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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0fc6 | Problem:
Sean $a$ y $b$ dos números positivos primos entre sí. Se dice que un entero positivo $n$ es débil si no puede ser escrito en la forma $n=a x+b y$, para algunos enteros $x$ e $y$ no negativos. Prueba que si $n$ es débil y $n<\frac{a b}{6}$, entonces existe un entero $k \geq 2$, tal que $k n$ es débil. | [
"Solution:\n\nTrivialmente se observa que la suma de enteros positivos no débiles es no débil. Esto motiva considerar para cada entero positivo $n$ los enteros $2 n$ y $3 n$. Si ambos no son débiles, entonces $k n$ no es débil para cada $k \\geq 2$, ya que $k n$ puede ser escrito en la forma $2 n r+3 n s$ para algu... | Spain | LIV Olimpiada matemática Española (Concurso Final) | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
04jg | Let $n$ be a positive integer and let $S_n = \sum_{k=1}^{n} k!(k^2 + k + 1)$. Determine $\frac{S_n + 1}{(n+1)!}$. | [] | Croatia | Croatia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n+1 | |
06pb | Let $a_{1}, a_{2}, \ldots, a_{100}$ be nonnegative real numbers such that $a_{1}^{2}+a_{2}^{2}+\ldots+a_{100}^{2}=1$. Prove that
$$
a_{1}^{2} a_{2}+a_{2}^{2} a_{3}+\ldots+a_{100}^{2} a_{1}<\frac{12}{25}
$$ | [
"Let $S=\\sum_{k=1}^{100} a_{k}^{2} a_{k+1}$. (As usual, we consider the indices modulo $100$, e.g. we set $a_{101}=a_{1}$ and $a_{102}=a_{2}$.)\nApplying the Cauchy-Schwarz inequality to sequences $\\left(a_{k+1}\\right)$ and $\\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\\right)$, and then the AM-GM inequality to numbers $a... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0hoc | Problem:
Some soldiers are standing in a line in the east-west direction, each of them facing north. Their officer commands, "Right face!" They should now all be facing east, but, as they are at the very beginning of their military career, some of them get the order wrong and turn to the west. Every soldier who is the... | [
"Solution:\n\nGive the soldiers ID numbers from $1$ upwards from west to east. At any moment, define the confusion index to be the sum of the ID numbers of the soldiers who are facing west. Note that whenever a pair of soldiers turn around, soldier $n+1$ turns from west to east (decreasing the confusion index by $n... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0che | Let $A \in \mathcal{M}_n(\mathbb{C})$ be a matrix with the property $A^T = -A$, where $A^T$ is the transpose of $A$.
a) If $A \in \mathcal{M}_n(\mathbb{R})$ and $A^2 = O_n$, prove that $A = O_n$.
b) If $n$ is an odd natural number and there is a matrix $B \in \mathcal{M}_n(\mathbb{C})$ such that $A$ is the adjoint of... | [
"a) Assume $A = (a_{ij})_{1 \\le i,j \\le n}$ and $A^2 = (m_{ij})_{1 \\le i,j \\le n}$. The property $A^T = -A$ leads to the relations $a_{ji} = -a_{ij}$, for $i, j = 1, \\dots, n$, that is $A$ is antisymmetric. Then\n$$\nm_{ii} = \\sum_{j=1}^{n} a_{ij} a_{ji} = - \\sum_{j=1}^{n} a_{ij}^2, \\quad i = 1, \\dots, n.\... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
01vq | Consider the expression $M(n, m) = |n\sqrt{n^2 + a} - bm|$, where $n$ and $m$ are arbitrary positive integers and the numbers $a$ and $b$ are fixed, moreover $a$ is an odd positive integer, and $b$ is a rational number with an odd denominator of its representation as an irreducible fraction.
Prove that there is
a) no ... | [
"a) The solution of part **a)** is almost obvious and its statement holds for any $b \\in \\mathbb{Q}$ and $a \\in \\mathbb{N}$. Indeed, if $n\\sqrt{n^2+a} = bm$, then, since $b \\in \\mathbb{Q}$, the number $\\sqrt{n^2+a}$ must be rational and therefore integer. Hence $n^2+a = k^2$ for some positive integer $k$. B... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
06w8 | For every integer $n \geqslant 1$ consider the $n \times n$ table with entry $\left\lfloor\frac{ij}{n+1}\right\rfloor$ at the intersection of row $i$ and column $j$, for every $i=1, \ldots, n$ and $j=1, \ldots, n$. Determine all integers $n \geqslant 1$ for which the sum of the $n^{2}$ entries in the table is equal to ... | [
"Answer: All integers $n$ for which $n+1$ is a prime.\n\nFirst, observe that every pair $x, y$ of real numbers for which the sum $x+y$ is integer satisfies\n$$\n\\lfloor x\\rfloor+\\lfloor y\\rfloor \\geqslant x+y-1 \\tag{1}\n$$\nThe inequality is strict if $x$ and $y$ are integers, and it holds with equality other... | IMO | IMO 2021 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | All integers n such that n+1 is prime. | |
09co | $\Delta ABC$ өгчээ. $A$ оройд харгалах гадаад багтсан тойгийн төв нь $J$ байг. Энэ тойрог нь $BC$ хэрмийг $M$ цэгт, харин $AB$ ба $AC$ ба $AC$ шулуунуудыг $K$ ба $L$ цэгүүдэд шүргэнэ. $LM$ ба $BJ$ шулуунууд $F$ цэгт, $KM$ ба $CJ$ шулуунууд $G$ цэгт огтоллддог. $AF$ ба $BC$ шулуунуудын огтлоодлын цэг $S$, $AG$ ба $BC$ ш... | [
"Let $\\alpha = \\angle CAB$, $\\beta = \\angle ABC$ and $\\gamma = \\angle BCA$. The line $AJ$ is the bisector of $\\angle CAB$, so $\\angle JAK = \\angle JAL = \\frac{\\alpha}{2}$. By $\\angle AKJ = \\angle ALJ = 90^\\circ$ the points $K$ and $L$ lie on the circle $\\omega$ with diameter $AJ$.\n\nThe triangle $KB... | Mongolia | ОУМО-53 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Mongolian | proof only | null | |
05c9 | On a plane, 5 points are chosen arbitrarily. Find the largest possible number of distinct right triangles with all vertices in the chosen points. | [
"A square $ABCD$ and its centre $E$ determine 8 distinct right triangles: $ABC$, $BCD$, $CDA$, $DAB$, $AEB$, $BEC$, $CED$, $DEA$.\n\n\nFig. 48\n\nWe show that more than 8 right triangles is impossible. Firstly, note that among any 5 points, one can choose 4 points that are vertices of a rec... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | English | proof and answer | 8 | |
0i84 | Determine all pairs of positive integers $(a, b)$ such that
$$
\frac{a^2}{2ab^2 - b^3 + 1}
$$
is a positive integer. | [
"**First Solution.** (Based on work by Anders Kaseorg) Rewrite equation $(*)$ as $a^2 - 2ab^2k = -b^3k + k$. Adding $b^4k^2$ to both sides completes the square on the left-hand side and gives\n$$\n(kb^2 - a)^2 = b^4k^2 - b^3k + k,\n$$\nor\n$$\n(2kb^2 - 2a)^2 = (2b^2k)^2 - 2b(2b^2k) + 4k.\n$$\nCompleting the square ... | United States | USA IMO 2003 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | (a, b) = (2t, 1), (t, 2t), (8t^4 - t, 2t) for all positive integers t | |
0ju8 | Problem:
In a $17 \times 17$ matrix $M$, all entries are $\pm 1$. The maximum possible value of $|\operatorname{det} M|$ is $N$. Estimate $N$.
An estimate of $E>0$ earns $\left\lfloor 20 \min (N / E, E / N)^{2}\right\rfloor$ points. | [
"Solution:\n\nThis is Hadamard's maximal determinant problem. There's an upper bound of $n^{\\frac{1}{2} n}$ which empirically seems to give reasonably good estimates, but in fact this is open for general $n$."
] | United States | HMMT February | [
"Algebra > Linear Algebra > Determinants",
"Algebra > Linear Algebra > Matrices"
] | null | final answer only | null | |
0jqa | Problem:
Let $S$ be a subset of the set $\{1,2,3, \ldots, 2015\}$ such that for any two elements $a, b \in S$, the difference $a-b$ does not divide the sum $a+b$. Find the maximum possible size of $S$. | [
"Solution:\nAnswer: $672$\nFrom each of the sets $\\{1,2,3\\}, \\{4,5,6\\}, \\{7,8,9\\}, \\ldots$ at most $1$ element can be in $S$. This leads to an upper bound of $\\left\\lceil\\frac{2015}{3}\\right\\rceil = 672$ which we can obtain with the set $\\{1,4,7, \\ldots, 2014\\}$."
] | United States | HMMT November 2015 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 672 | |
083e | Problem:
Sia $B$ un punto interno al segmento $AC$ con $AB$ di lunghezza $2$ e $BC$ di lunghezza $3$. Costruiti i triangoli equilateri $ABA'$ e $CBC'$, dalla stessa parte rispetto al segmento $AC$, quanto misura l'area del triangolo $A'BC'$?
(A) $\frac{3}{2} \sqrt{3}$
(B) $3$
(C) $3 \sqrt{3}$
(D) $5$
(E) $\frac{25}{8... | [] | Italy | UNIONE MATEMATICA ITALIANA Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | MCQ | A | |
0ka3 | Problem:
Tessa the hyper-ant has a 2019-dimensional hypercube. For a real number $k$, she calls a placement of nonzero real numbers on the $2^{2019}$ vertices of the hypercube $k$-harmonic if for any vertex, the sum of all 2019 numbers that are edge-adjacent to this vertex is equal to $k$ times the number on this vert... | [
"Solution:\n\nBy adding up all the equations on each vertex, we get $2019 S = k S$ where $S$ is the sum of all entries, so $k = 2019$ unless $S = 0$. In the latter case, by adding up all the equations on a half of the cube, we get $2018 S - S = k S$ where $S$ is the sum of all entries on that half of the cube, so $... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Linear Algebra > Vectors",
"Discrete Mathematics > Other"
] | null | proof and answer | 2040200 | |
0f3o | Problem:
$x_0$ is a real number in the interval $(0, 1)$ with decimal representation $0.d_1 d_2 d_3 \ldots$. We obtain the sequence $x_n$ as follows. $x_{n+1}$ is obtained from $x_n$ by rearranging the 5 digits $d_{n+1}$, $d_{n+2}$, $d_{n+3}$, $d_{n+4}$, $d_{n+5}$. Show that the sequence $x_n$ converges. Can the limit... | [] | Soviet Union | ASU | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | The sequence always converges because each decimal position is altered only finitely many times. Yes, the limit can be irrational even when the start is rational; for example, with x0 = 1/99999 (decimal repeating 00001), one can choose the digit fixed at each position to create a nonperiodic limit. For the last part, a... | |
0avw | Problem:
Find all positive real numbers $a, b, c, d$ such that for all $x \in \mathbb{R}$,
$$
(a x+b)^{2016}+\left(x^{2}+c x+d\right)^{1008}=8(x-2)^{2016}
$$ | [
"Solution:\nCompare coefficients of $x^{2016}$ in the equation to obtain $a^{2016}+1=8$, i.e. $a=7^{1/2016}$. Then, take $x=2$ to obtain\n$$\n(2 a+b)^{2016}+(4+2 c+d)^{1008}=0\n$$\nSince the LHS is a sum of even-exponent powers, the equation will be solved in $\\mathbb{R}$ if and only if both addends are zero. In p... | Philippines | 18th PMO National Stage Oral Phase | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a = 7^{1/2016}, b = -2·7^{1/2016}, c = -4, d = 4 | |
0aj7 | Let $d(n)$ denote the number of positive divisors of $n$. For positive integer $n$ we define $f(n)$ as
$$
f(n) = d(k_1) + d(k_2) + d(k_3) + \dots + d(k_m),
$$
where $1 = k_1 < k_2 < \dots < k_m = n$ are all divisors of the number $n$. We call an integer $n > 1$ almost perfect if $f(n) = n$. Find all almost perfect numb... | [
"Alternative way to define $f(n)$ is\n$$\nf(n) = \\sum_{k|n, k \\ge 1} d(k).\n$$\nLet $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_r^{\\alpha_r}$ be the prime factorization of $n$. We have $d(n) = \\prod_{i=1}^r (\\alpha_i + 1)$.\nWe prove the function $f$ is multiplicative, in particular, given coprime $n, m$ we ... | North Macedonia | European Mathematical Cup | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 3, 18, 36 | |
02m5 | For each positive integer $n$ let $f(n)$ be the number of products of integers bigger than $1$ whose result is at most $n$, i.e. $f(n)$ is the number of $k$-uples $(a_1, a_2, \dots, a_k)$ where $k$ is a natural number, $a_i \ge 2$ is an integer for all $i$ and $a_1 \cdot a_2 \cdot \dots a_k \le n$ (include, by conventi... | [
"Extend the definition of $f$ to real numbers, that is, $f(x)$ is the number of $k$-uples whose product is at most $x$, $x \\in \\mathbb{R}$. If the last number in a $k$-uple is $m$, then we obtain a product that doesn't exceed $x/m$. Conversely, given a number $m$ and a product that doesn't exceed $x/m$ we obtain ... | Brazil | XXXI Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | English | proof only | null | |
03yo | Given that $0 < x, y < 1$, determine, with proof, the maximum value of $\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)}$. | [
"When $x = y = \\frac{1}{3}$, the value of the expression is $\\frac{1}{8}$.\n\nWe will prove that $\\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)} \\le \\frac{1}{8}$ for any $0 < x, y < 1$ as follows.\n\nIf $x + y \\ge 1$, then $\\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)} \\le 0 < \\frac{1}{8}$.\n\nIf $x + y < 1$, then let $1 - x - y =... | China | China Western Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 1/8 | |
0bd9 | Problem:
Adottak az $a, b \in \mathbb{R}$ és $z \in \mathbb{C} \setminus \mathbb{R}$ számok úgy, hogy fennálljon az $|a-b|=|a+b-2z|$ egyenlőség.
a)
Igazold, hogy a $|z-a|^{x} + |\bar{z}-b|^{x} = |a-b|^{x}$ egyenletnek egy és csak egy $x \in \mathbb{R}$ megoldása van!
b)
Határozd meg azokat az $x \in \mathbb{R}$ szám... | [] | Romania | Matematika tantárgyverseny Megyei szakasz | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | a) x = 2. b) All real x with x ≥ 2. | |
0cnq | Given a positive integer $n > 1$. An integer $a > n^2$ is chosen so that for each $i = 1, 2, \ldots, n$, the set $\{a + 1, a + 2, \ldots, a + n\}$ contains a multiple of the number $n^2 + i$. Prove that $a > n^4 - n^3$. (A. Golovanov) | [
"Заметим, что разность между любыми двумя числами вида $a+i$ ($i = 1, \\ldots, n$) не превосходит $n-1$.\nПусть кратное числу $n^2 + i$, содержащееся среди наших чисел — это $a_i(n^2 + i)$. Ясно, что $a_1 > 1$. Тогда найдется такое $1 \\le i \\le n-1$, что $a_i > a_{i+1}$ (в противном случае $a_1 \\le a_2 \\le \\do... | Russia | Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof only | null | |
0029 | Utilizando triangulitos equiláteros de cartón de lado $1$ se forma un triángulo equilátero de lado $2^{2004}$. A este triángulo se le extrae el triangulito de lado $1$ cuyo centro coincide con el centro del triángulo mayor.
Determinar si es posible cubrir totalmente la superficie restante, sin superposiciones ni hueco... | [] | Argentina | 15ª Olimpiada Matemática del Cono Sur | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | español | proof and answer | Yes | |
0l8s | Let be given an integer $n \ge 1$. Consider a permutation $(a_1, a_2, ..., a_{2n})$ of the first $2n$ positive integers such that the numbers $|a_{i+1} - a_i|$ $(i = 1, 2, ..., 2n-1)$ are distinct each from others.
Prove that $a_1 - a_{2n} = n$ if and only if $1 \le a_{2k} \le n$ for every $k = 1, 2, ..., n$. | [
"a) Sufficient condition. As $1 \\le a_{2k} \\le n$ ($\\forall k = 1, 2, \\dots, n$), we have:\n$$\nT = \\sum_{i=1}^{2n-1} |a_{i+1} - a_i| = 2(a_1 + a_3 + \\dots + a_{2n-1}) - 2(a_2 + a_4 + \\dots + a_{2n}) + a_{2n} - a_1 = 2n^2 + a_{2n} - a_1.\n$$\nOn the other hand, since $1 \\le |a_{i+1} - a_i| \\le 2n - 1$ $\\f... | Vietnam | VIETNAMESE MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0jbz | Problem:
Triangle $ABC$ is an equilateral triangle with side length $1$. Let $X_{0}, X_{1}, \ldots$ be an infinite sequence of points such that the following conditions hold:
- $X_{0}$ is the center of $ABC$.
- For all $i \geq 0$, $X_{2i+1}$ lies on segment $AB$ and $X_{2i+2}$ lies on segment $AC$.
- For all $i \geq ... | [
"Solution:\n\n$\\boxed{\\sqrt{\\dfrac{\\sqrt{6}}{3}}}$\n\nLet $Y$ be the foot of the perpendicular from $A$ to $X_{0} X_{1}$: note that the sum we wish to maximize is simply $X_{0}Y + YA$. However, it is not difficult to check (for example, by AM-GM) that $AY + YX_{0} \\geq \\sqrt{2} \\cdot AX_{0} = \\dfrac{\\sqrt{... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(6)/3 | |
05p7 | Problem:
Combien de tableaux $3 \times 3$ peut-on construire en les remplissant avec les nombres de 1 à 3 tels qu'il n'y ait pas deux fois le même nombre dans une ligne ni dans une colonne. Et de tableaux $4 \times 4$ avec les nombres de 1 à 4 ? | [
"Solution:\nNous présentons ci-dessous une preuve complète de l'exercice. Il n'était pas nécessaire d'être aussi précis pour avoir tous les points mais il est bon, lorsque que l'on compte des objets, d'avoir une idée de la preuve pour démontrer que nous n'avons rien oublié ni compté en double.\nNous avons six façon... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 3x3: 12; 4x4: 576 | |
0b0e | Problem:
Last December 7, a computer owned by Patrick Laroche from Florida, USA determined that the number $2^{82,589,933}-1$ is a prime number. This number had a whopping $24,862,048$ digits, and is currently the largest known prime number. The computer used software provided by the GIMPS, which is a distributed comp... | [
"Solution:\n\nMersenne primes"
] | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | Mersenne primes | |
0jvc | Problem:
What is the minimum value of the product
$$
\prod_{i=1}^{6} \frac{a_{i}-a_{i+1}}{a_{i+2}-a_{i+3}}
$$
given that $\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\right)$ is a permutation of $(1,2,3,4,5,6)$? (note $a_{7}=a_{1}, a_{8}=a_{2}$, etc.) | [
"Solution:\nAnswer: $1$\nThe product always evaluates to $1$."
] | United States | HMMT November | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 1 | |
0bl0 | Consider numbers $7^{n+7}, 11^{n+11}, 15^{n+15}, \dots, 4007^{n+4007}$, where $n$ is a positive integer. Prove that at least ten non-nil differences of the given numbers are divisible by $500$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0j17 | Problem:
How many ways are there to choose 2010 functions $f_{1}, \ldots, f_{2010}$ from $\{0,1\}$ to $\{0,1\}$ such that $f_{2010} \circ f_{2009} \circ \cdots \circ f_{1}$ is constant? Note: a function $g$ is constant if $g(a)=g(b)$ for all $a, b$ in the domain of $g$. | [
"Solution:\n\n$4^{2010}-2^{2010}$\n\nIf all 2010 functions are bijective, then the composition $f_{2010} \\circ f_{2009} \\circ \\cdots \\circ f_{1}$ will be bijective also, and therefore not constant. If, however, one of $f_{1}, \\ldots, f_{2010}$ is not bijective, say $f_{k}$, then $f_{k}(0)=f_{k}(1)=q$, so $f_{2... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 4^{2010}-2^{2010} | |
0kpb | Problem:
Define the annoyingness of a permutation of the first $n$ integers to be the minimum number of copies of the permutation that are needed to be placed next to each other so that the subsequence $1,2, \ldots, n$ appears. For instance, the annoyingness of $3,2,1$ is 3, and the annoyingness of $1,3,4,2$ is 2.
A ra... | [
"Solution:\nFor a given permutation $p_{1}, \\ldots, p_{n}$, let $f_{k}(p)$ be the smallest number of copies of $p$ that need to be placed next to each other to have $1, \\ldots, k$ appear as a subsequence. We are interested in finding the expectation of $f_{n}(p)$.\nNotice that if $k$ appears before $k+1$ in $p$, ... | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 2023/2 | |
0bex | Given a positive integer $n$ and the function $f: \mathbb{N} \to \mathbb{N}$ described by
$$
f(x) = \begin{cases} x/2 & , \text{if } x \text{ is even} \\ (x-1)/2 + 2^{n-1} & , \text{if } x \text{ is odd} \end{cases}.
$$
Determine the set $A = \{x \in \mathbb{N} \mid (\underbrace{f \circ f \circ \dots \circ f}_{n \text{... | [
"As for $x \\in \\{0, 1, \\dots, 2^n - 1\\}$ we have $f(x) \\in \\{0, 1, \\dots, 2^n - 1\\}$, and for $x \\ge 2^n$ we have $f(x) < x$, we get $f(x) \\le \\max(x, 2^n - 1)$. It follows that if $x \\in A$, then\n$$\nx = f^{[n]}(x) \\le \\max(f^{[n-1]}(x), 2^n - 1) \\le \\dots \\le \\max(f(x), 2^n - 1).\n$$\nIf $x \\g... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | A = {0, 1, ..., 2^n - 1} | |
090r | Let $ABC$ be an acute triangle with circumcenter $O$ and let $D$ be the foot of the perpendicular from $A$ to $BC$. Assume $\angle AOD = 90^\circ$ and $OD = 4\sqrt{7}$ hold. Let $E$ and $F$ be the feet of perpendiculars from $D$ to $AB$ and $AC$ respectively, and let the lines $AO$ and $EF$ meet at $P$. If $AP = 11$, f... | [
"Without loss of generality assume $AB \\ge AC$. First, since $\\angle AED = \\angle ADB = 90^\\circ$, triangles $AED$ and $ADB$ are similar, so $AE : AD = AD : AB$, that is $AD^2 = AB \\cdot AE$. Similarly $AD^2 = AC \\cdot AF$, hence $AB \\cdot AE = AC \\cdot AF$ and points $E$, $B$, $C$, $F$ are concyclic. Now,\... | Japan | The 35th Japanese Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 2√61 | |
075q | Find all functions $f$ from the set of real numbers to itself satisfying
$$
f(x(1+y)) = f(x)(1+f(y))
$$
for all real numbers $x, y$. | [
"If $f$ is not identically zero, then by standard substitutions we get that $f(x) = x$ for $x = 0, \\pm 1$. Using these it is easy to see that $f$ is additive and multiplicative on the set of real numbers. It then follows by induction and continuity that $f(x) = x$ for all real $x$. $\\square$"
] | India | Indija TS 2013 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 for all real x, and f(x) = x for all real x | |
065g | Let $\triangle ABC$ a triangle with $90^\circ \neq \hat{A} \neq 135^\circ$. Let $D$ and $E$ be external points to the triangle such that $DAB$ and $EAC$ are isosceles triangles with right angles in $D$ and $E$, respectively. Let $F = BE \cap CD$, and $M, N$ the midpoints of $BC, DE$, respectively.
Prove that, if three ... | [
"a) If $M, N, F$ are collinear, we must have $DE \\parallel BC$, hence the distances from $D$ and $E$ to $BC$ be equal, and this is equivalent to $b = c$.\n\nb) If $A, M, F$ are collinear, $\\tan \\vec{BAM} = \\frac{b \\sin A}{c - b \\cos A}$, $\\tan \\vec{DAN} = \\frac{b \\cos A}{c - b \\sin A}$, and $\\vec{DAN} =... | Greece | Mediterranean Mathematical Competition | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians"
] | English | proof only | null | |
0b4o | Problem:
A game is played on the number line. Initially, there is a token placed at the number $0$. In each move, the player can move the token from its current position $x$, to either $x+2023$ or $x-59$. The goal of the game is to move the token to either $1$ or $-1$. What is the minimum number of moves required to a... | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 247 | |
06zl | Problem:
Given three non-collinear points $M$, $N$, $H$ show how to construct a triangle which has $H$ as orthocenter and $M$ and $N$ as the midpoints of two sides. | [
"Solution:\n\nTake $H'$ so that $M$ is the midpoint of $HH'$. The circle diameter $NH'$ meets the line through $H$ perpendicular to $MN$ in two points (in general), either of which we may take as $A$. Then $B$ is the reflection of $A$ in $M$, and $C$ is the reflection of $A$ in $N$.\n\nTo s... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04lp | Find the smallest multiple of $84$ whose decimal representation contains only digits $6$ and $7$. | [] | Croatia | Mathematical competitions in Croatia | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 76776 | |
062n | Problem:
Auf einer Tafel stehe am Anfang eine positive ganze Zahl. Wenn eine Zahl $x$ auf der Tafel steht, darf man die Zahlen $2x+1$ und $\frac{x}{x+2}$ dazuschreiben. Irgendwann stehe auch die Zahl 2008 auf der Tafel. Man beweise, dass sie von Anfang an dastand. | [
"Solution:\n\nLösungsskizze: Anfangs stehe die Zahl $a$ auf der Tafel. Der Übergang von $x$ zu $2x+1$ oder $\\frac{x}{x+2}$ werde als Transformation bezeichnet. Alle Zahlen auf der Tafel sind positiv.\n\n1. Variante:\nAus der Zahl $a$ entstehen durch $k$ Transformationen stets Zahlen der Form $\\frac{ma+m-1}{(2^{k}... | Germany | 1. IMO-Auswahlklausur | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
0bfm | Prove that the sum between a number $n$ and its *reverse* is a multiple of $81$ if and only if the sum of the digits of $n$ is a multiple of $81$. | [
"Consider $n = \\overline{a_1a_2\\dots a_{m-1}a_m}$ and $r(n) = \\overline{a_m a_{m-1} \\dots a_2 a_1}$, its *reverse*.\n\n$$\nn + r(n) = \\sum_{j=0}^{m} (a_j + a_{m-j}) \\cdot 10^j = \\sum_{j=0}^{m} a_j (10^j + 10^{m-j}).\n$$\n\nNotice that $10^i + 10^{j+1} \\equiv 10^j + 10^{i+1} \\pmod{81}$, $\\forall i, j \\in ... | Romania | 64th NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0brx | Let $ABCD$ be a cyclic quadrilateral, and let $E$ and $F$ denote the midpoints of diagonals $[AC]$ and $[BD]$, respectively. If $\{G\} = AB \cap CD$, $\{H\} = AD \cap BC$, prove that:
a) the intersection points of the angle bisectors of $\angle AHB$ and $\angle AGD$ with the sides of the quadrilateral $ABCD$ are the v... | [
"We are going to treat only the case when $C \\in (GD)$ and $C \\in (BH)$, all the other cases being similar.\n\na) Let $M$ and $N$ be the intersection points of the angle bisector of angle $\\angle G$ with the sides $BC$ and $AD$, respectively. Consider $I, K$ the intersection points of the angle bisector of angle... | Romania | 67th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals"
] | English | proof only | null | |
0hvk | Problem:
Determine if it is possible to color each of the rational numbers either red or blue such that the following three conditions are all satisfied:
(i) $x$ and $-x$ are opposite colors, for all rational $x \neq 0$;
(ii) $x$ and $1-x$ are opposite colors, for all rational $x \neq 1 / 2$;
(iii) $x$ and $1 / x$ are ... | [
"Solution:\nThe answer is yes.\nWe will prove the following statement by induction: It is possible to color the rational numbers with denominator at most $n$ red and blue such that the conditions (i)-(iii) hold whenever the two rational numbers in question both have denominator at most $n$, and such that no number ... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Yes | |
09xs | The number $1$ is written on the blackboard. A *turn* consists of wiping out the number on the board and replacing it by the double of the number, or by the number one smaller. For example, we can replace $1$ by $2$ (the double) or $0$ (one smaller), and if $5$ is on the board, we can replace it by $10$ or $4$.
What is... | [
"B) $15$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | B | |
03k0 | Problem:
Show that the equation $x^{2} + y^{5} = z^{3}$ has infinitely many solutions in integers $x$, $y$, $z$ for which $x y z \neq 0$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
078x | At an IMOTC party, all people have pairwise distinct ages. Some pairs of people are friends and friendship is mutual. Call a person *junior* if they are younger than all their friends, and *senior* if they are older than all their friends. A person with no friends is both *junior* and *senior*. A sequence of pairwise d... | [
"Consider obvious graph theory interpretation, with vertices being labelled by the ages. Whenever we say an increasing path, we refer to the labels being monotonically increasing. For any vertex $w$, let $S(w)$ be the set of all $m \\pmod k$ such that there exists an increasing path $v_1, v_2, \\dots, v_m = w$ with... | India | IMO TST | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
0exe | Problem:
A spy-plane circles point $A$ at a distance $10\mathrm{km}$ with speed $1000\mathrm{km/h}$. A missile is fired towards the plane from $A$ at the same speed and moves so that it is always on the line between $A$ and the plane. How long does it take to hit? | [
"Solution:\nAnswer: $18\\pi$ sec.\n\nLet $C$ be the position of the spy-plane at the moment the missile is fired. Let $B$ be the point a quarter of the way around the circle from $C$ (in the direction the spy-plane is moving). Then the missile moves along the semi-circle on diameter $AB$ and hits the plane at $B$.\... | Soviet Union | 5th ASU | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 18π seconds | |
02zb | Problem:
Joana deve colocar três fichas em um tabuleiro $5 \times 5$, no qual as casas são numeradas de 1 a 25, sendo uma em cada casa. De quantas maneiras ela pode fazer isso, se:
a) As 3 fichas são de cores diferentes?
b) As 3 fichas são idênticas?
c) As fichas são de cores diferentes e não podem estar duas a dua... | [
"Solution:\n\na) Como são 25 casas para a primeira ficha, temos 25 possibilidades, para a segunda ficha, temos 24 possibilidades e para a terceira ficha, temos 23 possibilidades. Portanto, são $25 \\cdot 24 \\cdot 23 = 13.800$ possibilidades.\n\nb) Para as peças da mesma cor, devemos descontar o número de situações... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | a) 13800; b) 2300; c) 600 | |
01l7 | Let $M$ be the midpoint of the side $AB$ of the acute-angled non-isosceles triangle $ABC$, $H$ be the orthocenter of $ABC$, and $I$ be the incenter of $ABC$.
Prove that if $M$, $I$, and $H$ are collinear, then the length of the segment $CH$ is equal to the length of the radius of the incircle of the triangle $ABC$.
(Fo... | [
"Let $\\Gamma$ be incircle of the triangle $ABC$. Let $K$ be the point of tangency of $\\Gamma$ and the side $AB$. Let the line $CL$ meet the side $AB$ at $N$. It is easy to see that $AK = BN$ (it suffices to consider the homothety with the center $C$ which transform $\\Gamma$ into excircle touching $AB$ at $N$, an... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | English | proof only | null | |
07x9 | Let $\mathbb{Z}_+ = \{1, 2, 3, 4, \dots\}$ be the set of all positive integers. Find, with proof, all functions $f : \mathbb{Z}_+ \to \mathbb{Z}_+$ with the property that
$$
f(x + f(y) + f(f(z))) = z + f(y) + f(f(x))
$$
for all positive integers $x, y, z$. | [
"For simplicity write $f^2(x) = f(f(x))$, $f^3(x) = f(f(f(x)))$, etc. We first show\n$$\nf^2(x) = x \\quad \\text{for all } x > 0\n$$\nin two different ways.\n\n**Method 1, using injectivity.** Suppose $f(z_1) = f(z_2)$. Replacing $z$ by $z_1$ or by $z_2$ leads to the same LHS of (2), hence the RHS must agree in bo... | Ireland | IRL_ABooklet_2024 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | Exactly two functions: (1) f(n) = n for all positive integers n; (2) f(1) = 2, f(2) = 1, and f(n) = n for all n ≥ 3. | |
0g70 | $AD$, $BC$ 為圓 $O$ 的兩弦交於圓內一點 $P$，在線段 $AP$, $PC$ 與圓 $O$ 所圍區域內的一圓 $K$，與弦 $AD$, $BC$ 分別切於 $E$, $F$，且與圓 $O$ 相切於 $T$，$TF$ 交圓 $O$ 於第二點 $G$，$AG$ 與 $EF$ 交於點 $I$。
試證：
(1) $A$, $E$, $T$, $I$ 四點共圓。
(2) $GB = GI$。 | [
"延長 $BC$ 交兩圓公切線於 $R$。延長 $TE$ 交圓 $O$ 於第二點 $H$\n首先，由於 $TR$ 為兩圓公切線，故 $\\angle GHT = \\angle RTF = \\angle FET$，從而 $EF \\parallel GH$。\n接著再注意到 $\\angle RTC = \\angle TBC$，且 $RF$ 與 $RT$ 皆為對圓 $K$ 的切線，故\n$$\n\\angle RTF = \\angle RFT = \\angle TBC + \\angle BTG = \\angle RTC + \\angle BTG,\n$$\n$$\n\\angle BTG = \\angle R... | Taiwan | 二〇一二數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0h9b | Determine all possible real pairs $(x, y)$ that satisfy the following:
$$
4x + 3y = 2x \cdot \left[ \frac{x^2 + y^2}{x^2} \right].
$$ | [
"It follows from the given condition that $x \\neq 0$, therefore, equation can be rewritten as $2 + \\frac{3y}{2x} = \\left[ 1 + \\frac{y^2}{x^2} \\right]$. Let $k = \\frac{3y}{2x}$ is an integer. Since $[1 + \\frac{y^2}{x^2}] = 1 + \\frac{y^2}{x^2}$, the following holds:\n$$\n1 + k = \\left[ \\frac{4k^2}{9} \\righ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | All real pairs with x ≠ 0 and either y = -2x/3 or y = 2x. | |
0cut | In the plane, several lines in a general position are drawn. These lines partition the plane into regions. Prove that one can put into each region a positive number so that the sums of numbers on both sides of each line will be equal. | [
"Обозначим проведённые прямые $l_1, l_2, \\dots, l_n$, упорядочив их направления по часовой стрелке (см. рис. 10). Формально это означает следующее. Рассмотрим произвольную точку плёскости $O$. Проведем через неё прямые, параллельные нашим, заумеруем их по часовой стрелке, а потом присвоим нашим прямым те же номера... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English; Russian | proof only | null | |
0c17 | Let $\triangle ABC$ be a right triangle in $A$ and points $D$ and $E$ on $AB$ such that $\angle ACD = \angle DCE = \angle ECB$. Show that if $3\overrightarrow{AD} = 2\overrightarrow{DE}$ and $\overrightarrow{CD} + \overrightarrow{CE} = 2\overrightarrow{CM}$, then $\overrightarrow{AB} = 4\overrightarrow{AM}$.
Gabriel Po... | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05lh | Problem:
Soit $ABC$ un triangle non isocèle inscrit dans un cercle $\Gamma$ de rayon $R$. Le cercle passant par $A$ et tangent en $C$ à $[BC]$ recoupe le cercle passant par $B$ et tangent en $C$ à $[AC]$ au point $D$.
a) Montrer que $CD \leqslant R$.
b) Montrer que lorsque $C$ se déplace sur $\Gamma$, la droite $(CD... | [
"Solution:\n\n\n\na) On observe d'abord que $(AD, AC) = (CD, CB)$ et $(BC, BD) = (CA, CD)$, donc $DAC$ et $DCB$ sont directement semblables. On en déduit que $D^2 = DA \\cdot DB$.\n\nDe plus, en notant $\\gamma = (\\overrightarrow{CA}, \\overrightarrow{CB})$, on a $(\\overrightarrow{DC}, \\... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bg7 | Given a triangle $ABC$, a circle centered at some point $O$ meets the segments $BC, CA, AB$ in the pairs of points $X$ and $X'$, $Y$ and $Y'$, $Z$ and $Z'$, respectively, labeled in circular order: $X, X', Y, Y', Z, Z'$. Let $M$ be the Miquel point of the triangle $XYZ$ (i.e., the point of concurrence of the circles $A... | [
"We begin by reviewing some basic facts on conics. For an ellipse $\\Sigma$ with center $N$, foci $M$ and $M'$, semiaxes $a$ and $b$, it is known that the orthogonal projections $P$ and $P'$ of $M$ and $M'$ on any line $t$ tangent to $\\Sigma$ lie on the major auxiliary circle of $\\Sigma$, so that $NP = a = NP'$. ... | Romania | The Tenth IMAR Mathematical Competition | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry ... | null | proof only | null | |
08ju | Problem:
Prove that for all real $x, y$
$$
\frac{x+y}{x^{2}-x y+y^{2}} \leq \frac{2 \sqrt{2}}{\sqrt{x^{2}+y^{2}}}
$$ | [
"Solution:\nThe inequality rewrites as\n$$\n\\frac{x+y}{x^{2}-x y+y^{2}} \\leq \\frac{\\sqrt{2\\left(x^{2}+y^{2}\\right)}}{\\frac{x^{2}+y^{2}}{2}}\n$$\nNow it is enough to prove the next two simple inequalities:\n$$\nx+y \\leq \\sqrt{2\\left(x^{2}+y^{2}\\right)}, \\quad x^{2}-x y+y^{2} \\geq \\frac{x^{2}+y^{2}}{2}\... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0hym | Problem:
Bildert works in a cubicle in an office which consists of 27 cubicles arranged in a $3 \times 3 \times 3$ cube. Any two cubicles sharing a wall have a connecting door on this wall; for example, the corner cubicles have exactly 3 doors, while the center cubicle has 6 doors: one on each wall, one on the floor, ... | [
"Solution:\n\nSolution I. The answer is no. Denote the central cubicle by $C$, and denote the vertex, edge and face cubicles by $V, E$ and $F$, respectively. The trip must start with $C$ and include every one of the $8 V$'s, $6 F$'s, and $12 E$'s. The sequence must begin with $C F E$. Each cubicle $V$ is adjacent o... | United States | BAMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
0l9k | Let $F$ be the set of all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ satisfying the condition
$f(3x) \geq f(f(2x)) + x$ for every real positive number $x$.
Find the greatest real number $\alpha$ such that for all $f \in F$, we have
$$
f(x) \geq \alpha x
$$
for every real positive number $x$. | [
"• It is clear that the function $f(x) = x/2$, $x \\in \\mathbb{R}^+$, is a function belonging to $F$. Thus $\\alpha \\leq 1/2$.\n• Let $f$ be an arbitrary function in $F$. It is easy to see that\n$$\nf(x) \\geq x/3 \\quad \\forall x \\in \\mathbb{R}^+. \\qquad (1)\n$$\nConsider the sequence of numbers $\\{\\alpha_... | Vietnam | 2003 Vietnamese Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 1/2 | |
048q | How many positive integers less than $2011$ are divisible by either $2$ or $7$, but are not divisible by $5$? | [
"Let $N = 2011$.\n\nLet $A$ be the set of positive integers less than $2011$ divisible by $2$ or $7$.\nLet $B$ be the set of positive integers less than $2011$ divisible by $5$.\nWe are to find $|A \\setminus B|$.\n\nFirst, count the number of positive integers less than $2011$ divisible by $2$ or $7$.\n\nLet $S_2$... | Croatia | CroatianCompetitions2011 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 919 | |
0af7 | Дадени се 21 плочка во облик на квадратче, со иста димензија. На четири плочки е запишан бројот 1; на две плочки е запишан бројот 2; на седум плочки е запишан бројот 3; на 8 плочки е запишан бројот 4. Користејќи 20 од тие плочки, Димитар формирал правоаголник со димензии 4 на 5. За формираниот правоаголник збирот на бр... | [
"Да го означиме со $S$ збирот на сите броеви запишани на плочките кои го формираат правоаголникот. Од условот на задачата, имаме дека 4 е делител на $S$ и дека 5 е делител на $S$. Значи 20 е делител на $S$. Збирот на сите броеви запишани на 21-ната плочка е точно 61. Заклучуваме дека на неискористената плочка мора ... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Macedonian, English | proof and answer | 1 | |
02dj | Any positive integer $n$ can be written in the form $n = 2^b(2c + 1)$. We call $2c + 1$ the *odd part of* $n$. Given an odd integer $n > 0$, define the sequence $a_0, a_1, a_2, \ldots$ as follows: $a_0 = 2^n - 1$, $a_{k+1}$ is the odd part of $3a_k + 1$. Find $a_n$. | [
"An induction shows that $a_k = 3^k 2^{n-k} - 1$ for $k \\le n-1$. It is certainly true for $k=0$. Suppose it is true for $k < n-1$. Then $3a_k + 1 = 3^{k+1} 2^{n-k} - 2$. Since $n-k > 1$, the odd part is $3^{k+1} 2^{n-(k+1)} - 1$, so the result is true for $k+1$. That gets us as far as $a_{n-1} = 3^{n-1} 2 - 1$.\n... | Brazil | IV OBM | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (3^n - 1)/2 | |
037u | Problem:
Two real numbers $a$ and $b$ satisfy the inequality $b^{3} + b \leq a - a^{3}$. Find the maximum possible value of $a + b$. | [
"Solution:\n\nLet $a + b = c$. Thus,\n$$\n(c - a)^{3} + c - a \\leq a - a^{3} \\Longleftrightarrow 3 c a^{2} - (3 c^{2} + 2) a + c^{3} + c \\leq 0\n$$\nIf $c > 0$ then\n$$\n0 \\leq D = (3 c^{2} + 2)^{2} - 12 c (c^{3} + c) = 4 - 3 c^{4}\n$$\nHence $c \\leq \\sqrt[4]{\\frac{4}{3}}$ with equality when $a$ is the doubl... | Bulgaria | Team selection test for 23. BMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (4/3)^(1/4) | |
02r4 | Problem:
a) Severina escreveu um número inteiro positivo em cada lado de um quadrado. Em seguida, escreveu em cada vértice o produto dos números escritos nos lados que se encontram nesse vértice. A soma dos números escritos em dois lados opostos é 60 e a soma dos números escritos nos outros lados é 85. Qual é a soma do... | [
"Solution:\na) $1^{a}$ solução: Sejam $a, b, c$ e $d$ os números escritos nos lados do quadrado no sentido horário. Os números associados aos vértices são, portanto, $ab$, $bc$, $cd$ e $da$, e sua soma é\n$$\nab + bc + cd + da = b(a + c) + d(a + c) = (a + c)(b + d) = 85 \\times 60 = 5100\n$$\n![](attached_image_1.p... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Solid Geometry > 3D Shapes",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) 5100; b) 15 | |
0iq6 | Problem:
Let $f(n)$ be the number of times you have to hit the $\sqrt{\ }$ key on a calculator to get a number less than $2$ starting from $n$. For instance, $f(2)=1$, $f(5)=2$. For how many $1 < m < 2008$ is $f(m)$ odd? | [
"Solution:\nAnswer: $242$ This is $[2^{1}, 2^{2}) \\cup [2^{4}, 2^{8}) \\cup [2^{16}, 2^{32}) \\ldots$, and $2^{8} < 2008 < 2^{16}$ so we have exactly the first two intervals."
] | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 242 | |
04fn | On the playground there are $2014$ athletes with the numbers from $1$ to $2014$ on their shirts (each number is on exactly one shirt). At the beginning they are all standing. In certain time intervals the coach shouts out all positive integers from $1$ to $2014$ in the increasing order. All athletes having a multiple o... | [
"Every athlete will change his position as many times as the number on his shirt has divisors. Hence, at the end in the position of crouch will be those athletes whose shirt numbers have an odd number of divisors.\nAll divisors of the number $n$ can be grouped into two element sets $\\{d, \\frac{n}{d}\\}$, unless $... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 44 | |
077x | Euler marks $n$ different points in the Euclidean plane. For each pair of marked points, Gauss writes down the number $\lfloor \log_2 d \rfloor$ where $d$ is the distance between the two points. Prove that Gauss writes down less than $2n$ distinct values.
*Note:* For any $d > 0$, $\lfloor \log_2 d \rfloor$ is the uniq... | [
"Let the $n$ points be $P_1, P_2, \\ldots, P_n$.\n\nLet $D$ be the set of all pairwise distances between the $n$ points. For each $d \\in D$, Gauss writes down $\\lfloor \\log_2 d \\rfloor$.\n\nLet $d_{\\min}$ and $d_{\\max}$ be the minimal and maximal pairwise distances among the $n$ points.\n\nAll values written ... | India | INMO_2023 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0elu | Let $a_1, a_2, ..., a_n, b_1, b_2, ..., b_n$ be the numbers $1, 2, ..., 2n$ in some order. Suppose that $a_1 < a_2 < ... < a_n$ and that $b_1 > b_2 > ... > b_n$. Prove that
$$
\sum_{i=1}^{n} |a_i - b_i| = n^2.
$$ | [
"Note that for a given $i$, $a_i$ and $b_i$ are never both in one of $\\{1, 2, ..., n\\}$ or $\\{n+1, ..., 2n\\}$. Suppose, without loss of generality, that $a_i, b_i \\le n$. Then $a_1, a_2, ..., a_i \\le n$, and $b_i, b_{i+1}, ..., b_n \\le n$, giving $n+1$ elements less than or equal to $n$, which is impossible.... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
02a6 | Problem:
Jogos de futebol - Os doze alunos de uma turma de olimpíada saíam para jogar futebol todos os dias após a aula de matemática, formando dois times de 6 jogadores cada e jogando entre si. A cada dia eles formavam dois times diferentes dos times formados em dias anteriores. Ao final do ano, eles verificaram que ... | [
"Solution:\n\nPara cada grupo de 5 alunos, existe um único time formado que os contém. Logo, contamos $C_{12}^{5} = \\frac{12 \\cdot 11 \\cdot 10 \\cdot 9 \\cdot 8}{5!} = 792$ times para cada 5 alunos escolhidos. Por outro lado, em cada time de 6 jogadores, temos $C_{6}^{5} = 6$ modos de escolhermos cinco jogadores... | Brazil | Nível 3 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 132 | |
0gdl | 設實數 $a, b, c, d$ 滿足
$$
(a + c)(b + d) = \sqrt{2}(ac - 2bd - 1).
$$
試證：
$$
(ab - 1)^2 + (bc - 1)^2 + (cd - 1)^2 + (da - 1)^2 + (ac - 1)^2 + (2bd + 1)^2 \geq 4.
$$
Let $a, b, c, d$ be real numbers satisfying
$$
(a + c)(b + d) = \sqrt{2}(ac - 2bd - 1).
$$
Show that
$$
(ab - 1)^2 + (bc - 1)^2 + (cd - 1)^2 + (da - 1)^2 + (... | [
"令 $A = (a+c)(b+d) = \\sqrt{2}(ac-2bd-1)$. 注意到\n$$\n\\begin{align*}\n\\sum_{\\text{cyc}} (ab-1)^2 &\\ge \\sum_{\\text{cyc}} (ab-1)^2 - \\left(\\sum_{\\text{cyc}} ab-1\\right)^2 \\\\\n&= -2 \\left(\\sum_{\\text{cyc}} ab^2 c\\right) - 4abcd + 3 \\\\\n&= 3 - 4abcd - 2(ab + cd)(ad + bc) \\\\\n&= 3 - 4abcd - 2(ab + cd)(... | Taiwan | 2020 Taiwan IMO 1J | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0ej6 | Problem:
Podana imamo tri števila $A=26^{351}$, $B=5^{702}$ in $C=3^{1053}$. Števila uredi po velikosti. Kateri spodnji zapis je pravilen?
(A) $A<B<C$
(B) $A<C<B$
(C) $B<C<A$
(D) $B<A<C$
(E) $C<A<B$ | [
"Solution:\n\n$B=5^{702}=(5^{2})^{351}=25^{351}$, $C=(3^{3})^{351}=27^{351}$. Ker je $25^{351}<26^{351}<27^{351}$, dobimo $B<A<C$. Pravilen je odgovor $\\mathrm{D}$."
] | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | D | |
06nx | Let $n \ge 4$ be a positive integer. Consider any set $A$ formed by $n$ distinct real numbers such that the following condition holds: for every $a \in A$, there exist distinct elements $x, y, z \in A$ such that $|x - a|, |y - a|, |z - a| \ge 1$. For each $n$, find the greatest real number $M$ such that
$$
\sum_{a \in ... | [
"The greatest $M$ is $4$ if $n = 4$, and is $3$ if $n \\ge 5$.\n\nLet $S$ be the sum $\\sum_{a \\in A} |a|$.\n\nFor $n = 4$, by considering $A = \\{-1, 0, 1, 2\\}$, we need $M \\le 4$. Let $a < b < c < d$ be the elements in $A$. From the condition, we must have $b - a, c - b, d - c \\ge 1$. This implies\n$$\nS = (|... | Hong Kong | IMO HK TST | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | M = 4 for n = 4; M = 3 for n ≥ 5 | |
07pk | Find all real numbers $x$ for which
$$
\frac{x\sqrt{14}}{\sqrt{x+1}+\sqrt{1-x}} > \sqrt{2-x}.
$$ | [
"For the square roots to exist, we require $-1 \\le x \\le 1$. The inequality is false when $x \\le 0$, so we assume $0 < x \\le 1$. By multiplying above and below by $\\sqrt{x+1} - \\sqrt{1-x}$, the inequality becomes\n$$\n\\sqrt{x+1} - \\sqrt{1-x} > \\sqrt{\\frac{4-2x}{7}}.\n$$\nThe left side is positive, since $... | Ireland | Ireland | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 3/5 < x <= 1 | |
0fqw | Problem:
Consideramos un triángulo $ABC$ y un punto $D$ en el lado $AC$. Si $\overline{AB} = \overline{DC} = 1$, $\angle DBC = 30^{\circ}$ y $\angle ABD = 90^{\circ}$, calcula el valor de $\overline{AD}$. | [
"Solution:\n\nLlamando $\\angle ADB = \\alpha$, tendremos que $\\angle BDC = 180 - \\alpha$. Utilizando el teorema de los senos en el triángulo $ADB$ tenemos que\n$$\n\\frac{x}{1} = \\frac{1}{\\sin \\alpha} = \\frac{BD}{\\sin(90 - \\alpha)}\n$$\ny en el triángulo $DBC$ tendremos que\n$$\n\\frac{1}{\\sin 30} = \\fra... | Spain | OME fase local | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 2^(1/3) | |
06ne | In a country there are only four types of coins, of denominations $74$, $87$, $111$ and $124$ dollars respectively. In how many different ways can one pay exactly $2023$ dollars using these coins? | [
"Answer: $14$\n\nNote that $74$ and $111$ have a common factor of $37$. Also, $87$ and $124$ are $13$ greater than $74$ and $111$ respectively. We call the coins with denominations $87$ and $124$ 'bad coins'.\n\nSince $2023 \\equiv 25 \\pmod{37}$, we must use $k$ bad coins such that $13k \\equiv 25 \\pmod{37}$. The... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 14 | |
005f | Se tienen $48$ enteros positivos menores que $70$ cuya suma es $140$. Demuestre que es posible elegir algunos de estos números tales que su suma sea exactamente $70$. Dé un contraejemplo con $47$ enteros positivos. | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | Counterexample for 47 integers: take forty-six ones and ninety-four; their sum is one hundred forty, and no subset sums to seventy. | |
0f73 | Problem:
Let $S$ be the set of all numbers which can be written as $1/mn$, where $m$ and $n$ are positive integers not exceeding $1986$. Show that the sum of the elements of $S$ is not an integer. | [] | Soviet Union | 20th ASU | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0hin | What is the maximum possible number of edges in a graph with $2n$ vertices, if there is exactly one way to divide its vertices into $n$ pairs such that in each pair the vertices are connected by an edge? | [
"Consider this partition into pairs, denote the vertices $A_1, A_2, ..., A_{2n}$ where the vertices $A_{2i-1}, A_{2i}$ are connected by an edge for each $i$ from $1$ to $n$. Note that for every two pairs $(A_{2i-1}, A_{2i})$, $(A_{2j-1}, A_{2j})$, there are at most two edges between them: there cannot be two edges ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | n^2 | |
01wf | Does there exist a positive integer $n$ which can be represented both as $n = a^2 - b$ and $n = b^2 - c$, where $a$, $b$, $c$ are three distinct divisors of $n$? | [
"Answer: no, such $n$ doesn't exist."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | No, such n doesn't exist. | |
0427 | Take randomly five different numbers from $1, 2, \ldots, 20$. Then the probability that there are at least two adjacent numbers among them is ______. | [
"Suppose $a_1 < a_2 < a_3 < a_4 < a_5$ are taken from $1, 2, \\ldots, 20$. If $a_1, a_2, a_3, a_4, a_5$ are not adjacent to each other, then we have\n$$\n1 \\leq a_1 < a_2 - 1 < a_3 - 2 < a_4 - 3 < a_5 - 4 \\leq 16,\n$$\nfrom which we know that the number of ways to select five numbers not adjacent to each other fr... | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 232/323 | |
0db1 | Let $ABCD$ be a cyclic quadrilateral with $AB = BC$ and $AD = CD$. A point $M$ lies on the minor arc $CD$ of its circumcircle. The lines $BM$ and $CD$ meet at point $P$, the lines $AM$ and $BD$ meet at point $Q$. Prove that $PQ \parallel AC$. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
045p | Let $n$ be a positive integer. Let $x_1, x_2, \dots, x_{2n}$ be $2n$ nonnegative real numbers such that $x_1 + x_2 + \dots + x_{2n} = 4$. Prove that there exist nonnegative integers $p, q$ such that $q \le n - 1$, and that
$$
\sum_{i=1}^{q} x_{p+2i-1} \le 1, \quad \sum_{i=q+1}^{n-1} x_{p+2i} \le 1.
$$
*Remark 1 : the ... | [
"**Proof:** Set $A = x_1 + x_3 + \\dots + x_{2n-1}$ and $B = x_2 + x_4 + \\dots + x_{2n}$.\nIf one of $A, B$ is less than or equal to 1, the problem is obvious. If $A > 1$ and $B > 1$, for $0 \\le k \\le n-1$, let $m(k) \\in \\{1, 2, \\dots, n-1\\}$ be the unique integer such that\n$$\n\\sum_{i=0}^{m(k)} x_{2k+2i+1... | China | 2022 China Team Selection Test | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0dsr | Let $a, b, c, d$ be positive integers such that $a + c = 20$ and $\frac{a}{b} + \frac{c}{d} < 1$. Find the maximum possible value of $\frac{a}{b} + \frac{c}{d}$. | [
"Therefore the maximum value of the sum of the fractions is attained when $y$ is minimum, which is $1$. When $y = 1$, since $ac < xy$, the optimum value for $x$ is $ac + 1$ (we want to choose $x$ so that it is as small as possible). Thus for fixed $a, c$, the maximum value is\n$$\n\\frac{a}{a+ac+1} + \\frac{c}{c+1}... | Singapore | Singapore Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1385/1386 | |
0erp | Determine all pairs of real numbers $a$ and $b$, $b > 0$, such that the solutions to the two equations
$$
x^2 + a x + a = b
$$
and
$$
x^2 + a x + a = -b
$$
are four consecutive integers. | [
"The quadratic formula gives us\n$$\n\\frac{-a \\pm \\sqrt{a^2 - 4a + 4b}}{2}\n$$\nand\n$$\n\\frac{-a \\pm \\sqrt{a^2 - 4a - 4b}}{2}\n$$\n\nSuppose that the four consecutive numbers are $n - 1, n, n + 1, n + 2$. The parabola $y = x^2 + a x + a$ reaches its minimum at $x = -a/2$, and the line $x = -a/2$ is its axis ... | South Africa | South African Mathematics Olympiad Third Round | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | a = -1, b = 1 and a = 5, b = 1 | |
09xe | Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
$$
f(x + y f(x + y)) = y^2 + f(x) f(y)
$$ | [
"\n2. Let $M, N, R, S$ be the midpoints of line segments $BC, CA, BD, AD$. Let $Z$ be the centroid of $\\triangle ABC$. Quadrilateral $QFMC$ is cyclic as $\\angle QFC = 90^\\circ = \\angle QMC$. Note that therefore $CQ$ is a diameter of the circumcircle of $QFMC$. Analogously, we see that $... | Netherlands | IMO Team Selection Test 3 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = x + 1 and f(x) = 1 - x | |
0fbp | Problem:
En el plano tenemos una recta $r$ y dos puntos $A$ y $B$ exteriores a la recta y en el mismo semiplano. Determinar un punto $M$ de la recta tal que el ángulo de $r$ con $A M$ sea doble del de $r$ con $B M$. (Considérese como ángulo de dos rectas el menor de los ángulos que forman). | [
"Solution:\n\n\n\nSe traza con centro $B$ la circunferencia tangente a $r$. Se trazan por $A$ las tangentes a la circunferencia anterior que cortan a $r$ en $M$ y $M'$. $B$ es el incentro del triángulo $A M' M$."
] | Spain | OME 12 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof and answer | null | |
00fy | Let $n$, $k$ be given positive integers with $n > k$. Prove that
$$
\frac{1}{n+1} \cdot \frac{n^{n}}{k^{k}(n-k)^{n-k}} < \frac{n!}{k!(n-k)!} < \frac{n^{n}}{k^{k}(n-k)^{n-k}}.
$$ | [
"The inequality is equivalent to\n$$\n\\frac{n^{n}}{n+1} < \\binom{n}{k} k^{k}(n-k)^{n-k} < n^{n}\n$$\nwhich suggests investigating the binomial expansion of\n$$\nn^{n} = ((n-k) + k)^{n} = \\sum_{i=0}^{n} \\binom{n}{i} (n-k)^{n-i} k^{i}.\n$$\nThe $(k+1)$th term $T_{k+1}$ of the expansion is $\\binom{n}{k} k^{k} (n-... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0ky0 | Problem:
There are 100 people standing in a line from left to right. Half of them are randomly chosen to face right (with all $\binom{100}{50}$ possible choices being equally likely), and the others face left. Then, while there is a pair of people who are facing each other and have no one between them, the leftmost su... | [
"Solution:\n\nNotice that the order in which the people leave the line is irrelevant. Give each right-facing person a weight of $1$, and each left-facing person a weight of $-1$. We claim the answer for some arrangement of these $2n$ people is $-2$ times the minimum prefix sum. For instance:\n\n$$\n\\begin{gathered... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2^100 / binom(100,50) - 1 | |
0avq | Problem:
Suppose that $S_{k}$ is the sum of the first $k$ terms of an arithmetic sequence with common difference $3$. If the value of $\frac{S_{3n}}{S_{n}}$ does not depend on $n$, what is the $100$th term of the sequence? | [] | Philippines | 19th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 597/2 | |
0ecm | The figure shows three concentric semicircles with radii $1$, $2$, and $4$. Regions denoted by $X$ have area $x$ and regions denoted by $Y$ have area $y$. What is the ratio between areas $x$ and $y$?
(A) $1 : 3$ (B) $1 : 2$ (C) $2 : 3$ (D) $3 : 8$ (E) $4 : 9$ | [
"Areas of the three semicircles are equal to $\\frac{\\pi}{2}$, $\\frac{4\\pi}{2}$, and $\\frac{16\\pi}{2}$. Therefore, the area $x$ is equal to $\\frac{4\\pi}{2} - \\frac{\\pi}{2} = \\frac{3\\pi}{2}$ and the area $y$ is equal to $\\frac{16\\pi}{2} - \\frac{4\\pi}{2} = 6\\pi$. The ratio we are looking for is $\\fra... | Slovenia | National Math Olympiad 2015 – First Round | [
"Geometry > Plane Geometry > Circles > Coaxal circles"
] | null | MCQ | D | |
0iiz | Problem:
Points $A$, $C$, and $B$ lie on a line in that order such that $AC = 4$ and $BC = 2$. Circles $\omega_{1}$, $\omega_{2}$, and $\omega_{3}$ have $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$ as diameters. Circle $\Gamma$ is externally tangent to $\omega_{1}$ and $\omega_{2}$ at $D$ and $E$ respectively... | [
"Solution:\n\nLet the center of $\\omega_{i}$ be $O_{i}$ for $i=1,2,3$ and let $O$ denote the center of $\\Gamma$. Then $O$, $D$, and $O_{1}$ are collinear, as are $O$, $E$, and $O_{2}$. Denote by $F$ the point of tangency between $\\Gamma$ and $\\omega_{3}$; then $F$, $O$, and $O_{3}$ are collinear. Writing $r$ fo... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2/3 | |
0dha | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that
$$
2f(x)f(x + y) - f(x^2) = \frac{x}{2}(f(2x) + 4f(f(y)))
$$
for all $x, y \in \mathbb{R}$. | [
"Put $x = 0$, we get $2f(0)f(y) - f(0) = 0$, if $f(0) \\neq 0$ then $f(y) = 1/2$ for all $y$, which does not satisfy. Thus $f(0) = 0$.\n\nPut $y = 0$, we get $2f(x)^2 - f(x^2) = x/2 \\cdot f(2x)$ then plugging back $2f(x) \\cdot f(x+y) = 2f(x)^2 + 2x \\cdot f(f(y))$ so\n$$\nf(x) \\cdot f(x+y) = f(x)^2 + x \\cdot f(... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0 and f(x) = x | |
0iiw | Problem:
Let $f(x)$ be a degree $2006$ polynomial with complex roots $c_{1}, c_{2}, \ldots, c_{2006}$, such that the set
$$
\left\{\left|c_{1}\right|,\left|c_{2}\right|, \ldots,\left|c_{2006}\right|\right\}
$$
consists of exactly $1006$ distinct values. What is the minimum number of real roots of $f(x)$? | [
"Solution:\nThe complex roots of the polynomial must come in pairs, $c_{i}$ and $\\overline{c_{i}}$, both of which have the same absolute value. If $n$ is the number of distinct absolute values $\\left|c_{i}\\right|$ corresponding to those of non-real roots, then there are at least $2n$ non-real roots of $f(x)$. Th... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 6 | |
0hvi | Problem:
Show that the polynomial $\left(x^{2}+x\right)^{2^{1000}}+1$ cannot be factored as the product of two nonconstant polynomials with integer coefficients. | [
"Solution:\nAssume for contradiction this is not the case, and the polynomial can be written as\n$$\n\\left(x^{2}+x\\right)^{2^{1000}}+1=f(x) g(x)\n$$\nfor some nonconstant $f$ and $g$ with integer coefficients. Clearly we may assume $f$ and $g$ have leading coefficient one. Taking modulo 2 we obtain that\n$$\nf(x)... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0aop | Problem:
Simplify: $\left(\frac{2^{-1}+3^{-1}}{2^{-1}-3^{-1}}\right)^{-1}$. | [
"Solution:\n\n$$\n\\left(\\frac{2^{-1}+3^{-1}}{2^{-1}-3^{-1}}\\right)^{-1} = \\frac{2^{-1}-3^{-1}}{2^{-1}+3^{-1}} = \\frac{\\frac{1}{2}-\\frac{1}{3}}{\\frac{1}{2}+\\frac{1}{3}} = \\frac{\\frac{1}{6}}{\\frac{5}{6}} = \\frac{1}{5}\n$$"
] | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 1/5 | |
0ihb | Problem:
In how many ways can the cells of a $4 \times 4$ table be filled in with the digits $1,2, \ldots, 9$ so that each of the 4-digit numbers formed by the columns is divisible by each of the 4-digit numbers formed by the rows? | [
"Solution:\nIf $a$ and $b$ are 4-digit numbers with the same first digit, and $a$ divides $b$, then since $b < a + 1000 \\leq 2a$, $b$ must equal $a$. In particular, since the number formed by the first row of the table divides the number in the first column (and both have the same first digit), these numbers must ... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 9 | |
07t9 | Suppose $a, b, c$ are the side lengths of a triangle. Prove that
$$
a^2 + b^2 + c^2 \leq a \max(b, c) + b \max(c, a) + c \max(a, b)
$$
with equality iff $a = b = c$. | [
"Since $\\max(x, y) = \\frac{x+y+|x-y|}{2}$, twice the expression\n$$\na \\max(b, c) + b \\max(c, a) + c \\max(a, b)\n$$\nis equal to\n$$\na(b+c) + b(c+a) + c(a+b) + a|b-c| + b|c-a| + c|a-b|.\n$$\nTherefore,\n$$\n\\begin{aligned}\n& 2(a \\max(b,c) + b \\max(c,a) + c \\max(a,b)) - 2(a^2 + b^2 + c^2) \\\\\n&= 2 \\sum... | Ireland | IRL_ABooklet_2020 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0c3v | Problem:
Fie $k$ un număr real, cu $k>2$.
a) Arătaţi că pentru orice numere pozitive $x, y$ şi $z$ are loc inegalitatea
$$
\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}>2 \sqrt{\frac{(x+y)(y+z)(z+x)}{x y+y z+z x}}
$$
b) Demonstraţi că există numere pozitive $x, y$ şi $z$ pentru care
$$
\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}<k \sqrt{\f... | [
"Solution:\n\na) Avem $\\sqrt{x+y}+\\sqrt{y+z}+\\sqrt{z+x}>2 \\sqrt{\\frac{(x+y)(y+z)(z+x)}{x y+y z+z x}} \\Leftrightarrow$\n$x+y+z+\\sqrt{x^{2}+x y+y z+z x}+\\sqrt{y^{2}+x y+y z+z x}+\\sqrt{z^{2}+x y+y z+z x}>$\n$2 \\cdot \\frac{(x+y)(y+z)(z+x)}{x y+y z+z x}$. Dar\n$\\sqrt{x^{2}+x y+y z+z x}>x, \\sqrt{y^{2}+x y+y ... | Romania | Al cincilea test de selecţie pentru OBMJ | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0c27 | Consider a positive integer $n \ge 2$ and a $n \times n$ square (see the figure). The main diagonal of this square consists of the hatched squares. We write 0 into the $1 \times 1$ squares situated below the main diagonal and nonzero natural numbers into the other $1 \times 1$ squares (including the hatched ones). Afte... | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 6 | |
0db3 | Let 6 pairwise different digits are given and all of them are different from $0$. Prove that there exist $2$ six-digit integers, such that their difference is equal to $9$ and each of them contains all given $6$ digits. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | English | proof only | null | |
0c2u | Let $a$, $b$, $c$ be positive real numbers, with $abc(a + b + c) = 3$. Prove that
$$
\frac{1}{a^2 + b^2 + 1} + \frac{1}{b^2 + c^2 + 1} + \frac{1}{c^2 + a^2 + 1} \le 1.
$$ | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
019q | Prove that for positive $a$, $b$, $c$ and $l > m$
$$
\frac{a^{3l} + a^{3m} + 1}{a^l + b^{l-m}a^m + b^l} + \frac{b^{3l} + b^{3m} + 1}{b^l + c^{l-m}b^m + c^l} + \frac{c^{3l} + c^{3m} + 1}{c^l + a^{l-m}c^m + a^l} \ge a^m + b^m + c^m.
$$ | [
"Observe that Cauchy–Schwarz inequality can be written in the form\n$$\n\\frac{x}{a} + \\frac{y}{b} + \\frac{z}{c} \\geq \\frac{(\\sqrt{x} + \\sqrt{y} + \\sqrt{z})^2}{a + b + c}.\n$$\nAll sums in the following inequalities are cyclic\n$$\n\\begin{align*}\n\\sum \\frac{a^{3l} + a^{3m} + 1}{a^l + b^{l-m}a^m + b^l} &\... | Baltic Way | Baltic Way 2013 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null |
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