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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0kik | Problem:
Integers $0 \leq a, b, c, d \leq 9$ satisfy
$$
\begin{gathered}
6a + 9b + 3c + d = 88 \\
a - b + c - d = -6 \\
a - 9b + 3c - d = -46
\end{gathered}
$$
Find $1000a + 100b + 10c + d$. | [
"Solution:\n\nLet $N = \\overline{abcd}$ be an at most-four digit number. Note that the first equation gives us $N \\equiv 4 \\pmod{7}$. The second equation gives us $N \\equiv 6 \\pmod{11}$. The third equation gives us $N \\equiv 7 \\pmod{13}$. Using CRT, we get $N \\equiv \\frac{1}{2} \\equiv 501 \\pmod{1001}$. T... | United States | HMMT November 2021 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | final answer only | 6507 | |
0cw5 | 10,000 kids came to a camp; each has exactly 11 friends at the camp (friendship is always mutual). Each kid wears a T-shirt of one of the seven fixed colors, so that every two friends wear T-shirts of distinct colors. The guides requested that some kids (at least one) change their T-shirt colors (with some others among... | [
"Перейдём к графу, вершины которого соответствуют детям, а рёбра — дружбам. Напомним, что раскраска вершин называется правильной, если цвета любых двух вершин, соединённые ребром, различны. Таким образом, граф правильно раскрашен в 7 цветов, в нём выделено 100 стабильных вершин, и требуется перекрасить часть осталь... | Russia | Final round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English; Russian | proof only | null | |
0k7a | Problem:
For a positive integer $n$, let $\tau(n)$ be the number of positive integer divisors of $n$. How many integers $1 \leq n \leq 50$ are there such that $\tau(\tau(n))$ is odd? | [
"Solution:\n\nNote that $\\tau(n)$ is odd if and only if $n$ is a perfect square. Thus, it suffices to find the number of integers $n$ in the given range such that $\\tau(n) = k^{2}$ for some positive integer $k$.\n\nIf $k = 1$, then we obtain $n = 1$ as our only solution.\n\nIf $k = 2$, we see that $n$ is either i... | United States | HMMT November 2019 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 17 | |
09ot | Determine the number of ways to color cells in a $3 \times 61$ grid such that every $2 \times 2$ subgrid contains exactly one colored cell. | [] | Mongolia | MMO2025 Round 2 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 3*2^30 + 2 | |
0f0k | Problem:
Given 4 points which do not lie in a plane, how many parallelepipeds have all 4 points as vertices? | [] | Soviet Union | ASU | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | final answer only | 8 | |
00ql | Prove that there exist infinitely many non isosceles triangles with rational side lengths, rational lengths of altitudes, and perimeter equal to $3$. | [
"If the lengths $a$, $b$ and $c$ of the sides are rational, since $a h_a = b h_b = c h_c = 2A$, where by $h_a$, $h_b$ and $h_c$ we denote the lengths of the altitudes of the triangle, it is enough to find an infinite number of triangles with rational area. From Heron's formula we have\n$$\nA = \\sqrt{\\frac{3}{2} \... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Number Theory > Diophantine Equations > Pythagorean triples"
] | English | proof only | null | |
0hqk | Problem:
Let $ABC$ be a triangle with incenter $I$. A line through $I$ parallel to $BC$ intersects sides $AB$ and $AC$ at $D$ and $E$ respectively. Prove that the perimeter of $\triangle ADE$ is equal to $AB + AC$. | [
"Solution:\n\nBecause $DE \\parallel BC$, $\\angle BID = \\angle IBC$ which is the same as $\\angle DBI$ since $I$ is on the bisector of $\\angle ABC$. Thus $\\triangle BDI$ is isosceles, implying $BD = DI$. Similarly $CE = EI$. Thus the perimeter of $\\triangle ADE$ is\n$$\nAD + AE + DE = AD + AE + DI + EI = AD + ... | United States | Berkeley Math Circle Monthly Contest 7 | [
"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 | |
08mz | Problem:
Consider a rectangle whose lengths of sides are natural numbers. If someone places as many squares as possible, each with area $3$, inside of the given rectangle, such that the sides of the squares are parallel to the rectangle sides, then the maximal number of these squares fill exactly half of the area of t... | [
"Solution:\n\nLet $ABCD$ be a rectangle with $AB = m$ and $AD = n$ where $m, n$ are natural numbers such that $m \\geq n \\geq 2$. Suppose that inside of the rectangle $ABCD$ is placed a rectangular lattice consisting of some identical squares whose areas are equal to $3$, where $k$ of them are placed along the sid... | JBMO | JBMO Shortlist | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Rectangles with side lengths (2, 3), (3, 4), (3, 6), (3, 8), (3, 10), (3, 12), up to ordering of the sides. | |
08ri | A $3 \times 3$ grid is given. We color each square by red or blue so that no red $2 \times 2$ square nor blue $2 \times 2$ square appears. How many such colorings are there?
We consider two colorings different even if they correspond by rotation and/or reversal. | [
"We call the squares corners, edges or the center, according to their position. We first consider the case that the center is colored red. Call the edges $A$, $B$, $C$ and $D$ clockwise. We divide the cases by how many edges are colored red. Note that any $2 \\times 2$ square must include the center.\n\n* If there ... | Japan | The 16th Japanese Mathematical Olympiad - The First Round | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 322 | |
0k5x | Problem:
An artist paints identical dragons on two circular discs of the same size. On the first disc, the dragon covers the center, but on the second it is not. Show that it is possible to cut the second disc into two pieces that can be reassembled so the dragon covers the center. | [
"Solution:\n\nLay the second disc on top of the first disc so that the dragons are lined up, and cut off the piece of the second disc that hangs over the edge of the first disc. This piece is congruent to the piece of the first disc that is not covered by the second disc, so the two pieces can be glued together to ... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
00e8 | Let $ABCD$ be a convex quadrilateral such that $AB = CD$, $\angle BCD = 2\angle BAD$, $\angle ABC = 2\angle ADC$, and $\angle BAD \neq \angle ADC$. Find the angle between diagonals $AC$ and $BD$. | [
"Name $\\angle BAD = \\alpha$, $\\angle BCD = 2\\alpha$, $\\angle ADC = \\beta$, $\\angle ABC = 2\\beta$. The interior angles of $ABCD$ add up to $3\\alpha + 3\\beta = 360^\\circ$. Hence, $\\alpha + \\beta = 120^\\circ$. As $\\alpha \\neq \\beta$, we may assume without loss of generality that $\\beta < 60^\\circ < ... | Argentina | Rioplatense Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 60 degrees | |
01gm | A mason has bricks with dimensions $2 \times 5 \times 8$ and other bricks with dimensions $2 \times 3 \times 7$. She also has a box whose interior has dimensions $10 \times 11 \times 14$. The bricks and the interior of the box are all rectangular parallelepipeds. The mason packs bricks into the box filling the entire v... | [
"Answer: 24.\nLet the number of $2 \\times 5 \\times 8$ bricks in the box be $x$, and the number of $2 \\times 3 \\times 7$ bricks $y$. We must figure out the sum $x + y$. The volume of the box is divisible by $7$, and so is the volume of any $2 \\times 3 \\times 7$ brick. The volume of a $2 \\times 5 \\times 8$ br... | Baltic Way | Baltic Way 2020 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 24 | |
0cze | In an acute triangle $ABC$ the angle bisector $AL$, $L \in BC$, intersects its circumcircle at $N$. Let $K$ and $M$ be the projections of $L$ onto sides $AB$ and $AC$. Prove that triangle $ABC$ and quadrilateral $AKNM$ have equal areas. | [
"Let $P$ be the second point of intersection of segment $BC$ and the circle circumscribed about quadrilateral $AKLM$. Denote by $E$ the intersection point of the lines $KN$ and $BC$ and by $F$ the intersection point of the lines $MN$ and $BC$.\n\nThen $\\widehat{BCN} = \\widehat{BAN}$ and $\\widehat{MAL} = \\wideha... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0f3w | Problem:
A chess board is placed on top of an identical board and rotated through $45$ degrees about its center. What is the area which is black in both boards? | [] | Soviet Union | 15th ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | one quarter of the board area | |
03td | For nonnegative real numbers $a$, $b$, $c$ with $a + b + c = 1$, prove that
$$
\sqrt{a + \frac{(b-c)^2}{4}} + \sqrt{b} + \sqrt{c} \le \sqrt{3}.
$$ | [
"**Proof I** Without loss of generality, we may assume that $b \\ge c$.\nWe set $\\sqrt{b} = x + y$ and $\\sqrt{c} = x - y$ for some nonnegative real numbers $x$ and $y$. Hence $b - c = 4xy$ and $a = 1 - 2x^2 - 2y^2$. It follows that\n$$\n\\sqrt{a + \\frac{(b-c)^2}{4}} + \\sqrt{b} + \\sqrt{c} = \\sqrt{1 - 2x^2 - 2y... | China | China Girls' Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0b1d | Problem:
Find the number of ordered pairs $(x, y)$ of positive integers satisfying $x y + x = y + 92$. | [
"Solution:\n\nWe manipulate the equation to form $x y + x - y - 1 = 91$ or $(x - 1)(y + 1) = 91$.\n\nSince $x$ and $y$ must be positive integers, and $91 = 7 \\times 13$, we must have\n\n$\\left\\{\\begin{array}{l}x - 1 = 1 \\\\ y + 1 = 91\\end{array}\\right.$\n\n$\\left\\{\\begin{array}{l}x - 1 = 91 \\\\ y + 1 = 1... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Number Theory > Diophantine Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 3 | |
0fkz | Problem:
Determina justificadamente todos los pares de números enteros $(x, y)$ que verifican la ecuación $x^{2}-y^{4}=2009$. | [
"Solution:\n\nDada una solución $(x, y)$ cualquiera, es claro que también son soluciones $(x,-y),(-x, y)$ y $(-x,-y)$, con lo que se puede asumir sin pérdida de generalidad que $x, y \\geq 0$. Supongamos entonces que es así. Es claro que $(x-y^{2})(x+y^{2})=7^{2} \\cdot 41$.\n\nSi $x-y^{2}$ y $x+y^{2}$ no son primo... | Spain | Fase Nacional de la XLV Olimpiada Matemática Española | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (45, 2), (45, -2), (-45, 2), (-45, -2) | |
042x | Let the 9-element set $A = \{a + bi \mid a, b \in \{1, 2, 3\}\}$, with $i$ being the imaginary unit. $\alpha = (z_1, z_2, \dots, z_9)$ is a permutation of all the elements in $A$, satisfying $|z_1| \le |z_2| \le \dots \le |z_9|$. The number of such permutations $\alpha$ is ______. | [
"Since\n$$\n\\begin{align*}\n|1+i| < |2+i| = |1+2i| < |2+2i| < |3+i| \\\\\n= |1+3i| < |3+2i| = |2+3i| < |3+3i|,\n\\end{align*}\n$$\nit follows that\n$$\n\\begin{align*}\nz_1 &= 1 + i, \\\\\n\\{z_2, z_3\\} = \\{2 + i, 1 + 2i\\}, \\\\\nz_4 &= 2 + 2i, \\\\\n\\{z_5, z_6\\} = \\{3 + i, 1 + 3i\\}, \\\\\n\\{z_7, z_8\\} = ... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | 8 | |
0ego | Problem:
Koliko je takih trimestnih naravnih števil, pri katerih se poljubni dve števki razlikujeta za vsaj 4?
(A) 10
(B) 15
(C) 18
(D) 20
(E) 21 | [
"Solution:\n\nObravnavajmo vse možnosti glede na prvo števko števila.\nČe je prva števka števila enaka $1$, tedaj moramo drugo in tretjo števko izbrati iz množice $\\{5,6,7,8,9\\}$. Imamo le dve možnosti, $59$ in $95$.\nPodobno, če je prva števka enaka $8$, morata biti preostali dve iz množice $\\{0,1,2,3,4\\}$. Sp... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | C | |
07ur | The equation $AB \times CD = EFGH$, where each of the letters $A, B, C, D, E, F, G, H$ represents a different digit and the values of $A, C$ and $E$ are all non-zero, has many solutions, e.g. $46 \times 85 = 3910$. Prove that the largest value of $EFGH$ for which there is a solution, is $72 \times 95 = 6840$. | [
"We cannot have $B = 0$ or $D = 0$, as this would imply $H = 0$. Similarly, we cannot have $B = 1$ or $D = 1$, as this would imply $H = D$ or $H = B$. For a solution $EFGH$ to be bigger than $6840$ one of the numbers needs to be greater than $86$ because otherwise the product would not exceed $79 \\cdot 86 = 6794$.... | Ireland | IRL_ABooklet | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 6840 | |
061p | Problem:
Es sei $B$ ein beliebiger Punkt auf einem Kreis $k_{1}$ und es sei $A$ ein von $B$ verschiedener Punkt auf der Tangente an $k_{1}$ in $B$. Ferner sei $C$ ein Punkt außerhalb von $k_{1}$ mit der Eigenschaft, dass die Strecke $A C$ den Kreis $k_{1}$ in zwei verschiedenen Punkten schneidet. Schließlich sei $k_{2... | [
"Solution:\n\nWir bezeichnen die Mittelpunkte der Kreise $k_{1}$ und $k_{2}$ mit $O_{1}$ bzw. $O_{2}$. Die Lösung setzt nun voraus, dass $\\angle C D B$ stumpf ist (siehe Fig.); andernfalls sind lediglich einige Winkel modulo $180^{\\circ}$ zu vertauschen.\nEs sei $P$ der Umkreismittelpunkt des Dreiecks $BDC$; dann... | Germany | Auswahlwettbewerb zur IMO | [
"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 | |
0d60 | Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} .
$$
Prove that
$$
\sum_{1 \leq i<j \leq n} a_{i} a_{j}\left(1-a_{i} a_{j}\right) \geq 0 .
$$ | [
"The original inequality is equivalent to\n$$\n2 \\sum_{1 \\leq i<j \\leq n} a_{i} a_{j} \\geq 2 \\sum_{1 \\leq i<j \\leq n} a_{i}^{2} a_{j}^{2},\n$$\nor\n$$\n\\left(\\sum_{i=1}^{n} a_{i}\\right)^{2}-\\sum_{i=1}^{n} a_{i}^{2} \\geq\\left(\\sum_{i=1}^{n} a_{i}^{2}\\right)^{2}-\\sum_{i=1}^{n} a_{i}^{4} .\n$$\nUsing t... | Saudi Arabia | SAMC 2015 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English, Arabic | proof only | null | |
0l0b | Problem:
Jerry and Neil have a 3-sided die that rolls the numbers $1$, $2$, and $3$, each with probability $\frac{1}{3}$. Jerry rolls first, then Neil rolls the die repeatedly until his number is at least as large as Jerry's. Compute the probability that Neil's final number is $3$. | [
"Solution:\n\nIf Jerry rolls $k$, then there is a $\\frac{1}{4-k}$ probability that Neil's number is $3$, since Neil has an equal chance of rolling any of the $4-k$ integers not less than $k$. Thus, the answer is\n$$\n\\frac{1}{3}\\left(1+\\frac{1}{2}+\\frac{1}{3}\\right)=\\frac{11}{18}\n$$"
] | United States | HMMT February 2024 Guts Round | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 11/18 | |
0io5 | Problem:
Convex quadrilateral $A B C D$ has sides $A B = B C = 7$, $C D = 5$, and $A D = 3$. Given additionally that $m \angle A B C = 60^{\circ}$, find $B D$. | [
"Solution:\n\nAnswer: $8$. Triangle $A B C$ is equilateral, so $A C = 7$ as well. Now the law of cosines shows that $m \\angle C D A = 120^{\\circ}$; i.e., $A B C D$ is cyclic. Ptolemy's theorem now gives $A C \\cdot B D = A B \\cdot C D + A D \\cdot B C$, or simply $B D = C D + A D = 8$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 8 | |
06rx | Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$. The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$. Prove that $D$, $H$, $F$, $G$ are concyclic. | [
"We show first that the triangles $FDG$ and $FBE$ are similar. Since $ABCD$ is cyclic, the triangles $EAB$ and $EDC$ are similar, as well as $FAB$ and $FCD$. The parallelogram $ECGD$ yields $GD = EC$ and $\\angle CDG = \\angle DCE$; also $\\angle DCE = \\angle DCA = \\angle DBA$ by inscribed angles. Therefore\n$$\n... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0h03 | For what minimum integer number $N$ one can change "/" to "+" and "-" in the following expression: $1 \ast 2 \ast 3 \ast \dots \ast N$ in order to get:
a) $2010$; b) $2011$? | [
"a) If we have all $+$, then: $1 + 2 + 3 + \\dots + 62 = 1953 < 2010$, $1 + 2 + 3 + \\dots + 63 = 2016 > 2011$. Hence, $N \\ge 63$. From the other hand, $N = 63$ will do: $1 + 2 - 3 + \\dots + 62 + 63 = 2016 - 6 = 2010$.\n\nb) $N = 63, 64$ are not the solutions, because change of sign before any number does not cha... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010) | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a) 63; b) 65 | |
076u | Suppose $n \ge 0$ is an integer and all the roots of $x^3 + \alpha x + 4 - (2 \times 2016^n) = 0$ are integers. Find all possible values of $\alpha$. | [] | India | INMO-2017 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Other"
] | English | proof and answer | -3 | |
0dsi | Let $O$ be a point inside a triangle $ABC$ such that $\angle BOC = 90^\circ$ and $\angle BAO = \angle BCO$. Prove that $\angle OMN = 90^\circ$, where $M$, $N$ are the midpoints of $AC$ and $BC$ respectively. | [
"Extend $CO$ to a point $D$ such that $CO = OD$. Since $\\angle BOC = 90^\\circ$, the triangles $BOC$ and $BOD$ are congruent. Let $\\omega_1$ be the circumcircle of the triangle $BOD$. Since $\\angle BDO = \\angle BCO = \\angle BAO$, the point $A$ lies on $\\omega_1$. As $M$, $O$, $N$ are the midpoints of $CA$, $C... | Singapore | Singapore Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jcl | Problem:
Let $x_{1} = y_{1} = x_{2} = y_{2} = 1$, then for $n \geq 3$ let $x_{n} = x_{n-1} y_{n-2} + x_{n-2} y_{n-1}$ and $y_{n} = y_{n-1} y_{n-2} - x_{n-1} x_{n-2}$. What are the last two digits of $\left|x_{2012}\right|$? | [
"Solution:\n\nLet $z_{n} = y_{n} + x_{n} i$. Then the recursion implies that:\n$$\n\\begin{aligned}\n& z_{1} = z_{2} = 1 + i \\\\\n& z_{n} = z_{n-1} z_{n-2}\n\\end{aligned}\n$$\nThis implies that\n$$\nz_{n} = \\left(z_{1}\\right)^{F_{n}}\n$$\nwhere $F_{n}$ is the $n^{\\text{th}}$ Fibonacci number $\\left(F_{1} = F_... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | final answer only | 84 | |
0aux | Problem:
Let $m$ be the product of all positive integer divisors of $360{,}000$. Suppose the prime factors of $m$ are $p_{1}, p_{2}, \ldots, p_{k}$, for some positive integer $k$, and $m = p_{1}^{e_{1}} p_{2}^{e_{2}} \cdot \ldots \cdot p_{k}^{e_{k}}$, for some positive integers $e_{1}, e_{2}, \ldots, e_{k}$. Find $e_{... | [
"Solution:\n\nLet $d(m)$ be the number of positive divisors of $m$. Since $360{,}000 = 2^{6} \\cdot 3^{2} \\cdot 5^{4}$, we have $m = 360{,}000^{\\frac{d(m)}{2}}$. Thus,\n$$\ne_{1} + e_{2} + e_{3} = \\frac{(6 + 2 + 4)}{2} \\cdot [(6 + 1)(2 + 1)(4 + 1)] = 630\n$$"
] | Philippines | 18th PMO National Stage Oral Phase | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 630 | |
07sd | The triple $(1, 5, 7)$ is such that the squares $(1, 25, 49)$ are in arithmetic progression. Show that there are infinitely many triples $(a, b, c)$ of relatively prime positive integers such that $a^2$, $b^2$ and $c^2$ are in arithmetic progression. | [
"Let $b^2 = u^2 + v^2$ be a Pythagorean triple with integers $u < v < b$. Then\n$$\n(v-u)^2 + (v+u)^2 = v^2 - 2uv + u^2 + v^2 + 2uv + u^2 = 2u^2 + 2v^2 = 2b^2.\n$$\nRearranging,\n$$\nb^2 - (v-u)^2 = (v+u)^2 - b^2.\n$$\nThus, setting $(a, b, c) = (v-u, b, v+u)$ produces squares in arithmetic progression. It is clear... | Ireland | IRL_ABooklet_2020 | [
"Number Theory > Diophantine Equations > Pythagorean triples"
] | null | proof only | null | |
0jmf | Problem:
Consider the parabola consisting of the points $(x, y)$ in the real plane satisfying
$$
(y+x) = (y-x)^2 + 3(y-x) + 3.
$$
Find the minimum possible value of $y$. | [
"Solution:\n\nLet $w = y - x$. Adding $w$ to both sides and dividing by two gives\n$$\ny = \\frac{w^2 + 4w + 3}{2} = \\frac{(w+2)^2 - 1}{2}\n$$\nwhich is minimized when $w = -2$. This yields $y = -\\frac{1}{2}$."
] | United States | HMMT November 2014 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | -1/2 | |
0e4u | Problem:
V šestkotniku $A B C D E F$ velja $\angle B A F=150^\circ$, $\angle A C B=\angle A D C=90^\circ$, $|A C|=|B C|$, trikotnik $A B C$ je podoben trikotniku $A D E$ in trikotnik $B C D$ je podoben trikotniku $D E F$. Izračunaj razmerje dolžin daljic $A B$ in $A F$. | [
"Solution:\n\nOznačimo $|A B|=a$. Ker je $A B C$ enakokrak pravokotni trikotnik z vrhom $C$, je $|A C|=\\frac{a}{\\sqrt{2}}=\\frac{a \\sqrt{2}}{2}$ in $\\angle A B C=\\angle B A C=\\frac{\\pi}{4}$. Zaradi podobnosti trikotnikov $A D E$ in $A B C$ sledi $\\angle A E D=\\angle A C B=\\frac{\\pi}{2}$ in $|D E|=|A E|$.... | Slovenia | 55. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 8/3 | |
0aiv | $A = \{a, b, c\}$ is a set containing three positive integers. Prove that we can find a set $B \subset A$, $B = \{x, y\}$ such that for all odd positive integers $m, n$ we have
$$
10 \mid x^m y^n - x^n y^m
$$ | [
"Let $f(x, y) = x^m y^n - x^n y^m$. If $n = m$, the problem statement will be fulfilled no matter how we choose $B$ so from now on, without loss of generality, we consider $n > m$.\nSince $m$ and $n$ are both odd, we have that $n - m$ is even and we get\n$$\n\\begin{align*}\nf(x, y) &= x^m y^m (y^{n-m} - x^{n-m}) \... | North Macedonia | European Mathematical Cup | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
00fl | Is there an infinite set of points in the plane such that no three points are collinear, and the distance between any two points is rational? | [
"The answer is yes and we present the following construction: the idea is considering points in the unit circle of the form $P_{n}=(\\cos (2 n \\theta), \\sin (2 n \\theta))$ for an appropriate $\\theta$. Then the distance $P_{m} P_{n}$ is the length of the chord with central angle $(2 m-2 n) \\theta \\bmod \\pi$, ... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1994 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Number Theory > Diophantine Equations > Pythagorean triples"
] | null | proof and answer | Yes | |
02b6 | Problem:
Seja $ABC$ um triângulo com $AB = 13$, $BC = 15$ e $AC = 9$. Seja $r$ a reta paralela a $BC$ traçada por $A$. A bissetriz do ângulo $\angle BCA$ corta a reta $r$ em $E$ e a bissetriz do ângulo $\angle CBA$ corta $r$ em $F$. Calcular a medida do segmento $EF$. | [
"Solution:\n\n\n\nComo a reta $EF$ é paralela ao lado $BC$, os ângulos alternos internos gerados pela transversal $CF$ são iguais, isto é, $\\angle FCB = \\angle CFA$. Por outro lado, como $CF$ é bissetriz, temos $\\angle FCE = \\angle FCA$ e assim, $\\angle FCA = \\angle CFA$, donde o triâ... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 22 | |
0jvt | Problem:
In cyclic quadrilateral $ABCD$ with $AB = AD = 49$ and $AC = 73$, let $I$ and $J$ denote the incenters of triangles $ABD$ and $CBD$. If diagonal $\overline{BD}$ bisects $\overline{IJ}$, find the length of $IJ$. | [
"Solution:\n\nLet $O$ be circumcenter, $R$ the circumradius and $r$ the common inradius. We have $IO^{2} = JO^{2} = R(R-2r)$ by a result of Euler; denote $x$ for the common value of $IO$ and $JO$. Additionally, we know $AJ = AB = AD = 49$ (angle chase to find that $\\angle BJA = \\angle JBA$). Since $A$ is the midp... | United States | HMMT February 2016 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneo... | null | proof and answer | 28*sqrt(69)/5 | |
0hh0 | Let $I$ be the incenter of the triangle $ABC$, and $P$ be any point on the arc $BAC$ of the circumscribed circle. On the tangent to the circumscribed circle $\omega$ of the triangle $API$ at the point $I$ points $K$ and $L$ were selected so that $BK = KI$ and $CL = LI$. Prove that the circumcircle of the triangle $PKL$... | [
"Mark the midpoints of the arcs $AB$, $AC$, $BC$ that do not contain other points, by the points $W_C$, $W_B$, $W_A$ respectively (fig. 13). It is clear that $K$ and $L$ lie on $W_A W_C$ and $W_A W_B$ respectively. Then.\n\nFig. 13\n$$\n\\begin{aligned}\n\\angle(KL, PI) &= \\angle(AI, AP) =... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03pl | Determine all pairs of positive integers $(a, b)$ such that
$$
\frac{a^2}{2ab^2 - b^3 + 1}
$$
is a positive integer. | [
"**Solution I** Let $(a, b)$ be a pair of positive integers satisfying the condition. Since $k = \\frac{a^2}{2ab^2 - b^3 + 1} > 0$, we have $2ab^2 - b^3 + 1 > 0$ or $a > \\frac{b}{2} - \\frac{1}{2b^2}$, and hence $a \\ge \\frac{b}{2}$. Using this, we infer from $k \\ge 1$, or $a^2 \\ge b^2(2a-b)+1$, that $a^2 > b^2... | China | International Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (a, b) = (2l, 1), (l, 2l), or (8l^4 - l, 2l) for any positive integer l | |
08xa | Let $n$ be a positive integer. Points $P_1, P_2, \dots, P_{4n}$ are placed in a plane in such a way that no 3 points among them lie on any straight line. Furthermore, for each $i = 1, 2, \dots, 4n$ if we rotate the half-line $P_i P_{i-1}$ starting at $P_i$ around the point $P_i$ by $90^\circ$ clockwise, then the half l... | [
"Let for $k = 1, 2, \\dots, n$ $A_k = P_{4k-3}$, $B_k = P_{4k-2}$, $C_k = P_{4k-1}$, $D_k = P_{4k}$. Also, let $A_{n+1} = A_1$ and $B_{n+1} = B_1$. From now on let us say that the directed line segments $A_iB_i, B_iC_i, C_iD_i, D_iA_{i+1}$ are leftward, downward, rightward, upward segments, respectively. Furthermor... | Japan | Japan 2013 Final Round | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof and answer | (2n-1)(n-1) | |
07pp | In triangle $\triangle ABC$, the angle $\angle BAC$ is less than $90^\circ$. The perpendiculars from $C$ on $AB$ and from $B$ on $AC$ intersect the circumcircle of $\triangle ABC$ again at $D$ and $E$ respectively. If $|DE| = |BC|$, find the measure of the angle $\angle BAC$. | [
"Let $CD$ and $BE$ intersect $AB$ and $AC$ at $F$ and $G$ respectively. Let $P$ be the intersection point of $CD$ and $BE$.\n\nBecause $\\angle DEB = \\angle DCB$ and $|DE| = |BC|$ it follows that $\\triangle DPE$ is congruent to $\\triangle BPC$. In particular, $|DP| = |BP|$. This implies $\\angle BDP = \\angle PB... | Ireland | Ireland | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 60° | |
0jdq | Problem:
Compute the prime factorization of $1007021035035021007001$. (You should write your answer in the form $p_{1}^{e_{1}} p_{2}^{e_{2}} \ldots p_{k}^{e_{k}}$, where $p_{1}, \ldots, p_{k}$ are distinct prime numbers and $e_{1}, \ldots, e_{k}$ are positive integers.) | [
"Solution:\n\nAnswer: $7^{7} \\cdot 11^{7} \\cdot 13^{7}$ The number in question is\n$$\n\\sum_{i=0}^{7}\\binom{7}{i} 1000^{i}=(1000+1)^{7}=1001^{7}=7^{7} \\cdot 11^{7} \\cdot 13^{7}\n$$"
] | United States | HMMT 2013 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 7^7 * 11^7 * 13^7 | |
02c9 | Problem:
Sinais e radicais - Quais dos números dados são negativos?
a. $10-3 \sqrt{11}$
b. $3 \sqrt{11}-10$
c. $18-5 \sqrt{13}$
d. $51-10 \sqrt{26}$
e. $10 \sqrt{26}-51$ | [
"Solution:\n\nTemos $3 \\sqrt{11}=\\sqrt{9 \\times 11}=\\sqrt{99}$. Como $100>99$, obtemos $10=\\sqrt{100}>\\sqrt{99}=3 \\sqrt{11}$, portanto, $10-3 \\sqrt{11}>0$ e $3 \\sqrt{11}-10<0$.\n\nAnalogamente, temos $10 \\sqrt{26}=\\sqrt{100 \\times 26}=\\sqrt{2600}$. Como $2601>2600$, obtemos\n$$\n51=\\sqrt{2601}>\\sqrt{... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | b, c, e | |
01ak | A circle $\omega$ is tangent to the side $BC$ of a triangle $ABC$ at point $T$. The side $AB$ intersects $\omega$ at points $P$ and $R$ ($A$ is closer to $P$ than $R$); the side $AC$ intersects $\omega$ at points $Q$ and $S$ ($A$ is closer to $Q$ than $S$). The lines $AT$, $BQ$ and $CP$ are concurrent. Prove that the l... | [
"By Ceva's theorem,\n$$\nAP \\cdot BT \\cdot CQ = BP \\cdot CT \\cdot AQ \\quad (2)\n$$\nBy the power of the point,\n$$\nBT^2 = BR \\cdot BP \\quad (3)\n$$\n$$\nCT^2 = CS \\cdot CQ \\quad (4)\n$$\n$$\nAP \\cdot AR = AQ \\cdot AS \\quad (5)\n$$\nCombining (2) with (3), (4), (5), we obtain\n$$\nAS \\cdot BT \\cdot \\... | Baltic Way | Baltic Way 2013 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof only | null | |
09zf | Consider the equilateral triangle $PQR$. Inside this triangle the regular hexagon $ABCDEF$ is drawn. Points $B$, $D$ and $F$ are the midpoints of the sides of the triangle $PQR$. The area of the pentagon $QBAFR$ is equal to $1$.
What is the area of the triangle $PQR$?
A) $\frac{11}{10}$
B) $\f... | [] | Netherlands | First Round | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | MCQ | null | |
01ta | We say that a diagonal of a convex hexagon is *good* if it divides the hexagon into two circumscribed quadrilaterals.
Find the greatest number of good diagonals in a convex hexagon. | [
"Answer: 1.\n\nShow that among any two intersecting main diagonals of the hexagon at most one diagonal can be good. Suppose, contrary to our claim, that there are two good intersecting main diagonals. Without loss of generality, we assume that $AD$ and $BE$ are good diagonals of the hexagon $ABCDEF$ (see Fig. 1). T... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 1 | |
0iq4 | Problem:
Explain how any unit fraction $\frac{1}{n}$ can be decomposed into other unit fractions. | [
"Solution:\n$\\frac{1}{n} = \\frac{1}{2n} + \\frac{1}{3n} + \\frac{1}{6n}$"
] | United States | Harvard-MIT November Tournament | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 1/n = 1/(2n) + 1/(3n) + 1/(6n) | |
0frm | Sea $ABC$ un triángulo con $AB \neq AC$, sea $I$ su incentro, $\gamma$ su circunferencia inscrita y $D$ el punto medio de $BC$. La tangente a $\gamma$ por $D$ diferente de $BC$ toca a $\gamma$ en $E$. Demuestra que $AE$ y $DI$ son paralelas. | [
"(Ver la figura 1) Sea $P$ el punto de tangencia de $\\gamma$ con $BC$. Sean $Q$ y $R$ los puntos simétricos de $P$ con respecto a $D$ e $I$, respectivamente. Tenemos que:\n\n* $ER$ es paralela a $DI$. Por un lado, $DI$ es perpendicular a $PE$ (es de hecho la mediatriz de $PE$). Por otro lado, por ser $R$ el punto ... | Spain | LVII Olimpiada Matemática Española Concurso Final Nacional | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Misce... | null | proof only | null | |
0an4 | Problem:
Let $s_{n}$ be the sum of the digits of a natural number $n$. Find the smallest value of $\frac{n}{s_{n}}$ if $n$ is a four-digit number. | [
"Solution:\n\nLet $n$ be a four-digit number, so $1000 \\leq n \\leq 9999$. Let $s_n$ denote the sum of the digits of $n$.\n\nWe are to find the smallest value of $\\frac{n}{s_n}$.\n\nLet $n = 1000a + 100b + 10c + d$, where $a, b, c, d$ are digits and $a \\geq 1$.\nThen $s_n = a + b + c + d$.\n\nWe want to minimize... | Philippines | 18th PMO Area Stage | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 1099/19 | |
06wj | Determine the largest $N$ for which there exists a table $T$ of integers with $N$ rows and 100 columns that has the following properties:
(i) Every row contains the numbers $1,2, \ldots, 100$ in some order.
(ii) For any two distinct rows $r$ and $s$, there is a column $c$ such that $|T(r, c)-T(s, c)| \geqslant 2$.
Here... | [
"Answer: The largest such integer is $N=100!/2^{50}$.\n\n## Solution 1.\n\nNon-existence of a larger table. Let us consider some fixed row in the table, and let us replace (for $k=1,2, \\ldots, 50$) each of two numbers $2k-1$ and $2k$ respectively by the symbol $x_{k}$. The resulting pattern is an arrangement of 50... | IMO | IMO 2021 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | 100!/2^{50} | |
0cqz | В языке племени АУ две буквы — «а» и «у». Некоторые последовательности этих букв являются словами, причём в каждом слове не меньше одной и не больше 13 букв. Известно, что если написать подряд любые два слова, то полученная последовательность букв не будет словом. Найдите максимальное возможное количество слов в таком ... | [
"**Ответ.** $2^{14} - 2^7 = 16256$.\n\n**Первое решение.** Если все последовательности, количество букв в которых не меньше 7 и не больше 13, являются словами, то, очевидно, условие задачи соблюдается; при этом количество таких слов равно $2^7 + \\dots + 2^{13} = 2^{14} - 2^7$. Осталось показать, что это количество... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 16256 | |
007z | Find all functions $f: \mathbb{N} \to \mathbb{R}$ that satisfy the equation
$$
f(x + y) = f(x) + f(y)
$$
for all $x, y \in \mathbb{N}$ such that $10^6 - 10^{-6} < \frac{x}{y} < 10^6 + 10^{-6}$. | [
"All functions of the form $f(x) = cx$ with $c \\in \\mathbb{R}$ are solutions; they are the only ones. More generally let $b > a > 0$, and let the open interval $\\Delta = (a, b)$ contain an integer (in our case $a = 10^6 - 10^{-6}$, $b = 10^6 + 10^{-6}$). Consider any function $f: \\mathbb{N} \\to \\mathbb{R}$ su... | Argentina | Mathematical Olympiad Rioplatense | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | All such functions are f(n) = c n for some real constant c. | |
06gu | Determine the number of positive integers $m$ for which there exist nonnegative integers $x_0, x_1, x_2, \dots, x_{2011}$ such that $m^{x_0} = m^{x_1} + m^{x_2} + \dots + m^{x_{2011}}$. | [
"There are 16 such positive integers $m$.\nLet $a_k = m^k$. We would like to represent $a_{x_0}$ as a sum of 2011 numbers of the form $a_{x_j}$. To get such a representation, we can do in the following way. Starting from the number $a_{x_0}$, each time we choose a number $a_{k+1}$ and break it into $m$ numbers $a_k... | Hong Kong | Year 2012 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 16 | |
032o | Problem:
Find all values of $a$ such that the equation
$$
4^{x} - (a^{2} + 3a - 2) 2^{x} + 3a^{3} - 2a^{2} = 0
$$
has a unique solution. | [
"Solution:\nSetting $y = 2^{x}$, we have to find all values of $a$ such that the equation\n$$\ny^{2} - (a^{2} + 3a - 2) y + 3a^{3} - 2a^{2} = 0 \\Longleftrightarrow (y - a^{2})(y - 3a + 2) = 0\n$$\nhas exactly one positive root.\n\nObviously $a = 0$ is not a solution. For $a \\neq 0$ the equation has a positive roo... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (-∞, 0) ∪ (0, 2/3] ∪ {1, 2} | |
0f25 | Problem:
You have $1000$ tickets numbered $000$, $001$, $\ldots$, $999$ and $100$ boxes numbered $00$, $01$, $\ldots$, $99$. You may put each ticket into any box whose number can be obtained from the ticket number by deleting one digit. Show that you can put every ticket into $50$ boxes, but not into less than $50$. S... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Three-digit tickets (delete one): minimum 50 boxes. Four-digit tickets (delete two): minimum 34 boxes. In general, for n+2‑digit tickets deleting n digits, write 10 = q(n+1) + r with 0 ≤ r ≤ n; the minimum number of boxes is r(q+1)^2 + (n+1 − r)q^2 (equivalently, partition the ten digits into n+1 groups as equally as p... | |
03r6 | Find all positive integers $n$ such that $n$ has a multiple which is alternating.
We call a positive integer *alternating* if every two consecutive digits in its decimal representation are of different parity. | [
"**Solution I**\n\nLemma 1: If $k$ is a positive integer, then there exist $0 \\le a_1, a_2, \\dots, a_{2k} \\le 9$ such that $a_1, a_3, \\dots, a_{2k-1}$ are odd integers, $a_2, a_4, \\dots, a_{2k}$ are even integers, and\n$$\n2^{2k+1} \\mid \\overline{a_1 a_2 \\cdots a_{2k}}\n$$\nProof of Lemma 1 by mathematical ... | China | International Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | All positive integers not divisible by 20. | |
00n3 | Alice and Bob play a game that allows the playing numbers $19$ and $20$ and the two possible starting numbers $9$ and $10$. Alice chooses her playing number and assigns the remaining playing number to Bob while Bob independently chooses the starting number.
Alice adds her playing number to the starting number, Bob add... | [
"Let $s$ be the starting number and $a$ Alice's playing number and $b$ Bob's playing number. Furthermore, let $n$ be the number of rounds of the game (i.e., the number of times that Bob adds his number).\n\nIn order for Bob to win, the equation $s + 39n = 2019$ must have an integer solution for $n$. But neither $s ... | Austria | Austria2019 | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Bob cannot win; Bob should choose starting number 9. | |
0gkl | Let $a$, $b$ and $c$ be positive real numbers such that $abc = 1$. Prove that
$$
\begin{aligned}
& \frac{1}{\sqrt{a^3 + 2b^3 + 6}} + \frac{1}{\sqrt{b^3 + 2c^3 + 6}} + \frac{1}{\sqrt{c^3 + 2a^3 + 6}} \\
\le & \frac{a^2}{\sqrt{a^4 + 4b + 4c}} + \frac{b^2}{\sqrt{b^4 + 4c + 4a}} + \frac{c^2}{\sqrt{c^4 + 4a + 4b}}
\end{alig... | [
"First we prove that\n$$\n\\frac{a^2}{\\sqrt{a^4 + 4b + 4c}} + \\frac{b^2}{\\sqrt{b^4 + 4c + 4a}} + \\frac{c^2}{\\sqrt{c^4 + 4a + 4b}} \\ge 1\n$$\nApplying AM-GM inequality, we conclude that\n$$\n\\begin{aligned} \\frac{a^2}{\\sqrt{a^4 + 4b + 4c}} &= \\frac{a^2}{\\sqrt{a^4 + 4ab^2c + 4abc^2}} \\\\ &\\ge \\frac{a^2}... | Thailand | Thailand Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
09y4 | Problem:
Laat $m$ en $n$ natuurlijke getallen zijn met $mn$ even. Jetze gaat een $m \times n$ bord (dus bestaande uit $m$ rijen en $n$ kolommen) bedekken met dominostenen, zodat elke dominosteen precies twee vakjes bedekt, dominostenen niet uitsteken of overlappen, en alle vakjes bedekt worden door een dominosteen. Me... | [
"Solution:\n\nStel eerst dat $n$ oneven is. Dan is het duidelijk dat $V \\geq 1$; het verschil is immers altijd oneven. We laten zien dat $V=1$ altijd mogelijk is. Kleur hiertoe de verticale dominostenen in oneven kolommen rood en de verticale dominostenen in even kolommen blauw. Omdat in elke rij elke horizontale ... | Netherlands | IMO-selectietoets III | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | V = 1 if n is odd; V = 2 if n ≡ 2 mod 4; V = 0 if n ≡ 0 mod 4 (assuming mn is even). | |
0gs4 | The sequence $a_1, a_2, \dots$ satisfies the equality $\sum_{i=1}^{n} a_{\lfloor \frac{n}{i} \rfloor} = n^{10}$ for each positive integer $n$. Given an arbitrary positive integer $c$, prove that $\frac{c^{a_n} - c^{a_{n-1}}}{n}$ is an integer for each $n > 1$. | [
"For any $n > 1$, we can write the given equality for $n$ and $n-1$, then subtract them side by side to obtain the following:\n$$\n\\sum_{i=1}^{n-1} \\left( a_{\\lfloor \\frac{n}{i} \\rfloor} - a_{\\lfloor \\frac{n-1}{i} \\rfloor} \\right) + a_1 = n^{10} - (n-1)^{10}.\n$$\nDefining a new sequence by $b_1 = a_1 = 1$... | Turkey | 26th Turkish Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > Möbius inversion",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
05rc | Problem:
Soit $\Gamma_{1}$ et $\Gamma_{2}$ deux cercles s'intersectant en $M$ et $N$. Soit $A$ et $B$ les points de contact de ces deux cercles avec leur tangente extérieure commune la plus proche de $M$. Soient $C$ et $D$ les symétriques respectifs de $A$ et $B$ par rapport à $M$, et $E$ et $F$ les points d'intersecti... | [
"Solution:\n\nL'énoncé ainsi formulé peut sembler curieux, mais on remarque qu'il suffit de prouver que $\\widehat{EMF}+\\widehat{ENF}=180^\\circ$ pour obtenir le résultat voulu.\n\nSoit $L$ l'intersection de $(MN)$ et $(AB)$. $L$ est sur l'axe radical de $\\Gamma_{1}$ et $\\Gamma_{2}$, don... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 5 : Pot-POURRI | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01h5 | For a positive integer $n$, we let $s(n)$ denote the sum of the proper divisors of $n$ (excluding $n$ itself). Does there exist a positive integer $a$ for which the equation
$$
s(n) = a + n
$$
has infinitely many solutions? | [
"Answer: Yes, for example $a = 12$.\nWhenever $p$ is prime,\n$$\ns(6p) = 1 + 2 + 3 + 6 + p + 2p + 3p = 12 + 6p.\n$$"
] | Baltic Way | Baltic Way 2020 | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | Yes; a = 12 | |
0883 | Problem:
Dimostrare che tutte le potenze di $3$ hanno la cifra delle decine pari. | [
"Solution:\nSia $r(k)$ il resto di $3^{k}$ nella divisione per $20$. Al fine di provare la tesi, è sufficiente mostrare che per ogni numero naturale $k$ si ha $r(k)<10$. È evidente che $3 \\cdot r(k)$ ed $r(k+1)$ hanno lo stesso resto nella divisione per $20$: si ha perciò $r(0)=1$, $r(1)=3$, $r(2)=9$, $r(3)=7$, $r... | Italy | Progetto Olimpiadi di Matematica | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
09qg | Problem:
Vind alle paren $(x, y)$ van gehele getallen die voldoen aan
$$
x^{2}+y^{2}+3^{3}=456 \sqrt{x-y} .
$$ | [
"Solution:\nOmdat links een geheel getal staat, moet rechts ook een geheel getal staan. De wortel uit een geheel getal is ofwel geheel ofwel irrationaal (maar nooit een niet-gehele breuk), dus moet $\\sqrt{x-y}$ wel geheel zijn. De rechterkant van de vergelijking is deelbaar door $3$, dus de linkerkant moet dat ook... | Netherlands | Dutch TST toets 8 juni | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (30, 21) and (-21, -30) | |
0iws | Problem:
A torus (donut) having inner radius $2$ and outer radius $4$ sits on a flat table. What is the radius of the largest spherical ball that can be placed on top of the center torus so that the ball still touches the horizontal plane? (If the $x$-$y$ plane is the table, the torus is formed by revolving the circle ... | [
"Solution:\nLet $r$ be the radius of the sphere. One can see that it satisfies $(r+1)^2 = (r-1)^2 + 3^2$ by the Pythagorean Theorem, so $r = 9/4$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 9/4 | |
098j | Problem:
Fie numerele reale pozitive $a_{1}, \cdots, a_{n}$ ($n \in \mathbf{N}$), astfel încât $a_{1}+\cdots+a_{n}<1$. Demonstrați inegalitatea
$$
\frac{a_{1}+a_{2}+\cdots+a_{n}}{a_{1} a_{2} \cdots a_{n}} \cdot \frac{\left(1-a_{1}\right) \cdots\left(1-a_{n}\right)}{1-a_{1}-\cdots-a_{n}} \geq n^{n+1}
$$ | [
"Solution:\nNotăm:\n$$\na_{n+1}=1-\\left(a_{1}+\\cdots+a_{n}\\right)\n$$\nAtunci, conform condiției problemei, avem $a_{n+1}>0$ și inegalitatea cerută capătă forma\n$$\n\\frac{\\left(1-a_{n+1}\\right) \\cdot\\left(1-a_{1}\\right) \\cdot\\left(1-a_{2}\\right) \\cdots\\left(1-a_{n}\\right)}{a_{1} a_{2} \\cdots a_{n} ... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
08zi | Let $A$ be a set of functions defined for integers from $1$ to $2023$ and taking integer values from $1$ to $2023$. Suppose that $A$ satisfies the following two conditions:
* For any function $f$ belonging to $A$ and any integers $x$ and $y$ with $1 \le x < y \le 2023$, we have $f(x) \ge f(y)$.
* For any functions $f$ ... | [
"$$\n\\boxed{\\binom{2022}{1011}}\n$$\nWe say that $x$ is a fixed point of $f$ if we have $f(x) = x$. Suppose that a function $f$ belonging to $\\mathcal{A}$ has two fixed points $a$ and $b$ with $a < b$. Then, it follows from the first condition that $a = f(a) \\ge f(b) = b$, which is a contradiction. Therefore, e... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | 2022 choose 1011 | |
0hje | Problem:
Given $n$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$ such that $a_{1} a_{2} \cdots a_{n}=1$, prove that
$$
\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right) \geq 2^{n}
$$
When does the equality hold? | [
"Solution:\n\nBy the inequality $a+b \\geq 2 \\sqrt{a b}$ which holds for positive numbers $a, b$ (and equality is if and only if $a=b$), we see that $1+a_{1} \\geq 2 \\sqrt{a_{1}}$, $1+a_{2} \\geq 2 \\sqrt{a_{2}}$, $\\ldots$, $1+a_{n} \\geq 2 \\sqrt{a_{n}}$. Multiplying these inequalities we get\n$$\n\\left(1+a_{1... | United States | Berkeley Math Circle | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | At least 2^n; equality holds if and only if all the numbers are 1. | |
07id | Consider the sequence $(a_n)_{n \in \mathbb{N}}$ where $a_1 = 2$ and for $n \ge 1$ we have
$$
a_{n+1} = \left(1 + \frac{1}{n}\right)^n a_n.
$$
Prove that there are infinitely many $n$ such that $\frac{1}{n+1}a_n a_{n-1} \dots a_1$ is a perfect square of a natural number. | [
"By induction, we prove $a_n = \\frac{2n^n}{n!}$. The base is clear, if the hypothesis holds true for $n$, we have\n$$\na_{n+1} = \\left(1 + \\frac{1}{n}\\right)^n a_n = 2 \\cdot \\left(\\frac{n+1}{n}\\right)^n \\frac{n^n}{n!} = 2 \\cdot \\frac{(n+1)^n}{n!} = 2 \\cdot \\frac{(n+1)^{n+1}}{(n+1)!},\n$$\nnow we have:\... | Iran | 40th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization t... | null | proof only | null | |
094r | Problem:
There is a sheet of paper (like this one) on an infinite blackboard. Marvin secretly chooses a convex 2024-gon $P$ that lies fully on the piece of paper. Tigerin wants to find the vertices of $P$. In each step, Tigerin can draw a line $g$ on the blackboard that is fully outside the piece of paper, then Marvin... | [
"Solution:\n\nOne of the key observations is the following. If 3 answer lines intersect at a common point $X$, then $X$ must be a vertex of $P$.\n\nLet us start by querying the four sides of the paper. This determines a rectangle on the paper which contains $P$ and each of the sides of the rectangle contain at leas... | Middle European Mathematical Olympiad (MEMO) | MEMO Szeged | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
02f8 | Let $P_1P_2...P_n$ be a polygon inscribed in a circle and $Q$ a point outside the plane of the circle. Let $\beta_i$ be the plane perpendicular to $QP_i$ and passing through $P_i$, $i = 1, 2, ..., n$. Prove that the intersection of all planes $\beta_i$ is a point. | [
"Let $S$ be the sphere that contains the circumcircle of the polygon and $Q$ and let $Q'$ be the point diametrically opposite to $Q$ in $S$. Then $\\angle Q'P_iQ = 90^\\circ$, that is, $Q'P_i$ is perpendicular to $QP_i$ and since $\\beta_i$ contains all lines perpendicular to $QP_i$ passing through $P_i$, $Q'$ belo... | Brazil | XV OBM | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0g9k | 設三角形 $ABC$ 的內切圓為 $\omega$, 內心為 $I$, 外接圓為 $\Gamma$。令 $D$ 為 $\omega$ 在 $BC$ 邊的切點, 並令 $M$ 是 $ID$ 的中點。設 $A'$ 為 $A$ 在 $\Gamma$ 上的對徑點 (即 $AA'$ 為 $\Gamma$ 的一條直徑)。設 $X$ 為直線 $A'M$ 與圓 $\Gamma$ 的另一個交點。
證明：三角形 $AXD$ 的外接圓與直線 $BC$ 相切。
Let $ABC$ be a triangle with incircle $\omega$, incentre $I$ and circumcircle $\Gamma$.
Let $D$ be... | [
"考慮一個過 $A$ 點、並與 $BC$ 直線切於 $D$ 點的圓；設此圓與 $\\Gamma$ 的另一個交點為 $Y$。我們將證明 $Y, M, A'$ 共線。故 $X = Y$，而本題就得證了。\n設 $\\omega$ 分別切 $AC, AB$ 邊於 $E, F$ 點; 令 $EF$ 交 $BC$ 於 $P$ 點, 而 $AX$ 交 $BC$ 於 $J$ 點。\n\n\n\n因為 $JB \\cdot JC = JA \\cdot JY = JD^2$ 且 $(B, C, D, P) = -1$, 所以 $J$ 是 $DP$ 的中點。注意到 $AD$ 是 $P$ 點對圓... | Taiwan | 2015 Math Olympiad Second Stage Training Camp | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry >... | null | proof only | null | |
0fuu | Problem:
Seien $a$, $b$, $c$ positive reelle Zahlen mit $a b c = 1$. Bestimme alle möglichen Werte, die der Ausdruck
$$
\frac{1+a}{1+a+a b} + \frac{1+b}{1+b+b c} + \frac{1+c}{1+c+c a}
$$
annehmen kann. | [
"Solution:\n\nSei $A$ der gegebene Ausdruck. Es gilt wegen $a b c = 1$\n$$\n\\begin{aligned}\nA & = \\frac{1+a}{1+a+a b} + \\frac{a(1+b)}{a(1+b+b c)} + \\frac{a b(1+c)}{a b(1+c+c a)} \\\\\n& = \\frac{1+a}{1+a+a b} + \\frac{a+a b}{1+a+a b} + \\frac{a b+1}{1+a+a b} \\\\\n& = \\frac{2(1+a+a b)}{1+a+a b} = 2\n\\end{ali... | Switzerland | SMO Finalrunde | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 2 | |
061n | Problem:
Man bestimme alle Funktionen $f: \mathbb{R} \rightarrow \mathbb{R}$ mit der Eigenschaft
$$
f(f(x)+y)=2 x+f(f(y)-x)
$$
für alle $x, y \in \mathbb{R}$. | [
"Solution:\nFür eine beliebige reelle Zahl $z$ setzen wir $a=f(z)$, $b=z+f(0)$, $n=f\\left(\\frac{a+b}{2}\\right)$ und $m=f(0)-\\frac{a+b}{2}$. Wegen $\\mathbb{D}_f=\\mathbb{R}$ sind $a$, $b$, $n$ und $m$ wohlbestimmt. Einsetzen von $x=0$, $y=z$ in $(*)$ liefert $f(z+f(0))=f(f(z))$, also $f(a)=f(b)$ (I).\n\nEinsetz... | Germany | Auswahlwettbewerb zur IMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x + c for any real constant c | |
0bep | Problem:
Adott az $f:[0, \pi / 2] \rightarrow[0, \infty)$ növekvő függvény. Igazold, hogy:
a) $\int_{0}^{\pi / 2}(f(x)-f(\pi / 4))(\sin x-\cos x) \, \mathrm{d} x \geq 0$.
b) Létezik $a \in[\pi / 4, \pi / 2]$ úgy, hogy $\int_{0}^{a} f(x) \sin x \, \mathrm{d} x=\int_{0}^{a} f(x) \cos x \, \mathrm{d} x$. | [] | Romania | Matematika tantárgyverseny Megyei szakasz | [
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Functions",
"Precalculus > Trigonometric functions"
] | null | proof only | null | |
0h65 | Find all integer $n$, such that:
$$
(n-2013)(n-2014)(n-2016)(n-2017)=4.
$$ | [
"If integer number $n$ satisfies the given condition, then $4$ can be presented as product of four pairwise distinct integers. Since the integer divisors of this number are only $\\pm1$, $\\pm2$ and $\\pm4$, we have that sought-for divisors are $\\pm1$ and $\\pm2$. Indeed, if the absolute value of one of divisors i... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 2015 | |
0bjh | Given any integer $n \ge 2$, show that there exists a set of $n$ pairwise coprime composite integers in arithmetic progression. | [
"Fix a prime $p > n$ and an integer $N \\ge p + (n-1)n!$ and consider the arithmetic progression of length $n$ consisting of $N! + p + kn!$, $k = 0, 1, \\dots, n-1$.\n\nSuppose, if possible, that $q$ is a prime factor of two of these numbers. Then $q$ divides their difference which is of the form $kn!$, for some po... | Romania | THE 2014 DANUBE MATHEMATICAL COMPETITION | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
03dr | Find all values of the real parameter $a$ for which the solution of the system of simultaneous equations
$$
\begin{cases} |x-2| + |x-1| + |x+3| \le 6 \\ x^2 - a x + 2a \le 0 \end{cases}
$$
is a closed interval of length 1. | [] | Bulgaria | Bulgaria 2022 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | -1 | |
0b1c | Problem:
A positive integer $T$ is said to be triangular if $T = 1 + 2 + \cdots + n$ for some positive integer $n$. Find the smallest positive integer $k$ such that whenever $T$ is triangular, $81T + k$ is also triangular. | [
"Solution:\n\nClearly, taking $T = 1$, we must have $k \\geq 10$. We show that $k = 10$ indeed works. If $T$ is triangular, then $T = \\frac{n(n+1)}{2}$ for some positive integer $n$. Then, we have\n$$\n\\begin{aligned}\n81T + 10 & = \\frac{81n(n+1)}{2} + 10 \\\\\n& = \\frac{81n^2 + 81n + 20}{2} \\\\\n& = \\frac{(9... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 10 | |
0dyj | Prove that for all positive integers $n$ and all real numbers $x_1, x_2, \dots, x_n$ the inequality
$$
\sqrt{x_1^2 + 1} + 2\sqrt{x_2^2 + 1} + \dots + n\sqrt{x_n^2 + 1} \ge \sqrt{(x_1 + 2x_2 + \dots + nx_n)^2 + \frac{n^2(n+1)^2}{4}}
$$
holds. | [
"We prove the inequality by induction. Obviously, for $n=1$ the equality holds and for $n=2$ we have\n$$\n\\sqrt{x_1^2 + 1} + 2\\sqrt{x_2^2 + 1} \\ge \\sqrt{(x_1 + 2x_2)^2 + 9}.\n$$\nAfter squaring both sides we rewrite it as\n$$\n\\sqrt{(x_1^2 + 1)(x_2^2 + 1)} \\ge x_1 x_2 + 1.\n$$\nIf the right-hand side is negat... | Slovenia | Slovenija 2008 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0gr4 | Find the minimal possible value of $\frac{\sqrt{a-b}}{b-1}$ over all real numbers $a > b > 1$ satisfying
$$
(ab+1)^2 + (a+b)^2 \le 2(a+b)(a^2 - ab + b^2 + 1).
$$ | [
"Using the given equation, we get\n$$\n\\begin{aligned}\n0 &\\ge (ab+1)^2 + (a+b)^2 - 2(a+b)(a^2 - ab + b^2 + 1) \\\\\n&= a^2b^2 + 4ab + 1 - 2a^3 - 2b^3 + a^2 + b^2 - 2a - 2b \\\\\n&= (a^2 - 2b + 1)(b^2 - 2a + 1)\n\\end{aligned}\n$$\nSince $a > b$, we get $a^2 - 2b + 1 > b^2 - 2b + 1 = (b-1)^2 \\ge 0$ and $b^2 - 2a... | Turkey | 22nd Junior Turkish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 1/sqrt(2) | |
09yc | Problem:
Zij $p$ een priemgetal groter dan $2$. Patricia wil $7$ niet-noodzakelijk verschillende getallen uit $\{1,2, \ldots, p\}$ aan de zwarte stippen in onderstaande figuur toekennen, op zo'n manier dat het product van drie getallen op een lijn of cirkel altijd dezelfde rest bij deling door $p$ heeft.
![](attached... | [
"Solution:\n\na.\nMet 'lijnen' bedoelen we in het vervolg van dit onderdeel de zes lijnen en de cirkel. Zodra het getal $p$ ergens gebruikt wordt, is er een 'lijn' met product deelbaar door $p$, dus met rest $0$ na deling door $p$. Alle 'lijnen' moeten dan rest $0$ geven, dus alle 'lijnen' bevatten minstens één kee... | Netherlands | Selectietoets | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) 3; b) 8(p−1) | |
08nj | Problem:
Let $O_{1}$ be a point in the exterior of the circle $c(O, R)$ and let $O_{1}N$, $O_{1}D$ be the tangent segments from $O_{1}$ to the circle. On the segment $O_{1}N$ consider the point $B$ such that $BN = R$. Let the line from $B$ parallel to $ON$ intersect the segment $O_{1}D$ at $C$. If $A$ is a point on the... | [
"Solution:\nObviously, the segment $BC$ is tangent to the circle $c$. Let $M$ be the point of tangency (Figure 6).\n\n\n\nCall $Q$, $M$ the tangency points of $BA$, $BC$ with $c^{\\prime}$ and $c$ respectively, and call $H$ the midpoint of segment $AC$. It is well known that\n$$\nAQ = \\fra... | JBMO | JBMO Shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof and answer | (R + r) * sqrt(a^2 - (R + r)^2) | |
0dy9 | Problem:
Dana je družina linearnih funkcij: $f(x) = m x + r x - 2 m + r + 1$. Poišči vrednosti parametrov $m$ in $r$ tako, da bo graf izbrane linearne funkcije vzporeden premici $y = -x$ in bo potekal skozi točko $T(0, 3)$. | [
"Solution:\n\nOdčitamo smerni koeficient družine premic $f(x)$. Ker sta premici vzporedni, enačimo smerna koeficienta družine premic in simetrale sodih kvadrantov $m + r = -1$. Točka $T$ je presečišče premice z ordinatno osjo, zato iz družine funkcij $f(x)$ odčitamo začetno vrednost. Zapišemo enačbo $-2 m + r + 1 =... | Slovenia | Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | m = -1, r = 0 | |
038l | Solve the system
$$
\left|
\begin{array}{l}
\sqrt{x^2 + y^2 - 16(x + y) - 9y + 7} = y - 2 \\
x + 13\sqrt[4]{x - y} = y + 42
\end{array}
\right.
$$ | [
"Writing the second equation in the form\n$x - y + 13\\sqrt[4]{x - y} - 42 = 0$\nand setting $u = \\sqrt[4]{x-y}$, $u \\ge 0$, we get that $u^4 + 13u - 42 = 0$. Note that this equation has only one positive root $u = 2$ and hence $x = y + 16$. Then the first equation easily gives\n$y^2 - 5y + 3 = 0.$\nThe last equa... | Bulgaria | Spring Mathematical Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | x = (37 + sqrt(13)) / 2, y = (5 + sqrt(13)) / 2 | |
0b78 | Let $(a_n)_{n \ge 0}$ be a sequence of positive real numbers such that
$$
\sum_{k=0}^{n} C_{n}^{k} a_{k} a_{n-k} = a_{n}^{2}, \text{ for any } n \ge 0.
$$
Prove that $(a_n)_{n \ge 0}$ is a geometrical sequence. | [
"Let $a_0 = a$. It is obvious that $a_1 = 2a$, and, for $n = 2$,\n$$\na_2^2 - 2aa_2 - 8a^2 = 0,\n$$\nimplying $a_2 = 4a$, for $a_2 > 0$.\nUse induction on $n$ to prove that $a_n = 2^n a$. Assume that $a_k = 2^k a$, for all $k$, $0 \\le k \\le n$, to prove $a_{n+1} = 2^{n+1} a$. We have\n$$\n2aa_{n+1} + a^2 2^{n+1} ... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
... | English | proof only | null | |
0gws | What should be the value of the parameter $\alpha$ in order that equation $|x - \frac{1}{2}| + |x - \sin \alpha| = \cos 3\alpha$ has a single solution? Find this solution. | [
"\nFig.8\n\nThe graph of functions $y = |x - a| + |x - b|$ for $b > a$ is shown in **fig.8**. Thus it is clear that equation $|x - a| + |x - b| = c$ may have a single solution only if $a = b$ and $c = 0$ (**fig.9**).\n\nTherefore for our equation the following condition should be satisfied:... | Ukraine | Ukrajina 2008 | [
"Precalculus > Functions",
"Precalculus > Trigonometric functions"
] | English | proof and answer | All such parameters are α = (-1)^n·π/6 + πn for n ∈ ℤ, and the unique solution is x = 1/2. | |
02ym | Problem:
Vinte jogadores participaram de um torneio de xadrez. Cada jogador enfrentou todos outro jogador exatamente uma vez e cada partida terminou com a vitória de um dos jogadores ou em empate. Nesse torneio, notou-se que para cada partida que terminou em empate, cada um dos demais 18 jogadores venceu pelo menos um... | [
"Solution:\n\nInicialmente mostraremos que cada jogador participou de no máximo um empate. Suponha, por absurdo, que o jogador $A$ empatou com os jogadores $B$ e $C$. Como $A$ empatou com $B$, pelas regras mencionadas, $C$ ganhou de $A$ ou de $B$. Dado que $C$ empatou com $A$, a única possibilidade é $C$ ter ganho ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0im6 | Problem:
How many 5-digit numbers $\overline{a b c d e}$ exist such that digits $b$ and $d$ are each the sum of the digits to their immediate left and right? (That is, $b = a + c$ and $d = c + e$.) | [
"Solution:\n\nAnswer: 330. Note that $a > 0$, so that $b > c$, and $e \\geq 0$ so that $d \\geq c$. Conversely, for each choice of $(b, c, d)$ with $b > c$ and $d \\geq c$, there exists a unique pair $(a, e)$ such that $\\overline{a b c d e}$ is a number having the desired property. Thus, we compute\n$$\n\\sum_{c=0... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 330 | |
079u | Let $p$ be an odd prime number. We say a polynomial
$$
f(x) = \sum_{j=0}^{n} a_{j}x^{j} \text{ is } i\text{-residue if } \sum_{p-1 < j, j > 0} a_{j} \equiv i \pmod{p}.
$$
Show that $\{f(1), \dots, f(p-1)\}$ is a complete residue system modulo $p$ if and only if polynomials $f(x), \dots, f(x)^{p-2}$ are 0-residue and ... | [
"**Lemma.** Let $p$ be an odd prime number and $k$ a positive integer then we have\n$$\n\\sum_{i=0}^{p-1} i^k \\equiv \\begin{cases} 0 & p-1 \\nmid k \\\\ -1 & p-1 \\mid k \\end{cases} \\pmod{p}.\n$$\nProof. If $p-1 \\nmid k$ the statement is obvious by *Fermat Little Theorem*. For $p-1 \\nmid k$\nconsider $g$ a pr... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0hza | Problem:
Find an ordered pair $(a, b)$ of real numbers for which $x^{2}+a x+b$ has a non-real root whose cube is 343. | [
"Solution:\n\nThe cube roots of $343$ are the roots of $x^{3}-343$, which is $(x-7)(x^{2}+7x+49)$. Therefore the ordered pair we want is $(7, 49)$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | final answer only | (7, 49) | |
02t7 | Problem:
Considere dois tambores de capacidade suficientemente grande, um deles vazio e o outro cheio de líquido.
a) Determine se é possível colocar exatamente um litro do líquido do tambor cheio, no vazio, usando dois baldes, um com capacidade de 5 litros e o outro com capacidade de 7 litros.
b) Determine se é possíve... | [
"Solution:\na) Basta encher o tambor vazio com 15 litros ($3 \\times 5$ litros) usando três vezes o balde de 5 litros e, em seguida, retirar 14 litros ($2 \\times 7$ litros) usando o balde de 7 litros duas vezes. Dessa forma, transportamos $3 \\times 5 - 2 \\times 7 = 1$ litro.\n\nb) A quantidade $a$ que podemos tr... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | a) Yes, it is possible. b) No, it is not possible. | |
0f50 | Problem:
$3n$ points divide a circle into $3n$ arcs. One third of the arcs have length $1$, one third have length $2$ and one third have length $3$. Show that two of the points are at opposite ends of a diameter. | [] | Soviet Union | 16th ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0921 | Problem:
Let $x, y, z, w \in \mathbb{R} \setminus \{0\}$ such that $x+y \neq 0$, $z+w \neq 0$, and $x y+z w \geq 0$. Prove the inequality
$$
\left(\frac{x+y}{z+w}+\frac{z+w}{x+y}\right)^{-1}+\frac{1}{2} \geq \left(\frac{x}{z}+\frac{z}{x}\right)^{-1}+\left(\frac{y}{w}+\frac{w}{y}\right)^{-1}
$$ | [
"Solution:\nWe first subtract $1$ on both sides and rewrite the inequality as\n$$\n\\left(\\frac{(x+y)(z+w)}{(x+y)^{2}+(z+w)^{2}}-\\frac{1}{2}\\right) \\geq \\left(\\frac{x z}{x^{2}+z^{2}}-\\frac{1}{2}\\right)+\\left(\\frac{y w}{y^{2}+w^{2}}-\\frac{1}{2}\\right)\n$$\nwhich is equivalent to\n$$\n\\frac{(x-z)^{2}}{x^... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0l2p | Problem:
Albert writes down all of the multiples of $9$ between $9$ and $999$, inclusive. Compute the sum of the digits he wrote. | [
"Solution:\n\nIf $x$ is a multiple of $9$, so is $999 - x$, and their digits always sum to $27$. The $112$ multiples from $0$ to $999$ form $56$ such pairs, so the sum of their digits is $56 \\cdot 27 = 1512$."
] | United States | HMMT November | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 1512 | |
00us | Find all positive integers $n$ such that there exist positive integers $a$, $b$, $c$ satisfying $\gcd(a, b, c) = 1$ and $a + b + c = \gcd(ab + c, ac - b) = n$. | [
"Answer: All positive integers with all prime divisors of the form $4k + 1$.\n\nIn the proof we will use the following well-known Lemma.\n**Lemma.** Let $m$ be an odd positive integer. The equation $x^2 \\equiv -1 \\pmod m$ has an integer solution if and only if all prime divisors of $m$ are of the form $4k + 1$.\n... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All positive integers with all prime divisors of the form 4k + 1. | |
0g4y | Problem:
Während der Weltmeisterschaft gibt es $n$ unterschiedliche Panini-Sticker zu sammeln. Marcos Freunde versuchen alle ihre Sammlungen zu vervollständigen, jedoch hat bis jetzt noch keiner alle Sticker! Wir nennen ein Paar zweier seiner Freunde komplett, falls ihre kombinierte Sammlung jeden Sticker mindestens e... | [
"Solution:\n\nUm zu zeigen, dass mindestens $n$ Tische nötig sind, nehmen wir an, wir haben $n$ Freunde, wobei jedem genau ein Bild fehlt, und sodass jedem ein anderes Bild fehlt. Jedes Paar dieser Freunde ist komplett, also müssen sie an $n$ unterschiedlichen Tischen sitzen.\n\nUm zu zeigen, dass $n$ Tische immer ... | Switzerland | Zweite Runde 2023 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0g0p | Problem:
On considère 2016 points distincts dans le plan. Montrer qu'on peut trouver au moins 45 distances différentes entre ces points. | [
"Solution:\nSoit $E$ l'enveloppe convexe des 2016 points (le plus petit ensemble convexe qui contient tous les points) et soit $P$ un point de notre ensemble qui est sur le bord de $E$. Si depuis $P$ il y a au moins 45 distances, l'exercice est prouvé.\n\nSinon on peut tracer 44 cercles centrés en $P$ de telle mani... | Switzerland | SMO - Finalrunde | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Circles"
] | null | proof only | null | |
0jyb | Problem:
The length of a rectangle is three times its width. Given that its perimeter and area are both numerically equal to $k > 0$, find $k$. | [
"Solution:\n\nLet $a$ be the width of the rectangle. Then the length of the rectangle is $3a$, so the perimeter is $2(a + 3a) = 8a$, and the area is $3a^2$. Since the perimeter and area are both numerically equal to $k$, we know that\n$$\n8a = 3a^2 = k\n$$\nBecause $k > 0$, the rectangle is non-degenerate. It follo... | United States | HMMT November | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 64/3 | |
0ja6 | Problem:
Let $a_{0}, a_{1}, a_{2}, \ldots$ denote the sequence of real numbers such that $a_{0}=2$ and $a_{n+1}=\frac{a_{n}}{1+a_{n}}$ for $n \geq 0$. Compute $a_{2012}$. | [
"Solution:\nAnswer: $\\frac{2}{4025}$\n\nCalculating out the first few terms, note that they follow the pattern $a_{n}=\\frac{2}{2n+1}$. Plugging this back into the recursion shows that it indeed works."
] | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 2/4025 |
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