proof only | null | |
0doc | Problem:
Дат је конвексан полиедар са бар 5 темена у чијем се сваком темену састају тачно по три ивице. Доказати да је могуће доделити сваком темену тог полиедра неки рационалан број тако да буду задовољени следећи услови:
(i) бар један од додељених бројева је једнак 2020;
(ii) за сваку страну полиедра, производ броје... | [
"Solution:\n\nОзначимо $c = 2020$. Прво размотримо случај када постоји страна $\\mathcal{F}$ са парним бројем темена. Тада је довољно доделити теменима стране $\\mathcal{F}$ наизменично бројеве $c$ и $\\frac{1}{c}$, а свим осталим теменима полиедра број $1$. Заиста, производ бројева на страни $\\mathcal{F}$ је једн... | Serbia | 14. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0kih | Problem:
A disk of radius $1$ rolls all the way around the inside of a square of side length $s > 4$ and sweeps out a region of area $A$. A second disk of radius $1$ rolls all the way around the outside of the same square and sweeps out a region of area $2A$. The value of $s$ can be written as $a + \frac{b\pi}{c}$, whe... | [] | United States | AMC 10 A | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | 13 | |
096o | Problem:
Fie $I_{n}=n \int_{0}^{1} \frac{\cos x}{1+e^{n x}} d x$, unde $n \in \mathbb{N}^{*}$. Calculați: $\lim _{n \rightarrow \infty} I_{n}$. | [
"Solution:\n\nUtilizăm metoda integrării prin părți și obținem\n$$\n\\begin{gathered}\nI_{n}=n \\int_{0}^{1} \\frac{\\cos x}{1+e^{n x}} d x=n \\int_{0}^{1} \\cos x \\frac{e^{-n x}}{1+e^{-n x}} d x=-\\left.\\cos x \\cdot \\ln \\left(1+e^{-n x}\\right)\\right|_{0} ^{1}- \\\\\n-\\int_{0}^{1} \\sin x \\ln \\left(1+e^{-... | Moldova | Olimpiada Republicană la Matematică | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Limits"
] | null | proof and answer | ln 2 | |
0bgh | Problem:
Se consideră triunghiul $ABC$ şi punctele $D, E \in (BC)$, $F, G \in (CA)$, $H, I \in (AB)$ astfel încât $BD = CE$, $CF = AG$ şi $AH = BI$. Notăm cu $M, N, P$ mijloacele segmentelor $[GH]$, $[DI]$, respectiv $[EF]$ şi cu $M'$ intersecţia dreptelor $AM$ şi $BC$.
a) Arătaţi că
$$
\frac{BM'}{CM'} = \frac{AG}{AH}... | [] | Romania | Olimpiada Naţională de Matematică Etapa Judeţeană | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0707 | Problem:
$ABCD$ is an $n \times n$ board. We call a diagonal row of cells a positive diagonal if it is parallel to $AC$. How many coins must be placed on an $n \times n$ board such that every cell either has a coin or is in the same row, column or positive diagonal as a coin? | [
"Solution:\n\nThere must be at least $n - k$ rows without a coin and at least $n - k$ columns without a coin. Let $r_1, r_2, \\ldots, r_{n-k}$ be cells in the top row without a coin which are also in a column without a coin. Let $r_1, c_2, c_3, \\ldots, c_{n-k}$ be cells in the first column without a coin which are... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | ceil((2n - 1)/3) | |
00sj | A town-planner has built an isolated city whose road network consists of $2N$ roundabouts, each connecting exactly three roads. A series of tunnels and bridges ensure that all roads in the town meet only at roundabouts. All roads are two-way, and each roundabout is oriented clockwise.
Vlad has recently passed his drivi... | [
"$N$ odd. In fact, the number of trajectories has the same parity as $N$.\nThe setting is a (multi)graph where every vertex has degree three. Each vertex has an *orientation*, an ordering of its incident edges. We call Vlad's possible paths *trajectories*, and a *complete trajectory* if he traverses every edge in b... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof and answer | N odd | |
088o | Problem:
Una scacchiera $8$ per $8$ viene riempita con le lettere $A$, $B$, $C$, $D$ in modo che due caselle con un lato o un vertice in comune contengano lettere diverse, e in modo che le lettere $A$ e le lettere $B$ abbiano la proprietà seguente: ogni qual volta una $A$ o una $B$ ha una certa lettera $X$ adiacente i... | [
"Solution:\n\nLa risposta è (A). Numeriamo le righe da $1$ a $8$ e le colonne da $1$ a $8$; consideriamo il quadrato $2 \\times 2$ centrale, quello formato dalle colonne $4$ e $5$ intersecate con le righe $4$ e $5$. Esaminiamo separatamente due casi:\n\na. Nel quadrato mettiamo una $A$ e una $B$ in caselle che s'in... | Italy | Olimpiadi della Matematica - Gara di Febbraio | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | MCQ | A | |
0ddk | Find all positive integers $n$, such that $n$ is a perfect number and $\varphi(n)$ is power of 2. Note: a positive integer $n$ is called perfect if the sum of all its positive divisors is equal to $2n$. | [
"Suppose that $\\varphi(n) = 2^m$ for some $m \\in \\mathbb{Z}^+$. Based on the formula of $\\varphi(n)$, we have\n$$\n\\varphi(n) = \\prod_{p|n} p^{t-1}(p-1) = 2^m\n$$\nwith $v_p(n) = t$. This implies that $t = 1$ for all odd prime divisors, since otherwise, $p \\mid 2^m$, a contradiction. Thus $p-1$ is a power of... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmet... | null | proof and answer | 6 | |
06dv | Denote by $\sigma(n)$ the sum of divisors of $n$. A positive integer $n$ ($n \ge 2$) is redundant if for any integer $k$, with $k < n$, we have $\frac{\sigma(k)}{k} < \frac{\sigma(n)}{n}$. (For instance if $n = 1, 2, 3, 4, 5$, then $\frac{\sigma(n)}{n} = 1, \frac{3}{2}, \frac{4}{3}, \frac{7}{4}, \frac{6}{5}$ respective... | [
"(Superabundant numbers/IMO Shortlist 1983) When $n = m!$, we have\n$$\n\\frac{\\sigma(m!)}{m!} \\ge \\frac{1}{m!} \\left( m! + \\frac{m!}{2} + \\frac{m!}{3} + \\dots + \\frac{m!}{m} \\right) = 1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{m}.\n$$\nSince the harmonic series diverges, the sequence $\\left\\{ ... | Hong Kong | IMO HK TST | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
08vq | A square with side length $3$ is partitioned into a $3 \times 3$ grid with $9$ unit squares. In each of the $9$ unit squares a distinct positive integer greater than or equal to $1$ and less than or equal to $9$ is inserted. For each of the $3$ columns the second largest of the $3$ numbers appearing on that column is m... | [
"First we show that if one of the marked numbers is $5$, then $5$ is the second largest of the three marked numbers. For this purpose, let us note that if $5$ is marked for one of the columns, then there is exactly one number less than or equal to $4$ in this column, and therefore, there must be a column which cont... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 207360 | |
0cr1 | The language of Au tribe contains only two letters $a$ and $u$. Some sequences of these letters are words, and each word contains not less than $1$ and not more than $13$ letters. Moreover, if one writes down two words consecutively (with no space between them) then the resulting sequence is never a word. Find the maxi... | [
"**Answer.** $2^{14} - 2^7 = 16256$.\n\n**First solution.** If all sequences whose number of letters is at least $7$ and at most $13$ are words, then, obviously, the condition of the problem is satisfied; in this case, the number of such words is $2^7 + \\dots + 2^{13} = 2^{14} - 2^7$. It remains to show that this ... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 2^14 - 2^7 = 16256 | |
0kel | Problem:
Let $S$ be a set of intervals defined recursively as follows:
- Initially, $[1,1000]$ is the only interval in $S$.
- If $l \neq r$ and $[l, r] \in S$, then both $\left[l,\left\lfloor\frac{l+r}{2}\right\rfloor\right],\left[\left\lfloor\frac{l+r}{2}\right\rfloor+1, r\right] \in S$.
(Note that $S$ can contain int... | [
"Solution:\nThe answer is given by computing the sum of the lengths of all intervals in $S$ and dividing this value by $1000$, where the length of an interval $[i, j]$ is given by $j-i+1$. An interval may be categorized based on how many times $[1,1000]$ must be split to attain it. An interval that is derived from ... | United States | HMMT February 2020 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 10.976 | |
09ef | Draw a circle $\gamma$ through point $M$ which is the middle point of the arc $BC$ not containing vertex $A$ and the center $I$ of incircle of the triangle $ABC$. If the circle $\gamma$ intersects side $BC$ in points $D$, $E$ and the lines $MD$, $ME$ intersect the circle $\omega$ in points $P$, $Q$ which are different ... | [
"Let's prove that all possible lines $PQ$ pass through the point $I$.\n\n\n\nIn order to prove this we need to prove that $\\angle PIM + \\angle QIM = 180^\\circ$. It is easy see $MI = MB = MC$. It follows from $\\angle IBM = \\angle BIM = \\frac{\\alpha+\\beta}{2}$, $\\angle ICM = \\angle ... | Mongolia | Mongolian 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 > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00j9 | Determine all functions $f$ mapping the integers to the integers with the following property: For any two (not necessarily different) numbers $m$ and $n$, $\text{gcd}(m, n)$ is a divisor of $f(m) + f(n)$. (Note that $\text{gcd}(m, n) = \text{gcd}(|m|, |n|)$ and $\text{gcd}(m, 0) = |m|$ holds for all integers $m$ and $n... | [
"If $t$ is an odd number and we set $m = n = t$, we see that $t|2f(t)$ must hold, which means that $t|f(t)$ must hold for all odd values of $t$.\n\nIf we now set $m = 0$ and $n = t$ (with $t$ still odd), we further see that $t|f(0) + f(t)$ must also hold, which means that $f(0)$ must be divisible by all odd numbers... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All functions f: Z -> Z with n | f(n) for every integer n (equivalently, f(n) = n · g(n) for an arbitrary integer-valued function g, and in particular f(0) = 0). | |
0770 | Let $ABC$ be a triangle with $\angle A = 90^{\circ}$ and $AB < AC$. Let $AD$ be the altitude from $A$ to $BC$. Let $P$, $Q$ and $I$ denote respectively the incentres of triangles $ABD$, $ACD$ and $ABC$. Prove that $AI$ is perpendicular to $PQ$ and $AI = PQ$. | [] | India | INMO-2017 | [
"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 | |
0bnu | Problem:
Határozd meg az összes olyan $k \geq 1$ és $n \geq 2$ természetes számot, amelyre létezik $A, B \in \mathcal{M}_{n}(\mathbb{Z})$ úgy, hogy $A^{3}=O_{n}$ és $A^{k} B+B A=I_{n}$.
Problem:
Determinaţi toate numerele naturale $k \geq 1$ şi $n \geq 2$ cu proprietatea că există $A, B \in \mathcal{M}_{n}(\mathbb{Z... | [
"Solution:\n\nFie $A, B \\in \\mathcal{M}_{n}(\\mathbb{Z})$ astfel încât $A^{3}=O_{n}$ şi $A^{k} B+B A=I_{n}$.\n\nDacă $k \\geq 3$, atunci $B A=I_{n}$ (deoarece $A^{k}=O_{n}$ ), deci $A$ este inversabilă, în contradicţie cu $A^{3}=O_{n}$. \n\nDacă $k=2$ atunci din $A^{2} B+B A=I_{n}$, prin înmulţire la stânga cu $A... | Romania | Olimpiada Naţională de Matematică, Etapa Judeţeană şi a Municipiului Bucureşti | [
"Algebra > Linear Algebra > Matrices"
] | null | proof and answer | k = 1 and n is an even integer at least 2. | |
0d4c | Let $ABC$ be a triangle and let $P$ be a point on $BC$. Points $M$ and $N$ lie on $AB$ and $AC$, respectively, such that $MN$ is not parallel to $BC$ and $AMPN$ is a parallelogram. Line $MN$ meets the circumcircle of $ABC$ at $R$ and $S$. Prove that the circumcircle of triangle $RPS$ is tangent to $BC$. | [
"Because $BM$ and $NP$ are parallel, and $AN$ and $MP$ are parallel, we have\n$$\n\\frac{QP}{QB} = \\frac{NP}{MB} = \\frac{AM}{MB} = \\frac{CP}{PB},\n$$\nand\n$$\n\\frac{QC}{QP} = \\frac{CN}{PM} = \\frac{CN}{NA} = \\frac{CP}{PB}.\n$$\n\n\n\nWe deduce that\n$$\nQP^2 = QB \\cdot QC.\n$$\nBut ... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | English, Arabic | proof only | null | |
0l41 | Problem:
Derek is bored in math class and is drawing a flower. He first draws 8 points $A_{1}, A_{2}, \ldots, A_{8}$ equally spaced around an enormous circle. He then draws 8 arcs outside the circle where the $i$th arc for $i=1,2, \ldots, 8$ has endpoints $A_{i}, A_{i+1}$ with $A_{9}=A_{1}$, such that all of the arcs ... | [
"Solution:\n\n\n\nDraw the centers $O_{1}, \\ldots, O_{8}$ of the arcs, and connect these centers to form a regular octagon as shown. For each $i=1, \\ldots, 8$, extend line $A_{i} O_{i}$ to hit each arc again at $X_{i}$.\n\nThe blue arcs $\\left(\\operatorname{arc} A_{i} X_{i}\\right.$ for... | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 10π | |
0693 | In the triangle $ABC$, in which $\angle A = 60^\circ$, $D \in (BC)$ is such that $AD$ is the internal bisector of angle $\angle A$. Let it be $r_B, r_C$ and $r$, respectively, the inradius of the triangles $ABD$, $ADC$ and $ABC$. Show that $\frac{1}{r_B} + \frac{1}{r_C} = 2\left(\frac{1}{r} + \frac{1}{b} + \frac{1}{c}\... | [
"It is well known that $AD = \\frac{2bc}{b+c} \\cos \\frac{A}{2} = \\frac{bc\\sqrt{3}}{b+c}$. Let $h = AM$ be the length of the altitude from $A$ in the triangle $ABC$. From the theorem of the internal bisector we get\n$$\nBD = \\frac{ac}{b+c}, \\quad CD = \\frac{ab}{b+c}.\n$$\nLet $p_{ABD} = \\frac{BD + DA + AB}{2... | Greece | 21st Mediterranean Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0jdg | Problem:
What is the smallest non-square positive integer that is the product of four prime numbers (not necessarily distinct)? | [
"Solution:\n\nAnswer: $24$\n\nThe smallest two integers that are the product of four primes are $2^{4} = 16$ and $2^{3} \\cdot 3 = 24$. Since $16$ is a perfect square and $24$ is not, the answer is $24$."
] | United States | HMMT November 2013 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 24 | |
0elp | Determine all positive integers $n$ such that $5^n - 1$ can be written as a product of an even number of consecutive integers. | [
"$5^n - 1$ cannot be a product of more than five consecutive integers: since one of the factors would have to be divisible by $5$, this would have to be the case for the product as well, but $5^n - 1$ is not divisible by $5$.\n\nNow we have to consider two cases:\n\nCase 1: If $5^n - 1$ is a product of two consecut... | South Africa | The South African Mathematical Olympiad Third Round | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 2 | |
0jnx | Problem:
Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be a function such that for any integers $x, y$, we have
$$
f\left(x^{2}-3 y^{2}\right)+f\left(x^{2}+y^{2}\right)=2(x+y) f(x-y)
$$
Suppose that $f(n)>0$ for all $n>0$ and that $f(2015) \cdot f(2016)$ is a perfect square. Find the minimum possible value of $f(1)+f(2)$. | [
"Solution:\nPlugging in $-y$ in place of $y$ in the equation and comparing the result with the original equation gives\n$$\n(x-y) f(x+y)=(x+y) f(x-y)\n$$\nThis shows that whenever $a, b \\in \\mathbb{Z}-\\{0\\}$ with $a \\equiv b (\\bmod 2)$, we have\n$$\n\\frac{f(a)}{a}=\\frac{f(b)}{b}\n$$\nwhich implies that ther... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 246 | |
0568 | On a plane a finite number of points are marked of which no three are collinear. Assume that there exists a non-convex polygon with all of its vertices located at some of those points. Prove that there exists a non-convex quadrilateral such that all its vertices lie at the marked points. | [
"There has to be a point inside the convex hull of the set of marked points, otherwise any subset of points would form a convex polygon. Let us partition the convex hull into triangles by drawing a necessary amount of diagonals. As no three points are on the same line, the marked point inside the convex hull must b... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0hnw | Problem:
For any positive integer $n \geq 3$, show that you can write $1$ as a sum of $n$ fractions with numerator $1$ and different denominators. | [
"Solution:\n\nFor $n=3$, we can write\n$$\n1 = \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{6}.\n$$\nFor $n=4$, we can write\n$$\n1 = \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{9} + \\frac{1}{18}.\n$$\nWe claim that we can continue this pattern by breaking the smallest fraction into two smaller fractions, such that the sma... | United States | Berkeley Math Circle | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Other"
] | null | proof only | null | |
080o | Problem:
Sia $ABC$ un triangolo tale che l'angolo $\widehat{ACB} = 60^\circ$. Sia $M$ il punto medio del lato $AB$ e siano $H$ e $K$ i piedi delle altezze che partono da $B$ e da $A$ rispettivamente. Dimostrare che il triangolo $HMK$ è equilatero. | [] | Italy | Progetto Olimpiadi di Matematica 2000 GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07jk | Find all sequences $(a_n)_{n \in \mathbb{N}}$ of positive integers such that for every $n \ge 3$ we have:
$$
\frac{1}{a_1 a_3} + \frac{1}{a_2 a_4} + \frac{1}{a_3 a_5} + \dots + \frac{1}{a_{n-2} a_n} = 1 - \frac{1}{a_1^2 + a_2^2 + \dots + a_{n-1}^2}
$$ | [
"Let $S_{n-2} = \\sum_{k=1}^{n-2} \\frac{1}{a_k a_{k+2}}$ and $Q_m = \\sum_{k=1}^m a_k^2$. The given relation is $S_{n-2} = 1 - \\frac{1}{Q_{n-1}}$ for $n \\ge 3$.\n\nFor $n = 3$:\n$$\n\\frac{1}{a_1 a_3} = 1 - \\frac{1}{a_1^2 + a_2^2} \\quad (1)\n$$\nThis implies $a_1 = 1$. If $a_1 \\ge 2$, then since $a_i \\in \\m... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | a_n is the Fibonacci sequence with a_1 = 1, a_2 = 1 (i.e., a_n = F_n with F_1 = 1, F_2 = 1) | |
045a | In a convex quadrilateral $ABCD$, let $I$ and $J$ be the incenter of $\triangle ABC$ and $\triangle ADC$, respectively. Assume that $IJ$, $AC$, $BD$ meet at point $P$. The line through $P$ perpendicular to $BD$ meets the exterior angle bisectors of $\angle BAD$ at $E$, and meets the exterior angle bisector of $\angle B... | [
"**Proof 1:** If $AB \\parallel CD$ and $AD \\parallel BC$, then $ABCD$ is a parallelogram. In this case, $P$ is the midpoint of $AC$ and $AE \\parallel CF$. So $PE = PF$.\n\nWe may assume that $AB$ is not parallel to $CD$ from now on. We first show that $AB + AD = CB + CD$. As shown in the following picture, we ma... | China | 2022 China Team Selection Test | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0gid | 我們稱一個正整數為傑出數, 若其等於 $1, 2, \ldots, n$ 的最小公倍數, 其中 $n$ 為正整數。找出所有符合 $x + y = z$ 的傑出數 $x$、$y$、$z$。
A positive integer is *superb* if it is the least common multiple of $1, 2, \ldots, n$ for some positive integer $n$. Find all superb $x, y, z$ such that $x + y = z$. | [
"所有可能的 $(x, y, z)$ 為 $x = y = [1, 2, \\ldots, 2^k - 1]$ and $z = [1, 2, \\ldots, 2^k]$, 其中 $k$ 為任意正整數而 $[\\ldots]$ 表最小公倍數。\n\n顯然 $z > x, y$。假設 $x = [1, \\ldots, a]$, $y = [1, \\ldots, b]$ 且 $z = [1, \\ldots, c]$, 並不失一般性假設 $a \\le b$。進一步地, 不失一般性讓 $a$ 最大(也就是對於任何 $a' > a$ 有 $x \\ne [1, \\ldots, a']$), $b$ 最大且 $c$ 最小。這... | Taiwan | Taiwan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | Chinese; English | proof and answer | All solutions are given by: for any positive integer k, x = y = lcm(1,2,\ldots,2^k - 1) and z = lcm(1,2,\ldots,2^k). | |
04h5 | Find all positive integers $n$ such that
$$
a! + b! + c! = 2^n
$$
for some positive integers $a, b, c$. | [
"Without loss of generality, assume $a \\leq b \\leq c$.\n\nIf $c \\geq 6$, then $c!$ is divisible by $2^4 = 16$, and $a!$ and $b!$ are also divisible by $2$ for $a, b \\geq 2$. Thus, $a! + b! + c!$ is divisible by $2$, but for $c \\geq 6$, $c!$ is divisible by $8$ and higher powers of $2$.\n\nLet us check small va... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | [2, 3, 5, 7] | |
0g1o | Problem:
Sei $ABC$ ein Dreieck mit $AB \neq AC$ und Umkreis $k$. Die Tangente an $k$ durch $A$ schneide $BC$ in $P$. Die Winkelhalbierende von $\angle APB$ schneide $AB$ in $D$ und $AC$ in $E$. Zeige, dass das Dreieck $ADE$ gleichschenklig ist. | [
"Solution:\n\nSei $\\alpha := \\angle BAC$ und $\\beta := \\angle CAB$. Mit dem Tangentenwinkelsatz folgt $\\angle CAP = \\beta$. Wir benutzen nun die Innenwinkelsumme im Dreieck $ABP$ und erhalten $\\angle APB = 180^{\\circ} - \\alpha - 2\\beta$. Mit der Winkelhalbierenden folgt $\\angle APD = 90^{\\circ} - \\beta... | Switzerland | SMO - Vorrunde | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0g1f | Problem:
Déterminer la plus petite constante réelle $C$ telle que pour tous $a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \in \mathbb{R}_{>0}$, pas nécessairement distincts, il existe toujours quatre indices distincts $i, j, k, l$ tels que:
$$
\left|\frac{a_{i}}{a_{j}}-\frac{a_{k}}{a_{l}}\right| \leq C
$$ | [
"Solution:\n\nAprès quelques essais, on remarque que $C=1/2$ est optimal.\n\n- $C \\geq 1/2$ : en substituant $(1/2, 1, 1, 1, n)$ et en laissant $n \\rightarrow \\infty$, on obtient $C \\geq 1/2$.\n\n- $C \\leq 1/2$ : supposons que $a_{5} \\geq a_{4} \\geq \\ldots \\geq a_{1}$. On découpe $[0,1]$ en deux sous inter... | Switzerland | IMO-Selektion | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 1/2 | |
07ox | A square $ABCD$ is inscribed in a circle. Let $E$ be the midpoint of $AD$. The line $CE$ meets the circle again at $F$. The lines $FB$ and $AC$ meet at $G$ and the line $GE$ meets the arc $AFD$ of the circle at $K$.
Find, with proof, the measure of $\angle KAD$. | [
"Let $O$ be the intersection point of the diagonals $AC$ and $BD$, which also is the centre of the circle. Because $A$, $B$, $C$, $F$ are on a circle, we have $\\angle BFC = \\angle BAC = 45^\\circ$. Since $\\angle CAD = 45^\\circ$, we obtain $\\angle BFC = \\angle CAD$ and hence $\\angle AEGF$ is cyclic.\n![](atta... | Ireland | Irska 2014 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 15° | |
04az | Find all pairs $(x, y)$ of integers such that $x^3 + x^2 + x = y^2 + y$. | [
"If $y = 0$ or $y = -1$, the right hand side of the given equation is equal to zero so $x^3 + x^2 + x = x(x^2 + x + 1) = 0$, and hence $x = 0$. That gives us two solutions: $(0, 0)$ and $(0, -1)$.\n\nWe shall prove that no other solutions exist.\n\nAssume that $y \\in \\mathbb{Z} \\setminus \\{-1, 0\\}$. Then $y^2 ... | Croatia | Hrvatska 2011 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | [(0, 0), (0, -1)] | |
07ed | Suppose that there are $n$ distinct real numbers on the board. We write all pairwise differences of these numbers and clear all the previous numbers. Prove that if $n$ is odd, it is possible to divide these $\binom{n}{2}$ obtained numbers into two sets with equal sum. | [
"Let $n$ be an odd number and $a_1 < a_2 < \\dots < a_n$ be the given real numbers. Put $a_i - a_j$ (where $i > j$) into the first set if $i$ and $j$ have the same parity and put them into the second set if they have different parity. We claim that\n\nthe coefficient of $a_i$'s in both sets are the same. Suppose $i... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
07fp | $P(x)$ is a non-constant monic polynomial with integer coefficients. Assume that $P_1(x), P_2(x), \dots, P_n(x)$ are monic polynomials with integer coefficients such that for all $1 \le i \le n$, $\deg(P_i) \ge \deg(P)$. We know that for any natural number $x$, there exists a natural number $y$ and an index $i$ ($1 \le... | [
"We firstly prove that there exists an index $i$ such that $\\deg P_i(x) = \\deg P(x)$ and $P(x) = P_i(y)$ for infinitely many $x, y \\in \\mathbb{N}$.\nAssume to the contrary. Let the degree of $P_{t+1}(x), \\dots, P_n(x)$ be equal to $\\deg P(x)$ and $\\deg P_i(x) > \\deg P(x)$ for all $i \\in \\{1, 2, \\dots, t\... | Iran | 37th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
087f | Problem:
Per rubare un prezioso gioiello, un ladro deve scoprire il codice che permette di aprire la porta della cassaforte. Le informazioni che è riuscito a carpire sono le seguenti:
- il codice è un numero
- qualsiasi sottosequenza di cifre consecutive del codice (dunque sia ogni cifra presa singolarmente, che ogni ... | [
"Solution:\n\nLa risposta è $373$.\n\nTroviamo tutti i numeri che verificano le condizioni del testo.\n\nInnanzi tutto notiamo che se abbiamo trovato tutti i numeri accettabili di $n$ cifre, allora un numero accettabile di $n+1$ cifre dovrà necessariamente contenere uno dei numeri di $n$ cifre. In particolare, se s... | Italy | Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 373 | |
0fyh | Problem:
Betrachte die Eckpunkte eines regulären $n$-Ecks und verbinde diese mit Seiten oder Diagonalen irgendwie zu einem geschlossenen Streckenzug, der jede Ecke genau einmal durchläuft. Ein paralleles Paar ist eine Menge von zwei verschiedenen parallelen Strecken in diesem Streckenzug. Zeige:
a. Ist $n$ gerade, da... | [
"Solution:\n\nNummeriere die Ecken der Reihe nach von $1$ bis $n$. Jeder solche Streckenzug entspricht einer Permutation $(a_{1}, \\ldots, a_{n})$ der Menge $\\{1,2, \\ldots, n\\}$. Zwei Strecken $a_{i} a_{i+1}$ und $a_{j} a_{j+1}$ sind dabei genau dann parallel, wenn\n$$\na_{i}+a_{i+1} \\equiv a_{j}+a_{j+1} \\quad... | Switzerland | SMO Finalrunde | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0fu9 | Problem:
Bestimme alle $k$, für die eine natürliche Zahl $n$ existiert, sodass $1^{n}+2^{n}+3^{n}+4^{n}$ mit genau $k$ Nullen endet. | [
"Solution:\n\nSei $A(n)=1^{n}+2^{n}+3^{n}+4^{n}$. Die Beispiele $A(1)=10$, $A(3)=100$ und $A(4)=354$ zeigen, dass $k=0,1,2$ möglich ist. Nehme nun an, dass $A(n)$ mit mindestens drei Nullen endet, dann gilt $8 \\mid A(n)$. Für $n \\geq 3$ ist aber $A(n) \\equiv 1+0+3^{n}+0=4$ oder $2$ $\\pmod{8}$, Widerspruch."
] | Switzerland | SMO Finalrunde | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | k = 0, 1, 2 | |
04cw | Find all integers greater than $1$ whose all divisors written in an increasing sequence make a geometric sequence. | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Exactly the prime powers, i.e., all integers of the form p^k with p prime and k a positive integer. | |
0hed | Let $ABCD$ be a square with the side-length $10$. Let $E$ and $F$ be the points on the sides $BC$ and $AD$ respectively, so that $ABEF$ is a rectangle. The rectangle $KLMN$ is such that its vertices $K$, $L$, $M$ and $N$ belong to the segments $CD$, $DF$, $FE$ and $EC$ respectively. It turns out that the rectangles $AB... | [
"Consider the diagonals $AE$ and $LN$ of the equal rectangles $ABEF$ and $KLMN$. Clearly, $AE = LN$. Let $NX$ be the perpendicular to the segment $AD$ (Fig. 1). Clearly, $NX = AB = 10$. Then $\\triangle AEF = \\triangle NXL$, since they are right triangles with equal hypotenuses and sides. Thus, $AF = LX$. Let $LY ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 5 | |
05dd | Problem:
There are infinitely many people registered on the social network Mugbook. Some pairs of (different) users are registered as friends, but each person has only finitely many friends. Every user has at least one friend. (Friendship is symmetric; that is, if $A$ is a friend of $B$, then $B$ is a friend of $A$.)
... | [
"Solution:\n\nFor any person $A$, let $f^{0}(x)=x$, let $f(A)$ be $A$'s best friend, and define $f^{k+1}(A)=f\\left(f^{k}(A)\\right)$, so any person who is a $k$-best friend is $f^{k}(A)$ for some person $A$; clearly a $k$-best friend is also an $\\ell$-best friend for all $\\ell<k$. Let $X$ be a popular person. Fo... | European Girls' Mathematical Olympiad (EGMO) | European Girls' Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0a3e | Problem:
Laat driehoek $\triangle A B C$ gegeven zijn. Op lijnstuk $B C$ ligt punt $P$, zodanig dat de cirkel met middellijn $B P$ door het middelpunt van de ingeschreven cirkel van $\triangle A B C$ gaat. Bewijs dat
$$
\frac{|B P|}{|P C|}=\frac{c}{s-c}
$$
waarbij $c$ de lengte is van lijnstuk $A B$, en $s$ de helft v... | [
"Solution:\n\nWe introduceren het middelpunt $I$ van de ingeschreven cirkel, en noemen de ingeschreven cirkel zelf $\\omega$. Wegens het gegeven geldt met Thales dat $\\angle B I P=90^{\\circ}$.\n\n\n\na.\nWe introduceren de tweede raaklijn $\\ell$ door $P$ aan $\\omega$ naast $B C$. Merk o... | Netherlands | IMO-selectietoets III | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
... | null | proof only | null | |
09gu | Solve the following system of equations in real numbers:
$$
\begin{cases}
a^2 = b^5 + c^5 \\
b^2 = c^5 + a^5 \\
c^2 = a^5 + b^5.
\end{cases}
$$ | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | a = b = c = 0 or a = b = c = (1/2)^{1/3} | |
0c23 | Problem:
Fie $f:[a, b] \longrightarrow \mathbb{R}$ o funcţie integrabilă, iar $(a_{n})_{n \geq 1}$ un şir de numere reale strict pozitive cu proprietatea că $\lim_{n \rightarrow \infty} a_{n}=0$.
a) Dacă $A=\{m \cdot a_{n} \mid m, n \in \mathbb{N}^{*}\}$, arătaţi că orice interval deschis de numere strict pozitive co... | [
"Solution:\n\na) Deoarece $\\lim_{n \\rightarrow \\infty} a_{n}=0$, pentru orice $c, d>0$, cu $c<d$, există $n \\in \\mathbb{N}^{*}$ cu $a_{n}<d-c$. Pentru $m=\\left\\lfloor\\frac{c}{a_{n}}\\right\\rfloor+1$ rezultă atunci că $m \\cdot a_{n} \\in (c, d) \\cap A$.\n\nb) Funcţia $f$ fiind integrabilă, este mărginită.... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0jnv | Problem:
Unit squares $ABCD$ and $EFGH$ have centers $O_1$ and $O_2$ respectively, and are originally situated such that $B$ and $E$ are at the same position and $C$ and $H$ are at the same position. The squares then rotate clockwise about their centers at the rate of one revolution per hour. After 5 minutes, what is ... | [
"Solution:\n\nThe answer is $\\frac{2-\\sqrt{3}}{4}$.\n\nNote that $AE = BF = CG = DH = 1$ at all times. Suppose that the squares have rotated $\\theta$ radians. Then $\\angle O_1O_2H = \\frac{\\pi}{4} - \\theta = \\angle O_1DH$, so $\\angle HDC = \\frac{\\pi}{4} - \\angle O_1DH = \\theta$. Let $P$ be the intersect... | United States | HMMT November 2015 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | (2 - sqrt(3)) / 4 | |
01js | Find all possible values of the expression
$$
\frac{1}{a}\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{b+c}\right)+\frac{1}{b}\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{c+a}\right)+\frac{1}{c}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{a+b}\right)-\frac{1}{a+b+c}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b}+\frac{1}{b+c}+\fr... | [
"Answer: $1$.\nBy condition, $1/a + 1/b + 1/c = 1$. Let $x = 1/a$, $y = 1/b$, $z = 1/c$. Then\n$$\nx + y + z = 1. \\quad (1)\n$$\nWe rewrite the initial expression as\n$$\n\\begin{align*}\n&+ \\left( \\frac{xyz}{x+y} - \\frac{xyz}{xy+yz+zx} \\cdot \\frac{xy}{x+y} \\right) - \\frac{xyz}{xy+yz+zx} = \\\\\n&= 1 + \\fr... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 1 | |
0bbu | Find all the functions $f: \mathbb{Z} \to \mathbb{N}$ fulfilling the conditions:
i) $f(m+n) = f(m)+f(n)+2mn$, for every integers $m, n$;
ii) $f(f(x))-f(1)$ is a perfect square for each integer $x$. | [
"Let us try to find all functions $f: \\mathbb{Z} \\to \\mathbb{N}$ such that:\n\ni) $f(m+n) = f(m) + f(n) + 2mn$ for all $m, n \\in \\mathbb{Z}$;\nii) $f(f(x)) - f(1)$ is a perfect square for all $x \\in \\mathbb{Z}$.\n\nLet us first solve the functional equation (i).\n\nLet $P(m, n)$ denote the assertion $f(m+n) ... | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | No such function exists. | |
0h87 | Triangle **ABC** is such that $\angle ABC = \angle ACB = 30^\circ$. On the side **BC** point **D** is selected. The point **K** is such that **D** is the midpoint of **AK**. It turned out that $\angle BKA > 60^\circ$. Prove that $3AD < CB$. | [
"Let us select the points $X$ and $Y$ on the side $BC$ such that $BX = AX$ and $CY = YA$. Then $\\angle AXY = \\angle ABC + \\angle BAX = 2\\angle ABC = 60^\\circ$ (fig. 39).\n\nAnalogously, $\\angle XYA = 60^\\circ$ and then $\\triangle XYA$ is equilateral. Then $BX = AX = XY = AY = YC$, i.e. the points $X$, $Y$ d... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
08g4 | Problem:
Su una circonferenza di centro $O$ e diametro $A B$, fissiamo un punto $C$ distinto da $A$ e $B$. Facciamo poi variare un punto $D$, anch'esso distinto da $A$ e $B$, sull'arco $A B$ della circonferenza a cui non appartiene $C$, e definiamo $E$ come il punto del segmento $C D$ tale che le rette $B E$ e $C D$ s... | [
"Solution:\n\nCon riferimento alla figura a sinistra, indichiamo con $\\theta$ l'ampiezza dell'angolo $\\angle B C D$.\n\n\n\nLa soluzione del problema si può suddividere nei seguenti sei passi.\n\n- I punti $B, O, E, D$ stanno su una stessa circonferenza se e solo se $D$ è il punto medio d... | Italy | Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Misce... | null | proof only | null | |
04uc | Consider positive integers $a, b, c$ that are side-lengths of a non-degenerate triangle and such that $\text{GCD}(a, b, c) = 1$ and the fractions
$$
\frac{a^2 + b^2 - c^2}{a + b - c}, \quad \frac{b^2 + c^2 - a^2}{b + c - a}, \quad \frac{c^2 + a^2 - b^2}{c + a - b}
$$
are all integers. Prove that the product of the deno... | [
"Let $z = a + b - c$, $x = b + c - a$, $y = c + a - b$ be the (positive) denominators. Then $a = (y + z)/2$, $b = (x + z)/2$, $c = (x + y)/2$ and\n$$\na^2 + b^2 - c^2 = \\frac{1}{4}((y + z)^2 + (x + z)^2 - (x + y)^2) = \\frac{1}{2}(z(z + x + y) - x y),\n$$\nhence $z \\mid x y$ and likewise $y \\mid x z$ and $x \\mi... | Czech Republic | 67th Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0kca | Problem:
Three distinct vertices of a regular $2020$-gon are chosen uniformly at random. The probability that the triangle they form is isosceles can be expressed as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Compute $100 a + b$. | [
"Solution:\nThe number of isosceles triangles that share vertices with the $2020$-gon is $2020 \\cdot 1009$, since there are $2020$ ways to choose the apex of the triangle and then $1009$ ways to choose the other two vertices. (Since $2020$ is not divisible by $3$, there are no equilateral triangles, so no triangle... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 773 | |
01i9 | Distinct positive integers $a$, $b$, $c$, $d$ satisfy
$$
\begin{cases}
a|b^2 + c^2 + d^2 \\
b|a^2 + c^2 + d^2 \\
c|a^2 + b^2 + d^2 \\
d|a^2 + b^2 + c^2
\end{cases} .
$$
Also it is known that none of them is larger than the product of three others. What is the largest possible number of primes among them? | [
"At first we note that the given condition is equivalent to $a$, $b$, $c$, $d$ dividing $a^2 + b^2 + c^2 + d^2$.\nIt is possible that three of the given numbers are primes, for example for $a = 2$, $b = 3$, $c = 13$ and $d = 26$. In this case $2^2 + 3^2 + 13^2 + 26^2 = 13 \\cdot 66$ which is divisible by all four g... | Baltic Way | Baltic Way 2021 Shortlist | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 3 | |
0iw9 | Problem:
Consider a square, inside which is inscribed a circle, inside which is inscribed a square, inside which is inscribed a circle, and so on, with the outermost square having side length $1$. Find the difference between the sum of the areas of the squares and the sum of the areas of the circles. | [
"Solution:\n\nAnswer: $2 - \\frac{\\pi}{2}$\n\n\n\nThe ratio of the area of each square and the circle immediately inside it is $\\frac{4}{\\pi}$. The total sum of the areas of the squares is $1 + \\frac{1}{2} + \\frac{1}{4} + \\ldots = 2$. Difference in area is then $2 - \\frac{\\pi}{2}$."... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2 - pi/2 | |
00l8 | We are given an acute triangle $ABC$ with $AB > AC$ and orthocenter $H$. The point $E$ lies symmetric to $C$ with respect to the altitude $AH$. Let $F$ be the intersection of the lines $EH$ and $AC$.
Prove that the circumcenter of the triangle $AEF$ lies on the line $AB$. | [
"Let $\\theta$ be the angle between $AF$ and the tangent $t$ at $A$ to the circumcircle of $AEF$. By the inscribed angle theorem, we have $\\angle FEA = \\theta$. Due to the reflection, we have $\\angle ACH = \\angle FEA = \\theta$. Because of $\\angle ACH = \\theta$, the tangent $t$ is parallel to $CH$ and thus or... | Austria | National Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
001n | Sobre una mesa se tienen $n \geq 2$ bolsas de plástico, todas de diferente color. Cada una de ellas está en contacto con la mesa o está adentro de una de las otras bolsas. La operación permitida es elegir una de las bolsas que está en contacto con la mesa y realizar el siguiente intercambio: todas las bolsas que tenga ... | [] | Argentina | XII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | español | proof and answer | (n+1)^{n-1} | |
03ce | Let $m > 2017$ be positive integer and $N = m^{2017} + 1$. The numbers $N, N-m, N-2m, \dots, m+1, 1$ are written (in that order) on the blackboard. On every move the leftmost number is deleted together with all its divisors (if any). Find the last deleted number. | [
"Let $a$ be the smallest number on the board such that $(m+1)a > N$. It is easy to see that $a = \\frac{m^{2017}+m^2+m+1}{m+1}$. We will prove that $a$ is the last deleted number.\n\nThe numbers $2a, 3a, \\dots, ma$ are not on the board since they are not congruent $1$ modulo $m$. Therefore $a$ can not be deleted a... | Bulgaria | BULGARIAN NATIONAL MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | (m^{2017} + m^2 + m + 1) / (m + 1) | |
0c94 | Let $n$ be a non-negative integer, and let $f: [-1, 1] \to \mathbb{R}$ be a twice differentiable function vanishing at the origin. Assuming $f''$ continuous, show that
$$
(2n + 3) \int_{-1}^{1} x^{2n} f(x) \, dx = f''(c), \quad \text{for some } c \text{ in the closed interval } [-1, 1].
$$ | [
"$$\n\\int_{-1}^{1} x^{2n} f(x) \\, dx = \\int_{-1}^{1} x^{2n} g(x) \\, dx + \\int_{-1}^{1} x^{2n} h(x) \\, dx = 2 \\int_{0}^{1} x^{2n} g(x) \\, dx.\n$$\nSince $f(0) = 0$ and $g'(x) = \\frac{1}{2}(f'(x)-f'(-x))$, it follows that $g(0) = g'(0) = 0$; and since $g''(x) = \\frac{1}{2}(f''(x)+f''(-x))$ and $f''$ is cont... | Romania | Romanian Mathematical Olympiad | [
"Calculus > Differential Calculus > Applications",
"Calculus > Differential Calculus > Derivatives",
"Calculus > Integral Calculus > Applications"
] | English | proof only | null | |
074l | Let $a$, $b$, $c$ be integers such that $b$ is even. Suppose the equation $x^3 + a x^2 + b x + c = 0$ has roots $\alpha$, $\beta$, $\gamma$ such that $\alpha^2 = \beta + \gamma$. Prove that $\alpha$ is an integer and $\beta \neq \gamma$. | [
"Let $\\alpha$, $\\beta$, $\\gamma$ be the roots of the cubic. Then we have\n$$\n\\alpha + \\beta + \\gamma = -a, \\quad \\alpha\\beta + \\beta\\gamma + \\gamma\\alpha = b, \\quad \\alpha\\beta\\gamma = -c.\n$$\nThus $\\alpha^2 = \\beta + \\gamma = -a - \\alpha$. We also observe that $\\alpha \\neq 0$; otherwise $c... | India | Indija TS 2010 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
03zm | Line $x - 2y - 1 = 0$ and parabola $y^2 = 4x$ intersect at points $A$, $B$, point $C$ is on the parabola, and $\angle ACB = 90^\circ$. Then the coordinate of $C$ is ______. | [
"Let $A(x_1, y_1)$, $B(x_2, y_2)$, $C(t^2, 2t)$. From\n$$\n\\begin{cases} x - 2y - 1 = 0, \\\\ y^2 = 4x, \\end{cases}\n$$\nwe get $y^2 - 8y - 4 = 0$, which means $y_1 + y_2 = 8$, $y_1 \\cdot y_2 = -4$.\nSince $x_1 = 2y_1 + 1$, $x_2 = 2y_2 + 1$, we have\n$$\n\\begin{aligned}\nx_1 + x_2 &= 2(y_1 + y_2) + 2 = 18, \\\\... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | final answer only | (1, -2) or (9, -6) | |
05h3 | Problem:
On désigne par $K$ la valeur maximale de
$$
\left|x_{1}-x_{2}\right| \cdot\left|x_{1}-x_{3}\right| \cdot\left|x_{1}-x_{4}\right| \cdot\left|x_{2}-x_{3}\right| \cdot\left|x_{2}-x_{4}\right| \cdot\left|x_{3}-x_{4}\right|
$$
ou $x_{1}, x_{2}, x_{3}, x_{4} \in [0 ; 1]$.
a. Prouver que $\frac{4}{243} < K < \frac{... | [
"Solution:\n\na. Soit $x_{1}, x_{2}, x_{3}, x_{4} \\in [0 ; 1]$.\nOn pose\n$$\nf\\left(x_{1}, x_{2}, x_{3}, x_{4}\\right) = \\left|x_{1}-x_{2}\\right| \\cdot \\left|x_{1}-x_{3}\\right| \\cdot \\left|x_{1}-x_{4}\\right| \\cdot \\left|x_{2}-x_{3}\\right| \\cdot \\left|x_{2}-x_{4}\\right| \\cdot \\left|x_{3}-x_{4}\\ri... | France | Olympiades Françaises de Mathématiques, Envoi No. 6 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | sqrt(5)/125 | |
0635 | Problem:
Es seien $a$, $b$ und $c$ drei positive reelle Zahlen mit $a^{2}+b^{2}+c^{2} \geq 3$. Man beweise, dass
$$
\frac{(a+1)(b+2)}{(b+1)(b+5)}+\frac{(b+1)(c+2)}{(c+1)(c+5)}+\frac{(c+1)(a+2)}{(a+1)(a+5)} \geq \frac{3}{2}
$$ | [
"Solution:\n\nDie Funktion $f:(0, \\infty) \\rightarrow \\mathbb{R}, f(x)=\\frac{x+2}{(x+1)(x+5)}$ ist streng monoton fallend und konvex, denn die erste Ableitung\n$$\nf^{\\prime}(x)=-\\frac{x^{2}+4 x+7}{(x+1)^{2}(x+5)^{2}}\n$$\nist negativ, und die zweite Ableitung\n$$\nf^{\\prime \\prime}(x)=\\frac{2\\left(x^{3}+... | Germany | 1. Auswahlklausur | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0jok | Problem:
Let $A_{1}, A_{2}, \ldots, A_{2015}$ be distinct points on the unit circle with center $O$. For every two distinct integers $i, j$, let $P_{i j}$ be the midpoint of $A_{i}$ and $A_{j}$. Find the smallest possible value of
$$
\sum_{1 \leq i<j \leq 2015} O P_{i j}^{2}
$$ | [
"Solution:\nAnswer: $\\frac{2015 \\cdot 2013}{4}$ OR $\\frac{4056195}{4}$\n\nUse vectors. $\\sum\\left|a_{i}+a_{j}\\right|^{2} / 4=\\sum\\left(2+2 a_{i} \\cdot a_{j}\\right) / 4=\\frac{1}{2}\\binom{2015}{2}+\\frac{1}{4}\\left(\\left|\\sum a_{i}\\right|^{2}-\\sum\\left|a_{i}\\right|^{2}\\right) \\geq 2015 \\cdot \\f... | United States | HMMT February | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 4056195/4 | |
0i5j | Problem:
Massachusetts Avenue is ten blocks long. One boy and one girl live on each block. They want to form friendships such that each boy is friends with exactly one girl and vice versa. Nobody wants a friend living more than one block away (but they may be on the same block). How many pairings are possible? | [
"Solution:\n\n89\n\nLet $a_n$ be the number of pairings if there are $n$ blocks; we have $a_1 = 1$, $a_2 = 2$, and we claim the Fibonacci recurrence is satisfied. Indeed, if there are $n$ blocks, either the boy on block 1 is friends with the girl on block 1, leaving $a_{n-1}$ possible pairings for the people on the... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof and answer | 89 | |
0grp | Show that for all positive real numbers $x$, $y$ and $z$
$$
\frac{x^2+1}{(x+y)^2+4(z+1)} + \frac{y^2+1}{(y+z)^2+4(x+1)} + \frac{z^2+1}{(z+x)^2+4(y+1)} \ge \frac{1}{2}
$$ | [
"Note that $(x+y)^2 \\le 2(x^2+y^2)$ and $4z+4 \\le 2(z^2+3)$. Therefore,\n$$\n\\frac{x^2+1}{(x+y)^2+4(z+1)} \\ge \\frac{x^2+1}{2(x^2+y^2+z^2+3)}\n$$\nSimilarly we can write analogous inequalities for pairs $(y, z)$ and $(z, x)$. The sum of these three inequalities yields\n$$\n\\frac{x^2+1}{(x+y)^2+4(z+1)} + \\frac... | Turkey | Team Selection Test | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
01ka | Given a trapezium $ABCD$ ($AD \parallel BC$) with $AD = 3BC$. A circle $\Gamma_1$ with the center $B$ passes through the midpoint of $BD$, and a circle $\Gamma_2$ with the center $C$ passes through the midpoint of $AC$.
Prove that the line through the points of intersection of $\Gamma_1$ and $\Gamma_2$ meets the side $... | [
"Let $R$ be the midpoint of $BD$, $M$ be the midpoint of $AD$, $P$ be the point of intersection of $AC$ and $BD$. We show that $LD = 2LP$.\n\n\n\nBy condition, $BP : PD = 1 : 3$. So $BP = PR = 0.5RD$, hence $LP$ is the median of the triangle $BLP$, and $LR$ is the median of the triangle $BL... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
02i2 | Problem:
Usando apenas os dígitos $1,2,3,4$ e $5$, Peri construiu uma sequência da seguinte forma: um $1$, dois $2$, três $3$, quatro $4$, cinco $5$, seis $1$, sete $2$ e assim por diante; abaixo vemos os primeiros termos desta sequência:
$1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,1,1,1,1,1,1,2,2,2, \ldots$
Qual é o $100^{\circ... | [
"Solution:\n\nAgrupamos a sequência em blocos numerados consecutivamente, cada bloco formado pelos termos iguais consecutivos, como mostrado a seguir.\n$$\n\\underbrace{1}_{\\text{bloco 1}}, \\underbrace{2,2}_{\\text{bloco 2}}, \\underbrace{3,3,3}_{\\text{bloco 3}}, \\underbrace{4,4,4,4}_{\\text{bloco 4}}, \\underb... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 4 | |
0kkp | Problem:
The function $f: \mathbb{Z}^2 \rightarrow \mathbb{Z}$ satisfies
- $f(x, 0) = f(0, y) = 0$, and
- $f(x, y) = f(x-1, y) + f(x, y-1) + x + y$
for all nonnegative integers $x$ and $y$. Find $f(6,12)$. | [
"Solution:\n\nWe claim $f(x, y) = \\binom{x+y+2}{x+1} - (x+y+2)$. Indeed, the hypothesis holds true for our base cases $f(x, 0)$ and $f(0, y)$, and moreover,\n$$\nf(x-1, y) + f(x, y-1) + x + y = \\binom{x+y+1}{x} + \\binom{x+y+1}{x+1} - 2(x+y+1) + x + y = \\binom{x+y+2}{x+1} - (x+y+2).\n$$\nThus, the final answer i... | United States | HMMT November 2021 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 77500 | |
048p | The sum of lengths of legs of the trapezoid is $4\sqrt{10}$, and its height is $6$. The area of the trapezoid is $72$. If a circle can be circumscribed to the trapezoid, determine its radius. | [
"Solution:\n\nLet the trapezoid have bases $a$ and $b$ ($a > b$), and legs $c$ and $d$. Since a circle can be circumscribed about the trapezoid, it must be tangential, so $a + b = c + d$.\n\nGiven:\n- $c + d = 4\\sqrt{10}$\n- $a + b = 4\\sqrt{10}$\n- Height $h = 6$\n- Area $S = 72$\n\nThe area of a trapezoid is $S ... | Croatia | Hrvatska 2011 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 5√2 | |
0is6 | Problem:
Joe B. is frustrated with chess. He breaks the board, leaving a $4 \times 4$ board, and throws 3 black knights and 3 white kings at the board. Miraculously, they all land in distinct squares! What is the expected number of checks in the resulting position? (Note that a knight can administer multiple checks an... | [
"Solution:\n\nWe first compute the expected number of checks between a single knight-king pair. If the king is located at any of the 4 corners, the knight has 2 possible checks. If the king is located in one of the 8 squares on the side of the board but not in the corner, the knight has 3 possible checks. If the ki... | United States | 1st Annual Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 9/5 | |
00kx | Let $k_1$ and $k_2$ be internally tangent circles with common point $X$. Let $P$ be a point lying neither on one of the two circles nor on the line through the two centers. Let $N_1$ be the point on $k_1$ closest to $P$ and $F_1$ be the point on $k_1$ that is farthest from $P$. Analogously, let $N_2$ be the point on $k... | [
"The line segment $N_1F_1$ is a diameter of $k_1$ passing through $P$. Similarly, $N_2F_2$ is a diameter of $k_2$ passing through $P$.\nDue to Thales's theorem, we have $\\angle N_1XF_1 = 90^\\circ$ and $\\angle N_2XF_2 = 90^\\circ$.\nLet $\\angle N_2XF_1 = \\alpha$, we obtain\n$$\n\\angle N_1XN_2 = 90^\\circ - \\a... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0l8n | For every couple of real numbers $(a, b)$, consider the sequence of numbers $\{x_n\}$, $n = 0, 1, 2, \dots$, defined by:
$$
x_0 = a \quad \text{and} \quad x_{n+1} = x_n + b \sin x_n \quad \text{for every } n = 0, 1, 2, \dots
$$
Prove that:
1) for every real number $a$, the sequence $\{x_n\}$ corresponding to $(a, b)$ h... | [
"1).\n+ For $a = k\\pi$ ($k \\in \\mathbb{Z}$), we have $x_n = k\\pi \\quad \\forall n \\in \\mathbb{N}$, therefore $\\lim_{n \\to \\infty} x_n = k\\pi$.\n\n+ For $a \\neq k\\pi$ ($k \\in \\mathbb{Z}$), consider the function $f(x) = x + \\sin x$ defined on $\\mathbb{R}$. We have $f'(x) = 1 + \\cos x \\ge 0 \\quad \... | Vietnam | VIETNAMESE MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | 1) For b = 1, the sequence converges for every real a. Its limit is the odd multiple of pi determined by the interval containing a: if a ∈ (2kπ, (2k+1)π), the limit is (2k+1)π; if a ∈ ((2k−1)π, 2kπ), the limit is (2k−1)π; if a = kπ, the limit is kπ. Equivalently, lim_{n→∞} x_n = (2⌊a/(2π)⌋ + sign(a/(2π)))π.
2) For any ... | |
0fl7 | Problem:
Sea $ABCD$ un cuadrilátero convexo. Sea $P$ la intersección de $AC$ y $BD$. El ángulo $\angle APD = 60^\circ$. Sean $E, F, G$ y $H$ los puntos medios de los lados $AB, BC, CD$ y $DA$ respectivamente. Halla el mayor número real positivo $k$ tal que
$$
EG + 3HF \geq k d + (1-k) s
$$
siendo $s$ el semiperímetro ... | [
"Solution:\n\nProbaremos que $k = 1 + \\sqrt{3}$ y que la igualdad se da si, y sólo si, $ABCD$ es un rectángulo.\n\nSean $W, X, Y$ y $Z$ cuatro puntos exteriores a $ABCD$ de modo que los triángulos $ABW$ y $DCY$ sean equiláteros, el triángulo $BCX$ sea isósceles en $X$, el triángulo $AZD$ s... | Spain | XLVI Olimpiada Matemática Española Fase nacional | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | k = 1 + sqrt(3), with equality if and only if the quadrilateral is a rectangle. | |
0ga1 | 設 $\langle f_n \rangle$ 為費氏數列, 亦即: $f_0 = 0$、$f_1 = 1$; 且對所有非負整數 $n$, $f_{n+2} = f_{n+1} + f_n$ 均成立。
試找出所有的正整數對 $(a, b)$ 滿足 $a < b$, 並且對任意的正整數 $n$, $f_n - 2n \cdot a^n$ 總能被 $b$ 整除。
Let $\langle f_n \rangle$ be the Fibonacci sequence, that is, $f_0 = 0$, $f_1 = 1$, and that $f_{n+2} = f_{n+1} + f_n$ holds for all nonne... | [
"由題設，$b \\mid f_1 - 2a$，即 $b \\mid 1 - 2a$。但因 $b > a$，所以 $b = 2a - 1$。而對任意正整數 $n$，都有\n$$\nb \\mid f_n - 2n a^n, \\quad b \\mid f_{n+1} - 2(n+1)a^{n+1}, \\quad b \\mid f_{n+2} - 2(n+2)a^{n+2}. \\quad (1)\n$$\n由此三式，加上 $f_{n+2} = f_{n+1} + f_n$ 以及 $b = 2a-1$ 等條件，知\n$$\nb \\mid (n + 2)a^{n+2} - (n + 1)a^{n+1} - n a^n.\... | Taiwan | 二〇一六數學奧林匹亞競賽第二階段選訓營 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | (3, 5) | |
0gm6 | Let $p$ be a prime number. Determine the maximal degree of a polynomial $T(x)$ whose coefficients belong to $\{0,1,...,p-1\}$ whose degree is less than $p$, and which satisfies
$$
T(n) \equiv T(m) \pmod{p} \implies n \equiv m \pmod{p}
$$
for all integers $n, m$. | [] | Turkey | VIIth NATIONAL MATHEMATICAL OLYMPIAD | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof and answer | p−2 for p≥3; for p=2, the maximal degree is 1 | |
0gnc | Find all positive odd integers $n$ for which there exist odd integers $x_1, x_2, \dots, x_n$ such that
$$
x_1^2 + x_2^2 + \dots + x_n^2 = n^4.
$$ | [
"Since $n$ is odd, $n^4 \\equiv 1 \\pmod{8}$. Since $x_i$ is odd, $x_i^2 \\equiv 1 \\pmod{8}$ for $1 \\le i \\le n$. Hence $n = x_1^2 + x_2^2 + \\dots + x_n^2 \\equiv n^4 \\equiv 1 \\pmod{8}$.\n\nOn the other hand, if $n \\equiv 1 \\pmod{8}$, then odd numbers $x_1, x_2, \\dots, x_n$ satisfying the required equality... | Turkey | Team Selection Test for IMO 2007 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | All positive odd integers congruent to 1 modulo 8. | |
0kwd | Problem:
Let $L O V E R$ be a convex pentagon such that $L O V E$ is a rectangle. Given that $O V = 20$ and $L O = V E = R E = R L = 23$, compute the radius of the circle passing through $R$, $O$, and $V$. | [
"Solution:\n\n\n\nLet $X$ be the point such that $R X O L$ is a rhombus. Note that line $R X$ defines a line of symmetry on the pentagon $L O V E R$. Then by symmetry $R X V E$ is also a rhombus, so $R X = O X = V X = 23$. This makes $X$ the center of the circle, and the radius is $23$."
] | United States | HMMT November | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 23 | |
025z | Problem:
Brincando com números - A soma $1+1+4$ dos algarismos do número $114$ divide o próprio número. Qual é o maior número, menor do que $900$, que satisfaz esta propriedade? | [
"Solution:\n\nComo queremos encontrar o maior número possível, menor do que $900$, iniciaremos com o algarismo $8$ na casa da centena. Observemos que o número $800$ satisfaz a propriedade. Logo, o número procurado é maior que ou igual a $800$.\n\nDevemos então encontrar $a$ e $b$ tais que $8+a+b$ divida $8ab = 800 ... | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 888 | |
0kpu | Problem:
Alice is thinking of a positive real number $x$, and Bob is thinking of a positive real number $y$. Given that $x^{\sqrt{y}}=27$ and $(\sqrt{x})^{y}=9$, compute $x y$. | [
"Solution:\n\nNote that\n$$\n27^{\\sqrt{y}}=\\left(x^{\\sqrt{y}}\\right)^{\\sqrt{y}}=x^{y}=(\\sqrt{x})^{2 y}=81,\n$$\nso $\\sqrt{y}=4 / 3$ or $y=16 / 9$. It follows that $x^{4 / 3}=27$ or $x=9 \\sqrt[4]{3}$. The final answer is $9 \\sqrt[4]{3} \\cdot 16 / 9=16 \\sqrt[4]{3}$."
] | United States | HMMT November 2022 | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 16*3^(1/4) | |
0d9e | Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = 6$. Prove that
$$
\frac{1}{a^2 b + 16} + \frac{1}{b^2 c + 16} + \frac{1}{c^2 a + 16} \geq \frac{1}{8}
$$ | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null |
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