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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0ez1 | Problem:
What is the smallest positive integer $a$ such that we can find integers $b$ and $c$ so that $a x^2 + b x + c$ has two distinct positive roots less than $1$? | [
"Solution:\n$4x^2 - 4x + 1 = (2x - 1)^2$, which has the double root $1/2$. So it remains to consider $a = 1,2,3$.\n\n- $b/a$ is the sum of the roots, so $b$ is negative. $c/a$ is the product of the roots, so $c$ is positive. If $a = 1$, then the product of the roots is $c$, which is at least $1$, so both roots cann... | Soviet Union | 3rd ASU | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 4 | |
04ut | Find all triples $(x, y, z)$ of positive real numbers such that
$$
\begin{aligned}
x^2 + 2y^2 &= x + 2y + 3z, \\
y^2 + 2z^2 &= 2x + 3y + 4z, \\
z^2 + 2x^2 &= 3x + 4y + 5z.
\end{aligned}
$$
(Patrik Bak) | [] | Czech Republic | Second Round | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | (10/3, 5/3, 10/3) | |
0514 | Find all triples $(a, b, c)$ of positive integers such that
$$
a^{bc} + b^{ca} + c^{ab} = 3abc.
$$ | [
"First assume $a \\ge 2$, $b \\ge 2$, $c \\ge 2$. W.l.o.g., let $c$ be the greatest among the three numbers. Then $a^{bc} + b^{ca} + c^{ab} \\ge a^4 + b^4 + c^4 > b^4 + c^4 \\ge 2b^2c^2 = 2b \\cdot c \\cdot bc > 3 \\cdot a \\cdot bc$. Thus there are no solutions in this case.\n\nIt remains to study triples that con... | Estonia | Estonian Math Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (1,1,1) and all permutations of (1,2,3) | |
0en0 | Let $P$ be a point inside triangle $ABC$. Construct the points $P_1$, $P_2$, $P_3$ such that $PP_1 \perp BC$, $PP_2 \perp CA$, $PP_3 \perp AB$ and $BP_3 = BP_1$, $CP_2 = CP_1$. Prove that $AP_3 = AP_2$. | [
"Place the points $A$, $B$ and $C$ in the complex plane such that $P$ is at the origin. Then we have the following equations:\n$$\nPP_1 \\perp BC \\implies \\frac{B-C}{P_1} = \\frac{\\bar{C} - \\bar{B}}{\\bar{P}_1} \\qquad (3)\n$$\n$$\nPP_2 \\perp AC \\implies \\frac{A-C}{P_2} = \\frac{\\bar{C} - \\bar{A}}{\\bar{P}... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
05gv | Problem:
Dans un club d'échecs de 20 personnes, 14 parties sont jouées et chaque joueur a joué au moins une partie. Montrer qu'il y a 6 parties auxquelles 12 joueurs différents ont participé. | [
"Solution:\n\nSoit $r$ le plus grand entier tel qu'il existe un ensemble $M$ de $r$ parties faisant intervenir $2r$ joueurs différents.\n\nNotons $J$ l'ensemble des $2r$ personnes qui ont joué les parties de $M$. Par maximalité de $M$, si $a$ et $b$ sont deux joueurs n'appartenant pas à $J$, alors $a$ et $b$ n'ont ... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof only | null | |
01fm | In a directed graph $G$ outgoing degree of each vertex equals $3$ (loops and bidirectional edges are allowed and considered as cycles). Prove that the graph contains $2$ non-intersecting directed cycles. | [
"We proceed by induction on $n$, the number of vertices.\n\nFor the base case we consider $n = 4$ where the only possible digraph is the complete digraph which contains $2$ disjoint cycles of length $2$.\n\nWe now assume that any $3$-out digraph, with $n - 1 \\ge 4$ vertices, contains $2$ disjoint cycles.\n\nLet $G... | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0g4g | Problem:
Let $S$ be a non-empty set of positive integers such that for any $n \in S$, all positive divisors of $2^{n}+1$ are also in $S$. Prove that $S$ contains an integer of the form
$$
\left(p_{1} p_{2} \ldots p_{2023}\right)^{2023}
$$
where $p_{1}, p_{2}, \ldots, p_{2023}$ are distinct prime numbers, all greater th... | [
"Solution:\nSince $S$ is non-empty, it contains some integer $a$. Since $1 \\mid 2^{a}+1$, it follows that $1 \\in S$ and thus $2^{1}+1=3 \\in S$. We call a pair $(m, n)$ of positive integers $k$-valid if we have $m, n \\in S$, $n|m| 2^{n}+1$ and $m / n$ is divisible by $k$ distinct primes. We will show inductively... | Switzerland | Switzerland Selection Solution | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0jka | Problem:
In triangle $ABC$, let the parabola with focus $A$ and directrix $BC$ intersect sides $AB$ and $AC$ at $A_1$ and $A_2$, respectively. Similarly, let the parabola with focus $B$ and directrix $CA$ intersect sides $BC$ and $BA$ at $B_1$ and $B_2$, respectively. Finally, let the parabola with focus $C$ and direc... | [
"Solution:\n\nBy the definition of a parabola, we get $AA_1 = A_1B \\sin B$ and similarly for the other points. So $\\frac{AB_2}{AB} = \\frac{AC_1}{AC}$, giving $B_2C_1 \\parallel BC$, and similarly for the other sides. So $DEF$ (WLOG, in that order) is similar to $ABC$. It suffices to scale after finding the lengt... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous >... | null | proof and answer | 6728/3375 | |
03ta | Let complex number $z = (a + \cos \theta) + (2a - \sin \theta)i$. If $|z| \le 2$ for any $\theta \in \mathbb{R}$, then the range of real number $a$ is ______. | [
"By the definition given above, we have, for any $\\theta \\in \\mathbb{R}$,\n$$\n\\begin{align*}\n|z| \\le 2 & \\Leftrightarrow (a + \\cos \\theta)^2 + (2a - \\sin \\theta)^2 \\le 4 \\\\\n& \\Leftrightarrow 2a(\\cos \\theta - 2\\sin \\theta) \\le 3 - 5a^2 \\\\\n& \\Leftrightarrow -2\\sqrt{5}a \\sin(\\theta - \\var... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | [-√5/5, √5/5] | |
0ice | Problem:
A plane $P$ slices through a cube of volume $1$ with a cross-section in the shape of a regular hexagon. This cube also has an inscribed sphere, whose intersection with $P$ is a circle. What is the area of the region inside the regular hexagon but outside the circle? | [
"Solution:\nOne can show that the hexagon must have as its vertices the midpoints of six edges of the cube, as illustrated; for example, this readily follows from the fact that opposite sides of the hexagons and the medians between them are parallel. We then conclude that the side of the hexagon is $\\sqrt{2} / 2$ ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | final answer only | (3*sqrt(3)-pi)/4 | |
0f8d | Problem:
A polygonal line with a finite number of segments has all its vertices on a parabola. Any two adjacent segments make equal angles with the tangent to the parabola at their point of intersection. One end of the polygonal line is also on the axis of the parabola. Show that the other vertices of the polygonal lin... | [] | Soviet Union | 22nd ASU | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
05au | Do there exist positive integers $x$ and $y$ such that
$$
11x^5 + 33y = 13y^5 + 31x + 2024?
$$ | [
"The given equation is equivalent to\n$$\n(11x^5 - 31x) + (33y - 13y^5) = 2024.\n$$\nThe fifth power of any positive integer ends with the same digit as the number itself. Therefore, $11x^5$ ends with the same digit as $11x$, which in turn ends with the same digit as $x$. Since $31x$ also ends with the same digit, ... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | No | |
05o5 | Problem:
Une cellule peut se diviser en 42 ou en 44 petites cellules. Combien de divisions faut-il pour obtenir, à partir d'une cellule, exactement 2017 cellules? | [
"Solution:\n\nUne division augmente de $41$ ou $43$ le nombre de cellules. Soit $a$ le nombre de divisions en $42$ cellules, et $b$ le nombre de divisions en $44$ cellules. On veut trouver $a+b$, sachant que\n$$\n2017 = 1 + 41a + 43b = 1 + 41(a+b) + 2b = 1 + 43(a+b) - 2b\n$$\nOn voit que $41(a+b) \\leq 2016$, donc ... | France | Envoi 1: Arithmétique | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 48 | |
0061 | Hallar el menor número entero positivo $N$ que cumple las siguientes dos condiciones:
* $N$ tiene por lo menos dos factores primos distintos.
* Para cualesquiera $p$ y $q$ factores primos de $N$, con $p$ distinto de $q$, la suma $p+q$ divide a $N$. | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Spanish | proof and answer | 2520 | |
0ctj | In the space, three segments $A_1A_2$, $B_1B_2$, and $C_1C_2$ are chosen.
They are not coplanar, and they meet at point $P$. Denote by $O_{ijk}$ the center of the sphere passing through $A_i, B_j, C_k$, and $P$. Prove that the four lines $O_{111}O_{222}, O_{112}O_{221}, O_{121}O_{212}$, and $O_{211}O_{122}$ have a com... | [
"For a fixed $i$, all the points $O_{ijk}$ belong to the perpendicular bisector $\\alpha_i$ of $PA_i$; notice that $\\alpha_1 \\parallel \\alpha_2$. By similar reasons, the points $O_{ijk}$ appear to be eight vertices of some parallelepiped.\n\nFor any segment $XY$, the perpendicular bisector to this segment is the... | Russia | Russian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English; Russian | proof only | null | |
05t9 | Problem:
Andréa, Baptiste et Camille jouent au foot à trois. Un des joueurs est aux cages, les deux autres sont sur le terrain et essaient de marquer. Le joueur qui marque devient ensuite gardien pour le tir suivant.
Durant l'après-midi, Andréa a été sur le terrain 12 fois, Baptiste l'a été 21 fois et Camille a été a... | [
"Solution:\n\nSi on note $n$ le nombre de parties jouées, alors au total il y a eu $n$ fois un gardien. On utilise à présent les hypothèses données par l'énoncé. Si Andréa a été 12 fois sur le terrain, il a donc été $n-12$ fois aux cages. Si Baptiste a été 21 fois sur le terrain, il a été $n-21$ fois aux cages. Ain... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Andréa | |
0kty | Problem:
Suppose $P(x)$ is a cubic polynomial with integer coefficients such that $P(\sqrt{5})=5$ and $P(\sqrt[3]{5})=5 \sqrt[3]{5}$. Compute $P(5)$. | [
"Solution:\n\nWrite $P(x)=a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are integers. Then we have that\n$$\n\\begin{aligned}\nP(\\sqrt{5})-5 & =(5 a+c) \\sqrt{5}+(5 b+d-5)=0 \\\\\nP(\\sqrt[3]{5})-5 \\sqrt[3]{5} & =(5 a+d)+(c-5) \\sqrt[3]{5}+b \\sqrt[3]{25}=0\n\\end{aligned}\n$$\nRecall that $\\sqrt{5}$ is irrational.... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | -95 | |
0k80 | Problem:
The coefficients of the polynomial $P(x)$ are nonnegative integers, each less than $100$. Given that $P(10) = 331633$ and $P(-10) = 273373$, compute $P(1)$. | [
"Solution:\nLet\n$$\nP(x) = a_{0} + a_{1} x + a_{2} x^{2} + \\ldots\n$$\nThen\n$$\n\\frac{1}{2}(P(10) + P(-10)) = a_{0} + 100 a_{2} + \\ldots\n$$\nand\n$$\n\\frac{1}{2}(P(10) - P(-10)) = 10 a_{1} + 1000 a_{3} + \\ldots\n$$\nSince all the coefficients are nonnegative integers, these expressions give us each of the c... | United States | HMMT November 2019 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 100 | |
0a8p | Problem:
The point $D$ inside the equilateral triangle $\triangle ABC$ satisfies $\angle ADC = 150^\circ$. Prove that a triangle with side lengths $|AD|, |BD|, |CD|$ is necessarily a right-angled triangle. | [
"Solution:\n\n(See Figure 14.) We rotate the figure counterclockwise $60^\\circ$ around $C$. Because $ABC$ is an equilateral triangle, $\\angle BAC = 60^\\circ$, so $A$ is mapped on $B$. Assume $D$ maps to $E$. The properties of rotation imply $AD = BE$ and $\\angle BEC = 150^\\circ$. Because the triangle $DEC$ is ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 17 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0eed | Problem:
Poišči vse četverice celih števil $a, b, c$ in $d$, ki rešijo sistem enačb
$$
\begin{aligned}
& a^{2}+b^{2}+c^{2}=d+13 \\
& a+2 b+3 c=\frac{d}{2}+13
\end{aligned}
$$ | [
"Solution:\nIz druge enačbe izrazimo $d=2 a+4 b+6 c-26$ in vstavimo v prvo enačbo, da dobimo\n$$\na^{2}+b^{2}+c^{2}=2 a+4 b+6 c-13\n$$\nVse člene postavimo na levo stran $a^{2}+b^{2}+c^{2}-2 a-4 b-6 c+13=0$ in dopolnimo do popolnih kvadratov\n$$\n(a-1)^{2}+(b-2)^{2}+(c-3)^{2}=1\n$$\nKer so vsi trije popolni kvadrat... | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | [[0, 2, 3, 0], [2, 2, 3, 4], [1, 1, 3, -2], [1, 3, 3, 6], [1, 2, 2, -4], [1, 2, 4, 8]] | |
0ibm | Problem:
What is the largest whole number that is equal to the product of its digits? | [
"Solution:\n\nSuppose the number $n$ has $k+1$ digits, the first of which is $d$. Then the number is at least $d \\cdot 10^{k}$. On the other hand, each of the digits after the first is at most $9$, so the product of the digits is at most $d \\cdot 9^{k}$. Thus, if $n$ equals the product of its digits, then\n$$\nd ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 9 | |
093t | Problem:
Let $m$ and $n$ be positive integers. Some squares of an $m \times n$ board are coloured red. A sequence $a_{1}, a_{2}, \ldots, a_{2 r}$ of $2 r \geqslant 4$ pairwise distinct red squares is called a bishop circuit if for every $k \in\{1, \ldots, 2 r\}$, the squares $a_{k}$ and $a_{k+1}$ lie on a diagonal, bu... | [
"Solution:\n\nObviously, for the tables $1 \\times n$ and $n \\times 1$, the largest number of black cells is $n$. Therefore, we assume that $m \\geq 2$ and $n \\geq 2$ for the rest of the solution. In the table $m \\times n$, we can color the first two rows, the first column and the last column, which is $2 m + 2 ... | Middle European Mathematical Olympiad (MEMO) | 15th Middle European Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | If min(m, n) = 1, the maximum is mn. If m, n ≥ 2, the maximum is 2m + 2n − 4. | |
0fs7 | Problem:
Beweise für positive reelle Zahlen $x, y, z$ mit $x+y+z=1$ die folgende Ungleichung:
$$
\frac{x^{2}+y^{2}}{z}+\frac{y^{2}+z^{2}}{x}+\frac{z^{2}+x^{2}}{y} \geq 2
$$ | [
"Solution:\n\nEs gibt sehr viele Wege, diese Ungleichung zu beweisen. Wir geben 10 verschiedene Lösungen als Auswahl. Wir verwenden folgende Abkürzungen:\nHM, GM, AM, QM: harmonisches, geometrisches, arithmetisches, quadratisches Mittel. C.S.: Ungleichung von Cauchy-Schwarz.\nSetze\n$$\nA=\\frac{x^{2}+y^{2}}{z}+\\f... | Switzerland | Vorselektionsprüfung | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof only | null | |
0c64 | Let $ABC$ be a triangle, and let $I$ and $O$ be its incenter and circumcenter, respectively. The $A$-excircle touches the lines $AB$, $AC$, $BC$ at $K$, $L$, $M$, respectively. Show that, if the midpoint of the segment $KL$ lies on the circle $ABC$, then $I$, $M$, $O$ are collinear.
Pavel Kozhevnikov, Russian Olympiad... | [
"Leaving the trivial case $AB = AC$ aside, we show that $I$, $M$, $O$ all lie on the Euler line of the triangle formed by the three excenters $I_A$, $I_B$, $I_C$. Recall that the Euler line of a triangle is the line through the orthocenter, the center of the nine-point circle and the circumcenter of that triangle. ... | Romania | SELECTION TESTS FOR THE 2019 BMO AND IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0dyk | Find all prime numbers $p$ such that the polynomial
$$
q(x) = 2x^3 - 2p x^2 + (1-p)x + p
$$
has at least one rational root. | [
"If $p = 2$, we have $q(x) = 2x^3 - 4x^2 - x + 2 = (x-2)(2x^2 - 1)$ and $x = 2$ is a rational root. Now, let $p$ be an odd prime. The only possible candidates for rational roots are $\\pm 1, \\pm p, \\pm \\frac{1}{2}$ and $\\pm \\frac{p}{2}$. Let us consider all possible cases.\nSince $q(1) = 3 - 2p$ and $3 - 2p$ i... | Slovenia | Slovenija 2008 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 2 | |
045g | Let $I$ be the incenter of a triangle $ABC$ with $AB > AC$, and let $AM$ be the median. The line through $I$ perpendicular to $BC$ meets the line $AM$ at point $L$. Let $J$ be the reflection of $I$ with respect to $A$. Prove that $\angle ABJ = \angle LBI$. | [
"*Proof.* We need the following lemma.\n**Lemma:** In triangle *ABC*, let *I* be the incenter. The incircle $\\odot I$ is tangent to sides $BC$, $CA$, $AB$ at points $A_1$, $B_1$, $C_1$, respectively. Let $A_1I$ intersect $B_1C_1$ at $L$. Then, $AL$ passes through the midpoint $M$ of $BC$.\n![](attached_image_1.png... | China | 2022 CGMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
09bj | Зөв $2n$ өнцөгтийн бүх диагоналуудыг татав. Нэг оройгоос гарсан $2n - 1$ хэрчмийн аль нэгийг нь үлдээж бусдыг нь арчсан гэе. Энэ үйлдлийг бүх оройн хувьд гүйцэтгэе. Ингэхэд үүсэх $n$ хэрчим дотор уртаараа тэнцүү 2 хэрчим ямагт олддог байх $n$-ийн бүх утгыг ол. | [
"Зөв $2n$ өнцөгтийн $A_1, A_2, \\dots, A_{2n}$ оройнуудыг хоёр оройгоос тогтох $n$ ширхэг хосуудад хуваав. Нэг хосын 2 оройг холбоход үүсэх $n$ хэрчим дотор урт нь тэнцүү 2 хэрчим олддог бол $n = 4m + 2$ эсвэл $n = 4m + 3$ хэлбэртэй гэж батлая. Эсрэгээс нь бүх хосын оройнуудыг холбосон хэрчмийн уртууд ялгаатай гэе.... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Mongolian | proof and answer | n = 4m + 2 or n = 4m + 3 | |
0kky | Problem:
Geoff walks on the number line for 40 minutes, starting at the point 0. On the $n$th minute, he flips a fair coin. If it comes up heads he walks $\frac{1}{n}$ in the positive direction and if it comes up tails he walks $\frac{1}{n}$ in the negative direction. Let $p$ be the probability that he never leaves th... | [
"Solution:\n\nTo estimate it by hand, we'll do casework on the most likely ways that Geoff will go past $+2$, and double the answer. If Geoff starts with one of the three sequences below, he will be past 2 or very close to 2:\n$$\n(+,+,+,+),\\ (+,+,+,-,+,+),\\ (+,+,-,+,+,+)\n$$\nThe probability of one of these happ... | United States | HMMT Spring 2021 Guts Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 8101 | |
07s1 | Prove that
$$
\sum_{k=0}^{n} (n + 1 - k) \sin((2k + 1)x) \geq 0, \quad \forall x \in [0, \pi].
$$ | [
"Since\n$$\n2 \\sin(x) \\sin((2k + 1)x) = \\cos(2kx) - \\cos((2k + 2)x), \\quad k = 0, 1, 2, \\dots\n$$\nit follows that\n$$\n\\begin{aligned}\n& 2 \\sin(x) \\sum_{k=0}^{n} (n + 1 - k) \\sin((2k + 1)x) \\\\\n&= \\sum_{k=0}^{n} (n + 1 - k) (\\cos(2kx) - \\cos((2k + 2)x)) \\\\\n&= \\sum_{k=0}^{n} (n + 1 - k) \\cos(2k... | Ireland | Irish | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
0c49 | Problem:
Fie $ABC$ un triunghi oarecare astfel încât $AB > AC$. Punctul $P \in (AB)$ are proprietatea că $\angle ACP \equiv \angle ABC$. Fie $D$ simetricul lui $P$ faţă de $AC$ şi $E$ punctul în care cercul circumscris triunghiului $BCD$ taie a doua oară dreapta $AC$. Demonstraţi că $AE = AC$. | [
"Solution:\n\nFie $Q$ punctul în care cercul circumscris triunghiului $BCD$ intersectează a doua oară dreapta $AB$. Atunci $\\angle QEA \\equiv \\angle QBC \\equiv \\angle ECP$, deci $EQ \\parallel PC$.\n\nÎn plus, $\\angle ECP \\equiv \\angle ECD$ implică $QD \\parallel EC$, deci $EQ = CD = CP$.\n\nDeducem că $EQC... | Romania | Olimpiada Naţională de Matematică Primul baraj pentru Olimpiada Balcanică de Matematică pentru Juniori | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
016p | In a rectangle $ABCD$ where $AB = 2BC$ the diagonals intersect in a point $E$, the angle bisector of the angle $\angle CAD$ intersects the side $CD$ in a point $F$ and the diagonal $BD$ in a point $G$, and $EG = 25$. Determine the length of $FC$. | [
"Let $x = BC$. Then $BD = \\sqrt{5}x$ and $DE = \\frac{\\sqrt{5}}{2}x$. Since the angle bisector divides the opposite side in the ratio of the adjacent sides, $DG = \\frac{2}{\\sqrt{5}}GE$ and $DF = \\frac{1}{\\sqrt{5}}FC$. Hence\n$$\n\\frac{\\sqrt{5}}{2}x = DE = DG + GE = \\frac{2}{\\sqrt{5}}GE + GE = 10\\sqrt{5} ... | Baltic Way | BALTIC WAY | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 25 + 15√5 | |
03t9 | Let $p$ be a prime number greater than $3$. Prove that there exist integers $a_1, a_2, \dots, a_t$ that satisfy the following conditions:
a.
$$
-\frac{p}{2} < a_1 < a_2 < \cdots < a_t \le \frac{p}{2},
$$
b.
$$
\frac{p-a_1}{|a_1|} \cdot \frac{p-a_2}{|a_2|} \cdot \cdots \cdot \frac{p-a_t}{|a_t|} = 3^m,
$$
where $m$ is ... | [
"**Proof** By the Division Algorithm, there exist unique integers $q$ and $r$ such that $p = 3q + r$, where $0 < r < 3$.\nTaking $b_0 = r$, then\n$$\n\\frac{p-b_0}{|b_0|} = \\frac{3^{c_0} \\cdot b_1^*}{|b_0|}, \\text{ where } 3 \\nmid b_1^* \\text{ and } 0 < b_1^* < \\frac{p}{2}.\n$$\nTaking $b_1 = \\pm b_1^*$ such... | China | China Girls' Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
08ly | Problem:
Determine all integers $a, b, c$ satisfying the identities:
$$
\begin{gathered}
a+b+c=15 \\
(a-3)^{3}+(b-5)^{3}+(c-7)^{3}=540
\end{gathered}
$$ | [
"Solution:\nWe will use the following fact:\n\nLemma: If $x, y, z$ are integers such that\n$$\nx+y+z=0\n$$\nthen\n$$\nx^{3}+y^{3}+z^{3}=3 x y z\n$$\nProof: Let\n$$\nx+y+z=0.\n$$\nThen we have\n$$\nx^{3}+y^{3}+z^{3}=x^{3}+y^{3}+(-x-y)^{3}=x^{3}+y^{3}-x^{3}-y^{3}-3 x y(x+y)=3 x y z\n$$\nNow, from\n$$\na+b+c=15\n$$\nw... | JBMO | 2009 Shortlist JBMO | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | [(-1, 0, 16), (-2, 1, 16), (-1, 14, 2), (-2, 14, 3), (12, 0, 3), (12, 1, 2)] | |
0cdm | The triangle $ABC$ has $\angle BAC = 90^\circ$ and $\angle ACB = 54^\circ$. Take the bisector $BD$ ($D \in AC$) of the angle $ABC$ and the point $E$ on the segment $BD$ so that $DE = DC$. Prove that $BE = 2 \cdot AD$. | [
"A straightforward angle chasing gives $\\angle ABC = 90^\\circ - 54^\\circ = 36^\\circ$, then $\\angle ABD = \\angle CBD = \\frac{1}{2} \\angle ABC = 18^\\circ$. So $\\angle BDA = 90^\\circ - \\angle ABD = 72^\\circ$, $\\angle BDC = 180^\\circ - \\angle BDA = 108^\\circ$.\n\nThe isosceles triangle $CDE$ gives $\\a... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
05bs | Prove that the value of the expression
$$
x^4 - 2x^3 - 88x^2 + 90x + 2025
$$
is positive regardless of the value of $x$ on the real line. | [
"Using equalities $2025 = 45^2$ and $90 = 2 \\cdot 45$ we get\n$$\n\\begin{aligned}\nx^4 - 2x^3 - 88x^2 + 90x + 2025 &= x^4 - 2x^3 - 89x^2 + (x + 45)^2 \\\\\n&= (x^2)^2 - 2x^2(x + 45) + (x + 45)^2 + x^2 \\\\\n&= (x^2 - x - 45)^2 + x^2.\n\\end{aligned}\n$$\nBut $(x^2 - x - 45)^2 + x^2 \\ge 0$. Equality is impossible... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
08dx | Problem:
a. Supponiamo che $n = k^{2}$ sia un quadrato perfetto. Dimostrare che il numero di divisori positivi di $n$ strettamente minori di $k$ è uguale al numero di divisori di $n$ strettamente maggiori di $k$.
b. Supponiamo che $n = k^{2}$ sia un quadrato perfetto. Dimostrare che $n$ ha al massimo $2k-1$ divisori ... | [
"Solution:\n\na. Se $d$ è un divisore di $n$ allora anche $n / d$ è un divisore di $n$. Ne segue che per ogni divisore $d$ di $n = k^{2}$ con $0 < d < k$ esiste un corrispondente divisore $n / d$ con $n / d = k^{2} / d > k$, e viceversa per ogni divisore $d$ di $k^{2}$ con $d > k$ esiste un corrispondente divisore ... | Italy | Olimpiadi della Matematica | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | k = 1 and k = 2 | |
059e | We call positive integers $n, m$ an *interesting pair* if $n < m$ and the greatest prime factor of $n$ is equal to the greatest prime factor of $m$.
a. For an interesting pair $n, m$, will there always exist a prime $p$ such that $n < p < m$?
b. Among the first 25 positive integers, how many don't form an interesting... | [
"(a) The pair $24, 27$ is interesting, as the greatest prime factor of them both is $3$. However, there are no primes between them.\n\n(b) The number $1$ and prime numbers cannot form a pair with any smaller integers. However, each composite number forms a pair with its greatest prime factor, which is clearly small... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | a) No; for example, 24 and 27. b) 10 | |
0ho3 | Problem:
Find all pairs of integers $(x, y)$ for which $x^{2} + x y = y^{2}$. | [
"Solution:\n\nThe only such pair is $(0, 0)$. If $x = 0$, we easily get $y = 0$ which satisfies the equation. Otherwise, dividing through by $x^{2}$ we get\n$$\n1 + \\frac{y}{x} = \\left(\\frac{y}{x}\\right)^{2}.\n$$\nThis implies $\\left(\\frac{y}{x}\\right)^{2} - \\frac{y}{x} - 1 = 0$ and the quadratic formula gi... | United States | Berkeley Math Circle | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (0, 0) | |
03bh | The incircle of $\triangle ABC$ touches the sides $BC$ and $AC$ at points $A_1$ and $B_1$, respectively. The lines $B_1A_1$ and $AB$ are concurrent at $X$ such that $A$ lies between $X$ and $B$. If $\angle CXB = 90^\circ$ and $BC^2 = AB^2 + BC \cdot AC$ find the angles of $\triangle ABC$. | [
"Let $Y \\in BC$ be such that $AY \\parallel XA_1$. Thus $BY = a-b$ and since $\\frac{a}{c} = \\frac{c}{a-b}$ we have that $\\triangle ABC \\sim \\triangle YBA$. Therefore $\\angle AXA_1 = \\angle BAY = \\gamma$ which implies that quadrilateral $XAA_1C$ is cyclic. It follows from $\\angle XAC = \\angle XA_1C = 90^\... | Bulgaria | Team selection test for the 54th IMO | [
"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... | English | proof and answer | A = 108°, B = 36°, C = 36° | |
00aw | Given is a table with $n$ rows and $12$ columns. Each cell in it contains a $0$ or a $1$. The table has the following properties:
a. Every two rows are different.
b. Every row contains exactly $4$ entries equal to $1$.
c. For every $3$ rows there is a column that intersects them at three entries equal to $0$.
Find ... | [
"The answer is $\\binom{11}{4} = 330$; here and later on $\\binom{n}{k}$ denotes a binomial coefficient.\n\nHere is an example with $n=330$. Form the $\\binom{11}{4} = 330$ (unordered) quadruples $i, j, k, l$ with elements from $\\{1, 2, \\dots, 11\\}$. To every such quadruple assign a row of length $12$ in which t... | Argentina | Argentina_2017 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 330 | |
04ll | Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(x + f(y)) = f(f(y)) + 2x f(y) + x^2
$$
holds for all real numbers $x$ and $y$. | [
"Considering $x = -f(y)$ we get $f(0) = f(f(y)) - (f(y))^2$. Denoting $f(0)$ by $a$ and substituting $y = 0$ we arrive at $f(a) = a^2 + a$. Now we return to the original equality and plug in $y = 0$:\n$$\nf(x + a) = f(a) + 2x a + x^2 = x^2 + 2x a + a^2 + a = (x + a)^2 + a.\n$$\nThus, all the possible solutions are ... | Croatia | Mathematical competitions in Croatia | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x^2 + a for any real constant a | |
0j6m | Problem:
Determine the value of
$$
\sum_{k=1}^{2011} \frac{k-1}{k!(2011-k)!} .
$$ | [
"Solution:\nAnswer: $\\frac{2009\\left(2^{2010}\\right)+1}{2011!}$ We note that\n$$\n\\begin{aligned}\n(2011!) \\sum_{k=1}^{2011} \\frac{k-1}{k!(2011-k)!} & =\\sum_{k=1}^{2011} \\frac{(2011!)(k-1)}{k!(2011-k)!} \\\\\n& =\\sum_{k=1}^{2011} \\frac{k(2011)!}{k!(2011-k)!}-\\sum_{k=1}^{2011} \\frac{2011!}{k!(2011-k)!} \... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | (2009*2^{2010} + 1)/(2011!) | |
0fzl | Problem:
Sei $ABC$ ein Dreieck, in welchem $\alpha = \angle BAC$ der strikt kleinste Winkel ist, und $P$ ein Punkt auf der Seite $BC$. Weiter sei $D$ ein Punkt auf der Geraden $AB$, sodass $B$ zwischen $A$ und $D$ liegt und $\angle BPD = \alpha$ gilt, und $E$ ein Punkt auf der Geraden $AC$, sodass $C$ zwischen $A$ und... | [
"Solution:\n\nSei $F$ der Schnittpunkt der Geraden $EP$ und $AD$ und sei $G$ der Schnittpunkt der Geraden $DP$ und $AE$. Weiter sei $S$ der Schnittpunkt der Geraden $BE$ und $CD$ und $T$ der Schnittpunkt der Geraden $BG$ und $CF$. Da $E, C, G$ und $D, B, F$ jeweils auf einer Geraden liegen, können wir Pappus anwend... | Switzerland | IMO-Selektionsprüfung | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometr... | null | proof only | null | |
0kop | Let $a$, $b$, $x$, and $y$ be real numbers with $a > 4$ and $b > 1$ such that
$$
\frac{x^2}{a^2} + \frac{y^2}{a^2 - 16} = \frac{(x - 20)^2}{b^2 - 1} + \frac{(y - 11)^2}{b^2} = 1.
$$
Find the least possible value for $a + b$. | [
"The graph of\n$$\n\\frac{x^2}{a^2} + \\frac{y^2}{a^2 - 16} = 1\n$$\nis an ellipse centered at $(0, 0)$ with major axis parallel to the $x$-axis of length $2a$. The distance from the center to the foci is $\\sqrt{a^2 - (a^2 - 16)} = 4$, so the foci are $F_1 = (-4, 0)$ and $F_2 = (4, 0)$.\nSimilarly, the graph of\n$... | United States | 2022 AIME II | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loc... | null | proof and answer | 23 | |
06zu | Problem:
Find all functions $f$ on the positive integers with positive integer values such that
(1) if $x < y$, then $f(x) < f(y)$, and
(2) $f(y f(x)) = x^{2} f(x y)$. | [
"Solution:\n\nPut $y = 1$. Then $f(f(x)) = x^{2} f(x)$.\n\nPut $y = f(z)$, then $f(f(z) f(x)) = x^{2} f(x f(z)) = x^{2} z^{2} f(x z) = f(f(x z))$.\n\nBut $f$ is $(1,1)$ so $f(x z) = f(x) f(z)$.\n\nNow suppose $f(m) > m^{2}$ for some $m$. Then by (1), $f(f(m)) > f(m^{2}) = f(m \\cdot m) = f(m)^{2}$.\n\nBut $f(f(m)) ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | f(n) = n^2 | |
047m | There are 100 students in a summer camp, with exactly 2024 pairs of mutual friends. We need to divide them into 50 groups of 2 students each. Prove that:
1. It's possible to group them such that at most 20 groups contain mutual friends;
2. It's possible to group them such that at least 23 groups contain mutual friends;... | [
"*Proof.* Using the language of graph theory to describe this problem, we represent each student as a vertex and connect two vertices with an edge if the corresponding students are friends, forming a simple graph $G$ of order $100$. By the given condition, $G$ has exactly $2024$ edges.\n\nPair the $100$ vertices in... | China | 2024 CGMO | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0erf | A box contains $5$ red pens, $6$ blue pens and $4$ green pens. Three pens are consecutively taken from the box, at random and without replacement. The probability that the first pen is red, the second pen blue and the third pen green, is
(A) $\frac{7}{225}$ (B) $\frac{1}{5}$ (C) $\frac{9}{82}$ (D) $\frac{4}{91}$ (E) $\... | [
"There are $5 + 6 + 4 = 15$ pens in the box, so the probability of first picking a red pen is $\\frac{5}{15} = \\frac{1}{3}$. This leaves $14$ pens (since the red pen is not replaced), and the probability of next picking a blue pen is $\\frac{6}{14} = \\frac{3}{7}$. Finally, the probability of picking a green pen i... | South Africa | South African Mathematics Olympiad First Round | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | English | MCQ | D | |
0ad9 | Let $a$ and $b$ be integers bigger than $2$. Prove that there is a positive integer $k$ and a finite sequence $n_1, n_2, \dots, n_k$ of positive integers, such that $n_1 = a$, $n_k = b$ and $(n_i + n_{i+1}) \mid n_i n_{i+1}$ for every $i = 1, 2, \dots, k$. | [
"We'll write $a \\leftrightarrow b$ if there exists such a sequence. It is easy to see that $\\leftrightarrow$ is an equivalence relation. Notice that $n \\leftrightarrow 2n$ for every integer $n$, $n \\ge 3$ because in that case the desired finite sequence is\n$$\nn_1 = n,\\ n_2 = n(n-1),\\ n_3 = n(n-1)(n-2),\\ n_... | North Macedonia | Macedonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
03sj | Given an integer $n \ge 2$, define $M_0(x_0, y_0)$ to be an intersection point of the parabola $y^2 = nx - 1$ and the line $y = x$. Prove that for any positive integer $m$, there exists an integer $k \ge 2$ such that $(x_0^m, y_0^m)$ is an intersection point of $y^2 = kx - 1$ and $y = x$. | [
"Since $M_0(x_0, y_0)$ is an intersection point of $y^2 = nx - 1$ and $y = x$, we get $x_0 = y_0 = \\frac{n \\pm \\sqrt{n^2 - 4}}{2}$. Then obviously $x_0 + \\frac{1}{x_0} = n$.\n\nLet $(x_0^m, y_0^m)$ be an intersection point of $y^2 = kx - 1$ and $y = x$. Then we get\n$$\nk = x_0^m + \\frac{1}{x_0^m}.\n$$\nWe den... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials",
"Algebra > Intermediate Algebra > Quadratic functions",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
056l | Find all functions $f : \mathbb{R} \to \mathbb{R}$ that satisfy $f(xy + f(xy)) = 2x f(y)$ for all $x, y \in \mathbb{R}$. | [
"By choosing $(x, y) = (z, 1)$ in the original equation, where $z$ is any real number, we get $f(z + f(z)) = 2z f(1)$. However, by choosing $(x, y) = (1, z)$ in the original equation, we get $f(z + f(z)) = 2 f(z)$. From these equations together we obtain $f(z) = f(1) z$.\n\nBy choosing $z = 1$ in equation $f(z + f(... | Estonia | IMO Team Selection Contest | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0 for all x; f(x) = x for all x | |
0jj8 | Problem:
Let $ABCD$ be a trapezoid with $AB \parallel CD$ and $\angle D = 90^{\circ}$. Suppose that there is a point $E$ on $CD$ such that $AE = BE$ and that triangles $AED$ and $CEB$ are similar, but not congruent. Given that $\frac{CD}{AB} = 2014$, find $\frac{BC}{AD}$. | [
"Solution:\n\nLet $M$ be the midpoint of $AB$. Let $AM = MB = ED = a$, $ME = AD = b$, and $AE = BE = c$. Since $\\triangle BEC \\sim \\triangle DAE$, but $\\triangle BEC$ is not congruent to $\\triangle DAE$, we must have $\\triangle BEC \\sim \\triangle DAE$. Thus, $BC / BE = AD / DE = b / a$, so $BC = b c / a$, a... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(4027) | |
0gmd | Show that there is no continuous function $f: \mathbb{R} \to \mathbb{R}$ satisfying
$$
f(x - f(x)) = \frac{x}{2} \quad \text{for all real numbers } x.
$$ | [] | Turkey | TEAM SELECTION EXAMINATION FOR THE 42nd INTERNATIONAL MATH- EMATICAL OLYMPIAD. TURKEY. | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
0b3a | Problem:
Three real numbers $a_{1}, a_{2}, a_{3}$ form an arithmetic sequence. After $a_{1}$ is increased by $1$, the three numbers now form a geometric sequence. If $a_{1}$ is a positive integer, what is the smallest positive value of the common difference?
(a) $1$
(b) $\sqrt{2}+1$
(c) $3$
(d) $\sqrt{5}+2$ | [] | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | b | |
0fg9 | Problem:
Tenemos en el plano un punto $P_{0}$ y un triángulo $A_{1} A_{2} A_{3}$. Definimos $A_{s}=A_{s-3}$ para todo $s \geq 4$. Construimos una sucesión de puntos $P_{1}, P_{2}, P_{3}, \ldots$, de forma que $P_{k+1}$ es la imagen de $P_{k}$ por la rotación de centro $A_{k+1}$ y ángulo $120^{\circ}$ en sentido horari... | [] | Spain | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0amq | Problem:
The amount $4.5$ is split into two nonnegative real numbers uniformly at random. Then each number is rounded to its nearest integer. For instance, if $4.5$ is split into $\sqrt{2}$ and $4.5-\sqrt{2}$, then the resulting integers are $1$ and $3$, respectively. What is the probability that the two integers sum ... | [] | Philippines | 18th PMO Area Stage | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 4/9 | |
0kxz | Problem:
Pentagon $S P E A K$ is inscribed in triangle $N O W$ such that $S$ and $P$ lie on segment $N O$, $K$ and $A$ lie on segment $N W$, and $E$ lies on segment $O W$. Suppose that $N S = S P = P O$ and $N K = K A = A W$. Given that $E P = E K = 5$ and $E A = E S = 6$, compute $O W$. | [
"Solution:\n\n\n\nNote that $[E S K] = [E P A]$, since one has half the base but double the height. Since the sides are the same, we must have $\\sin \\angle S E K = \\sin \\angle P E A$, so $\\angle S E K + \\angle P E A = 180^{\\circ}$.\n\nLet $O W = 3x$, so $S K = x$ and $P A = 2x$. Then... | United States | HMMT November | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 3 sqrt(610)/5 | |
0bmm | Define a sequence of integers by $a_0 = 1$, and $a_n = \sum_{k=0}^{n-1} \binom{n}{k} a_k$, $n \ge 1$. Let $m$ be a positive integer, let $p$ be a prime, and let $q$ and $r$ be non-negative integers. Prove that the difference $a_{p^m q+r} - a_{p^{m-1} q+r}$ is divisible by $p^m$. | [
"Consider the $\\mathbb{R}$-vector space $\\mathbb{R}[X]$ of all polynomials with real coefficients and define an $\\mathbb{R}$-linear functional $L: \\mathbb{R}[X] \\to \\mathbb{R}$ by $LX^n = a_n$, $n = 0, 1, 2, \\dots$. Thus, if $f = \\sum_k \\alpha_k X^k$, then $Lf = \\sum_k \\alpha_k a_k$. Since $(X+1)^n = \\s... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson ... | null | proof only | null | |
0e1b | Problem:
Dan je ostrokotni trikotnik $ABC$, v katerem je $|AB| > |BC| > |AC|$. Naj bo $D$ od $C$ različna točka na daljici $BC$, da je $|AC| = |AD|$. Označimo s $H$ višinsko točko trikotnika $ABC$, z $A_1$ in $B_1$ pa nožišči višin iz točk $A$ in $B$. Premici $A_1B_1$ in $DH$ se sekata v točki $E$. Dokaži, da so točke... | [
"Solution:\n\nNaj bo $C_1$ nožišče višine na stranico $AB$. Trikotnik $CAD$ je zaradi $|AC| = |AD|$ enakokrak. Premica $AA_1$ je višina tudi v tem enakokrakem trikotniku, zato je $\\angle CDH = \\angle HCD = C_1CB = \\frac{\\pi}{2} - \\angle CBA$. Od tod sledi\n$$\n\\begin{aligned}\n\\angle EDB & = \\pi - \\angle C... | Slovenia | Slovenian Secondary School Mathematical 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 > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07qr | A point $A$ moves on the circumference of a circle centre $O$ and radius $r$. Let $BC$ be a fixed line segment in the plane. Prove that the locus of the centroid of $\triangle ABC$ is a circle whose radius is $r/3$ and whose centre is the centroid of $\triangle OBC$. | [
"Let $D$ be the midpoint of $BC$ and $M$ the centroid of $\\triangle OBC$. For any given point $A$ on the circle centre $O$, radius $r$, let $G$ be the centroid of the triangle $ABC$.\n\n\n\nWe know that $|OD| = 3|M\\bar{D}|$ and $|AD| = 3|G\\bar{D}|$, hence $AO$ is parallel to $GM$. The in... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
033g | Problem:
Find the maximum possible value of the product of different positive integers with sum $2004$. | [
"Solution:\nLet $x_{1} + x_{2} + \\cdots + x_{k} = 2004$, $x_{1}, x_{2}, \\ldots, x_{k} \\in \\mathbb{N}$, $x_{1} < x_{2} < \\cdots < x_{k}$ and the product $x_{1} x_{2} \\ldots x_{k}$ is maximal. Assume that for some $i, j$, $1 \\leq i < j \\leq k$ one has that $x_{i} \\leq x_{i+1} - 2$ and $x_{j} \\leq x_{j+1} - ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 63!/11 | |
0gjw | Genji the ninja is to jump along the real axis, starting at the point $0$. In doing so, he alternates between the following two types of jumps, his first jump being type 1.
**Type 1:** Genji jumps from the current position $x$ to a point $y$ in the set
$$
\{x, x+1, x+2, x+3, x+4, x+5, x+6\}.
$$
**Type 2:** Genji jump... | [] | Thailand | Selected Problems from Thailand Training Camp | [
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
05pl | Problem:
a) Soit $x_{1} \geqslant x_{2}>0$ et $y_{1} \geqslant y_{2}>0$ des réels tels que
$$
x_{1} \geqslant y_{1} \text{ et } x_{1} x_{2} \geqslant y_{1} y_{2}
$$
Prouver que
$$
x_{1}+x_{2} \geqslant y_{1}+y_{2}
$$
b) Soit $x_{1} \geqslant x_{2} \geqslant \ldots \geqslant x_{n}>0$ et $y_{1} \geqslant y_{2} \geqslant ... | [
"Solution:\na) Procédons par l'absurde: supposons que $y_{1}+y_{2}>x_{1}+x_{2}$. Alors $y_{2}>x_{1}+x_{2}-y_{1}$ et on a\n$$\n0 \\geqslant y_{1} y_{2}-x_{1} x_{2}>y_{1}\\left(x_{1}+x_{2}-y_{1}\\right)-x_{1} x_{2}=\\left(x_{1}-y_{1}\\right)\\left(y_{1}-x_{2}\\right)\n$$\nPuisque $x_{1} \\geqslant y_{1}$, alors $y_{1... | France | Olympiades Françaises de Mathématiques - Envoi 2 (Algèbre) | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
04rb | Let $x$, $y$ and $z$ be any positive real numbers. Prove the inequality
$$
(x + y + z) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \le m^2, \quad \text{where } m = \min \left( \frac{x}{y} + \frac{y}{z} + \frac{z}{x}, \frac{y}{x} + \frac{z}{y} + \frac{x}{z} \right).
$$
Find when the equality holds. | [
"Since the inequality involves the minimum of two positive numbers and since the function $y = x^2$ is increasing on the set $\\mathbb{R}^+$, our task is to verify\n$$\n(x+y+z)\\left(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\right) \\le \\left(\\frac{x}{y}+\\frac{y}{z}+\\frac{z}{x}\\right)^2 \\quad \\text{and} \\quad... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Equality holds if and only if x = y = z. | |
0a3i | Problem:
Gegeven zijn reële getallen $a$, $b$ en $c$ waarvoor geldt dat $0 \leq a \leq b \leq c$ en $a+b+c=1$. Bewijs dat
$$
a b \sqrt{b-a}+b c \sqrt{c-b}+a c \sqrt{c-a}<\frac{1}{4}
$$ | [
"Solution:\n\nVoor elke $x \\leq y$ hebben we wegens de rekenkundig-meetkundig gemiddelde ongelijkheid dat $\\sqrt{y-x} \\leq \\frac{y-x+1}{2}$ met gelijkheid dan en slechts dan als $y-x=1$. Verder merken we op dat $1-a=b+c$ en de symmetrische varianten, omdat $a+b+c=1$. Als we dit toepassen op de wortels aan de li... | Netherlands | IMO-selectietoets II | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
07pd | Prove that $||1-4t^2|-3|t|| \le \frac{3}{2}$ for real numbers $t$ that satisfy $-1 \le t \le 1$. | [
"By letting $s = |t|$ we see that it is enough to show $|1 - 4s^2 - 3s| \\le \\frac{3}{2}$, for $0 \\le s \\le 1$. Now $|1 - 4s^2 - 3s| = |f(s)|$, where\n$$\n\\begin{aligned}\nf(s) &= |1 - 4s^2| - 3s \\\\\n&= \\begin{cases} 1 - 4s^2 - 3s, & \\text{if } 0 \\le s \\le \\frac{1}{2} \\\\ 4s^2 - 3s - 1, & \\text{if } \\... | Ireland | Ireland | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0ccj | We say that 13 positive integers form a *special* group if the numbers of the group are consecutive.
a) Determine the number of special groups for which the sum of their elements is a three-digit perfect square.
b) Find the maximum number of primes in a special group. | [
"a) Denote by $a, a+1, \\dots, a+12$ the elements of a special group. Their sum is $13a + 78 = 13(a + 6)$ and it must be a three-digit perfect square, therefore $13(a + 6) \\in \\{169, 676\\}$, thus $a \\in \\{7, 46\\}$. There are only 2 special groups with the given property.\n\nb) If $a = 1$ or $a = 2$, there are... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a) 2; b) 6 | |
02s8 | Problem:
Na cidade de Trocalândia, $20\%$ dos gatos pensam que são cachorros e $25\%$ dos cachorros pensam que são gatos. Certo dia, um psicólogo veterinário resolve testar todos os gatos e cachorros de Trocalândia, verificando que $30\%$ do total pensava ser gato. Que proporção dos animais testados era de cães? | [
"Solution:\n\nSejam $C$ e $G$, respectivamente, o número de cães e gatos de Trocalândia. O número de gatos que pensam que são gatos é\n$$\n\\frac{80 G}{100}\n$$\nO número de cachorros que pensam que são gatos é\n$$\n\\frac{25 C}{100}\n$$\nLogo, o número total de animais que pensam que são gatos é\n$$\n\\frac{80 G+2... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 10/11 | |
09jl | Find the smallest positive number $0 < c < 1$ satisfying the following property:
Any simple, triangle-free graph with $n \ge 3$ vertices, where the degree of any vertex is strictly bigger than $cn$, is bipartite.
**Explanation:** A simple graph is an undirected graph with no loops and no multiple edges. The degree of ... | [
"Answer: $c = \\frac{2}{5}$.\nWe say that a simple graph with $n$ vertices is *c-good* if it is triangle-free and the degree of any vertex is strictly bigger than $cn$. A length 5 cycle is triangle-free and the degree of any vertex is 2. For $c < \\frac{2}{5}$, this graph is *c*-good, but not bipartite.\n\nNow we s... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2/5 | |
011r | Problem:
The real-valued function $f$ is defined for all positive integers. For any integers $a > 1$, $b > 1$ with $d = \operatorname{gcd}(a, b)$, we have
$$
f(ab) = f(d) \cdot \left(f\left(\frac{a}{d}\right) + f\left(\frac{b}{d}\right)\right),
$$
Determine all possible values of $f(2001)$. | [
"Solution:\n\n0 and $\\frac{1}{2}$.\n\nObviously the constant functions $f(n) = 0$ and $f(n) = \\frac{1}{2}$ provide solutions.\n\nWe show that there are no other solutions. Assume $f(2001) \\neq 0$. Since $2001 = 3 \\cdot 667$ and $\\operatorname{gcd}(3, 667) = 1$, then\n$$\nf(2001) = f(1) \\cdot (f(3) + f(667)),\... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 0 and 1/2 | |
03d8 | Solve the equation $x^3 - 3x - 1 = 0$. | [
"Let $x = 2 \\cos \\theta$. Then\n$$\nx^3 - 3x = (2 \\cos \\theta)^3 - 3 \\cdot 2 \\cos \\theta = 8 \\cos^3 \\theta - 6 \\cos \\theta.\n$$\nRecall that $\\cos 3\\theta = 4 \\cos^3 \\theta - 3 \\cos \\theta$, so\n$$\n8 \\cos^3 \\theta - 6 \\cos \\theta = 2(4 \\cos^3 \\theta - 3 \\cos \\theta) = 2 \\cos 3\\theta.\n$$... | Bulgaria | Bulgaria 2022 | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | x = 2 cos(π/9), x = 2 cos(7π/9), x = 2 cos(13π/9) | |
0af3 | Даден е правоаголник $ABCD$ и точка $S$ (точката $S$ не мора да лежи во рамнината на правоаголникот). Дали растојанијата од точката $S$ до темињата на правоаголникот може во некој редослед да бидат еднакви на 1, 3, 5 и 7. | [
"Нека за правоаголникот $ABCD$ точката $S$ е таква да растојанијата на точката до темињата на правоаголникот по некој редослед се еднакви на 1, 3, 5, 7. Ќе ја пресликаме точката $S$, во однос на централна симетрија со центар пресекот на дијагоналите на правоаголникот $E$, во точка $F$. Тогаш четириаголниците $AFCS$... | North Macedonia | Републички натпревар по математика за средно образование | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | Macedonian, English | proof only | null | |
0i90 | Problem:
A cylinder of base radius $1$ is cut into two equal parts along a plane passing through the center of the cylinder and tangent to the two base circles. Suppose that each piece's surface area is $m$ times its volume. Find the greatest lower bound for all possible values of $m$ as the height of the cylinder vari... | [
"Solution:\nLet $h$ be the height of the cylinder. Then the volume of each piece is half the volume of the cylinder, so it is $\\frac{1}{2} \\pi h$. The base of the piece has area $\\pi$, and the ellipse formed by the cut has area $\\pi \\cdot 1 \\cdot \\sqrt{1+\\frac{h^{2}}{4}}$ because its area is the product of ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 3 | |
0fo6 | Se considera un polígono regular de $90$ vértices, numerados del $1$ al $90$ de manera aleatoria. Probar que siempre podemos encontrar dos vértices consecutivos cuyo producto es mayor o igual que $2014$. | [
"Consideremos el primer par de números consecutivos cuyo producto es mayor o igual que $2014$, que son el $45$ y el $46$. Por lo tanto, para que no se cumpliera el enunciado, los números que deben ir a izquierda y derecha de los vértices numerados del $46$ al $90$ tendrían que ser menores o iguales que $44$. Sin em... | Spain | L Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof only | null | |
081f | Problem:
Chiamiamo numeri monotòni gli interi positivi tali che
- si scrivono usando almeno due cifre;
- nessuna cifra è zero;
- le cifre compaiono in ordine strettamente crescente o strettamente decrescente.
(Ad esempio $127$ e $9742$ sono numeri monotoni, mentre $172$, $1224$ e $7320$ non lo sono.)
a. Calcolare la ... | [
"Solution:\n\na. Chiamiamo crescenti i numeri monotoni le cui cifre compaiono in ordine crescente, e analogamente chiamiamo decrescenti i numeri monotoni le cui cifre compaiono in ordine decrescente. Chiamiamo inoltre gemello di un numero monotono $N$, l'intero che si scrive sostituendo ogni cifra di $N$ con il suo... | Italy | Gara Nazionale di Matematica | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | a: 13999860; b: 3 | |
09yh | Problem:
Op een rechthoekig bord met $m \times n$ vakjes ($m, n \geq 3$) liggen domino's ($2 \times 1$ of $1 \times 2$-tegels), die elkaar niet overlappen en niet uitsteken buiten het bord. Elke domino bedekt precies twee vakjes van het bord. Neem aan dat de bedekking met domino's de eigenschap heeft dat er geen enkel... | [
"Solution:\n\nKoppel elk leeg vakje aan de domino die direct rechts van dit vakje ligt (tenzij het vakje aan de rechterrand van het bord ligt). Stel dat er nu twee lege vakjes aan dezelfde domino gekoppeld worden, dan moet deze domino verticaal liggen en zijn beide vakjes links van hem leeg. Echter, dan zou er daar... | Netherlands | IMO-selectietoets | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0c1c | Let $a, b, c \in [1, \infty)$. Prove that:
$$
\frac{a\sqrt{b}}{a+b} + \frac{b\sqrt{c}}{b+c} + \frac{c\sqrt{a}}{c+a} + \frac{3}{2} \le a+b+c.
$$ | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0a6m | Problem:
Let $ABC$ be an acute scalene triangle with $AC > BC > AB$. Let the orthocentre be $H$ and circumcentre be $O$. Suppose that lines $BO$ and $CH$ intersect at a point $D$. Point $E$ (where $E \neq C$) lies on side $AC$ so that $OECD$ is cyclic. Point $F$ (where $F \neq C$) lies on side $BC$ such that $CE = FE$... | [
"Solution:\n\nLet $\\alpha = \\angle BAC$. Let $BB'$ and $CC'$ be altitudes in triangle $ABC$, as shown.\n\n\n\nClaim. Triangle $CDE$ is isosceles with $CE = DE$.\n\nProof.\n\n$$\n\\angle DEC = \\angle DOC \\\\\n\\qquad = \\angle BOC \\\\\n\\qquad = 2\\angle BAC \\\\\n\\qquad = 2\\alpha\n$$... | New Zealand | NZMO Round Two | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0eci | Problem:
V vrsti stoji $n$ luči, $n \geq 3$, ki so oštevilčene s števili od $1$ do $n$. Na začetku je vsaka liha luč v vrsti prižgana, vsaka soda luč pa ugasnjena. V vsaki potezi lahko hkrati zamenjamo stanje treh zaporednih luči v vrsti (ugasnjene prižgemo, prižgane pa ugasnemo).
a. Dokaži, da vrstni red izvajanja p... | [
"Solution:\n\na.\nOglejmo si, kaj se dogaja s stanjem ene luči pri izvajanju potez. Vsaka poteza stanje luči bodisi zamenja bodisi ohrani. Končno stanje neke luči je zato odvisno le od števila izvedenih potez, ki stanje te luči zamenjajo, ni pa odvisno od vrstnega izvajanja potez. Torej tudi končno stanje vseh luči... | Slovenia | 59. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Exactly those n that are multiples of 3 | |
0e8t | Function $f$ satisfies $f(x) = x^2 + 1$. What is the value of $\frac{f(f(x)+x)}{f(x)}$?
(A) $x^2 + x + 1$
(B) $x^2 + 2x + 2$
(C) $x^2 + 1$
(D) $x^2 + 2x + 1$
(E) $x^2 + x$ | [
"From the definition of $f$ we get\n$$\n\\frac{f(f(x)+x)}{f(x)} = \\frac{(x^2+1+x)^2+1}{x^2+1} = \\frac{x^4+2x^3+3x^2+2x+2}{x^2+1}.\n$$\nDivision of polynomials gives us the result, $x^2+2x+2$. Hence, the correct answer is $B$."
] | Slovenia | National Math Olympiad 2013 - First Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | B | |
0ck1 | The positive integers $a < b < c$ are such that $a + b + 2c$ is a common multiple of $b$ and $c$ (*).
a) Prove that the greatest common divisor of $a + b$ and $c$ is $c$.
b) Find the positive integers $k < 1000$, so that $abc = k^2$ and $a, b, c$ fulfill (*). | [
"a) Since $c \\mid a + b + 2c$ and $c \\mid 2c$, $c \\mid a + b$. Now $a < b < c$ implies $a + b < 2c$, hence $a + b = c$. Then $(a + b, c) = c$.\n\nb) From $b \\mid a + b + 2c$ and (a), $b \\mid 3a + 3b$. Since $b \\mid 3b$, $b \\mid 3a$. From $3a < 3b$ follows $3a \\in \\{b, 2b\\}$.\n\nCase I: $3a = b$. Then $a =... | Romania | 75th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | {18, 144, 486, 900} | |
01ua | In an isosceles triangle $ABC$ with $AB = BC$, points $K$ and $M$ are the midpoints of the sides $AB$ and $AC$, respectively. The circumscribed circle of the triangle $CKB$ meets the line $BM$ at point $N$ different from $M$. The line passing through $N$ parallel to the side $AC$ meets the circumscribed circle of the t... | [
"Let $O$ be the center of the circumscribed circle of the triangle $ABC$. Since $KBCN$ is an inscribed quadrilateral and $NB$ is the bisector of the angle $KBC$, we have $KN = NC$ (see Fig. 1). Moreover, since $N$ lies on the perpendicular bisector of the segment $AC$, we have $AN = NC$, therefore, $N$ is the circu... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
04nx | Let $d$ be a positive integer, and let $(a_n)$ be an arithmetic sequence of positive integers with the common difference $d$. If $d \le 2018$, prove that at most 11 consecutive terms of that sequence can be prime numbers. | [] | Croatia | Croatia_2018 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0fp8 | Cada 20 minutos durante una semana se travasa una cantidad exacta de litros de agua (siempre la misma cantidad) desde un tanque con $25000$ litros a otro depósito inicialmente vacío. Desde este segundo depósito, a intervalos regulares de tiempo, se extrae primero $1$ litro, luego $2$, luego $3$, etc. Justo al final de ... | [
"Sea $n$ el número de extracciones de agua realizadas durante la semana. En total habrán extraído $T_n = 1 + 2 + \\dots + n = n(n+1)/2$ litros. Por otro lado, si el caudal que se trasvasa cada $20$ minutos al segundo depósito es de $k$ litros, el total de litros que ha entrado es $7 \\times 24 \\times 3 \\times k =... | Spain | LII Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Spanish | proof and answer | 2016 | |
02uj | Problem:
a) José aprendeu um método para calcular produtos de dois números de uma forma mais rápida baseado na fatoração:
$$
(n-k)(n+k)=n^{2}-k^{2}
$$
Para calcular $23 \cdot 17$, ele escolhe $n=20, k=3$ e calcula:
$$
23 \cdot 17=20^{2}-3^{2}=400-9=391
$$
Determine, sem usar a calculadora, o valor de $\sqrt{1001 \cdot... | [
"Solution:\n\na) Basta escolher $n=1002$ e $k=1$, pois\n$$\n\\begin{aligned}\n\\sqrt{1001 \\cdot 1003+1} & =\\sqrt{1002^{2}-1^{2}+1} \\\\\n& =\\sqrt{1002^{2}} \\\\\n& =1002\n\\end{aligned}\n$$\n\nb)\n$$\n\\begin{aligned}\n(n(n+3)+1)^{2} & =n^{2}(n+3)^{2}+2 n(n+3)+1 \\\\\n& =n(n+3)[n(n+3)+2]+1 \\\\\n& =n(n+3)\\left[... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a) 1002; b) The identity holds: n(n+1)(n+2)(n+3)+1 equals (n(n+3)+1) squared; c) 4062239 | |
0k8s | Problem:
Let $a_{1}, a_{2}, \ldots$ be an arithmetic sequence and $b_{1}, b_{2}, \ldots$ be a geometric sequence. Suppose that $a_{1} b_{1}=20$, $a_{2} b_{2}=19$, and $a_{3} b_{3}=14$. Find the greatest possible value of $a_{4} b_{4}$. | [
"Solution:\nLet $\\{a_{n}\\}$ have common difference $d$ and $\\{b_{n}\\}$ have common ratio $r$; for brevity, let $a_{1}=a$ and $b_{1}=b$. Then we have the equations $a b=20$, $(a+d) b r=19$, and $(a+2 d) b r^{2}=14$, and we want to maximize $(a+3 d) b r^{3}$.\n\nThe equation $(a+d) b r=19$ expands as $a b r+d b r... | United States | HMMT February 2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 37/4 | |
0hca | A circle passes through vertices **A**, **B** of a parallelogram $ABCD$, and intersects diagonals $BD$ and $AC$ in points $X$ and $Y$, respectively. Circumscribed circle of $\angle ADX$ intersects diagonal $AC$ in point $Z$. Prove that $AY = CZ$. | [
"Since $A$, $B$, $Z$, $D$ lie on one circle, and $A$, $B$, $X$, $Y$ lie on another circle (fig. 26),\n\n$$\n\\angle AYB = \\angle AXB = 180^\\circ - \\angle AXD = 180^\\circ - \\angle AZD = \\angle DZC.\n$$\n\nSince $ABCD$ is a parallelogram, $\\angle BAY = \\angle DCZ$, hence, $\\angle ABY = \\angle CDZ$, which yi... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ac2 | In the isosceles triangle $ABC$ let $D$ be the base point for the altitude from $C$ to the base $AB$ of $ABC$. Calculate the length of the segment that connects the middle point of $AD$ and the middle point of $AC$ if it is known that the perimeter of the triangle $ABC$ is $36\, ext{cm}$ and the perimeter of the triang... | [
"$2 \\cdot L_{\\Delta ADC} = L_{\\Delta ABC} + 2 \\cdot \\overline{CD}$. From here, we obtain $2 \\cdot 29 = 36 + 2 \\cdot \\overline{CD}$ or $\\overline{CD} = (2 \\cdot 29 - 36) : 2 = 11\\,\\text{cm}$. Let $M$ be the middle point of $AD$ and $N$ be the middle point of $AC$. Then $MN$ is the middle line of the tria... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 5.5 cm | |
05t5 | Problem:
Soit $n$ un entier naturel non nul. On dit qu'une fonction $f: \mathbb{R} \rightarrow \mathbb{R}$ est $n$-positive si, pour tous les réels $x_{1}, \ldots, x_{n}$ tels que $x_{1}+\ldots+x_{n}=0$, on a $f\left(x_{1}\right)+\ldots+f\left(x_{n}\right) \geqslant 0$.
a) Toute fonction 2020-positive est-elle nécess... | [
"Solution:\n\nDans les deux solutions, on pose $n=1010$.\n\na. Soit $f$ une fonction $2n$-positive, et soit $x_{1}, \\ldots, x_{n}$ des réels tels que $x_{1}+\\ldots+x_{n}=0$. En notant que $x_{1}+\\ldots+x_{n}+x_{1}+\\ldots+x_{n}=0$ et que $f$ est $2n$-positive, on constate que\n$$\n0 \\leqslant f\\left(x_{1}\\rig... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | a) yes; b) no | |
08bx | Problem:
Sia $n$ un intero positivo tale che la rappresentazione decimale di $2^{n}$ inizia con la cifra 7 (ovvero la cifra non nulla più a sinistra è 7). Con che cifra inizia la rappresentazione decimale di $5^{n}$?
(A) 1
(B) 2
(C) 5
(D) 7
(E) Dipende da $n$ | [
"Solution:\n\nLa risposta è (A). Essa può essere trovata sfruttando l'uguaglianza $5^{n} = \\frac{10^{n}}{2^{n}}$. Sia $c$ il numero di cifre decimali di $2^{n}$; l'algoritmo di divisione in colonna comporta che la prima cifra del quoziente $\\frac{10^{c}}{2^{n}}$ sia il risultato della divisione intera di $10^{c}$... | Italy | Gara di Febbraio | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | A | |
0acr | In a square shaped area on the ground with side length $12\text{ m}$ a cylindrical hole is dug with diameter $8\text{ m}$. The dug soil is equally allotted on the square shaped area, outside of the hole, and then is pressed as much as was before the digging. How deep should be the dug in order the hole to be deep? | [
"The area of the land where the dug soil is allotted is $(12^2 - 4^2\\pi)\\text{ m}^2 = (144 - 16\\pi)\\text{ m}^2$. Let $x$ be the height of the hole before the dug soil is allotted. Then the thickness of the layer of allotted and then pressed soil is $(3-x)\\text{ m}$. Because the dug soil is pressed as much as i... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Solid Geometry > Volume",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 3 − π/3 meters (approximately 1.95 meters) | |
0a39 | Problem:
Vind alle paren priemgetallen $(p, q)$ waarvoor er positieve gehele $(m, n)$ bestaan zodat
$$
(p+q)^{m} = (p-q)^{n}
$$ | [
"Solution:\n\nDe enige deler die $p-q$ en $p+q$ gemeen kunnen hebben, is $2$, want $p$ en $q$ zijn verschillende priemgetallen. Inderdaad, een deler $d$ van $p+q$ en $p-q$ is ook een deler van $(p+q)+(p-q)=2p$ en van $(p+q)-(p-q)=2q$. En we weten dat $\\operatorname{ggd}(2p, 2q)=2$, dus $d$ moet een deler zijn van ... | Netherlands | Maarttoets | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (p, q) = (3, 5) or (5, 3) | |
09p3 | Let $\alpha, \beta, \gamma$ be the angles opposite the sides $a, b, c$ of a triangle, respectively. Prove that if the lengths of $a, b, c$ form an arithmetic progression in this order, then the values $\cos \alpha, 1 - \cos \beta, \cos \gamma$ also form an arithmetic progression in that order.
(Otgonbayar Uuye) | [
"By the Law of Sines, the values $\\sin \\alpha, \\sin \\beta, \\sin \\gamma$ form an arithmetic progression. Therefore\n$$\n2 \\cos \\frac{\\beta}{2} \\cos \\frac{\\gamma - \\alpha}{2} = 2 \\sin \\frac{\\alpha + \\gamma}{2} \\cos \\frac{\\gamma - \\alpha}{2} = \\sin \\alpha + \\sin \\gamma = 2 \\sin \\beta = 4 \\s... | Mongolia | MMO2025 Round 4 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
06vw | Let $a$ and $b$ be two positive integers. Prove that the integer
$$
a^{2} + \left\lceil \frac{4 a^{2}}{b} \right\rceil
$$
is not a square. (Here $\lceil z\rceil$ denotes the least integer greater than or equal to $z$.) | [
"Assume, for contradiction, that\n$$\na^{2} + \\left\\lceil \\frac{4 a^{2}}{b} \\right\\rceil = (a + k)^{2}, \\quad \\text{or} \\quad \\left\\lceil \\frac{(2a)^{2}}{b} \\right\\rceil = (2a + k)k.\n$$\nClearly, $k \\geqslant 1$. In other words, the equation\n$$\n\\left\\lceil \\frac{c^{2}}{b} \\right\\rceil = (c + k... | IMO | IMO 2019 Shortlisted Problems | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof only | null | |
0jr9 | Problem:
Let $ABCD$ be a quadrilateral with an inscribed circle $\omega$ that has center $I$. If $IA = 5$, $IB = 7$, $IC = 4$, $ID = 9$, find the value of $\frac{AB}{CD}$. | [
"Solution:\n\nThe $I$-altitudes of triangles $AIB$ and $CID$ are both equal to the radius of $\\omega$, hence have equal length. Therefore $\\frac{[AIB]}{[CID]} = \\frac{AB}{CD}$. Also note that $[AIB] = IA \\cdot IB \\cdot \\sin AIB$ and $[CID] = IC \\cdot ID \\cdot \\sin CID$, but since lines $IA, IB, IC, ID$ bis... | United States | HMMT November 2015 | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 35/36 | |
035f | Problem:
Two circles $k_{1}$ and $k_{2}$ are externally tangent at point $T$. A line meets $k_{1}$ at points $A$ and $B$ and is tangent to $k_{2}$ at point $X$. The line $X T$ meets $k_{1}$ at point $S$ and let $C$ be a point on the arc $\overparen{T S}$ which does not contain $A$ and $B$. Let $C Y$ be the tangent lin... | [
"Solution:\n\na)\nSince the circles $k_{1}$ and $k_{2}$ are tangent at the point $T$, we have\n$$\n\\Varangle B X T = \\frac{\\widehat{X T}}{2} = \\frac{\\widehat{T S}}{2} = \\Varangle T A S.\n$$\nThen it follows easily that $S$ is the midpoint of the arc $\\widehat{A B}$, i.e. $S A = S B$. Hence $\\Varangle T C I ... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ... | null | proof only | null | |
029n | Problem:
Família Sétimo - O Sr. e Sra. Sétimo têm 7 filhos, todos nascidos em 1º de abril, na verdade em seis 1º de abril consecutivos. Este ano, para seus aniversários, a Sra. Sétimo fez um bolo com velinhas para cada um - o número de velas igual ao número de anos de cada um. João Sétimo, o filho que mais gosta de Ma... | [
"Solution:\n\nOs nascimentos ocorreram em seis 1º de abril, logo existem irmãos gêmeos. Como nesse ano temos 2 bolos a mais que há 2 anos atrás, então há 2 anos atrás o mais jovem ainda não tinha nascido, o penúltimo filho tinha acabado de nascer, e os gêmeos já tinham nascido. Atualmente o mais jovem tem 1 ano e o... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 26 | |
0cl0 | The sequence $(a_n)$ of positive integers is defined by $a_1 = 1$ and $a_n = (a_{n-1}+1)^2$ for $n \ge 2$. Let $p$ be an odd prime. Prove that $a_{2p} - a_p$ has at least $p$ pairwise distinct prime divisors. | [
"Let $b_0 = 1$ and write $b_n = a_n + 1$, $n \\ge 1$. Clearly, $b_{n+1} = b_n^2 + 1$, $n \\ge 0$. If $n > m \\ge 1$, it then follows that $a_n - a_m = b_n - b_m = b_{n-1}^2 - b_{m-1}^2 = (b_{n-1} - b_{m-1})(b_{n-1} - b_{m-1})$, so\n\n$$\na_n - a_m = b_n - b_m = (b_{n-m} - b_0) \\prod_{k=0}^{m-1} (b_{n-m+k} + b_k). ... | Romania | 75th NMO Selection Tests | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
08jg | Problem:
Prove that each positive integer is equal to a difference of two positive integers with the same number of the prime divisors. | [] | JBMO | The second selection test for JMBO 2003 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Other"
] | null | proof only | null | |
0fno | Tenemos 50 fichas numeradas del 1 al 50, y hay que colorearlas de rojo o azul. Sabemos que la ficha 5 es de color azul. Para la coloración del resto de fichas se siguen las siguientes reglas:
a) Si la ficha con el número $x$ y la ficha con el número $y$ son de distinto color, entonces la ficha con el número $|x - y|$ s... | [
"Observemos que dos números que se diferencian en $5$ tienen el mismo color. En efecto, si fueran de distinto color, su diferencia debería ser de color rojo, por la regla a). Pero su diferencia es $5$, que es de color azul. Por tanto basta con saber el color de los $4$ primeros números. Aquí, distinguimos dos casos... | Spain | L Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | 2 | |
08ox | Problem:
Let $c \equiv c(O, K)$ be a circle with center $O$ and radius $R$ and $A, B$ be two points on it, not belonging to the same diameter. The bisector of the angle $A \hat{B} O$ intersects the circle $c$ at point $C$, the circumcircle of the triangle $A O B$, say $^{c_{1}}$ at point $K$ and the circumcircle of th... | [
"Solution:\n\nThe segments $O B, O C$ are equal, as radii of the circle $^{c}$. Hence $O B C$ is an isosceles triangle and\n$$\n\\hat{B}_{1}=\\hat{C}_{1}=\\hat{x}\n$$\n\n\n\nThe chord $B C$ is the bisector of the angle $O \\hat{B} A$, and hence\n$$\n\\hat{B}_{1}=\\hat{B}_{2}=\\hat{x}\n$$\nT... | JBMO | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.