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
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values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0l94 | Consider the equation
$$
\frac{1}{x-1} + \frac{1}{4x-1} + \dots + \frac{1}{k^2x-1} + \dots + \frac{1}{n^2x-1} = \frac{1}{2}
$$
where $n$ is a positive integer parameter.
1/ Prove that for every positive integer $n$, the considered equation has a unique root greater than $1$, which is denoted by $x_n$.
2/ Prove that the... | [
"The given equation can be written in the form:\n$$\nf_n(x) = \\frac{-1}{2} + \\frac{1}{x-1} + \\frac{1}{4x-1} + \\dots + \\frac{1}{k^2x-1} + \\dots + \\frac{1}{n^2x-1} = 0 \\quad (1)\n$$\n1/ It is easily seen that for every $n \\in \\mathbb{N}^*$, the function $f_n(x)$ is continuous and decreasing on the interval ... | Vietnam | THE 2002 VIETNAMESE MATHEMATICAL OLYMPIAD | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications",
"Precalculus > Limits"
] | English | proof and answer | 4 | |
0j7q | Problem:
Let $S$ be a finite, nonempty set of real numbers such that the distance between any two distinct points in $S$ is an element of $S$. In other words, $|x-y|$ is in $S$ whenever $x \neq y$ and $x$ and $y$ are both in $S$.
Prove that the elements of $S$ may be arranged in an arithmetic progression. This means th... | [
"Solution:\nIf $S$ has just one element, then $S=\\{a\\}$ satisfies the given condition.\n\nLet the elements of $S$ be $a_1 < a_2 < \\cdots < a_n$.\n\nIf $a_1 < 0$, then $|a_n - a_1| \\geq a_n - a_1 > a_n$, so $|a_n - a_1|$ cannot be an element of $S$—contradiction.\n\nNow suppose $a_1 > 0$. The numbers\n$$\n\\begi... | United States | 13th Bay Area Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ij1 | Problem:
Let $N$ denote the number of subsets of $\{1,2,3, \ldots, 100\}$ that contain more prime numbers than multiples of $4$. Compute the largest integer $k$ such that $2^{k}$ divides $N$. | [
"Solution:\nLet $S$ denote a subset with the said property. Note that there are $25$ multiples of $4$ and $25$ primes in the set $\\{1,2,3, \\ldots, 100\\}$, with no overlap between the two. Let $T$ denote the subset of $50$ numbers that are neither prime nor a multiple of $4$, and let $U$ denote the $50$ other num... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 52 | |
05yv | Problem:
Soient $ABC$ un triangle isocèle en $A$, $O$ le centre de son cercle circonscrit et $I$ le centre de son cercle inscrit. La parallèle à $(AB)$ passant par $I$ coupe $(AC)$ en $D$. Montrer que $(CI)$ et $(DO)$ sont perpendiculaires. | [
"Solution:\n\n\n\nPremièrement, comme $AB = AC$, $A$, $O$, $I$ sont alignés et la droite $(AI)$ coupe $[BC]$ en son milieu $M$. On pose $P$ l'intersection des droites $(DO)$ et $(CI)$.\n\nPar chasse aux angles, on montre que $I$, $O$, $D$, $C$ sont cocycliques, en montrant que $\\widehat{IO... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"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 | |
06r2 | Given six positive numbers $a, b, c, d, e, f$ such that $a < b < c < d < e < f$. Let $a + c + e = S$ and $b + d + f = T$. Prove that
$$
2 S T > \sqrt{3(S + T)(S(b d + b f + d f) + T(a c + a e + c e))}
$$ | [
"We define also $\\sigma = a c + c e + a e$, $\\tau = b d + b f + d f$. The idea of the solution is to interpret (1) as a natural inequality on the roots of an appropriate polynomial.\n\nActually, consider the polynomial\n$$\nP(x) = (b + d + f)(x - a)(x - c)(x - e) + (a + c + e)(x - b)(x - d)(x - f)\n$$\n$$\n= T\\l... | IMO | 51st IMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities ... | English | proof only | null | |
01w9 | Is it true that for any nonzero rational numbers $a$ and $b$ one can find integers $m$ and $n$ such that the number $(am + b)^2 + (a + nb)^2$ is integer? | [
"**Answer: no.**\nLet $b = 1$ and let $a$ be equal to some irreducible fraction $p/q$ with the denominator $q$ such that the squares of integers are never congruent to $q-1$ modulo $q$ (i.e., $-1$ is a quadratic nonresidue modulo $q$). For example, let $a = 1/3$. Transform the expression from the problem condition:... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | No | |
0kjs | Problem:
Let $P$ be a point selected uniformly at random in the cube $[0,1]^3$. The plane parallel to $x+y+z=0$ passing through $P$ intersects the cube in a two-dimensional region $\mathcal{R}$. Let $t$ be the expected value of the perimeter of $\mathcal{R}$. If $t^2$ can be written as $\frac{a}{b}$, where $a$ and $b$... | [
"Solution:\n\nWe can divide the cube into 3 regions based on the value of $x+y+z$ which defines the plane: $x+y+z<1$, $1 \\leq x+y+z \\leq 2$, and $x+y+z>2$. The two regions on the ends create tetrahedra, each of which has volume $1/6$. The middle region is a triangular antiprism with volume $2/3$.\n\nIf our point ... | United States | HMMT Spring 2021 Guts Round | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 12108 | |
06et | Let $\triangle ABC$ be given with $R$ radius of its circumcircle. Let $AB = c$, $BC = a$ and $CA = b$. Let $K_1$ and $K_2$ be the circles which pass through $C$ and are tangent to $AB$ at $A$ and $B$ respectively. Let $K$ be the circle of radius $r$ which is externally tangent to $K_1$, $K_2$ and tangent to the line $A... | [
"a.\nThe answer is $r = \\frac{abc^2}{4R(a+b)^2}$ or any equivalent expression.\n\nLet $O$, $O_1$, $O_2$ be the centres of $K$, $K_1$, $K_2$ respectively. Let $R_1$ and $R_2$ be the radii of $K_1$ and $K_2$ respectively. We have\n$$\nR_1 = AO_1 = \\frac{AC}{2 \\cos \\angle CAO_1} = \\frac{b}{2 \\sin A} = \\frac{bR}... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | r = abc^2 / (4R(a+b)^2); angle C = 90 degrees | |
0c1m | Consider the rectangular cuboid $ABCD A'B'C'D'$. Denote by $M, N, P$ the midpoints of $[AB]$, $[BC]$ and $[BB']$, respectively. Let $\{O\}$ be the intersection point of lines $A'N$ and $C'M$.
a) Prove that the points $D$, $O$ and $P$ are collinear.
b) Show that $MC' \perp (A'PN)$ if and only if $ABCDA'B'C'D'$ is a cube... | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
0ipu | Problem:
Point $A$ lies at $(0,4)$ and point $B$ lies at $(3,8)$. Find the $x$-coordinate of the point $X$ on the $x$-axis maximizing $\angle A X B$. | [
"Solution:\n\nAnswer: $5 \\sqrt{2}-3$\n\nLet $X$ be a point on the $x$-axis and let $\\theta=\\angle A X B$. We can easily see that the circle with diameter $A B$ does not meet the $x$-axis, so $\\theta \\leq \\pi$. Thus, maximizing $\\theta$ is equivalent to maximizing $\\sin \\theta$. By the Law of Sines, this in... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | final answer only | 5 sqrt(2)-3 | |
0ixz | Find all triples $(x, y, z)$ of real numbers that satisfy the system of equations
$$
\begin{cases} x^3 = 3x - 12y + 50, \\ y^3 = 12y + 3z - 2, \\ z^3 = 27z + 27x. \end{cases}
$$ | [
"**Solution 1.** Rewrite the system as\n$$\n\\begin{cases} x^3 - 3x - 2 = -12(y - 4), \\\\ y^3 - 12y - 16 = 3(z - 6), \\\\ z^3 - 27z - 54 = 27(x - 2). \\end{cases}\n$$\nThen factor the left sides to obtain\n$$\n\\begin{cases} (x+1)^2(x-2) = -12(y-4), \\\\ (y+2)^2(y-4) = 3(z-6), \\\\ (z+3)^2(z-6) = 27(x-2). \\end{ca... | United States | Team Selection Test 2009 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (2, 4, 6) | |
01ok | Three points $A$, $B$, $C$, are marked on the hyperbola $y = 1/x$ so that the triangle $ABC$ is equilateral.
Find all possible values of the product of the sum of abscissae and the sum of ordinates of the vertices of $ABC$. | [
"Answer 9.\n\nLet $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$ be the marked points. Then the required product is equal to\n$$\nT = (a+b+c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right).\n$$\n\n\n\nSince any vertical and any horizontal line meets the hyperbola $y = 1/x$ at most at on... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 9 | |
0dd8 | Given a positive integer $k$, show that there exists a prime $p$ such that one can choose distinct integers $a_1, a_2, \dots, a_{k+3} \in \{1, 2, \dots, p-1\}$ such that $p$ divides $a_i a_{i+1} a_{i+2} a_{i+3} - i$ for all $i = 1, 2, 3, \dots, k$. | [
"First we choose distinct positive rational numbers $r_1, \\dots, r_{k+3}$ such that\n$$\nr_i r_{i+1} r_{i+2} r_{i+3} = i \\quad \\text{for } 1 \\le i \\le k\n$$\nLet $r_1 = x, r_2 = y, r_3 = z$ be some distinct primes greater than $k$; the remaining terms satisfy $r_4 = \\frac{1}{r_1 r_2 r_3}$ and $r_{i+4} = \\fra... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0gr7 | There are finite number of vertical sticks each of length $2017$ fixed on some plate. There is a bead on each stick which slides freely on the stick. Some bead pairs are connected by elasticated ropes. The Young Ant freely moves on all ropes. The Old Ant freely moves on all ropes for which the difference between height... | [
"Let us reformulate the problem in terms of the graph theory. Let $G$ be a connected graph. Suppose that each vertex of $G$ can be properly coloured into one of colors $0, 1, \\ldots, 2017$: endpoints of each edge are differently coloured. Let $f(x)$ be a color of the vertex $x$. Prove that there is a proper recolo... | Turkey | 25th Turkish Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
02cc | Problem:
Ordenando frações - Qual dos números fica entre $2/5$ e $3/4$?
(a) $1/6$
(b) $4/3$
(c) $5/2$
(d) $4/7$
(e) $1/4$ | [
"Solution:\n\nA opção correta é (d).\n\nLembre que a ordem entre frações constituídas de inteiros positivos é determinada pelo produto cruzado dos inteiros, ou seja, $\\frac{a}{b} < \\frac{c}{d}$ equivale à afirmação $a \\times d < b \\times c$. Desse modo, temos\n$$\n\\frac{1}{6} < \\frac{1}{4} < \\frac{2}{5} < \\... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | d | |
0dlm | Let a grid $2m \times 2n$ is given. Laminates colors each cell of the grid either black or white such that the cells of each color form a polygon. Then Yamal covers the whole grid by $2mn$ dominoes of size $1 \times 2$. The domino is called combined if it covers 1 black and 1 white cell. Find the maximal number of comb... | [] | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | m + n - 1 | |
0cs8 | По кругу расставлены $99$ натуральных чисел. Известно, что любые два соседних числа отличаются или на $1$, или на $2$, или в два раза. Докажите, что хотя бы одно из этих чисел делится на $3$. (С. Берлов) | [
"Пусть ни одно из чисел не делится на $3$. Тогда каждое число даёт остаток $1$ или $2$ при делении на $3$. Но числа, дающие одинаковые ненулевые остатки при делении на $3$, не могут отличаться на $1$ или на $2$; не могут они и отличаться в два раза. Значит, соседние числа дают различные остатки при делении на $3$, ... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0j0a | Problem:
A rectangular piece of paper is folded along its diagonal (as depicted below) to form a non-convex pentagon that has an area of $\frac{7}{10}$ of the area of the original rectangle. Find the ratio of the longer side of the rectangle to the shorter side of the rectangle.
 | [
"Solution:\n\n\nGiven a polygon $P_{1} P_{2} \\cdots P_{k}$, let $\\left[P_{1} P_{2} \\cdots P_{k}\\right]$ denote its area. Let $A B C D$ be the rectangle. Suppose we fold $B$ across $\\overline{A C}$, and let $E$ be the intersection of $\\overline{A D}$ and $\\overline{B^{\\prime} C}$. Th... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | sqrt(5) | |
01s9 | Given integers $a$, $b$, $c$, with
$$
\frac{a^2}{a^2 + b^2} + \frac{c^2}{a^2 + c^2} = \frac{2c}{b+c},
$$
prove that the product $bc$ is a perfect square. | [
"If $b = 0$ or $c = 0$, then $bc$ is a perfect square. If $a = 0$, then from\n$$\n\\frac{a^2}{a^2 + b^2} + \\frac{c^2}{a^2 + c^2} = \\frac{2c}{b+c} \\quad (1)\n$$\nit follows that $b = c$, so $bc$ is a perfect square. Therefore we can assume that all $a$, $b$, $c$ are different from $0$. Then we can rewrite (1) as\... | Belarus | FINAL ROUND | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | English | proof only | null | |
0k9a | Problem:
Compute the sum of all positive integers $n < 2048$ such that $n$ has an even number of $1$'s in its binary representation. | [
"Solution:\n\nNote that the positive integers less than $2048$ are those with at most $11$ binary digits. Consider the contribution from any one of those digits. If we set that digit to $1$, then the remaining $10$ digits can be set in $2^{9} = 512$ ways so that the number of $1$'s is even. Therefore the answer is\... | United States | HMMT November 2019 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1048064 | |
0bm6 | Determine all positive integers expressible, for every integer $n \ge 3$, in the form $\frac{(a_1 + 1)(a_2 + 1)\cdots(a_n + 1) - 1}{a_1a_2\cdots a_n}$, where $a_1, a_2, \ldots, a_n$ are pairwise distinct positive integers. | [
"The integers greater than $3$ are ruled out by noticing that if $a_1, a_2, a_3$ are pairwise distinct positive integers, then\n$$\n\\begin{aligned}\n\\frac{(a_1+1)(a_2+1)(a_3+1)-1}{a_1 a_2 a_3} &= 1 + \\frac{1}{a_1} + \\frac{1}{a_2} + \\frac{1}{a_3} + \\frac{1}{a_1 a_2} + \\frac{1}{a_1 a_3} + \\frac{1}{a_2 a_3} \\... | Romania | 2015 Thirteenth IMAR Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Other"
] | English | proof and answer | 2 and 3 | |
0030 | Dado un entero positivo $n$, en un plano se consideran $2n$ puntos alineados $A_1, A_2, \dots, A_{2n}$. Cada punto se colorea de azul o rojo mediante el siguiente procedimiento: En el plano dado se trazan $n$ circunferencias con diámetros de extremos $A_i$ y $A_j$, disyuntas dos a dos. Cada $A_k$, $1 \le k \le 2n$, per... | [] | Argentina | XX Olimpiada Iberoamericana de Matemáticas | [
"Geometry > Plane Geometry > Circles",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Español | proof and answer | 2^n | |
06pg | In the Cartesian coordinate plane define the strip $S_{n}=\{(x, y) \mid n \leq x<n+1\}$ for every integer $n$. Assume that each strip $S_{n}$ is colored either red or blue, and let $a$ and $b$ be two distinct positive integers. Prove that there exists a rectangle with side lengths $a$ and $b$ such that its vertices hav... | [
"If $S_{n}$ and $S_{n+a}$ have the same color for some integer $n$, then we can choose the rectangle with vertices $(n, 0) \\in S_{n}, (n, b) \\in S_{n}, (n+a, 0) \\in S_{n+a}$, and $(n+a, b) \\in S_{n+a}$, and we are done. So it can be assumed that $S_{n}$ and $S_{n+a}$ have opposite colors for each $n$.\n\nSimila... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discr... | English | proof only | null | |
0cfo | Let $f : [0, 1] \to [0, \infty)$ be a convex function with $f(0) = 0$. Show that
$$
\left( \int_{0}^{1} f(x) \, dx \right)^{2} \leq \frac{3}{4} \int_{0}^{1} f^{2}(x) \, dx .
$$ | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
067u | Let $AB\Gamma$ be an acute angled triangle with $AB \le A\Gamma$ and $c(O, R)$ its circumcircle. The line perpendicular from $A$ to the tangent of $c(O, R)$ at $\Gamma$ intersects it at $\Delta$.
(α) If the triangle $AB\Gamma$ is isosceles with $AB = A\Gamma$, prove that: $\Gamma\Delta = \frac{B\Gamma}{2}$.
(β) If $\... | [
"(α) If $Z$ is the midpoint of $B\\Gamma$, then $AZ$ is altitude and median of the isosceles triangle $AB\\Gamma$. Then the right angled triangles $A\\Gamma Z$ and $A\\Gamma\\Delta$ are equal, because they have: $A\\Gamma$ common hypotenuse and $A\\Gamma\\Delta = A\\Gamma Z$, (the last equality arises from the equa... | Greece | Hellenic Mathematical Olympiad ARCHIMEDES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
0haq | Rounded tower has 16 doors, behind each there is a chest with gold of captain Flint. These doors are at equal distances to the neighboring ones, and are numbered clockwise from 1 to 16. 16 pirates come to the tower, each having a key, all keys are numbered from 1 to 16. It is known that key with number $n$ opens doors ... | [
"Consider some arrangement of keys. Then for every column there is a corresponding key. Paint gray all the cells of the column that can be opened by the corresponding key. In the column where there is key 2, 8 cells will be painted, where there is key 5 – 3 cells. In total there will be painted\n\n$$\n16 + 8 + 5 + ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 3 | |
0di7 | Given is an equilateral triangle $ABC$ with circumcenter $O$. Let $D$ be a point on the minor arc $BC$ of its circumcircle such that $DB > DC$. The perpendicular bisector of $OD$ meets the circumcircle at $E$, $F$, with $E$ lying on the minor arc $BC$. The lines $BE$ and $CF$ meet at $P$. Prove that $PD \perp BC$. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"G... | English | proof only | null | |
0fwy | Problem:
Zwei Kreise $k_{1}$ und $k_{2}$ schneiden sich in $A$ und $B$. Sei $r$ eine Gerade durch $B$, die $k_{1}$ in $C$ und $k_{2}$ in $D$ schneidet, so dass $B$ zwischen $C$ und $D$ liegt. Sei $s$ die Gerade parallel zu $A D$, die $k_{1}$ in $E$ berührt und zu $A D$ den kleinstmöglichen Abstand hat. Die Gerade $A E$... | [
"Solution:\n(a) Wir zeigen zuerst, dass $C E$ parallel zu $D F$ ist. Es gilt\n$$\n\\angle C E F=\\angle C E A=\\angle A B D=\\angle A F D=\\angle E F D\n$$\nalso ist $C E$ parallel zu $D F$. Sei $X$ der Schnittpunkt von $s$ und $t$. Um die Parallelität von $A C$ und $t$ zu zeigen, genügt es $\\angle D F X=\\angle E... | Switzerland | IMO Selektion 2008 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00h9 | Let $n$ be a positive integer. Consider $2n$ distinct lines on the plane, no two of which are parallel. Of the $2n$ lines, $n$ are colored blue, the other $n$ are colored red. Let $\mathcal{B}$ be the set of all points on the plane that lie on at least one blue line, and $\mathcal{R}$ the set of all points on the plane... | [
"Consider a line $\\ell$ on the plane and a point $P$ on it such that $\\ell$ is not parallel to any of the $2n$ lines. Rotate $\\ell$ about $P$ counterclockwise until it is parallel to one of the $2n$ lines. Take note of that line and keep rotating until all the $2n$ lines are met. The $2n$ lines are now ordered a... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > ... | English | proof only | null | |
0fyc | Problem:
Eine natürliche Zahl $x$ heisst gut, falls $x$ das Produkt einer geraden Anzahl (nicht notwendig verschiedener) Primzahlen ist. Seien $a, b$ natürliche Zahlen und definiere $m(x) = (x+a)(x+b)$.
a. Beweise, dass natürliche Zahlen $a, b$ existieren, sodass $m(1), m(2), \ldots, m(2010)$ alles gute Zahlen sind.
... | [
"Solution:\n\nFür eine natürliche Zahl $x$ setzen wir $u(x) = 0$ falls $x$ gut ist und $u(x) = 1$ sonst. Es gilt dann offensichtlich $u(xy) \\equiv u(x) + u(y) \\pmod{2}$.\n\na. In der binären Folge\n$$\nv = (u(2), u(3), u(4), \\ldots)\n$$\nbetrachte man die ersten $n = 2^{2010} + 1$ Teilfolgen der Länge 2010, d.h.... | Switzerland | IMO Selektion | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
01am | Four circles with a common center are drawn in a plane, the distances between adjacent circles are equal. Prove that it is not possible to draw a square with each vertex lying on a different circle. | [
"**Lemma 1.** If $ABCD$ is a square then for arbitrary point $E$\n$$\nEA^2 + EC^2 = EB^2 + ED^2.\n$$\n*Proof*. Let $A'$, $C'$ be projections of $E$ on sides $AB$, and $CD$ respectively. Then\n$$\nEA^2 + EC^2 = (EA'^2 + AA'^2) + (EC'^2 + CC'^2)\n$$\n$$\nEB^2 + ED^2 = (EA'^2 + A'B^2) + (EC'^2 + C'D^2)\n$$\nAs $A'B = ... | Baltic Way | Baltic Way 2013 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0e72 | Find all real numbers $x$ that solve the equation
$$
\cos(\pi \sin^2 x) + \sin(\pi \cos^2 x) = 1.
$$ | [
"Denote $y = \\pi \\sin^2 x$. Because $\\sin(\\pi \\cos^2 x) = \\sin(\\pi(1 - \\sin^2 x)) = \\sin(\\pi - \\pi \\sin^2 x) = \\sin(\\pi \\sin^2 x)$, we get the equation $\\cos y + \\sin y = 1$.\n\nBecause $\\cos y + \\sin y = \\sin y + \\sin\\left(\\frac{\\pi}{2} - y\\right) = 2\\sin\\frac{\\pi}{4}\\cos\\frac{2y-\\fr... | Slovenia | National Math Olympiad 2012 | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | x = n·π or x = π/4 + n·(π/2), for any integer n | |
0bh4 | A $4 \times 4$ magic square has in its first line the non-nil digits $a, b, c, d$, written in this order (from left to right), and each of the other lines contains the same numbers, written in different orders. It is known that the sum of the eight four-digit numbers obtained by reading the columns downwards and by rea... | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Smallest = 1899, Largest = 9981 | |
09ks | Consider a class of students. Within any group of six students, there are always two students who are not friends. Furthermore, if we select two of these non-friends, there will always be a student among the remaining four who is friends with both of the chosen students. How many students are there in the class altoget... | [
"Answer: 25.\nLet's denote the number of students in the class as $N$. To begin, we show that it is possible for $N$ to equal 25. The class is divided into five groups, with each group consisting of five students who are not friends with each other. Furthermore, any two different groups of two children are friends.... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 25 | |
02o5 | Let $O$ be the intersection point of the diagonals of the cyclic quadrilateral $ABCD$. The circumcircles of triangles $AOB$ and $COD$ meet lines $BC$ and $AD$ again at $M$, $N$, $P$ and $Q$. Prove that the quadrilateral $MNPQ$ is inscribed in a circle with center $O$. | [
"Consider angles oriented modulo $180^\\circ$. Since $\\angle PCO = \\angle BCA = \\angle BDA = \\angle ODN$ are inscribed in the circumcircle of the triangle $OCD$, $OP = ON$. Analogously, $OM = OQ$.\n\n\n\nNow, $\\angle QOP = \\angle CPO = \\angle CDO = \\angle CDB = \\angle CAB = \\angle... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0968 | Problem:
Fie $a$, $b$ şi $c$ lungimile laturilor unui triunghi arbitrar cu perimetrul egal cu $1$.
Arătaţi că are loc relaţia
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} > \frac{9}{\sqrt{1-a} + \sqrt{1-b} + \sqrt{1-c}}.
$$ | [
"Solution:\n\nDin ipoteză avem $a + b + c = 1$.\n\nÎntrucât $a$, $b$ şi $c$ sunt lungimile laturilor unui triunghi, are loc $a + b > c$, $c \\in (0 ; 1)$, atunci\n$$\n\\sqrt{a + b} > \\sqrt{c} > c \\Leftrightarrow \\sqrt{1 - c} > c \\Leftrightarrow \\frac{1}{\\sqrt{1 - c}} < \\frac{1}{c}.\n$$\nIdem $\\frac{1}{\\sqr... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0g6v | 令 $Z$ 表示所有整數所成的集合且 $n \ge 1$ 為一奇數。試求所有的函數 $f: Z \to Z$ 使得對任意的整數 $x, y$, $(f(x) - f(y))$ 能整除 $(x^n - y^n)$. | [
"答：$f(x) = \\epsilon x^d + c$, 其中 $\\epsilon \\in \\{-1, 1\\}$, 正整數 $d$ 為 $n$ 的因數且 $c$ 為一整數。\n滿足題設之所有函數為上述之 $f$.\n\n令函數 $f$ 為滿足題設之解。對任意整數 $n$, 定義函數 $g(x) = f(x) + n$ 也滿足題設。我們可假設 $f(0) = 0$.\n\n對任意質數 $p$, 取 $(x, y) = (p, 0)$ 則 $f(p)|p^n$. 因為質數有無窮多個, 所以存在整數 $d$ 與 $\\epsilon$ 且 $0 \\le d \\le n$, $\\epsilon \\in \\{-1... | Taiwan | 二〇一二數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | All functions of the form f(x) = ε x^d + c, where ε ∈ {−1, 1}, d is a positive divisor of n, and c is any integer. | |
03du | Find the number of all positive integers $4 \le n \le 2022$ that are not primes such that for any positive integer $k$ in the interval $[1, \sqrt{n}-1]$ the following holds: the number of ways to choose $k$ persons from a group of $n$ people is divisible by $n$. | [
"The condition of the problem requires $\\binom{n}{k}$ to be divisible by $n$ for all $1 \\le k \\le \\sqrt{n}-1$. Assume there exists a prime divisor $p$ of $n$ such that $p \\le \\sqrt{n}-1$ and set $k = p$. Since\n$$\np! \\binom{n}{p} = n(n-1)(n-2)\\cdots(n-p+1)\n$$\nand $\\binom{n}{p}$ is divisible by $n$, we h... | Bulgaria | Bulgaria 2022 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 22 | |
012i | Problem:
Let $n$ be a positive integer. Prove that the equation
$$
x + y + \frac{1}{x} + \frac{1}{y} = 3n
$$
does not have solutions in positive rational numbers. | [
"Solution:\nSuppose $x = \\frac{p}{q}$ and $y = \\frac{r}{s}$ satisfy the given equation, where $p, q, r, s$ are positive integers and $\\operatorname{gcd}(p, q) = 1$, $\\operatorname{gcd}(r, s) = 1$. We have\n$$\n\\frac{p}{q} + \\frac{r}{s} + \\frac{q}{p} + \\frac{s}{r} = 3n\n$$\nor\n$$\n\\left(p^2 + q^2\\right) r... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
00so | Let $\triangle ABC$ ($BC > AC$) be an acute triangle with circumcircle $k$ centered at $O$. The tangent to $k$ at $C$ intersects the line $AB$ at the point $D$. The circumcircles of triangles $BCD$, $OCD$ and $AOB$ intersect the ray $CA$ (beyond $A$) at the points $Q$, $P$ and $K$, respectively, such that $P \in (AK)$ ... | [
"As $DC$ is tangent to $k$ at $C$ then $\\angle OCD = 90^\\circ$. Denote by $X$ the midpoint of $AB$. Then $\\angle OXA = 90^\\circ$ because $OX$ is the perpendicular bisector of the side $AB$. The pentagon $PXOCD$ is inscribed in the circle with diameter $OD$, hence $\\angle PXA = \\angle PXD = \\angle PCD = \\ang... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0264 | Problem:
Mostre que $M=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$ é um número inteiro. | [
"Solution:\n\nSejam $a=\\sqrt[3]{\\sqrt{5}+2}$ e $b=\\sqrt[3]{\\sqrt{5}-2}$. Assim, $M=a-b$ e temos:\n$$\nM^{3}=(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)\n$$\nSabemos que $a^{3}-b^{3}=4$ e $a b=1$. Assim, $M^{3}+3 M-4=0$, ou seja, o número $M$ é raiz do polinômio $x^{3}+3 x-4$.\nPor sua vez, o número 1 é uma raiz do polinôm... | Brazil | Nível 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
05ug | Problem:
Soit $ABC$ un triangle isocèle en $A$, puis $D$ un point du segment $[BC]$ tel que $BD \neq CD$. Soit $P$ et $Q$ les projetés orthogonaux de $D$ sur $(AB)$ et $(AC)$. Enfin, soit $E$ le point d'intersection, autre que $A$, entre les cercles circonscrits à $ABC$ et $APQ$.
Démontrer que, si les droites $(EP)$,... | [
"Solution:\n\nUn premier réflexe est de dessiner une figure où le triangle $ABC$ est isocèle rectangle, mais sans indiquer qu'il l'est. De nombreuses symétries apparaissent alors. Notons $X$ le point d'intersection des droites $(AC)$ et $(EP)$, $\\ell$ la médiatrice de $[PQ]$, $O$ le centre du cercle circonscrit à ... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gc6 | 平面上, 有不等邊三角形 $ABC$ 與其外一點 $P$。在射線 $AB$, $AC$ 上分別取點 $B'$ 與 $C'$, 使得 $AB' = AC$, $AC' = AB$。令點 $Q$ 為 $P$ 對直線 $BC$ 的對稱點。設 $\triangle BB'P$ 與 $\triangle CC'P$ 的兩外接圓再交於點 $P'$, $\triangle BB'Q$ 與 $\triangle CC'Q$ 的兩外接圓再交於點 $Q'$。令點 $O$, $O'$ 分別為 $\triangle ABC$ 與 $\triangle AB'C'$ 的外心。證明
(1) $O'$, $P'$, $Q'$ 三點共線。
(2) $O'P' \c... | [
"1. 顯然 $B$, $C$, $B'$, $C'$ 共圓, 因為它們形成等腰梯形。對這圓及 $\\triangle BB'P$, $\\triangle CC'P$ 的外接圓使用三根軸定理, 得 $A$, $P$, $P'$ 共線。同理 $A$, $Q$, $Q'$ 共線。因 $AP \\cdot AP' = AB \\cdot AB' = AQ \\cdot AQ'$, 所以 $P$, $Q$, $P'$, $Q'$ 共圓。\n\n2. 令 $O_1$ 為 $\\triangle BB'O'$, $\\triangle CC'O'$ 的外接圓的另一交點。如 1., $A$, $O'$, $O_1$ 共線, 且 $Q$,... | Taiwan | 二〇一八數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Reflection",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroi... | null | proof only | null | |
0hm8 | Problem:
Determine whether there exists a number that begins with $2$ having the property that, when the $2$ is moved to the end, the number is
(a) doubled;
(b) tripled. | [
"Solution.\nThere is no such number. Suppose that it existed; represent it by $2 \\cdots ab$ where $a$ and $b$ are the last two digits of the number. Let us write the digit-by-digit multiplication:\n\n\n\nFrom the last column it is clear that there are only two choices for $b$: $1$ and $6$.... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) No such number exists. b) 285714. | |
0hcr | Some natural numbers are placed around a circle in such a way that the product of any two neighboring numbers is a perfect square. Prove that the product of any (not necessarily neighboring) two numbers is also a perfect square.
(Arseniy Nikolaev) | [
"Let us denote numbers as $a_1, a_2, \\ldots, a_k$. From the problem statement it follows that for any $1 \\le i < j \\le k$ the product $(a_i a_{i+1})(a_{i+1} a_{i+2}) \\ldots (a_{j-1} a_j) = n^2$ is a perfect square. Hence the product $a_i a_j = \\frac{n^2}{a_{i+1}^2 a_{i+2}^2 \\ldots a_{j-1}^2}$ is also a perfec... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null | |
0i2j | Let $n$ be an odd integer greater than $1$ and let $c_1, c_2, \dots, c_n$ be integers. For each permutation $a = (a_1, a_2, \dots, a_n)$ of $\{1, 2, \dots, n\}$, define $S(a) = \sum_{i=1}^n c_i a_i$. Prove that there exist permutations $b$ and $c$, $b \neq c$, such that $n!$ divides $S(b) - S(c)$. | [
"**First Solution.** Let $\\sum_a$ denote the sum over all $n!$ permutations $a = (a_1, a_2, \\dots, a_n)$. We compute $\\sum_a S(a) \\mod n!$ in two ways, one of which assuming that the desired conclusion is false, and reach a contradiction.\n\nSuppose, for the sake of contradiction, that the claim is false. Then ... | United States | USA IMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Generating functions",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0k9j | Problem:
Find the value of
$$
\sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \frac{a b(3 a+c)}{4^{a+b+c}(a+b)(b+c)(c+a)}
$$ | [
"Solution:\nLet $S$ denote the given sum. By summing over all six permutations of the variables $a, b, c$ we obtain\n$$\n\\begin{aligned}\n6 S & =\\sum_{a=1}^{\\infty} \\sum_{b=1}^{\\infty} \\sum_{c=1}^{\\infty} \\frac{3\\left(a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b\\right)+6 a b c}{4^{a+b+c}(a+b)(b+c)(c+a)... | United States | HMMT February 2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 1/54 | |
0eue | Let a quadrilateral $ABCD$ be inscribed in a circle $O$ with $\angle B$ and $\angle C$ obtuse. Let $E$ be the point of intersection of the lines $AB$ and $CD$. Let $P$ and $R$ be the feet of the perpendicular lines from $E$ to the lines $BC$ and $AD$, respectively. Let $Q$ be the point of intersection of the lines $EP$... | [
"Let $O_1$, $O_2$ be the circumcenters of the triangles $AED$ and $EBC$, respectively.\n\n**Lemma** Let $H$ be the foot of the altitude of the triangle $ABC$. The line symmetric to the line $AH$ with respect to the bisector of the angle $A$ passes the circumcenter $O$ of the triangle $ABC$.\n\n**Proof** Since the t... | South Korea | Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0hg0 | Let $P$ be a polynomial with integer coefficients of degree $d$. For the set $A = \{a_1, a_2, \dots, a_k\}$ of positive integers denote $S(A) = P(a_1) + P(a_2) + \dots + P(a_k)$. The positive integers $m, n$ are such that $m^{d+1} \mid n$. Prove that the set $\{1, 2, \dots, n\}$ can be split into $m$ disjoint subsets $... | [
"Let $n = k m^{d+1}$. We will construct the desired splitting in the following way. To determine which of the subsets each number $x \\in \\{1; 2; \\dots n\\}$ belongs to, let us write $x - 1$ in the form:\n$$\nx - 1 = c m^{d+1} + x_0 + x_1 m + x_2 m^2 + \\dots + x_d m^d,\n$$\nwhere $c \\ge 0$ is an integer, $x_i \... | Ukraine | Problems from Ukrainian Authors | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other"
] | English | proof only | null | |
00ep | A sequence of integers is defined as follows: $a_1 = 1$, $a_2 = 2$, and for each $n \ge 2$, $a_{n+1}$ is equal to the greatest prime divisor of $a_1 + a_2 + \dots + a_n$. Compute $a_{100}$. | [
"Let $p_1 < p_2 < p_3 < \\dots$ be the sequence of all prime numbers, and denote $s_n = a_1 + a_2 + \\dots + a_n$. Notice that $a_3 = 3$, and consequently $s_3 = 1 + 2 + 3 = 6 = 2 \\cdot 3 = p_1p_2$.\nSuppose that for some $n$ we have $s_n = p_k p_{k+1}$. Then, the greatest prime divisor of $s_n$ is $p_{k+1}$, so $... | Argentina | Cono Sur Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 53 | |
0kfn | Problem:
Circles $\omega_{a}, \omega_{b}, \omega_{c}$ have centers $A, B, C$, respectively and are pairwise externally tangent at points $D, E, F$ (with $D \in BC, E \in CA, F \in AB$). Lines $BE$ and $CF$ meet at $T$. Given that $\omega_{a}$ has radius $341$, there exists a line $\ell$ tangent to all three circles, a... | [
"Solution:\n\nWe will use the following notation: let $\\omega$ be the circle of radius $49$ tangent to each of $\\omega_{a}, \\omega_{b}, \\omega_{c}$. Let $\\omega_{a}, \\omega_{b}, \\omega_{c}$ have radii $r_{a}, r_{b}, r_{c}$ respectively. Let $\\gamma$ be the incircle of $ABC$, with center $I$ and radius $r$. ... | United States | HMMT February | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry >... | null | proof and answer | 294 | |
0kn5 | When a certain unfair die is rolled, an even number is 3 times as likely to appear as an odd number. The die is rolled twice. What is the probability that the sum of the numbers rolled is even?
(A) $\frac{3}{8}$ (B) $\frac{4}{9}$ (C) $\frac{5}{9}$ (D) $\frac{9}{16}$ (E) $\frac{5}{8}$ | [] | United States | AMC 10 A | [
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | E | |
0eg4 | Problem:
Kot med nosilkama diagonal štirikotnika $ABCD$ je velik $60^{\circ}$. Vsako oglišče štirikotnika prezrcalimo čez nosilko diagonale, ki poteka skozi njemu sosednji oglišči. Dokaži, da zrcalne slike oglišč ležijo na isti premici. | [
"\n\nZrcalne slike oglišč $A, B, C$ in $D$ pri zrcaljenju čez nosilke ustreznih diagonal označimo po vrsti z $A'$, $B'$, $C'$ in $D'$. Označimo presečišče nosilk diagonal s $P$.\n\nDenimo najprej, da $P$ leži v notranjosti štirikotnika. Zaradi simetrije lahko predpostavimo, da je $\\angle B... | Slovenia | Slovenian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Transformations",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cti | Positive real numbers $x$, $y$, and $z$ satisfy the inequality $xyz \ge xy + yz + zx$. Prove that
$$
\sqrt{xyz} \ge \sqrt{x} + \sqrt{y} + \sqrt{z}.
$$
Положительные числа $x$, $y$ и $z$ удовлетворяют условию $xyz \ge xy + yz + zx$. Докажите неравенство $\sqrt{xyz} \ge \sqrt{x} + \sqrt{y} + \sqrt{z}$. | [
"Rewrite the given and the required inequalities as $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} \\le 1$ and $\\frac{1}{\\sqrt{xy}} + \\frac{1}{\\sqrt{xz}} + \\frac{1}{\\sqrt{yz}} \\le 1$.\n\nBy the inequality of means, we have\n$$\nxy + xz \\geq 2\\sqrt{xy \\cdot xz}, \\quad xy + yz \\geq 2\\sqrt{xy \\cdot yz}, \\q... | Russia | Russian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English; Russian | proof only | null | |
0j4g | Problem:
Let $G_{1} G_{2} G_{3}$ be a triangle with $G_{1} G_{2}=7$, $G_{2} G_{3}=13$, and $G_{3} G_{1}=15$. Let $G_{4}$ be a point outside triangle $G_{1} G_{2} G_{3}$ so that ray $\overrightarrow{G_{1} G_{4}}$ cuts through the interior of the triangle, $G_{3} G_{4}=G_{4} G_{2}$, and $\angle G_{3} G_{1} G_{4}=30^{\ci... | [
"Solution:\n\nAnswer: $\\frac{169}{23}$\n\n\n\nWe first show that quadrilateral $G_{1} G_{2} G_{4} G_{3}$ is cyclic. Note that by the law of cosines,\n$$\n\\cos \\angle G_{2} G_{1} G_{3}=\\frac{7^{2}+15^{2}-13^{2}}{2 \\cdot 7 \\cdot 15}=\\frac{1}{2}\n$$\nso $\\angle G_{2} G_{1} G_{3}=60^{\\... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 169/23 | |
0db7 | Real nonzero numbers $x$, $y$, $z$ are such that $x + y + z = 0$. Moreover, it is known that
$$
A = \frac{x}{y} + \frac{y}{z} + \frac{z}{x} = \frac{x}{z} + \frac{z}{y} + \frac{y}{x} + 1.
$$
Determine $A$. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | -1 | |
03ji | Problem:
For every positive integer $n$ show that
$$
[\sqrt{n} + \sqrt{n+1}] = [\sqrt{4n+1}] = [\sqrt{4n+2}] = [\sqrt{4n+3}]
$$
where $[x]$ is the greatest integer less than or equal to $x$ (for example $[2.3]=2$, $[\pi]=3$, $[5]=5$). | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0l3f | Problem:
In triangle $ABC$, a circle $\omega$ with center $O$ passes through $B$ and $C$ and intersects segments $\overline{AB}$ and $\overline{AC}$ again at $B'$ and $C'$, respectively. Suppose that the circles with diameters $BB'$ and $CC'$ are externally tangent to each other at $T$. If $AB = 18$, $AC = 36$, and $A... | [
"Solution:\n\n\n\nBy Radical Axis Theorem, we know that $AT$ is tangent to both circles. Moreover, consider power of a point $A$ with respect to these three circles, we have $AB \\cdot AB' = AT^2 = AC \\cdot AC'$. Thus $AB' = \\frac{12^2}{18} = 8$, and $AC' = \\frac{12^2}{36} = 4$.\n\nConsi... | United States | HMMT February 2024 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 65/3 | |
0kmi | Problem:
Let
- $P$ be a point inside a triangle $\triangle ABC$,
- $\triangle DEF$ be the pedal triangle of $P$, i.e., let $D, E, F$ be the feet of the altitudes from $P$ to $BC, CA, AB$, respectively,
- $I$ be the incenter of $\triangle ABC$, and
- $\triangle XYZ$ be the Cevian triangle of $I$, i.e., $X, Y, Z$ be the... | [
"Solution:\n\nWhile there is a synthetic solution, we present a solution using (unnormalized) barycentric coordinates.\nLet $P = (x, y, z)$ in barycentric coordinates. We know that $PD = \\frac{2[\\triangle BCD]}{a} = 2[\\triangle ABC] \\frac{x}{a}$, so by SSS similarity if $PD, PE, PF$ forms a triangle, so do $x/a... | United States | Berkeley Math Circle: Monthly Contest 3 | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle ... | null | proof only | null | |
0htb | Problem:
Prove that any prime which is the difference of two cubes is also the sum of a square and three times a square. | [
"Solution:\nSuppose that\n$$\np = a^{3} - b^{3}\n$$\nwhere $p$ is prime and $a$ and $b$ are natural numbers. The right side can be factored:\n$$\np = (a - b)\\left(a^{2} + a b + b^{2}\\right)\n$$\nSince $p$ is prime, one of the factors must be $\\pm 1$. We note that $a^{2} + a b + b^{2} \\geq 1 + 1 + 1 = 3$, so $a ... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | null | proof only | null | |
0in5 | Determine whether or not there exist positive integers $a$ and $b$ such that $a$ does not divide $b^n - n$ for all positive integers $n$. | [
"**Solution 1.** For all pairs of positive integers $a$ and $b$, we claim that there exist infinitely many $n$ such that $a$ divides $b^n - n$.\nWe establish our claim by strong induction on $a$. The base case of $a = 1$ holds trivially. Now, suppose that the claim holds for all $a < a_0$. Since $\\varphi(a) < a$, ... | United States | Team Selection Test | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Residues and Primitive Roots > Multi... | null | proof and answer | No. For any positive integers a and b, there exist infinitely many positive integers n such that a divides b^n − n. | |
0awx | Problem:
The symbol $\div$ is well-known nowadays to indicate division between two numbers. But during the late medieval period, it had a completely different purpose - it was used to mark words or passages which were spurious, corrupt, or doubtful. The actual name for this "division sign" comes from the ancient Greek... | [
"Solution:\n\nObelus"
] | Philippines | Philippine Mathematical Olympiad | [
"Math Word Problems"
] | null | final answer only | Obelus | |
0fxy | Problem:
In einem spitzwinkligen Dreieck $ABC$ seien $BE$ und $CF$ Höhen. Zwei Kreise durch $A$ und $F$ berühren die Gerade $BC$ bei $P$ und $Q$, so dass $B$ zwischen $C$ und $Q$ liegt. Beweise, dass sich die Geraden $PE$ und $QF$ auf dem Umkreis von $AEF$ schneiden. | [
"Solution:\n\nSeien wie immer $\\angle BAC=\\alpha$ und $\\angle ABC=\\beta$. Der Schnittpunkt von $QF$ mit $PE$ sei $S$. Es genügt die Gleichung $\\angle FQB+\\angle EPB=180^{\\circ}-\\alpha$ zu zeigen, denn daraus folgt $\\angle FSE=\\alpha$ und $AFES$ ist ein Sehnenviereck. Wir kümmern uns zuerst um den Winkel $... | Switzerland | IMO Selektion | [
"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 > Circles > Radical axis ... | null | proof only | null | |
0d1c | Let $ABCDE$ be a pentagon with $\widehat{A} = \widehat{B} = \widehat{C} = \widehat{D} = 120^\circ$. Prove that
$$
4AC \cdot BD \ge 3AE \cdot ED.
$$ | [
"\nIt is clear that $AB \\parallel ED$ and $AE \\parallel CD$. Let $F$ be the intersection point of lines $AB$ and $CD$. The inequality is equivalent to\n$$\n\\frac{AC}{AF} \\cdot \\frac{BD}{FD} \\ge \\frac{3}{4}. \\quad (1)\n$$\nUsing the Law of Sines in triangles $ACF$ and $BDF$, we get t... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof only | null | |
0i60 | Problem:
The real function $f$ has the property that, whenever $a$, $b$, $n$ are positive integers such that $a + b = 2^{n}$, the equation $f(a) + f(b) = n^{2}$ holds. What is $f(2002)$? | [
"Solution:\n\nWe know $f(a) = n^{2} - f\\left(2^{n} - a\\right)$ for any $a$, $n$ with $2^{n} > a$; repeated application gives\n$$\n\\begin{gathered}\nf(2002) = 11^{2} - f(46) = 11^{2} - \\left(6^{2} - f(18)\\right) = 11^{2} - \\left(6^{2} - \\left(5^{2} - f(14)\\right)\\right) \\\\\n= 11^{2} - \\left(6^{2} - \\lef... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 96 | |
0149 | Problem:
A rectangle is divided into $200 \times 3$ unit squares. Prove that the number of ways of splitting this rectangle into rectangles of size $1 \times 2$ is divisible by $3$. | [
"Solution:\n\nLet us denote the number of ways to split some figure into dominos by a small picture of this figure with a sign \\#. For example, $\\# \\boxplus = 2$.\nLet $N_{n} = \\#$ ( $n$ rows) and $\\gamma_{n} = \\#$ ( $n-2$ full rows and one row with two cells).\nWe are going to find a recurrence relation for ... | Baltic Way | Baltic Way 2005 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
03oj | As shown in the diagram, there is a sequence of the curves $P_0, P_1, P_2, \dots$. It is known that the region enclosed by $P_0$ has area $1$ and $P_0$ is an equilateral triangle. We obtain $P_{k+1}$ from $P_k$ by operating as follows: Trisecting every side of $P_k$, then we construct an equilateral triangle outwardly ... | [
"(1) We perform the operation on $P_0$. It is easy to see that each side of $P_0$ becomes $4$ sides of $P_1$. So the number of sides of $P_1$ is $3 \\cdot 4$. In the same way, we operate on $P_1$. Each side of $P_1$ becomes $4$ sides of $P_2$. So the number of sides of $P_2$ is $3 \\cdot 4^2$. Consequently, it is n... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | S_n = 8/5 − (3/5)(4/9)^n; lim_{n→∞} S_n = 8/5 | |
0gch | 令 $N$ 與 $Z$ 分別表示所有正整數與整數之集合。試求所有函數 $f: N \to Z$ 滿足:
$n$ 整除 $f(m)$ 的充要條件為 $m$ 能整除 $\sum_{d|n} f(d)$, 對於所有的正整數 $n$ 和 $m$ 皆成立。 | [
"答：$f(n) = 0, \\forall n \\in N$\n\n若 $f(1) \\neq 0$, 則取 $m = 1, n = |f(1)| + 1$, 因為 $1|\\sum_{d|n} f(d)$, 所以 $n|f(1)$ 但 $0 < |f(1)| < n$, 矛盾。故 $f(1) = 0$.\n\n底下證明對所有正整數 $k$ 有 $f(2^k) = 0$.\n利用反證法, 假設存在正整數 $k$ 使得 $f(2^k) \\neq 0$, 假設 $t$ 是最小的那一個, 那麼對所有 $m > |f(2^t)|$ 都有\n$m|\\sum_{d|2^{t-1}} f(d)$ 但 $m$ 無法整除 $\\sum... | Taiwan | 二〇一八數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | f(n) = 0 for all positive integers n | |
0fq1 | Problem:
Un trazador de puntos medios es un instrumento que dibuja el punto medio exacto de dos puntos previamente señalados. Partiendo de dos puntos a distancia $1$ y utilizando sólo el trazador de puntos medios, debes obtener dos puntos a una distancia estrictamente comprendida entre $\frac{1}{2017}$ y $\frac{1}{201... | [
"Solution:\n\nSin pérdida de generalidad podemos trabajar con la recta real y considerar que uno de los puntos iniciales es el $0$ y el otro el $1$. Es fácil comprobar que después de $k$ aplicaciones del trazador, todos los puntos hallados son de la forma $\\frac{n}{2^{k}}$ con $0 \\leq n \\leq 2^{k}$ siendo la fra... | Spain | LIII Olimpiada matemática Española | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra... | null | proof and answer | 17 | |
0iec | Problem:
You start out with a big pile of $3^{2004}$ cards, with the numbers $1,2,3, \ldots, 3^{2004}$ written on them. You arrange the cards into groups of three any way you like; from each group, you keep the card with the largest number and discard the other two. You now again arrange these $3^{2003}$ remaining card... | [
"Solution:\n$3^{2004}-2 \\cdot 3^{1002}+2$\nWe claim that if you have cards numbered $1,2, \\ldots, 3^{2 n}$ and perform $2 n$ successive grouping operations, then $c$ is a possible value for your last remaining card if and only if\n$$\n3^{n} \\leq c \\leq 3^{2 n}-3^{n}+1\n$$\nThis gives $3^{2 n}-2 \\cdot 3^{n}+2$ ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3^{2004} - 2*3^{1002} + 2 | |
0785 | Let $S(k)$ denote the sum of digits (in base 10) of any positive integer $k$. Does there exist a positive integer $N$ so that $S(2^n) \le S(2^{n+1})$ for all integers $n > N$? | [] | India | IMOTC Practice Test 1 | [
"Number Theory > Other"
] | null | proof and answer | No | |
03su | Suppose $A = \{x \mid 5x - a \le 0\}$, $B = \{x \mid 6x - b > 0\}$, $a, b \in \mathbb{N}$, and $A \cap B \cap \mathbb{N} = \{2, 3, 4\}$. The number of such pairs $(a, b)$ is ( ). | [
"Since $5x - a \\le 0 \\Rightarrow x \\le \\frac{a}{5}$, $6x - b > 0 \\Rightarrow x > \\frac{b}{6}$. In order to satisfy $A \\cap B \\cap \\mathbb{N} = \\{2, 3, 4\\}$, we have\n$$\n\\begin{cases}\n1 \\le \\frac{b}{6} < 2, \\\\\n4 \\le \\frac{a}{5} < 5,\n\\end{cases}\n$$\nor\n$$\n\\begin{cases}\n6 \\le b < 12, \\\\\... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | C | |
0ck4 | Let $n \ge 2$ be an integer and $a$ and $b$ complex numbers such that $a \neq 0$ and $b^k \neq 1$, for any $k \in \{1, 2, \dots, n\}$. Suppose that the matrices $A, B \in \mathcal{M}_n(\mathbb{C})$ are such that $BA = aI_n + bAB$. Prove that $A$ and $B$ are invertible.
Sorin Rădulescu and Mihai Piticari | [
"If $b = 0$, then $BA = aI_n$, so $A$ and $B$ are invertible. We will suppose $b \\neq 0$.\n\nDenote by $\\sigma(X)$ the set of eigenvalues of a matrix $X \\in \\mathcal{M}_n(\\mathbb{C})$. Let $\\lambda \\in \\sigma(AB)$. Then\n$$\n\\det(BA - (b\\lambda + a)I_n) = \\det(bAB - b\\lambda I_n) = b^n \\det(AB - \\lamb... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
0ib6 | Problem:
Let $f(x) = x^{2} + x^{4} + x^{6} + x^{8} + \cdots$, for all real $x$ such that the sum converges. For how many real numbers $x$ does $f(x) = x$? | [
"Solution:\nClearly $x = 0$ works. Otherwise, we want $x = \\dfrac{x^{2}}{1 - x^{2}}$, or $x^{2} + x - 1 = 0$. Discard the negative root (since the sum doesn't converge there), but $\\dfrac{-1 + \\sqrt{5}}{2}$ works, for a total of 2 values."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 2 | |
08ai | Problem:
Per ogni intero positivo $n$, sia $D_{n}$ il massimo comune divisore di tutti i numeri della forma $a^{n} + (a+1)^{n} + (a+2)^{n}$ al variare di $a$ fra tutti gli interi positivi.
a. Dimostrare che, per ogni $n$, $D_{n}$ è della forma $3^{k}$ per qualche intero $k \geq 0$.
b. Dimostrare che, per ogni $k \ge... | [] | Italy | Olimpiade Italiana di Matematica | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
0jev | Problem:
Let $S$ be a subset of $\{1,2,3, \ldots, 12\}$ such that it is impossible to partition $S$ into $k$ disjoint subsets, each of whose elements sum to the same value, for any integer $k \geq 2$. Find the maximum possible sum of the elements of $S$. | [
"Solution:\n77\n\nWe note that the maximum possible sum is 78 (the entire set). However, this could be partitioned into 2 subsets with sum 39: $\\{1,2,3,10,11,12\\}$ and $\\{4,5,6,7,8,9\\}$. The next largest possible sum is 77 (the entire set except 1). If $k \\geq 2$ subsets each had equal sum, then they would hav... | United States | HMMT November 2013 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 77 | |
09wk | One hundred brownies (girl scouts) are sitting in a big circle around the camp fire. Each brownie has one or more chestnuts and no two brownies have the same number of chestnuts. Each brownie divides her number of chestnuts by the number of chestnuts of her right neighbour and writes down the remainder on a green piece... | [
"$100$"
] | Netherlands | Second Round | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 100 | |
0f5t | Problem:
Each weight in a set of $n$ has integral weight and the total weight of the set is $2n$. A balance is initially empty. We then place the weights onto a pan of the balance one at a time. Each time we place the heaviest weight not yet placed. If the pans balance, then we place the weight onto the left pan. Othe... | [] | Soviet Union | 18th ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0jwu | Problem:
The game of Penta is played with teams of five players each, and there are five roles the players can play. Each of the five players chooses two of five roles they wish to play. If each player chooses their roles randomly, what is the probability that each role will have exactly two players? | [
"Solution:\n\nConsider a graph with five vertices corresponding to the roles, and draw an edge between two vertices if a player picks both roles. Thus there are exactly $5$ edges in the graph, and we want to find the probability that each vertex has degree $2$. In particular, we want to find the probability that th... | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 51/2500 | |
0i6w | Problem:
Denote by $\langle x\rangle$ the fractional part of the real number $x$ (for instance, $\langle 3.2\rangle=0.2$ ). A positive integer $N$ is selected randomly from the set $\{1,2,3, \ldots, M\}$, with each integer having the same probability of being picked, and $\left\langle\frac{87}{303} N\right\rangle$ is c... | [
"Solution:\nThis method of picking $N$ is equivalent to uniformly randomly selecting a positive integer. Call this the average value of $\\left\\langle\\frac{87}{303} N\\right\\rangle$ for $N$ a positive integer. In lowest terms, $\\frac{87}{303}=\\frac{29}{101}$, so the answer is the same as the average value of $... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 50/101 | |
0gpl | In a convex quadrilateral $ABCD$, the diagonals intersect at the point $E$ and $\angle AEB = 90^\circ$. A point $P$ is chosen on the side $[AD]$ other than $A$ so that $PE = EC$. The circumcircle of the triangle $BCD$ intersects the side $[AD]$ at the point $Q$ other than $A$. The circle passing through $A$ and tangent... | [
"Let $\\angle EAP = \\alpha$ and $\\angle ECD = \\beta$. Note that $\\angle RPE = \\alpha$. Choose a point $X$ on $[ED]$ such that $\\angle RPX = 90^\\circ$. Observe that $R, P, X, E$ are cyclic and hence $\\angle RXE = \\alpha$ which implies that $A, R, X, D$ are concyclic. Therefore $ER \\cdot EA = EX \\cdot ED$.... | Turkey | 17th Junior Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
... | English | proof only | null | |
03y4 | Let $f(x)$ and $g(x)$ be strictly increasing linear functions from $\mathbf{R}$ to $\mathbf{R}$ such that $f(x)$ is an integer if and only if $g(x)$ is an integer. Prove that for any real number $x$, $f(x) - g(x)$ is an integer. | [
"By symmetry, we may assume that $a \\ge c$.\nWe claim that $a = c$. Assume on the contrary that $a > c$. Because $a > c > 0$, the ranges of $f$ and $g$ are both $\\mathbf{R}$. There is a $x_0$ such that $f(x_0) = a x_0 + b$ is an integer. Hence $g(x_0) = c x_0 + d$ is also an integer. But then,\n$$\nf\\left(x_0 + ... | China | China Girls' Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0j2c | Problem:
A polynomial $P$ is of the form $\pm x^{6} \pm x^{5} \pm x^{4} \pm x^{3} \pm x^{2} \pm x \pm 1$. Given that $P(2)=27$, what is $P(3)$? | [
"Solution:\nAnswer: 439\nWe use the following lemma:\n\nLemma. The sign of $\\pm 2^{n} \\pm 2^{n-1} \\pm \\cdots \\pm 2 \\pm 1$ is the same as the sign of the $2^{n}$ term.\n\nProof. Without loss of generality, let $2^{n}$ be positive. (We can flip all signs.) Notice that $2^{n} \\pm 2^{n-1} \\pm 2^{n-2} \\pm \\cdo... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 439 | |
0dg5 | In the cells of the grid $10 \times 10$ are written positive integers, all of them less than $11$. It is known that the sum of $2$ numbers written in the cells having common vertex is a prime number. Prove that there are $17$ cells containing the same number. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0evs | Let $O$ be the circumcircle of a triangle $ABC$. Choose a point $D (\neq A)$ on the extension of segment $BA$ towards $A$. Points $E$ and $F$ are two distinct points on circle $O$ such that lines $DE$ and $DF$ are both tangent to circle $O$. Segment $EF$ intersects side $CA$ at point $T (\neq C)$. Choose a point $P (\n... | [
"Let $Z$ be the intersection point of line $CD$ and $O$. For cyclic quadrilateral $BAZC$, $D$ is the intersection point of two sides $BA$ and $CZ$, and $EF$ is the polar line of $D$ with respect to the circle $O$. Therefore, the intersection point of $AC$ and $BZ$ should lie on $EF$, which means the points $B, T, Z... | South Korea | The 37th Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | English | proof only | null | |
0kxs | Problem:
Find, with proof, all nonconstant polynomials $P(x)$ with real coefficients such that, for all nonzero real numbers $z$ with $P(z) \neq 0$ and $P\left(\frac{1}{z}\right) \neq 0$, we have
$$
\frac{1}{P(z)}+\frac{1}{P\left(\frac{1}{z}\right)}=z+\frac{1}{z}
$$ | [
"Solution:\n\nIt is straightforward to plug in and verify the above answers. Hence, we focus on showing that these are all possible solutions. The key claim is the following.\n\nClaim: If $r \\neq 0$ is a root of $P(z)$ with multiplicity $n$, then $1 / r$ is also a root of $P(z)$ with multiplicity $n$.\n\nProof 1 (... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | All nonconstant real polynomials are exactly those of the form P(x) = x(1 + ε x^k)/(1 + x^2), where ε = 1 with k ≡ 2 mod 4 (k = 2, 6, 10, …), or ε = −1 with k ≡ 0 mod 4 (k = 4, 8, 12, …). | |
05x2 | Problem:
Est-il possible de trouver un bloc de 1000 nombres entiers strictement positifs consécutifs qui contient exactement 5 nombres premiers? | [
"Solution:\n\nA première vue, il semble difficile de garantir exactement 5 nombre premier dans un bloc de 1000 entiers consécutifs : aucun théorème d'arithmétique élémentaire permet de s'assurer d'avoir des nombres premiers rapprochés, mais pas d'autres nombres premier entre eux. On peut donc essayer de se demander... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | null | proof only | null | |
0g03 | Problem:
Sei $ABC$ ein spitzwinkliges Dreieck und $W$ der Schnittpunkt der Winkelhalbierenden von $\angle ACB$ und der Seite $AB$. Weiter seien $I_{A}$ und $I_{B}$ die Inkreismittelpunkte der Dreiecke $AWC$ respektive $WBC$. Die Geraden $I_{A}W$ und $I_{B}B$ schneiden sich im Punkt $D$. Sei $M$ der Mittelpunkt der Str... | [
"Solution:\n\nDie Geraden $WI_{A}$ und $WI_{B}$ halbieren die Winkel $\\angle CWA$ und $\\angle BWC$, folglich gilt:\n$$\n\\angle I_{B}WD = \\angle I_{B}WC + \\angle CWI_{A} = \\frac{1}{2}(\\angle BWC + \\angle CWA) = 90^{\\circ}\n$$\nSomit liegt $W$ auf dem Thaleskreis über $DI_{B}$, dessen Mittelpunkt $M$ ist.\nM... | Switzerland | SMO - Vorrunde | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07qi | Prove that the sequence
$$
x_n = \sqrt[n]{9^{n-1}(n+9)}, \quad n = 2, 3, 4, \dots
$$
is strictly decreasing, and bounded below by $9$. | [
"To prove\n$$\nx_{n+1} = \\sqrt[n+1]{9^n(n+10)} < \\sqrt[n]{9^{n-1}(n+9)} = x_n \\quad \\text{for all } n \\ge 2\n$$\nis equivalent to showing\n$$\n(9^n(n+10))^n < (9^{n-1}(n+9))^{n+1} \\quad \\text{for } n=2,3,\\dots\n$$\nSimplifying this, the claim is that\n$$\n9(n + 10)^n < (n + 9)^{n+1}, \\quad n = 2, 3, \\dots... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof only | null | |
07ma | A row of $n$ lamps is labelled from left to right with the numbers $1$ to $n$, where $n$ is an *odd* positive integer. Each lamp has a switch which, if pressed, turns it from OFF to ON or from ON to OFF; however the switches may only be pressed according to the following rules:
(a) Switch 1 may be pressed at any time;
... | [
"For any $n \\ge 1$, let $L_n$ denote the minimum number of times the switches must be pressed to turn all the lamps OFF. It is easy to check that $L_1 = 1$ and $L_2 = 2$.\n\nConsider the problem of turning lamps $1, 2, \\dots, k$ from all ON to all OFF, for some $k \\ge 3$. First, lamp $k$ must be turned OFF at so... | Ireland | Irish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | (2^{n+1} - 1)/3 | |
00ko | Let $a, b, c$ be integers with $a^3 + b^3 + c^3$ divisible by $18$. Prove that $abc$ is divisible by $6$. | [
"We need to prove that $abc$ is divisible by $2$ and by $3$. We will give proofs by contradiction.\n\nSuppose $abc$ odd. This implies that $a$, $b$ and $c$ are odd. Therefore, $a^3 + b^3 + c^3$ is odd and certainly not divisible by $18$. This contradiction shows that $abc$ is even.\n\nSuppose that $abc$ is not divi... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0a0m | Find all functions $f: \mathbb{R} \to \mathbb{R}$ for which
$$
f(a - b)f(c - d) + f(a - d)f(b - c) \le (a - c)f(b - d)
$$
for all real numbers $a$, $b$, $c$ and $d$.
*Note that there is only one occurrence of $f$ on the right hand side!* | [
"The solutions to the given functional inequality are $f(x) = 0$ for all $x$ and $f(x) = x$ for all $x$. For $f(x) = 0$, we easily find that equality always holds. For $f(x) = x$ we check that\n$$\n\\begin{aligned}\n(a-b)(c-d) + (a-d)(b-c) &= ac - ad - bc + bd + ab - ac - bd + cd \\\\\n&= -ad - bc + ab + cd \\\\\n&... | Netherlands | BxMO Team Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0 for all real x; f(x) = x for all real x | |
012c | Problem:
Let $ABC$ be an acute triangle with $\angle BAC > \angle BCA$, and let $D$ be a point on side $AC$ such that $|AB| = |BD|$. Furthermore, let $F$ be a point on the circumcircle of triangle $ABC$ such that line $FD$ is perpendicular to side $BC$ and points $F, B$ lie on different sides of line $AC$. Prove that ... | [
"Solution:\n\nLet $E$ be the other point on the circumcircle of triangle $ABC$ such that $|AB| = |EB|$. Let $D'$ be the point of intersection of side $AC$ and the line perpendicular to side $BC$, passing through $E$. Then $\\angle ECB = \\angle BCA$ and the triangle $ECD'$ is isosceles. As $ED' \\perp BC$, the tria... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
0eww | Problem:
A circle center $O$ is inscribed in $ABCD$ (touching every side). Prove that $\angle AOB + \angle COD$ equals $180$ degrees. | [
"Solution:\nLet $AB$ touch the circle at $W$, $BC$ at $X$, $CD$ at $Y$, and $DA$ at $Z$. Then $AO$ bisects angle $ZOW$ and $BO$ bisects angle $XOW$. So $\\angle AOB$ is half angle $ZOX$. Similarly $\\angle COD$ is half angle $XOZ$ and hence $\\angle AOB + \\angle COD$ equals $180$."
] | Soviet Union | 4th ASU | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a5p | Problem:
On a table, there is an empty bag and a chessboard containing exactly one token on each square. Next to the table is a large pile that contains an unlimited supply of tokens. Using only the following types of moves what is the maximum possible number of tokens that can be in the bag?
- Type 1: Choose a non-emp... | [
"Solution:\nLet $a_{i,j}$ be the number of tokens in the square in the $i^{\\mathrm{th}}$ row and $j^{\\mathrm{th}}$ column, where the first row is the topmost row and the first column is the leftmost column. Furthermore let $b$ denote the number of tokens in the bag. We define a monovariant as follows.\n$$m = \\fr... | New Zealand | New Zealand Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 43350 | |
03nq | Problem:
Nina and Tadashi play the following game. Initially, a triple $(a, b, c)$ of nonnegative integers with $a+b+c=2021$ is written on a blackboard. Nina and Tadashi then take moves in turn, with Nina first. A player making a move chooses a positive integer $k$ and one of the three entries on the board; then the p... | [
"Solution:\n\nThe answer is $3^{\\text{number of 1's in binary expansion of } 2021}=3^{8}=6561$.\n\nThroughout this solution, we say two nonnegative integers overlap in the $2^{\\ell}$ position if their binary representations both have a 1 in that position. We say that two nonnegative integers overlap if they overl... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 6561 | |
02ad | Problem:
Dizemos que um número inteiro positivo de três dígitos é três estrelas se ele for o resultado do produto de três números primos distintos. Por exemplo, $286=2 \cdot 11 \cdot 13$ é um número três estrelas, mas $30=2 \cdot 3 \cdot 5$ e $275=5 \cdot 5 \cdot 13$ não são números três estrelas, pois o primeiro só p... | [
"Solution:\n\n(a) Os dois primeiros números de três dígitos são $100=2 \\cdot 2 \\cdot 5 \\cdot 5$ e $101=101$ (que é primo). Ao testar $102$, temos $102=2 \\cdot 3 \\cdot 17$, que é o menor número três estrelas.\n\n(b) Basta mostrar que todo número três estrelas possui pelo menos um dos fatores primos do conjunto ... | Brazil | null | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 102 | |
0kqw | Problem:
Let $\left(x_{1}, y_{1}\right), \ldots,\left(x_{k}, y_{k}\right)$ be the distinct real solutions to the equation
$$
\left(x^{2}+y^{2}\right)^{6}=\left(x^{2}-y^{2}\right)^{4}=\left(2 x^{3}-6 x y^{2}\right)^{3}
$$
Then $\sum_{i=1}^{k}\left(x_{i}+y_{i}\right)$ can be expressed as $\frac{a}{b}$, where $a$ and $b$ ... | [
"Solution:\nUsing polar coordinates, we can transform the problem to finding the intersections between $r=\\cos 2 \\theta$ and $r=2 \\cos 3 \\theta$. Drawing this out gives us a four-leaf clover and a large 3-leaf clover, which intersect at 7 points (one point being the origin). Note that since this graph is symmet... | United States | HMMT February | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | 516 | |
06pt | Let $k$ be a positive integer. Prove that the number $\left(4 k^{2}-1\right)^{2}$ has a positive divisor of the form $8 k n-1$ if and only if $k$ is even. | [
"The statement follows from the following fact.\n\nLemma. For arbitrary positive integers $x$ and $y$, the number $4 x y-1$ divides $\\left(4 x^{2}-1\\right)^{2}$ if and only if $x=y$.\n\nProof. If $x=y$ then $4 x y-1=4 x^{2}-1$ obviously divides $\\left(4 x^{2}-1\\right)^{2}$ so it is sufficient to consider the op... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
02bd | Problem:
A professora Jussara gosta de fazer cópias reduzidas. Ela começa desenhando o triângulo retângulo $A B C$ abaixo:
Em seguida, a professora Jussara traça um segmento $A P_{1}$ de forma que $A P_{1}$ seja perpendicular a $B C$, conforme mostra a figura abaixo.
... | [
"Solution:\na) Usaremos aqui que a soma dos ângulos internos de um triângulo é igual a $180^{\\circ}$. Considere a seguinte figura:\n\nUsando os ângulos internos do triângulo $A B C$, concluímos que $\\alpha+\\beta+90^{\\circ}=180^{\\circ}$. Logo temos que $\\alpha+\\beta=90^{\\circ}$. Um m... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | a) The claim is correct: triangles APC and ABC are similar. b) AP1 = 12/5. c) Area(ABC) : Area(AP1C) = 25/16. d) Area(P1P2C) : Area(AP1C) = 16/25. e) Area(P5P6C) = 6 × (16/25)^6. |
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