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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0bns | Let $a \ge 0$ and let $(x_n)_{n \ge 1}$ be a sequence of real numbers. Given that $\left(\frac{x_1+\dots+x_n}{n^a}\right)_{n \ge 1}$ is a bounded sequence, prove that the sequence $(y_n)_{n \ge 1}$, defined by
$$
y_n = \frac{x_1}{1^b} + \frac{x_2}{2^b} + \dots + \frac{x_n}{n^b},
$$
is a convergent sequence for all $b >... | [
"Let $S_n = \\sum_{k=1}^{n} x_k$, $n \\in \\mathbb{N}^*$. Using the hypothesis, one can find a constant $c > 0$ such that $|S_n| \\le c n^a$, $\\forall n \\in \\mathbb{N}^*$. Let $n, p \\in \\mathbb{N}^*$; we have:\n$$\n\\begin{align*}\n|y_{n+p} - y_n| &= \\left| \\sum_{k=n+1}^{n+p} \\frac{x_k}{k^b} \\right| = \\le... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
08j8 | Problem:
Is there a convex quadrilateral, whose diagonals divide it into four triangles, such that their areas are four distinct prime integers?
Problem:
Există un patrulater convex pe care diagonalele să-l împartă în patru triunghiuri cu ariile numere prime distincte? | [
"Solution:\n\nNo. Let the areas of those triangles be the prime numbers $p$, $q$, $r$ and $t$. But for the areas of the triangles we have $pq = rt$, where the triangles with areas $p$ and $q$ have only a common vertex. This is not possible for distinct primes."
] | JBMO | 7th JBMO | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | No | |
0k82 | Problem:
Three points are chosen inside a unit cube uniformly and independently at random. What is the probability that there exists a cube with side length $\frac{1}{2}$ and edges parallel to those of the unit cube that contains all three points? | [
"Solution:\n\nLet the unit cube be placed on a $x y z$-coordinate system, with edges parallel to the $x$, $y$, $z$ axes. Suppose the three points are labeled $A$, $B$, $C$. If there exists a cube with side length $\\frac{1}{2}$ and edges parallel to the edges of the unit cube that contain all three points, then the... | United States | HMMT February 2019 | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | final answer only | 1/8 | |
0ilt | Problem:
Compute
$$
\int_{0}^{\infty} \frac{e^{-x} \sin (x)}{x} d x
$$ | [
"Solution:\nAnswer: $\\frac{\\pi}{4}$. We can compute the integral by introducing a parameter and exchanging the order of integration:\n$$\n\\begin{aligned}\n\\int_{0}^{\\infty} e^{-x}\\left(\\frac{\\sin (x)}{x}\\right) \\mathrm{d} x & =\\int_{0}^{\\infty} e^{-x}\\left(\\int_{0}^{1} \\cos (a x) \\mathrm{d} a\\right... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | pi/4 | |
0kmd | A school has 100 students and 5 teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are 50, 20, 20, 5, and 5. Let $t$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let... | [] | United States | AMC 12 A | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | MCQ | B | |
0byt | Let $ABCD A'B'C'D'$ be a cube with side length $1$. An ant walks on the cube's faces, starting from $A$ and ending at $C'$. The ant moves only on the cube's edges or on the diagonals of its faces. Knowing that the ant never passes through the same point twice, find the maximal length of such a walk. | [
"We claim that the maximal length of a walk is $3 + 4\\sqrt{2}$. An example of such a walk is $A \\to A' \\to B \\to D \\to D' \\to C \\to B' \\to C'$. Since the cube has $8$ vertices, the ant's walk consists of at most $7$ steps, each of length either $1$ or $\\sqrt{2}$.\n\nThe length of a $6$ steps walk is at mos... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Solid Geometry > 3D Shapes",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 3 + 4√2 | |
0l6b | Problem:
On the perimeter of a unit circle, 12 points are chosen uniformly and independently at random. Estimate the expected value of the area of the convex 12-gon formed by these points.
Submit a positive number $E$ written in decimal. If the correct answer is $A$, you will receive round $\left(20e^{-15|E - A|}\righ... | [
"Solution:\nWe compute the exact answer as given above. Let $n = 12$, and $\\theta_{1}, \\theta_{2}, \\ldots, \\theta_{n}$ be uniformly randomly generated such that $\\theta_{1} + \\cdots + \\theta_{n} = 2\\pi$. We are trying to estimate\n\n$$\n\\mathbb{E}\\left[\\frac{1}{2}\\sum_{i = 1}^{n}\\sin (\\theta_{i})\\rig... | United States | HMMT February | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 33/π − 1485/(2π^3) + 10395/π^5 − 155925/(2π^7) + 467775/(2π^9) ≈ 2.559 | |
04bj | Determine all pairs of prime numbers $p$ and $q$ for which there exists an integer $a$ such that $a^4 = p a^3 + q$. | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (p, q) = (2, 3) | |
0hbn | Let $ABC$ be an acute angle triangle, where $AB < AC$, and let $M$ be the midpoint of $BC$, and $K$ be the midpoint of the polygonal chain $BAC$. Show that $\sqrt{2} KM > AB$.
(Heorhii Naumenko)
 | [
"Let $N$ be the midpoint of $AC$. Since $K$ is the midpoint of polygonal chain $BAC$, the following holds (Fig. 15):\n$$\n\\frac{1}{2}(AB + AC) = KC = KN + NC = KN + \\frac{1}{2}AC \\Rightarrow KN = \\frac{1}{2}AB = NM.\n$$\nBy the cosine theorem for $\\triangle KNM$:\n$$\n\\begin{aligned}\n& KM^2 = KN^2 + NM^2 - 2... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0k0s | Problem:
For some real number $c$, the graphs of the equation $y = |x - 20| + |x + 18|$ and the line $y = x + c$ intersect at exactly one point. What is $c$? | [
"Solution:\n\nWe want to know the value of $c$ so that the graph $|x-20| + |x+18| - x = c$ has one solution. The graph of the function $|x-20| + |x+18| - x$ consists of an infinite section of slope $-3$ for $x \\in (-\\infty, -18]$, then a finite section of slope $-1$ for $x \\in [-18, 20]$, then an infinite sectio... | United States | HMMT February | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 18 | |
0634 | Problem:
Es sei $\Gamma$ der Umkreis des bei $C$ gleichschenkligen Dreiecks $A B C$. Im Inneren der Seite $\overline{B C}$ liege der Punkt $M$. Es gebe einen Punkt $N$ auf dem Strahl $A M$, für den $M$ zwischen $A$ und $N$ liegt und der $|A N|=|A C|$ erfüllt. Der Umkreis des Dreiecks $C M N$ schneide $\Gamma$ in den b... | [
"Solution:\n\nDie Idee zur Lösung geht von der Beobachtung aus, dass die Forderung $|A N|=|A C|=|B C|$ unverändert bleibt, wenn man $A$ und $C$ sowie $N$ und $B$ miteinander vertauscht. Dies legt nahe, dass man auch den Umkreis des Dreiecks $A M B$ betrachten sollte. Der zweite Schnittpunkt der Umkreise von $A M B$... | Germany | 1. Auswahlklausur | [
"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 > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
02b9 | Problem:
Asterisco $-\mathrm{Se}$ $\frac{*}{24}-\frac{3}{8}-\frac{2}{3}=\frac{1}{6}$, quanto vale $*$ ?
(a) 20
(b) 21
(c) 23
(d) 25
(e) 29 | [
"Solution:\n\nA opção correta é (e).\n$$\n\\frac{1}{6}=\\frac{*}{24}-\\frac{3}{8}-\\frac{2}{3}=\\frac{*}{24}-\\left(\\frac{3}{8}+\\frac{2}{3}\\right)=\\frac{*}{24}-\\frac{25}{24}=\\frac{*-25}{24}\n$$\nLogo, $\\frac{*-25}{24}=\\frac{1}{6}=\\frac{4}{24}$, donde $*-25=4$, ou seja, $*=29$."
] | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | e | |
0fl5 | Problem:
Determina los lados del triángulo rectángulo del que se conocen el perímetro, $p=96$, y la altura sobre la hipotenusa, $h=\frac{96}{5}$. | [
"Solution:\nConsideramos el triángulo rectángulo de la figura.\n\nBuscamos relaciones entre estos segmentos.\nEl área del triángulo es: $\\frac{b c}{2}=\\frac{a h}{2}$, de aquí se deduce que\n$$\nb c=a h\n$$\nPor ser $p=a+b+c$, se tiene $b+c=p-a$, luego $(b+c)^{2}=(p-a)^{2}$, y de aquí, uti... | Spain | Spain | [
"Geometry > Plane Geometry > Triangles",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 40, 32, 24 | |
0klp | Problem:
In an $n \times n$ square grid, $n$ squares are marked so that every rectangle composed of exactly $n$ grid squares contains at least one marked square. Determine all possible values of $n$. | [
"Solution:\n\nIn this solution, we will reference cells by row and column, measured from left to right and top to bottom. The cell at $(0,0)$ is then in the top-left corner; the cell at $(1,3)$ is in the fourth cell of the second row.\n\n| $(0,0)$ | | | |\n| :--- | :--- | :--- | :--- |\n| | | | $(1,3)$ ... | United States | HMIC 2021 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1, primes, and prime squares | |
0euu | Find the maximum value of
$$
\frac{1}{a^2 - 4a + 9} + \frac{1}{b^2 - 4b + 9} + \frac{1}{c^2 - 4c + 9}
$$
where $a$, $b$, $c$ are non-negative real numbers satisfying $a + b + c = 1$. | [
"Note that for $0 \\le x \\le 1$ the inequality\n$$\n\\frac{1}{x^2 - 4x + 9} \\le \\frac{x + 2}{18}\n$$\nholds, where the equality holds if and only if $x = 0$ or $x = 1$. Hence,\n$$\n\\frac{1}{a^2 - 4a + 9} + \\frac{1}{b^2 - 4b + 9} + \\frac{1}{c^2 - 4c + 9} \\le \\frac{1}{18}(a + b + c + 6) = \\frac{7}{18}\n$$\nS... | South Korea | 24th Korean Mathematical Olympiad Final Round | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 7/18 | |
07j4 | We call a sequence $(P_n)_{n=1,...}$ of polynomials an *arithmetic sequence with common difference* $Q(x)$ if $P_{n+1} = P_n + Q$, $n = 1, \dots$. Suppose that we have an arithmetic sequence of polynomials with the common difference $Q(x)$ and the first term $P(x)$ such that $P$, $Q$ are *monic* polynomials with intege... | [
"We shall firstly prove that the set of integer roots of the $(P + nQ)_{n=1,...}$ would be unbounded. For sake of this, notice that if $m \\ne n$ the polynomials $P + mQ$, $P + nQ$ have no common integer root; otherwise $P$ and $Q$ would have. Now, let $r_n$ be an integer root of $P + nQ$. It follows that $P(r_n)/Q... | Iran | 41th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
05ko | Problem:
On considère un échiquier classique $8 \times 8$. Un zigzag est un chemin sur les cases blanches, qui part d'une case (quelconque) de la ligne du bas, et monte de ligne en ligne jusqu'à atteindre celle du haut (sur n'importe quelle case) : à chaque étape, on monte d'une case en diagonale. Combien y a-t-il de ... | [
"Solution:\n\nOn écrit un 1 dans chacune des quatre cases blanches du bas. Puis dans la deuxième ligne, on écrit la somme des cases blanches qui y mènent : c'est le nombre de manières d'atteindre ces cases. On recommence dans la troisième ligne, etc. jusqu'à la huitième ligne : sur chacune des quatre cases, on a le... | France | Envoi de combinatoire | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | 296 | |
09ip | Prove that $\underbrace{99\ldots99}_{2997}$ is divisible by $998001$ and find the first four digits of the quotient. | [] | Mongolia | Mongolian Mathematical Olympiad Round 1 | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1002 | |
07md | Determine all pairs of real numbers $(m, c)$ such that for all $x \ge 0$
$$
mx + c \le x^3.
$$ | [
"Suppose $(m, c)$ is an allowable pair. Since the inequality must hold when $x = 0$, we see immediately that $c$ cannot be positive. Moreover,\n$$\nm \\le \\frac{x^3 - c}{x}, \\quad \\forall x > 0,\n$$\nand so, if $c = 0$, then $m \\le 0$. Otherwise, $c < 0$, and so, by the AM-GM inequality,\n$$\n\\begin{aligned}\n... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | {(m, c) : c <= 0 and 27 c^2 - 4 m^3 >= 0} (equivalently, c = -2 s^3 and m <= 3 s^2 for some s >= 0) | |
0ews | Problem:
a. A $6 \times 6$ board is tiled with $2 \times 1$ dominos. Prove that we can always divide the board into two rectangles each of which is tiled separately (with no domino crossing the dividing line).
b. Is this true for an $8 \times 8$ board? | [
"Solution:\n\na.\nWe say a domino bridges two columns if half the domino is in each column. We show that for $0 < n < 6$ the number of dominoes bridging columns $n$ and $n+1$ must be at least $2$ and even.\n\nConsider first $n = 1$. There cannot be $3$ dominos entirely in column $1$, or it would be separately tiled... | Soviet Union | 3rd ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | a: yes; b: no | |
00el | We have 10 bottles, each with 1-liter capacity. Initially, 9 of them are empty and the other is completely filled with orange juice. A move consists of picking a non-empty bottle, dividing its content into 3 equal parts, and placing these 3 parts in any 3 bottles. Is it possible, after a sequence of moves, that all 10 ... | [
"The answer is no. We will show that, at all times, the amount of liters of orange juice in any bottle can be represented as $\\frac{m}{3^k}$ for some nonnegative integers $m, k$. This is clearly true at the beginning of the process. On each move, we pick one bottle that has $\\frac{m_1}{3^{k_1}}$ and add $\\frac{m... | Argentina | Rioplatense Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No | |
00mc | Es sei $S = \{1, 2, \dots, 2017\}$.
Man bestimme die größtmögliche natürliche Zahl $n$, für die es $n$ verschiedene Teilmengen von $S$ gibt, sodass für keine zwei dieser Teilmengen ihre Vereinigung gleich $S$ ist.
(Gerhard Woeginger) | [
"Es gibt $2^{2016}$ Teilmengen von $S$, die das Element $2017$ nicht enthalten. Die Vereinigung von je zwei dieser Teilmengen enthält $2017$ ebenfalls nicht und ist daher ungleich $S$. Daher ist das gesuchte $n$ mindestens $2^{2016}$.\nWenn wir jede Teilmenge von $S$ mit ihrem Komplement zu einem Paar zusammenfasse... | Austria | 48. Österreichische Mathematik-Olympiade | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | German | proof and answer | 2^{2016} | |
0hrk | Problem:
If $n$ is a natural number such that $2n+1$ and $3n+1$ are perfect squares, prove that $5n+3$ can't be a prime number. | [
"Solution:\nSuppose that $5n+3$ is a prime number. Let $x$ and $y$ be natural numbers such that $x^2 = 2n+1$ and $y^2 = 3n+1$. Then\n$$\n5n+3 = 4(2n+1) - (3n+1) = 4x^2 - y^2 = (2x - y)(2x + y).\n$$\nSince $5n+3$ is a prime number and $x$ and $y$ are natural numbers, we must have $2x - y = 1$, implying that $y = 2x ... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0472 | Let $\mathbb{N} = \{0,1,2,\dots\}$ be the set of all non-negative integers. For each $n \in \mathbb{N}$, define the Catalan number
$$
C_n = \frac{1}{n+1} \binom{2n}{n} = \frac{(2n)!}{n!(n+1)!}.
$$
Prove that for any positive integer $m$, we have
$$
\sum_{\substack{i,j,k \in \mathbb{N} \\ i+j+k=m}} C_{i+j}C_{i+k}C_{j+k}... | [
"*Proof 1.* Consider the set of ordered triples:\n$$\nA = \\{ (u, v, w) \\in \\mathbb{N}^3 \\mid u, v, w \\le m; u + v + w = 2m \\},\n$$\n$$\nB = \\{(u, v, w) \\in \\mathbb{N}^3 \\mid u + v + w = 2m\\}, \\quad B_1 = \\{(u, v, w) \\in B \\mid u \\ge m + 1\\},\n$$\n$$\nB_2 = \\{(u, v, w) \\in B \\mid v \\ge m + 1\\},... | China | The 65th IMO China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0563 | A slider on the scrolling bar of Paul's mail client shows the proportion of emails which are preceding the email which is currently open. Paul noticed that before deleting some emails the slider was on 10%. Paul deleted some consecutive emails, starting from the one which was currently open. After that the slider was o... | [
"Let $n$ be the number of emails before deletion. As the slider showed $10\\%$ before deletion there were $0.1n$ emails preceding the email that was open when Paul started the deletion. After the end of the deletion those emails accounted for $50\\%$ from the remaining emails which means that $0.2n$ emails remained... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 80% | |
0goe | Let $S$ denote the set of polynomials $P(x) = a x^2 + b x$ where $0 \leq a, b < 2010^{18}$ are integers. Determine the number of polynomials $P$ in $S$ for which there exists a polynomial $Q$ in $S$ such that $Q(P(n)) \equiv n \pmod{2010^{18}}$ for all integers $n$. | [
"We will show that there exists $Q(x) = c x^2 + d x$ for $P(x) = a x^2 + b x$ if and only if $2^8 1005^9 \\mid a$ and $(2010, b) = 1$. Then it follows that the answer is\n$$\n2 \\cdot 2010^9 \\cdot 2010^{18} \\left(1 - \\frac{1}{2}\\right) \\left(1 - \\frac{1}{3}\\right) \\left(1 - \\frac{1}{5}\\right) \\left(1 - \... | Turkey | 18th Turkish Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Algebra > Algebraic Expressions > Pol... | English | proof and answer | 2^5 * 3 * 11 * 2010^{26} | |
0i43 | Problem:
A man, standing on a lawn, is wearing a circular sombrero of radius $3$ feet. Unfortunately, the hat blocks the sunlight so effectively that the grass directly under it dies instantly. If the man walks in a circle of radius $5$ feet, what area of dead grass will result? | [
"Solution:\n$60\\pi\\ \\mathrm{ft}^2$\n\nLet $O$ be the center of the man's circular trajectory. The sombrero kills all the grass that is within $3$ feet of any point that is $5$ feet away from $O$—i.e., all the grass at points $P$ with $2 \\leq OP \\leq 8$. The area of this annulus is then $\\pi\\left(8^2 - 2^2\\r... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 60π ft^2 | |
0j6u | Problem:
Let $x$ be a real number such that $2^{x} = 3$. Determine the value of $4^{3x+2}$. | [
"Solution:\n\nWe have\n$$\n4^{3x+2} = 4^{3x} \\cdot 4^{2} = \\left(2^{2}\\right)^{3x} \\cdot 16 = 2^{6x} \\cdot 16 = \\left(2^{x}\\right)^{6} \\cdot 16 = 3^{6} \\cdot 16 = 11664\n$$"
] | United States | Harvard-MIT November Tournament | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 11664 | |
08dr | Problem:
Se si taglia un foglio A4 precisamente a metà lungo una retta parallela al lato più corto, si creano due fogli che hanno la stessa forma di quello originale, cioè che si ottengono da esso tramite una rotazione e una riduzione di scala. Quest'anno diremo che un foglio rettangolare è "contemporaneo" se, taglian... | [
"Solution:\n\nLa risposta è $(A)$. Indichiamo con $x$ la lunghezza del lato lungo del foglio contemporaneo. I rettangoli ritagliati hanno lo stesso rapporto tra i lati, ma sappiamo che il loro lato lungo misura $1$, mentre il loro lato corto misura $\\frac{x}{2020}$. Scrivendo l'uguaglianza tra i rapporti, cioè la ... | Italy | Olimpiadi della Matematica | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | A | |
0945 | Problem:
Let $k$ be a positive integer and $a_{1}, a_{2}, \ldots, a_{k}$ be nonnegative real numbers. Initially, there is a sequence of $n \geq k$ zeros written on a blackboard. At each step, Nicole chooses $k$ consecutive numbers written on the blackboard and increases the first number by $a_{1}$, the second one by $... | [
"Solution:\n\nDenote by $L_{i}, 0 \\leq i < n$, the number of tiles that John puts in such a way that $a_{0}$ is added at position $i$. Note that $L_{n-k} = L_{n-k+1} = \\cdots = L_{n-1} = 0$. Analogously, we define $R_{i}, 0 \\leq i < n$ for the number of times $a_{k-1}$ was added at position $i$.\n\nFirst, note t... | Middle European Mathematical Olympiad (MEMO) | MEMO Team Competition | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0kk0 | Problem:
Let $x$, $y$, $z$ be real numbers satisfying
$$
\begin{aligned}
2x + y + 4xy + 6xz &= -6 \\
y + 2z + 2xy + 6yz &= 4 \\
x - z + 2xz - 4yz &= -3
\end{aligned}
$$
Find $x^{2} + y^{2} + z^{2}$. | [
"Solution:\nWe multiply the first, second, and third equations by $\\frac{1}{2}$, $-\\frac{1}{2}$, and $-1$, respectively, then add the three resulting equations. This gives $xy + xz + yz = -2$.\n\nDoing the same with the coefficients $-1$, $2$, and $3$ gives $x + y + z = 5$, from which $(x + y + z)^2 = 25$.\n\nSo ... | United States | HMMT November 2021 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 29 | |
09su | Problem:
Zij $\triangle ABC$ een rechthoekige driehoek met $\angle A = 90^\circ$ en omgeschreven cirkel $\Gamma$. De ingeschreven cirkel raakt aan $BC$ in een punt $D$. Zij $E$ het midden van de boog $AB$ van $\Gamma$ waar $C$ niet op ligt en zij $F$ het midden van de boog $AC$ van $\Gamma$ waar $B$ niet op ligt.
a) ... | [
"Solution:\n\na)\nHet midden $E$ van boog $AB$ waar $C$ niet op ligt, ligt op de bissectrice $CI$. Net zo ligt $F$ op $BI$. Er geldt $\\angle IFC = \\angle BFC = \\angle BAC = 90^\\circ$ omdat $ABCF$ een koordenvierhoek is en $\\angle IDC = 90^\\circ$ omdat $D$ het raakpunt van de ingeschreven cirkel aan $BC$ is. D... | Netherlands | Selectietoets | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0dfh | Prove that for any positive integer $n$ at least one coefficient of the polynomial
$$
(x^4 + x^3 - 3x^2 + x + 2)^n
$$
is negative. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0for | Problem:
Encuentra todas las aplicaciones $f: \mathbb{Z} \rightarrow \mathbb{Z}$ que verifican $f(n)+f(n+1)=2 n+1$ para cualquier entero $n$ y además
$$
\sum_{i=1}^{63} f(i)=2015
$$ | [
"Solution:\n\nNótese que $f(n+1)=2 n+1-f(n)$, con lo que podemos hallar sucesivamente\n$$\nf(1)=1-f(0), \\quad f(2)=3-f(1)=2+f(0), \\quad f(3)=5-f(2)=3-f(0), \\ldots\n$$\n\nEsto nos permite conjeturar que $f(n)=n+(-1)^{n} f(0)$ para todo $n \\geq 0$, cosa que podemos demostrar por inducción, siendo cierto como ya s... | Spain | null | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | f(n) = n + (-1)^n | |
0a63 | Problem:
Let triangle $ABC$ be right-angled at $A$. Let $D$ be the point on $AC$ such that $BD$ bisects angle $\angle ABC$. Prove that $BC - BD = 2AB$ if and only if $\frac{1}{BD} - \frac{1}{BC} = \frac{1}{2AB}$. | [
"Solution:\n\nWlog let $AB = 1$ and $BC = a$. Also let $BD = x$. We will try to find all the lengths in the diagram in terms of $a$.\n\n\n\nBy Pythagoras in $\\triangle ABC$ we get $AC = \\sqrt{a^2 - 1}$. By the angle-bisector theorem we get $\\frac{AD}{DC} = \\frac{AB}{BC} = \\frac{1}{a}$, and so $DC = a \\times A... | New Zealand | NZMO Round One | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0bav | For each positive integer $n$ consider the set $A_n$ of all the numbers obtained by choosing signs in $\pm 1 \pm 2 \pm \cdots \pm n$; for instance, $A_2 = \{-3, -1, 1, 3\}$ and $A_3 = \{-6, -4, -2, 0, 2, 4, 6\}$. Find the cardinal of the set $A_n$. | [
"The difference of any two elements of the set is even, implying that all elements have the same parity.\nWe claim that all numbers between $-\\frac{n(n+1)}{2}$ and $\\frac{n(n+1)}{2}$, sharing the same parity with them belong to the set $A_n$ – and only them. Indeed, let $x \\in A_n$, $x < \\frac{n(n+1)}{2}$. If (... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n(n+1)/2 + 1 | |
0ftk | Problem:
Beweise für jede positive reelle Zahl $a$ und jedes ganze $n \geq 1$ die Ungleichung
$$
a^{n}+\frac{1}{a^{n}}-2 \geq n^{2}\left(a+\frac{1}{a}-2\right)
$$
und bestimme alle Fälle, in denen das Gleichheitszeichen gilt. | [
"Solution:\n\nMultiplikation mit $a^{n}$ und Anwenden der Binomischen Formeln ergibt die äquivalente Ungleichung\n$$\n\\left(a^{n}-1\\right)^{2} \\geq n^{2} a^{n-1}(a-1)^{2}\n$$\nNach AM-GM gilt nun aber\n$$\n\\begin{aligned}\n\\left(a^{n}-1\\right)^{2} & =(a-1)^{2}\\left(1+a+a^{2}+\\ldots+a^{n-1}\\right)^{2} \\\\\... | Switzerland | IMO Selektion | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds exactly when the exponent equals one or the base equals one. | |
0j8x | Problem:
The integers from $1$ to $n$ are written in increasing order from left to right on a blackboard. David and Goliath play the following game: starting with David, the two players alternate erasing any two consecutive numbers and replacing them with their sum or product. Play continues until only one number on t... | [
"Solution:\n\nAnswer: $4022$\n\nIf $n$ is odd and greater than $1$, then Goliath makes the last move. No matter what two numbers are on the board, Goliath can combine them to make an even number. Hence Goliath has a winning strategy in this case.\n\nNow suppose $n$ is even. We can replace all numbers on the board b... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 4022 | |
0bqg | Problem:
Fie $ABC$ un triunghi și punctele $M$, $N$ pe laturile $[AB]$ respectiv $[BC]$ astfel încât $\frac{AM}{MB}=\frac{m}{n}$ și $\frac{BN}{NC}=\frac{n}{p}$, unde $m, n, p$ sunt numere reale pozitive cu proprietatea $p^{2}=mn$. Notăm cu $P$ intersecția dreptelor $CM$ și $AN$. Arătați că $n \overrightarrow{PA}+m \ov... | [] | Romania | OLIMPIADA NATIONALĂ DE MATEMATICĂ ETAPA LOCALĂ | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0hqz | Problem:
Let $a$, $b$, $c$, $x$ be real numbers such that
$$
a x^{2}-b x-c = b x^{2}-c x-a = c x^{2}-a x-b.
$$
Prove that $a = b = c$. | [
"Solution:\nLet $u = a - b$, $v = b - c$, and $w = c - a$. Then $u + v + w = 0$, and we would like to prove that $u = v = w = 0$. We have\n$$\n\\begin{aligned}\na x^{2} - b x - c &= b x^{2} - c x - a \\\\\n(a - b) x^{2} + (-b + c) x + (-c + a) &= 0 \\\\\nu x^{2} - v x - w &= 0.\n\\end{aligned}\n$$\n\nIn a similar m... | United States | Berkeley Math Circle | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof only | null | |
01ss | Given real numbers $a, b, c, d$ such that $\sin a + b > \sin c + d$, $a + \sin b > c + \sin d$, prove that $a + b > c + d$. | [
"By condition, we have\n$$\na + \\sin b > c + \\sin d, \\qquad (1)\n$$\n$$\n\\sin a + b > \\sin c + d. \\qquad (2)\n$$\nSuppose, contrary to our claim, that\n$$\nc+d \\geq a+b. \\tag{3}\n$$\nSumming (1) and (3), we get\n$$\na + \\sin b + c + d > c + \\sin d + a + b \\implies d - \\sin d > b - \\sin b. \\quad (4)\n$... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Functions",
"Precalculus > Trigonometric functions"
] | English | proof only | null | |
0d5f | Find the number of strictly increasing sequences of nonnegative integers with the first term $0$ and the last term $15$, and among any two consecutive terms, exactly one of them is even. | [
"Let $A_{n}$ be the set of such sequences with the last term is $n$ instead of $15$, and $a_{n}=|A_{n}|$. We will show that $(a_{n})$ is in fact the Fibonacci sequence and deduce that $a_{15}=610$.\n\nWe can check easily that $a_{1}=a_{2}=1$. For $n \\geq 1$, we consider the second-last term of each sequence in $A_... | Saudi Arabia | SAMC 2015 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English, Arabic | proof and answer | 610 | |
0bos | Find all positive integers $N$ having an even number of digits (no leading zeroes) such that, if we insert a multiplication sign after the first $n$ digits of $N$, the result of the multiplication is a divisor of $N$. | [
"Let $N = \\overline{a_1a_2\\dots a_n b_1b_2\\dots b_n}$ be the $2n$-digit number, with $a_1 \\ge 1$. Also let $A = \\overline{a_1a_2\\dots a_n}$ and $B = \\overline{b_1b_2\\dots b_n}$, so by hypothesis we are given $10^n A + B = N = kAB$ for some positive integer $k$. Multiplying with $k$ and adding $10^n$, we arr... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integer... | null | proof and answer | All solutions N are:
- Two-digit: N ∈ {11, 12, 15, 24, 36}.
- Four-digit: N = 1352.
- Infinite family (k = 3): For any n ≥ 1, let A = (5·10^{n−1} + 1)/3 and B = (10^n + 2)/3; then N is the concatenation AB (examples: n = 2 gives N = 1734).
- Infinite family (k = 7): For n = 6t + 3 (t ≥ 0), let A = B = (10^n + 1)/7; the... | |
06wn | Determine all integers $n \geqslant 3$ satisfying the following property: every convex $n$-gon whose sides all have length $1$ contains an equilateral triangle of side length $1$.
(Every polygon is assumed to contain its boundary.) | [
"Answer: All odd $n \\geqslant 3$.\n\nFirst we show that for every even $n \\geqslant 4$ there exists a polygon violating the required statement. Consider a regular $k$-gon $A_{0} A_{1}, \\ldots A_{k-1}$ with side length $1$. Let $B_{1}, B_{2}, \\ldots, B_{n / 2-1}$ be the points symmetric to $A_{1}, A_{2}, \\ldots... | IMO | IMO 2021 Shortlisted Problems | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | All odd n ≥ 3 | |
03y0 | Let $x_1, x_2, \dots, x_n$ (where $n \ge 2$) be real numbers with
$$x_1^2 + x_2^2 + \dots + x_n^2 = 1.$$
Prove that
$$
\sum_{k=1}^{n} \left( 1 - \frac{k}{\sum_{i=1}^{n} ix_i^2} \right)^2 \cdot \frac{x_k^2}{k} \le \left( \frac{n-1}{n+1} \right)^2 \sum_{k=1}^{n} \frac{x_k^2}{k}.
$$
Determine when the equality holds. | [
"**Comment:** Expanding the left-hand side of the desired inequality gives\n$$\n\\begin{align*}\n& \\sum_{k=1}^{n} \\left( 1 - \\frac{k}{\\sum_{i=1}^{n} ix_i^2} \\right)^2 \\cdot \\frac{x_k^2}{k} \\\\\n&= \\sum_{k=1}^{n} \\frac{x_k^2}{k} - \\sum_{k=1}^{n} \\frac{2x_k^2}{\\sum_{i=1}^{n} ix_i^2} + \\sum_{k=1}^{n} \\f... | China | China Girls' Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | Equality holds if and only if x2 = x3 = ... = x_{n-1} = 0 and x1^2 = xn^2 = 1/2. | |
0bhb | Find all differentiable functions $f: \mathbb{R} \to \mathbb{R}$, whose derivative is bounded in a neighborhood of the origin and fulfill the condition
$$
x f(x) - y f(y) = (x^2 - y^2) \max(f'(x), f'(y)),
$$
for every real numbers $x$ and $y$. | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All solutions are f(x) = c x for a real constant c. | |
01qb | Let $AA_1$, $BB_1$ be the altitudes of an acute non-isosceles triangle $ABC$. The circumcircle of triangle $ABC$ meets that of triangle $A_1B_1C$ at point $N$ (different from $C$). Let $M$ be the midpoint of $AB$ and $K$ be the intersection point of $CN$ and $AB$.
Prove that the line of centers of the circumcircles of... | [
"Let $O$ and $L$ be the circumcenters of the triangles $ABC$ and $KMC$, respectively, $H$ the orthocenter of the triangle $ABC$. Let $S$ and $P$ be the midpoints of the segments $KM$ and $CH$, respectively. Let $R$ denote the circumradius of the triangle $ABC$. Since $A$, $B$, $A_1$, $B_1$ lie on the circle with $A... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > ... | English | proof only | null | |
0dhk | Given a quadrilateral $ABCD$, the external angle bisectors of $\angle CAD$, $\angle CBD$ intersect at $P$. Show that if $AD + AC = BC + BD$, then $\angle APD = \angle BPC$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | English | proof only | null | |
0gy6 | Find all solutions in positive integer $m, n$ to the equation
$$
m n^2 = 2009(n + 1).
$$ | [
"Obviously, $n$ and $(n+1)$ are coprime numbers. Consequently, for this equation we have that $m$ is divisible by $(n+1)$ and $2009$ is divisible by $n^2$. Since $2009 = 7^2 \\cdot 41$, this can be achieved in two ways: $n=1$ or $n=7$.\n\nIf $n=1$, then $m = 2009 \\cdot 2 = 4018$.\n\nIf $n=7$, then $49m = 2009 \\cd... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (m, n) = (4018, 1) and (328, 7) | |
00xi | Problem:
Find all integers satisfying the equation $2^{x} \cdot (4 - x) = 2x + 4$. | [
"Solution:\nSince $2^{x}$ must be positive, we have $\\frac{2x + 4}{4 - x} > 0$ yielding $-2 < x < 4$. Thus it suffices to check the points $-1, 0, 1, 2, 3$. The three solutions are $x = 0, 1, 2$."
] | Baltic Way | Baltic Way 1992 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 0, 1, 2 | |
03i9 | Problem:
Sketch the graph of $x^{3} + x y + y^{3} = 3$. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
05te | Problem:
Soit $ABC$ un triangle et $\Gamma$ un cercle passant par $A$. On suppose que $\Gamma$ recoupe les segments $[AB]$ et $[AC]$ en deux points, que l'on appelle respectivement $D$ et $E$, et qu'il coupe le segment $[BC]$ en deux points, que l'on appelle $F$ et $G$, de sorte que $F$ se trouve entre $B$ et $G$. Soi... | [
"Solution:\n\nCommençons par tracer une figure.\n\n\nUne première remarque frappante est que $T$ semble être situé sur le cercle $\\Gamma$. Après avoir vérifié qu'il l'était bien sur une deuxième figure, on s'empresse donc de démontrer ce premier résultat. Pour ce faire, on entame donc une ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0l4u | The 27 cells of a $3 \times 9$ grid are filled in using the numbers 1 through 9 so that each row contains 9 different numbers, and each of the three $3 \times 3$ blocks heavily outlined in the example below contains 9 different numbers, as in the first three rows of a Sudoku puzzle.
The numb... | [
"Call 1, 2, 3 *small numbers*, 4, 5, 6 *medium numbers*, and 7, 8, 9 *large numbers*. There are $9!$ ways to fill in the first $3 \\times 3$ block of the grid to contain each of the numbers 1 through 9. Without loss of generality suppose that the first row is 123, the second row is 456, and the third row is 789. No... | United States | AIME I | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 81 | |
0333 | Problem:
The incircle of $\triangle ABC$ is tangent to the sides $AC$ and $BC$, $AC \neq BC$, at points $P$ and $Q$, respectively. The excircles to the sides $AC$ and $BC$ are tangent to the line $AB$ at points $M$ and $N$. Find $\Varangle ACB$ if the points $M, N, P$ and $Q$ are concyclic. | [
"Solution:\n\nThe perpendicular bisector of the segment $AB$ and the bisector of $\\Varangle ACB$ meet at the midpoint $D$ of the arc $\\overparen{AB}$ of the circumcircle of $\\triangle ABC$ which does not contain $C$.\n\n\n\nNext we use the standard notation for the elements of $\\triangl... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof and answer | 90° | |
0dqh | An infinite sequence of integers, $a_0, a_1, a_2, \dots$, with $a_0 > 0$, has the property that for any $n \ge 0$, $a_{n+1} = a_n - b_n$, where $b_n$ is the number having the same sign as $a_n$, but having the digits written in the reverse order. For example if $a_0 = 1210$, $a_1 = 1089$ and $a_2 = -8712$, etc. Find th... | [
"If $a_0$ has a single digit, then $a_1 = 0$. Thus $a_0$ has at least 2 digits. If $a_0 = \\overline{ab} = 10a + b$, then $a_1 = 9(a-b)$ which is divisible by 9. It follows that all subsequent terms are divisible by 9. Checking all 2-digit multiples of 9 shows that eventually 9 appears (Note that $\\overline{ab}$ a... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof and answer | 1012 | |
06n3 | $ABCD$ is a square with side length $1$. $P$ and $Q$ are points on $AB$ and $BC$ respectively such that $BP = BQ = \frac{1}{\sqrt{2}}$. $N$ is the foot of perpendicular from $B$ to $CP$. Find $NQ^2$. | [
"Answer: $\\frac{5 - 2\\sqrt{2}}{6}$\n\nSet the square in the first quadrant of the coordinate plane with $B$ at the origin and $C$ at $(1, 0)$. Then the coordinates of $P$ and $Q$ are $(0, \\frac{1}{\\sqrt{2}})$ and $(\\frac{1}{\\sqrt{2}}, 0)$ respectively.\n\nAs $PC$ has slope $-\\frac{1}{\\sqrt{2}}$, the slope o... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | (5 - 2√2)/6 | |
0dg9 | Let the sequence $a_i$ be defined in the following way: $a_1 = m \in \mathbb{Z}_+$ and inductively $a_{i+1} = a_i + \lfloor \sqrt{a_i} \rfloor$. Prove that the sequence $a_i$ contains infinitely many perfect squares. | [
"Let us analyze the sequence $a_i$ defined by $a_1 = m$ and $a_{i+1} = a_i + \\lfloor \\sqrt{a_i} \\rfloor$.\n\nSuppose at some step $a_k$ is a perfect square, say $a_k = n^2$ for some integer $n \\geq 1$. Then:\n$$\na_{k+1} = a_k + \\lfloor \\sqrt{a_k} \\rfloor = n^2 + \\lfloor n \\rfloor = n^2 + n = n(n+1)\n$$\n\... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
0efc | Problem:
Polona je razdelila pravokotnik na 9 manjših pravokotnikov in na tri izmed štirih vogalnih pravokotnikov zapisala njihove ploščine (glej sliko). Koliko je ploščina četrtega vogalnega pravokotnika?
(A) 9
(B) 13.5
(C) 14
(D) 15
(E) 16
 | [
"Solution:\nOznačimo višini prve in tretje vrstice pravokotnikov z $x$ in $y$, širini prvega in tretjega stolpca pravokotnikov pa $z$ in $w$. Tedaj so ploščine treh vogalnih pravokotnikov s številkami enake $x z = 8$, $x w = 12$, $y z = 10$, ploščina četrtega vogalnega pravokotnika pa je $y w = \\frac{y z \\cdot x ... | Slovenia | Slovenian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
07a8 | Is it possible to write $\binom{n}{2}$ consecutive natural numbers on the edges of a complete graph with $n$ vertices such that for every path (or cycle) of length $3$ with edges $a, b, c$ ($b$ lies between $a, c$) the greatest common divisor of the numbers of edges $a$ and $c$ divides the number of edge $b$? | [
"First we claim that for every positive integer $k$, the edges which their numbers are divisible by $k$ form a cluster. Indeed, suppose on the contrary that $S$ is the largest cluster such that the number of its edges is divisible by $k$ and $v \\notin S$ is another such edge. Applying problem statement on edge $v$... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Other",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | Possible if and only if the graph has three vertices (n = 3); impossible for larger n. | |
07ng | $P$ is a point on a diameter $AB$ of a circle, centre $O$. The points $C$ and $D$ are on the circumference of the circle, on the same side of $AB$, such that $\angle APC = \angle BPD$. Prove that the quadrilateral $PODC$ is cyclic. | [
"Extend $CP$ to meet the circumference at $E$. Then, using the assumption, we get $\\angle BPE = \\angle APC = \\angle DPB$, hence $D$ is the reflection of $E$ in the line $AB$. Therefore, $ED$ is perpendicular to $AB$ and so $\\angle BPE + \\angle PED = 90^\\circ$.\n\n\n\nWe also have $\\a... | Ireland | Ireland | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0jjq | Problem:
2014 triangles have non-overlapping interiors contained in a circle of radius $1$. What is the largest possible value of the sum of their areas? | [
"Answer: N/A This problem turned out to be much trickier than we expected. We have yet to see a complete solution, but let us know if you find one!"
] | United States | HMMT 2014 HMIC | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 1008 * sin(pi / 1008) | |
0chq | Consider a field $\mathbb{L}$ with $q$ elements. Prove that:
a) if $q \equiv 3 \pmod 4$ and $n \in \mathbb{N}$, $n \ge 2$, is a positive integer divisible by $q-1$, then $x^n = (x^2+1)^n$ for any $x \in \mathbb{L}^*$;
b) if there is an integer $n \in \mathbb{N}^*, n \ge 2$, such that $x^n = (x^2+1)^n$ for any $x \in \m... | [
"a) As the multiplicative group $\\mathbb{L}^*$ has order $q-1$, we have for any $x \\in \\mathbb{L}^*$, $x^{q-1} = 1$, and as $n$ is a multiple of $q-1$, we get $x^n = 1$, for any $x \\in \\mathbb{L}^*$.\nBecause $4$ is not a divisor of the order $q-1$ of the group $\\mathbb{L}^*$, there are no elements of order $... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Abstract Algebra > Field Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0bij | We call a composite positive integer $n$ nice if it is possible to arrange its factors that are larger than $1$ on a circle such that two neighboring numbers are not coprime. How many of the elements of the set $\{1, 2, 3, \ldots, 100\}$ are nice? | [
"If $n = pq$, where $p, q$ are distinct primes, it is clear that we can not arrange $p, q$ and $pq$ without $p$ and $q$ being neighbors, therefore $pq$ is not nice.\n\nIf $n$ is not a product of two distinct primes, then $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$, where $k \\ge 1$ and $\\alpha_i \... | Romania | 65th NMO Selection Tests for JBMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 44 | |
080k | Problem:
Qual è il minimo numero di lanci di un dado a 6 facce che si devono effettuare per avere una probabilità superiore al $50\%$ che la somma di tutti i punteggi ottenuti sia maggiore od uguale a $48$? | [] | Italy | Progetto Olimpiadi di Matematica 2000 GARA di SECONDO LIVELLO | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 14 | |
0cxz | How many integers in the set $\{1,2, \ldots, 2010\}$ divide $5^{2010!}-3^{2010!}$? | [
"Let $k \\in \\{1,2, \\ldots, 2010\\}$. If $3 \\mid k$, then $k \\nmid 5^{2010!}-3^{2010!}$. Also, if $5 \\mid k$, then $k \\nmid 5^{2010!}-3^{2010!}$. It follows that any multiple of $3$ or $5$ in the set $\\{1,2, \\ldots, 2010\\}$ is not a divisor of $5^{2010!}-3^{2010!}$. Any number $k$ in $\\{1,2, \\ldots, 2010... | Saudi Arabia | SAMC | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | 1072 | |
0ayc | Problem:
In triangle $ABC$, $AB = 6$, $BC = 10$, and $CA = 14$. If $D$, $E$, and $F$ are the midpoints of sides $BC$, $CA$, and $AB$, respectively, find $AD^{2} + BE^{2} + CF^{2}$. | [
"Solution:\n\nWe use Stewart's Theorem with the median $AD$ (note that $BD = CD = \\frac{BC}{2}$):\n$$\nCA^{2}(BD) + AB^{2}(CD) = BC\\left(AD^{2} + (BD)(CD)\\right) \\rightarrow \\frac{1}{2}\\left(CA^{2} + AB^{2}\\right) = AD^{2} + \\left(\\frac{BC}{2}\\right)^{2}\n$$\nSimilarly, by using Stewart's Theorem with the... | Philippines | 20th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 249 | |
08y7 | There are $4030$ numbers consisting of two of each number lying in between (and including) $1$ and $2015$. Suppose we line up these numbers from left to right. An ordered sequence of $2015$ numbers chosen from this line-up of $4030$ numbers and considered with the order inherited from the original line-up is called a *... | [
"$\\binom{4030}{2015} - 2015$\n\nIn the following, by a subsequence of the given sequence of positive integers we mean a sequence obtained by choosing entries from the original sequence and retaining the order of the choices.\n\nThere are $\\binom{4030}{2015}$ ways of choosing $2015$ numbers from the given sequence... | Japan | Japan 2015 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | binom(4030,2015) - 2015 | |
05ke | Problem:
Un cercle de centre $O$ est inscrit dans un quadrilatère $A B C D$ dont les côtés ne sont pas parallèles. Montrer que le point $O$ coïncide avec le point d'intersection des lignes médianes du quadrilatère si et seulement si $O A \cdot O C=O B \cdot O D$. (Une ligne médiane du quadrilatère est une droite relia... | [
"Solution:\n\nSoient $I$ et $J$ les milieux des côtés $[A B]$ et $[C D]$ respectivement. Les droites $(A B)$ et $(C D)$ se coupent au point $P$. Sans perte de généralité, on peut supposer que $P$ est le point d'intersection des demi-droites $[A B)$ et $[C D)$.\n\n\n\n1) Supposons que $O$ es... | France | OFM | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kcr | Problem:
We call a positive integer $t$ good if there is a sequence $a_{0}, a_{1}, \ldots$ of positive integers satisfying $a_{0}=15$, $a_{1}=t$, and
$$
a_{n-1} a_{n+1} = (a_{n} - 1)(a_{n} + 1)
$$
for all positive integers $n$. Find the sum of all good numbers. | [
"Solution:\n\nBy the condition of the problem statement, we have\n$$\na_{n}^{2} - a_{n-1} a_{n+1} = 1 = a_{n-1}^{2} - a_{n-2} a_{n}\n$$\nThis is equivalent to\n$$\n\\frac{a_{n-2} + a_{n}}{a_{n-1}} = \\frac{a_{n-1} + a_{n+1}}{a_{n}}\n$$\nLet $k = \\frac{a_{0} + a_{2}}{a_{1}}$. Then we have\n$$\n\\frac{a_{n-1} + a_{n... | United States | HMMT February 2020 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 296 | |
0j6o | Problem:
Let $ABC$ be a triangle such that $AB = AC = 182$ and $BC = 140$. Let $X_{1}$ lie on $AC$ such that $CX_{1} = 130$. Let the line through $X_{1}$ perpendicular to $BX_{1}$ at $X_{1}$ meet $AB$ at $X_{2}$. Define $X_{2}, X_{3}, \ldots$, as follows: for $n$ odd and $n \geq 1$, let $X_{n+1}$ be the intersection of... | [
"Solution:\nAnswer: $\\frac{1106}{5}$\nLet $M$ and $N$ denote the perpendiculars from $X_{1}$ and $A$ to $BC$, respectively. Since triangle $ABC$ is isosceles, we have $M$ is the midpoint of $BC$. Moreover, since $AM$ is parallel to $X_{1}N$, we have $\\frac{NC}{X_{1}C} = \\frac{MC}{AC} \\Leftrightarrow \\frac{X_{1... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1106/5 | |
0gi4 | 平面上設點 $O$ 為圓 $\Gamma$ 的圓心, 另在 $\Gamma$ 上有兩點 $A$, $B$ 滿足 $O, A, B$ 不共線。設點 $M$ 為線段 $AB$ 的中點, 並分別在直線 $OA$, $OB$ 上取點 $P$, $Q$ 使得 $P \neq A$ 且 $P, M, Q$ 三點共線。令過 $P$ 與 $AB$ 平行的直線, 和過 $Q$ 與 $OM$ 平行的直線交於點 $X$; 又令過 $X$ 與 $OA$ 平行的直線, 和過 $B$ 與 $OX$ 垂直的直線交於點 $Y$。證明: 若點 $X$ 在圓 $\Gamma$ 上, 則點 $Y$ 也在圓 $\Gamma$ 上。 | [
"不失一般性, 可設 $OA = OB = OX = 1$。設有向角 $\\angle AOB = \\theta$、$\\angle BOX = \\varphi$。又設 $A = \\varphi + \\theta/2$, $B = \\theta/2$。在 $\\triangle OPX$ 上使用正弦定律, 可知有向長度 $OP$ 等於 $\\cos A/ \\cos B$。又在 $\\triangle OQX$ 上使用正弦定律, 可知有向長度 $OQ$ 等於 $\\sin A/ \\sin B$。於是由于孟氏定理,\n$$\n\\frac{\\frac{\\cos A}{\\cos B}}{1 - \\frac{\... | Taiwan | 2023 年台灣數學奧林匹亞考試 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Chinese (Traditional) | proof only | null | |
0f3l | Problem:
Show that there are infinitely many positive integers $n$ such that $[a^{3/2}] + [b^{3/2}] = n$ has at least $1980$ integer solutions. | [
"Solution:\n\nConsider all $a, b$ in the range $1, 2, 3, \\ldots, N^{2}$. There are $N^{4}$ possible pairs of values. But $[a^{3/2}]$ and $[b^{3/2}]$ are in the range $1, 2, \\ldots, N^{3}$, so their sum is in the range $1, 2, \\ldots, 2N^{3}$. Hence one of these values has at least $N/2$ solutions. By taking $N$ s... | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0kjo | Convex quadrilateral $ABCD$ has $AB = 18$, $\angle A = 60^\circ$, and $\overline{AB} \parallel \overline{CD}$. In some order, the lengths of the four sides form an arithmetic progression, and side $AB$ is a side of maximum length. The length of another side is $a$. What is the sum of all possible values of $a$?
(A) 24 ... | [] | United States | AMC 12 A | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | D | |
0iv6 | Problem:
The corner of a unit cube is chopped off such that the cut runs through the three vertices adjacent to the vertex of the chosen corner. What is the height of the cube when the freshly-cut face is placed on a table? | [
"Solution:\n\nThe major diagonal has a length of $\\sqrt{3}$. The volume of the pyramid is $1/6$, and so its height $h$ satisfies $\\frac{1}{3} \\cdot h \\cdot \\frac{\\sqrt{3}}{4}(\\sqrt{2})^{2} = 1/6$ since the freshly cut face is an equilateral triangle of side length $\\sqrt{2}$. Thus $h = \\sqrt{3}/3$, and the... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 2/√3 | |
0cm4 | Problem:
Amy and Bob play the game. At the beginning, Amy writes down a positive integer on the board. Then the players take moves in turn, Bob moves first. On any move of his, Bob replaces the number $n$ on the blackboard with a number of the form $n-a^{2}$, where $a$ is a positive integer. On any move of hers, Amy r... | [
"Solution:\n\nThe answer is in the negative. For a positive integer $n$, we define its square-free part $S(n)$ to be the smallest positive integer $a$ such that $n / a$ is a square of an integer. In other words, $S(n)$ is the product of all primes having odd exponents in the prime expansion of $n$. We also agree th... | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition Day 1 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | No; Amy cannot prevent Bob's win. Bob can always force a win (in at most S(N) moves). | |
0dz7 | Find the smallest three-digit integer with the property that its triple has only even digits. | [
"Denote the three-digit number by $\\overline{abc}$. Its triple is equal to\n$$\n3 \\cdot \\overline{abc} = (3a) \\cdot 100 + (3b) \\cdot 10 + 3c.\n$$\nObviously, $a$ has to be at least 1. If $a = 1$ and we want the digit at the hundreds in $3 \\cdot \\overline{abc}$ to be even, we require $3b \\cdot 10 + 3c \\ge 1... | Slovenia | Slovenija 2008 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 134 | |
0ck5 | Find the real numbers $x$ such that $3^x + 3^{[x]} + 3^{\{x\}} = 4$. | [
"There are no numbers $x \\in [1, \\infty)$ with the property in the statement: if $x \\ge 1$, then $[x] \\ge 1$ and, since $\\{x\\} \\in [0, 1)$, we obtain\n$$\n4 = 3^x + 3^{[x]} + 3^{\\{x\\}} \\ge 3 + 3 + 1 = 7,\n$$\ncontradiction.\n\nThere are no numbers $x \\in (-\\infty, -1)$ with the property in the statement... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | x = 1 - log_3 2 or x = log_3(11/12) | |
080c | Problem:
La tela di un dipinto rettangolare è circondata da un passepartout (cioè un riquadro) largo $10~\mathrm{cm}$. Attorno a quest'ultimo vi è poi una cornice, anch'essa larga $10~\mathrm{cm}$ (nella figura, il rettangolo bianco rappresenta la tela, la superficie tratteggiata il passepartout, la superficie nera la... | [
"Solution:\n\nLa risposta è (D). Siano infatti $a$ e $b$ le dimensioni della tela.\nQuelle del passepartout sono allora $(a+20)$ e $(b+20)$ e quelle della cornice sono $(a+40)$ e $(b+40)$. L'ipotesi dice che si ha $(a+40)(b+40)=2(a+20)(b+20)$ da cui $a b=800$, condizione ovviamente equivalente all'ipotesi stessa.\n... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | D | |
05pn | Problem:
Soit $ABC$ un triangle dont l'orthocentre $H$ est distinct des sommets ainsi que du centre du cercle circonscrit $O$. On désigne par $M, N, P$ les centres des cercles circonscrits aux triangles $HBC$, $HCA$ et $HAB$.
Montrer que les droites $(AM)$, $(BN)$, $(CP)$ et $(OH)$ sont concourantes.
![](attached_im... | [
"Solution:\n\nSoit $H'$ le symétrique de $H$ par rapport à $(BC)$. On sait que $H'$ est sur le cercle circonscrit à $ABC$ (il est facile de vérifier $\\widehat{B H' C} = \\widehat{B H C} = 180^{\\circ} - \\widehat{BAC}$). Le centre du cercle circonscrit à $H'BC$ est donc $O$. Par symétrie par rapport à $(BC)$, le c... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / C... | null | proof only | null | |
0i74 | Problem:
Draw a square of side length $1$. Connect its sides' midpoints to form a second square. Connect the midpoints of the sides of the second square to form a third square. Connect the midpoints of the sides of the third square to form a fourth square. And so forth. What is the sum of the areas of all the squares ... | [
"Solution:\n\nThe area of the first square is $1$, the area of the second is $\\frac{1}{2}$, the area of the third is $\\frac{1}{4}$, etc., so the answer is $1 + \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\cdots = 2$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 2 | |
00ob | Let $h$ be a semicircle with diameter $AB$. The two circles $k_1$ and $k_2$, $k_1 \neq k_2$, touch the segment $AB$ at the points $C$ and $D$, respectively, and the semicircle $h$ from the inside at the points $E$ and $F$, respectively. Prove that the four points $C$, $D$, $E$ and $F$ lie on a circle. | [
"We first consider the case where $C$ and $D$ are both not the center of $AB$, so that the tangents in $C$ and $D$ are both not parallel to $AB$.\n\nThe tangent in $E$ intersects $AB$ in $X$, the tangent in $F$ intersects $AB$ in $Y$ and the two tangents intersect each other in $Z$. Let $I$ now be the intersection ... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
018k | In Greifswald, there are three schools called $A$, $B$, and $C$, each of which is attended by at least one student. Among any three students $a$ from $A$, $b$ from $B$, and $c$ from $C$ there are two knowing each other and two others not knowing each other. Prove that either some student from $A$ knows all students fro... | [
"Assume the contrary and let $a$ be a student from $A$ knowing as many students from $B$ as possible. As $a$ does not know all students from $B$, there is a student $b$ from $B$ not known to $a$. Similarly, we may pick a student $c$ from $C$ not known to $b$ and then a student $a'$ from $A$ not known to $c$. Applyi... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0i26 | Let $a, b$, and $c$ be nonnegative real numbers such that
$$
a^2 + b^2 + c^2 + abc = 4.
$$
Prove that
$$
0 \le ab + bc + ca - abc \le 2.
$$ | [
"**First Solution.** (By Richard Stong) From the condition, at least one of $a, b$, and $c$ does not exceed $1$, say $a \\le 1$. Then\n$$\nab + bc + ca - abc = a(b + c) + bc(1 - a) \\ge 0.\n$$\nTo obtain equality, we have $a(b + c) = bc(1 - a) = 0$. If $a = 1$, then $b + c = 0$ or $b = c = 0$, which contradicts the... | United States | USA IMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Intermediate Algebra > Quadratic functions",
"Geometry > Plane Geometry > Triangles > Triangl... | English | proof only | null | |
0kd6 | Problem:
In acute triangle $A B C$, let $H$ be the orthocenter and $D$ the foot of the altitude from $A$. The circumcircle of triangle $B H C$ intersects $A C$ at $E \neq C$, and $A B$ at $F \neq B$. If $B D = 3$, $C D = 7$, and $\frac{A H}{H D} = \frac{5}{7}$, the area of triangle $A E F$ can be expressed as $\frac{a... | [
"Solution:\n\n\n\nLet $A H$ intersect the circumcircle of $\\triangle A B C$ again at $P$, and the circumcircle of $\\triangle B H C$ again at $Q$. Because $\\angle B H C = 180 - \\angle A = \\angle B P C$, $P$ is the reflection of $H$ over $D$. Thus, we know that $P D = H D$. From power of... | United States | HMMO 2020 | [
"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 > Miscellaneous > Distance chasing"
] | null | final answer only | 12017 | |
0l1e | Problem:
Let $ABC$ be an equilateral triangle. A regular hexagon $P X Q Y R Z$ of side length $2$ is placed so that $P$, $Q$, and $R$ lie on segments $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. If points $A$, $X$, and $Y$ are collinear, compute $BC$. | [
"\n\nNotice that $\\angle QAR = 60^{\\circ}$, and $\\triangle YAR$ is isosceles with base angle $120^{\\circ}$. This implies that $Y$ is the circumcenter of $\\triangle AQR$. Thus, $YA = YR = YQ = 2$. We have $\\angle AYR = 90^{\\circ}$, so $AR = 2\\sqrt{2}$. Moreover, $\\angle AYQ = 150^{\... | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Mi... | null | proof and answer | sqrt(6) + 3sqrt(2) | |
08gr | Problem:
Sono dati, nello spazio, tre punti distinti $X, Y$ e $Z$; ci si chiede se esista un punto $P$ diverso da $X, Y$ e $Z$ tale che le rette $P X, P Y$ e $P Z$ siano a due a due perpendicolari. Quattro amici fanno le seguenti affermazioni:
Alberto: "Esistono $X, Y, Z$ e $P$ appartenenti allo stesso piano che sodd... | [
"Solution:\n\nLa risposta è $(\\mathbf{C})$. L'unico che ha ragione è Carlo.\n\nAlberto ha torto, infatti se esistessero $X, Y, Z, P$ appartenenti allo stesso piano tali che $P X, P Y$ e $P Z$ sono a due a due perpendicolari allora $P Z \\perp P X, P Z \\perp P Y$ e quindi $P Y \\parallel P X$, che è assurdo.\n\nBa... | Italy | Olimpiadi di Matematica | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | MCQ | C | |
08tm | $$
\frac{3}{2} < \frac{4a+b}{a+4b} + \frac{4b+c}{b+4c} + \frac{4c+a}{c+4a} < 9
$$
for positive real numbers $a, b, c$. | [
"Let\n$$\nS = \\frac{4a+b}{a+4b} + \\frac{4b+c}{b+4c} + \\frac{4c+a}{c+4a}\n$$\nWe will show that $\\frac{3}{2} < S < 9$.\n\nFirst, let us show that $\\frac{3}{2} < S$. Let us suppose that $a$ is greater than or equal to $b$ and $c$. Then,\n$$\n\\begin{aligned}\n\\frac{4a+b}{a+4b} &= \\frac{(a+b)+3a}{(a+b)+3b} \\ge... | Japan | Japan Junior Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0csi | В выпуклом $n$-угольнике проведено несколько диагоналей. Проведённая диагональ называется хорошей, если она пересекается (по внутренним точкам) ровно с одной из других проведённых диагоналей. Найдите наибольшее возможное количество хороших диагоналей.
(С. Берлов) | [
"**Ответ.** $n - 2$ при чётных $n$, $n - 3$ при нечётных $n$.\n\nМы будем пользоваться следующей известной леммой.\n\n**Лемма.** В выпуклом $n$-угольнике нельзя провести более $n - 3$ диагоналей, не имеющих общих внутренних точек.\n\nСначала докажем индукцией по $n$, что количество хороших диагоналей не превосходит... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n − 2 for even n; n − 3 for odd n | |
03q2 | Suppose in the tetrahedron $ABCD$, $AB = 1$, $CD = \sqrt{3}$, the distance and angle between the lines $AB$ and $CD$ are $2$ and $\frac{\pi}{3}$ respectively. Then the volume of the tetrahedron equals ( ).
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{1}{2}$
(C) $\frac{1}{3}$
(D) $\frac{\sqrt{3}}{3}$ | [
"As in the diagram, from point $C$ draw a line $CE$ such that it is equal and parallel to $AB$. Construct a prism $ABF$-$ECD$ with $\\triangle CDE$ as base and $BC$ as a lateral edge. Denote $V_1$ as the volume of the tetrahedron and $V_2$ the volume of the prism, then $V_1 = \\frac{1}{3} V_2$.\n\n | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Other 3D problems"
] | English | MCQ | B | |
0le1 | Consider a rectangle board $ABCD$ of size $m \times n$ with $(m + 1) \times (n + 1)$ intersections. Some engineers want to build a route from $A$ which goes along the segments parallel to the sides of the board, passes through each intersection exactly once and finally turns back to $A$.
a. Prove that they can build t... | [
"We number the rows $1$ to $m+1$ from left to right, the columns $1$ to $n+1$ from top to bottom and suppose that the point $A$ is $(1, 1)$ (at the top left of the board).\n\na. *Necessary condition:* The route can be written as a letter sequence consisting of $L, R, U$ and $D$, which respectively represents left, ... | Vietnam | VN IMO Booklet | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) A route exists if and only if at least one of the side lengths is odd. b) The minimum number of turns is: if both side lengths are odd, then two times the smaller plus one; if one is even and the other is odd, then two times the odd one plus one. | |
0bg3 | Find all natural numbers $m, n$ so that $85^m - n^4 = 4$. | [
"$$(n - 1)^2 = 5^m - 1 \\quad \\text{and} \\quad (n + 1)^2 = 17^m - 1.$$\nFor $m > 1$ there are many (more than one) perfect squares between $5^m - 1$ and $17^m - 1$ (e.g. $9^m$ and $16^m$), therefore $m = 1$ and $n = 3$."
] | Romania | The Danube Mathematical Competition | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | m = 1, n = 3 | |
0i1y | Problem:
Call three sides of an opaque cube adjacent if someone can see them all at once. Draw a plane through the centers of each triple of adjacent sides of a cube with edge length $1$. Find the volume of the closed figure bounded by the resulting planes. | [
"Solution:\n\nThe volume of the figure is half the volume of the cube (which can be seen by cutting the cube into $8$ equal cubes and realizing that the planes cut each of these cubes in half), namely $\\frac{1}{2}$."
] | United States | Harvard-MIT Math Tournament | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 1/2 | |
0bol | a) Show that, if $I \subset \mathbb{R}$ is a closed bounded interval, and $f: I \to \mathbb{R}$ is a non-constant monic polynomial function such that $\max_{x \in I} |f(x)| < 2$, then there exists a non-constant monic polynomial function $g: I \to \mathbb{R}$ such that $\max_{x \in I} |g(x)| < 1$.
b) Show that there e... | [
"a) Let $I \\subset \\mathbb{R}$ be a closed bounded interval, let $P(I)$ be the set of all polynomial functions $f: I \\to \\mathbb{R}$, and let $\\|f\\| = \\max_{x \\in I} |f(x)|$. Define $A: P(I) \\to P(I)$ by $Af(x) = (f(x))^2 - \\frac{1}{2}\\|f\\|^2$, $x \\in I$. If $f$ is monic (respectively, non-constant), t... | Romania | 2015 Thirteenth IMAR Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0ahq | Determine all positive integers $x$, $y$ and $z$ such that
$$
x^5 + 4^y = 2013^z.
$$ | [
"Note that $2013$ is divisible by $11$, and that $x^5$ is congruent with $0$, $1$ or $-1$ modulo $11$. Since $4^y$ is congruent $4$, $5$, $9$, $3$ or $1$ modulo $11$, we get that $x^5 = -1 \\pmod{11}$, and $4^y = 1 \\pmod{11}$. Therefore $5|y$, and the given equation is of the form $a^5 + b^5 = 2013^z = 3^z \\cdot ... | North Macedonia | Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | no positive integer solutions | |
00wg | Problem:
Let $*$ denote an operation, assigning a real number $a * b$ to each pair of real numbers $(a, b)$ (e.g., $a * b = a + b^{2} - 17$). Devise an equation which is true (for all possible values of variables) provided the operation $*$ is commutative or associative and which can be false otherwise. | [
"Solution:\n\nA suitable equation is $x * (x * x) = (x * x) * x$ which is obviously true if $*$ is any commutative or associative operation but does not hold in general, e.g., $1 - (1 - 1) \\neq (1 - 1) - 1$."
] | Baltic Way | Baltic Way | [
"Algebra > Abstract Algebra > Other"
] | null | proof and answer | x * (x * x) = (x * x) * x | |
04te | Let $ABC$ be an acute triangle with altitudes $AK$, $BL$, $CM$. Prove that triangle $ABC$ is isosceles if and only if
$$
AM + BK + CL = AL + BM + CK.
$$ | [
"Points $K$, $L$, $M$ are defined as feet of altitudes and we need to express perpendicularity. Since the statement involves lengths and not angles, we characterise the perpendicularity by lengths of segments.\nComparing Pythagorean theorems in triangles $AMC$, $BMC$ we learn $AC^2 - BC^2 = AM^2 - BM^2$. Denoting t... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0924 | Problem:
Let $A B C$ be an acute triangle. Construct a triangle $P Q R$ such that $A B=2 P Q$, $B C=2 Q R$, $C A=2 R P$, and the lines $P Q$, $Q R$, and $R P$ pass through the points $A$, $B$, and $C$, respectively. (All six points $A, B, C, P, Q$, and $R$ are distinct.) | [
"\nSince the angles of triangles $P Q R$ and $A B C$ are equal, $\\angle Q A B=\\angle R B C=\\angle P C A$. Let $S$ denote the Brocard point of $A B C$, the one for which $\\angle S A B=\\angle S B C=\\angle S C A$.\nThe circle $A P C$ is tangent to $A B$ by the reverse tangent-chord theor... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellan... | null | proof only | null | |
07mf | The functions $f(n) = n^2$ and $f(n) = -n^2$ both satisfy the following equations:
$$
f(f(n) + 1) - f(f(n) - 1) = 4n^2
$$
Either show that there is no other function $f : \mathbb{Z} \to \mathbb{Z}$ satisfying the above equation for all $n \in \mathbb{Z}$, or prove that another such function exists. | [
"There are infinitely many solutions. We first examine some constraints.\nIf $f(n) = f(m)$ for $n, m \\ge 0$, then $4n^2 = 4m^2$ and so $n = m$. Similarly $n = m$ if $f(n) = f(m)$ for $n, m \\le 0$. Now let $a = f(0)$ and substitute $n = 0$ in the given functional equation. We obtain $f(a+1) = f(a-1)$, hence $a+1$ ... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
04ry | Let $n$ be a natural number whose all positive divisors are denoted as $d_1, d_2, \dots, d_k$ in such a way that $d_1 < d_2 < \dots < d_k$ (thus $d_1 = 1$ and $d_k = n$). Determine all the values of $n$ for which both equalities $d_5 - d_3 = 50$ and $11d_5 + 8d_7 = 3n$ hold. (Matúš Harminc) | [
"We distinguish whether $n$ is odd or even.\n\ni) The case of $n$ odd. Since all the $d_i$'s are odd too, it follows from $11d_5+8d_7 = 3n$ that $d_7 \\mid 11d_5$ as well as $d_5 \\mid 8d_7$, hence $d_5 \\mid d_7$. In view of $d_7 > d_5$, the relations $d_5 \\mid d_7 \\mid 11d_5$ imply that $d_7 = 11d_5$. Substitut... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 2013 | |
08v9 | Suppose a positive integer $n$ is a perfect square. Consider the set of numbers that can be represented as the product of 2 numbers, both of which are greater than or equal to $n$ (2 numbers may be the same). Express the number, which is the $n$-th smallest in this set, in terms of $n$. | [
"We will show that the desired answer is $(n + \\sqrt{n} - 1)^2$.\n\nFor a positive integer $c$ greater than or equal to $2n$, let us call a number a *c*-product if it can be represented as a product of 2 numbers greater than or equal to $n$ whose sum equals $c$. We also denote for any real number $x$ the greatest ... | Japan | Japan Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | (n + sqrt(n) - 1)^2 |
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