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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
04w2 | We shall say that an odd prime $p$ is *kooky* if the sum of all primes smaller than $p$ is a multiple of $p$. Can two consecutive primes be kooky? | [
"We shall prove that there are no consecutive kooky primes.\nOrder all the primes in an increasing sequence $p_1 = 2 < p_2 = 3 < p_3 < \\dots$ and, for the sake of contradiction, suppose that $p_n$ and $p_{n+1}$ are both kooky for some $n > 1$. This means that there exist positive integers $a$ and $b$ such that\n$$... | Czech Republic | First Round of the 73rd Czech and Slovak Mathematical Olympiad (take-home part) | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | No | |
0dx8 | Problem:
Poišči vsa realna števila $x$, ki zadoščajo enačbi $\sqrt{x^{2}-2 x+17}=3 \sqrt{3}-1$, in jih zapiši v obliki $x=m+n \sqrt{3}$, kjer sta $m$ in $n$ celi števili. | [
"Solution:\n1. Ker sta obe strani enačbe pozitivni, lahko enačbo kvadriramo in dobimo\n$$\nx^{2}-2 x+17=27-6 \\sqrt{3}+1=28-6 \\sqrt{3}\n$$\nNato levo stran dopolnimo do popolnega kvadrata\n$$\n(x-1)^{2}+16=28-6 \\sqrt{3}\n$$\nin enačbo preuredimo do\n$$\n(x-1)^{2}=12-6 \\sqrt{3}\n$$\nOd tod po korenjenju dobimo $x... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Equations and Inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | x = 4 - sqrt(3) or x = -2 + sqrt(3) | |
03js | Problem:
Given the numbers $1, 2, 2^{2}, \ldots, 2^{n-1}$. For a specific permutation $\sigma = X_{1}, X_{2}, \ldots, X_{n}$ of these numbers we define $S_{1}(\sigma) = X_{1}$, $S_{2}(\sigma) = X_{1} + X_{2}$, $S_{3}(\sigma) = X_{1} + X_{2} + X_{3}$, $\ldots$ and $Q(\sigma) = S_{1}(\sigma) S_{2}(\sigma) \cdots S_{n}(\s... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2^{-n(n-1)/2} | |
0av8 | Problem:
Last January 7, 2016, a team from the University of Central Missouri headed by Curtis Cooper discovered the largest prime number known so far:
$$
2^{74,207,281}-1
$$
which contains over 22.3 million digits. Curtis Cooper is part of a large collaborative project called GIMPS, where mathematicians use their comp... | [
"Solution:\nGreat Internet Mersenne Prime Search"
] | Philippines | 18th PMO National Stage Oral Phase | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | Great Internet Mersenne Prime Search | |
0bft | Given six points on a circle, $A$, $a$, $B$, $b$, $C$, $c$, show that the Pascal lines of the hexagrams $AaBbCc$, $AbBcCa$, $AcBaCb$ are concurrent.
 | [
"The lines $Aa$ and $bC$ meet at $D$, and the lines $Bb$ and $cA$ meet at $D'$ to determine the Pascal line of the hexagram $AaBbCc$; similarly, the lines $Bc$ and $aA$ meet at $E$, and the lines $Ca$ and $bB$ meet at $E'$ to determine the Pascal line of the hexagram $AbBcCa$; finally, the lines $Cb$ and $cB$ meet ... | Romania | The Danube Mathematical Competition | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem"
] | null | proof only | null | |
0jaa | Problem:
For each positive integer $n$, there is a circle around the origin with radius $n$. Rainbow Dash starts off somewhere on the plane, but not on a circle. She takes off in some direction in a straight path. She moves $\frac{\sqrt{5}}{5}$ units before crossing a circle, then $\sqrt{5}$ units, then $\frac{3 \sqrt... | [
"Solution:\n\nAnswer: $\\frac{2 \\sqrt{170}-9 \\sqrt{5}}{5}$ Note that the distance from Rainbow Dash's starting point to the first place in which she hits a circle is irrelevant, except in checking that this distance is small enough that she does not hit another circle beforehand. It will be clear at the end that ... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (2\sqrt{170}-9\sqrt{5})/5 | |
08gm | Problem:
Gli aventi diritto di voto sull'isola di Cavalfurfandia sono $5000$; ciascuno è un cavaliere, nel qual caso rilascia solo dichiarazioni vere, o un furfante, nel qual caso rilascia solo dichiarazioni false. Tremila abitanti dichiarano che voteranno Carla, duemila che voteranno Flavia. Ciascuno dei tremila che ... | [
"Solution:\n\nLa risposta è (C). Osserviamo che abbiamo due casi distinti a seconda che l'affermazione \"Tutti quelli che dichiarano di votare per Flavia sono dei furfanti\" sia vera o falsa:\n\n- se l'affermazione è vera, allora tutti coloro che l'hanno rilasciata (che sono le $3000$ persone che hanno dichiarato d... | Italy | Olimpiadi di Matematica | [
"Discrete Mathematics > Logic"
] | null | MCQ | C | |
0bik | Given an odd prime $p$, determine all polynomials $f$ and $g$ with integral coefficients satisfying the condition $f(g(X)) = \sum_{k=0}^{p-1} X^k$. | [
"More generally, let $k$ be an integer greater than $1$ and let $P$ be a polynomial of degree at most $k-2$ with integral coefficients. The polynomials $f$ and $g$ with integral coefficients satisfying the condition $f(g(X)) = X^k + X^{k-1} + P(X)$ are: $f(X) = (\\pm X \\mp a)^k + (\\pm X \\mp a)^{k-1} + P(\\pm X \... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Abstract Algebra > Field Theory",
"Number Theor... | null | proof and answer | All integer‑coefficient solutions are exactly the two families with consistent signs:
1) f(X) = ± X ∓ a and g(X) = ± Φ_p(X) + a, where a is any integer;
2) f(X) = Φ_p(± X ∓ b) and g(X) = ± X + b, where b is any integer.
Here Φ_p(X) = 1 + X + X^2 + … + X^{p−1} is the p‑th cyclotomic polynomial, and the ± signs in each p... | |
05ld | Problem:
Soit $n > 0$ un entier, et $a, b, c$ des entiers strictement positifs tels que
$$
(a + b c)(b + a c) = 19^{n}
$$
Prouver que $n$ est pair. | [
"Solution:\nProcédons par l'absurde, et supposons qu'il existe un triplet $(a, b, c)$ tel que $(a + b c)(b + a c) = 19^{n}$, où $n$ est un entier impair. Sans perte de généralité, on peut supposer que $a \\geqslant b$ et que la somme $a + b + c$ est minimale.\n\nPuisque $19$ est premier, on sait que $b + a c \\geqs... | France | Olympiades Françaises de Mathématiques - Test de Février | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysi... | null | proof only | null | |
0dzo | Problem:
Igralca imata kup enakih žetonov, s katerega izmenično jemljeta po enega in ga postavljata na poljubno prazno polje kvadratne tabele velikosti $2008 \times 2008$. Zmaga tisti, ki prvi postavi žeton tako, da skupaj s tremi drugimi tvori oglišča pravokotnika, ki ima stranice vzporedne stranicam tabele. Kateri i... | [
"Solution:\n\nZmagovalno strategijo ima drugi igralec. Stolpce razdeli v $1004$ parov in sicer sta v prvem paru stolpca $1$ in $2$, v drugem stolpca $3$ in $4$, in tako naprej, do zadnjega para, v katerem sta stolpca $2007$ in $2008$. Kadarkoli da prvi igralec žeton v enega izmed stolpcev, da drugi igralec žeton v ... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | second player | |
0cmt | Given integers $x, y \in [2, 100]$, prove that there exists a positive integer $n$ such that the number $x^{2^n} + y^{2^n}$ is composite.
(S. Berlov, A. Kanel-Belov) | [
"If $x = y$, then $n = 1$ works, since $x^2 + y^2$ is an even number greater than $2$.\n\nNow assume $x \\neq y$. In this case, we will show that for some $n$, the number $x^{2^n} + y^{2^n}$ is divisible by $257$ and not equal to $257$. Then it will be composite, as required.\n\nSuppose $x^{2^n} + y^{2^n} = 257$. L... | Russia | Russian mathematical olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English; Russian | proof only | null | |
0km5 | Problem:
Suppose that $m$ and $n$ are positive integers with $m < n$ such that the interval $[m, n)$ contains more multiples of $2021$ than multiples of $2000$. Compute the maximum possible value of $n - m$. | [
"Solution:\nLet $a = 2021$ and $b = 2000$. It is clear that we may increase $y - x$ unless both $x - 1$ and $y + 1$ are multiples of $b$, so we may assume that our interval is of length $b(k + 1) - 1$, where there are $k$ multiples of $b$ in our interval. There are at least $k + 1$ multiples of $a$, and so it is of... | United States | HMMT Spring 2021 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 191999 | |
0cq1 | Let $\triangle ABC$ be an isosceles triangle with $AB = AC$. Point $D$ is chosen on the smaller arc $AB$ of the circumcircle of $ABC$. Point $E$ is chosen on the ray $AD$ outside the segment $AD$ so that points $A$ and $E$ lie on the same side of line $BC$. The circumcircle of triangle $BDE$ intersects side $AB$ at poi... | [
"Обозначим через $\\Omega$ и $\\omega$ описанные окружности треугольников $ABC$ и $BDE$. Положим $\\angle ACB = \\angle ABC = \\alpha$. Четырёхугольник $BDAC$ вписан в $\\Omega$, значит, $\\angle ADB = 180^\\circ - \\angle ACB = 180^\\circ - \\alpha$. Углы $ADB$ и $EDB$ — смежные, откуда $\\angle EDB = 180^\\circ -... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Russian | proof only | null | |
05a9 | In an unopened pack, there are red, green and blue candies that weigh $2$ g, $5$ g and $25$ g respectively. There are the same number of candies of each color. After Mary eats some candies from that pack, the remaining candies weigh exactly $787$ g in total. Find the least number of candies that Mary could have eaten. | [
"Let there be $n$ candies of each color in the unopened pack. Since one red, one green and one blue candy weigh $32$ grams in total, the total weight of all candies in the unopened pack is $32n$ grams. Therefore Mary eats $32n - 787$ or $32(n - 25) + 13$ grams of candies.\n\nNote that if $n = 27$, then Mary could h... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 4 | |
01wx | Let $a_0, a_1, a_2, \dots$ be a sequence of real numbers such that $a_0 = 0$, $a_1 = 1$, and for every $n \ge 2$ there exists $1 \le k \le n$ satisfying
$$
a_n = \frac{a_{n-1} + \dots + a_{n-k}}{k}.
$$
Find the maximal possible $a_{2018} - a_{2017}$. | [
"See IMO-2018 Shortlist, Problem A4."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 1/4034 | |
043y | Given positive integers $a$, $b$, $c$ which are pairwise coprime. Let $f(n)$ represent the number of nonnegative integer solutions $(x, y, z)$ of the equation $a x + b y + c z = n$. Prove: there exist real constants $\alpha$, $\beta$, $\gamma$, such that for every nonnegative real number $n$,
$$
|f(n) - (\alpha n^2 + \... | [
"Consider the generating function of $\\{f(n)\\}$:\n$$\n\\begin{align*}\nG(t) &= \\sum_{n=0}^{\\infty} f(n) t^n \\\\\n&= (1 + t^a + t^{2a} + \\dots)(1 + t^b + t^{2b} + \\dots)(1 + t^c + t^{2c} + \\dots) \\\\\n&= \\frac{1}{(1 - t^a)(1 - t^b)(1 - t^c)}.\n\\end{align*}\n$$\nSince $a$, $b$, $c$ are pairwise coprime, th... | China | China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Complex numbers",
"Number Theory > Diophantine Equations > Techniques: modulo... | null | proof only | null | |
0gcg | 令 $\mathbb{N}$ 表示所有正整數之集合。試求所有函數 $f : \mathbb{N} \rightarrow \mathbb{N}$ 滿足
$$
f(x + y(f(x))) = x + f(x)f(y)
$$
對於所有正整數 $x, y$ 皆成立。
令 $\mathbb{N}$ 表示所有正整數之集合。試求所有函數 $f : \mathbb{N} \to \mathbb{N}$ 滿足
$$
f(x + y(f(x))) = x + f(x)f(y)
$$
對於所有正整數 $x, y$ 皆成立。 | [
"首先，當 $x \\neq z \\equiv x \\pmod{f(x)}$ 時，我們有 $f(z) \\equiv x \\pmod{f(x)}$ 那麼在原式中用 $x + f(x)$ 跟 $f(x)$ 取代 $x, y$ 會得到\n$$\nf(x + f(x) + f(x)f(x + f(x))) = x + f(x) + f(x + f(x))f(f(x))\n$$\n兩邊 mod $f(x)$ 可知一個重要的關係式\n$$\nf(x) \\mid x f(f(x)) \\quad (1)\n$$\n特別地，我們有\n$$\nf(x + y f(x)) \\mid (x + y f(x)) f(f(x + y f(... | Taiwan | 二〇一八數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | f(x) = x for all positive integers x | |
0496 | Given nine positive integers, none of which has a prime divisor greater than $6$, prove that there are two of them whose product is a perfect square. | [] | Croatia | Hrvatska 2011 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
031n | Problem:
Find all real numbers $a$ such that $4[a n]=n+[a[a n]]$ for any positive integer $n$ ($[x]$ denotes the largest integer less than or equal to $x$). | [
"Solution:\nIt follows by the given condition that $4(a n-1)<n+a(a n)$ and $4 a n>n+a(a n-1)-1$, i.e.,\n$$\n1+a^2-\\frac{a+1}{n}<4 a<1+a^2+\\frac{4}{n}\n$$\nLetting $n \\rightarrow \\infty$ gives $1+a^2=4 a$, so $a=2-\\sqrt{3}$ or $a=2+\\sqrt{3}$.\nThe given equality for $n=1$ yields that the first case is impossib... | Bulgaria | 52. Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2+\sqrt{3} | |
0a0z | In a square, we draw the inscribed circle, into which we draw an inscribed square whose sides are parallel to the sides of the original square. We colour the area between these squares dark grey. In the smaller square, we again draw the inscribed circle containing another inscribed square with parallel sides; the area ... | [
"$\\frac{1}{2}$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 1/2 | |
0gzx | Find all the value of parameter $a$, at which the equation $x^2 - 3x[x] + 2x = a$ has two positive roots. | [
"Let's build the graph of the equation, namely the function $y = x^2 - 3x[x] + 2x = a$. It is easy to understand that we are interested only when $x > 0$.\n$$\n\\begin{align*}\nx \\in [0, 1) &\\Rightarrow [x] = 0 \\Rightarrow y = x^2 + 2x \\\\\nx \\in [1, 2) &\\Rightarrow [x] = 1 \\Rightarrow y = x^2 - x \\\\\nx \\... | Ukraine | The Problems of Ukrainian Authors | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (0, 2) ∪ (-49/4, -12) | |
0e5z | Problem:
Naj bo $p$ polinom stopnje $2$, katerega vsaj en koeficient ni celo število. Denimo, da je za vsako celo število $n$ tudi število $p(n)$ celo. Dokaži, da ima tedaj polinom $q(x) = p(x) - \frac{1}{2} x(x+1)$ same celoštevilske koeficiente. | [
"Solution:\n\nOznačimo $P(x) = \\alpha x^{2} + \\beta x + \\gamma$. Tedaj morajo biti števila $P(0) = \\gamma$, $P(1) - P(0) = (\\alpha + \\beta + \\gamma) - \\gamma = \\alpha + \\beta$ in $P(2) - 2 P(1) + P(0) = (4\\alpha + 2\\beta + \\gamma) - 2(\\alpha + \\beta + \\gamma) + \\gamma = 2\\alpha$ cela.\n\nOznačimo ... | Slovenia | 56. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0ca3 | Problem:
Se consideră patrulaterul convex $ABCD$ în care unghiurile $A$ şi $C$ nu sunt ascuţite. Pe laturile $AB$, $BC$, $CD$ şi $DA$ se iau punctele $K$, $L$, $M$, respectiv $N$. Demonstraţi că perimetrul patrulaterului $KLMN$ este cel puţin egal cu dublul lungimii diagonalei $AC$. | [] | Romania | Al doilea test de selecţie pentru OBMJ | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
07nd | $ABC$ is an equilateral triangle. $D$ is the midpoint of $BC$, $E$ is a point on $AC$ between $A$ and $C$ and $F$ is a point on $AB$ between $A$ and $B$. The area of $\triangle EDC$ is $24\sqrt{3}$, the area of $\triangle AFE$ is $24\sqrt{3}$ and the area of $\triangle FBD$ is $54\sqrt{3}$. Find the length of the sides... | [
"Let $x = |BD| = |DC|$. Because $\\triangle ABC$ is equilateral, considering the areas of $\\triangle BDF$ and $\\triangle DCE$ gives\n$$\n54\\sqrt{3} = \\frac{1}{2}x \\cdot |BF| \\cdot \\sin(60^\\circ) = \\frac{1}{4}\\sqrt{3} \\cdot x \\cdot |BF| \\quad \\Rightarrow \\quad x \\cdot |BF| = 216\n$$\n$$\n24\\sqrt{3} ... | Ireland | Ireland | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 24 | |
062e | Problem:
Man bestimme alle positiven ganzen Zahlen $k$ und $n$ mit folgender Eigenschaft:
Die Zahl $k^{4}+n^{2}$ ist ohne Rest durch die Zahl $7^{k}-3^{n}$ teilbar. | [
"Solution:\n\nDas einzige Lösungspaar ist $(k ; n)=(2 ; 4)$. Für jedes Lösungspaar sind $k^{4}+n^{2}$ und $7^{k}-3^{n}$ ganzzahlig, wobei $7^{k}-3^{n}$ nicht notwendig positiv sein muss. Weil $7^{k}$ und $3^{n}$ ungerade sind, gilt $2 \\mid 7^{k}-3^{n}$ und daher $2 \\mid k^{4}+n^{2}$. Deshalb müssen $k$ und $n$ en... | Germany | 2. IMO-Auswahlklausur | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2, 4) | |
0go6 | Let $P$ be a point in the interior of a triangle $ABC$ that is not on the median belonging to $A$ and satisfying $\angle CAP = \angle BCP$. Let $BP \cap CA = \{B'\}$ and $CP \cap AB = \{C'\}$. Let $Q$ be the second point of intersection of the line $AP$ and the circumcircle of $ABC$, $R$ be the point of intersection of... | [
"Since $PS \\parallel AC$ and $\\angle CAP = \\angle BCP$ we have $\\angle QPC = \\angle ACB = \\angle AQB = \\angle PQB$. Therefore $BQ \\parallel PC$.\n\nWe will first show that $SQ = RB'$ if and only if $AB \\parallel B'Q$. Since $PS$ and $AB'$ are parallel, we have $\\frac{PQ}{PA} = \\frac{SQ}{SB'}$. If $SQ = R... | Turkey | 18th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
0abv | The vertices $A$ and $B$ of an equilateral triangle $ABC$ lie on a circle $k$ of radius $1$, and the vertex $C$ is in the interior of circle $k$. A point $D$, different from $B$, lies on $k$ so that $AD = AB$. The line $DC$ intersects $k$ for the second time at point $E$. Find the length of the line segment $CE$. | [
"As $AD = AC$, $\\triangle CDA$ is isosceles. If $\\angle ADC = \\angle ACD = \\alpha$ and $\\angle BCE = \\beta$ then $\\beta = 120^\\circ - \\alpha$. The quadrilateral $ABED$ is cyclic, so $\\angle ABE = 180^\\circ - \\alpha$. Then $\\angle CBE = 120^\\circ - \\alpha$ so $\\angle CBE = \\beta$. Thus $\\triangle C... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 1 | |
06m5 | Let $n$ be an integer, and let $A$ be a subset of $\{0, 1, 2, 3, \dots, 5^n\}$ consisting of $4n+2$ numbers. Prove that there exist $a, b, c \in A$ such that $a < b < c$ and $c + 2a > 3b$. | [
"(IMO Shortlist 2021 A1) See the official solution."
] | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0gl7 | Let $ABC$ be a triangle with $AB \neq AC$. Let the angle bisector of $\angle BAC$ intersect $BC$ at $P$, and intersect the perpendicular bisector of segment $BC$ at $Q$. Prove that $\frac{PQ}{AQ} = \left(\frac{BC}{AB+AC}\right)^2$. | [
"We first show that $Q$ lies on the circumcircle of triangle $ABC$.\n\n\n\nLet the perpendicular bisector of $BC$ intersect the circumcircle of $ABC$ at $Q'$. Since $Q'$ bisects the arc $BC$ we have $\\angle BAQ' = \\angle CAQ'$. Thus $Q'$ lies on the angle bisector of $\\angle BAC$. Hence ... | Thailand | The 13th Thailand Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0irm | Problem:
Cyclic pentagon $ABCD E$ has side lengths $AB = BC = 5$, $CD = DE = 12$, and $AE = 14$. Determine the radius of its circumcircle. | [
"Solution:\n\nLet $C'$ be the point on minor arc $BCD$ such that $BC' = 12$ and $C'D = 5$, and write $AC' = BD = C'E = x$, $AD = y$, and $BD = z$. Ptolemy applied to quadrilaterals $ABC'D$, $BC'DE$, and $ABDE$ gives\n$$\n\\begin{aligned}\n& x^2 = 12y + 5^2 \\\\\n& x^2 = 5z + 12^2 \\\\\n& yz = 14x + 5 \\cdot 12\n\\e... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"... | null | proof and answer | 225 sqrt(11) / 88 | |
0irv | Let $\mathcal{P}$ be a convex polygon with $n$ sides, $n \ge 3$. Any set of $n-3$ diagonals of $\mathcal{P}$ that do not intersect in the interior of the polygon determine a *triangulation* of $\mathcal{P}$ into $n-2$ triangles. If $\mathcal{P}$ is regular and there is a triangulation of $\mathcal{P}$ consisting of onl... | [
"**Solution.** The answer is $n = 2^{m+1} + 2^k$, where $m$ and $k$ are nonnegative integers. In other words, $n$ is either a power of 2 (when $m+1=k$) or the sum of two nonequal powers of 2 (with $1=2^0$ being considered as a power of 2).\nWe start with the following observation.\n**Lemma.** Let $Q = Q_0Q_1\\dots ... | United States | USAMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | n = 2^{m+1} + 2^k for nonnegative integers m and k (i.e., n is a power of two or the sum of two distinct powers of two, counting 1 = 2^0). | |
06i2 | If $x$ and $y$ are real numbers, find the minimum value of $\sqrt{4+y^2}+\sqrt{x^2+y^2-4x-4y+8}+\sqrt{x^2-8x+17}$.
若 $x$、$y$ 為實數, 求 $\sqrt{4+y^2}+\sqrt{x^2+y^2-4x-4y+8}+\sqrt{x^2-8x+17}$ 的最小值。 | [] | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English; Chinese | proof and answer | 5 | |
07kl | Circles $S$ and $T$ intersect at $P$ and $Q$, with $S$ passing through the centre of $T$. Distinct points $A$ and $B$ lie on $S$, inside $T$, and are equidistant from the centre of $T$. The line $PA$ meets $T$ again at $D$. Prove that $|AD| = |PB|$. | [
"Let $O$ be the centre of the circle $T$. Extend $PB$ to meet the circle at $C$. Let $\\angle QPA = \\alpha$ and let $\\angle BPO = \\beta$. Since $|AO| = |OB|$ the join of the two centres is perpendicular to $AB$. The join of the two centres is also perpendicular to $PQ$. Thus $PQ \\parallel AB$. Hence $\\angle BA... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
075x | Let $h \ge 3$ be an integer and $X$ the set of all positive integers that are greater than or equal to $2h$. Let $S$ be a nonempty subset of $X$ such that the following two conditions hold:
* if $a + b \in S$ with $a \ge h, b \ge h$, then $ab \in S$;
* if $ab \in S$ with $a \ge h, b \ge h$, then $a+b \in S$.
Prove tha... | [
"Let $f: X \\to \\{0, 1\\}$ be such that $f(x) = 1$ if and only if $x \\in S$. Then $f(a+b) = f(ab)$ whenever $a \\ge h, b \\ge h$. If $a \\ge h+2$ then\n$$\nf(2a - 1) = f(a^2 - a) = f(a^3 - 2a^2) = f(a^2 + a - 2) = f(2a + 1).\n$$\nGiven $n \\ge 2h$, it is easy to see that there exists an $a \\ge h+2$ such that $f(... | India | Indija TS 2013 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
08qj | Problem:
Let $\triangle ABC$ be a right-angled triangle with $\angle BAC = 90^\circ$ and let $E$ be the foot of the perpendicular from $A$ on $BC$. Let $Z \neq A$ be a point on the line $AB$ with $AB = BZ$. Let $(c)$ be the circumcircle of the triangle $\triangle AEZ$. Let $D$ be the second point of intersection of $(... | [
"Solution:\n\nWe will first show that $PA$ is tangent to $(c)$ at $A$.\nSince $E, D, Z, A$ are concyclic, then $\\angle EDC = \\angle EAZ = \\angle EAB$. Since also the triangles $\\triangle ABC$ and $\\triangle EBA$ are similar, then $\\angle EAB = \\angle BCA$, therefore $\\angle EDC = \\angle BCA$.\nSince $\\ang... | JBMO | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ... | null | proof only | null | |
0ajs | Let $x$, $y$, $z$ be positive real numbers. Prove that
$$
\sqrt{\frac{xy}{x^2 + y^2 + 2z^2}} + \sqrt{\frac{yz}{y^2 + z^2 + 2x^2}} + \sqrt{\frac{zx}{z^2 + x^2 + 2y^2}} \le \frac{3}{2}.
$$ | [
"We have\n$$\n\\begin{aligned}\n& \\sqrt{\\frac{xy}{x^2 + y^2 + 2z^2}} + \\sqrt{\\frac{yz}{y^2 + z^2 + 2x^2}} + \\sqrt{\\frac{zx}{z^2 + x^2 + 2y^2}} \n\\le \\\\\n& \\sqrt{\\frac{xy}{xy + yz + zx + z^2}} + \\sqrt{\\frac{yz}{xy + yz + zx + x^2}} + \\sqrt{\\frac{zx}{xy + yz + zx + y^2}} \n= \\\\\n& \\sqrt{\\frac{xy}{(... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0c03 | Prove that the equation $x^2 + y^2 + z^2 = x + y + z + 1$ has no rational solutions. | [
"The equation can be written equivalently as $(2x - 1)^2 + (2y - 1)^2 + (2z - 1)^2 = 7$.\n\nIf $(x, y, z)$ would be a solution of this equation with rational components, denoting $2x - 1 = \\frac{a_1}{b_1}$, $2y - 1 = \\frac{a_2}{b_2}$, $2z - 1 = \\frac{a_3}{b_3}$, one would have integers $a_1, b_1, a_2, b_2, a_3, ... | Romania | 69th NMO Selection Tests for JBMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
04cb | A board of size $1000 \times 1000$ is given. Is it possible to colour exactly $125$ unit squares of the board in such a way that every coloured unit square has an odd number of coloured neighbouring unit squares? Two unit squares are *neighbouring* if they have a common side. | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No | |
038j | A quadrilateral $ABCD$ with $\angle BAD + \angle ADC > 180^{\circ}$ is circumscribed around a circle of center $I$. A line through $I$ meets $AB$ and $CD$ at points $X$ and $Y$, respectively. Prove that if $IX = IY$ then $AX \cdot DY = BX \cdot CY$. | [
"Denote by $M$ and $N$ the tangent points of the incircle of $ABCD$ with $AB$ and $CD$, respectively. It follows from $\\angle BAD + \\angle ADC > 180^{\\circ}$ that $AB \\parallel CD$ and $\\angle MIN < 180^{\\circ}$. Also, the equalities $IM = IN$, $\\angle IMX = \\angle INY$ and $IX = IY$ show that $\\triangle I... | Bulgaria | Bulgarian National Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0dzd | Problem:
Med funkcijami $f(x)=\frac{a x^{2}+b x+8}{(x+c)^{2}}$ izberi tisto, ki ima definicijsko območje $\mathrm{D}_{f}=(-\infty, 1) \cup(1, \infty)$, ničlo 2 in asimptoto $y=0$. | [
"Solution:\n\nIz definicijskega območja razberemo, da je pol $x=1$, zato zapišemo enačbo $(1+c)^{2}=0$.\nIzračunamo $c=-1$. Iz enačbe asimptote razberemo, da je stopnja števca manjša od stopnje\nimenovalca, zato je $a=0$. Upoštevamo, da je število 2 ničla funkcije: $a \\cdot 0+2 b+8=0$, od\nkoder izračunamo $b=-4$.... | Slovenia | Državno tekmovanje | [
"Precalculus > Functions"
] | null | proof and answer | f(x) = (-4x + 8)/(x + 1)^2 | |
018z | Let $\Gamma$ be a circle, and $A$ a point outside $\Gamma$. For a point $B$ on $\Gamma$, let $C$ be the third vertex of the equilateral triangle $ABC$ (with vertices $A$, $B$ and $C$ going clockwise). Find the path traced out by $C$ as $B$ moves around $\Gamma$. | [
"Let $\\Gamma$ have centre $O$ and radius $r$. Let $O'$ be the third vertex of the equilateral triangle $AOO'$ (clockwise). Then the locus of $C$ is a circle with centre $O'$ and radius $r$: for consider the triangles $AOB$ and $AO'C$. Then $AO = AO'$ and $BA = CA$ (by construction: equilateral triangles) and $\\an... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | A circle of the same radius as the original, centered at the third vertex of the clockwise equilateral triangle built on the external point and the original circle’s center. | |
0cca | Let $n$ and $k$ be natural numbers, where $1 \le k < n$. At each vertex of a regular polygon with $n$ sides, either $1$ or $-1$ is written. At each step, we choose $k$ consecutive vertices and change their signs. Is it possible, starting from any configuration and performing this transformation multiple times, to obtai... | [
"Let $d = (n, k)$. If $d > 1$ and $n = dm$, $k = dl$, then by performing the global move\n$$\ne_1 e_2 \\dots e_k e_{k+1} e_{2k+1} \\dots e_{m(k-k+1)} e_{2e_{k+2}} e_{2k+2} \\dots e_{m(k-k+2)}\n$$\nwe obtain the initial configuration. The same configuration can also be obtained by performing the null move. Thus, in ... | Romania | THE Eighteenth IMAR Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | Yes if and only if gcd(n, k) = 1 and k is odd; otherwise no. | |
07a0 | Determine all increasing sequence $a_1, a_2, a_3, \ldots$ of positive integers such that for every $i, j \in \mathbb{N}$, the number of positive divisors of $i + j$ and $a_i + a_j$ are equal (A sequence $a_1, a_2, a_3, \ldots$ is increasing if $i \le j$ implies $a_i \le a_j$). | [
"First it is easy to show that the sequence is strictly increasing. Assume that $a_i = a_{i+1}$ for some $i$. Let $j = p - i$ for a large prime $p$. Now $i + j$ is a prime number. So $a_i + a_j$ is prime too. But from $a_i = a_{i+1}$ we get $a_{i+1} + a_j$ and $i + 1 + j$ are prime numbers. So $i + j$ and $i + j + ... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | a_n = n for all n | |
0gdm | 證明：存在一個正實數 $c > 0$，會使得有無窮多組正整數 $(n, N)$ 滿足下列條件：
(a) $N > c \cdot 2^{k} n \log(n)$，其中 $k$ 為 $N$ 的質因數個數；
(b) 對 $N$ 的任一質因數 $p$，設 $e$ 為 $N$ 的標準質因數分解中 $p$ 的次方數，都有 $n$ 被 $\varphi(p^e)$ 整除，其中 $\varphi$ 為歐拉函數，即 $\varphi(s)$ 為不超過 $s$ 而又與 $s$ 互質的正整數個數。
Prove that there is a positive real number $c > 0$ such that there are infin... | [
"We construct arbitrarily large pairs $(n, N)$ of integers that meet the conditions of the problem. Fix an integer $x \\ge 3$, and let $N = \\prod_{p \\le x} p$ be the product of the primes at most $x$. We set $n = 2^{2-\\pi(x)} \\prod_{p \\le x} (p-1)$, where $\\pi(x)$ denotes the number of primes at most $x$. Thi... | Taiwan | 2020 Taiwan IMO 3J | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
03iu | Problem:
The area of a triangle is determined by the lengths of its sides. Is the volume of a tetrahedron determined by the areas of its faces? | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Solid Geometry > Volume"
] | null | proof only | null | |
0j4a | Problem:
Let $\Omega$ be a circle of radius $8$ centered at point $O$, and let $M$ be a point on $\Omega$. Let $S$ be the set of points $P$ such that $P$ is contained within $\Omega$, or such that there exists some rectangle $ABCD$ containing $P$ whose center is on $\Omega$ with $AB=4$, $BC=5$, and $BC \parallel OM$. ... | [
"Solution:\n\nAnswer: $164+64\\pi$\n\nWe wish to consider the union of all rectangles $ABCD$ with $AB=4$, $BC=5$, and $BC \\parallel \\overline{OM}$, with center $X$ on $\\Omega$. Consider translating rectangle $ABCD$ along the radius $XO$ to a rectangle $A'B'C'D'$ now centered at $O$. It is now clear that every po... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 164 + 64*pi | |
0fml | Sean $X$, $Y$ los extremos de un diámetro de una circunferencia $\Gamma$ y $N$ el punto medio de uno de los arcos $XY$ de $\Gamma$. Sean $A$ y $B$ dos puntos en el segmento $XY$. Las rectas $NA$ y $NB$ cortan nuevamente a $\Gamma$ en los puntos $C$ y $D$, respectivamente. Las tangentes a $\Gamma$ en $C$ y $D$ se cortan... | [
"**Solución redactada por Daniel Lasaosa Medarde, basada en la solución dada en la prueba por Marcos García Fierro, del equipo español.**\n\nTracemos la paralela a $XY$ por $P$, que corta a las rectas $NA$ en $E$, y $NB$ en $F$, y sea $O$ el centro de $\\Gamma$. Por ser $PC$ tangente a $\\Gamma$ en $C$, tenemos que... | Spain | Olimpiada Iberoamericana de Matemáticas | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Spanish | proof only | null | |
0bgm | Problem:
Fie $a \in (0,1)$ şi $\mathcal{C}$ mulţimea funcţiilor crescătoare $f:[0,1] \rightarrow [0, \infty)$, astfel încât $\int_{0}^{1} f(x) \, \mathrm{d}x = 1$. Să se determine:
a. $\max_{f \in \mathcal{C}} \int_{0}^{a} f(x) \, \mathrm{d}x$,
b. $\max_{f \in \mathcal{C}} \int_{0}^{a} (f(x))^{2} \, \mathrm{d}x$. | [
"Solution:\n\na.\nFie $f$ o funcţie din mulţimea $\\mathcal{C}$. Arătăm că\n$$\n\\int_{0}^{a} f(x) \\, \\mathrm{d}x \\leq a\n$$\nÎntr-adevăr,\n$$\n\\begin{aligned}\na - \\int_{0}^{a} f(x) \\, \\mathrm{d}x &= a \\int_{0}^{1} f(x) \\, \\mathrm{d}x - \\int_{0}^{a} f(x) \\, \\mathrm{d}x = a \\int_{a}^{1} f(x) \\, \\mat... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) a
b) a if a ≤ 1/2; 1/(4(1−a)) if a > 1/2 | |
0ag4 | Let $x_1, x_2, \dots, x_n$ be positive real numbers and $n \ge 3$. Prove the inequality
$$
\frac{x_1 x_3}{x_1 x_3 + x_2 x_4} + \frac{x_2 x_4}{x_2 x_4 + x_3 x_5} + \dots + \frac{x_{n-1} x_1}{x_{n-1} x_1 + x_n x_2} + \frac{x_n x_2}{x_n x_2 + x_1 x_3} \le n-1
$$ | [
"In every one of the fractions $\\frac{x_{i-1}x_{i+1}}{x_{i-1}x_{i+1} + x_i x_{i+2}}$ (where $i \\in \\{1, 2, \\dots, n\\}$ and $x_0 = x_n$ and $x_{n+1} = x_1$) we divide by $x_i x_{i+2}$ and if we denote $y_i = \\frac{x_{i-1}x_{i+1}}{x_i x_{i+2}}$ the required inequality is transformed to the form:\n$$\n\\frac{y_1... | North Macedonia | null | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0ils | Problem:
$ABCD$ is a convex quadrilateral such that $AB = 2$, $BC = 3$, $CD = 7$, and $AD = 6$. It also has an incircle. Given that $\angle ABC$ is right, determine the radius of this incircle. | [] | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | (1 + sqrt(13))/3 | |
019g | Find all integer solutions to the equation
$$
2x^3 - y^2 = 3.
$$ | [
"Consider the equation modulo $8$. Then $y^2 \\equiv 0, 1, 4 \\pmod{8}$. This means $2x^3 \\equiv 3, 4, 7 \\pmod{8}$ all of which are impossible. The equation lacks a solution."
] | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | No integer solutions | |
0lc7 | Find all natural number $n$ and prime number $p$ for which the polynomial
$$
x^n - p x + p^2
$$
can be written as product of two polynomials from $\mathbb{Z}[x]$ with degree at least $1$. | [] | Vietnam | Vietnamese Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | n=3, p=2 | |
0ia5 | Problem:
The numbers $112$, $121$, $123$, $153$, $243$, $313$, and $322$ are among the rows, columns, and diagonals of a $3 \times 3$ square grid of digits (rows and diagonals read left-to-right, and columns read top-to-bottom). What 3-digit number completes the list? | [
"Solution:\n\n| 1 | 1 | 2 |\n| :--- | :--- | :--- |\n| 5 | 2 | 4 |\n| 3 | 1 | 3 |\n\nThe center digit is the middle digit of 4 numbers (hence at least 3 members of the above list), so it must be $2$. The top-left digit begins at least 2 members of the above list, so it must be $1$ or $3$. If it is $3$, then after p... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Logic"
] | null | proof and answer | 524 | |
0567 | The midpoints of the sides $BC$, $CA$, and $AB$ of triangle $ABC$ are $D$, $E$, and $F$, respectively. The reflections of centroid $M$ of $ABC$ around points $D$, $E$, and $F$ are $X$, $Y$, and $Z$, respectively. Segments $XZ$ and $YZ$ intersect the side $AB$ in points $K$ and $L$, respectively. Prove that $AL = BK$. | [
"As $\\frac{MY}{MB} = \\frac{2ME}{2ME} = 1$ and analogously $\\frac{MZ}{MC} = 1$ we have $BC \\parallel YZ$. Let $G$ be the intersection of lines $YZ$ and $AD$ (Fig. 14).\n\nThen $\\frac{MG}{MD} = \\frac{MY}{MB} = 1$ from which $MG = MD = \\frac{1}{3}AD$ and $AG = AD - MD - MG = \\frac{1}{3}AD$. Hence $\\frac{AL}{A... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
0dyl | Let $\triangle ABC$ be a right triangle with the right angle at $C$ and let $D$ be a point on the segment $BC$. Denote the circumcircle of the triangle $ABD$ by $\mathcal{K}$. Let $E$ be a point on $\mathcal{K}$, such that the chord $DE$ is perpendicular to $AB$. Prove that the triangle $AEB$ is isosceles with the apex... | [
"Let $T$ be the intersection of the chords $DE$ and $AB$. We know that $DTB$ is a right triangle. First, assume that the triangle $ABE$ is isosceles and write $\\angle AEB = \\angle BAE = \\alpha$. Inscribed angles $\\angle EAB$ and $\\angle EDB$ over $BE$ are equal, so $\\angle EDB = \\alpha$. Since $DE$ is perpen... | Slovenia | Slovenija 2008 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0e33 | A teacher gives Matej four pieces of paper, each containing a non-zero digit. Matej arranges them in a line thus forming a four-digit number. He then exchanges two of the pieces (without flipping or rotating them in the process) to form another four-digit number. Can he always do this in such a way that the two numbers... | [
"The answer is yes. If one of the digits is even, then Matej can form two even numbers by putting the piece of paper with the even digit in the place of units and exchanging two of the other three pieces of paper. If one of the four digits is equal to $5$, then Matej can act similarly. He should put the piece of pa... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0a66 | Problem:
At each vertex of a regular $14$-gon, lies a coin. Initially $7$ coins are heads, and $7$ coins are tails. Determine the minimum number $t$ such that it's always possible to turn over at most $t$ of the coins so that in the resulting $14$-gon, no two adjacent coins are both heads and no two adjacent coins are... | [
"Solution:\n\nNumber the vertices from $1$ to $14$. We need to make sure that all the even-numbered vertices are tails and all the odd numbered vertices are heads or vice versa.\n\nLet's look at the $7$ odd numbered vertices.\n\nWithout loss of generality, assume that there are initially more heads than tails on th... | New Zealand | NZMO Round Two | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 6 | |
0g17 | Problem:
Seien $B = (-1, 0)$ und $C = (1, 0)$ fixe Punkte in der Ebene. Eine nichtleere, beschränkte Teilmenge $S$ der Ebene heißt nett, falls die folgenden Bedingungen gelten:
(i) Es gibt einen Punkt $T$ in $S$, sodass für jeden anderen Punkt $Q$ in $S$ die Strecke $TQ$ vollständig in $S$ liegt.
(ii) Für jedes Drei... | [
"Solution:\n\nSeien $k_{B}$ und $k_{C}$ zwei Kreise um $B$ bzw. $C$ mit Radius $2$. Wir definieren die zwei Gebiete $S$ und $S'$ folgendermassen:\n\n\n\nDie Gebiete sind in der Skizze eingezeichnet. Offensichtlich erfüllen beide Gebiete die Bedingung (i).\n\nUm Bedingung (ii) zu zeigen, neh... | Switzerland | IMO-Selektion | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0jcg | Let $x_0, x_1, \dots, x_{n_0-1}$ be integers, and let $d_1, d_2, \dots, d_k$ be positive integers with $n_0 = d_1 > d_2 > \dots > d_k$ and $\gcd(d_1, d_2, \dots, d_k) = 1$. For every integer $n \ge n_0$, define
$$
x_n = \left\lfloor \frac{x_{n-d_1} + x_{n-d_2} + \dots + x_{n-d_k}}{k} \right\rfloor.
$$
Show that the seq... | [
"**Solution** (By Adam Hesterberg). Note that $x_n \\le \\max\\{x_{n-1}, \\dots, x_{n-n_0}\\}$, so $\\{x_n\\}$ is a bounded sequence. Let $X$ be the largest integer that occurs infinitely often in the sequence $\\{x_n\\}$, and let $N$ be an integer such that for all $n > N$, $x_n \\le X$. We now have a lemma.\n\n**... | United States | Team Selection Test Selection Test | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0kma | Problem:
Let $P$ be a polynomial with integer coefficients such that $P(2020) = P(2021) = 2021$. Prove that $P$ has no integer roots. | [
"Solution:\nLet\n$$\nP(x) = a_n x^n + a_{n-1} x^{n-1} + \\cdots + a_1 x + a_0.\n$$\nIf we plug in $x = 2020$, all terms except $a_0$ will be even, so since $P(2020)$ is odd, $a_0$ must be odd. But then if we plug in any other even number for $x$, $P(x)$ will still be odd since all the terms except $a_0$ will still ... | United States | Berkeley Math Circle: Monthly Contest 3 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
00qn | Let $a, b, c, p, q, r$ be positive integers such that $a^p + b^q + c^r = a^q + b^r + c^p = a^r + b^p + c^q$.
Prove that $a = b = c$ or $p = q = r$. | [
"The proof is essentially a size argument. We split into three cases, the first two of which are quite straightforward.\n\nCase 1: two of $p, q, r$ are equal, wlog $q = r$. Subtract $a^q + b^q + c^q$ from the given equation to show that\n$$\na^p - a^q = b^p - b^q = c^p - c^q.\n$$\nNow if $p > q$ then $x^p - x^q = x... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0avk | Problem:
Let $\lfloor x\rfloor$ be the greatest integer not exceeding $x$. For instance, $\lfloor 3.4\rfloor=3,\lfloor 2\rfloor=2$, and $\lfloor-2.7\rfloor=-3$. Determine the value of the constant $\lambda>0$ so that $2\lfloor\lambda n\rfloor=1-n+\lfloor\lambda\lfloor\lambda n\rfloor\rfloor$ for all positive integers ... | [] | Philippines | 19th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1 + sqrt(2) | |
0jxu | Problem:
Points $A$, $B$, $C$, $D$ lie on a circle in that order such that $\frac{A B}{B C}=\frac{D A}{C D}$. If $A C=3$ and $B D=B C=4$, find $A D$. | [
"Solution:\n\nBy Ptolemy's theorem, we have $A B \\cdot C D + B C \\cdot D A = A C \\cdot B D = 3 \\cdot 4 = 12$.\n\nSince the condition implies $A B \\cdot C D = B C \\cdot D A$, we have $D A = \\frac{6}{B C} = \\frac{3}{2}$."
] | United States | HMMT November | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof and answer | 3/2 | |
09hs | For nonnegative real numbers $a, b, c \ge 0$ satisfying $a+b+c \le 1$, prove
$$ (b+ac)^4 + c^2 \ge (ab+c)^4 + c^2a^4. $$
(Proposed by Otgonbayar Uuye) | [] | Mongolia | Round 2 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0jmy | Problem:
Farmer Yang has a $2015 \times 2015$ square grid of corn plants. One day, the plant in the very center of the grid becomes diseased. Every day, every plant adjacent to a diseased plant becomes diseased. After how many days will all of Yang's corn plants be diseased? | [
"Solution:\n\nAnswer: $2014$\n\nAfter $k$ minutes, the diseased plants are the ones with taxicab distance at most $k$ from the center. The plants on the corner are the farthest from the center and have taxicab distance $2014$ from the center, so all the plants will be diseased after $2014$ minutes."
] | United States | HMMT November 2015 | [
"Discrete Mathematics > Other"
] | null | final answer only | 2014 | |
04to | Find all triplets of integers $(a, b, c)$ such that each of the fractions
$$
\frac{a}{b+c}, \quad \frac{b}{c+a}, \quad \frac{c}{a+b}
$$
is an integer. | [
"The considered fractions are symmetric in the sense that if we permute $(a, b, c)$ we get the same three fractions (possibly in different order). The same is true if we replace $(a, b, c)$ by $(-a, -b, -c)$. This simplifies the following casework.\n\nLet us suppose that numbers $a, b, c$ are such that all three fr... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All integer triples are either (i) nonzero triples with a + b + c = 0, or (ii) permutations of (0, t, t) with t ≠ 0. | |
080v | Problem:
Un treno lungo 500 metri attraversa a velocità costante una galleria lunga 3 chilometri. Sapendo che sono passati 50 secondi dal momento in cui l'ultima carrozza del treno è entrata nella galleria a quando il locomotore emerge dall'altra uscita, si può affermare che la velocità del treno è:
(A) $50 \mathrm{~k... | [
"Solution:\n\nLa risposta è $(\\mathbf{E})$. In 50 secondi il locomotore del treno percorre $3-0,5=2,5$ chilometri. Perciò la sua velocità è\n$$\n\\frac{2,5}{50} \\mathrm{~km} / \\mathrm{s}=0,05 \\cdot 3600 \\mathrm{~km} / \\mathrm{h}=180 \\mathrm{~km} / \\mathrm{h} .\n$$"
] | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | E | |
0czi | Let $n \geq 2$ be a positive integer and let $x_{n}$ be a positive real root to the equation $x(x+1) \ldots(x+n)=1$. Prove that
$$
x_{n}<\frac{1}{\sqrt{n!H_{n}}}
$$
where $H_{n}=1+\frac{1}{2}+\ldots+\frac{1}{n}$. | [
"Because $x_{n}$ satisfies the equation we have $x_{n}^{2} \\cdot S_{n-1}< 1$, where $S_{n-1}$ is the $n-1$ symmetric sum of $1,2, \\ldots, n$. We can write\n$$\nS_{n-1}=n!\\left(1+\\frac{1}{2}+\\ldots+\\frac{1}{n}\\right)=n!\\cdot H_{n}\n$$\nand the above inequality is equivalent to $x_{n}^{2}<\\frac{1}{n!H_{n}}$,... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0b3b | Problem:
Point $G$ lies on side $AB$ of square $ABCD$ and square $AEFG$ is drawn outwards $ABCD$, as shown in the figure below. Suppose that the area of triangle $EGC$ is $1/16$ of the area of pentagon $DEFBC$. What is the ratio of the areas of $AEFG$ and $ABCD$?
(a) $4:25$
(b) $9:49$
(c) $1... | [] | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | a | |
06r8 | The vertices $X$, $Y$, $Z$ of an equilateral triangle $XYZ$ lie respectively on the sides $BC$, $CA$, $AB$ of an acute-angled triangle $ABC$. Prove that the incenter of triangle $ABC$ lies inside triangle $XYZ$.
The vertices $X$, $Y$, $Z$ of an equilateral triangle $XYZ$ lie respectively on the sides $BC$, $CA$, $AB$ ... | [
"We will prove a stronger fact; namely, we will show that the incenter $I$ of triangle $ABC$ lies inside the incircle of triangle $XYZ$ (and hence surely inside triangle $XYZ$ itself). We denote by $d(U, VW)$ the distance between point $U$ and line $VW$.\n\nDenote by $O$ the incenter of $\\triangle XYZ$ and by $r$,... | IMO | 51st IMO Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
06c7 | Let $s, t$ be given nonzero integers, and let $(x, y)$ be any ordered pair of integers. A move changes $(x, y)$ to $(x + t, y - s)$. The pair $(x, y)$ is 'good' if after some (maybe zero) number of moves it describes a pair of integers that are not relatively prime.
a. Determine if $(s, t)$ is a good pair.
b. Show th... | [
"a. Yes. Since $s, t \\neq 0$, $s^2 + t^2 \\geq 2$. Let $p$ be any prime divisor of $s^2 + t^2$. If $p \\mid t$, then $p \\mid s$, and so $\\gcd(s, t) \\geq p$. This shows $(s, t)$ is good. If $p \\nmid t$, then there exists $k \\in \\mathbb{Z}^+$ such that $tk \\equiv -s \\pmod{p}$. Then we have $p \\mid s + kt$. ... | Hong Kong | CHKMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
03sc | Suppose that the real numbers $a_1, a_2, \dots, a_n$ satisfy $a_1 + a_2 + \dots + a_n = 0$. Prove
$$
\max_{1 \le i \le n} a_i^2 \le \frac{n}{3} \sum_{i=1}^{n-1} (a_i - a_{i+1})^2.
$$
(pose by Zhu Huawei) | [
"It is sufficient to prove that: For every $k \\in \\{1, 2, \\dots, n\\}$, we have\n$$\na_k^2 \\le \\frac{n}{3} \\sum_{i=1}^{n-1} (a_i - a_{i+1})^2.\n$$\nLet $d_k = a_k - a_{k+1}$, $k = 1, 2, \\dots, n-1$, then\n$$\na_k = a_k,\n$$\n$$\na_{k+1} = a_k - d_k,\\quad a_{k+2} = a_k - d_k - d_{k+1}, \\dots,\n$$\n$$\na_n =... | China | China Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0g4w | Problem:
Soit $\omega_{1}$ un cercle de diamètre $J K$. Soit $t$ la tangente à $\omega_{1}$ en $J$ et soit $U \neq J$ un autre point sur $t$. Soit $\omega_{2}$ le plus petit cercle centré en $U$ qui intersecte $\omega_{1}$ en un seul point $Y$. Soit $I$ la deuxième intersection de la droite $J K$ avec le cercle circon... | [
"Solution:\n\nOn dénote le centre de $\\omega_{1}$ par $O$ et le centre circonscrit au triangle $J Y U$ par $\\omega_{3}$. On affirme que $F$ se trouve sur $\\omega_{3}$. Pour prouver cette affirmation, on note d'abord que $U, Y$, et $O$ sont colinéaires comme $\\omega_{1}$ et $\\omega_{2}$ sont tangents en $Y$. Co... | Switzerland | Deuxième tour 2023 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
079h | We call a sequence $a_0, \ldots, a_{1389}$ of real numbers *concave* if for every $0 < i < 1389$ we have $a_i \ge \frac{a_{i-1} + a_{i+1}}{2}$. Find the maximum number $c$ such that for every *concave* sequence of nonnegative numbers, we have
$$
\sum_{i=0}^{1389} i a_i^2 \ge c \sum_{i=0}^{1389} a_i^2
$$ | [
"First consider the concave sequence $a_i = (1389 - i)d$, $0 \\le i \\le 1389$, $d > 0$. Note that the maximum possible value of $c$ for this sequence is $c_0 = \\frac{\\sum_{i=0}^{1389} i a_i^2}{\\sum_{i=0}^{1389} a_i^2} = \\frac{1389^2 - 5 \\times 1389}{4 \\times 1389 - 2}$.\n\nNow we claim that for every concave... | Iran | 27th Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (1389^2 - 5*1389)/(4*1389 - 2) | |
0gtn | Find the largest possible value of $k$, if in every graph on $2022$ vertices having no cycles, it is always possible to choose $k$ vertices such that any chosen vertex is adjacent to at most two chosen vertices. | [
"**Answer: 1517.**\nConsider a graph on $n$ vertices having no cycles. By induction over $n$, we will show that the maximum possible value of $k$ is $\\lfloor 3n/4 \\rfloor$.\n\nIt can be readily shown that the claim holds for $n \\le 4$. Let $G$ be an acyclic graph on $n \\ge 5$ vertices. Assume that the claim hol... | Turkey | Team Selection Test | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 1517 | |
01be | A square-shaped pizza with a side length $30$ cm is cut into pieces. All cuts are parallel to the sides, and the total length of them is $240$ cm. Show that there is a piece which has an area at least $36\ \text{cm}^2$. | [
"Let $s_1, \\dots, s_n$ be areas of the pieces, and $p_1, \\dots, p_n$ their perimeters. Then $\\sum p_i = 4 \\cdot 30 + 2 \\cdot 240 = 600$, and $\\sum s_i = 900$.\nLet the smallest rectangle that can be drawn around the $i$-th piece of pizza have sides $a_i$ and $b_i$. Then\n$$\n\\sqrt{s_i} \\le \\sqrt{a_i b_i} \... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
02l8 | Problem:
Uma soma alternada - Se $S_{n}=1-2+3-4+5-6+\ldots+(-1)^{n+1} n$, onde $n$ é um inteiro positivo, então $S_{1992}+S_{1993}$ é:
(a) -2
(b) -1
(c) 0
(d) 1
(e) 2 | [
"Solution:\n\nLembre que\n$$\n(-1)^{n+1}= \\begin{cases}1 & \\text{ se } n \\text{ é ímpar } \\\\ -1 & \\text{ se } n \\text{ é par }\\end{cases}\n$$\nObservemos que associando duas a duas parcelas consecutivas,\n$$\n(1-2)+(3-4)+(5-6)+\\cdots\n$$\nobtemos uma soma de $n$ parcelas todas iguais a $-1$. Logo,\n$$\nS_{... | Brazil | Nível 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | d | |
074v | For each integer $n \ge 1$, define $a_n = \lfloor \frac{n}{\sqrt{n}} \rfloor$, where $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$, for any real number $x$. Find the number of all $n$ in the set $\{1, 2, 3, \dots, 2010\}$ for which $a_n > a_{n+1}$. | [
"Let us examine the first few natural numbers: $1, 2, 3, 4, 5, 6, 7, 8, 9$. Here we see that $a_n = 1, 2, 3, 2, 2, 3, 3, 4, 3$. We observe that $a_n \\le a_{n+1}$ for all $n$ except when $n+1$ is a square in which case $a_n > a_{n+1}$. We prove that this observation is valid in general. Consider the range\n\n$$\nm^... | India | Indija mo 2011 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 43 | |
018c | Let $f$ be a real-valued function of a real variable such that
$$
f(f(x)) = x^2 - x + 1
$$
for all real numbers $x$. Determine $f(0)$. | [
"Let $f(0) = a$ and $f(1) = b$.\nThen $f(f(0)) = f(a)$.\nBut $f(f(0)) = 0^2 - 0 + 1 = 1$. So $f(a) = 1$.\nAlso $f(f(1)) = f(b)$.\nBut $f(f(1)) = 1^2 - 1 + 1 = 1$. So $f(b) = 1$.\nFrom (1), $f(f(a)) = f(1)$.\nBut $f(f(a)) = a^2 - a + 1$. So $a^2 - a + 1 = b$.\nFrom (2), $f(f(b)) = f(1)$, giving $b^2 - b + 1 = b$. So... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 1 | |
06zv | Problem:
$ABC$ is an equilateral triangle. $D$ is on the side $AB$ and $E$ is on the side $AC$ such that $DE$ touches the incircle. Show that $AD/DB + AE/EC = 1$. | [
"Solution:\n\nPut $BD = x$, $CE = y$, $BC = a$. Then since the two tangents from $B$ to the incircle are of equal length, and similarly the two tangents from $D$ and $E$, we have $ED + BC = BD + CE$, or $ED = x + y - a$.\n\nBy the cosine law, $ED^2 = AE^2 + AD^2 - AE \\cdot AD$. Substituting and simplifying, we get... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0j70 | Problem:
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f(0)=0$, $f(1)=1$, and $\left|f^{\prime}(x)\right| \leq 2$ for all real numbers $x$. If $a$ and $b$ are real numbers such that the set of possible values of $\int_{0}^{1} f(x) d x$ is the open interval $(a, b)$, determine $b-a$... | [
"Solution:\n\nAnswer: $\\frac{3}{4}$\n\nDraw lines of slope $\\pm 2$ passing through $(0,0)$ and $(1,1)$. These form a parallelogram with vertices $(0,0)$, $(.75,1.5)$, $(1,1)$, $(.25,-.5)$. By the mean value theorem, no point of $(x, f(x))$ lies outside this parallelogram, but we can construct functions arbitraril... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof and answer | 3/4 | |
0avu | Problem:
Two players, $A$ (first player) and $B$, take alternate turns in playing a game using $2016$ chips as follows: the player whose turn it is, must remove $s$ chips from the remaining pile of chips, where $s \in \{2,4,5\}$. No one can skip a turn. The player who at some point is unable to make a move (cannot rem... | [
"Solution:\n\nWe call the remaining number of chips a winning position if there exists at least one move such that the player (whose turn it is) can force a win. The remaining number of chips is a losing position if any move by the player will give the opponent a chance to force a win, or a winning position. Thus, ... | Philippines | 18th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Player B | |
0kuv | Problem:
Compute the number of ways to tile a $3 \times 5$ rectangle with one $1 \times 1$ tile, one $1 \times 2$ tile, one $1 \times 3$ tile, one $1 \times 4$ tile, and one $1 \times 5$ tile. (The tiles can be rotated, and tilings that differ by rotation or reflection are considered distinct.) | [
"Solution:\n\nOur strategy is to first place the $1 \\times 5$ and the $1 \\times 4$ tiles since their size restricts their location. We have three cases:\n\n- Case 1: first row. There are 4 ways to place the $1 \\times 4$ tile. There is an empty cell next to the $1 \\times 4$ tile, which can either be occupied by ... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics"
] | null | final answer only | 40 | |
01y1 | At each node of the checkered $n \times n$ board sat a beetle. At midnight, each beetle crawled into the center of a cell. It turned out that the distance between any two beetles sitting in the adjacent (along the side) nodes did not increase.
Prove that at least one beetle crawled into the center of a cell at the vert... | [
"**First solution.** We will prove the statement of the problem in the general case of rectangular boards $n \\times m$. Suppose in contrary that there exist such positive integers $n$ and $m$ and a movement of beetles, that none of the beetles crawled into the center of the cell at the vertex of which it initially... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0ez2 | Problem:
$n$ is an integer. Prove that the sum of all fractions $1/rs$, where $r$ and $s$ are relatively prime integers satisfying $0 < r < s \leq n$, $r + s > n$, is $1/2$. | [
"Solution:\n\nWe use induction on $n$. If $n = 2$, then the only such fraction is $r = 1$, $s = 2$, giving $1/rs = 1/2$, so the result holds.\n\nSuppose it holds for $n - 1$. As we move to $n$, we lose the fractions with $r + s = n$. The other fractions $1/rs$ which satisfy the conditions for $n - 1$ also satisfy t... | Soviet Union | 3rd ASU | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1/2 | |
09x5 | The sequence $a_0, a_1, a_2, \dots$ of integers is defined by $a_0 = 3$ and
$$
a_{n+1} - a_n = n(a_n - 1)
$$
for all $n \ge 0$. Determine all integers $m \ge 2$ for which $\gcd(m, a_n) = 1$ for all $n \ge 0$. | [
"The sequence is given by the formula $a_n = 2 \\cdot n! + 1$ for $n \\ge 0$. (We use the usual definition $0! = 1$, which satisfies $1! = 1 \\cdot 0!$, in the same way we have $n! = n \\cdot (n-1)!$ for other positive integers $n$.) We will prove the equality by induction. We have $a_0 = 3$, which equals $2 \\cdot... | Netherlands | IMO Team Selection Test 1 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | All m that are powers of two: m = 2^i for i ≥ 1. | |
09zn | Sil has a lot of cards, which are yellow on one side and blue on the other. Most cards have a number on both sides. If two cards have the same number on the yellow side, then they have the same number on the blue side. There are also cards with a $\times$ on the yellow side and a $+$ on the blue side. Finally, there ar... | [] | Netherlands | Junior Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | MCQ | E | |
0ld4 | Let $f_n(x)$ be a sequence of polynomials, where $f_0(x) = 2$, $f_1(x) = 3x$, and
$$
f_n(x) = 3x f_{n-1}(x) + (1 - x - 2x^2) f_{n-2}(x)
$$
for all $n \ge 2$. Determine all positive integers $n$ such that $f_n(x)$ is divisible by $x^3 - x^2 + x$. | [
"By direct calculation, one can obtain\n$$\nf_n(x) = (2x-1)^n + (x+1)^n\n$$\nfor all positive integers $n$. Let $Q(x) = x^3 - x^2 + x = x(x^2 - x + 1)$ and $n$ be the natural number such that $Q(x)$ is a divisor of $f_n(x)$. It is easy to see that\n$$\n(-1)^n + 1^n = f_n(0) = 0,\n$$\nthen $n$ is odd.\n\nLet $R(x) =... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Modular Arithmetic > Fe... | English | proof and answer | all positive integers n with n ≡ 3 (mod 6) | |
09gr | Let $a$ and $b$ be positive real numbers such that $a + b = 2$. Prove that
$$
\frac{a}{1 + b + b^2} + \frac{b}{1 + a + a^2} \ge \frac{2}{3}.
$$ | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
06fr | Find all nonnegative integers $m$ and $n$ that satisfy the equation:
$$
107^{56}(m^2 - 1) - 2m + 3 = \binom{113^{114}}{n}.
$$
(If $n$ and $r$ are nonnegative integers satisfying $r \le n$, then $\binom{n}{r} = C_r^n = \frac{n!}{r!(n-r)!}$ and $\binom{n}{r} = 0$ if $r > n$.) | [
"The solutions are $(m, n) = (1, 0), (1, 113^{114})$.\nWhen $n = 0$ or $n = 113^{114}$, the equation becomes $107^{56}(m^2 - 1) - 2m + 3 = 1$. This implies\n$$\n(m - 1)(107^{56}(m + 1) - 2) = 0.\n$$\nClearly, $107^{56}(m + 1) - 2 = 0$ has no solution. Thus, the only solution is $m = 1$.\n\nWhen $0 < n < 113^{114}$,... | Hong Kong | CHKMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Residues and Primitive Roots > Primitive root... | null | proof and answer | m = 1, n = 0 or n = 113^{114} | |
07dd | $2n-1$ distinct positive real numbers with sum $S$ are given. Prove that there are at least $\binom{2n-2}{n-1}$ $n$-tuples among these numbers such that the sum of each $n$-tuple is at least $\frac{S}{2}$. | [
"Let's place these $2n-1$ numbers around a circle and label them by $\\pi(1), \\pi(2), \\dots, \\pi(2n-1)$ respectively and let\n$$\nS_i = \\pi(i) + \\pi(i+1) + \\dots + \\pi(i + n - 1), \\quad i \\le 2n-1\n$$\n(Numbers are considered modulo $2n-1$.) We claim that at least $n$ of $S_i$'s have a value not less than ... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof only | null | |
0l51 | Problem:
Let $a$, $b$, and $c$ be pairwise distinct complex numbers such that
$$a^{2} = b + 6,\quad b^{2} = c + 6,\quad \mathrm{and}\quad c^{2} = a + 6.$$
Compute the two possible values of $a + b + c$. | [
"Solution:\n\nSolution 1: Notice that any of $a$, $b$, or $c$ being $3$ or $-2$ implies $a = b = c$, which is invalid. Thus, \n$$(a^{2} - 9)(b^{2} - 9)(c^{2} - 9) = (b - 3)(c - 3)(a - 3)\\implies (a + 3)(b + 3)(c + 3) = 1,$$ \n$$(a^{2} - 4)(b^{2} - 4)(c^{2} - 4) = (b + 2)(c + 2)(a + 2)\\implies (a - 2)(b - 2)(c - ... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | (-1 ± sqrt(17))/2 | |
00aq | Let $n \ge 2$ be a natural number. For each pair $a, b$ of relatively prime natural numbers let $d_{a,b}$ be the greatest common divisor of $na+b$ and $a+nb$. Find the maximum value of $d_{a,b}$. | [
"The maximum value of $d_{a,b}$ equals $n^2-1$.\n\nLet $a$ and $b$ be relatively prime. Since $d_{a,b}$ divides $na+b$ and $a+nb$, it also divides the numbers $u = (n+1)(a+b) = (na+b)+(a+nb)$ and $v = (n-1)(a-b) = (na+b)-(a+nb)$. Hence $d_{a,b}$ divides $(n-1)u + (n+1)v = 2(n^2-1)a$ and $(n-1)u - (n+1)v = 2(n^2-1)b... | Argentina | Argentine National Olympiad 2016 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | n^2 - 1 | |
0i47 | Problem:
The squares of a chessboard are numbered from left to right and top to bottom (so that the first row reads $1, 2, \ldots, 8$, the second reads $9, 10, \ldots, 16$, and so forth). The number $1$ is on a black square. How many black squares contain odd numbers? | [
"Solution:\n\n$16$. The black squares in the $n$th row contain odd numbers when $n$ is odd and even numbers when $n$ is even; thus there are four rows where the black squares contain odd numbers, and each such row contributes four black squares."
] | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 16 | |
0l1r | Problem:
Compute the sum of all two-digit positive integers $x$ such that for all three-digit (base 10) positive integers $\underline{a}\ \underline{b}\ \underline{c}$, if $\underline{a}\ \underline{b}\ \underline{c}$ is a multiple of $x$, then the three-digit (base 10) number $\underline{b}\ \underline{c}\ \underline... | [
"Solution:\n\nNote that $\\overline{a b c 0} - \\overline{b c a} = a\\left(10^{4}-1\\right)$ must also be a multiple of $x$. Choosing $a=1$ means that $x$ divides $10^{3}-1$, and this is clearly a necessary and sufficient condition. The only two-digit factors of $10^{3}-1$ are $27$ and $37$, so our answer is $27+37... | United States | HMMT February 2024 | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 64 | |
031x | Problem:
Find the number of real solutions of the system
$$
\left\lvert\,
\begin{aligned}
& x + y + z = 3 x y \\
& x^{2} + y^{2} + z^{2} = 3 x z \\
& x^{3} + y^{3} + z^{3} = 3 y z
\end{aligned}
\right.
$$ | [
"Solution:\nNote first that the triple $(0, 0, 0)$ is a solution of the system.\n\nIf $y = 0$, then it follows from the first equation that $x = -z$ and the second one gives that $x = z = 0$.\n\nIf $y \\neq 0$, set $a = \\frac{x}{y}$ and $b = \\frac{z}{y}$. Then the system becomes\n$$\n\\left\\lvert\\,\n\\begin{ali... | Bulgaria | 52. Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 4 | |
0l23 | Problem:
Positive integers $a$, $b$, and $c$ have the property that $\operatorname{lcm}(a, b)$, $\operatorname{lcm}(b, c)$, and $\operatorname{lcm}(c, a)$ end in 4, 6, and 7, respectively, when written in base 10. Compute the minimum possible value of $a+b+c$. | [
"Solution:\n\nNote that $a+b+c=28$ is achieved when $(a, b, c)=(19,6,3)$. To show we cannot do better, first observe we would need $a+c<27$ and $\\operatorname{lcm}(a, c) \\leq a c \\leq 13 \\cdot 13=169$, which is only possible when $\\operatorname{lcm}(a, c)$ is $7, 17, 57, 77$, or $117$. We do casework on each v... | United States | HMMT November 2024 | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Modular Arithmetic",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 28 | |
08lk | Problem:
Is it possible to cover a given square with a few congruent right-angled triangles with acute angle equal to $30^{\circ}$? (The triangles may not overlap and may not exceed the margins of the square.) | [
"Solution:\nWe will prove that desired covering is impossible.\nLet us assume the opposite, i.e., a square with side length $a$ can be tiled with $k$ congruent right-angled triangles, whose sides are of lengths $b$, $b \\sqrt{3}$, and $2b$.\n\nThen the area of such a triangle is $\\frac{b^{2} \\sqrt{3}}{2}$.\nAnd t... | JBMO | 2008 Shortlist JBMO | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Intermediate Algebra > Other"
] | null | proof only | null | |
0j2j | A blackboard contains 68 pairs of nonzero integers. Suppose that for each positive integer $k$ at most one of the pairs $(k, k)$ and $(-k, -k)$ is written on the blackboard. A student erases some of the 136 integers, subject to the condition that no two erased integers may add to 0. The student then scores one point fo... | [
"(By Zuming Feng and Paul Zeitz). The answer is 43.\n\nWe first show that we can always get 43 points. Without loss of generality, we assume that the value of $x$ is positive for every pair of the form $(x, x)$ (otherwise, replace every occurrence of $x$ on the blackboard by $-x$, and every occurrence of $-x$ by $x... | United States | USAMO 2010 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 43 | |
0892 | Problem:
Marco, Fabrizio e Giovanni, tre matematici, sfidano un gruppo di quattro fisici a un torneo di calcio balilla. Giocano un incontro per ogni possibile combinazione di due matematici (uno in attacco, uno in difesa) contro due fisici (uno in attacco, uno in difesa). Ciascun incontro ha la stessa durata, e in tot... | [
"Solution:\n\nLa risposta è (D). Per simmetria, ciascuno fra Marco, Fabrizio e Giovanni deve trascorrere lo stesso tempo alla difesa della squadra dei matematici, che dunque risulta essere un terzo delle 24 ore del torneo, cioè 8 ore. In alternativa possiamo osservare che ci saranno $3 \\cdot 2 \\cdot 4 \\cdot 3$ i... | Italy | Italian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | D |
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