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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0g16 | Problem:
Soient $x, y, z$ des nombres réels positifs ou nuls avec $x y + y z + z x = 1$. Montrer que
$$
\frac{4}{x+y+z} \leq (x+y)(\sqrt{3} z + 1)
$$ | [
"Solution:\n\nPremière solution: On réécrit l'équation sous la forme\n$$\n4 \\leq (x+y+z)(x+y)(\\sqrt{3} z+1) = \\left(x^{2} + 2 x y + y^{2} + x z + y z\\right)(\\sqrt{3} z+1) = \\sqrt{3} z\\left(x^{2} + 2 x y + y^{2} + x z + y z\\right) + x^{2} + x y + y^{2} + 1\n$$\nOn peut encore transformer pour obtenir $3 \\le... | Switzerland | SMO Finalrunde | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
015q | The parliament of some country consists of several persons, some of them are friends. Computer prints all the lists of the members of the parliament such that no two persons in one list are friends. The number of these lists (including empty list) equals $M$. The members of the parliament want to constitute two committ... | [
"Consider the graph $G$ in which the parliamentaries correspond to vertices and friends are depicted by edges. We will identify the specified sets of vertices and the subgraphs of $G$ that contain all the edges of graph $G$ between these vertices. With this assumption we see that for any possible committees $A$ and... | Baltic Way | Baltic Way SHL | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0fgv | Problem:
Probar que los binomios $25 x+31 y$ y $3 x+7 y$ son múltiplos de 41 para los mismos valores de $x$ e $y$. | [
"Solution:\n\nSi $a$ y $b$ son enteros, módulo 41 tenemos las siguientes equivalencias:\n$$\n\\begin{aligned}\n25 a+31 b & \\equiv 0 \\Leftrightarrow \\quad \\text{ (multiplicando por } 2) \\\\\n\\Leftrightarrow 50 a+62 b & \\equiv 0 \\Leftrightarrow \\\\\n\\Leftrightarrow 9 a+21 b & \\equiv 0 \\Leftrightarrow \\\\... | Spain | OME 24 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0boe | The isosceles triangle $ABE$, with $m(\angle ABE) = 120^\circ$ is constructed outside the square $ABCD$. Denote $M$ the orthogonal projection of $B$ onto the bisector of angle $EAB$, $N$ the orthogonal projection of $M$ onto $AB$ and $P$ the meeting point of the straight lines $CN$ and $MB$. Let $G$ be the centroid of ... | [
"$m(\\angle BAM) = 15^\\circ$ implies $MN = \\frac{1}{4}AB$. From $MN \\parallel BC$ follows $\\triangle PMN \\sim \\triangle PBC$, hence\n$$\n\\frac{PM}{PB} = \\frac{MN}{BC} = \\frac{1}{4}\n$$\nand, denoting $Q = BM \\cap AE$, we come to $\\frac{PM}{BQ} = \\frac{1}{6}$, whence\n$$\nPB = PM + \\frac{1}{2}BQ = \\fra... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
06bo | Prove that for positive real numbers $x, y, z$,
$$
\frac{xyz(x + y + z + \sqrt{x^2 + y^2 + z^2})}{(x^2 + y^2 + z^2)(xy + yz + zx)} \le \frac{3 + \sqrt{3}}{9}.
$$ | [
"By the QM-AM inequality, we have\n$$\nx + y + z \\le \\sqrt{3(x^2 + y^2 + z^2)}.\n$$\nThis implies\n$$\n\\frac{xyz(x + y + z + \\sqrt{x^2 + y^2 + z^2})}{(x^2 + y^2 + z^2)(xy + yz + zx)} \\le \\frac{(\\sqrt{3} + 1)xyz\\sqrt{x^2 + y^2 + z^2}}{(x^2 + y^2 + z^2)(xy + yz + zx)} = \\frac{(\\sqrt{3} + 1)xyz}{\\sqrt{x^2 +... | Hong Kong | IMO HK TST | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
00iu | Determine all triples $(x, y, z)$ of real numbers satisfying the following system of equations:
$$
\begin{aligned}
2^{\sqrt[3]{x^2}} \cdot 4^{\sqrt[3]{y^2}} \cdot 16^{\sqrt[3]{z^2}} &= 128 \\
(xy^2 + z^4)^2 &= 4 + (xy^2 - z^4)^2.
\end{aligned}
$$ | [
"The first equation can be transformed as follows:\n$$\n\\begin{aligned}\n& 2^{\\sqrt[3]{x^2}} \\cdot 4^{\\sqrt[3]{y^2}} \\cdot 16^{\\sqrt[3]{z^2}} = 128 \\\\\n\\Leftrightarrow & 2^{\\sqrt[3]{x^2} + 2 \\cdot \\sqrt[3]{y^2} + 4 \\cdot \\sqrt[3]{z^2}} = 2^7 \\\\\n\\Leftrightarrow & \\sqrt[3]{x^2} + 2\\sqrt[3]{y^2} + ... | Austria | AustriaMO2011 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1) | |
05rd | Problem:
Carine et Cyril jouent au jeu suivant. Tout d'abord, Cyril choisit un entier $n \geqslant 1$. Puis, dans chacune des cases d'une grille $3 \times 3$ (apparentée à un jeu de morpion), il écrit un entier.
Vient ensuite le tour de Carine. Elle peut, autant de fois qu'elle le souhaite, effectuer l'opération suiv... | [
"Solution:\n\nOn va montrer que Carine a une stratégie gagnante. Tout d'abord, adoptons quelques notations. On note $(i, j)$ la case située en ligne $i$ et en colonne $j$, et on note $k_{i, j}$ l'entier écrit sur cette case. Sans perte de généralité, on suppose que nos entiers sont écrits modulo $n$.\n\nOn va égale... | France | TEST DU 27 FÉVRIER 2019 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | Carine | |
08hh | Problem:
Find all integers $n$ for which the number $\log_{2n-1}(n^2+2)$ is rational. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 5 | |
0cac | Problem:
Fie $f:[0,1] \longrightarrow [0,1]$ o funcție continuă și bijectivă, cu proprietatea că $f(0)=0$. Arătați că pentru orice $\alpha \geq 0$ are loc inegalitatea
$$
(\alpha+2) \cdot \int_{0}^{1} x^{\alpha}\left(f(x)+f^{-1}(x)\right) \, \mathrm{d}x \leq 2
$$ | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof only | null | |
0epu | Several small villages are situated on the banks of a straight river. On one side, there are $20$ villages in a row, and on the other there are $15$ villages in a row. We would like to build bridges, each of which connects a village on the one side with a village on the other side. The bridges must be straight, must no... | [
"We show that the answer is generally $\\binom{a+b-2}{a-1} = \\frac{(a+b-2)!}{(a-1)!(b-1)!}$ if there are $a$ towns on one side and $b$ on the other. In our particular instance, there are thus $\\binom{33}{14} = 818\\,809\\,200$ ways to build the bridges. We prove our general formula by induction on the total numbe... | South Africa | South African Mathematics Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 818809200 | |
08j1 | Problem:
Let $n \geq 4$ be a positive integer. On the chessboard table with dimensions $n \times n$ we put coins. We consider the diagonals of the table, each diagonal with at least two unit squares. What is the smallest number of coins to put on the table so that on each horizontal, each vertical, and each diagonal t... | [] | JBMO | The second selection test for IMO 2003 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | For odd n, the minimum is 2n − 3; for even n, the minimum is 2n − 2. | |
005g | Demonstrar que, para cada entero positivo $n$, existe un entero positivo $k$ tal que la representación decimal de cada uno de los números $k, 2k, \ldots, nk$ contiene todos los dígitos $0,1,2,3,4,5,6,7,8,9$. | [] | Argentina | XVIII Olimpiada Matemática del Cono Sur | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | Español | proof only | null | |
07zq | Problem:
Qual è la cifra delle unità del numero $2^{(2^{1})} + 2^{(2^{2})} + 2^{(2^{3})} + 2^{(2^{4})} + \ldots + 2^{(2^{1999})}$?
(A) 0
(B) 2
(C) 4
(D) 6
(E) 8. | [
"Solution:\n\nConsideriamo la cifra delle unità di $2^{(2^{k})}$ per $k \\geq 1$.\n\nOsserviamo che la cifra delle unità di $2^n$ segue un ciclo di lunghezza 4:\n\n| $n$ | $2^n$ | cifra delle unità |\n|-----|-------|-------------------|\n| 1 | 2 | 2 |\n| 2 | 4 | 4 |\n| 3 ... | Italy | Italian Mathematical Olympiad - Febbraio Round | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | B | |
08zp | Let $a_1, a_2, a_3, a_4, a_5, a_6, a_7$ be distinct positive integers. Find the minimum possible value of $|a_7 - a_1|$ assuming that the sequence $a_1, 2a_2, 3a_3, 4a_4, 5a_5, 6a_6, 7a_7$ is an arithmetic progression. A sequence $x_1, x_2, \dots, x_7$ is called an arithmetic progression if $x_2 - x_1 = x_3 - x_2 = \do... | [
"360\nThe assumption implies that there holds $ia_i = a_1 + (i-1)(2a_2 - a_1)$ for any integer $2 \\le i \\le 7$. Subtracting $a_1$ from both sides and dividing by $i$, one obtains\n$$\na_i - a_1 = \\frac{ia_i - ia_1}{i} = \\frac{(i-1)(2a_2 - a_1) - (i-1)a_1}{i} = \\frac{2(i-1)(a_2 - a_1)}{i}.\n$$\nSince $i$ and $i... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 360 | |
0ixt | Problem:
How many sequences of $5$ positive integers $(a, b, c, d, e)$ satisfy $a b c d e \leq a+b+c+d+e \leq 10$? | [
"Solution:\nWe count based on how many $1$'s the sequence contains.\n\nIf $a = b = c = d = e = 1$ then this gives us $1$ possibility.\n\nIf $a = b = c = d = 1$ and $e \\neq 1$, $e$ can be $2, 3, 4, 5, 6$. Each such sequence $(1, 1, 1, 1, e)$ can be arranged in $5$ different ways, for a total of $5 \\cdot 5 = 25$ wa... | United States | $12^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 116 | |
0bwm | Let $ABC$ be a triangle and $D$ the midpoint of the side $BC$. Denote by $E$ the symmetric of $A$ with respect to $D$. Let $F$ be the foot of the perpendicular from $E$ to $AC$. It is known that $AC = 2CF$ and $2m(\angle CAD) + m(\angle BAD) = 90^\circ$. Show that the triangle $ABC$ is equilateral.
Vlad Robu | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Ang... | English | proof only | null | |
0enx | Find all positive integers $n$ for which there exist non-negative integers $a_1, a_2, \dots, a_n$ such that
$$
\frac{1}{2^{a_1}} + \frac{1}{2^{a_2}} + \dots + \frac{1}{2^{a_n}} = \frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \dots + \frac{n}{3^{a_n}} = 1.
$$ | [
"Let $M = \\max\\{a_1, \\dots, a_n\\}$. Then we have\n$$\n3^M = 1 \\cdot 3^{M-a_1} + 2 \\cdot 3^{M-a_2} + \\dots + n \\cdot 3^{M-a_n} \\equiv 1 + 2 + \\dots + n = \\frac{n(n+1)}{2} \\pmod n.\n$$\nTherefore, the number $\\frac{n(n+1)}{2}$ must be odd and hence $n \\equiv 1 \\pmod 4$ or $n \\equiv 2 \\pmod 4$.\n\nWe ... | South Africa | International Mathematical Olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All positive integers n with n ≡ 1 or 2 modulo 4 | |
08fa | Problem:
Sia $ABC$ un triangolo, sia $r$ la bisettrice interna dell'angolo acuto $\widehat{BAC}$ e siano $K$ la proiezione di $B$ su $r$, $L$ la proiezione di $K$ su $AB$ e $D$ il simmetrico di $B$ rispetto ad $L$. Chiamiamo infine $H$ il piede dell'altezza del triangolo $ABC$ uscente da $B$. Dimostrare che:
a. $BH =... | [
"Solution:\n\na.\nOsserviamo intanto che il quadrilatero $AHKB$ è inscrivibile in una circonferenza, in quanto gli angoli $\\widehat{AHB}$ e $\\widehat{AKB}$ sono uguali (in particolare, retti). Gli angoli alla circonferenza $\\widehat{KBH}$ e $\\widehat{KAH}$ insistono sul medesimo arco e sono quindi uguali. D'alt... | Italy | Gara di Febbraio | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscell... | null | proof only | null | |
0bw1 | Consider the sequence of integers $a_0, a_1, a_2, \dots$, where $a_n = n^6 - 2017$ if $n$ is divisible by $7$, and $a_n = \frac{1}{7}(n^6 - 2017)$ otherwise. Determine the largest length a string of consecutive terms sharing a common divisor greater than $1$ may have. | [
"To show it less than $3$, we let $b_n = n^6 - 2017$, $n = 0, 1, 2, \\dots$, and prove that\n$$\n\\text{gcd}(b_{n-1}, b_n, b_{n+1}) = \\begin{cases} 1 & \\text{if } n \\equiv 0, \\pm 1 \\pmod 7, \\\\ 7 & \\text{if } n \\equiv \\pm 2, \\pm 3 \\pmod 7. \\end{cases}\n$$\nFix an index $n$, and let $d$ be a positive int... | Romania | Eleventh STARS OF MATHEMATICS Competition | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | 2 | |
096h | Problem:
Calculați: $$\int_{0}^{\ln 2} \frac{e^{3 x}-e^{-3 x}}{\sqrt{e^{x}+e^{-x}}} d x.$$ | [
"Solution:\n$$\n\\begin{gathered}\n\\int_{0}^{\\ln 2} \\frac{e^{3 x}-e^{-3 x}}{\\sqrt{e^{x}+e^{-x}}} d x = \\int_{0}^{\\ln 2} \\frac{\\left(e^{x}-e^{-x}\\right)\\left[\\left(e^{x}+e^{-x}\\right)^{2}-1\\right]}{\\sqrt{e^{x}+e^{-x}}} d x =\n\\left|\\begin{array}{c}\n\\sqrt{e^{x}+e^{-x}}=t \\\\\ne^{x}+e^{-x}=t^{2} \\\... | Moldova | Olimpiada Republicană la Matematică | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | final answer only | (5√10 + 8√2)/20 | |
04bd | Let $\triangle ABC$ be a triangle with an obtuse angle at vertex $B$, let $D$ and $E$ be midpoints of the segments $\overline{AB}$ and $\overline{AC}$ respectively, let $F$ be a point on the segment $\overline{BC}$ such that $\angle BFE = 90^\circ$, and let $G$ be a point on the segment $\overline{DE}$ such that $\angl... | [
"Obviously, $BFEG$ is a rectangle. The triangles $ADE$ and $ABC$ are similar with the coefficient of similarity $2$. The points $A$, $F$ and $G$ are collinear if and only if $|FB| : |GD| = 2 : 1$.\n\nLet $|BC| = a$ and $|BF| = x$. Then $|ED| = \\frac{1}{2}a$, $|EG| = |FB| = x$, $|GD| = |DE| - |EG| = \\frac{1}{2}a -... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
05dh | Problem:
Determine all integers $m$ for which the $m \times m$ square can be dissected into five rectangles, the side lengths of which are the integers $1,2,3, \ldots, 10$ in some order. | [
"Solution:\nThe solution naturally divides into three different parts: we first obtain some bounds on $m$. We then describe the structure of possible dissections, and finally, we deal with the few remaining cases.\n\nIn the first part of the solution, we get rid of the cases with $m \\leqslant 10$ or $m \\geqslant ... | European Girls' Mathematical Olympiad (EGMO) | EGMO | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 11 and 13 | |
0145 | Problem:
Find three different polynomials $P(x)$ with real coefficients such that $P\left(x^{2}+1\right)=P(x)^{2}+1$ for all real $x$. | [
"Solution:\nLet $Q(x)=x^{2}+1$. Then the equation that $P$ must satisfy can be written $P(Q(x))=Q(P(x))$, and it is clear that this will be satisfied for $P(x)=x$, $P(x)=Q(x)$ and $P(x)=Q(Q(x))$.",
"Solution:\nFor all reals $x$ we have $P(x)^{2}+1=P\\left(x^{2}+1\\right)=P(-x)^{2}+1$ and consequently, $(P(x)+P(-x... | Baltic Way | Baltic Way 2005 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | P(x) = x; P(x) = x^2 + 1; P(x) = x^4 + 2x^2 + 2 | |
0iu8 | Problem:
Compute
$$
\prod_{n=0}^{\infty}\left(1-\left(\frac{1}{2}\right)^{3^{n}}+\left(\frac{1}{4}\right)^{3^{n}}\right)
$$ | [
"Solution:\n\n$\\boxed{\\dfrac{2}{3}}$\n\nWe can rewrite each term as\n$$\n\\frac{1+\\left(\\frac{1}{2}\\right)^{3^{n+1}}}{1+\\left(\\frac{1}{2}\\right)^{3^{n}}}\n$$\nIn the infinite product, each term of the form $1+\\left(\\frac{1}{2}\\right)^{3^{n}}$ with $n>0$ appears once in the numerator and once in the denom... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2/3 | |
0ez3 | Problem:
Given $n$ points in space such that the triangle formed from any three of the points has an angle greater than $120$ degrees. Prove that the points can be labeled $1, 2, 3, \ldots, n$ so that the angle defined by $i$, $i+1$, $i+2$ is greater than $120$ degrees for $i = 1, 2, \ldots, n-2$. | [] | Soviet Union | 3rd ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | null | proof only | null | |
0io9 | Problem:
Forty two cards are labeled with the natural numbers $1$ through $42$ and randomly shuffled into a stack. One by one, cards are taken off of the top of the stack until a card labeled with a prime number is removed. How many cards are removed on average? | [
"Solution:\n\nAnswer: $\\frac{43}{14}$. Note that there are $13$ prime numbers amongst the cards. We may view these as separating the remaining $29$ cards into $14$ groups of nonprimes - those appearing before the first prime, between the first and second, etc. Each of these groups is equally likely to appear first... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 43/14 | |
0imz | Problem:
A sequence $\{a_{n}\}_{n \geq 0}$ of real numbers satisfies the recursion $a_{n+1} = a_{n}^{3} - 3 a_{n}^{2} + 3$ for all positive integers $n$. For how many values of $a_{0}$ does $a_{2007} = a_{0}$? | [
"Solution:\nAnswer: $\\mathbf{3}^{\\mathbf{2007}}$. If $x$ appears in the sequence, the next term $x^{3} - 3x^{2} + 3$ is the same if and only if $0 = x^{3} - 3x^{2} - x + 3 = (x-3)(x-1)(x+1)$. Moreover, that next term is strictly larger if $x > 3$ and strictly smaller if $x < -1$. It follows that no values of $a_{... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials"
] | null | proof and answer | 3^{2007} | |
05i0 | Problem:
Soit $t$ un entier naturel non-nul. Montrer qu'il existe un entier $n>1$ premier avec $t$ tel que pour tout entier $k \geq 1$, l'entier $n^{k}+t$ ne soit pas une puissance (c'est-à-dire qu'il ne soit pas de la forme $m^{r}$ avec $m \geq 1$ et $r \geq 2$ ). | [
"Solution:\n\nPour que $n$ soit premier avec $t$, on va le chercher sous la forme $1+t s$ où $s$ est entier. On aura alors $n^{k}+t \\equiv 1+t\\pmod{s}$. En particulier, si $s$ est divisible par $(t+1)$, alors $n^{k}+t$ l'est également.\n\nOn va d'abord traiter le cas où $t+1$ n'est pas une puissance. Dans ce cas,... | France | Envoi 1 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0dfb | Characterize all positive integers $n > 1$ for which the expression
$$
(n-1)! \cdot \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n-1}\right)
$$
is not divisible by $n$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | English | proof and answer | Exactly those n equal to 2p with p an odd prime, and n = 2, 4, 8. | |
08dd | Problem:
Diciamo che un numero intero positivo con un numero pari di cifre è "corretto" se, leggendo ad alta voce le singole cifre, otteniamo una corretta descrizione del numero stesso. O meglio, se ogni cifra in posizione dispari indica quante volte compare la cifra successiva in tutto il numero.
Ad esempio, $1210$ è... | [
"Solution:\n\na. Il numero $\\mathcal{A} = 12131415161718$ è corretto. Poiché una corretta descrizione del numero non dipende dall'ordine, spezzando $\\mathcal{A}$ in blocchi di due cifre consecutive\n\n| 12 | 13 | 14 | 15 | 16 | 17 | 18 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n\ne permutandoli tra lo... | Italy | Progetto Olimpiadi della Matematica | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) At least 5040 such numbers (hence more than 2019). b) There are finitely many. c) The largest has 162 digits. | |
096g | Problem:
Fie șirul $\left(a_{n}\right)_{n=1}^{\infty}$, $a_{n}=\underbrace{11 \ldots 1}_{n \text{ cifre }}\underbrace{88 \ldots 8}_{n \text{ cifre }}$, $\forall n \geq 1$. Calculați $\lim _{n \rightarrow \infty}\left\{\sqrt{a_{n}}\right\}$, unde $\{t\}$ reprezintă partea fracționară a numărului $t$. | [
"Solution:\n\nFie $b_{n}=\\underbrace{11 \\ldots 1}_{n \\text{ cifre }}$ și $c_{n}=3 b_{n}$, $\\forall n \\geq 1$. Observăm că $\\lim _{n \\rightarrow \\infty} b_{n}=\\lim _{n \\rightarrow \\infty} c_{n}=+\\infty$. Avem\n$$\n\\begin{aligned}\na_{n} & =\\underbrace{11 \\ldots 1}_{n \\text{ cifre }}\\underbrace{88 \\... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1/2 | |
01xk | Find all triples $(x; y; z)$ of real numbers satisfying the system
$$
\begin{cases} (x+1)(x^2+1) = y^3+1, \\ (y+1)(y^2+1) = z^3+1, \\ (z+1)(z^2+1) = x^3+1. \end{cases}
$$ | [
"It is easy to see that the triples $(0; 0; 0)$ and $(-1; -1; -1)$ satisfy the system. We will show that there are no other solutions.\nFrom the equations\n$$\n(x+1)(x^2+1) = y^3+1, \\quad (1)\n$$\n$$\n(y+1)(y^2+1) = z^3+1, \\quad (2)\n$$\n$$\n(z+1)(z^2+1) = x^3+1 \\quad (3)\n$$\nit follows that if $x = -1$, then, ... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | (0,0,0) and (-1,-1,-1) | |
0ixu | Problem:
Find the maximum value of $x+y$, given that $x^{2}+y^{2}-3y-1=0$. | [
"Solution:\n\nWe can rewrite $x^{2}+y^{2}-3y-1=0$ as\n$$\nx^{2} + \\left(y - \\frac{3}{2}\\right)^{2} = \\frac{13}{4}.\n$$\nWe then see that the set of solutions to $x^{2}+y^{2}-3y-1=0$ is the circle of radius $\\frac{\\sqrt{13}}{2}$ and center $\\left(0, \\frac{3}{2}\\right)$.\n\nThis can be written as\n$$\nx = \\... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | (sqrt(26) + 3)/2 | |
0fiv | Problem:
Encuentra el mayor número entero $N$ que cumpla las siguientes condiciones:
a. $E(N / 3)$ tiene sus tres cifras iguales.
b. $E(N / 3)$ es suma de números naturales consecutivos comenzando en 1, es decir, existe un natural $n$ tal que
$$
E(N / 3) = 1 + 2 + 3 + \cdots + n
$$
Nota: $E(x)$ es la parte entera d... | [
"Solution:\n\nDe la condición a) sale $z = E\\left(\\frac{N}{3}\\right) = 111 \\cdot k$ para todo $k \\in \\mathbb{N}$, $1 \\leq k \\leq 9$.\n\nDe la condición b) sale $z = E\\left(\\frac{N}{3}\\right) = 1 + 2 + 3 + \\cdots + n$, o bien, $z = \\frac{n(n+1)}{2}$, que equivale a $n^2 + n - 2z = 0$, es decir, $n = \\f... | Spain | Olimpiada Matemática Española | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 2000 | |
05tr | Problem:
Déterminer le plus petit entier $n \geqslant 2$ tel qu'il existe des entiers strictement positifs $a_{1}, \ldots, a_{n}$ tels que
$$
a_{1}^{2}+\ldots+a_{n}^{2} \mid \left(a_{1}+\ldots+a_{n}\right)^{2}-1
$$ | [
"Solution:\n\nAnalysons le problème. Dans ce problème, on cherche le plus petit entier $n$ satisfaisant une certaine propriété. Supposons que l'on veuille montrer que le plus petit entier recherché est l'entier $c$. Pour montrer que $c$ est bien le plus petit entier, on doit d'une part montrer que si un entier $n$ ... | France | Envoi 5: Pot Pourri | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 9 | |
0a04 | Problem:
Vind alle natuurlijke getallen $n$ waarvoor er een geheel getal $a>2$ bestaat zo dat $a^{d}+2^{d} \mid a^{n}-2^{n}$ voor alle positieve delers $d \neq n$ van $n$. | [
"Solution:\n\nAntwoord: $n$ is priem of een tweemacht (incl. $n=1$).\n\nInderdaad, als $n$ een oneven priemgetal is, dan is de enige deler $d=1$. Kies $a=2^{k}-2$ met $3 \\leq k \\leq n+1$, bijvoorbeeld $a=6$. Dan moeten we controleren dat $2^{k}-2+2=2^{k}$ een deler is $\\left(2^{k}-2\\right)^{n}-2^{n}$. Dit is he... | Netherlands | IMO-selectietoets III | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All n that are prime or a power of two (including n equal to one). | |
0bek | Let $f: [0, \pi/2] \to [0, \infty)$ be an increasing function. Prove that:
a. $$\int_0^{\pi/2} (f(x) - f(\pi/4))(\sin x - \cos x) dx \geq 0.$$
b. There exists $a \in [\pi/4, \pi/2]$ such that $\int_0^a f(x) \sin x dx = \int_0^a f(x) \cos x dx$. | [
"a. Let $I = \\int_0^{\\pi/4} (f(x) - f(\\pi/4))(\\sin x - \\cos x) dx$ and $J = \\int_{\\pi/4}^{\\pi/2} (f(x) - f(\\pi/4))(\\sin x - \\cos x) dx$ and notice that\n$$\n\\int_0^{\\pi/2} (f(x) - f(\\pi/4))(\\sin x - \\cos x) dx = I + J.\n$$\nFor $x \\in [0, \\pi/4]$ we have $f(x) - f(\\pi/4) \\leq 0$ and $\\sin x - \... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | null | proof only | null | |
0hlv | Problem:
Let $A B C D E$ be a convex pentagon. If $\alpha=\angle D A C$, $\beta=\angle E B D$, $\gamma=\angle A C E$, $\delta=\angle B D A$, and $\epsilon=\angle B E C$, as shown in the picture, calculate the sum $\alpha+\beta+\gamma+\delta+\epsilon$.
 | [
"Solution:\n\nLet $M, N, P, Q, R$ denote the intersections of the lines $A C$ and $B E$, $A C$ and $B D$, $C E$ and $B D$, $D A$ and $C E$, $E B$ and $A E$, respectively. Since the sum of the internal angles of a triangle is $180^{\\circ}$, from $\\triangle A M R$ we get $\\alpha=180^{\\circ}-\\angle A M R-\\angle ... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 180° | |
040v | In a $n \times n$ chart, two cells of the chart are called adjacent if they have a common side. At the origin, there is a number $+1$ in every cell. An operation means that: one can choose a cell, and change the signs of all the numbers which are in the adjacent cells of it (doesn't change the sign of the number in it)... | [] | China | China Western Invitational Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Linear Algebra > Matrices"
] | English | proof and answer | All n ≥ 2 with n not congruent to 1 modulo 4 | |
0c3c | Let $p$ be a prime number. Consider a commutative finite group $G$, with at least three elements, identity element $e$, and so that $x^p = e$, for every $x \in G$. Denote $Aut(G)$ the group of the automorphisms of $G$. Prove that:
a) $|G| = p^n$, where $n$ is the smallest cardinality of a set of generators of $G$;
b)... | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Linear Algebra > Linear transformations"
] | null | proof only | null | |
09zc | We construct a sequence of numbers starting with $2022$ and $21$. Each next number in the sequence is equal to the positive difference of the two previous numbers. So the third and fourth number in the sequence are $2001$ and $1980$.
At which place in the sequence do we find the number $0$ for the first time? | [] | Netherlands | Junior Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 154 | |
0f8w | Problem:
$S$ and $S'$ are two intersecting spheres. The line $BXB'$ is parallel to the line of centers, where $B$ is a point on $S$, $B'$ is a point on $S'$, and $X$ lies on both spheres. $A$ is another point on $S$, and $A'$ is another point on $S'$ such that the line $AA'$ has a point on both spheres. Show that the ... | [] | Soviet Union | 23rd ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0h9o | Given a triangle $ABC$ with $\angle ACB > 90^\circ$, $\angle CBA > 45^\circ$. Points $P$ and $T$ belong to the sides $AC$ and $AB$, respectively, so that $PT = BC$ and $PT \perp BC$. Points $P_1$ and $T_1$ belong to the sides $AC$ and $AB$ so that $AP = CP_1$ and $AT = BT_1$. Show that $\angle CBA - \angle P_1T_1A = 45... | [
"Let $S$ be such that $\\triangle ABC$ and $\\triangle BCS$ are in a different half planes with respect to $BC$, and $\\triangle BCS$ is isosceles and has right angle $CBS$. Then segments $BS$ and $PT$ are equal and parallel ($TP = BC = BS$, $PS \\parallel AB$), hence $TPSB$ is a parallelogram (Fig. 21). Thus, $PS ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gdt | 有 $N$ 隻怪獸，每一隻的體重都是一個正實數。每次我們把其中兩隻怪獸融合成一隻，新怪獸的體重是之前兩隻怪獸的體重和。經過一系列的操作後，我們最終將所有怪獸融合成一隻。在這個過程中，如果有某一次融合，被融合的兩隻怪獸中有一隻的體重大於另一隻的 2.020 倍，則我們稱這一次融合是危險的。一個融合順序的危險程度，是其過程中危險融合的次數。
試證：不論起始怪獸的體重如何分配，“每一次都將最輕的兩隻怪獸融合”都是所有融合順序中，讓危險程度達到最低的一個順序。
There are $N$ monsters, each with a positive weight. On each step, two of the monsters are... | [
"原題中的 2.020 可以改成任意 $k > 2$，故以下證明一般性的 $k$ 的狀況。\n讓我們將怪獸標記，使得第 $i$ 隻怪獸的體重 $x_i$ 滿足 $x_1 \\le x_2 \\le \\dots \\le x_N$。我們定義一隻怪獸 $i$ 是**危險的**，若且唯若 $x_i > k \\sum_{j<i} x_j$。\n首先，注意到起始危險怪獸數是危險程度的下界。假設 $x_i$ 是危險的，並令 $M = \\{i-1\\}$。當第 $i-1$ 號怪獸與 $j$ 號怪獸融合時，將 $j$ 加入集合 $M$ 中。則易知在此過程中，當第一次 $M \\cap \\{j: j \\ge i\\}$ 時，必為一危... | Taiwan | 2020 Taiwan IMO 3J | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0frl | Disponemos de $2n$ bombillas colocadas en dos filas (A y B) y numeradas de 1 a $n$ en cada fila. Algunas (o ninguna) de las bombillas están encendidas y el resto apagadas; decimos que eso es un “estado”. Dos estados son distintos si hay una bombilla que está encendida en uno de ellos y apagada en el otro. Diremos que u... | [
"Es obvio que $ET = 2^{2n}$, puesto que cada una de las $2n$ bombillas puede estar apagada o encendida. El número de “estados buenos” con $k$ bombillas encendidas en cada fila es $\\binom{n}{k}^2$, ya que hay $\\binom{n}{k}$ formas de elegir las $k$ bombillas encendidas de la fila A y otras tantas de elegir las $k$... | Spain | LVII Olimpiada Matemática Española Concurso Final Nacional | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
04mt | Let $ABCD$ be a rectangle. A certain number of red lines are drawn between the lines $AB$ and $CD$, parallel to them, and a certain number of blue lines are drawn between the lines $AD$ and $BC$, parallel to them. This divides the rectangle into $775$ smaller rectangles, while the red and the blue lines intersect at $7... | [] | Croatia | Croatia_2018 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 24 red lines and 30 blue lines, or 30 red lines and 24 blue lines | |
0ks4 | Problem:
Given a positive integer $k$, let $\|k\|$ denote the absolute difference between $k$ and the nearest perfect square. For example, $\|13\|=3$ since the nearest perfect square to $13$ is $16$. Compute the smallest positive integer $n$ such that
$$
\frac{\|1\|+\|2\|+\cdots+\|n\|}{n}=100.
$$
Proposed by: Carl Schi... | [
"Solution:\nNote that from $n = m^2$ to $n = (m+1)^2$, $\\|n\\|$ increases from $0$ to a peak of $m$ (which is repeated twice), and then goes back down to $0$. Therefore\n$$\n\\sum_{n=1}^{m^2} \\|n\\| = \\sum_{k=1}^{m-1} 2(1+2+\\cdots+k) = \\sum_{k=1}^{m-1} 2\\binom{k+1}{2} = 2\\binom{m+1}{3} = \\frac{m}{3}\\left(m... | United States | HMMT February 2022 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 89800 | |
00wl | Problem:
Find the smallest positive integer $n$ having the property: for any set of $n$ distinct integers $a_{1}, a_{2}, \ldots, a_{n}$ the product of all differences $a_{i}-a_{j}$, $i<j$, is divisible by $1991$. | [
"Solution:\n\nLet $S = \\prod_{1 \\leq i < j \\leq n} (a_{i} - a_{j})$. Note that $1991 = 11 \\cdot 181$. Therefore $S$ is divisible by $1991$ if and only if it is divisible by both $11$ and $181$.\n\nIf $n \\leq 181$ then we can take the numbers $a_{1}, \\ldots, a_{n}$ from distinct congruence classes modulo $181$... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 182 | |
00t3 | A number of $N$ children are at a party, and they sit in a circle to play a game of Pass the Parcel. Because the host has no other form of entertainment, the parcel has infinitely many layers. On turn $i$, starting with $i = 1$, the following two things happen in order:
(1) The parcel is passed $i^2$ positions clockwis... | [
"Every child receives a prize if and only if $N = 2^a3^b$ for some non-negative integers $a$ and $b$. For convenience, say $N$ is *good* if every child receives a prize.\n\nNumber the children $0, \\ldots, N-1$ clockwise around the circle, child number $0$ starting with the parcel. After $n$ turns, the parcel will ... | Balkan Mathematical Olympiad | BMO Short List | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | All N of the form 2^a 3^b for non-negative integers a and b | |
0cat | Solve in real numbers the equation $2(5^x + 6^x - 3^x) = 7^x + 9^x$. | [
"We claim that the only solutions are $x = 0$ and $x = 1$.\n\nThe given equation can be rewritten as follows:\n$$\n5^x \\left( \\left(\\frac{7}{5}\\right)^x + \\left(\\frac{3}{5}\\right)^x - 2 \\right) + 6^x \\left( \\left(\\frac{9}{6}\\right)^x + \\left(\\frac{3}{6}\\right)^x - 2 \\right) = 0. \\ (\\star)\n$$\nCon... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | x = 0 and x = 1 | |
0cr4 | Треугольник $ABC$ ($AB > BC$) вписан в окружность $\Omega$. На сторонах $AB$ и $BC$ выбраны точки $M$ и $N$ соответственно так, что $AM = CN$. Прямые $MN$ и $AC$ пересекаются в точке $K$. Пусть $P$ — центр вписанной окружности треугольника $AMK$, а $Q$ — центр вневписанной окружности треугольника $CNK$, касающейся стор... | [
"**Первое решение.** Мы будем использовать следующую известную лемму о трезубце.\n**Лемма.** Пусть $L$ — середина дуги $YZ$ (не содержащей точку $X$) описанной окружности треугольника $XYZ$. Пусть $I$ — центр вписанной окружности треугольника $XYZ$, а $I_x$ — центр вневписанной окружности этого треугольника, касающ... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tange... | null | proof only | null | |
0cdq | Let $f : [0, 1] \to \mathbb{R}$ be a monotone increasing differentiable function, with a continuous derivative, such that $f(0) = 0$. Let $g : [0, 1] \to \mathbb{R}$ be the function defined by
$$
g(x) = f(x) + (x-1)f'(x), \quad \text{for any } x \in [0, 1].
$$
a) Show that
$$
\int_{0}^{1} g(x) \, dx = 0.
$$ | [
"a) Because $g(x) = f(x) + (x-1)f'(x) = ((x-1)f(x))'$, it follows that\n$$\n\\int_{0}^{1} g(x) \\, dx = (x-1)f(x)\\Big|_{0}^{1} = 0.\n$$"
] | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Derivatives"
] | null | proof and answer | 0 | |
0lbq | Let be given two positive real numbers $a_0$ and $a_1$. The sequence $(a_n)$ is defined by
$$
a_{n+2} = 1 + \frac{a_{n+1}}{a_n}, \quad n = 0,1,2,\ldots
$$
Prove that $|a_{2012} - 2| < 10^{-200}$. | [] | Vietnam | Vietnamese Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
08pc | Problem:
Let $ABC$ be an acute angled triangle with orthocenter $H$ and circumcenter $O$. Assume the circumcenter $X$ of $BHC$ lies on the circumcircle of $ABC$. Reflect $O$ across $X$ to obtain $O'$, and let the lines $XH$ and $O'A$ meet at $K$. Let $L$, $M$ and $N$ be the midpoints of $[XB]$, $[XC]$ and $[BC]$, resp... | [
"Solution:\n\nThe circumcircles of $ABC$ and $BHC$ have the same radius. So, $XB = XC = XH = XO = r$ (where $r$ is the radius of the circle $ABC$) and $O'$ lies on $C(X, r)$. We conclude that $OX$ is the perpendicular bisector for $[BC]$. So, $BOX$ and $COX$ are equilateral triangles.\n\nIt is known that $AH = 2ON ... | JBMO | Junior Balkan Mathematics Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > D... | null | proof only | null | |
00mr | Determine all digits $z$ such that for each integer $k \ge 1$ there exists an integer $n \ge 1$ with the property that the decimal representation of $n^9$ ends with at least $k$ digits $z$. | [
"For $z=0$ we easily find $10^l$ with any sufficiently large integer $l$ such that $9l \\ge k$.\n\nFor $z \\in \\{2, 4, 6, 8\\}$ the number $n^9$ is even and therefore also $n$ must be even, and hence $n^9$ must be divisible by $2^9$. However, numbers ending with 222, 444 or 666 are already not divisible by 8, and ... | Austria | 49th Austrian Mathematical Olympiad, National Competition (Final Round, part 2) | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | {0, 1, 3, 7, 9} | |
00cn | Se tiene un conjunto $M$ de 2019 números reales tales que para todo par $a, b$ de números de $M$ se verifica que $a^2 + b\sqrt{2}$ es un número racional. Demostrar que para todo $a$ de $M$ vale que $a\sqrt{2}$ es un número racional. | [] | Argentina | Nacional OMA 2019 | [
"Algebra > Intermediate Algebra > Other"
] | Spanish | proof only | null | |
0bag | Let $G$ be the set of matrices
$$
\begin{pmatrix} a & b \\ \hat{0} & \hat{1} \end{pmatrix}, \quad a, b \in \mathbb{Z}_7, a \neq \hat{0}.
$$
a) Show that $G$ is a group with respect to the matrix multiplication.
b) Prove that there exists no proper homomorphism from $G$ to $\mathbb{Z}_7$. | [
"a. Consider $A = \\begin{pmatrix} a & b \\\\ \\hat{0} & \\hat{1} \\end{pmatrix}$ and $B = \\begin{pmatrix} x & y \\\\ \\hat{0} & \\hat{1} \\end{pmatrix}$ two elements of $G$. Then\n$$ AB = \\begin{pmatrix} ax & ay + b \\\\ \\hat{0} & \\hat{1} \\end{pmatrix} \\in G, \\text{ for } ax \\neq \\hat{0}. $$\nTo end the p... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Linear Algebra > Matrices",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
06tz | Let $A_{1}$, $B_{1}$ and $C_{1}$ be points on sides $BC$, $CA$ and $AB$ of an acute triangle $ABC$ respectively, such that $AA_{1}$, $BB_{1}$ and $CC_{1}$ are the internal angle bisectors of triangle $ABC$. Let $I$ be the incentre of triangle $ABC$, and $H$ be the orthocentre of triangle $A_{1}B_{1}C_{1}$. Show that
$$... | [
"Without loss of generality, assume $\\alpha = \\angle BAC \\leqslant \\beta = \\angle CBA \\leqslant \\gamma = \\angle ACB$. Denote by $a$, $b$, $c$ the lengths of $BC$, $CA$, $AB$ respectively. We first show that triangle $A_{1}B_{1}C_{1}$ is acute.\nChoose points $D$ and $E$ on side $BC$ such that $B_{1}D \\para... | IMO | IMO 2016 Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities >... | English | proof only | null | |
0his | Problem:
If $P_{1} P_{2} \ldots P_{100}$ is a regular 100-gon, what is the measure of the angle $\angle P_{20} P_{2} P_{1}$ in degrees? | [
"Solution:\nIntroduce the circumcircle of the polygon. The aforementioned angle cuts out an arc which is $\\frac{81}{100}$ of that of the entire circle, i.e. which measures $\\frac{81}{100} \\cdot 360^{\\circ}$. By the inscribed angle theorem, this means an answer of $\\frac{81}{100} \\cdot 180^{\\circ} = 145.8^{\\... | United States | Berkeley Math Circle: Monthly Contest 8 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 145.8° | |
0dom | Let $s = (AB + BC + AC)/2$ be the semiperimeter of triangle $ABC$. We choose two points $L$ and $N$ lying on the rays $AB$ and $CB$ satisfying $AL = CN = s$. Let $K$ be the point symmetric to $B$ with respect to the center of the circumcircle of triangle $ABC$. Prove that the perpendicular drawn from the point $K$ to t... | [
"Let $BC = a$, $AC = b$, $AB = c$ and $I$ be the incenter of the triangle $ABC$.\n\n\n\nFrom the point $I$ let us draw parallel lines to $AB$ and $BC$, intersecting $AK$ and $KC$ at $P$ and $Q$, respectively. In the triangle $IPQ$ we have $\\angle PIQ = \\angle ABC$, $IQ = s - c$, $IP = s -... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous >... | English | proof only | null | |
06d2 | $ABC$ is a triangle with $BC > CA > AB$. $D$ is a point on side $BC$, and $E$ is a point on $BA$ produced beyond $A$ so that $BD = BE = CA$. Let $P$ be a point on side $AC$ such that $E, B, D, P$ are concyclic, and let $Q$ be the second intersection point of $BP$ with the circumcircle of $\triangle ABC$. Prove that $AQ... | [
"Using the concyclic points, we have\n$$\n\\angle CAQ = \\angle CBQ = \\angle DBP = \\angle DEP\n$$\nand\n$$\n\\angle QCA = \\angle QBA = \\angle PBE = \\angle PDE.\n$$\nThis implies $\\triangle QAC \\sim \\triangle PED$. Let\n$$\n\\frac{QA}{PE} = \\frac{QC}{PD} = \\frac{AC}{ED} = k.\n$$\n\nApplying Ptolemy's theor... | Hong Kong | HKG TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ey9 | Problem:
499 white rooks and a black king are placed on a $1000 \times 1000$ chess board. The rook and king moves are the same as in ordinary chess, except that taking is not allowed and the king is allowed to remain in check. No matter what the initial situation and no matter how white moves, the black king can alway... | [
"Solution:\n\na. True. Black moves to one end of a main diagonal and then moves along the diagonal to the opposite end. Each of the 499 rooks is in some row. Since black moves through each row, every rook must change row. But each of the rooks is also in some column and so every rook must also change column. A rook... | Soviet Union | 1st ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) True; b) False; c) False | |
0lfg | Given the sequence $\{a_n\}_{n=1}^{\infty}$ defined by
$$
a_n = \frac{1}{4[1 - \log_4 n]}
$$
for all positive integers $n$. Put
$$
b_n = \frac{1}{n^2} \left( a_1 + a_2 + \cdots + a_n - \frac{1}{a_1 + a_2} \right), \quad \forall n \in \mathbb{Z}^+
$$
a) Find a polynomial $P(x)$ with real coefficients such that $b_n = P\... | [
"a) We will prove that the polynomial $P(x) = -\\frac{1}{5}x^2 + x$ satisfies the required properties. Obviously $b_1 = \\frac{4}{5} = P(1) = P\\left(\\frac{1}{1}\\right)$, so we only need to consider the case $n > 1$.\nNotice that, for each non-integer real number $x$, then $[-x] = -[x] - 1$. Thus, for each positi... | Vietnam | Vietnamese MO | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a) P(x) = -1/5 x^2 + x. b) There exists a strictly increasing subsequence with limit 2024/2025. | |
09ol | Two cells of a $2 \times 64$ grid are to be colored. A subrectangle is defined as a rectangle whose sides are aligned with the grid lines and whose vertices lie on grid points. Determine which two cells should be colored so that the number of subrectangles containing at least one of the two colored cells is maximized. ... | [] | Mongolia | MMO2025 Round 3 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Color one cell in each row at columns 26 and 39; the maximum number of subrectangles is 3380. | |
0h0k | What is the least possible value of
$$
(x_1 - x_2)^2 + (x_2 - x_3)^2 + \dots + (x_{n-1} - x_n)^2 + (x_n - x_1)^2,
$$
if $x_1, x_2, \dots, x_n$ are distinct integer numbers. | [
"**Answer:** $4n - 6$.\n\nWe prove by induction that\n$$\n(x_1 - x_2)^2 + (x_2 - x_3)^2 + \\dots + (x_{n-1} - x_n)^2 + (x_n - x_1)^2 \\geq 4n - 6,\n$$\nif $x_1, x_2, \\dots, x_n$ are distinct integer numbers.\n\nThe base is trivial. Indeed, $S_2 = (x_1 - x_2)^2 + (x_2 - x_1)^2 \\geq 2$, because all numbers are inte... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010) | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 4n - 6 | |
0hgm | For any non-negative real numbers $x$ and $y$ show the inequality:
$$
x^2 y^2 + x^2 y + xy^2 \leq x^4 y + x + y^4.
$$ | [
"If $x = 0$ or $y = 0$, the inequality is trivial. Now assume $x, y > 0$. Divide both parts of the inequality by $xy$. We get:\n$$\nxy + x + y \\leq x^3 + \\frac{1}{y} + \\frac{y^3}{x}.\n$$\nApply the famous inequality, which is a consequence of the Cauchy-Schwarz inequality:\n$$\n\\frac{a_1^2}{b_1} + \\frac{a_2^2}... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
07dj | In triangle $ABC$ let $M$ be the midpoint of $BC$. Let $\omega$ be a circle inside of $ABC$ and tangent to $AB, AC$ at $E, F$, respectively. The tangents from $M$ to $\omega$ meet this circle at $P, Q$ such that $P$ and $B$ lie on the same side of $AM$. Lines $PM, BF$ cut each other at $X$, and $Y$ is the intersection ... | [
"Assume that $CP, BQ$ cut each other at point $Z$. Since $2PM = BC$, we conclude that points $B, P, Q, C$ lie on the circle $\\Omega$ with center $M$.\n\n\n\nWe have\n$$\n\\widehat{PZQ} = \\widehat{BPZ} + \\widehat{PBZ} = 90^\\circ + \\widehat{PBQ} = 180^\\circ - \\widehat{MPQ} = \\frac{\\w... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0e36 | Each point on the sides of the triangle $\mathcal{T}$ is either red or blue. Prove that one can find points $A$, $B$, $C$ and $D$ on the sides of the triangle $\mathcal{T}$, such that all four points have the same colour and the quadrilateral $ABCD$ is a trapezoid. | [
"Denote the vertices of the triangle by $A$, $B$ and $C$ and let $A'$, $B'$ and $C'$ be the midpoints of the sides $BC$, $CA$ and $AB$. At least two of the points $A'$, $B'$ and $C'$ are the same colour. We may assume that the points $A'$ and $B'$ are both red. The segment $A'B'$ is parallel to the segment $AB$. If... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
03tk | Given $f(x) = \frac{\sin(\pi x) - \cos(\pi x) + 2}{\sqrt{x}}$ for $\frac{1}{4} \le x \le \frac{5}{4}$, the minimum of $f(x)$ is ______. | [
"By rewriting $f(x)$, we have\n$$\nf(x) = \\frac{\\sqrt{2}\\sin\\left(\\pi x - \\frac{\\pi}{4}\\right) + 2}{\\sqrt{x}}\n$$\nfor $\\frac{1}{4} \\le x \\le \\frac{5}{4}$. Define $g(x) = \\sqrt{2}\\sin(\\pi x - \\frac{\\pi}{4})$, where $\\frac{1}{4} \\le x \\le \\frac{5}{4}$. Then $g(x) \\ge 0$, and $g(x)$ is monotone... | China | China Mathematical Competition | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | English | proof and answer | 4√5/5 | |
04pi | Let $a$ be a positive real number such that $\log_{4a} 40\sqrt{3} = \log_{3a} 45$. Prove that $a^3$ is an integer and find its value. | [] | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | 75 | |
06kj | Let $ABC$ be a triangle with $AB = AC$. A circle $\Gamma$ lies outside triangle $ABC$ and is tangent to line $AC$ at $C$. Point $D$ lies on $\Gamma$ such that the circumcircle of triangle $ABD$ is internally tangent to $\Gamma$. Segment $AD$ meets $\Gamma$ again at $E$. Prove that $BE$ is tangent to $\Gamma$. | [
"Note that $AB^2 = AC^2 = AE \\times AD$. This implies $\\triangle ABE \\sim \\triangle ADB$. It follows that $\\angle AEB = \\angle ABD$. Let $BD$ meet $\\Gamma$ again at $F$. Consider the homothety with centre $D$ taking $\\Gamma$ to the circumcircle of $\\triangle ABD$. Since $EF$ is mapped to $AB$, the lines $E... | Hong Kong | null | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01vm | The point $X$ is marked inside the triangle $ABC$. The circumcircles of the triangles $AXB$ and $AXC$ intersect the side $BC$ again at $D$ and $E$ respectively. The line $DX$ intersects the side $AC$ at $K$, and the line $EX$ intersects the side $AB$ at $L$.
Prove that $LK \parallel BC$. | [
"First we prove that the points $A$, $L$, $X$ and $K$ lie on the same circle. Since the quadrilateral $ABDX$ is cyclic, $\\angle AXK = \\angle ABC$. Similarly $\\angle LXA = \\angle BCA$. Now from the equality $\\angle KAL + \\angle LXK = \\angle CAB + \\angle ABC + \\angle BCA = 180^\\circ$ it follows that the qua... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
02t6 | Problem:
O mágico Magimático diz para uma pessoa da plateia escolher uma peça qualquer de um dominó comum. Tal peça é formada por um par de números de 0 a 6. Em seguida, ele diz para a pessoa escolher um dos números da peça e realizar a seguinte sequência de operações:
1. multiplicá-lo por 5;
2. somar o resultado ante... | [
"Solution:\n\na) Vamos revelar o segredo do mágico. Suponha que o par de números escritos no dominó é $(x, y)$ e que o número escolhido para a sequência de operações foi o $x$. Assim, as operações realizadas pelo membro da plateia foram:\n$$\nx \\rightarrow 5x \\rightarrow 5x+15 \\rightarrow 2(5x+15) \\rightarrow 2... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) The tile is (3, 2).
b) In general, if the announced result is n, then y is the units digit of n and x = (n − 30 − y)/10 (equivalently, x is the tens digit of n minus 3). | |
0cej | Find the real solutions of the equation
$$
\frac{4x - 1}{x^2 - 2x + 2} + \frac{4x + 7}{x^2 + 2x + 2} = \frac{4(2x + 3)}{x^2 + 2}.
$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | x = ±√2, x = (4 + √34)/3, x = (4 - √34)/3 | |
00j8 | Determine the largest number $m$ such that the inequality
$$
(a^2 + 4(b^2 + c^2))(b^2 + 4(c^2 + a^2))(c^2 + 4(a^2 + b^2)) \geq m
$$
holds for all real numbers $a, b$ and $c$ not equal to $0$ and satisfying the condition $\left|\frac{1}{a}\right| + \left|\frac{1}{b}\right| + \left|\frac{1}{c}\right| \le 3$. | [
"We first note that we can consider only positive values of $a, b$ and $c$, since the absolute values of the variables are calculated in all instances (both as the absolute values of their reciprocals and as the squares of the variables). So for now, let $a, b, c > 0$.\n\nBy the geometric-harmonic means inequality,... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 729 | |
02w1 | Problem:
Um quadrado $9 \times 9$ foi dividido em 81 quadrados $1 \times 1$ e 8 deles foram pintados de preto e o restante de branco.
a) Mostre que independente de onde eles tenham sido pintados, sempre é possível encontrar um quadrado $3 \times 3$, formado pela justaposição de quadradinhos e com lados paralelos aos l... | [
"Solution:\n\na) Divida o tabuleiro $9 \\times 9$ em $3 \\times 3 = 9$ subtabuleiros de tamanho $3 \\times 3$ como indicado na figura abaixo:\n\n\n\nComo foram pintados apenas 8 quadradinhos e $8 < 9$, pelo menos um dos quadrados $3 \\times 3$ terá todos os seus quadrados não pintados.\n\nb... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | null | |
0b6m | Find the largest constant $K \ge 0$ such that for any $0 \le k \le K$, and for any non-negative real numbers $a, b, c$, satisfying $a^2 + b^2 + c^2 + kabc = k + 3$, to have $a + b + c \le 3$. | [
"Let us work from first principles. Whenever (at least) one variable is zero, say $c$, it follows $a^2 + b^2 = k + 3$, hence $a + b + c \\le \\sqrt{2(k+3)} \\le 3$ for $k \\le 3/2$, with equality holding for $a = b = 3/2$. We thus have a starting tentative limiting bound of $K = 3/2$.\n\nLet also notice that $a = b... | Romania | 2010 Fourth STARS OF MATHEMATICS COMPETITION | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof and answer | 3/2 | |
0erv | For two positive real numbers $a$ and $b$, which may be equal, what is the smallest possible value of $\frac{a}{b} + \frac{b}{a}$? | [
"It is easy to see that if $a = b$, then the expression is equal to $2$, and a few trials will suggest that $2$ is the smallest value. A proof is that\n\nLet $x = \\frac{a}{b}$, so $x > 0$. Then $\\frac{a}{b} + \\frac{b}{a} = x + \\frac{1}{x}$.\n\nBy the AM-GM inequality:\n\n$$\nx + \\frac{1}{x} \\geq 2\\sqrt{x \\c... | South Africa | South African Mathematics Olympiad Second Round | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 2 | |
00jj | In a triangle $ABC$, $H_a$, $H_b$ and $H_c$ are the feet of the altitudes on the sides $BC$, $CA$ and $AB$ respectively. For which triangles are two of the line segments $H_aH_b$, $H_bH_c$ and $H_cH_a$ of equal length? | [
"We first consider the situation in which no angle in $ABC$ is obtuse. If $ABC$ is right-angled with hypotenuse $AB$, we have $H_a = H_b = C$, and therefore certainly $H_bH_c = H_cH_a$. Any right-angled triangle $ABC$ therefore certainly has the required property. If $ABC$ is not right-angled, the triangles $AH_bB$... | Austria | Austrian 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"
] | null | proof and answer | Exactly the right-angled triangles, the isosceles triangles, and those triangles in which two angles satisfy alpha plus twice beta equals ninety degrees. | |
0bcw | Find all continuous functions $f: \mathbb{R} \to \mathbb{R}$ having the property that $(a^2 + ab + b^2) \int_a^b f(x) \, dx = 3 \int_a^b x^2 f(x) \, dx$, for all $a, b \in \mathbb{R}$. | [
"Choose $a = 0$ and $b = t > 0$. Then $t^2 F(t) = 3 \\int_0^t x^2 f(x) \\, dx$, where $F(t) = \\int_0^t f(x) \\, dx$. Taking the derivative we get $2tF(t) + t^2 f(t) = 3t^2 f(t)$, so $2t(tf(t) - F(t)) = 0$, that is $\\left(\\frac{F(t)}{t}\\right)' = 0$, for any $t \\in (0, \\infty)$. It follows that the function $g... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All constant functions: f(x) = k for some real constant k. | |
0hms | Problem:
Given an $n \times n$ matrix whose entries $a_{i j}$ satisfy $a_{i j} = \frac{1}{i + j - 1}$, $n$ numbers are chosen from the matrix no two of which are from the same row or the same column. Prove that the sum of these $n$ numbers is at least $1$. | [
"Solution:\n\nSuppose that $a_{i j}$ and $a_{k l}$ are among the chosen numbers and suppose that $i < k$ and $j < l$. It is straightforward to show that $a_{i j} + a_{k l} \\geqslant a_{i l} + a_{k j}$. Hence, whenever $a_{i j}$ and $a_{k l}$ with $i < k$ and $j < l$ are among chosen numbers, we can lower the sum b... | United States | Berkeley Math Circle | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0jov | Problem:
Let $w$, $x$, $y$, and $z$ be positive real numbers such that
$$
\begin{aligned}
0 & \neq \cos w \cos x \cos y \cos z \\
2 \pi & = w + x + y + z \\
3 \tan w & = k(1 + \sec w) \\
4 \tan x & = k(1 + \sec x) \\
5 \tan y & = k(1 + \sec y) \\
6 \tan z & = k(1 + \sec z)
\end{aligned}
$$
(Here $\sec t$ denotes $\frac... | [
"Solution:\nAnswer: $\\sqrt{19}$\n\nFrom the identity $\\tan \\frac{u}{2} = \\frac{\\sin u}{1 + \\cos u}$, the conditions work out to $3 \\tan \\frac{w}{2} = 4 \\tan \\frac{x}{2} = 5 \\tan \\frac{y}{2} = 6 \\tan \\frac{z}{2} = k$. Let $a = \\tan \\frac{w}{2}$, $b = \\tan \\frac{x}{2}$, $c = \\tan \\frac{y}{2}$, and... | United States | HMMT February | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | sqrt(19) | |
090b | A quadrilateral $ABCD$ is inscribed in a circle, with $AB = 7$ and $BC = 18$. The bisector of $\angle CDA$ intersects side $BC$ at point $E$, and a point $F$ on segment $DE$ satisfies $\angle AED = \angle FCD$. When $BE = 5$ and $EF = 3$, find the length of segment $DF$. | [
"<table><tr><td>17</td></tr><tr><td>3</td></tr></table>\nSince $\\angle ADE = \\angle FDC$ and $\\angle DEA = \\angle DCF$, triangles $DAE$ and $DFC$ are similar, hence $DA : DE = DF : DC$. Additionally we have $\\angle ADF = \\angle EDC$ and thus triangles $DAF$ and $DEC$ are similar in the same orientation. This ... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 17/3 | |
06h0 | Let $a, b$ be real numbers. It is given that the equation $x^4 + a x^3 + b x^2 + a x + 1 = 0$ has at least one real root. Find the minimum value of $a^2 + b^2$. | [
"The minimum value of $a^2 + b^2$ is $\\frac{4}{5}$.\n\nWLOG assume $a \\ge 0$. (Indeed, by flipping the sign of $a$, we only need to flip the sign of the root $x$.) Clearly, $x \\ne 0$. Rewrite the equation as\n$$\nx^2 + a x + b + \\frac{a}{x} + \\frac{1}{x^2} = 0.\n$$\nThis is the same as\n$$\n\\left(x + \\frac{1... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 4/5 | |
065x | Let $ABCD$ be a quadrilateral inscribed in the circle $c(O, R)$ and let $K, L, M, N, S, T$ be the midpoints of the segments $AB, BC, CD, AD, AC$ and $BD$, respectively. Prove that the centers of the circumcircles of the triangles $KLS, LMT, MNS$ and $NKT$ form a cyclic quadrilateral similar to $ABCD$. | [
"Let $c_1, c_2, c_3, c_4$ be the circumcircles of the triangles $KLS, LMT, MNS$ and $NKT$, respectively. It is easy to prove that the quadrilaterals $KLMN, TLSN$ and $KSMT$ are parallelograms and so $KM, NL, TS$ will pass through the same point $G$ and therefore the points $K, S, L$ are symmetric of the points $M, ... | Greece | SELECTION EXAMINATION | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations... | English | proof only | null | |
0h5q | There are 9 weights with labels $1$ g, $2$ g, $\ldots$, $9$ g respectively. It's known that one of the weights is lighter than the label says, while the other eight labels are correct. Is it possible to detect the counterfeit weight using scales with no additional weights no more than twice? | [
"At the first move, put $1+4+9$ on the left side and $2+5+7$ on the right side. If the total weights are equal, the counterfeit weight is among the other three. Then put on scales the combinations $3+4$ and $1+6$.\n\n* If $3+4=1+6$, they are genuine, hence, $8$ is counterfeit.\n\n* If $3+4<1+6$, then $3$ is counter... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Discrete Mathematics > Algorithms"
] | English | proof and answer | Yes | |
00vd | Determine all real polynomials $P(x)$ such that
$$
P^2(x) + P^2(y) + P^2(x + y) = 2P(x^2 + xy + y^2)
$$
for every $x, y \in \mathbb{R}$. | [
"Setting $y = 0$ we get\n$$\n2P^2(x) + P^2(0) = 2P(x^2) \\qquad (1)\n$$\nfor every $x \\in \\mathbb{R}$. We claim that $P(x)$ is a monomial. Indeed, if this is not the case, let $ax^n$ and $bx^m$, with $n > m$ be the two non-zero terms with the largest powers of $x$. Comparing the coefficients of $x^{m+n}$ in (1) w... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | P(x) = 0, P(x) = 2/3, P(x) = x, P(x) = x^2 | |
02vh | Problem:
a) Considere o tabuleiro $4 \times 4$ da figura a seguir. As letras e os números foram colocados para ajudar a localizar os quadradinhos. Por exemplo, o quadradinho $A 1$ é o do canto superior esquerdo e o $D 4$ é o do canto inferior direito.
Mostre uma maneira de escolhermos 12 quadr... | [
"Solution:\na) Existem várias maneiras de fazer isso (veremos exatamente quantas no próximo item). No exemplo da figura da esquerda abaixo, apenas os quadradinhos $A 1, B 2, C 3$ e $D 4$ não são escolhidos.\n\n\nb) Vamos usar o princípio fundamental da contagem. Na primeira linha, temos 4 m... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 24 ways on the 4×4 board; 576 ways on the 8×8 board. | |
0kcw | Problem:
In $\triangle ABC$, $\omega$ is the circumcircle, $I$ is the incenter and $I_{A}$ is the $A$-excenter. Let $M$ be the midpoint of arc $\widehat{BAC}$ on $\omega$, and suppose that $X, Y$ are the projections of $I$ onto $MI_{A}$ and $I_{A}$ onto $MI$, respectively. If $\triangle XYI_{A}$ is an equilateral tria... | [
"Solution:\n\nUsing Fact 5, we know that $II_{A}$ intersects the circle $(ABC)$ at $M_{A}$, which is the center of $\\left(II_{A}BCXY\\right)$. Let $R$ be the radius of the latter circle. We have $R=\\frac{1}{\\sqrt{3}}$.\n\nWe have $\\angle AIM=\\angle YII_{A}=\\angle YIX=\\frac{\\pi}{3}$. Also, $\\angle II_{A}M=\... | United States | HMMT February 2020 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof and answer | sqrt(6)/7 | |
08bn | Problem:
Dato un numero reale $x$ il simbolo $\lfloor x\rfloor$ indica la sua parte intera (cioè il più grande intero minore o uguale ad $x$) e $\{x\}$ la sua parte frazionaria (cioè $x-\lfloor x\rfloor$). Siano $x$, $y$, $z$ tre numeri reali positivi che soddisfano il seguente sistema:
$$
\left\{\begin{array}{l}
3\l... | [
"Solution:\n\nLa risposta è (D). Iniziamo dalla prima equazione del sistema: abbiamo $19 < 20,3 + \\{y\\} - \\{z\\} < 22$, quindi il valore di $3\\lfloor x\\rfloor$, che è un intero multiplo di $3$, deve essere necessariamente $21$, da cui $\\lfloor x\\rfloor = 7$.\n\nConsideriamo ora la seconda equazione: poiché $... | Italy | Gara di Febbraio | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
0dy8 | Find all integral solutions of the equation
$$
\frac{x^2}{2} + \frac{5}{y} = 7.
$$ | [
"If $(x, y)$ satisfies the equation then so does $(-x, y)$. We can thus limit ourselves to only non-negative values of $x$.\n\nMultiply the equation by $2y$ to get $x^2y + 10 = 14y$. This implies $y = \\frac{10}{14 - x^2}$.\n\nWe see that $14 - x^2$ has to be a divisor of $10$, so $-10 \\leq 14 - x^2 \\leq 10$, $-2... | Slovenia | Slovenija 2008 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (2, 1), (3, 2), (4, -5), (-2, 1), (-3, 2), (-4, -5) | |
091m | Problem:
In a plane the circles $\mathcal{K}_{1}$ and $\mathcal{K}_{2}$ with centers $I_{1}$ and $I_{2}$, respectively, intersect in two points $A$ and $B$. Assume that $\varangle I_{1} A I_{2}$ is obtuse. The tangent to $\mathcal{K}_{1}$ in $A$ intersects $\mathcal{K}_{2}$ again in $C$ and the tangent to $\mathcal{K}... | [
"Solution:\n\n\nSince $A D$ is tangent to $\\mathcal{K}_{2}$, it follows that $\\varangle A C B=\\varangle D A B$. Similarly, $\\varangle A D B=\\varangle B A C$.\nFrom this we have $\\varangle D B C=(\\varangle A D B+\\varangle D A B)+(\\varangle B A C+\\varangle A C B)=2(\\varangle D A B+... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
... | null | proof only | null | |
02jd | Problem:
O perímetro de um retângulo é $100~\mathrm{cm}$ e a diagonal mede $x~\mathrm{cm}$. Qual é a área do retângulo em função de $x$?
A) $625-x^{2}$
B) $625-\frac{x^{2}}{2}$
C) $1250-\frac{x^{2}}{2}$
D) $250-\frac{x^{2}}{2}$
E) $2500-\frac{x^{2}}{2}$ | [
"Solution:\n\nSolução 1:\nComo o perímetro do retângulo é $100$, seu semiperímetro é $50$. Como o semiperímetro de um retângulo é a soma do comprimento com a largura, concluímos que esses são da forma $a$ e $50-a$. A área de um retângulo é o produto do comprimento pela largura. No nosso caso, esta área é $(50-a) \\... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
08zh | Alice and Bob are playing a game using a blackboard. Initially, each of $2, 3, \dots, 50$ is written on the blackboard once. Also, a non-empty subset $S$ of $\{2, 3, \dots, 50\}$ is given. In the first turn, Alice erases all the elements of $S$ from the blackboard. After that, the two players, starting with Bob, take t... | [
"For $k$ primes $p_1, p_2, \\dots, p_k$, denote by $X(p_1, p_2, \\dots, p_k)$ the set of all integers in $\\{2, 3, \\dots, 50\\}$ that do not have any prime factor other than $p_1, p_2, \\dots, p_k$. The blackboard is said to be in $(p_1, p_2, \\dots, p_k)$-good situation or simply good, if the set of all integers ... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 32767 | |
0fcf | Problem:
Hallar todas las funciones reales continuas $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ que cumplen, para todo $x$ real positivo, la condición
$$
x+\frac{1}{x}=f(x)+\frac{1}{f(x)}
$$ | [
"Solution:\nEscribimos\n$$\nx-f=\\frac{1}{f}-\\frac{1}{x}=\\frac{x-f}{x f}\n$$\nde donde sale\n$$\n(x-f)\\left(1-\\frac{1}{x f}\\right)=0\n$$\nDe aquí resulta que, para cada $a>0$, será, o bien $f(a)=a$, o bien $f(a)=\\frac{1}{a}$.\nLas funciones de $\\mathbb{R}^{+}$ en $\\mathbb{R}^{+}$ definidas por $f(x)=x$ y po... | Spain | null | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x; f(x) = 1/x; f(x) = max(x, 1/x); f(x) = min(x, 1/x) | |
0k9m | Problem:
Carl only eats food in the shape of equilateral pentagons. Unfortunately, for dinner he receives a piece of steak in the shape of an equilateral triangle. So that he can eat it, he cuts off two corners with straight cuts to form an equilateral pentagon. The set of possible perimeters of the pentagon he obtain... | [
"Solution:\n\nAssume that the triangle has side length $1$. We will show the pentagon side length $x$ is in $\\left[2 \\sqrt{3}-3, \\frac{1}{2}\\right)$. Call the triangle $A B C$ and let corners $B, C$ be cut. Choose $P$ on $A B$, $Q, R$ on $B C$, and $S$ on $A C$ such that $A P Q R S$ is equilateral. If $x \\geq ... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4\sqrt{3}-6 | |
0cfe | Find all the positive integers $m$ and $n$ so that $n = 3^m + s(n) + 35$, where $s(n)$ is the sum of the digits of $n$. | [] | Romania | 74th NMO Shortlisted Problems | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | No solutions | |
092q | Problem:
A tract of land in the shape of an $8 \times 8$ square, whose sides are oriented north-south and east-west, consists of 64 smaller $1 \times 1$ square plots. There can be at most one house on each of the individual plots. A house can only occupy a single $1 \times 1$ square plot.
A house is said to be blocked ... | [
"Solution:\nLet us represent the tract as an $8 \\times 8$-chessboard, with cells colored black if the corresponding parcel is occupied, and white otherwise. We denote by $(i, j)$ the cell in the $i$-th row and $j$-th column (with the first row being the northernmost and the first column being the westernmost). We ... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 50 | |
0dc2 | Let $E$ be a point lies inside the parallelogram $ABCD$ such that $\angle BCE = \angle BAE$. Prove that the circumcenters of triangles $ABE$, $BCE$, $CDE$, $DAE$ are concyclic. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0d1i | For every real numbers $x_1, \dots, x_n$, prove that
$$
\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2+x_2^2} + \dots + \frac{x_n}{1+x_1^2+\dots+x_n^2} < \sqrt{n}.
$$ | [
"Using Cauchy-Schwarz inequality we have\n$$\n\\begin{aligned}\n& \\left( \\frac{x_1}{1+x_1^2} + \\dots + \\frac{x_n}{1+x_1^2 + \\dots + x_n^2} \\right) \\\\\n& \\le n \\left[ \\frac{x_1^2}{(1+x_1^2)^2} + \\dots + \\frac{x_n^2}{(1+x_1^2 + \\dots + x_n^2)^2} \\right] \\\\\n& \\le n \\left[ \\frac{x_1^2}{1 \\cdot (1+... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null |
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