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values | competition stringlengths 3 108 ⌀ | topics_flat listlengths 0 12 | language stringclasses 18
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values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0576 | Jüri writes on blackboard some consecutive integers. It is known that the total number of these integers is greater than one and the least of them is greater than $2$. Mari writes on blackboard consecutive integers, too, with the same total number of them as Jüri, but the least of them equals $1$. Is it possible that t... | [
"For example, the product $8 \\cdot 9$ of two consecutive integers divided by the product $1 \\cdot 2$ of the first two positive integers equals $6^2$."
] | Estonia | Open Contests | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Yes | |
08s7 | 3 rectangles are placed on a plane, and any 2 rectangles have their edges that are parallel to each other. They divide the plane into several parts. Determine the maximum possible number of such parts. (We consider the area that are contained in no rectangles as one part. For example, if we have only 1 rectangle, the p... | [
"We will show that in case of $n$ rectangles the answer is $2n^2 - 2n + 2$. We call each divided part region.\n\n$n$ rectangles $R_{1,n}, \\dots, R_{i,n+1-i}, \\dots, R_{n,1}$ divide plane into $2n^2 - 2n + 2$ regions, where $R_{x,y}$ is a rectangle with corners at $(x, y)$, $(x, -y)$, $(-x, y)$, $(-x, -y)$. Theref... | Japan | Japanese Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 14 | |
00wd | Problem:
The squares of a squared paper are enumerated as follows:
Devise a polynomial $p(m, n)$ of two variables $m, n$ such that for any positive integers $m$ and $n$ the number written in the square with coordinates $(m, n)$ will be equal to $p(m, n)$. | [
"Solution:\nSince the square with the coordinates $(m, n)$ is the $n$th on the $(n+m-1)$-th diagonal, it contains the number\n$$\np(m, n) = \\sum_{i=1}^{n+m-2} i + n = \\frac{(n+m-1)(n+m-2)}{2} + n\n$$"
] | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | p(m, n) = ((n+m-1)(n+m-2))/2 + n | |
0djb | Let $ABC$ be an acute, non-isosceles triangle. A circle $(O)$ passes through $B$, $C$ and meets $AB$, $AC$ at $D$, $E$. Lines $BE$, $CD$ meet at $I$. Denote $M$, $N$ as midpoints of $BE$, $CD$ and $MN$ meets $AB$, $AC$ at $P$, $Q$. Prove that $(OI)$ is tangent to $(APQ)$. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
016j | Problem:
In a party of eight persons, each pair of persons either know each other or do not know each other. Each person knows exactly three of the others. Determine whether the following two conditions can be satisfied simultaneously:
- for any three persons, at least two do not know each other;
- for any four person... | [] | Baltic Way | Baltic Way | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | null | proof and answer | Yes | |
0kay | Let $n$ be a positive integer. Tasty and Stacy are given a circular necklace with $3n$ sapphire beads and $3n$ turquoise beads, such that no three consecutive beads have the same color. They play a cooperative game where they alternate turns removing three consecutive beads, subject to the following conditions:
* Tasty... | [
"Observe that the condition about not having three consecutive matching beads is equivalent to saying there are no zigzag of lengths 1.\n\n\nThe main claim is that the game is winnable (for either player going first) if and only if there are at most $2n$ dividers. We prove this in two parts... | United States | USA IMO TST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0f09 | Problem:
a, $m$, $n$ are positive integers and $a > 1$. Show that if $a^{m} + 1$ divides $a^{n} + 1$, then $m$ divides $n$.
The positive integer $b$ is relatively prime to $a$, show that if $a^{m} + b^{m}$ divides $a^{n} + b^{n}$ then $m$ divides $n$. | [] | Soviet Union | ASU | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
066l | Determine all positive integers $x, y$ and $z$ such that
$$
x^5 + 4y^2 = 2013^2.
$$ | [] | Greece | Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | no solutions | |
069a | Find all triples $(x, y, z)$ of real numbers which are solutions of the system:
$$
\begin{cases} x^2 + y^2 + 25z^2 = 6xz + 8yz \\ 3x^2 + 2y^2 + z^2 = 240 \end{cases}
$$ | [
"The first equation can be written in the form:\n$$\n\\begin{aligned} x^2 - 6xz + 9z^2 + y^2 - 8yz + 16z^2 &= 0 \n& \\Leftrightarrow (x - 3z)^2 + (y - 4z)^2 &= 0 \\\\\n\\Leftrightarrow x - 3z &= 0 \\quad \\text{and} \\quad y - 4z &= 0 \n& \\Leftrightarrow x = 3z \\quad \\text{and} \\quad y = 4z. \\end{aligned}\n$$\... | Greece | 36th Hellenic Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (6, 8, 2) and (-6, -8, -2) | |
0kt0 | Problem:
"Very Frustrating Game" has six levels. When a level is attempted, the player goes to the next level if they succeed, but back to the previous level if they fail (or if they are on level 1 they restart).
a) Aerith has a $\frac{1}{2}$ success rate on all levels. How many level attempts on average would it tak... | [
"Solution:\n\na.\nLet $A(x) = 42 - (x + x^{2})$. One can check that $A(x) = 1 + \\frac{1}{2} A(x+1) + \\frac{1}{2} A(x-1)$, that $A(0) = A(-1)$, and that $A(6) = 0$. The answer when starting on level $n+1$ is therefore $A(n)$. Plugging in $n=0$ gives $42$.\n\nb.\nLet $B(x) = 360 - 3\\left(2(2^{x} - 1) - x\\right)$.... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | a) 42; b) 360 | |
0hss | Problem:
We have four bowls labeled $A, B, C, D$ in a row, and we have four indistinguishable marbles which we can initially distribute among the bowls any way we like. A move consists of transferring one marble from a bowl to one of the adjacent bowls in the row. Is it possible to perform a succession of moves in whi... | [
"Solution:\n\nThe answer is no. Call a position of the marbles \"even\" if bowls $A$ and $C$ have an even total number of marbles and \"odd\" otherwise. It is easy to see that any move changes an even position to an odd position and vice versa. However, note that, for any nonnegative integer $k$, the number of ways... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
05xd | Problem:
Soit $k \geqslant 1$ un entier. Quel est le plus petit entier $n$ tel que, quelque soit la manière de placer $n$ points dans le plan, il est possible de choisir un sous-ensemble $S$ constitué de $k$ de ces points qui vérifie "pour toute paire $P, Q$ de points de $S$, la distance entre $P$ et $Q$ est inférieur... | [
"Solution:\n\nCommençons par montrer que $n = (k-1)^2 + 1$ est possible. Pour cela, considérons le disque de rayon $1$ centré sur un point $P_1$ de l'ensemble. S'il y a au moins $k$ points dans ce disque, on peut prendre pour $S$ ces $k$ points (les distances entre eux sont inférieures ou égales à $2$). On peut don... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (k-1)^2 + 1 | |
07f1 | Let $k \in \mathbb{N}$ and $\{a_n\}_{n \ge 0}$ and $\{b_n\}_{n \ge 0}$ be two sequences of positive integers such that $a_i, b_i \in \{1, 2, \dots, 9\}$. For all $n \ge 0$
$$
\overline{a_n \dots a_1 a_0} + k \mid \overline{b_n \dots b_1 b_0} + k.
$$
Prove that there is a number $t$ with $1 \le t \le 9$ and $N \in \math... | [
"Let $A_n = \\overline{a_n \\dots a_1 a_0}$, $B_n = \\overline{b_n \\dots b_1 b_0}$, and $C_n = \\frac{B_n + k}{A_n + k}$. Obviously $C_n < 10$, so $C_n \\in \\{1, 2, \\dots, 9\\}$.\n\n**Claim.** If the sequence $\\{a_i\\}$ doesn't eventually become constant $9$, there exists $N \\in \\mathbb{N}$ such that $k + A_n... | Iran | 37th Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0kep | Problem:
Consider $\triangle A B C$. Choose a point $M$ on its side $B C$ and let $O$ be the center of the circle passing through the vertices of $\triangle A B M$. Let $k$ be the circle that passes through $A$ and $M$ and whose center lies on line $B C$. Let line $M O$ intersect $k$ again in point $K$. Prove that the... | [
"Solution:\n\nLet $D$ be the reflection of $A$ across side $B C$, which clearly lies on $k$. Let $K'$ be the point where lines $B D$ and $M O$ intersect. We will eventually show that $K' = K$. Then $K$ lies on line $B D$, which is therefore the same as line $B K$. Since $B$ and $D$ don't depend on the choice of $M$... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
0bey | Let $n$ be a positive integer and $M = \{1, 2, \dots, 2n + 1\}$. Find out in how many ways we can split the set $M$ into three mutually disjoint nonempty sets $A, B, C$ so that both the following are true:
(i) for each $a \in A$ and $b \in B$, the remainder of the division of $a$ by $b$ belongs to $C$,
(ii) for each $c... | [
"We notice that $a > b$, for all $a \\in A$ and $b \\in B$. Indeed, the contrary would imply that $a < b$, so the remainder of the division of $a$ by $b$ is $a$ and $a$ belongs to both $B$ and $C$. This shows that $A$ is made by consecutive numbers and $2n+1 \\in A$.\n\nTake $c \\in C$. From the assumption there ex... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Other"
] | null | proof and answer | n | |
0dld | Let $a_1 \le a_2 \le \dots \le a_{50}$ be real numbers such that
(i) $a_1 + a_2 + \dots + a_{50} = 0,$
(ii) $|a_1| + |a_2| + \dots + |a_{50}| = 624.$
Find the minimum value for $S = a_{50} - a_1$ and maximum value of $T = a_1a_2 \dots a_{50}$. | [
"Suppose that $k$ is the maximum index such that $a_k \\le 0$ then $a_k \\le 0 < a_{k+1}$. Thus $a_{k+1} + \\dots + a_{50} = -(a_1 + a_2 + \\dots + a_k)$ and then\n$$\n|a_1| + |a_2| + \\dots + |a_k| = |a_1 + a_2 + \\dots + a_k| = a_{k+1} + a_{k+2} + \\dots + a_{50} = \\frac{624}{2} = 312.\n$$\nWe have\n$$\n-312 = a... | Saudi Arabia | Saudi Booklet | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | Minimum S = 624/25. Maximum T = 13^24 * 12^26. | |
0hjx | Problem:
On Predictable Island, everyone is either a liar (who always lies) or a truth-teller (who always tells the truth). You encounter Aerith who says, "Bob and I are both liars." What are they actually? | [
"Solution:\n\nIf Aerith was a truth-teller, she would be lying, because it wouldn't be the case that they're both liars. Thus, she must be a liar. Therefore, the statement must be false, which means Bob has to be a truth-teller."
] | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Discrete Mathematics > Logic"
] | null | proof and answer | Aerith is a liar and Bob is a truth-teller. | |
0js3 | Problem:
Determine the largest integer $n$ such that there exist monic quadratic polynomials $p_{1}(x), p_{2}(x), p_{3}(x)$ with integer coefficients so that for all integers $i \in [1, n]$ there exists some $j \in [1,3]$ and $m \in \mathbb{Z}$ such that $p_{j}(m) = i$. | [
"Solution:\n\nThe construction for $n=9$ can be achieved with the polynomials $x^{2}+x+1$, $x^{2}+x+2$, and $x^{2}+5$.\n\nFirst we consider what kinds of polynomials we can have. Let $p(x) = (x+h)^{2} + k$. $h$ is either an integer or half an integer. Let $k=0$. If $h$ is an integer then $p(x)$ hits the perfect squ... | United States | HMMT November 2016 | [
"Algebra > Algebraic Expressions > Polynomials",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 9 | |
07z2 | Problem:
Si consideri il quadrante infinito in figura, dove tutti i quadratini hanno lato $1$. È possibile colorare di nero alcuni dei quadratini in modo che siano soddisfatte entrambe le seguenti proprietà?
- Per ogni numero naturale $n$, il quadrato con vertice in $O$ e di lato $n$ (con i lati paralleli agli assi) h... | [
"Solution:\n\nSì, è possibile. Per esempio, si colorino di nero:\n- il primo quadratino avente vertice in $O$;\n- i primi $2$ quadratini sulle due diagonali adiacenti alla diagonale centrale;\n- i primi $3$ quadratini sulle due diagonali adiacenti alle precedenti, e così via.\n\nÈ chiaro che ogni diagonale a $45^{\... | Italy | XIII GARA NAZIONALE di MATEMATICA | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | Yes | |
0e7c | Problem:
Največ koliko je lahko največji skupni delitelj števil $11 n+4$ in $7 n+2$, če je $n$ naravno število? | [
"Solution:\n\nNajvečji skupni delitelj števil $11 n+4$ in $7 n+2$ deli tudi $7(11 n+4)-11(7 n+2)=6$, torej je lahko največ $6$. Če je $n=4$, je $11 n+4=48$ in $7 n+2=30$, največji skupni delitelj teh dveh števil pa je natanko $6$. Odgovor je torej $6$.\n\n\n2. način. Največji skupni delitelj števil $11 n+4$ in $7 n... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 6 | |
00d3 | Juli tiene un mazo de 54 cartas y le propone a Bruno el siguiente juego. Juli ubica las cartas en una fila, algunas boca arriba y las demás boca abajo. Bruno puede hacer repetidas veces el siguiente movimiento: elige una de las cartas y da vuelta esa carta y sus dos vecinas (las que estaban boca arriba las pone boca ab... | [
"Demostraremos que Bruno tiene estrategia ganadora.\n\nConsideramos el caso en el que todas las cartas están boca arriba salvo la primera de la izquierda. Notamos con $\\overline{a_i}$ si la carta del lugar $i$ está boca arriba y $\\underline{a_i}$ si la carta del lugar $i$ está boca abajo. En este caso tenemos: $\... | Argentina | Nacional OMA | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Spanish | proof and answer | Bruno tiene estrategia ganadora. | |
0asg | Problem:
How many times does the graph of $y+1=\left|\log_{1/2}|x|\right|$ cross the $x$-axis? | [
"Solution:\n4"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 4 | |
03nz | Problem:
Let $f(x)$ be a non-constant polynomial with integer coefficients such that $f(1) \neq 1$. For a positive integer $n$, define $\operatorname{divs}(n)$ to be the set of positive divisors of $n$.
A positive integer $m$ is $f$-cool if there exists a positive integer $n$ for which
$$
f[\operatorname{divs}(m)]=\ope... | [
"Solution:\nAssume for the sake of contradiction that there are infinitely many $f$-cool integers.\nIf $f(x)$ has a negative leading coefficient, then a sufficiently large $f$-cool integer $m$ will have $f(m)<0$. But this implies $m$ is not $f$-cool, contradiction.\nThus $f(x)$ has a positive leading coefficient, s... | Canada | CMO 2023 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
04ir | In a triangle $ABC$ we have $|BC| + |AC| = 2|AB|$ and $\angle BAC - \angle CBA = 90^\circ$.
Determine the cosine of the angle $\angle ACB$. | [
"Let $D$ be the point on the side $\\overline{BC}$ such that $\\angle DAC = 90^\\circ$. Let us denote $\\varphi = \\angle CDA$ and $x = |CD|$. Then $\\cos \\angle ACB = \\sin \\varphi$.\n\n\n\nThen we have $|AC| = x \\sin \\varphi$ and $|BD| = |AD| = x \\cos \\varphi$. We also have $\\angle... | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 3/4 | |
01vv | A positive integer $n$ is fixed. Numbers $0$ and $1$ are placed in all cells (exactly one number in any cell) of a $k \times n$ table ($k$ is a number of the rows in the table, $n$ is a number of the columns in it). We call a table *nice* if the following property is fulfilled: for any partition of the set of the rows ... | [
"Answer: $k = n + 1$.\nIn total there are $2^{k-1} - 1$ ways to split the rows into two groups and $2^n - 1$ ways to select some set of columns. Since there is at least one column set for any row split and the column set uniquely defines the row split, $2^{k-1} - 1 \\le 2^n - 1$, and therefore $k \\le n + 1$. We wi... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | k = n + 1 | |
0hoq | Problem:
The set $S$ is a finite subset of $[0,1]$ with the following property: for all $x \in S$, there exist $a, b \in S \cup \{0,1\}$ with $a, b \neq x$ such that $x = (a + b)/2$. Prove that all the numbers in $S$ are rational. | [
"Solution:\n\nLet $x_{1}, x_{2}, \\ldots, x_{n}$ be the elements of $S$. They satisfy a system of $n$ linear equations of the form $x_{i} = (x_{a} + x_{b}) / 2$, where some of the $x_{a}, x_{b}$ may be replaced by $0$ or $1$. If the system is not redundant (its solution is unique), then it can be solved by the usua... | United States | Berkeley Math Circle Take-Home Contest | [
"Algebra > Linear Algebra > Vectors",
"Algebra > Linear Algebra > Matrices",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0646 | Problem:
Gegeben sei ein Dreieck $ABC$. Ein Kreis $k$ geht durch $A$, schneidet die Seiten $\overline{AB}$ und $\overline{AC}$ nochmals in den Punkten $D$ bzw. $E$ und schneidet die Seite $\overline{BC}$ in den Punkten $F$ und $G$, wobei $F$ zwischen $B$ und $G$ liegt. Die Tangente an den Kreis durch $B$, $D$ und $F$ ... | [
"Solution:\n\nNach dem Sehnen-Tangentenwinkelsatz für die Sehne $\\overline{FB}$ gilt (siehe Figur) $\\angle TFB = 180^{\\circ} - \\angle BDF = \\angle FDA$. Im Sehnenviereck $ADFG$ gilt daher $\\angle FDA = 180^{\\circ} - \\angle AGF = \\angle CGA$ (kleine Bögen), analog ist $\\angle TGB = 180^{\\circ} - \\angle C... | Germany | Auswahlklausur | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0exq | Problem:
Given 100 points on the plane. Prove that you can cover them with a collection of circles whose diameters total less than 100 and the distance between any two of which is more than 1. [The distance between circles radii $r$ and $s$ with centers a distance $d$ apart is the greater of 0 and $d - r - s$.] | [
"Solution:\n\nIf we have two circles diameters $d$ and $d'$, the distance between which is less than $1$, then they are contained in a circle diameter $d + d' + 1$. [If the line through the centers cuts the circles in $A$, $B$, $A'$, $B'$, then take a circle diameter $AB'$.] So start with $100$ circles of diameter ... | Soviet Union | 6th ASU | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
07g3 | a) Prove that every graph with $2n$ vertices and $n+2$ edges has an independent set of size $n$.
b) Find the number of graphs with $2n$ vertices and $n+3$ edges having no independent set of size $n$. | [
"a) Let $A$ be a maximal independent subset of vertices and $B$ be its complement. Assume to the contrary that $A$ has at most $n-1$ elements and so $B$ consists of at least $n+1$ vertices. Since $A$ is maximal, there should be an edge from every vertex in $B$ to $A$, so there are at least $n+1$ edges between $A$ a... | Iran | 38th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | a) Every graph with 2n vertices and n+2 edges has an independent set of size n.
b) Such graphs are exactly those consisting of two disjoint triangles and n−3 disjoint edges. For labeled graphs on 2n vertices, the number is C(2n,3)·C(2n−3,3)/2 · (2n−6)! / (2^{n−3}(n−3)!). | |
09i5 | We say a rational number is good if it can be written as $\frac{a}{b} + \frac{b}{a}$ for some positive integers $a$ and $b$.
(i) Show that any integer $n \ge 4$ can be written as the sum of some good numbers.
(ii) What is the minimum number of good numbers that $n = 57$ can be written as the sum? | [
"Answer: (ii) Three.\n(i) $2 = \\frac{1}{1} + \\frac{1}{1}$ and $\\frac{5}{2} = \\frac{2}{1} + \\frac{1}{2}$ are good, thus $4 = 2 + 2$ and $5 = \\frac{5}{2} + \\frac{5}{2}$ are sums of two good numbers. For $n \\ge 4$, we can write $n = 2(k+2)$ or $n = 2k+5$ with $k \\ge 0$, thus $n$ is a sum of good numbers.\n\n(... | Mongolia | Round 3 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 3 | |
076v | Problem:
In the given figure, $ABCD$ is a square paper. It is folded along $EF$ such that $A$ goes to a point $A'$ ($A' \neq C, B$) on the side $BC$ and $D$ goes to $D'$. The line $A'D'$ cuts $CD$ in $G$. Show that the inradius of the triangle $GCA'$ is the sum of the inradii of the triangles $GD'F$ and $A'BE$.
![](a... | [
"Solution:\n\nObserve that the triangles $GCA'$ and $A'BE$ are similar to the triangle $GD'F$. If $GF = u$, $GD' = v$ and $D'F = w$, then we have\n$$\nA'G = p u, \\quad CG = p v, \\quad A'C = p w, \\quad A'E = q u, \\quad BE = q w, \\quad A'B = q v\n$$\nIf $r$ is the inradius of $\\triangle GD'F$, then $p r$ and $q... | India | Indian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
02r2 | Problem:
Para qualquer número positivo $x$, dizemos que os números $x+1$ e $\frac{x}{x+1}$ são filhos de $x$ e que os dois são irmãos. Por exemplo, $\frac{3}{2}$ e $\frac{1}{3}$ são irmãos, pois são filhos de $\frac{1}{2}$; de fato, $\frac{3}{2}=\frac{1}{2}+1$ e $\frac{1}{3}=\frac{\frac{1}{2}}{\frac{1}{2}+1}$.
a) Enc... | [
"Solution:\n\nObservação: para facilitar a escrita da solução, vamos dizer que $x$ é pai de $y$ se $y$ é filho de $x$.\n\na) Suponhamos que $\\frac{5}{7}$ seja filho de um número positivo $x$. Então $\\frac{5}{7}=x+1$ ou $\\frac{5}{7}=\\frac{x}{x+1}$. A primeira equação leva a\n$$\nx=\\frac{5}{7}-1=-\\frac{2}{7}\n$... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) 7/2. b) No; a child has a unique positive parent. c) Yes; repeatedly taking the smaller child from one yields one over two thousand eight. | |
0gtb | In a triangle $ABC$ with $90^\circ > \hat{A} > \hat{B} > \hat{C}$ let $H$ be the orthocenter and $O$ be the circumcenter. The lines $HO$ and $BC$ intersect at point $T$. Let $X$ be the circumcenter of the triangle $AHO$. Prove that the reflection of $H$ over the line $TX$ lies on the circumcircle of $ABC$. | [
"Let $S$ and $M$ be the second intersections of the circumcircle with the line $AH$ and the circle $(AHO)$, respectively. Since $OA = OM$, one has $\\angle MHO = \\angle OHS$, hence $M$ and $S$ are reflections of one another over the line $OH$. Therefore, the perpendicular bisector of $[MH]$, the perpendicular bise... | Turkey | 30th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0c6k | Let $f_0$, $f_1$, $f_2$ and $f_3$ be polynomials in $\mathbb{R}[X]$ such that $f_k(1) = f_{k+1}(0)$, $k = 0, 1, 2, 3$ (indices are reduced modulo $4$). Show that there exists a polynomial $f$ in $\mathbb{R}[X, Y]$ such that $f(X, 0) = f_0(X)$, $f(1, Y) = f_1(Y)$, $f(1 - X, 1) = f_2(X)$, and $f(0, 1 - Y) = f_3(Y)$. | [
"The idea is to consider a suitable $\\mathbb{R}[Y]$-linear combination of $f_0(X)$ and $f_2(1 - X)$, namely, $(1 - Y)f_0(X) + Yf_2(1 - X)$, and a suitable $\\mathbb{R}[X]$-linear combination of $f_1(Y)$ and $f_3(1-Y)$, namely, $Xf_1(Y) + (1-X)f_3(1-Y)$, along with a suitable degree $2$ corrective term, $a_{11}XY +... | Romania | IMAR Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | English | proof only | null | |
08aq | Problem:
Cinque amici devono scendere da una seggiovia a cinque posti e possono farlo andando in tre direzioni differenti: a sinistra, dritto oppure a destra. Scendendo da una seggiovia è facile scontrarsi con i propri compagni di risalita. Per esempio: se io decido di andare dritto e qualcuno alla mia sinistra di and... | [
"Solution:\n\nLa risposta è $\\mathbf{( E )}$. Cominciamo contando i casi totali, cioè le stringhe di lunghezza 5 composte con le lettere S, A e D (Sinistra, Avanti e Destra). Ciascuno dei cinque amici può prendere indipendentemente dagli altri una tra le tre direzioni, quindi queste stringhe sono $3^{5}=243$.\n\nP... | Italy | Progetto Olimpiadi della Matematica - Gara di Febbraio | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | MCQ | E | |
02jy | Problem:
Artur quer desenhar uma "espiral" de 4 metros de comprimento formada de segmentos de reta. Ele já traçou 7 segmentos, como mostra a figura. Quantos segmentos ainda faltam traçar?
(A) 28
(B) 30
(C) 24
(D) 32
(E) 36
 | [
"Solution:\n\nA figura mostra que a \"espiral\" é formada de segmentos cujos comprimentos formam uma sequência finita da forma $1, 1, 2, 2, 3, 3, 4, 4, \\ldots, n, n$ (se os dois últimos segmentos da espiral têm o mesmo comprimento) ou da forma $1, 1, 2, 2, 3, 3, 4, 4, \\ldots, n, n, n+1$ (se os dois últimos segmen... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | D | |
05uy | Problem:
Déterminer tous les entiers $a$ tels que $a-3$ divise $a^{3}-3$. | [
"Solution:\nOn sait que $a-3 \\mid a^{3}-3^{3}$, donc il divise la différence\n$$\na-3 \\mid a^{3}-3-\\left(a^{3}-3^{3}\\right)=27-3=24\n$$\nRéciproquement, il suffit que $a-3$ divise $24$ pour qu'il divise $24+a^{3}-3^{3}=a^{3}-3$. Les solutions sont donc exactement les diviseurs de $24$ auxquels on ajoute $3$. On... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a ∈ {−21, −9, −5, −3, −1, 0, 1, 2, 4, 5, 6, 7, 9, 11, 15, 27} | |
07u3 | A sequence whose first term is positive has the property that any given term is the area of an equilateral triangle whose perimeter is the preceding term. If the first three terms form an arithmetic progression, determine all possible values of the first term. | [
"An equilateral triangle of side length $x$ has perimeter $3x$ and area $\\sqrt{3}x^2/4$. If the preceding term of the sequence is $T_{n-1}$, then its successor is\n$$\nT_n = \\frac{\\sqrt{3}(T_{n-1}/3)^2}{4} = \\frac{T_{n-1}^2 \\sqrt{3}}{36}.\n$$\nThus the first three terms are\n$$\na, \\frac{a^2\\sqrt{3}}{36}, \\... | Ireland | IRL_ABooklet | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 12*sqrt(3) and 6*sqrt(3)*(sqrt(5)-1) | |
07cq | In triangle $ABC$ denote by $O$ and $H$ be the circumcenter and the orthocenter. The point $P$ is the reflection of $A$ with respect to $OH$. Assume that $P$ is not on the same side of $BC$ as $A$. Points $E$ and $F$ lie on sides $AB$ and $AC$, respectively, such that $BE = PC$ and $CF = PB$. Let $K$ be the intersectio... | [
"Let $M$ be the midpoint of $BC$ and $S$ be the reflection of $P$ over $M$. First we show that $S$ lies on $OH$.\n\n\n\nWe have\n$$\n\\begin{cases} PM = MS \\\\ PK = KA \\end{cases} \\Rightarrow MK \\parallel AS, MK = \\frac{AS}{2}.\n$$\n\n$$\n\\triangle MKO \\sim \\triangle AHS.\n$$\nBecau... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
0d2m | Let $ABC$ be an acute triangle, $M$ be the midpoint of $BC$ and $P$ be a point on line segment $AM$. Lines $BP$ and $CP$ meet the circumcircle of $ABC$ again at $X$ and $Y$, respectively, and sides $AC$ at $D$ and $AB$ at $E$, respectively. Prove that the circumcircles of $AXD$ and $AYE$ have a common point $T \neq A$ ... | [
"By applying Ceva to the concurrent cevians $AM$, $BD$ and $CE$, we obtain\n$$\n\\frac{AE}{EB} = \\frac{AD}{DC} \\cdot \\frac{CM}{MB} = \\frac{AD}{DC}.\n$$\nWe deduce from Thales' theorem that segments $ED$ and $BC$ are parallel.\n\n\n\nSince quadrilateral $BCXY$ is cyclic, we have $\\angle... | Saudi Arabia | Selection tests for the International Mathematical Olympiad 2013 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angl... | English | proof only | null | |
0kc7 | Problem:
A vertex-induced subgraph is a subset of the vertices of a graph together with any edges whose endpoints are both in this subset.
An undirected graph contains 10 nodes and $m$ edges, with no loops or multiple edges. What is the minimum possible value of $m$ such that this graph must contain a nonempty vertex-i... | [
"Solution:\nSuppose that we want to find the vertex-induced subgraph of maximum size where each vertex has degree at least 5. To do so, we start with the entire graph and repeatedly remove any vertex with degree less than 5.\n\nIf there are vertices left after this process terminates, then the subgraph induced by t... | United States | HMMT February 2020 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 31 | |
0e0s | Find all positive integers $n$ for which $\lfloor \frac{n^2}{4} \rfloor$ is a perfect square. | [
"Let $n$ be even, $n = 2k$. Then $\\frac{n^2}{4} = k^2$ is even, so $\\lfloor \\frac{n^2}{4} \\rfloor = k^2$ is a perfect square.\n\nIf $n$ is odd it can be written as $n = 2k + 1$, $k \\ge 0$. In this case\n$$\n\\lfloor \\frac{n^2}{4} \\rfloor = \\lfloor \\frac{4k^2 + 4k + 1}{4} \\rfloor = \\lfloor k^2 + k + \\fra... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | n = 1 or n is even | |
0037 | En el triángulo escaleno $ABC$, con $\angle BAC = 90^\circ$, se consideran las circunferencias inscrita y circunscrita. La recta tangente en $A$ a la circunferencia circunscrita corta a la recta $BC$ en $M$. Sean $S$ y $R$ los puntos de tangencia de la circunferencia inscrita con los catetos $AC$ y $AB$, respectivament... | [] | Argentina | 21° OLIMPIADA IBEROAMERICANA DE MATEMÁTICA | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Ge... | Español | proof only | null | |
0hd4 | Let $ABC$ be an acute triangle with circumcircle $w$ with center $O$. Let $A_0$ and $C_0$ be the second intersection points of $w$ with the continuation of the altitudes from $A$ and $C$. Let $A_1$ and $C_1$ be the intersection points of the line $A_0C_0$ with the sides of the triangle $AB$ and $BC$ respectively. Let $... | [
"For now let's consider the points $A_0$, $A_1$ and $A_2$. Since $\\triangle ABC$ is an acute triangle, then the point $O$ lies inside $\\triangle ABC$, hence the point $A_2$ lies on the line $AC$ (Fig. 40). Let's prove that the quadrilateral $OA_0CA_2$ is inscribed. Indeed, since $A_2O \\parallel BC$ we have\n$$\n... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0bdp | A rook starts moving on an infinite chessboard, alternating horizontal and vertical moves. The length of the first move is one square, of the second – two squares, of the third – three squares and so on.
a) Is it possible for the rook to arrive at its starting point after exactly $2013$ moves?
b) Find all $n$ for which... | [
"We will attach integer coordinates to each square of the board, the starting point being $(0,0)$. Then, after each move one coordinate will be of the form $\\pm 1 \\pm 3 \\pm 5 \\pm \\dots$, and the other will be of the form $\\pm 2 \\pm 4 \\pm 6 \\pm \\dots$.\n\na) If $(0,0)$ could be reached in exactly $2013$ mo... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | a) No. b) All n of the form 8k or 8k−1 for k ≥ 1. | |
02i8 | Problem:
O preço de uma corrida de táxi é $R\$ 2,50$ fixos ("bandeirada"), mais $R\$ 0,10$ por cada 100 metros rodados. Tenho apenas $R\$ 10,00$ no bolso. Logo, tenho dinheiro para uma corrida de até:
A) $2,5~\mathrm{km}$
B) $5,0~\mathrm{km}$
C) $7,5~\mathrm{km}$
D) $10,0~\mathrm{km}$
E) $12,5~\mathrm{km}$ | [
"Solution:\n\nComo a bandeirada é fixa, temos $10,00 - 2,50 = 7,50$ reais a serem gastos apenas com os metros rodados. Cada trecho de 100 metros rodado custa $R\\$ 0,10$, então com $R\\$ 7,50$ posso fazer uma corrida de $\\frac{7,50}{0,10} = \\frac{750}{10} = 75$ trechos de 100 metros cada um, ou seja $75 \\times 1... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | C | |
06hh | Let $[x]$ denote the greatest integer not exceeding $x$. Find the last two digits of $\lfloor \frac{1}{3} \rfloor + \lfloor \frac{2}{3} \rfloor + \lfloor \frac{2^2}{3} \rfloor + \dots + \lfloor \frac{2^{2014}}{3} \rfloor$.
設 $[x]$ 表示不超過 $x$ 的最大整數。求 $\lfloor \frac{1}{3} \rfloor + \lfloor \frac{2}{3} \rfloor + \lfloor \... | [] | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English; Chinese | proof and answer | 15 | |
0f9c | Problem:
Two grasshoppers sit at opposite ends of the interval $[0, 1]$. A finite number of points (greater than zero) in the interval are marked. A move is for a grasshopper to select a marked point and jump over it to the equidistant point the other side. This point must lie in the interval for the move to be allowe... | [] | Soviet Union | 24th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | n equals the number of marked points | |
00wp | Problem:
A castle has a number of halls and $n$ doors. Every door leads into another hall or outside. Every hall has at least two doors. A knight enters the castle. In any hall, he can choose any door for exit except the one he just used to enter that hall. Find a strategy allowing the knight to get outside after visi... | [
"Solution:\n\nThe knight can use the following strategy: exit from any hall through the door immediately to the right of the one he used to enter that hall. Then, knowing which door was passed last and in which direction we can uniquely restore the whole path of the knight up to that point. Therefore, he will not b... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
07y8 | Problem:
Siano $a < b < c < d < e$ numeri reali. Si calcolano tutte le possibili somme a due a due di questi 5 numeri. Di queste 10 somme, le tre più piccole sono $32, 36, 37$, mentre le due più grandi sono $48$ e $51$. Si determinino tutti i possibili valori che può assumere $e$. | [
"Solution:\n\nÈ innanzitutto evidente che le due somme più piccole, $32$ e $36$, sono rispettivamente $a+b$ e $a+c$. Allo stesso modo, le due somme più grandi, $48$ e $51$, sono rispettivamente $c+e$ e $d+e$. Dunque $a+b=32$, $a+c=36$; sottraendo la prima equazione dalla seconda si ricava\n$$\nc-b=4.\n$$\nAllo stes... | Italy | null | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 27.5 | |
02a8 | Problem:
O salta-ficha - Temos dez fichas numeradas colocadas em linha reta, como na figura dada.
## (1) (2) (3) (4) (5) (6) (7) (8) (9) (10)
Queremos arrumá-las em cinco pilhas, com duas fichas em cada pilha. A regra para isso é que só podemos movimentar uma ficha fazendo-a saltar sobre uma ou mais fichas, ou sobre u... | [
"Solution:\n\n(a) A ficha 7 salta sobre as fichas 8 e 9 formando uma pilha com a ficha 10;\n\n(b) a ficha 4 salta sobre as fichas 5 e 6 formando uma pilha com a ficha 8 ;\n\n(c) a ficha 6 salta sobre as fichas 3 e 5 formando uma pilha com a ficha 2 ;\n\n(d) a ficha 5 salta sobre a pilha $(4,8)$ formando uma pilha c... | Brazil | Desafios | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | final answer only | A valid sequence of five moves is:
1) Move 7 over 8 and 9 to stack with 10.
2) Move 4 over 5 and 6 to stack with 8.
3) Move 6 over 3 and 5 to stack with 2.
4) Move 5 over the stack (4,8) to stack with 9.
5) Move 1 over the stack (6,2) to stack with 3. | |
0370 | Problem:
Forty thieves are to distribute 4000 euro amongst them. A group of five thieves is called poor if they have no more than 500 euro all together. What is the minimum number of poor groups amongst all possible groups of five thieves? | [
"Solution:\n\nIf 39 thieves take 101 euro each and the last one takes 61 euro, then the only poor groups are those having as a member the last thief. So this distribution of the money gives $\\binom{39}{4}$ poor groups. We shall prove that this is the required minimum.\n\nLet $r$ be the number of all possible divis... | Bulgaria | Spring Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 82251 | |
0c4h | Problem:
Se consideră $n$ greutăţi, $n \geq 2$, având masele $m_{1}, m_{2}, \ldots, m_{n}$, unde $m_{k} \in \mathbb{N}$, $1 \leq m_{k} \leq k$ pentru orice $k \in \{1,2, \ldots, n\}$. Demonstraţi că putem aşeza greutăţile pe talerele unei balanţe astfel încât aceasta să stea în echilibru dacă şi numai dacă $m_{1}+m_{2... | [
"Solution:\n\nEste evident că pentru ca greutăţile să poată echilibra balanţa suma maselor trebuie să fie pară (şi anume dublul sumei maselor greutăţilor dintr-un taler).\n\nReciproc, demonstrăm prin inducţie \"tare\" după $n$ că dacă suma maselor este pară atunci greutăţile pot echilibra o balanţă. Presupunem aşad... | Romania | Al treilea test de selecţie pentru OBMJ | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
025h | Problem:
Uma mediana de um triângulo é o segmento que liga um vértice ao ponto médio do lado oposto. Considere o triângulo $ABC$ na figura a seguir e sejam $M$, $N$ e $P$ os pontos médios dos lados $BC$, $CA$ e $AB$, respectivamente. As medianas $BN$ e $CP$ se cortam no ponto $G$. Seja $X$ o ponto médio do segmento $A... | [
"Solution:\n\n(a) Observe que os triângulos $AXN$ e $AGC$ são semelhantes, pois $\\frac{AX}{AG}=\\frac{AN}{AC}=\\frac{1}{2}$ e o ângulo $A$ é comum aos dois triângulos. Com isto, $XN=\\frac{GC}{2}$ e $XN$ é paralelo a $GC$. De maneira análoga, podemos provar que $PX$ é paralelo a $BN$. Assim, o quadrilátero $GPXN$ ... | Brazil | NÍVEL 3 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04u0 | Let $ABCD$ be a convex quadrilateral such that $\angle ABC = \angle ACD$ and $\angle ACB = \angle ADC$. Suppose that the circumcenter $O$ of triangle $BCD$ is different from $A$. Prove that the angle $OAC$ is right. (Patrik Bak) | [
"Since $\\angle ABC + \\angle CDA < 180^\\circ$, point $A$ lies inside the circumcircle $\\omega$ of triangle $BCD$. Denote by $C'$, $D'$ the second intersection of $\\omega$ with rays $CA$, $DA$, respectively (Fig. 3). We angle chase:\n$$\n\\angle D'C'C = \\angle D'DC = \\angle ADC = \\angle ACB.\n$$\nHence $BCC'D... | Czech Republic | 67th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fi7 | Problem:
Un cuadrado de lado $5$ se divide en $25$ cuadrados unidad por medio de rectas paralelas a los lados. Sea $A$ el conjunto de los $16$ puntos interiores, que son vértices de los cuadrados unidad, pero que no están en los lados del cuadrado inicial.
¿Cuál es el mayor número de puntos de $A$ que se pueden elegi... | [
"Solution:\n\nNumeremos los $16$ puntos como indica la tabla siguiente.\n\n| 13 | 14 | 15 | 16 |\n| :---: | :---: | :---: | :---: |\n| 9 | 10 | 11 | 12 |\n| 5 | 6 | 7 | 8 |\n| 1 | 2 | 3 | 4 |\n\nPor simple tanteo se obtiene un conjunto de seis puntos que cumple la condición del enunciado, por ejemplo $\\{1,2,3,8,12... | Spain | Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | 6 | |
0ddp | Let $ABCD$ be a cyclic quadrilateral with $O$ is circumcenter and $AC$ meets $BD$ at $I$. Suppose that rays $DA$, $CD$ meet at $E$ and rays $BA$, $CD$ meet at $F$. The Gauss line of $ABCD$ meets $AB$, $BC$, $CD$, $DA$ at points $M$, $N$, $P$, $Q$ respectively. Prove that the circle of diameter $OI$ is tangent to two ci... | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geome... | null | proof only | null | |
065t | Determine prime positive integers $p$ and $q$ satisfying the equation
$$
p^4 + p^3 + p^2 + p = q^2 + q.
$$ | [
"The given equation can be written as\n$$\np^4 + p^3 + p^2 + p = q^2 + q \\\\ \\Leftrightarrow p(p+1)(p^2+1) = q(q+1) \\quad (1)\n$$\n$$\n\\Leftrightarrow p(p^2-1)(p^2+1) = q(q+1)(p-1) \\quad (2)\n$$\nFor $q \\le p$ is not possible. Hence we should have $q > p$.\nTherefore from (2) we conclude that:\n$$\nq|(p^2-1)(... | Greece | SELECTION EXAMINATION | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | p = 2, q = 5 | |
0au3 | Problem:
In $\triangle ABC$, $\angle A = 80^\circ$, $\angle B = 30^\circ$, and $\angle C = 70^\circ$. Let $BH$ be an altitude of the triangle. Extend $BH$ to a point $D$ on the other side of $AC$ so that $BD = BC$. Find $\angle BDA$. | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 70° | |
02wq | Problem:
a) Qual dos números é maior: $2^{100}+3^{100}$ ou $4^{100}$ ?
b) Sejam $x$ e $y$ números naturais tais que
$$
2^{x} \cdot 3^{y}=\left(24^{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{60}}\right) \cdot\left(24^{\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots+\frac{1}{60}}\right)^{2} \cdot\left(24^{\frac{1}{4... | [
"Solution:\na) Perceba que\n$$\n4^{100}=4^{99}+4^{99}+4^{99}+4^{99}=3 \\cdot 4^{99}+4^{99}\n$$\nComparemos as parcelas separadamente:\n$$\n\\begin{aligned}\n4 & >3 \\\\\n4^{99} & >3^{99} \\\\\n3 \\cdot 4^{99} & >3 \\cdot 3^{99} \\\\\n3 \\cdot 4^{99} & >3^{100}\n\\end{aligned}\n$$\ne\n$$\n4^{99}=2^{198}>2^{100}\n$$\... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a) 4^100 is greater than 2^100 + 3^100. b) 3540 | |
0l9j | Let be given a real number $\alpha \neq 0$. Consider the sequence of real numbers $\{x_n\}$, $n = 1, 2, 3, \dots$, defined by:
$$
x_1 = 0 \quad \text{and} \quad x_{n+1}(x_n + \alpha) = \alpha + 1 \quad \text{for every } n = 1, 2, 3, \dots
$$
1/ Find the general term of the sequence $\{x_n\}$.
2/ Prove that the sequence... | [
"1/ • For $\\alpha = -1$, it is easily seen that $x_n = 0 \\quad \\forall n = 1, 2, 3, \\dots$.\n• For $\\alpha \\neq -1$, we have $x_n \\neq -\\alpha \\quad \\forall n = 1, 2, 3, \\dots$, so we can write the inductive relation defining the sequence $\\{x_n\\}$ in the equivalent form:\n$$\nx_{n+1} = \\frac{\\alpha ... | Vietnam | 2003 Vietnamese Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | General term:
- If alpha = -2: x_n = (n - 1)/n for all n.
- If alpha != -2: x_n = (alpha + 1)[(-1)^{n-2} + (alpha + 1)^{n-1}] / [(-1)^{n-1} + (alpha + 1)^n] for all n (this includes the case alpha = -1, giving x_n = 0).
Limit as n -> infinity:
- If |alpha + 1| > 1 or alpha = -2, then lim x_n = 1.
- If |alpha + 1| < 1,... | |
03yb | Let $N^*$ be the set of positive integers. Define $a_1 = 2$, and for $n = 1, 2, \dots$,
$$
a_{n+1} = \min\{\lambda \mid \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n} + \frac{1}{\lambda} < 1, \lambda \in N^*\}.
$$
Prove that $a_{n+1} = a_n^2 - a_n + 1$ for $n = 1, 2, \dots$. | [
"By $a_1 = 2$, $a_2 = \\min\\{\\lambda \\mid \\frac{1}{a_1} + \\frac{1}{\\lambda} < 1, \\lambda \\in N^*\\}$, consider $\\frac{1}{a_1} + \\frac{1}{\\lambda} < 1$, then $\\frac{1}{\\lambda} < 1 - \\frac{1}{2} = \\frac{1}{2}$, $\\lambda > 2$, hence $a_2 = 3$. So the conclusion is true for $n = 1$.\n\nSuppose that the... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null | |
06qh | For any integer $n \geq 2$, let $N(n)$ be the maximal number of triples $(a_{i}, b_{i}, c_{i})$, $i=1, \ldots, N(n)$, consisting of nonnegative integers $a_{i}, b_{i}$ and $c_{i}$ such that the following two conditions are satisfied:
(1) $a_{i}+b_{i}+c_{i}=n$ for all $i=1, \ldots, N(n)$,
(2) If $i \neq j$, then $a_{i} ... | [
"Let $n \\geq 2$ be an integer and let $\\{T_{1}, \\ldots, T_{N}\\}$ be any set of triples of nonnegative integers satisfying the conditions (1) and (2). Since the $a$-coordinates are pairwise distinct we have\n$$\n\\sum_{i=1}^{N} a_{i} \\geq \\sum_{i=1}^{N}(i-1)=\\frac{N(N-1)}{2}\n$$\nAnalogously,\n$$\n\\sum_{i=1}... | IMO | IMO Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | N(n) = floor(2n/3) + 1 | |
063k | Problem:
Eine positive ganze Zahl $n$ heißt neckisch, wenn man sie in der Form $n = a^{b} + b$ mit zwei ganzen Zahlen $a, b \geq 2$ schreiben kann.
Man entscheide, ob es 102 aufeinanderfolgende positive ganze Zahlen gibt, von denen genau 100 neckisch sind. | [
"Solution:\n\nEs gibt solche Zahlen. Für eine positive ganze Zahl $m$ sei $f(m)$ die Anzahl der neckischen Zahlen unter den 102 aufeinander folgenden Zahlen $m, m+1, m+2, \\ldots, m+101$. Es sei $N$ das kleinste gemeinsame Vielfache der Zahlen $2,3,4, \\ldots, 101$. Dann ist $f\\left(2^{N}\\right) \\geq 100$, da fü... | Germany | 1. Auswahlklausur 2014/2015 | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
04uu | Find all functions $f: (0, \infty) \to (0, \infty)$ such that
$$
f\left(f(x) + \frac{y+1}{f(y)}\right) = \frac{1}{f(y)} + x + 1
$$
for all $x, y > 0$.
(Dominik Burek) | [] | Czech Republic | Czech-Polish-Slovak Match | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x for all x > 0 | |
02jt | Problem:
Veja as promoções de dois supermercados:
| Supermercado A | Supermercado B |
| :---: | :---: |
| 6 latas de 3 litros do
sorvete QUENTE | Sorvete QUENTE - lata de 3
litros |
| $R\$ 24,00$ | 4 latas - só $R\$ 14,00$ |
Joana quer comprar 12 latas de sorvete para a festa de seu aniversário. Em qual superm... | [
"Solution:\n\n(D) Se comprar no supermercado $A$, Joana gastará $2 \\times R\\$ 24,00 = R\\$ 48,00$.\nSe comprar no supermercado $B$, ela gastará $3 \\times R\\$ 14,00 = R\\$ 42,00$."
] | Brazil | Brazilian Mathematical Olympiad | [
"Math Word Problems"
] | null | MCQ | D | |
0ef7 | Problem:
Naj bo $a_{1}, \frac{1}{5}, a_{3}, \frac{16}{125}, \ldots$ geometrijsko zaporedje s samimi pozitivnimi členi. Kolikšna je vsota $a_{1}+a_{3}+a_{5}+\ldots+a_{99}$?
(A) $\frac{5}{4}\left(1-0,8^{50}\right)$
(B) $\frac{5}{4} \cdot 0,2^{50}$
(C) $\frac{5}{4}-0,2^{50}$
(D) $\frac{5}{4}\left(1-8 \cdot 10^{-50}\righ... | [
"Solution:\n\nIz $\\frac{a_{3}}{\\frac{1}{5}}=\\frac{\\frac{16}{125}}{a_{3}}$ sledi, da je $a_{3}=\\frac{4}{25}$. Potem je $q=\\frac{4}{5}$, $a_{1}=\\frac{1}{4}$. Vsota je $S=a_{1} \\cdot \\frac{q^{50}-1}{q-1}=\\frac{5}{4}\\left(1-0,8^{50}\\right)$. Pravilen odgovor je (A)."
] | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | A | |
09ym | Eleonora has a piece of paper in the shape of an equilateral triangle with an area of $1$. She folds the piece several times and puts it flat on the table. It turns out that the figure on the table is not more than four layers thick anywhere.
What is the minimum area of the figure lying on the table? | [] | Netherlands | Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 1/4 | |
0b4w | Problem:
Let $\mathcal{S}$ be the set of all points in the plane. Find all functions $f: \mathcal{S} \rightarrow \mathbb{R}$ such that for all nondegenerate triangles $A B C$ with orthocenter $H$, if $f(A) \leq f(B) \leq f(C)$, then
$$
f(A)+f(C)=f(B)+f(H)
$$ | [
"Solution:\nLet $P(A, B, C)$ be the problem assertion. First consider a non-right triangle $A B C$ with orthocenter $H$. Note that in the set $\\{A, B, C, H\\}$, the last point is the orthocenter of the other three. Thus, we can assume WLOG $f(A) \\leq f(B) \\leq f(C) \\leq f(H)$.\nBy considering $P(A, B, C)$, this... | Philippines | 25th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All constant functions f(P) = c for any real constant c | |
06ui | Let $n \geqslant 3$ be an integer. Two regular $n$-gons $\mathcal{A}$ and $\mathcal{B}$ are given in the plane. Prove that the vertices of $\mathcal{A}$ that lie inside $\mathcal{B}$ or on its boundary are consecutive.
(That is, prove that there exists a line separating those vertices of $\mathcal{A}$ that lie inside $... | [
"We start with finding a regular $n$-gon $\\mathcal{C}$ which (i) is inscribed into $\\mathcal{B}$ (that is, all vertices of $\\mathcal{C}$ lie on the perimeter of $\\mathcal{B}$); and (ii) is either a translation of $\\mathcal{A}$, or a homothetic image of $\\mathcal{A}$ with a positive factor.\nSuch a polygon may... | IMO | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coord... | English | proof only | null | |
061t | Problem:
Gegeben seien jeweils $n$ reelle Zahlen $x_{1}, x_{2}, \ldots, x_{n}$ bzw. $y_{1}, y_{2}, \ldots, y_{n}$. Die Elemente einer $n \times n$-Matrix $A$ seien folgendermaßen definiert: ($1 \leq i, j \leq n$)
$$
a_{ij}= \begin{cases}1 & \text{ wenn } x_{i}+y_{j} \geq 0 \\ 0 & \text{ wenn } x_{i}+y_{j}<0\end{cases}... | [
"Solution:\n\nWir nehmen an, dass es eine Matrix $B$ der geforderten Art gebe mit $B \\neq A$. Nun betrachten wir in $A$ nur noch diejenigen Elemente $a_{ij}$, die sich von den entsprechenden Elementen $b_{ij}$ unterscheiden. Es muss mindestens ein solches Element geben. Alle anderen Elemente von $A$ werden gestric... | Germany | Auswahlwettbewerb zur IMO | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
07em | Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that for each pair of positive integers $a$ and $b$,
$$
f(a) + 2ab + f(b)
$$
is a perfect square. | [
"$\\boxed{f(n) = (n + 2\\ell)^2 - 2\\ell^2,\\ \\forall n \\in \\mathbb{Z}^+,\\ \\ell \\ge 0}$.\n\nWe set $F(a, b) = f(a) + f(b) + 2ab$ and we know that $\\forall a, b \\in \\mathbb{Z}^+$, $F(a, b)$ is a perfect square.\n\n$F(a, 2) = f(a) + f(2) + 4a = x^2$\n$F(a, 1) = f(a) + f(1) + 2a = y^2$\n$\\implies 2a + f(2) -... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | For all positive integers n, f(n) = (n + 2ℓ)^2 − 2ℓ^2 with ℓ ≥ 0; values of f on non-integer arguments are unconstrained by the condition. | |
0iqj | Problem:
Call a number overweight if it has at least three positive integer divisors (including $1$ and the number), and call a number obese if it has at least four positive integer divisors (including $1$ and the number). How many positive integers between $1$ and $200$ are overweight, but not obese? | [
"Solution:\n\nA positive integer is overweight, but not obese, if it has exactly $3$ factors. This can only happen if that integer is the square of a prime. (If two primes, $p$ and $q$, divide the number, then $p$, $q$, $pq$, and $1$ all divide it, making it at least obese.)\n\nSo, the integers less than $200$ whic... | United States | 1st Annual Harvard-MIT November Tournament | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | 6 | |
03hz | Problem:
Let
$$
p(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + \cdots + a_{1} x + a_{0}
$$
and
$$
q(x) = b_{m} x^{m} + b_{m-1} x^{m-1} + \cdots + b_{1} x + b_{0}
$$
be two polynomials with integer coefficients. Suppose that all the coefficients of the product $p(x) \cdot q(x)$ are even but not all of them are divisible by $4$.... | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
03r3 | Prove that every positive integer $n$, except a finite number of them, can be represented as a sum of $2004$ positive integers: $n = a_1 + a_2 + \dots + a_{2004}$, where $1 \le a_1 < a_2 < \dots < a_{2004}$, and $a_i \mid a_{i+1}$, $i = 1, 2, \dots, 2003$. (posed by Chen Yonggao) | [
"We are going to prove a more general result: For any positive integer $r \\ge 2$, there exists $N(r) \\in \\mathbb{N}$ such that for every $n \\ge N(r)$, there are positive integers $a_1, a_2, \\dots, a_r$ satisfying\n$$\n\\begin{aligned}\nn &= a_1 + a_2 + \\dots + a_r, \\quad 1 \\le a_1 < a_2 < \\dots \\\\\n&< a_... | China | China Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
007w | In the acute-angled triangle $ABP$ ($AB > BP$) the altitudes are $BH$, $PQ$ and $AS$. The extension of $QS$ intersects line $AP$ at $C$. The extension of $HS$ intersects $BC$ at $L$. If $HS = SL$ and $HL$ is perpendicular to $BC$, compute $\frac{SL}{SC}$. | [
"Since $SH = SL$, we compute the ratio $\\frac{SH}{SC}$. Note that $AB > BP$ implies that $P$ is between $A$ and $C$. Denote $\\angle BAP = \\alpha$ and observe that $\\angle PSC = \\angle PSH = \\angle BSL = \\angle BSQ = \\alpha$. Indeed we have $\\angle HSB = \\angle QSP = 180^\\circ - \\alpha$ from the cyclic q... | Argentina | Mathematical Olympiad Rioplatense | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | 1/3 | |
09fu | Find all positive integers $n$ such that there exists a positive integer with $n^2$ divisors (including $1$ and itself) and an arrangement of those divisors (each divisor is used once) in an $n \times n$ grid, where the sum of the divisors in each row and column is the same.
(Battsengel B., Bayarmagnai G.) | [
"*Answer: $n=1$.*\n\nSuppose there exists $n \\ge 2$ satisfying both requirements. Let $a_j$ be the largest number in the $j$-th column for $1 \\le j \\le n$. Without loss of generality, we may assume that $a_1 > a_2 > \\dots > a_n$. Thus $j a_1 \\le a_j$ since $a_j$ divides $a_1$. Hence it is clear that the sum of... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n=1 | |
0907 | Find the smallest integer $n \ge 10$ such that
$$
\left[ \frac{n}{1} \right] \left[ \frac{n}{2} \right] \cdots \left[ \frac{n}{10} \right] = \binom{n}{10}
$$
Here denote by $[r]$ the largest integer less than or equal to $r$ for a real number $r$. For example, $[3.14] = 3$ and $[5] = 5$. | [
"2519\nNote that $[x] > x - 1$ for any real number $x$. In particular, for an integer $1 \\le k \\le 10$ we have $\\lfloor \\frac{n}{k} \\rfloor > \\frac{n}{k} - 1$, thus $k \\lfloor \\frac{n}{k} \\rfloor > n - k$. Since both sides are integers, we have $k \\lfloor \\frac{n}{k} \\rfloor \\ge n - k + 1$, thus $\\lfl... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof and answer | 2519 | |
0ad7 | Find all triples of positive integers $(x, y, z)$ such that
$$
xyz + xy + yz + zx + x + y + z = 243.
$$ | [
"We add $1$ on both sides of the given equality and we get\n$$\nxyz + xy + yz + zx + x + y + z + 1 = 244\n$$\nor\n$$\nxy(z+1) + x(z+1) + y(z+1) + z + 1 = 244\n$$\nor\n$$\n(z+1)(xy + x + y + 1) = 244.\n$$\nFrom where we obtain\n$$\n(x+1)(y+1)(z+1) = 244.\n$$\nBecause $244 = 2 \\cdot 2 \\cdot 61$ we get that $(x+1, y... | North Macedonia | Macedonian Mathematical Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | [(1, 1, 60), (1, 60, 1), (60, 1, 1)] | |
031c | Problem:
The incircle of $\triangle ABC$ has radius $r$ and is tangent to the sides $AB$, $BC$ and $CA$ at points $C_1$, $A_1$ and $B_1$, respectively. If $N = BC \cap B_1C_1$ and $AA_1 = 2A_1N = 2r\sqrt{3}$, find $\Varangle ANC$. | [
"Solution:\n\nWe shall use the standard notation for the elements of $\\triangle ABC$. We may assume that $b > c$. Denote by $I$ the incenter of $\\triangle ABC$. Then the condition $A_1N = r\\sqrt{3}$ implies that $INA_1$ is a right-angled triangle with $\\Varangle NIA_1 = 60^\\circ$. We shall prove that $AA_1 \\p... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > T... | null | proof and answer | 90° | |
0e52 | Find all real numbers $x$ that solve the equation
$$
\log_2(10x) + \log_4(100x) + \log_8(1000x) - 2\log_{64} x = 9.
$$
Write the result as a reduced fraction. | [
"Rewrite the equation as\n$$\n\\begin{align*}\n9 &= \\log_2(10x) + \\log_4(100x) + \\log_8(1000x) - 2\\log_{64} x \\\\\n&= \\log_2(10x) + \\frac{\\log(100x)}{\\log 4} + \\frac{\\log(1000x)}{\\log 8} - \\frac{2\\log x}{\\log 64} \\\\\n&= \\log_2(10x) + \\frac{\\log(100x)}{2\\log 2} + \\frac{\\log(1000x)}{3\\log 2} -... | Slovenia | National Math Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 16/25 | |
0f5h | Problem:
Find all real solutions $x$, $y$ to
$$
y^2 = x^3 - 3x^2 + 2x, \quad x^2 = y^3 - 3y^2 + 2y.
$$ | [] | Soviet Union | 17th ASU | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | (0, 0), (2 - sqrt(2), 2 - sqrt(2)), (2 + sqrt(2), 2 + sqrt(2)) | |
09qb | Problem:
Bekijk rijen $a_{1}, a_{2}, a_{3}, \ldots$ van positieve gehele getallen. Bepaal de kleinst mogelijke waarde van $a_{2010}$ als gegeven is:
(i) $a_{n}<a_{n+1}$ voor alle $n \geq 1$,
(ii) $a_{i}+a_{l}>a_{j}+a_{k}$ voor alle viertallen $(i, j, k, l)$ met $1 \leq i<j \leq k<l$. | [
"Solution:\n\nWe bewijzen met inductie dat $a_{n}-a_{1} \\geq 2^{n-1}-1$ voor alle $n \\geq 2$. Voor $n=2$ staat hier $a_{2}-a_{1} \\geq 1$ en dat volgt uit voorwaarde (i). Zij nu $m \\geq 2$ en stel dat $a_{m}-a_{1} \\geq 2^{m-1}-1$. We passen voorwaarde (ii) toe met $i=1, j=k=m$ en $l=m+1$. We vinden dat $a_{1}+a... | Netherlands | toets | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2^2009 | |
0kf6 | Problem:
Compute the maximum number of sides of a polygon that is the cross-section of a regular hexagonal prism. | [
"Solution:\n\nNote that since there are 8 faces to a regular hexagonal prism and a cross-section may only intersect a face once, the upper bound for our answer is 8.\n\nIndeed, we can construct a cross-section of the prism with 8 sides. Let $A B C D E F$ and $A' B' C' D' E' F'$ be the two bases of the prism, with $... | United States | HMMO 2020 | [
"Geometry > Solid Geometry > 3D Shapes"
] | null | final answer only | 8 | |
04ef | If $(1 + \sin t)(1 + \cos t) = \frac{5}{4}$, determine $\sin t + \cos t$. | [
"Let $S = \\sin t + \\cos t$.\n\nWe have:\n$$(1 + \\sin t)(1 + \\cos t) = 1 + \\sin t + \\cos t + \\sin t \\cos t = \\frac{5}{4}$$\n\nSo:\n$$1 + S + \\sin t \\cos t = \\frac{5}{4}$$\n$$S + \\sin t \\cos t = \\frac{1}{4}$$\n\nRecall that:\n$$(\\sin t + \\cos t)^2 = \\sin^2 t + 2 \\sin t \\cos t + \\cos^2 t = 1 + 2 \... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | -1 + sqrt(10)/2 | |
0hma | Problem:
On a $5 \times 5$ chessboard, a king moves according to the following rules:
- It can move one square at a time, horizontally, vertically, or diagonally. (These are the usual moves of the king in chess.)
- It can move in each of the eight allowable directions at most three times in its entire route.
The king c... | [
"Solution:\n\na. The answer is no. To visit all 25 squares, the king must make his maximum of 24 moves and thus must move in each of the eight allowable directions exactly 3 times. Three of these directions (or 9 moves) take the king from a row to the next higher row; three directions (or 9 moves) go to the next lo... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | (a) No; (b) Yes. | |
06il | Let $a_1, a_2, \dots, a_n$ be a sequence of real numbers lying between $1$ and $-1$, i.e. $-1 < a_i < 1$, for $1 \le i \le n$, and such that
(i) $a_1 + a_2 + \dots + a_n = 0$;
(ii) $a_1^2 + a_2^2 + \dots + a_n^2 = 40$.
Determine the smallest possible value of $n$. | [
"The smallest possible value of $n$ is $42$.\nFirstly, it is obvious that $40 = a_1^2 + a_2^2 + \\cdots + a_n^2 < 1 + 1 + \\cdots + 1 = n$. Suppose $n = 41$. WLOG assume $a_1 \\le a_2 \\le \\cdots \\le a_n$ and $a_j \\ne 0$ for each $j$. Note that $a_1 < 0 < a_{41}$ since $a_1 + a_2 + \\cdots + a_{41} = 0$. Let $k$... | Hong Kong | CHKMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 42 | |
0aem | 2. Учениците во IV одделение решавале тест по математика кој содржи 20 задачи. За секоја точно решена задача се добиваат 5 поени, а за секоја неточно решена или нерешена задача се губат по 3 поени.
а) Колку задачи решил Иван, ако освоил 76 поени?
б) Колку најмногу задачи треба да погреши ученик ако сака сигурно да доби... | [
"а) Нека е $x$-број на решени задачи. Според условите на задачата, се добива следната равенка:\n$$5x - 3(20 - x) = 76$$\n$$5x - 60 + 3x = 76$$\n$$8x = 76 + 60$$\n$$8x = 136$$\n$$x = 136 : 8$$\n$$x = 17$$\nИван решил $17$ задачи.\n\nб) $8245:x=97$, $x=8245:97$, $x=85$. Бидејќи за секоја нерешена или неточно решена з... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | Macedonian, English | final answer only | a) 17; b) 1 | |
0eds | Problem:
Naj bo $f(x)=\frac{1}{\sqrt{a^{x}}}$, $a>0$ in $a \neq 1$. Koliko je $f\left(\log _{a} 4\right)$ ?
(A) 2
(B) $-\frac{1}{2}$
(C) $\frac{1}{2}$
(D) -4
(E) -2 | [
"Solution:\n\nIzračunamo $f\\left(\\log _{a} 4\\right)=\\frac{1}{\\sqrt{a^{\\log _{a} 4}}}=\\frac{1}{\\sqrt{4}}=\\frac{1}{2}$."
] | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | MCQ | C | |
0941 | Problem:
Prove that there are infinitely many positive integers $n$ such that $n^{2}$ written in base 4 contains only digits 1 and 2. | [
"Solution:\nWe prove that there are infinitely many $n$'s such that $n^{2}$ written in base 4 contains only 1 and 2, with the first and last digit being 1. One example is $n=5$, for which $n^{2}=25=121_{4}$.\n\nNow we describe how for a given such $n$ we can obtain another, bigger one, which satisfies these require... | Middle European Mathematical Olympiad (MEMO) | 15th Middle European Mathematical Olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
073a | Problem:
Let $ABC$ be a triangle in which $AB = AC$. Let $D$ be the mid-point of $BC$ and $P$ be a point on $AD$. Suppose $E$ is the foot of perpendicular from $P$ on $AC$. If $\frac{AP}{PD} = \frac{BP}{PE} = \lambda$, $\frac{BD}{AD} = m$ and $z = m^2(1+\lambda)$, prove that
$$
z^2 - (\lambda^3 - \lambda^2 - 2)z + 1 = ... | [
"Solution:\nLet $AD = h$, $PD = y$ and $BD = DC = a$. We\n\nobserve that $BP^2 = a^2 + y^2$. Moreover,\n$PE = PA \\sin \\angle DAC = (h - y) \\frac{DC}{AC} = \\frac{a(h - y)}{b}$,\nwhere $b = AC = AB$. Using $AP / PD = (h - y) / y$, we obtain $y = h / (1 + \\lambda)$. Thus\n$$\n\\lambda^2 =... | India | INMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
01xb | The sum of several (not necessary different) positive integers not exceeding $10$ is equal to $S$.
Find all possible values of $S$ such that these numbers can always be partitioned into two groups with the sum of the numbers in each group not exceeding $70$. | [
"**Answer:** $S \\le 133$.\n\nClearly $S \\le 140$. Suppose that $S \\ge 134$ and let $S = 134 + m$, where $0 \\le m \\le 6$. Consider the next collection of numbers: one number equals $8$, $m$ numbers equal $10$ and $14-m$ numbers equal $9$ (the total sum equals $134+m$). At least eight of these $15$ numbers will ... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | S ≤ 133 | |
05v5 | Problem:
Soient $p$ un nombre premier impair et $x_{1}, \ldots, x_{p}$ des entiers relatifs.
On suppose que pour tout $k \geqslant 1$ entier, on a
$$
p \mid x_{1}^{k}+\cdots+x_{p}^{k}
$$
Montrer que les entiers $x_{1}, \ldots, x_{p}$ sont tous congrus modulo $p$. | [
"Solution:\n\nNotons que la relation est également vraie pour $k=0$ car $\\underbrace{1+\\ldots+1}_{p \\text{ fois }} \\equiv 0$ $\\bmod p$\n\nSoit $Q(X)=\\sum_{k=0}^{n} a_{k} X^{k}$ un polynôme à coefficients entiers. On a\n$$\nQ\\left(x_{1}\\right)+\\ldots+Q\\left(x_{p}\\right)=\\sum_{\\ell=1}^{p} \\sum_{k=0}^{n}... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Discrete Mathematics > Combinatorics ... | null | proof only | null | |
0ioo | Problem:
Cyclic quadrilateral $ABCD$ has side lengths $AB = 1$, $BC = 2$, $CD = 3$, and $AD = 4$. Determine $AC / BD$.
 | [
"Solution:\n\nAnswer: $\\dfrac{5}{7}$. Let the diagonals intersect at $P$. Note that triangles $ABP$ and $DCP$ are similar, so that $3AP = DP$ and $3BP = CP$. Additionally, triangles $BCP$ and $ADP$ are similar, so that $2BP = AP$. It follows that\n$$\n\\frac{AC}{BD} = \\frac{AP + PC}{BP + PD} = \\frac{2BP + 3BP}{B... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 5/7 | |
0jvb | Problem:
Kelvin the Frog has a pair of standard fair 8-sided dice (each labelled from 1 to 8). Alex the sketchy Kat also has a pair of fair 8-sided dice, but whose faces are labelled differently (the integers on each of Alex's dice need not be distinct). To Alex's dismay, when both Kelvin and Alex roll their dice, the... | [
"Solution:\n\nDefine the generating function of an event $A$ as the polynomial\n$$\ng(A, x) = \\sum p_{i} x^{i}\n$$\nwhere $p_{i}$ denotes the probability that $i$ occurs during event $A$. We note that the generating function is multiplicative; i.e.\n$$\ng(A \\text{ AND } B, x) = g(A) g(B) = \\sum p_{i} q_{j} x^{i+... | United States | HMMT February 2016 | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Abstract Algebra > Ring Theory"
] | null | proof and answer | 24, 28, 32 | |
0df2 | The perpendicular bisector to the side $AC$ of triangle $ABC$ meets $BC$ and $AB$ at points $A_1$ and $C_1$ respectively. Let points $O$ and $O_1$ be the circumcenters of triangles $ABC$ and $A_1BC_1$ respectively. Prove that $C_1O_1$ is a tangent to the circumcircle of the triangle $ABC$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05mr | Problem:
Soient $0 \leqslant a, b, c, d, e \leqslant 1$ des nombres réels. Montrer que
$$
(1+a+b+c+d+e)^2 \geqslant 4\left(a^2+b^2+c^2+d^2+e^2\right)
$$ | [
"Solution:\n\nL'idée est d'utiliser l'inégalité $(x+y)^2 \\geqslant 4 x y$ pour $x=1$ et une valeur particulière de $y$.\n\nPlus précisément, pour tous $x, y \\geqslant 0$, on a $(x-y)^2 \\geqslant 0$, et donc $(x+y)^2 \\geqslant 4 x y$ (cette inégalité est aussi l'inégalité arithmético-géométrique à deux variables... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES - ENVOI No. 3 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0l18 | Problem:
A circle is tangent to both branches of the hyperbola $x^{2}-20 y^{2}=24$ as well as the $x$-axis. Compute the area of this circle. | [
"Solution:\n\nInvert about the unit circle centered at the origin. $\\omega$ turns into a horizontal line, and the hyperbola turns into the following:\n$$\n\\begin{aligned}\n\\frac{x^{2}}{\\left(x^{2}+y^{2}\\right)^{2}}-\\frac{20 y^{2}}{\\left(x^{2}+y^{2}\\right)^{2}}=24 & \\Longrightarrow ... | United States | HMMT February 2024 Guts Round | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 504π | |
08g8 | Problem:
Dedalo acquista un numero finito di stringhe (ciascuna di lunghezza finita) composte dalle cifre binarie 0 e 1. Per ognuna delle stringhe acquistate, Dedalo paga $(1 / 2)^L$ dracme, dove $L$ indica la lunghezza della stringa. Il Minotauro scappa se riesce a trovare una sequenza infinita di cifre binarie che n... | [
"Solution:\n\nPer ogni $c>0$ (anche non necessariamente intero), il lucchese Dedalo, di origini genovesi, può imprigionare il Minotauro spendendo meno di $c$ centesimi di dracma.\n\nLemma. Per ogni coppia di interi positivi $k$ ed $m$, la spesa per acquistare tutte le stringhe di lunghezza $mk$ in cui non compaiono... | Italy | XXXIX Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | all positive integers | |
0euz | Let $n$ be a positive integer. Prove that there are infinitely many triples of integers $(x, y, z)$ each of which satisfies $nx^2 + y^3 = z^4$, $(x, y) = (y, z) = (z, x) = 1$. | [
"$$\ny^3 = z^4 - nx^2 = (z^2 - \\sqrt{nx})(z^2 + \\sqrt{nx}). \\quad (1)\n$$\nFor $y = s^2 - nt^2$ with $(n, t) = 1$, we have\n$$\n\\begin{aligned}\ny^3 &= (s^3 - 3\\sqrt{n}s^2t + 3nst^2 - n\\sqrt{nt^3})(s^3 + 3\\sqrt{n}s^2t + 3nst^2 + n\\sqrt{nt^3}) \\\\\n&= (z^2 - \\sqrt{nx})(z^2 + \\sqrt{nx}),\n\\end{aligned}\n$... | South Korea | 25th Korean Mathematical Olympiad Final Round | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divi... | null | proof only | null |
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