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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
05pt | Problem:
Trouver tous les polynômes $P$ à coefficients entiers tels que $P(P(n)+n)$ est un nombre premier pour une infinité d'entiers $n$. | [
"Solution:\n\nSoit $P$ un polynôme solution qui n'est pas constant. On peut écrire $P(P(X)+X) = P(X) Q(X)$, avec $Q$ un polynôme à coefficients entiers. En effet, si $P(X) = \\sum_{i=0}^{k} a_{i} X^{i}$, avec $a_{i} \\in \\mathbb{Z}$, on a\n$$\nP(P(X)+X) = \\sum_{i=0}^{k} a_{i}(P(X)+X)^{i} = a_{0} + \\sum_{i=1}^{k}... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | All constant polynomials equal to a prime, and all polynomials P(x) = −2x + b with b an odd integer. | |
0j6a | Problem:
Let $ABCD$ be a quadrilateral inscribed in a circle with center $O$. Let $P$ denote the intersection of $AC$ and $BD$. Let $M$ and $N$ denote the midpoints of $AD$ and $BC$. If $AP=1$, $BP=3$, $DP=\sqrt{3}$, and $AC$ is perpendicular to $BD$, find the area of triangle $MON$. | [
"Solution:\n\nAnswer: $\\frac{3}{4}$\n\nWe first prove that $ONPM$ is a parallelogram. Note that $APD$ and $BPC$ are both $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangles. Let $M'$ denote the intersection of $MP$ and $BC$. Since $\\angle BPM' = \\angle MPD = 30^{\\circ}$, we have $MP \\perp BC$. Since $ON$ is the pe... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane G... | null | proof and answer | 3/4 | |
06nr | Let $u, v, w$ be positive integers, with $v$ and $w$ not exceeding $2023$, such that $u - v = 2w$ and $(u^2 + w^2)(v^2 + w^2)$ is divisible by $289$. How many sets of possible values of $(u, v, w)$ are there? | [
"Answer: $67473$\n\nIt suffices to choose $v$ and $w$ since there is a unique choice for $u = v + 2w$ after fixing $v$ and $w$. Note that $289 = 17^2$ and $2023 = 7 \\times 17^2$. We consider two cases.\n\nCase 1: $w$ is divisible by $17$\n\nIn this case $v$ is also divisible by $17$ (if not, then $u = v + 2w$ not ... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 67473 | |
0fc2 | Problem:
Halla las soluciones enteras de la ecuación:
siendo $p$ un número primo.
$$
p \cdot (x + y) = x \cdot y
$$ | [
"Solution:\nYa que $p$ es primo, $p \\neq 0$ y $p \\neq 1$. De la ecuación resulta que $p$ divide a $x$ o $p$ divide a $y$. Como la ecuación es simétrica respecto de $x$ e $y$, si $(\\alpha, \\beta)$ es solución, también lo será $(\\beta, \\alpha)$.\n\nSi $p$ divide a $x$, $x = p \\cdot a$, ($a \\in \\mathbb{Z}$) l... | Spain | null | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (x, y) in {(0, 0), (2p, 2p), (p(p+1), p+1), (p+1, p(p+1)), (p(1-p), p-1), (p-1, p(1-p))} for prime p | |
0j8u | Problem:
Segments $A A'$, $B B'$, and $C C'$, each of length $2$, all intersect at a point $O$. If $\angle A O C' = \angle B O A' = \angle C O B' = 60^{\circ}$, find the maximum possible value of the sum of the areas of triangles $A O C'$, $B O A'$, and $C O B'$. | [
"Solution:\n\nAnswer: $\\sqrt{3}$\n\nExtend $O A$ to $D$ and $O C'$ to $E$ such that $A D = O A'$ and $C' E = O C$. Since $O D = O E = 2$ and $\\angle D O E = 60^{\\circ}$, we have $O D E$ is an equilateral triangle. Let $F$ be the point on $D E$ such that $D F = O B$ and $E F = O B'$. Clearly we have $\\triangle D... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(3) | |
0kve | Problem:
The spikiness of a sequence $a_{1}, a_{2}, \ldots, a_{n}$ of at least two real numbers is the sum $\sum_{i=1}^{n-1}\left|a_{i+1}-a_{i}\right|$. Suppose $x_{1}, x_{2}, \ldots, x_{9}$ are chosen uniformly and randomly from the interval $[0,1]$. Let $M$ be the largest possible value of the spikiness of a permuta... | [
"Solution:\n\nOur job is to arrange the nine numbers in a way that maximizes the spikiness. Let an element be a peak if it is higher than its neighbor(s) and a valley if it is lower than its neighbor(s). It is not hard to show that an optimal arrangement has every element either a peak or a valley (if you have some... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 79/20 | |
0jtu | Problem:
Let $ABC$ be an acute triangle with incenter $I$ and circumcenter $O$. Assume that $\angle OIA = 90^\circ$. Given that $AI = 97$ and $BC = 144$, compute the area of $\triangle ABC$. | [
"Solution:\n\nWe present five different solutions and outline a sixth and seventh one. In what follows, let $a = BC$, $b = CA$, $c = AB$ as usual, and denote by $r$ and $R$ the inradius and circumradius. Let $s = \\frac{1}{2}(a + b + c)$. In the first five solutions we will only prove that\n$$\n\\angle AIO = 90^\\c... | United States | HMMT February 2016 | [
"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 > Radical axis theorem",
"Geometry > Plane Geometry > Circles > T... | null | proof and answer | 14040 | |
0j6z | Problem:
Let $x$ and $y$ be complex numbers such that $|x|=|y|=1$.
a. Determine the maximum value of $|1+x|+|1+y|-|1+xy|$.
b. Determine the maximum value of $|1+x|+|1+xy|+|1+xy^{2}|+\ldots+|1+xy^{2011}|-1006|1+y|$. | [
"Solution:\n\na. Answer: $2\\sqrt{2}$\n\nb. Answer: $2012\\sqrt{2}$ We divide the terms into 1006 sums of the form\n$$\n|1+xy^{2k}|+|1+xy^{2k+1}| - |1+y|\n$$\nFor each of these, we obtain, as in part a,\n$$\n\\begin{aligned}\n|1+xy^{2k}|+|1+xy^{2k+1}| - |1+y| &\\leq |1+xy^{2k}| + |xy^{2k+1} - y| \\\\\n&= |1+xy^{2k}... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | a) 2√2; b) 2012√2 | |
0d4n | Let $\triangle ABC$ be an acute triangle, with $\widehat{A} > \widehat{B} \geq \widehat{C}$. Let $D$, $E$ and $F$ be the tangency points between the incircle of the triangle and sides $BC$, $CA$, $AB$, respectively. Let $J$ be a point on $(BD)$, $K$ a point on $(DC)$, $L$ a point on $(EC)$ and $M$ a point on $(FB)$, su... | [
"Let $I$ be the incenter of triangle $ABC$. Because $ID = IE = IF$, $AF = FM = JD = DK = LE = EA$, and the angles at $D$, $E$, and $F$ are right, we have by Pythagoras $IA = IM = IJ = IK = IL$. Therefore, the pentagon $AMJKL$ is cyclic.\n\n\n\nOn the other hand, $AM = 2AF = 2EA = LA$. We de... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | English, Arabic | proof only | null | |
0289 | Problem:
Um barco motorizado solta uma boia em um rio de margens retilíneas e paralelas às 10:00 e começa a navegar, na direção determinada pelo rio, contra a correnteza até às 10:15. Depois disto, ele retorna, também na direção determinada pelo rio. Em que instante o barco encontrará novamente a boia? | [
"Solution:\n\nComo tanto o barco quanto a boia vão estar sujeitos aos mesmos efeitos da correnteza do rio, para efeitos práticos, podemos considerar apenas a velocidade relativa do barco em relação à boia e supor que a correnteza é nula. Neste caso, se o barco levou 15 minutos para ir em correnteza parada, ele tamb... | Brazil | null | [
"Math Word Problems"
] | null | final answer only | 10:30 | |
097g | Problem:
Fie paralelipipedul $A B C D A_{1} B_{1} C_{1} D_{1}$, în care $m\left(\angle A_{1} A D\right)=m\left(\angle A_{1} A B\right)=m(\angle D A B)=60^{\circ}$, iar $C_{1} A_{1}=\sqrt{7}~\mathrm{cm}$, $C_{1} B=\sqrt{13}~\mathrm{cm}$, $C_{1} D=\sqrt{19}~\mathrm{cm}$. Determinați distanța de la punctul $A$ la planul ... | [
"Solution:\n\nAplicând teorema cosinusurilor, obținem sistemul\n$$\n\\left\\{\\begin{array}{c}\nA B^{2}+A D^{2}+A B \\cdot A D=7 \\\\\nA D^{2}+A_{1} A^{2}+A D \\cdot A_{1} A=13 \\\\\nA B^{2}+A_{1} A^{2}+A B \\cdot A_{1} A=19\n\\end{array}\\right.\n$$\nRezolvând sistemul\n$$\n\\left\\{\\begin{array}{c}\nA B^{2}+A D^... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2√6/5 cm | |
09dq | $\triangle ABC$ хурц өнцөгт гурвалжинд багтсан $\omega$ тойрог $BC$ тэнхлэгийг $K$ цэгт шүргэнэ. $AD$ өндөр, $AD$-ийн дундаж $M$ байг. Хэрэв $N$ нь $\omega$ болон $KM$-ийн огтлолцолын хоёрдугаар цэг бол $BCN$ гурвалжныг багтаасан тойрог ба $\omega$ тойргууд $N$ цэгт шүргэнэ гэдгийг батал. | [
"$AB = AC$ бол батлах зүйл илт үнэн. $AB < AC$ гэх үнийг авъя. $BC$-ийн дундажийг $A'$ гээд, уг дундаж дээр босгосон $BC$-д перпендикуляр шулуун $NK$-тай $P$ цэгт огтлолцдог байг. Мөн $BCN$ гурвалжныг багтаасан тойргийн төвийг $S$ гэе. Хэрэв бид $N$, $I$, $S$ цэгүүд нэг шулуун дээр оршино гэдгийг баталчихвал бодлог... | Mongolia | ММО-48 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonomet... | Mongolian | proof only | null | |
01q7 | The graph of the parabola $y = x^2$ is drawn on the Cartesian plane $Oxy$. The vertices of a triangle $ABC$ belong to the parabola. The median $BM$ of the triangle is parallel to the ordinate axis and is equal to $2$.
Find the area of the triangle $ABC$. | [
"Answer: $2\\sqrt{2}$.\n\nLet $A(a; a^2)$, $B(b; b^2)$, $C(c; c^2)$. Without loss of generality we assume that $a < c$. Since $M$ is the midpoint of $AC$, we have $M\\left(\\frac{a+c}{2}; \\frac{a^2+c^2}{2}\\right)$. Since $BM \\parallel Oy$, we see that the abscissae of $M$ and $B$ are equal, i.e. $\\frac{a+c}{2} ... | Belarus | Final Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 2√2 | |
088z | Problem:
Sia $ABC$ un triangolo acutangolo; sia $O$ il suo circocentro e siano $P, Q$ i punti (diversi da $A$) in cui rispettivamente l'altezza uscente dal vertice $A$ e il prolungamento di $AO$ incontrano la circonferenza circoscritta ad $ABC$.
a. Si dimostri che gli angoli $B\widehat{A}P$ e $Q\widehat{A}C$ sono con... | [
"Solution:\n\na. Sia $K$ il punto d'intersezione tra $AP$ e $BC$ (ovvero il piede dell'altezza uscente da $A$). Dato che $\\widehat{AKB} = 90^\\circ$, abbiamo che $\\widehat{BAK} = 90^\\circ - \\widehat{ABC}$. Inoltre $\\widehat{ABC} = \\frac{1}{2} \\widehat{AOC}$ (angoli rispettivamente alla circonferenza e al cen... | Italy | Olimpiadi della Matematica - Gara di Febbraio | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00m9 | Man bestimme alle natürlichen Zahlen $n \ge 2$, für die
$$
n = a^2 + b^2
$$
gilt, wobei $a$ der kleinste von 1 verschiedene Teiler von $n$ und $b$ ein beliebiger Teiler von $n$ ist. | [
"Wir unterscheiden für $b$ drei Fälle.\n\n1. $b = 1$. Dann ist $n = a^2 + 1$. Aus $a \\mid n$, das heißt $a \\mid a^2 + 1$, folgt $a \\mid 1$, also der Widerspruch $a = 1$.\n\n2. $b = a$. Dann ist $n = 2a^2$ mit $a$ prim. Weil $n$ gerade ist, muss $a = 2$ sein, was auf $n = 8$ führt.\n\n3. $b > a$. Aus $n = a^2 + b... | Austria | 48. Österreichische Mathematik-Olympiade | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | German | proof and answer | 8, 20 | |
08ho | Problem:
In the circle with radius $R$ the distinct chords $[AB]$ and $[CD]$ are concurrent and form an acute angle with measure $\alpha$. Prove that $AB + CD > 2R \sin \alpha$. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distan... | null | proof only | null | |
04zg | Let the angles of a triangle be $\alpha$, $\beta$, and $\gamma$, the perimeter $2p$ and the radius of the circumcircle $R$. Prove the inequality
$$
cot^2 \alpha + cot^2 \beta + cot^2 \gamma \ge 3 \left( \frac{9R^2}{p^2} - 1 \right).
$$
When is the equality achieved? | [
"Let the opposite sides of the angles $\\alpha$, $\\beta$, and $\\gamma$ be correspondingly $a$, $b$, and $c$. Since $\\cot^2 \\alpha = 1/\\sin^2 \\alpha - 1$ and from the law of sines $1/\\sin \\alpha = 2R/a$, we have $\\cot^2 \\alpha = 4R^2/a^2 - 1$; similarly $\\cot^2 \\beta = 4R^2/b^2 - 1$ and $\\cot^2 \\gamma ... | Estonia | Estonija 2010 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds if and only if the triangle is equilateral. | |
05lg | Problem:
Montrer que si la somme de tous les diviseurs positifs d'un entier $n \geqslant 1$ est une puissance de deux, alors le nombre de diviseurs positifs de $n$ est une puissance de deux. | [
"Solution:\n\nDécomposons $n$ en produit de facteurs premiers : $n=\\prod_{i=1}^{k} p_{i}^{\\alpha_{i}}$. On voit aisément que la somme des diviseurs positifs de $n$, notée $\\sigma(n)$, vaut\n$$\n\\sigma(n)=\\prod_{i=1}^{k}\\left(1+p_{i}+p_{i}^{2}+\\cdots+p_{i}^{\\alpha_{i}}\\right)\n$$\nAinsi, si $1 \\leqslant i ... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
034z | Problem:
The bisectors of $\angle BAC$, $\angle ABC$ and $\angle ACB$ of $\triangle ABC$ meet its circumcircle at points $A_1$, $B_1$ and $C_1$, respectively. The side $AB$ meets the lines $C_1B_1$ and $C_1A_1$ at points $M$ and $N$, respectively, the side $BC$ meets the lines $A_1C_1$ and $A_1B_1$ at points $P$ and $... | [
"Solution:\n\na) If $RT$ is the altitude of $\\triangle CRQ$, then $RT = CR \\sin \\gamma$. Using the Sine theorem for $\\triangle B_1RC$ we get\n$$\nCR = \\frac{B_1C \\sin \\frac{\\alpha}{2}}{\\cos \\frac{\\gamma}{2}} = \\frac{2R \\sin \\frac{\\beta}{2} \\sin \\frac{\\alpha}{2}}{\\cos \\frac{\\gamma}{2}}\n$$\nwher... | Bulgaria | 54. Bulgarian Mathematical 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"
] | null | proof only | null | |
0j42 | Problem:
Let $ABC$ be a triangle with area $1$. Let points $D$ and $E$ lie on $AB$ and $AC$, respectively, such that $DE$ is parallel to $BC$ and $DE / BC = 1/3$. If $F$ is the reflection of $A$ across $DE$, find the area of triangle $FBC$. | [
"Solution:\nLet $AF$ intersect $BC$ at $H$. Since $DE / BC = 1/3$ and $F$ and $A$ are equidistant from $DE$, we have $AF = \\frac{2}{3} AH$ and $FH = AH - AF = \\frac{1}{3} AH$. Furthermore, since $AF$ is perpendicular to $DE$, we have $AH$ and $FH$ are the altitudes of triangles $ABC$ and $FBC$ respectively. There... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 1/3 | |
0goc | Prove that
$$
a^2 b^2 (a^2 + b^2 - 2) \geq (a + b)(ab - 1)
$$
for all positive real numbers $a$ and $b$. | [
"Let $p = ab$ and $s = a + b$. We want to show that $p^2(s^2 - 2p - 2) \\ge s(p-1)$, or equivalently, $p^2s^2 - (p-1)s - 2p^2(p+1) \\ge 0$ when $s^2 \\ge 4p > 0$.\nIf $0 < p \\le 1$, then $p^2s - (p-1) \\ge p^2 \\cdot 2\\sqrt{p} + (1-p) > 0$, and if $p \\ge 1$, then $p^2s - (p-1) \\ge (p^2 \\cdot 2\\sqrt{p} - p) + ... | Turkey | 15th Junior Turkish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0fb3 | Problem:
Show that for any real numbers $x, y > 1$, we have
$$
\frac{x^2}{y - 1} + \frac{y^2}{x - 1} \geq 8.
$$ | [
"Solution:\n\nWe have $(x - 2)^2 \\geq 0$, so $x^2 \\geq 4(x - 1)$. Hence,\n$$\n\\frac{x}{\\sqrt{x - 1}} \\geq 2.\n$$\nNow by AM/GM,\n$$\n\\frac{x^2}{y - 1} + \\frac{y^2}{x - 1} \\geq \\frac{2xy}{\\sqrt{(x - 1)(y - 1)}}.\n$$\nBut the right-hand side $\\geq 2 \\cdot 2 \\cdot 2 = 8$."
] | Soviet Union | 1st CIS | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
03zt | Suppose $f(x) = |\lg(x+1)|$ and real numbers $a, b$ ($a < b$) satisfy $f(a) = f\left(-\frac{b+1}{b+2}\right)$, $f(10a + 6b + 21) = 4\lg 2$. Find the values of $a, b$. | [
"$|\\lg(a+1)| = |\\lg\\left(-\\frac{b+1}{b+2}+1\\right)| = |\\lg\\left(\\frac{1}{b+2}\\right)| = |\\lg(b+2)|.$\nThen either $a+1 = b+2$ or $(a+1)(b+2) = 1$. Since $a < b$, so $a+1 \\neq b+2$. Therefore, $(a+1)(b+2) = 1$.\nFrom $f(a) = |\\lg(a+1)|$ we know $0 < a+1 < 1$. Then\n$$\n0 < a + 1 < b + 1 < b + 2,\n$$\nwhi... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | a = -2/5, b = -1/3 | |
05yz | Problem:
Trouver tous les entiers $n \geqslant 2$ pour lesquels il existe un nombre réel $r>0$ et $n$ nombres réels $a_{1}<a_{2}<\cdots<a_{n}$ tels que les $n(n-1) / 2$ différences $a_{j}-a_{i}$ obtenues lorsque $1 \leqslant i<j \leqslant n$ sont égales, à l'ordre près, aux nombres $r^{1}, r^{2}, \ldots, r^{n(n-1) / 2... | [
"Solution:\n\nDans un tel problème, il est indispensable de commencer par étudier les petits cas.\n\n$\\triangleright$ Lorsque $n=2$, il suffit de choisir $a_{1}=0$ et $a_{2}=r=1$. Par conséquent, $n=2$ convient.\n\n$\\triangleright$ Lorsque $n=3$, il suffit de choisir $a_{1}=0, a_{2}=r=\\phi$ et $a_{3}=\\phi^{3}$,... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2, 3, 4 | |
01ni | Find all possible values of real number $\alpha$ such that there exist a function $f: \mathbb{R} \rightarrow \mathbb{R}$, and real number $\alpha$ satisfying the equalities $f(\alpha) = 0$ and $f(f(x)) = x f(x) + \alpha$ for all real $x$. | [
"Answer: $a=0$.\n\nIndeed, if $a=0$, then the function $f \\equiv 0$ satisfies the condition.\n\nNow let $a \\ne 0$. Suppose that $f(\\alpha) = 0$ for some $\\alpha$. We have $f(0) = f(f(\\alpha)) = \\alpha \\cdot f(\\alpha) + a = a$. Then $f(a) = f(f(0)) = 0 \\cdot f(0) + a = a$. Therefore, $a = f(a) = f(f(a)) = a... | Belarus | 62nd Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | alpha = 0 | |
0ckf | Let $ABC$ be a scalene acute triangle with incentre $I$ and circumcentre $O$. Let $AI$ cross $BC$ at $D$. On circle $ABC$, let $X$ and $Y$ be the mid-arc points of $ABC$ and $BCA$, respectively. Let $DX$ cross $CI$ at $E$ and let $DY$ cross $BI$ at $F$. Prove that the lines $FX$, $EY$ and $IO$ are concurrent on the ext... | [
"The argument hinges on the claim below:\n\n**Claim.** The lines $AE$ and $BI$ are perpendicular; similarly, $AF$ and $CI$ are perpendicular\n\n*Proof.* Let $\\alpha = \\angle BAC$, $\\beta = \\angle CAB$ and $\\gamma = \\angle ACB$. Let $DX$ cross the circle $ADC$ again at $D'$. Note that $\\angle ECX = \\angle AC... | Romania | 75th NMO Selection Tests | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Colline... | English | proof only | null | |
04co | A triangle with the orthocentre $H$ and circumcentre $O$ is given. If one of the angles of the triangle is $60^\circ$, prove that the angle bisector of that angle is perpendicular to the line $OH$. | [
"Without loss of generality, let $\\angle BAC = 60^\\circ$ and $|AB| > |CA|$. Let $D$ be the intersection of the angle bisector of $\\angle BAC$ and the circumcircle of the triangle $ABC$. Point $D$ lies on the perpendicular bisector of $BC$. Let $P$ be the midpoint of $BC$. Since $|OC| = |OD|$ and $\\angle COD = 2... | Croatia | Mathematica competitions in Croatia | [
"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"
] | English | proof only | null | |
0hx1 | Problem:
In the pattern shown below, row 1 (the bottom row) consists of two $1$'s, and row $n$ is formed by taking row $n-1$ and inserting between each adjacent pair of numbers $a$ and $b$ their sum $a+b$:
| 1 | 5 | 4 | 7 | 3 | 8 | 5 | 7 | 2 | 7 | 5 | 8 | 3 | 7 | 4 | 5 | 1 |
| :--- | :--- | :--- | :--- | :--- | :--- ... | [
"Solution:\n\nThe number of copies of $n$ in row $n$ is $\\phi(n)$, where $\\phi$ is the totient function, or the number of positive integers less than or equal to $n$ that are relatively prime to $n$. To see this, consider the following pattern of fractions, where we start with $\\frac{0}{1}$ and $\\frac{1}{1}$ an... | United States | Berkeley Math Circle: Monthly Contest 8 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1344 | |
0er3 | Boris takes a taxi to his home but falls asleep when the taxi is halfway to his house. He wakes up when the remaining part of his journey is equal to half the distance covered by the taxi while he was asleep.
If the fraction of the journey for which he slept is $\frac{1}{n}$, what is the value of $n$? | [
"3 Assume that the remaining distance home is 1 unit. Then he slept for 2 units. Hence, the whole journey is 6 units and the fraction of the ride that he slept for is $\\frac{2}{6} = \\frac{1}{3}$."
] | South Africa | South African Mathematics Olympiad Second Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 3 | |
0jn0 | Problem:
Find the number of triples $(a, b, c)$ of positive integers such that $a + a b + a b c = 11$. | [
"Solution:\nWe can write $a + a b + a b c = a(1 + b + b c)$. Since $11$ is prime, $a = 11$ or $a = 1$. But since $b, c$ are both positive integers, we cannot have $a = 11$, and so $a = 1$.\n\nThen $1 + b + b c = 11 \\Longrightarrow b + b c = 10 \\Longrightarrow b(c + 1) = 10$, and since $c$ is a positive integer, o... | United States | HMMT November 2015 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 3 | |
0grt | Point $(a, b)$ on the plane is a *primitive point* if $a, b$ are integers with $\gcd(a, b) = 1$. A graph whose vertices are primitive points is constructed as follows: an edge is drawn between points $(a_1, b_1)$ and $(a_2, b_2)$ if and only if $2a_1 = 2a_2 \in \{b_1 - b_2, b_2 - b_1\}$ or $2b_1 = 2b_2 \in \{a_1 - a_2,... | [
"First, note that point $(a, b)$ is connected to points $(a, b \\pm 2a)$ and $(a \\pm 2b, b)$, so a generic vertex – except finitely many exceptions – of the graph has valency $4$. Now, we observe that for a generic primitive point – except finitely many exceptions – $(a, b)$, exactly $1$ of the $4$ points connecte... | Turkey | Team Selection Test | [
"Discrete Mathematics > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Remove 1 edge; the resulting forest has 5 trees. | |
0ba1 | Prove that
$$
\int_{0}^{1} e^{e^{x}} \, dx \ge \frac{e(e^{2} - 3)}{2}.
$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof only | null | |
0eux | Let $x$, $y$, and $z$ be positive real numbers. Prove that
$$
\frac{2x^2 + xy}{(y + \sqrt{zx} + z)^2} + \frac{2y^2 + yz}{(z + \sqrt{xy} + x)^2} + \frac{2z^2 + zx}{(x + \sqrt{yz} + y)^2} \geq 1.
$$ | [
"### First Solution\nBy the Cauchy-Schwarz Inequality we have\n$$\n(y + z + z)(y + z + x) \\geq (y + z + \\sqrt{zx})^2 = (y + \\sqrt{zx} + z)^2.\n$$\nTherefore,\n$$\n\\frac{2x^2 + xy}{(y + \\sqrt{zx} + z)^2} \\geq \\frac{2x^2 + xy}{(x + y + z)(y + 2z)} = \\frac{2x}{y + 2z} - \\frac{x}{x + y + z}.\n$$\nApply triples... | South Korea | 25th Korean Mathematical Olympiad Final Round | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0ip0 | Problem:
Determine the number of ways of walking from $(0,0)$ to $(5,5)$ using only up and right unit steps such that the path does not pass through any of the following points: $(1,1)$, $(1,4)$, $(4,1)$, $(4,4)$.
 | [
"Solution:\nIn the following figure, each lattice point (with the bottom-left-most point $(0,0)$) is labeled with the number of ways of reaching there from $(0,0)$. With the exception of the forbidden points, the labels satisfy the recursion formula $f(x, y) = f(x-1, y) + f(x, y-1)$. We see from the diagram that th... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 34 | |
0bz4 | a) Show that in a right triangle with an angle of $30^\circ$, the leg opposite to the $30^\circ$-angle has half of the length of the hypotenuse.
b) Inside the triangle $ABC$, with $m(\angle A) = 100^\circ$ and $m(\angle B) = 20^\circ$, we consider the point $D$, such that $m(\angle DAB) = 30^\circ$ and $m(\angle DBA) ... | [
"a) If $M$ is on the hypotenuse of the triangle $ABC$, with a right angle in $A$, with $m(\\angle B) = 30^\\circ$ such that $m(\\angle BAM) = 30^\\circ$, then the triangle $MAC$ is equilateral. Hence $MA = CM = CA = \\frac{1}{2}BC$, and $MB = \\frac{1}{2}BC$, also.\n\nb) We construct $D' \\in (AD)$, such that $CA =... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 40° | |
0exh | Problem:
Two players have a $3 \times 3$ board. $9$ cards, each with a different number, are placed face up in front of the players. Each player in turn takes a card and places it on the board until all the cards have been played. The first player wins if the sum of the numbers in the first and third rows is greater t... | [
"Solution:\n\nThe first player always wins."
] | Soviet Union | 5th ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | The first player always wins. | |
025i | Problem:
Denise e Antônio jogam uma série de 8 jogos no qual o vencedor da primeira partida ganha 1 ponto, o da segunda 2 pontos, o da terceira 4 pontos, o da quarta 8 pontos e assim por diante, multiplicando por 2 o número de pontos de uma partida para a outra. No final, Denise ganhou 31 pontos a mais que Antônio e n... | [
"Solution:\n\n$1^{a}$, $2^{a}$, $3^{a}$, $4^{a}$ e $8^{a}$"
] | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | Denise won the 1st, 2nd, 3rd, 4th, and 8th games. | |
07z9 | Problem:
Una sfera di raggio $r = 15~\mathrm{cm}$ è appoggiata su due binari distanti fra loro $24~\mathrm{cm}$ come in figura. Se la sfera fa una rotazione completa, di quanto avanza sui binari?
(A) $24~\mathrm{cm}$
(B) $30~\mathrm{cm}$
(C) $15\pi~\mathrm{cm}$
(D) $18\pi~\mathrm{cm}$
(E) $30\pi~\mathrm{cm}$
![](att... | [] | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | MCQ | D | |
093a | Problem:
a. Prove that for every positive integer $m$ there exists an integer $n \geq m$ such that
$$
\left\lfloor\frac{n}{1}\right\rfloor \cdot\left\lfloor\frac{n}{2}\right\rfloor \cdots\left\lfloor\frac{n}{m}\right\rfloor=\binom{n}{m}
$$
b. Denote by $p(m)$ the smallest integer $n \geq m$ such that the equation (*)... | [
"Solution:\n\nIt is clear that $p(1)=1$ and $p(2)=3$. From now on we assume that $m \\geq 3$.\n\nFirst, we prove that for all positive integers $n$ and $k$ with $1 \\leq k \\leq n$, it holds that\n$$\n\\left\\lfloor\\frac{n}{k}\\right\\rfloor \\geq \\frac{n-k+1}{k}\n$$\nIndeed, if we write $n$ as $i k+r$ where $0 \... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
05en | Problem:
Soit $ABCD$ un carré de côté $1$. À l'intérieur du carré, on trace les arcs de cercles de centres $A$, $B$, $C$, $D$ et de rayon $1$. Déterminer l'aire de chaque portion délimitée à l'intérieur du carré. | [
"Solution:\n\nConsidérons les aires $a$, $b$, $c$ des portions indiquées sur la figure.\n\n\n\nComme l'aire du carré est égale à $1$, on a\n$$\na + 4b + 4c = 1.\n$$\nComme l'aire du quart de cercle $ABC$ est égale à $\\pi / 4$, on a\n$$\na + 3b + 2c = \\frac{\\pi}{4}.\n$$\nConsidérons maint... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | a = 1 - sqrt(3) + pi/3; b = -1 + sqrt(3)/2 + pi/12; c = 1 - sqrt(3)/4 - pi/6 | |
0izm | Problem:
An ant starts at the origin of a coordinate plane. Each minute, it either walks one unit to the right or one unit up, but it will never move in the same direction more than twice in a row. In how many different ways can it get to the point $(5,5)$? | [
"Solution:\n\nAnswer: $84$\n\nWe can change the ant's sequence of moves to a sequence $a_{1}, a_{2}, \\ldots, a_{10}$, with $a_{i}=0$ if the $i$-th step is up, and $a_{i}=1$ if the $i$-th step is right. We define a subsequence of moves $a_{i}, a_{i+1}, \\ldots, a_{j}$, ($i \\leq j$) as an up run if all terms of the... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 84 | |
006e | Encuentre todas las sucesiones $x_1, x_2, \dots, x_{50}$ de 50 enteros positivos, con máximo común divisor igual a 1, tales que, para cada par de índices distintos $i, j$, el mínimo común múltiplo de $x_i$ y $x_j$ divide a la suma de los cuadrados de los restantes 48 términos. | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Modular Arithmetic"
] | Spanish | proof and answer | x1 = x2 = ... = x50 = 1 | |
06ph | Let $\alpha < \frac{3 - \sqrt{5}}{2}$ be a positive real number. Prove that there exist positive integers $n$ and $p > \alpha \cdot 2^{n}$ for which one can select $2p$ pairwise distinct subsets $S_{1}, \ldots, S_{p}, T_{1}, \ldots, T_{p}$ of the set $\{1, 2, \ldots, n\}$ such that $S_{i} \cap T_{j} \neq \varnothing$ f... | [
"Let $k$ and $m$ be positive integers (to be determined later) and set $n = k m$. Decompose the set $\\{1, 2, \\ldots, n\\}$ into $k$ disjoint subsets, each of size $m$; denote these subsets by $A_{1}, \\ldots, A_{k}$. Define the following families of sets:\n$$\n\\begin{aligned}\n\\mathcal{S} & = \\left\\{ S \\subs... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof only | null | |
04x4 | Prove that there exists a positive integer $n$, such that for all integers $k$ the number $k^2 + k + n$ has no prime divisors less than $2008$. | [
"Let $p < 2008$ be a fixed prime number. There exists $r = r(p)$ such that $k^2 + k \\neq r \\pmod{p}$ for any integer $k$; this follows, for example, from the fact that if $k \\equiv 0 \\pmod{p-1}$ (mod $p$), then $k^2 + k \\equiv 0 \\pmod{p}$.\n\nNow if $\\{p_1, p_2, \\dots, p_m\\}$ is the set of all prime number... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
0fg8 | Problem:
Sea $d$ un entero positivo distinto de $2$, $5$ y $13$. Demostrar que se pueden encontrar elementos distintos $a, b$ en el conjunto $\{2, 5, 13, d\}$, de manera que $ab - 1$ no sea un cuadrado perfecto. | [] | Spain | International Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0jpp | Problem:
Consider the following seven false conjectures with absurdly high counterexamples. Pick any subset of them, and list their labels in order of their smallest counterexample (the smallest $n$ for which the conjecture is false) from smallest to largest. For example, if you believe that the below list is already o... | [
"Solution:\nAnswer: ACGPRES\nThe smallest counterexamples are:\n- Polya's conjecture: $906,150,257$\n- Euler's sum of powers: $31,858,749,840,007,945,920,321$\n- Cyclotomic polynomials: $105$\n- Prime race: $23,338,590,792$\n- Seventeen conjecture: $8,424,432,925,592,889,329,288,197,322,308,900,672,459,420,460,792,... | United States | HMMT November 2015 | [
"Number Theory > Other"
] | null | final answer only | ACGPRES | |
00w5 | Problem:
Let $m$ and $n$ be positive integers. Prove that $25 m+3 n$ is divisible by $83$ if and only if $3 m+7 n$ is divisible by $83$. | [
"Solution:\n\nUse the equality $2 \\cdot (25 m + 3 n) + 11 \\cdot (3 m + 7 n) = 83 m + 83 n$."
] | Baltic Way | Baltic Way | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0dtg | Let $ABCD$ be a parallelogram and the points $E$, $F$ are in the exterior. If triangles $BCF$ and $DEC$ are similar, i.e., $\triangle BCF \sim \triangle DEC$, prove that triangle $AEF$ is similar to these two triangles. | [
"Let $AB = x$, $BF = x'$, $AF = x'$, $AD = y$, $DE = y'$, $AE = y'$, $CF = z$, $CE = z'$, $EF = z'$. Then $x : y' : z' = x : y : z'$.\nLet $\\angle CDE = \\angle FBC = a$, $\\angle DCE = \\angle BFC = b$, $\\angle DEC = \\angle BCF = c$ and $\\angle ADC = \\angle ABC = d$. Then $\\angle BCD = \\angle BAD = 180^\\ci... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kws | Problem:
If $a$ and $b$ are positive real numbers such that $a \cdot 2^{b} = 8$ and $a^{b} = 2$, compute $a^{\log_{2} a} 2^{b^{2}}$. | [
"Solution:\n\nTaking $\\log_{2}$ of both equations gives $\\log_{2} a + b = 3$ and $b \\log_{2} a = 1$.\n\nWe wish to find $a^{\\log_{2} a} 2^{b^{2}}$; taking $\\log_{2}$ of that gives $\\left(\\log_{2} a\\right)^{2} + b^{2}$, which is equal to $\\left(\\log_{2} a + b\\right)^{2} - 2 b \\log_{2} a = 3^{2} - 2 = 7$.... | United States | HMMT February 2023 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 128 | |
0gdu | 由於疫情持續擴大, 政府決定對交通實施管制, 於是在平面上建築了 $n$ 條無限長直線作為牆壁, 其中任兩線不平行, 任三線不共點; 這些牆壁的集合被記為 $W_n$。政府將 $W_n$ 中每一面牆的其中一面塗上藍色, 另一面則塗上綠色, 這讓任兩面牆的交叉點都分為四個角落: 兩面牆都是藍色的同色角落, 兩面牆都是綠色的同色角落, 以及兩面牆一藍一綠的異色角落。為了保持最起碼的交通, 政府在任兩牆的交叉點開了一扇門, 使得人們可以從其異色角落通到另一個異色角落。除此之外, 人們沒有其他方式可以跨越牆。
給定 $W_n$, 令 $k(W_n)$ 為最大的正整數 $k$, 使得不論政府如何塗色, 我們都可以在平面上放置 $k$ 個人, ... | [
"對於所有 $n$, $k(W_n)$ 的唯一可能值是 $k = n + 1$。\n\n由數學歸納法易知 $W_n$ 將平面分為 $C_2^{n+1} + 1$ 區。我們將此題轉化為一個圖 $G$，其中每一點對應一區，而可由門相通的兩區以邊連線。\n\n首先我們證明 $k(W_n) \\ge n + 1$。注意到牆的交點必為 $C_2^n$ 個，因此 $G$ 的邊數必為 $C_2^n$ 個。讓我們將 $G$ 中所有的邊先移除，然後一條一條加回去，則我們每加回一條邊時，$G$ 中的連通區域數至多減一，從而最終 $G$ 的連通區域數至少為 $C_2^{n+1} + 1 - C_2^n = n + 1$。若我們在每個連通區域各擺一... | Taiwan | 2020 Taiwan IMO 3J | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n + 1 | |
00yf | Problem:
An equilateral triangle is divided into $9000000$ congruent equilateral triangles by lines parallel to its sides. Each vertex of the small triangles is coloured in one of three colours. Prove that there exist three points of the same colour being the vertices of a triangle with its sides parallel to the sides... | [
"Solution:\n\nConsider the side $AB$ of the big triangle $ABC$ as \"horizontal\" and suppose the statement of the problem does not hold. The side $AB$ contains $3001$ vertices $A = A_{0}, A_{1}, \\ldots, A_{3000} = B$ of $3$ colours. Hence, there are at least $1001$ vertices of one colour, e.g., red. For any two re... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
09br | Ямар нэг $p \in \mathbb{P}$, $k, n \in \mathbb{N}$ тоонуудын хувьд $b^2 + a = p^k$, $a^2 + b = n p^k$ байх бүх $a, b$ гэсэн ялгаатай натурал тоонуудыг ол. | [
"$b = 1$ бол\n$$\n\\begin{gather*}\na + 1 \\mid a^2 + 1 = (a + 1)^2 - 2a \\Rightarrow \\\\\na + 1 \\mid 2a \\Rightarrow a + 1 \\mid 2 \\Rightarrow a + 1 = 2 \\quad a = 1\n\\end{gather*}\n$$\nболж $a \\neq b$ гэдэгт зөрчинө.\n$$\na^2 + b \\equiv 0 (\\mod (b^2 + a)) \\Rightarrow a^2 \\equiv -b (\\mod (b^2 + a))\n$$\n... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Mongolian | proof and answer | (a, b) = (5, 2) | |
07yk | Problem:
Sia $h$ un numero intero positivo e sia $a_{n}$ la successione definita per ricorrenza nel modo seguente:
$$
\begin{aligned}
a_{0} & =1 \\
a_{n+1} & = \begin{cases}\frac{a_{n}}{2} & \text{ se } a_{n} \text{ è pari } \\
a_{n}+h & \text{ se } a_{n} \text{ è dispari.}\end{cases}
\end{aligned}
$$
(Ad esempio, se... | [
"Solution:\n\nI valori cercati sono tutti e soli gli interi positivi dispari. Infatti, se $h$ è pari, allora $a_{1}=1+h$ è dispari, quindi $a_{2}=1+2 h$ è dispari e così via, quindi tutti i numeri $a_{n}=1+n h$ sono dispari e maggiori di 1 .\nSe invece $h$ è dispari, osserviamo che tutti i termini della successione... | Italy | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All positive odd integers | |
00a8 | We say that a natural number is of type 1 (respectively type 2) if each of its digits at even (respectively odd) position is greater than or equal to each of its adjacent digits. Positions are counted from left to right; leading zeros are not allowed (the first digit is assumed nonzero). One-digit numbers are considere... | [
"The answer is yes for part a) and no for part b).\n\nConsider a number $a > 1$ of type 1. If $a$ is a 1-digit number then $a = (a - 1) + 1$ is the desired representation since $a - 1$ and $1$ are 1-digit numbers, hence type 2 numbers by definition.\n\nLet $a$ have at least two digits. Write it in the form $a = u_1... | Argentina | Argentine National Olympiad 2015 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | a) Yes. b) No. | |
043d | Suppose that arithmetic sequence $\{a_n\}$ satisfies $a_{2021} = a_{20} + a_{21} = 1$. Then the value of $a_1$ is ______. | [
"Let the common difference of $\\{a_n\\}$ be $d$. By the given condition, it follows that\n$$\n\\begin{cases}\na_1 + 2020d = 1, \\\\\n2a_1 + 39d = 1.\n\\end{cases}\n$$\nThe solution is $a_1 = \\frac{1981}{4001}$.\n$\\square$"
] | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 1981/4001 | |
0c51 | Let $m, n$ be integers, $1 \le m \le n$, $A \in \mathcal{M}_{m,n}(\mathbb{C})$, $B \in \mathcal{M}_{n,m}(\mathbb{C})$ and $f_{A,B} : \mathcal{M}_n(\mathbb{C}) \to \mathcal{M}_m(\mathbb{C})$, $f_{A,B}(X) = AXB$; $f_{B,A} : \mathcal{M}_m(\mathbb{C}) \to \mathcal{M}_n(\mathbb{C})$, $f_{B,A}(Y) = BYA$.
Prove that the follo... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Linear transformations"
] | English | proof only | null | |
04a4 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x^2 + f(y)) = y - x^2$ for all $x, y \in \mathbb{R}$. | [
"Taking $x = 0$ in the given equation we get $f(f(y)) = y$ for every $y \\in \\mathbb{R}$.\nTherefore, we have\n$$\nf(y - x^2) = f(f(x^2 + f(y))) = x^2 + f(y). \\quad (3.1)\n$$\nfor every $x, y \\in \\mathbb{R}$.\nTaking $y = x^2$ in (3.1) we get $f(0) = x^2 + f(x^2)$, that is\n$$\nf(x^2) = -x^2 + f(0),\n$$\nwhile ... | Croatia | Hrvatska 2011 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | All solutions are f(x) = c − x for any real constant c. | |
03v1 | Given an integer $n \ge 3$, prove that the set $X = \{1, 2, 3, \dots, n^2 - n\}$ can be divided into two non-intersecting subsets such that neither of them contains $n$ elements $a_1, a_2, \dots, a_n$ with $a_1 < a_2 < \dots < a_n$ and $a_k \le \frac{a_{k-1} + a_{k+1}}{2}$ for all $k = 2, \dots, n-1$. | [
"Define\n$$\nS_k = \\{k^2 - k + 1, k^2 - k + 2, \\dots, k^2\\},\n$$\n$$\nT_k = \\{k^2 + 1, k^2 + 2, \\dots, k^2 + k\\}.\n$$\nLet $S = \\bigcup_{k=1}^{n-1} S_k$, $T = \\bigcup_{k=1}^{n-1} T_k$. We will prove that $S$, $T$ are the required subsets of $X$.\nFirstly it is easy to verify that $S \\cap T = \\emptyset$ an... | China | China Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0jpi | Problem:
How many ways are there to place four points in the plane such that the set of pairwise distances between the points consists of exactly 2 elements? (Two configurations are the same if one can be obtained from the other via rotation and scaling.) | [
"Solution:\nAnswer: 6\nLet $A$, $B$, $C$, $D$ be the four points. There are 6 pairwise distances, so at least three of them must be equal.\n\nCase 1: There is no equilateral triangle. Then WLOG we have $AB = BC = CD = 1$.\n\n- Subcase 1.1: $AD = 1$ as well. Then $AC = BD \\neq 1$, so $ABCD$ is a square.\n\n- Subcas... | United States | HMMT November 2015 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | null | final answer only | 6 | |
0c1z | Let $f: \mathbb{R} \to \mathbb{R}$ be a function which has the Darboux property. Prove that if $f$ is injective on the set of irrational numbers, then $f$ is continuous on $\mathbb{R}$. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof only | null | |
06pm | Determine the smallest positive real number $k$ with the following property.
Let $ABCD$ be a convex quadrilateral, and let points $A_{1}, B_{1}, C_{1}$ and $D_{1}$ lie on sides $AB, BC$, $CD$ and $DA$, respectively. Consider the areas of triangles $AA_{1}D_{1}, BB_{1}A_{1}, CC_{1}B_{1}$, and $DD_{1}C_{1}$; let $S$ be t... | [
"Throughout the solution, triangles $AA_{1}D_{1}, BB_{1}A_{1}, CC_{1}B_{1}$, and $DD_{1}C_{1}$ will be referred to as border triangles. We will denote by $[\\mathcal{R}]$ the area of a region $\\mathcal{R}$.\n\nFirst, we show that $k \\geq 1$. Consider a triangle $ABC$ with unit area; let $A_{1}, B_{1}, K$ be the m... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 1 | |
0htd | Problem:
For $x$ a positive real number with finitely many decimal places, denote by $r(x)$ the number formed by reversing the digits and decimal point of $x$. For instance, $r(98.6)=6.89$ and $r(740)=0.047$.
a. Prove that for all positive real numbers $x$ and $y$ with finitely many decimal places,
$$
r(x y) \leq 10 ... | [
"Solution:\n\na. Write\n$$\nx=\\sum_{i=i_{0}}^{i_{1}} x_{i} 10^{i} \\quad \\text{ and } \\quad y=\\sum_{j=j_{0}}^{j_{1}} y_{j} 10^{j}\n$$\nwhere the $x_{i}$ and $y_{j}$ are digits, and the limits $i_{0}, i_{1}, j_{0}, j_{1}$ are integers, possibly negative. Then\n$$\nr(x)=\\sum_{i=i_{0}}^{i_{1}} x_{i} 10^{-1-i} \\q... | United States | Berkeley Math Circle | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Yes. For example, let x be the integer whose decimal representation is two thousand fifteen consecutive ones, and let y be the integer whose decimal representation consists of two thousand fifteen ones, each pair of consecutive ones separated by two thousand fourteen zeros. For these, equality holds. | |
0j6g | Problem:
In how many ways may thirteen beads be placed on a circular necklace if each bead is either blue or yellow and no two yellow beads may be placed in adjacent positions? (Beads of the same color are considered to be identical, and two arrangements are considered to be the same if and only if each can be obtaine... | [
"Solution:\n\nAnswer: $41$\n\nLet $t_{n}$ be the number of arrangements of $n$ beads in a row such that bead $i$ and $i+1$ are not both yellow for $1 \\leq i < n$. Let $a_{n}$ and $b_{n}$ be the number of arrangements satisfying the additional condition that beads $n$ and $1$ are not both yellow, and that beads $n$... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 41 | |
0gsy | Each of 29 guests at the party wear a hat of one of three colours. A guest is a *lucky* if at least two of its friends in the party wear differently coloured hats. Show that it is always possible to choose a guest and to change its hat to hat of one of the remaining two colours such that the total number of lucky guest... | [
"Assume the contrary. Each lucky guest has at least two friends. Note that a lucky guest having exactly 2 friends can be made unlucky in two different ways: the hat colours of her friends should be changed from $(a, b)$ to either $(a, a)$ or $(b, b)$. A lucky guest having at least 3 friends can be made unlucky in j... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
00jd | Determine all integer solutions of the equation
$$
(x - 1) \cdot x + (x + 1) + (y - 1) \cdot y + (y + 1) = 24 - 9 \cdot xy.
$$ | [
"Since $(x - 1) \\cdot x \\cdot (x + 1) + (y - 1) \\cdot y \\cdot (y + 1) = x^3 + y^3 - x - y$, adding $3xy(x + y)$ to both sides of the equation yields the equivalent equation\n$$\n(x+y)^3 - (x+y) = 24 + 3xy(x+y-3) \\iff (x+y)^3 - 27 - (x+y-3) = 3xy(x+y-3).\n$$\nSince $(x + y)^3 - 27 = (x + y - 3)((x + y)^2 + 3(x ... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | All integer pairs with x + y = 3, together with the six pairs (-2, -2), (-3, -2), (-2, -3), (-4, -3), (-3, -4), (-4, -4). | |
0jli | Problem:
Prove that there exists a nonzero complex number $c$ and a real number $d$ such that
$$
\left|\left|\frac{1}{1+z+z^{2}}\right|-\left|\frac{1}{1+z+z^{2}}-c\right| \right|=d
$$
for all $z$ with $|z|=1$ and $1+z+z^{2} \neq 0$. (Here, $|z|$ denotes the absolute value of the complex number $z$, so that $|a+b i|=... | [
"Solution:\n\nLet $f(z) = \\left|\\frac{1}{1+z+z^{2}}\\right|$. Parametrize $z = e^{i t} = \\cos t + i \\sin t$ and let $g(t) = f\\left(e^{i t}\\right)$, $0 \\leq t < 2\\pi$. Writing out $\\frac{1}{1+z+z^{2}}$ in terms of $t$ and simplifying, we find that $g(t) = \\frac{\\cos t - i \\sin t}{1 + 2 \\cos t}$. Letting... | United States | HMMT 2014 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0bvp | Consider an acute triangle $ABC$ in which $A_1$, $B_1$, and $C_1$ are the feet of the altitudes dropped from $A$, $B$, and $C$, respectively, and $H$ is the orthocenter. The perpendiculars dropped from $H$ onto $A_1C_1$ and $A_1B_1$ intersect lines $AB$ and $AC$ at $P$ and $Q$, respectively. Prove that the line perpend... | [
"Point $H$ is the incenter of triangle $A_1B_1C_1$, while $A$ is the excenter of the same triangle, corresponding to the side $B_1C_1$. Let $D$, $E$, and $F$ be the orthogonal projections of points $P$, $Q$, and $H$, respectively, onto the line $B_1C_1$. Let $S$ be the intersection of lines $PH$ and $A_1C_1$, and l... | Romania | The Danube Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
076x | Let $ABCDE$ be a convex pentagon in which $\angle A = \angle B = \angle C = \angle D = 120^\circ$ and side lengths are five consecutive integers in some order. Find all possible values of $AB + BC + CD$. | [] | India | INMO-2017 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | 6, 7, 9 | |
0fp5 | Sea $n \ge 1$ un número entero. Probar que $\sum_{0 \le k < n/2} \binom{n}{2k+1} 13^k$ es divisible por $2^{n-1}$. | [
"Sea $a_n = \\sum_{0 \\le k < n/2} \\binom{n}{2k+1} 13^k$. Teniendo en cuenta que $\\binom{n}{j} = 0$ para $j < 0, j > n$ y la relación\n$$\n\\binom{n}{j} + \\binom{n}{j+1} = \\binom{n+1}{j+1}\n$$\nresulta $a_1 = 1$, $a_2 = 2$ y para $n \\ge 3$,\n$$\n\\begin{align*}\na_n - a_{n-1} &= \\sum_{0 \\le k < n/2} \\binom{... | Spain | PREPARACIÓN BARCELONA CONTEST | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | Spanish | proof only | null | |
0gye | Let be given positive numbers $a, b, c$. Prove that:
$$
\frac{a}{2a^2 + b^2 + c^2} + \frac{b}{2b^2 + c^2 + a^2} + \frac{c}{2c^2 + a^2 + b^2} \le \frac{9}{4(a+b+c)}
$$ | [
"Using the inequality $a^2 + b^2 + c^2 \\ge ab + bc + ca$ we have\n$$\n\\frac{a}{(a+b)(a+c)} + \\frac{b}{(b+c)(b+a)} + \\frac{c}{(c+a)(c+b)} \\le \\frac{9}{4(a+b+c)}\n$$\n, is equivalent to\n$$ 8(a+b+c)(ab+bc+ca) \\le 9(a+b)(b+c)(c+a) $$\nand this equivalent to inequality\n$$ 6abc \\le a^2b + a^2c + b^2a + b^2c + c... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
01ld | Prove that there is no function $f: \mathbb{R} \to \mathbb{R}$ such that $f(f(x)) = 1 - x f(x)$ for all real $x$.
(I. Voronovich) | [
"Suppose that there exists a function $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying\n$$\nf(f(x)) = 1 - x f(x) \\quad (*)\n$$\nfor any $x \\in \\mathbb{R}$.\n\nSet $c = f(0)$. Using $(*)$ for $x = 0$, we obtain $f(c) = f(f(0)) = 1 - 0 \\cdot f(0) = 1 - 0 \\cdot c = 1$.\nSince\n$$\n1 = f(c), \\quad (1)\n$$\nwe have $f... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
0dn1 | Problem:
Наћи све природне бројеве $a$ и $b$ такве да
$$
a \mid b^{2}, \quad b \mid a^{2} \quad \text{и} \quad a+1 \mid b^{2}+1.
$$ | [
"Solution:\n\nНека је $b^{2}=c a$. Услови задатка дају $b^{2}=c a \\mid a^{4}$ и $a+1 \\mid c a+1$, а то је еквивалентно са\n$$\nc \\mid a^{3} \\quad \\text{и} \\quad a+1 \\mid c-1.\n$$\nНека је $c=d(a+1)+1,\\ d \\in \\mathbb{N}_{0}$. Како је $a^{3} \\equiv -1 \\pmod{a+1}$, имамо $\\frac{a^{3}}{c} \\equiv -1 \\pmod... | Serbia | Serbian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All pairs (a, b) of the forms (t^2, t), (t^2, t^3), and (t^2(t^2 − 1), t(t^2 − 1)^2), where t is a natural number (with t ≥ 2 in the third family). | |
0267 | Problem:
Um sanduíche e um prato de refeição custam em média $R\$ 5,00$ e $R\$ 7,00$, respectivamente. De quantas maneiras pode-se comprar sanduíches e pratos de refeição com $R\$ 90,00$, sem deixar troco? | [
"Solution:\n\nSe $S$ corresponde ao número de sanduíches e $P$ ao número de pratos de refeição, então $5S + 7P = 90$. Precisamos encontrar soluções inteiras $S$ e $P$ para essa equação. Temos:\n\n$$\n5S + 7P = 90 \\Longrightarrow P = \\frac{90 - 5S}{7} = 5 \\times \\frac{18 - S}{7}\n$$\n\nComo $P$ é um número natur... | Brazil | Lista 2 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 3 | |
0kio | Problem:
Compute the number of ordered pairs of integers $(a, b)$, with $2 \leq a, b \leq 2021$, that satisfy the equation
$$
a^{\log_{b}\left(a^{-4}\right)} = b^{\log_{a}\left(b a^{-3}\right)}.
$$ | [
"Solution:\n\nTaking $\\log_{a}$ of both sides and simplifying gives\n$$\n-4 \\log_{b} a = \\left(\\log_{a} b\\right)^2 - 3 \\log_{a} b\n$$\nPlugging in $x = \\log_{a} b$ and using $\\log_{b} a = \\frac{1}{\\log_{a} b}$ gives\n$$\nx^3 - 3x^2 + 4 = 0\n$$\nWe can factor the polynomial as $(x-2)(x-2)(x+1)$, meaning $b... | United States | HMMT Spring 2021 | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 43 | |
0ah2 | Let $ABCDEF$ be a convex hexagon of area $1$ whose opposite sides are parallel. The lines $AB$, $CD$ and $EF$ meet in pairs to determine the vertices of a triangle. Similarly, the lines $BC$, $DA$ and $FA$ meet in pairs to determine the vertices of another triangle. Show that the area of the least one of these two tria... | [
"Unless otherwise stated, throughout the proof indices take on values from $0$ to $5$ and are reduced modulo $6$. Label the vertices of the hexagon in circular order, $A_0, A_1, A_2, A_3, A_4, A_5$, and let the lines of support of the alternate sides $A_iA_{i+1}$ and $A_{i+2}A_{i+3}$ meet at $B_i$. To show that the... | North Macedonia | XXVIII-th Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof only | null | |
0dea | Let $F$ be the set of all sequences $(a_1, a_2, \dots, a_{2020})$ with $a_i \in \{-1, 1\}$ for all $i = 1, 2, \dots, 2020$. Prove that there exists a set $S$, such that $S \subset F$, $|S| = 2020$ and for any $(a_1, a_2, \dots, a_{2020}) \in F$ there exists $(b_1, b_2, \dots, b_{2020}) \in S$, such that $\sum_{i=1}^{20... | [
"For each $i \\in \\{1, 2, \\dots, 2021\\}$, let $e_i = (\\underbrace{1, \\dots, 1}_{i-1}, \\underbrace{-1, \\dots, -1}_{2021-i})$.\nFor two sequences $a = (a_1, a_2, \\dots, a_{2020})$, $b = (b_1, b_2, \\dots, b_{2020})$, we define\n$$\na \\cdot b = a_1 b_1 + a_2 b_2 + \\dots + a_{2020} b_{2020}.\n$$\nIt's easy to... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Algebra > Linear Algebra > Vectors",
"Discrete Mathematics > Other"
] | null | proof only | null | |
04lu | On a $2017 \times 2017$ board, some of the squares are occupied by a single ladybird; the rest of the squares are empty. The ladybirds move, never leaving the board, according to the following principles. Every second, each ladybird moves to a neighbouring square. The moves are horizontal (to the square immediately to ... | [
"We claim that the required number is $2016^2 + 1$. Let us first show that we can find an arrangement of $2016^2$ ladybirds and choose their paths so that no two occupy the same square at the same moment in time. We place the ladybirds in the lower left $2016 \\times 2016$ square of the board and let them all move ... | Croatia | Mathematical competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 4064257 | |
0jd7 | Problem:
A standard 52-card deck contains cards of 4 suits and 13 numbers, with exactly one card for each pairing of suit and number. If Maya draws two cards with replacement from this deck, what is the probability that the two cards have the same suit or have the same number, but not both? | [
"Solution:\nAnswer: $\\quad \\frac{15}{52}$\n\nAfter drawing the first card, there are 12 other cards from the same suit and 3 other cards with the same number, so the probability is $\\frac{12+3}{52}$."
] | United States | HMMT | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 15/52 | |
0bbm | Given an integer number $n \ge 2$, evaluate the sum
$$
\sum_{\sigma \in S_n} (\operatorname{sgn} \sigma) n^{\ell(\sigma)},
$$
where $S_n$ is the set of all $n$-element permutations, and $\ell(\sigma)$ is the number of disjoint cycles in the standard decomposition of $\sigma$. | [
"The sum in question is $f_n(n) = n!$, where\n$$\nf_n(x) = \\sum_{\\sigma \\in S_n} (\\operatorname{sgn} \\sigma) x^{\\ell(\\sigma)} = x(x-1)\\cdots(x-n+1).\n$$\n\nThe latter is a straightforward consequence of the following recurrence formula:\n$$\nf_n(x) = x f_{n-1}(x) - (n-1) f_{n-1}(x) = (x-n+1) f_{n-1}(x), \\q... | Romania | 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | n! | |
0f4s | Problem:
An $n \times n$ array has a cross in $n - 1$ cells. A move consists of moving a row to a new position or moving a column to a new position. For example, one might move row 2 to row 5, so that row 1 remained in the same position, row 3 became row 2, row 4 became row 3, row 5 became row 4, row 2 became row 5 an... | [] | Soviet Union | 16th ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
009a | There are $1000$ balls distributed in $79$ identical boxes. There may be empty boxes, but all the balls are not in one box. There are two allowed operations:
* Pass exactly $13$ balls from one box to another.
* Pass exactly $66$ balls from one box to another.
The balls are distributed so that it is impossible to gather... | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | English | proof and answer | Exactly the distributions where every box has either 12, 25, 38, 51, or 64 balls, with the counts satisfying t + 2u + 3v + 4w = 4 (where t, u, v, w are the numbers of boxes with 25, 38, 51, 64 balls respectively) and all remaining boxes have 12 balls. Concretely, the five possibilities are:
- 78 boxes with 12 balls and... | |
02vb | Problem:
A mediatriz de um segmento $X Y$ é a reta perpendicular ao segmento passando por seu ponto médio. A principal propriedade da mediatriz é que um ponto está sobre ela se, e somente se, a distância desse ponto até $X$ é igual à distância desse ponto até $Y$. Uma afirmação formada usando "se, e somente se" é equi... | [
"Solution:\n\na) Como os triângulos $P M X$ e $P M Y$ são retângulos e $M X=M Y$, pelo Teorema de Pitágoras, temos:\n$$\nP X^{2}=P M^{2}+M X^{2}=P M^{2}+M Y^{2}=P Y^{2}\n$$\nPortanto, $P X=P Y$.\n\nb) Veja que as mediatrizes de $A B$ e $B C$ não são paralelas, pois esses segmentos não são paralelos. Logo elas se en... | Brazil | Brazilian 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 > Constructions and loci",
"Geometry > Plane Geometry > Mis... | null | proof only | null | |
0iht | Problem:
Suppose we have an $n$-gon such that each interior angle, measured in degrees, is a positive integer. Suppose further that all angles are less than $180^{\circ}$, and that all angles are different sizes. What is the maximum possible value of $n$? Prove your answer. | [
"Solution:\n\nLet's work with the exterior angles (each is $180$ minus the interior angle). Then the conditions on the exterior angles are identical to the conditions on the interior angles: each is a positive integer between $1$ and $179$ inclusive. The sum of the exterior angles is exactly $360$.\n\nHowever, the ... | United States | Harvard-MIT Mathematics Tournament, Team Round A | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 26 | |
05xg | Problem:
Déterminer toutes les fonctions $f$ injective de $\mathbb{N}$ dans $\mathbb{N}$ telles que tout polynôme $\sum_{i=0}^{n} a_{i} x^{i}$ a une racine réelle si et seulement si le polynôme $\sum_{i=0}^{n} a_{i} x^{f(i)}$ a une racine réelle. | [
"Solution:\n\nIl s'avère que la condition d'injectivité était inutile : nous allons prouver le résultat sans l'injectivité de $f$. Notons que le polynôme $1$ n'a pas de racine réelle, la condition de l'énoncé nous apprend que $X^{f(0)}$ n'en a pas non plus : la seule possibilité pour $f(0)$ est donc $f(0)=0$. De mê... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | All such functions are f(n) = a n for all natural numbers n, where a is an odd integer. | |
0baw | Let $ABC$ be a triangle such that $AB < AC$. The perpendicular bisector of the side $BC$ meets the side $AC$ at the point $D$, and the (interior) bisectrix of the angle $ADB$ meets the circumcircle $ABC$ at the point $E$. Prove that the (interior) bisectrix of the angle $AEB$ and the line through the incentres of the t... | [
"The lines $BC$ and $DE$ are parallel, so the angles $BED$ and $DAE$ are equal. Then so are the angles $AED$ and $DBE$. Let $I$ and $J$ be the incentres of the triangles $ADE$ and $BDE$, respectively. It follows that the triangles $DIE$ and $DJB$ are similar, so $DI/DE = DJ/DB$. Since the angles $IDJ$ and $EDB$ are... | Romania | 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"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 only | null | |
0itg | Problem:
Determine the number of ways to select a sequence of 8 sets $A_{1}, A_{2}, \ldots, A_{8}$, such that each is a subset (possibly empty) of $\{1,2\}$, and $A_{m}$ contains $A_{n}$ if $m$ divides $n$. | [
"Solution:\nConsider an arbitrary $x \\in \\{1,2\\}$, and let us consider the number of ways for $x$ to be in some of the sets so that the constraints are satisfied. We divide into a few cases:\n\n- Case: $x \\notin A_{1}$. Then $x$ cannot be in any of the sets. So there is one possibility.\n\n- Case: $x \\in A_{1}... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2025 | |
0flo | Determine whether the number $\lambda_n = \sqrt{3n^2 + 2n + 2}$ is irrational for every non-negative integer $n$. Justify your answer. | [] | Spain | Spanija 2012 | [
"Number Theory > Other",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | It is irrational for every non-negative integer n. | |
04fk | Find the least positive integer $a$ such that the expression
$$
\frac{n^8 + n^6 + n^4 + n^2 + a}{n^2 - 1}
$$
for $n = 2014$ is an integer divisible by $3$. (Mea Bombardelli) | [
"**Arranging the expression we get**\n$$\n\\begin{aligned}\n\\frac{n^8 + n^6 + n^4 + n^2 + a}{n^2 - 1} &= \\frac{n^8 - n^6 + 2(n^6 - n^4) + 3(n^4 - n^2) + 4(n^2 - 1) + a + 4}{n^2 - 1} \\\\\n&= n^6 + 2 \\cdot n^4 + 3 \\cdot n^2 + 4 + \\frac{a+4}{n^2-1}.\n\\end{aligned}\n$$\nSince $2014 \\equiv 1 \\pmod{3}$, every po... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | 8112386 | |
088k | Problem:
Sono dati due numeri reali $a$ e $b$ tali che $|3 a-b+1|=|b|$. Allora necessariamente si ha...
(A) $b=0$
(B) $a=-\frac{1}{3}$
(C) $a<0$
(D) $b=\frac{3 a+1}{2}$
(E) nessuna delle precedenti. | [
"Solution:\n\nLa risposta è (E). Infatti, nel piano cartesiano aventi assi $a$ e $b$, la regione descritta dall'equazione $|3 a-b+1|=|b|$ è data dall'unione della retta $3 a+1=0$ e dalla retta $3 a-2 b+1=0$. Quindi nessuna delle prime quattro risposte proposte è corretta."
] | Italy | Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | E | |
0czm | Let $P$ be a point in the interior of triangle $A B C$. Lines $A P, B P, C P$ intersect sides $B C, C A, A B$ at $L, M, N$, respectively. Prove that
$$
A P \cdot B P \cdot C P \geq 8 P L \cdot P M \cdot P N .
$$ | [
"Let $K_{A}, K_{B}, K_{C}$ be the areas of triangle $B P C$, $C P A$, $A P B$, respectively. Let $A'$ and $P'$ be the projections of $A$ and $P$ on side $B C$. Triangles $A L A'$ and $P L P'$ are similar, hence we have\n$$\n\\frac{A L}{P L}=\\frac{A A'}{P P'}=\\frac{A A' \\cdot B C}{P P' \\cdot B C}=\\frac{K}{K_{A}... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0fjy | Problem:
Un poliedro convexo tiene por caras 12 cuadrados, 8 hexágonos regulares y 6 octógonos regulares. En cada vértice del poliedro concurren exactamente un cuadrado, un hexágono y un octógono. ¿Cuántos segmentos que unen pares de vértices del poliedro son interiores al mismo, es decir, no son aristas ni están cont... | [
"Solution:\n\nSea $V$ el número de vértices, $A$ el número de aristas, $D$ el número de diagonales sobre las caras, e $I$ el número de diagonales interiores.\nPuesto que cada vértice del poliedro está exactamente en una cara cuadrada, debe haber\n$$\nV = 4 \\cdot 12 = 48 \\text{ vértices.}\n$$\n(Obtendríamos el mis... | Spain | Spanish Mathematical Olympiad - Local Stage | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 840 | |
012m | Problem:
Let $n$ be a positive integer. Prove that
$$
\sum_{i=1}^{n} x_{i}\left(1-x_{i}\right)^{2} \leqslant\left(1-\frac{1}{n}\right)^{2}
$$
for all nonnegative real numbers $x_{1}, x_{2}, \ldots, x_{n}$ such that $x_{1}+x_{2}+\cdots+x_{n}=1$. | [
"Solution:\nExpanding the expressions at both sides we obtain the equivalent inequality\n$$\n-\\sum_{i} x_{i}^{3}+2 \\sum_{i} x_{i}^{2}-\\frac{2}{n}+\\frac{1}{n^{2}} \\geqslant 0\n$$\nIt is easy to check that the left hand side is equal to\n$$\n\\sum_{i}\\left(2-\\frac{2}{n}-x_{i}\\right)\\left(x_{i}-\\frac{1}{n}\\... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0e0u | Find the smallest positive integer $n$ such that $n^2$ is a perfect cube, and $n^3$ is a perfect square. | [
"Since $n$ is divisible by $20$, we can write $n = 2^{2+a} \\cdot 5^{1+b} \\cdot k$, where $k$ is a positive integer divisible neither by $2$ nor $5$, while $a$ and $b$ are non-negative integers. If $n^2 = 2^{2(2+a)} \\cdot 5^{2(1+b)} \\cdot k^2$ is a perfect cube, then $3$ divides $2(2+a)$ and $3$ divides $2(1+b)$... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1 | |
02ax | Problem:
Encontre todos os pares de inteiros positivos $x, y$ tais que $x$ e $y$ são primos entre si, $x<y$ e $2000\left(\frac{x}{y}+\frac{y}{x}\right)$ é um inteiro ímpar. | [
"Solution:\n\nSolução 1: Temos:\n$$\n2000\\left(\\frac{x}{y}+\\frac{y}{x}\\right)=16 \\times 125\\left(\\frac{x^{2}+y^{2}}{x y}\\right)\n$$\nComo $x$ e $y$ são primos entre si, concluímos que $x y$ e $x^{2}+y^{2}$ não têm fatores em comum (prove isso). Logo, $x$ e $y$ são divisores de $2000=16 \\times 125=2^{4} \\t... | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (1,16), (1,80), (1,400), (1,2000), (5,16), (16,25), (16,125) | |
03w1 | The sequence $\{x_n\}$ is defined by $x_1 = 2$, $x_2 = 12$, $x_{n+2} = 6x_{n+1} - x_n$, $n = 1, 2, \dots$. Let $p$ be an odd prime number. Let $q$ be a prime number such that $q \mid x_p$. Prove that if $q \neq 2$, then $q \ge 2p - 1$. | [
"It is easy to see\n$$\nx_n = \\frac{1}{2\\sqrt{2}}\\left((3+2\\sqrt{2})^n - (3-2\\sqrt{2})^n\\right), \\quad n = 1, 2, \\dots\n$$\nLet $a_n, b_n$ be positive integers and $a_n + b_n\\sqrt{2} = (3+2\\sqrt{2})^n$. Then\n$$\na_n - b_n\\sqrt{2} = (3 - 2\\sqrt{2})^n,\n$$\nso $x_n = b_n$, and $a_n^2 - 2b_n^2 = 1$, $n = ... | China | China National Team Selection Test | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Pell's equations"
] | English | proof only | null | |
091x | Problem:
Let $n$ be a positive integer. On a board consisting of $4 n \times 4 n$ squares, exactly $4 n$ tokens are placed so that each row and each column contains one token. In a step, a token is moved horizontally or vertically to a neighbouring square. Several tokens may occupy the same square at the same time. The... | [
"Solution:\nWe shall prove that $k(n)=6 n^{2}$.\n\nWe define the distance from a given square to a given diagonal to be the minimal number of steps needed to get from the square to the diagonal. This equals the minimal number of horizontal steps needed to do that. It also equals the minimal number of vertical steps... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | k(n) = 6n^2 | |
0580 | Players $A$, $B$ and $C$ are playing the following game. Initially, the number $1$ is written on a blackboard. On their move, each player replaces the number $n$ currently on the blackboard with either $n+1$, $7n+7$, or $4n^3 + 3n + 4$ at their own choice, under the condition that the new number must not be larger than... | [
"The game lasts while the number on the blackboard stays less than $10^9$, because it is possible to make a move of the first kind (replace $n$ with $n+1$). As the number on the blackboard is increased by at least $1$ by every move, it cannot stay less than $10^9$ infinitely. When the number on the blackboard equal... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | Player C | |
02c8 | Problem:
O maior resultado - Qual o maior resultado que podemos encontrar quando dividimos um número de 2 algarismos pela soma de seus algarismos? | [
"Solution:\n\nO maior resultado - Estamos procurando o maior valor de $\\frac{10a+b}{a+b}$, onde $a$ e $b$ representam algarismos, pelo menos um diferente de $0$. Temos\n$$\n\\frac{10a+b}{a+b} = \\frac{10a+10b-9b}{a+b} = \\frac{10a+10b}{a+b} - \\frac{9b}{a+b} = 10 - \\frac{9b}{a+b} \\leq 10\n$$\nLogo, se conseguirm... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 10 | |
0gtx | Find all functions $f: \mathbb{Z}^+ \to \mathbb{Z}^+$ such that the following conditions hold:
(i) $a - b \mid f(a) - f(b)$ for all positive integers $a \neq b$,
(ii) $f(\varphi(a)) = \varphi(f(a))$ for all $a \in \mathbb{Z}^+$.
Note. $\varphi$ is the Euler's totient function: $\varphi(n)$ is the total number of positi... | [
"Answer: $f(x) \\equiv 1$ and $f(x) = x$.\nBy putting $a = 1$ to $f(\\varphi(a)) = \\varphi(f(a))$ we get $f(1) = \\varphi(f(1))$ and hence $f(1) = 1$. By putting $a = 2$ we get $f(1) = \\varphi(f(2))$ and since $f(1) = 1$ we get that either $f(2)$ is either 1 or 2.\n\nCase 1: $f(2) = 1$. Since $1 = f(2) = f(\\varp... | Turkey | Team Selection Test for EGMO 2024 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics ... | English | proof and answer | f(x) = 1 for all x, and f(x) = x for all x | |
0597 | The teacher writes the digits $20212022$ in a row on a blackboard. Juku must write each arithmetic operator $(+, -, \cdot, :)$ exactly once somewhere between these digits in such a way that the result is a correct mathematical expression with a real value, and find this value.
a. Can Juku obtain the number $0$ as the ... | [
"*Answer:* (a) Yes; (b) Yes.\n\n\na. One possibility is $2 + 0 \\cdot 2120 : 2 - 2 = 0$.\n\n\nb. If the teacher allowed using parentheses, Juku could write the expression $(2 - 0 + 2) : 1 \\cdot 2022$ whose value is $8088$. We show that it is impossible to achieve so big value without using parentheses. For that, w... | Estonia | Estonian Math Competitions | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | a: Yes; b: Yes |
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