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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
020r | Problem:
Two circles $\Gamma_{1}$ and $\Gamma_{2}$ intersect at points $A$ and $Z$ (with $A \neq Z$). Let $B$ be the centre of $\Gamma_{1}$ and let $C$ be the centre of $\Gamma_{2}$. The exterior angle bisector of $\angle B A C$ intersects $\Gamma_{1}$ again at $X$ and $\Gamma_{2}$ again at $Y$. Prove that the interio... | [
"Solution:\n\nSolution I. We first prove that $\\angle A Z X = \\angle Y Z A$. Since the triangles $\\triangle B A X$, $\\triangle C A Y$ are isosceles, and $X Y$ is the external bisector of $\\angle B A C$, we see that\n$$\n\\angle B X A = \\angle X A B = \\angle C A Y = \\angle A Y C.\n$$\nUsing these equalities,... | Benelux Mathematical Olympiad | Benelux 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"
] | null | proof only | null | |
02vp | Problem:
Um Quadrado Latino é um tabuleiro $n \times n$ preenchido com $n$ símbolos distintos de modo que em cada linha e em cada coluna não existam símbolos repetidos. Por exemplo, a figura abaixo mostra um exemplo de um Quadrado Latino de dimensões $3 \times 3$. O nome foi inspirado em trabalhos do matemático Leonha... | [
"Solution:\n\nObserve que dado um Quadrado Latino, quando trocamos todas as casas de um símbolo pelas casas de outro, ainda obtemos outro Quadrado Latino. No exemplo dado no enunciado, ao trocarmos as casas de número 1 e 2 de posição, obtemos:\n\n| 2 | 1 | 3 |\n| :--- | :--- | :--- |\n| 3 | 2 | 1 |\n| 1 | 3 | 2 |\n... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 48 | |
0im4 | Problem:
The points of the plane are colored in black and white so that whenever three vertices of a parallelogram are the same color, the fourth vertex is that color, too. Prove that all the points of the plane are the same color. | [
"Solution:\n\nSuppose not. Let $A$ be a white point and $B$ a black point. Their midpoint $C$ is one of the two colors; without loss of generality suppose $C$ is black. Now pick any point $D$ not collinear with $A, B, C$, and construct $E$ so that $C A D E$ is a parallelogram. If $D, E$ are both white, then $C A D ... | United States | 9th Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
07zl | Problem:
Dato un foglio rettangolare di lati $a$ e $b$, con $a > b$, determinare l'area del triangolo che risulta dalla sovrapposizione dei due lembi che si ottengono piegando il foglio lungo una diagonale (il triangolo colorato in grigio nella figura).
 | [
"Solution:\n\nIl triangolo $ABC$ è isoscele (gli angoli $\\widehat{BAC}$ e $\\widehat{ABC}$ corrispondono a due angoli alterni interni nel rettangolo diviso da una diagonale) e la sua area è $S = CB \\cdot AD / 2$ ($AD$ è l'altezza relativa a $CB$).\n\nPosto $DB = a$, $AD = b$, $CB = AC = x$ e $DC = a - x$, il teor... | Italy | XV GARA NAZIONALE DI MATEMATICA | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | b(a^2 + b^2)/(4a) | |
0i3a | Problem:
The graph of $x^{2}-(y-1)^{2}=1$ has one tangent line with positive slope that passes through $(x, y)=(0,0)$. If the point of tangency is $(a, b)$, find $\sin ^{-1}\left(\frac{a}{b}\right)$ in radians. | [
"Solution:\nDifferentiating both sides of the equation, we find that $2x - 2(y-1) \\frac{\\mathrm{d}y}{\\mathrm{d}x} = 0$, and so $\\frac{\\mathrm{d}y}{\\mathrm{d}x} = \\frac{x}{y-1} = \\frac{a}{b-1}$. The line passing through $(0,0)$ and $(a, b)$ has slope $\\frac{b}{a}$, so $\\frac{b}{a} = \\frac{a}{b-1}$. Solvin... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | pi/4 | |
0bfi | Let $p \ge 2$ be a prime, $(K, +, \cdot)$ be a field with $p^2$ elements and
$$
L = \{0, 1, \dots, \underbrace{1+1+\cdots+1}_{p-1 \text{ times}}\}.
$$
Prove that for every $x \in K \setminus L$ there exists $a, b \in L$ so that $x^{2013} + ax + b = 0$. | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Algebra > Abstract Algebra > Field Theory",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
0a01 | Problem:
Zij $n$ een positief geheel getal. Voor een reëel getal $x \geq 1$ geldt dat $\left\lfloor x^{n+1}\right\rfloor$, $\left\lfloor x^{n+2}\right\rfloor, \ldots,\left\lfloor x^{4 n}\right\rfloor$ allemaal kwadraten van positieve gehele getallen zijn. Bewijs dat $\lfloor x\rfloor$ ook het kwadraat van een positief... | [
"Solution:\n\nWe bewijzen dit eerst voor $n=1$. Schrijf $x=a+r$, met $a \\geq 1$ geheel en $0 \\leq r<1$. Stel $\\left\\lfloor x^{2}\\right\\rfloor$, $\\left\\lfloor x^{3}\\right\\rfloor$ en $\\left\\lfloor x^{4}\\right\\rfloor$ zijn kwadraten. Er geldt dus $a \\leq x<a+1$, waaruit volgt dat $a^{2} \\leq x^{2}<(a+1... | Netherlands | IMO-selectietoets II | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
04og | Let $n$ be a positive integer. Prove that there exists a positive integer $k$ such that
$$
51^k - 17
$$
is divisible by $2^n$. | [
"### Claim.\nFor every positive integer $n \\ge 3$ there exists a positive integer $k$ such that $51^{2^{n-2}} - 1 = 2^n \\cdot (2k + 1)$.\n\n*Proof.* We prove the claim by mathematical induction on $n$.\n\nFor $n=3$ we have $51^2 - 1 = 3^2 - 1 = 8$ (mod 16), so $51^2 - 1$ is divisible by $2^3$ but not by $2^4$, th... | Croatia | Croatian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
01j5 | In an acute scalene triangle $\triangle ABC$ points $P, Q$ lie on the smaller arcs $\widehat{AB}$ and $\widehat{AC}$, respectively, and are the intersection of the midline parallel to $BC$ with the circumcircle of $\triangle ABC$. Points $X$ and $Y$ are the intersection of the perpendicular bisectors of segments $AB$ a... | [
"\nShortlist: Keep confidential!\n\n**Step 1:** Points $X, P, O, Q, Y$ lie on a circle.\n*Proof.* Since point $X$ lies on the perpendicular bisector of $AB$, we have that $XA, XB$ are tangents to $\\odot(ABC)$. This means that $\\angle XAO = \\angle OBX = 90^{\\circ}$, therefore quadrilater... | Baltic Way | Baltic Way 2023 Shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasi... | English | proof only | null | |
0i1b | Problem:
Let $O$ be the center of a circle $k$. Points $A, B, C, D, E, F$ on $k$ are chosen such that the triangles $OAB$, $OCD$, $OEF$ are equilateral. Let $L, M, N$ be the midpoints of $BC$, $DE$, $FA$ respectively. Prove that triangle $LMN$ also is equilateral. | [
"Solution:\n\nThe simplest way to do this is using complex numbers. Let $\\omega = (-1 + i \\sqrt{3}) / 2$, a cube root of unity, and we have $1 + \\omega = -\\omega^{2}$. Then a triangle $XYZ$ is equilateral (with vertices labeled in counterclockwise order) iff segment $ZX$ is the rotation image of $YZ$ through an... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
01bl | Let $\Gamma$ be the circumcircle of acute triangle $ABC$. The perpendicular to $AB$ from $C$ meets $AB$ at $D$ and $\Gamma$ again at $E$. The bisector of angle $C$ meets $AB$ at $F$ and $\Gamma$ again at $G$. The line $GD$ meets $\Gamma$ again at $H$ and the line $HF$ meets $\Gamma$ again at $I$. Prove that $AI = EB$.
... | [
"Since $CG$ bisects $\\angle ACB$, we have $\\angle AHG = \\angle ACG = \\angle GCB$. Thus, from the triangle $ADH$ we find that $\\angle HDB = \\angle HAB + \\angle AHG = \\angle HCB + \\angle GCB = \\angle GCH$. It follows that a pair of opposite angles in the quadrilateral $CFDH$ are supplementary, whence $CFDH$... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0ak9 | Let $a_1, a_2, \ldots, a_n$ be $n \ge 2$ real numbers such that $0 \le a_i \le \frac{\pi}{2}$. Prove that
$$
\left( \frac{1}{n} \sum_{i=1}^{n} \frac{1}{1 + \sin a_i} \right) \left( 1 + \prod_{i=1}^{n} (\sin a_i)^{\frac{1}{n}} \right) \le 1.
$$ | [] | North Macedonia | Mediterranean Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
03lw | Problem:
A self-avoiding rook walk on a chessboard (a rectangular grid of squares) is a path traced by a sequence of rook moves parallel to an edge of the board from one unit square to another, such that each begins where the previous move ended and such that no move ever crosses a square that has previously been cros... | [
"Solution:\n\nLet $r_{n}=R(3, n)$. It can be checked directly that $r_{1}=1$ and $r_{2}=4$. Let $1 \\leq i \\leq 3$ and $1 \\leq j$; let $(i, j)$ denote the cell in the $i$\\text{th} row from the bottom and the $j$\\text{th} column from the left, so that the paths in question go from $(1,1)$ to $(3,1)$.\nSuppose th... | Canada | 40th Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (1/(2*sqrt(2)))*((1+sqrt(2))^(n+1) - (1-sqrt(2))^(n+1)) - 1 | |
0f5e | Problem:
$X$ is a union of $k$ disjoint intervals of the real line. It has the property that for any $h < 1$ we can find two points of $X$ which are a distance $h$ apart. Show that the sum of the lengths of the intervals in $X$ is at least $1 / k$. | [] | Soviet Union | 17th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
00lp | The nonnegative real numbers $a$ and $b$ satisfy $a + b = 1$. Prove that
$$
\frac{1}{2} \le \frac{a^3 + b^3}{a^2 + b^2} \le 1.
$$
When do we have equality in the right inequality and when in the left inequality? | [
"$$\n\\frac{a^3 + b^3}{a^2 + b^2} = (a + b) \\frac{a^2 - ab + b^2}{a^2 + b^2} = 1 - \\frac{ab}{a^2 + b^2}.\n$$\nFrom this the right inequality is evident with equality for $ab = 0$, i.e. for $a = 0$, $b = 1$ and for $a = 1$, $b = 0$.\n\nThe left inequality is equivalent to\n$$\n\\frac{1}{2} \\le 1 - \\frac{ab}{a^2 ... | Austria | 48th Austrian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Right inequality equality holds when (a, b) = (1, 0) or (0, 1). Left inequality equality holds when a = b = 1/2. | |
05ex | Problem:
Soit $x, y$ des réels. Montrer que:
$$
|x|+|y| \leqslant |x-y|+|x+y|.
$$ | [
"Solution:\n\nSi $x$ et $y$ sont positifs, $|x|+|y|=x+y=|x+y|$.\n\nSi $x$ et $y$ sont négatifs $|x|+|y|=-x-y=|x+y|$.\n\nSi $x$ est positif et $y$ négatif $|x|+|y|=x-y=|x-y|$.\n\nSi $x$ est négatif et $y$ positif $|x|+|y|=y-x=|x-y|$.\n\nDans tous les cas par positivité de la valeur absolue $|x|+|y| \\leqslant |x-y|+... | France | ENVOI 2 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
01dw | A grasshopper is jumping along the set $\mathbb{Z}$ of integers. He starts at the origin; and for each jump, he may decide whether to jump to the left or to the right. For each $n \in \mathbb{N}_0$, the $n$-th jump has length $n^2$.
Prove or disprove that for each $k \in \mathbb{Z}$ the grasshopper can arrive at $k$ st... | [
"Clearly, the grasshopper can arrive at the integers $+1$ and $-1$, in one jump. And also, he can arrive at the integer $14$ using three jumps to the right ($1 + 4 + 9 = 14$).\nNote the following: if the grasshopper can arrive at a number $a \\in \\mathbb{Z}$ using $n$ jumps, then he can also arrive at the numbers ... | Baltic Way | Baltic Way shortlist | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0a05 | Problem:
Voor reële getallen $x$ en $y$ definiëren we $M(x, y)$ als het maximum van de drie getallen $x y$, $(x-1)(y-1)$ en $x+y-2 x y$. Bepaal de kleinst mogelijke waarde van $M(x, y)$ over alle reële getallen $x$ en $y$ met $0 \leq x, y \leq 1$. | [
"Solution:\n\nWe laten zien dat de minimale waarde $\\frac{4}{9}$ is. Deze waarde kan bereikt worden door $x=y=\\frac{2}{3}$ te nemen. Dan geldt $x y=\\frac{4}{9}$, $(x-1)(y-1)=\\frac{1}{9}$ en $x+y-2 x y=\\frac{4}{9}$, dus dan is het maximum inderdaad $\\frac{4}{9}$.\n\nNu gaan we bewijzen dat $M(x, y) \\geq \\fra... | Netherlands | MO-selectietoets | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 4/9 | |
03q0 | Let $n \ge 2$ be an integer. Find the largest real number $\lambda$ such that the inequality
$$
a_n^2 \ge \lambda(a_1 + a_2 + \cdots + a_{n-1}) + 2a_n
$$
holds for any positive integers $a_1, a_2, \dots, a_n$ satisfying $a_1 < a_2 < \dots < a_n$. | [
"For $a_i = i$, $i = 1, 2, \\dots, n$, we have $\\lambda \\le (n-2) \\div \\frac{n-1}{2} = \\frac{2n-4}{n-1}$.\n\nSince $a_k \\le a_n - (n-k)$, $k = 1, 2, \\dots, n-1$, $a_n \\ge n$, then we have\n$$\n\\begin{aligned}\n\\frac{2n-4}{n-1} \\sum_{i=1}^{n-1} a_i &\\le \\frac{2n-4}{n-1} \\left( (n-1)a_n - \\frac{n(n-1)}... | China | China Girls' Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | (2n-4)/(n-1) | |
0f47 | Problem:
Each of the numbers from $100$ to $999$ inclusive is written on a separate card. The cards are arranged in a pile in random order. We take cards off the pile one at a time and stack them into $10$ piles according to the last digit. We then put the $1$ pile on top of the $0$ pile, the $2$ pile on top of the $1... | [] | Soviet Union | 15th ASU | [
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
01fr | Let $ABCD$ be a convex quadrilateral such that $|AD| < |AB|$ and $|CD| < |CB|$. Prove that $\angle ABC < \angle ADC$. | [
"Let $D'$ be the point symmetric to point $D$ with respect to line $AC$. As triangles $ADC$ and $AD'C$ are similar, it suffices to prove that the inequalities $\\angle CAB + \\angle CAD' < 180^\\circ$ and $\\angle ACB + \\angle ACD' < 180^\\circ$ together imply $\\angle ABC < \\angle AD'C$.\n\nSuppose the contrary,... | Baltic Way | Baltic Way 2019 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gwo | Let's consider all the increasing geometric progressions and select only those that have the maximal number $M$ of elements in the set $A = \{1,2,3,...,2008\}$. Find $M$. | [
"Let one of the progressions $\\{a_n\\}$ which has the required maximal number $M$ of common members have the first member $a_0$ and denominator $q_0 > 1$. The progression has some common elements with set $A$. Let $n$ be the least number in $A$, which can also be found in the progression. Then you can define a new... | Ukraine | Ukrajina 2008 | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 11 | |
0als | Problem:
Let $f$ be a polynomial function that satisfies $f(x-5) = -3x^{2} + 45x - 108$. Find the roots of $f(x)$. | [
"Solution:\nLet $y = x - 5$, so $x = y + 5$.\n\nThen $f(y) = -3(x)^{2} + 45x - 108 = -3(y+5)^{2} + 45(y+5) - 108$.\n\nExpand $(y+5)^{2}$:\n$(y+5)^{2} = y^{2} + 10y + 25$\n\nSo:\n$f(y) = -3(y^{2} + 10y + 25) + 45y + 225 - 108$\n$= -3y^{2} - 30y - 75 + 45y + 225 - 108$\n$= -3y^{2} + 15y + (225 - 108 - 75)$\n$= -3y^{2... | Philippines | AREA STAGE | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 7, -2 | |
0cbt | Find all the functions $f : \mathbb{Q} \to \mathbb{Q}$, so that $f(1 - xy) = (1 - f(x))(1 - f(y))$, for every rational numbers $x$ and $y$. | [
"Let us find all functions $f : \\mathbb{Q} \\to \\mathbb{Q}$ such that for all $x, y \\in \\mathbb{Q}$,\n$$\nf(1 - xy) = (1 - f(x))(1 - f(y)).\n$$\n\nLet us denote the given equation as $(*)$.\n\nStep 1. Plug $x = 0$ into $(*)$:\n$$\nf(1 - 0 \\cdot y) = (1 - f(0))(1 - f(y)) \\implies f(1) = (1 - f(0))(1 - f(y)).\n... | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1 - x | |
05od | Problem:
Trouver tous les entiers naturels $n$ pour lesquels $2^{n} + 12^{n} + 2011^{n}$ est un carré parfait. | [
"Solution:\n\nPosons $u_{n} = 2^{n} + 12^{n} + 2011^{n}$. On regarde modulo $3$ : $2 \\equiv -1 \\pmod{3}$, $12 \\equiv 0 \\pmod{3}$ et $2011 \\equiv 1 \\pmod{3}$, donc\n$$\nu_{n} \\equiv (-1)^{n} + 0^{n} + 1^{n} \\equiv (-1)^{n} + 1 \\pmod{3}$$\nSi $n$ est pair, $u_{n} \\equiv 2 \\pmod{3}$. Or aucun carré ne peut ... | France | OCympiades Françaises de Mathématiques | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1 | |
04rd | Let $A$, $B$ be sets of positive integers such that a sum of arbitrary two different numbers from $A$ is in $B$ and a ratio of arbitrary two different numbers from $B$ (greater one to smaller one) is in $A$. Find the maximum number of elements in $A \cup B$.
(Martin Panák) | [
"Initially we will prove that the set $A$ consists of at most two numbers. Suppose that three numbers $a < b < c$ belong to the set $A$. Then the numbers $a + b < a + c < b + c$ are in $B$ and therefore the number\n$$\n\\frac{b + c}{a + c} = 1 + \\frac{b - a}{a + c}\n$$\nhas to be in $A$. This is a contradiction be... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 5 | |
0fi6 | Problem:
Calcular la suma de los cuadrados de los cien primeros términos de una progresión aritmética, sabiendo que la suma de ellos vale $-1$, y la suma de los términos de lugar par vale $+1$. | [
"Solution:\n\nSea la progresión, $a, a+d, a+2d, \\ldots, a+99d$; tenemos que hallar\n$$\n\\begin{aligned}\nS = a^{2} + (a+d)^{2} + \\ldots + (a+99d)^{2} = \\\\\n= 100a^{2} + 2ad(1+2+\\ldots+99) + d^{2}\\left(1^{2} + 2^{2} + \\ldots + 99^{2}\\right)\n\\end{aligned}\n$$\nPara calcular $a$ y $d$ resolvemos el sistema\... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 299.98 | |
0clk | Let $(x_n)_{n \ge 1}$ be an increasing unbounded sequence of natural numbers such that $x_1 = 1$ and $x_{n+1} \le 2x_n$ for all $n \ge 1$. Prove that every nonzero natural number can be written as a finite sum of pairwise distinct terms of the sequence $(x_n)_{n \ge 1}$.
*Note:* Two terms $x_i$ and $x_j$ of the sequen... | [
"Let $k$ be a nonzero natural number such that $1 \\le k < 2x_n$ for some natural number $n \\ge 1$. We will prove by induction that we can write $k = \\sum_{i=1}^{n} \\varepsilon_i x_i$ with $\\varepsilon_i \\in \\{0, 1\\}$ for $i = 1, \\dots, n$. Since the sequence $(x_n)_{n \\ge 1}$ is unbounded, we can cover al... | Romania | 75th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0fui | Problem:
Sei $ABC$ ein spitzwinkliges Dreieck mit Höhenschnittpunkt $H$ und seien $M$ und $N$ zwei Punkte auf $BC$, so dass $\overrightarrow{MN}=\overrightarrow{BC}$. Seien $P$ und $Q$ die Projektionen von $M$ bzw. $N$ auf $AC$ bzw. $AB$. Zeige, dass $APHQ$ ein Sehnenviereck ist. | [
"Solution:\n\nDie Translation um den Vektor $\\overrightarrow{BM}$ bildet $B$ auf $M$ und $C$ auf $N$ ab. Sei $G$ das Bild von $H$ bei dieser Translation. Wir zeigen nun, dass $P, Q$ und $H$ auf dem Thaleskreis über der Strecke $AG$ liegen.\nDie Gerade $BH$ steht senkrecht auf $AC$, somit auch das Bild, die Gerade ... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > ... | null | proof only | null | |
02rv | Problem:
Patrícia quer escrever algarismos nas faces de dois cubos de madeira de tal modo que qualquer dia do mês possa ser representado juntando um algarismo de uma face de um dos cubos e outro algarismo de uma face do outro cubo. Por exemplo, para representar o dia primeiro, Patrícia junta os cubos de modo a mostrar... | [
"Solution:\n\na) Os dias do mês que são escritos com dois algarismos iguais são 11 e 22. Logo, o 1 e o 2 devem aparecer em ambos os cubos. Além disso, embora não exista um dia 00 no mês, o zero deve aparecer em ambos os cubos, vejamos o porquê disso. Se o zero estivesse em apenas um dos cubos, para formar os dias d... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | a) The digits 0, 1, and 2 must appear on both cubes.
b) One valid assignment is: Cube A: 0, 1, 2, 3, 4, 5; Cube B: 0, 1, 2, 6, 7, 8. | |
07i0 | Find all the functions $f : \mathbb{N} \to \mathbb{N}$ such that for $x, y \in \mathbb{N}$:
$$
0 \le y + f(x) + f^{f(y)}(x) \le 1
$$
where
$$
f^n(x) = \underbrace{f(f(\cdots(f(x))\cdots))}_{n}
$$ | [
"**Lemma 1.** For every natural number $x$, $\\{f^n(x)\\}$ is unbounded.\n*Proof.* For every natural number $y$,\n$$\nf^{f(y)}(x) \\ge y + f(x) - 1,\n$$\nSo, send $y$ to the infinity, we are done.\n\n**Lemma 2.** All the terms of the sequence $(f^n(x))$ are distinct.\n*Proof.* If $i < j$ and $f^i(x) = f^j(x)$ then ... | Iran | 40th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(n) = n + 1 for all natural numbers n | |
0jbe | At a certain orphanage, every pair of orphans are either friends or enemies of each other. For every three of an orphan's friends, an even number of pairs of them are enemies. Prove that it's possible to assign each orphan two parents such that every pair of friends shares exactly one parent, but no pair of enemies doe... | [
"**Solution:** Form a graph $G$ with the orphans as vertices and with an edge between two orphans if they are friends. Let $A_1, A_2, \\dots$ be the maximal cliques of $G$; that is, every pair of orphans in $A_i$ is friends, and every orphan outside $A_i$ is an enemy of some orphan in $A_i$. We now have a sequence ... | United States | Team Selection Test Selection Test | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
06ql | On a $999 \times 999$ board a limp rook can move in the following way: From any square it can move to any of its adjacent squares, i.e. a square having a common side with it, and every move must be a turn, i.e. the directions of any two consecutive moves must be perpendicular. A nonintersecting route of the limp rook c... | [
"The answer is $998^{2}-4=4 \\cdot\\left(499^{2}-1\\right)$ squares.\n\nFirst we show that this number is an upper bound for the number of cells a limp rook can visit. To do this we color the cells with four colors $A, B, C$ and $D$ in the following way: for $(i, j) \\equiv(0,0) \\bmod 2$ use $A$, for $(i, j) \\equ... | IMO | IMO Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 996000 | |
05mo | Problem:
Pour tout entier $n \in \mathbb{N}^{*}$, on note $v_{3}(n)$ la valuation 3-adique de $n$, c'est-à-dire le plus grand entier $k$ tel que $n$ est divisible par $3^{k}$. On pose $u_{1}=2$ et $u_{n}=4 v_{3}(n)+2-\frac{2}{u_{n-1}}$ pour tout $n \geqslant 2$ (si tant est que $u_{n-1}$ soit défini et non nul).
Mont... | [
"Solution:\n\nTout d'abord, notons que $u_{1}=2, u_{2}=1, u_{3}=3, u_{4}=\\frac{3}{2}, u_{5}=\\frac{2}{3}$ et $u_{6}=3$. On prouve par récurrence que, pour tout entier $n \\geq 2$, on a $0<u_{n}$ et\n$$\n0<u_{3 n-1}<1<u_{3 n-2}<2<u_{3 n}=2+u_{n}:\n$$\nc'est déjà vrai pour $n=2$.\nPuisque $u_{3 n}=u_{n}+2$, on montr... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0btm | A set of $n$ points in Euclidean 3-dimensional space, no four of which are coplanar, is partitioned into two subsets $A$ and $B$. An $\mathcal{A}\mathcal{B}$-tree is a configuration of $n-1$ segments, each of which has an endpoint in $A$ and the other in $B$, and such that no segments form a closed polyline. An $\mathc... | [
"The configurations of segments under consideration are all bipartite geometric trees on the points $n$ whose vertex-parts are $A$ and $B$, and transforming one into another preserves the degree of any vertex in $A$, but not necessarily that of a vertex in $B$.\n\nThe idea is to devise a strict semi-invariant of th... | Romania | 2016 Eighth Romanian Master of Mathematics | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof only | null | |
05v9 | Problem:
Déterminer tous les polynômes $P \in \mathbb{Z}[X]$ tels que :
(i) $P(n) \geqslant 1$ pour tout $n \geqslant 1$
(ii) $P(mn)$ et $P(m) P(n)$ ont le même nombre de diviseurs premiers pour tous $m, n \geqslant 1$. | [
"Solution:\n\nDans la suite, pour tout entier $n \\in \\mathbb{N}^{*}$, on note $\\mathcal{D}(n)$ l'ensemble des diviseurs premiers de $n$, et $\\delta(n)=|\\mathcal{D}(n)|$. Soit $P \\in \\mathbb{Z}[X]$ un polynôme vérifiant les deux conditions.\n\nOn commence par montrer le résultat suivant :\n\nLemme : Soit $n \... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | All polynomials of the form a·X^d with a a positive integer and d a nonnegative integer. | |
04xh | Let $ABCD$ be a convex quadrilateral such that
$$
|AB| + |CD| = \sqrt{2} \cdot |AC| \quad \text{and} \quad |BC| + |DA| = \sqrt{2} \cdot |BD|.
$$
Prove that $ABCD$ is a parallelogram. | [
"The result immediately follows from the following **LEMMA**. If $ABCD$ is any quadrilateral (convex or non-convex), then\n$$\n(|AB| + |CD|)^2 + (|BC| + |DA|)^2 \\geq 2|AC|^2 + 2|BD|^2,\n$$\nwith the equality if and only if $ABCD$ is a parallelogram.\n\n**PROOF OF LEMMA.** The vectors $\\mathbf{a} = \\overrightarro... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
0jga | In scalene triangle $ABC$, let the feet of the perpendiculars from $A$ to $BC$, $B$ to $CA$, $C$ to $AB$ be $A_1$, $B_1$, $C_1$, respectively. Denote by $A_2$ the intersection of lines $BC$ and $B_1C_1$. Define $B_2$ and $C_2$ analogously. Let $D$, $E$, $F$ be the respective midpoints of sides $BC$, $CA$, $AB$. Show th... | [
"We claim that the point of concurrency is $H$, the orthocenter of triangle $ABC$. By symmetry, it suffices to show that the perpendicular from $D$ to $AA_2$ passes through $H$.\n\nLet $A_3$ be the projection of $D$ onto $AA_2$. Because $\\angle AA_1D = \\angle AA_3D = 90^\\circ$ and $\\angle BC_1C = \\angle BB_1C ... | United States | TSTST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07xc | Find all positive integers $n$ and $m$ such that
$$
\binom{n}{1} + \binom{n}{3} = 2^m.
$$ | [
"Suppose $n$ and $m$ satisfy the equation $n + n(n-1)(n-2)/6 = 2^m$. Rearranging the equation gives\n$$\nn(n^2 - 3n + 8) = 3 \\cdot 2^{m+1}. \\qquad (2)\n$$\nFirst we note that $n^2-3n+8 > n$ is equivalent to $n^2-4n+8 = (n-2)^2+4 > 0$, which is true for all integers $n$. Moreover, $n^2 - 3n + 8 \\ge 3n$ is equival... | Ireland | IRL_ABooklet_2024 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Alge... | null | proof and answer | [(1, 2), (2, 3), (3, 4), (6, 8), (11, 24)] | |
053g | Find all natural numbers $n$ such that the equation $x^2 + y^2 + z^2 = nxyz$ has solutions in positive integers. | [
"For $n = 1$ one of the solutions is $x = y = z = 3$ and for $n = 3$ one of the solutions is $x = y = z = 1$.\n\nIf $n$ is even and there exists an integer solution, then the right-hand side of the equation is even. This is possible only if at least one of the numbers $x, y, z$ is even. Then the right-hand side is ... | Estonia | IMO Team Selection Contest | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 1 and 3 | |
0clh | Suppose $A, B \in \mathcal{M}_n(\mathbb{C})$ are such that $A + B = AB + BA$. Prove that:
a) if $n$ is odd, then $\det(AB - BA) = 0$;
b) if $\det(A) \ne \det(B)$, then $\det(AB - BA) = 0$. | [
"a) Define $C = 2A - I_n$ and $D = 2B - I_n$. We have $CD - I_n = -(DC - I_n) = 2(AB - BA)$. As $n$ is odd, $\\det(CD - I_n) = -\\det(DC - I_n)$.\nAs for any two square matrices $X, Y$, we have $\\det(XY - I_n) = \\det(YX - I_n)$, in our case we get $\\det(CD - I_n) = -\\det(CD - I_n)$, that is $\\det(CD - I_n) = 0... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
0ktw | Problem:
A tangent line to a circle is a line that intersects the circle in exactly one point. A common tangent line to two circles is a line that is tangent to both circles. As an example, in the figure to the right, line $a$ is a common tangent to both circles, but line $b$ is only tangent to the larger circle.
![]... | [
"Solution:\n\nThe possible values of $n$ are $0, 1, 2, 3$, and $4$. These cases are illustrated below.\n\n"
] | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | 0, 1, 2, 3, 4 | |
01pt | Given a decimal fraction $d$ such that there are only digits 0, 1, and 2 in its decimal representation. It is known that if each 0 in the decimal representation of $d$ are replaced by 1, then the obtained decimal fraction is periodic; if each 1 in the decimal representation of $d$ are replaced by 2, then the obtained d... | [
"Answer: yes, one can.\nSuppose that if we replace each 0 in the decimal fraction $d$ by 1 we obtain the decimal fraction $a$, and if we replace each 1 in the decimal fraction $d$ by 2 we obtain the decimal fraction $b$. Since both the fractions $a$ and $b$ are periodic, we see that both these numbers are rational ... | Belarus | BelarusMO 2013_s | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | Yes, the decimal fraction is periodic. | |
0cty | King Hiero has 11 metal pieces indistinguishable in appearance. The King knows that their weights (in some order) are $1, 2, \ldots, 11$ kg. Also the King has a bag that breaks if it contains more than $11$ kg. Archimedes knows the weight of each piece, and he wants to prove to Hiero that the first piece weighs $1$ kg.... | [
"The moves are $1+2+3+5$ and $1+4+6$.\n\nShow that Archimedes needs to use the bag twice. Let him first put into the bag the pieces weighing $1$, $2$, $3$, and $5$ kg, and then the pieces weighing $1$, $4$, and $6$ kg. In both cases, the bag does not break.\n\nLet us prove that this could only happen if the $1$ kg ... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English; Russian | proof and answer | 2 | |
0iry | Problem:
How many numbers between $1$ and $1,000,000$ are perfect squares but not perfect cubes? | [
"Solution:\n\nA number is both a perfect square and a perfect cube if and only if it is exactly a perfect sixth power. So, the answer is the number of perfect squares, minus the number of perfect sixth powers, which is $1000 - 10 = 990$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 990 | |
07ff | Let $n > 2$ be a natural number. Prove that the following equation has no solution $x_1, x_2, \dots, x_n$ in positive integers numbers greater than 1:
$$
(x_1x_2\dots x_n)^2 = x_1^3 + x_2^3 + \dots + x_n^3
$$ | [
"Assume that $(x_1, x_2, \\dots, x_n)$ is a solution to the equation. Without loss of generality, let $x_n \\ge x_{n-1} \\ge \\dots \\ge x_1$. Since $(x_1x_2\\dots x_n)^2$ is divisible by $x_n^2$, we have\n$$\nx_n^2 \\mid x_1^3 + x_2^3 + \\dots + x_{n-1}^3 \\implies x_n^2 \\le x_1^3 + x_2^3 + \\dots + x_{n-1}^3.\n$... | Iran | 37th Iranian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0k8q | Problem:
For which positive integers $n$ does the polynomial $P(X) = X^{n} + X^{n-1} + \cdots + 1$ have a real root? | [
"Solution:\nThe answer is odd $n$ only. For such odd $n$, one can take $X = -1$ as a real root.\n\nSuppose $n$ is even. We claim $P$ has no real roots. Indeed, note first $P(1) = n + 1$, so $1$ is not a real root. But for any $x \\neq 1$, we have\n$$\nP(x) = x^{n} + \\cdots + 1 = \\frac{x^{n+1} - 1}{x - 1} \\neq 0\... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | All odd positive integers n | |
0065 | Diremos que un grupo de tres personas es *simétrico* si cada una conoce a las otras dos o bien cada una no conoce a ninguna de las otras dos. En una fiesta hay 20 personas y cada una conoce a exactamente otras 9 personas de la fiesta. Halle el número de grupos simétricos de tres personas que hay en la fiesta. | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | Spanish | proof and answer | 240 | |
051a | Let $n$ be a positive integer and $a_1, \dots, a_{2n}$ be real numbers in $\left[-\frac{1}{2}, \frac{1}{2}\right]$. Leaving out any one of the numbers, the sum of the remaining $2n-1$ numbers is always an integer. Prove that $a_1 = \dots = a_{2n}$. | [
"Assume that there exist $a_i$ and $a_j$ which are not equal. Let $S = a_1 + \\dots + a_{2n}$. Since $S - a_i$ and $S - a_j$ are integers, their difference $(S - a_i) - (S - a_j) = a_j - a_i$ is also an integer. Since $a_j - a_i \\neq 0$, and they belong to $\\left[-\\frac{1}{2}, \\frac{1}{2}\\right]$, their differ... | Estonia | Estonian Math Competitions | [
"Number Theory > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0a2s | On a one-way road with two lanes, there are blue and red cars driving. The left lane contains only blue cars, the right lane only red. At a certain point, the cars have to merge to one lane. In doing so, the blue cars are a bit bolder than the red ones, sometimes causing several blue cars to end up behind each other, b... | [] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | MCQ | C | |
0hcg | Andrii chooses a sign in front of each number in the expression $\pm 1 \pm 2 \pm 3 \pm \dots \pm 2018$. How many different positive values can Andrii obtain as a result of the computed expression? | [
"Clearly, the expression cannot equal $0$, since it contains $1009$ odd numbers. We want to show that the expression can be any odd number between $1$ and $1+2+3+\\dots+2018$.\n\nTake some configuration. Find the first from the left consecutive numbers with signs \"-\" and \"+\". The switching of these signs decrea... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 1018586 | |
097p | Problem:
Fie triunghiul ascuţitunghic $ABC$ cu $AB > AC$. Punctul $F$, situat pe latura $(BC)$, este piciorul înălţimii duse din vârful $A$, iar $H$ este ortocentrul triunghiului $ABC$. Pe semidreapta $(BC)$ se ia un punct $D$ astfel încât $C \in (BD)$. Cercul circumscris triunghiului $DFH$ intersectează segmentul $(A... | [
"Solution:\n\nDeoarece $m(\\angle DFH) = 90^{\\circ} \\Rightarrow DH$ este diametrul cercului circumscris triunghiului $DFH$. Cercul circumscris triunghiului $DFH$ intersectează segmentul $(AD)$ în punctul $N$, prin urmare, $m(\\angle DNH) = 90^{\\circ} = m(\\angle ANH)$. De aici obţinem că dreapta $NH$ este perpen... | Moldova | Olimpiada Republicană la Matematică, Ziua a doua | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0aoz | Problem:
For what values of $k$ does the equation
$$
|x-2007|+|x+2007|=k
$$
have $(-\infty,-2007) \cup(2007,+\infty)$ as its solution set? | [
"Solution:\nIf $x \\in (-\\infty, -2007)$, then\n$$\n-(x-2007)-(x+2007)=k \\quad \\text{or} \\quad x=-\\frac{k}{2}.\n$$\nIf $x \\in [-2007, 2007]$, then\n$$\n-(x-2007)+(x+2007)=k \\quad \\text{or} \\quad k=4014.\n$$\nIf $x \\in (2007, +\\infty)$, then\n$$\n(x-2007)+(x+2007)=k \\quad \\text{or} \\quad x=\\frac{k}{2}... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | k > 4014 | |
0i91 | Problem:
If $x$, $y$, and $z$ are real numbers such that $2x^{2} + y^{2} + z^{2} = 2x - 4y + 2xz - 5$, find the maximum possible value of $x - y + z$. | [
"Solution:\nThe equation rearranges as $(x-1)^{2} + (y+2)^{2} + (x-z)^{2} = 0$, so we must have $x = 1$, $y = -2$, $z = 1$, giving us $4$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 4 | |
08ld | Problem:
If for the real numbers $x_{1}, x_{2}, \ldots, x_{n}$ it is $0 < x_{i} < 1$, for any $i$, show that
$$
1 + \sum_{1 \leq i < j \leq n} x_{i} x_{j} > \sum_{i=1}^{n} x_{i}
$$ | [
"Solution:\nWe'll prove it by induction.\nFor $n=1$ the desired result becomes $1 > x_{1}$ which is true.\nLet the result be true for some natural number $n \\geq 1$.\nWe'll prove it to be true for $n+1$ as well, and we'll be done.\nSo let $x_{1}, x_{2}, \\ldots, x_{n}, x_{n+1}$ be $n+1$ given real numbers with $0 ... | JBMO | 2008 Shortlist JBMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0hoi | Problem:
A point $M$ and a circle $k$ are given in the plane. If $ABCD$ is an arbitrary square inscribed in $k$, prove that the sum $MA^{4} + MB^{4} + MC^{4} + MD^{4}$ is independent of the positioning of the square. | [
"Solution:\n\nBy scaling, assume $k$ has radius $1$. Position the circle in the coordinate plane with its center $O$ at the origin, and points $A$, $B$, $C$, $D$ at $(1,0)$, $(0,1)$, $(-1,0)$, $(0,-1)$, respectively. Let $M$ have coordinates $(x, y)$. Then, using the Pythagorean theorem to expand $MA^{2}$, $MB^{2}$... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
00vc | Let $a, b$ be positive integers such that $a + 1, b + 1$ and $ab$ are all perfect squares. Prove that $\text{gcd}(a, b) + 1$ is also a perfect square. | [
"Let $a + 1 = A^2$, $b + 1 = B^2$ where $1 < A < B$ are positive integers. We prove the result by induction on $\\max\\{a, b\\} = b$.\nDefine $C := AB - \\sqrt{ab} = AB - \\sqrt{(A^2 - 1)(B^2 - 1)}$ which is an integer as $ab$ is a perfect square. Define $c := C^2 - 1$. We will now prove the following:\n* $1 < c < ... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Infinite descent / root flipping"
] | English | proof only | null | |
0ciz | Let $f : [\frac{1}{2}, 2] \to \mathbb{R}$ be a continuous and monotone function, for which the following equality holds:
$$
\int_{\frac{1}{2}}^{2} \frac{(1-x^2) \cdot f(x)}{1+x^2+x^4} \, dx = 0.
$$
Show that
$$
\int_{\frac{1}{2}}^{2} \frac{x f(x)}{1+x^2} \, dx = f(2) \cdot \ln 2.
$$ | [] | Romania | 75th NMO | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | English | proof and answer | f(2) · ln 2 | |
05rj | Problem:
Soit $ABCD$ un trapèze tel que $(AB)$ soit parallèle à $(CD)$. Soit $P$ un point de $[AC]$ et $Q$ un point de $[BD]$ tels que $\widehat{APD} = \widehat{BQC}$.
Démontrer que $\widehat{AQD} = \widehat{BPC}$. | [
"Solution:\n\nL'énoncé nous donne plein d'égalités d'angles : exploitons-les ! Tout d'abord, on remarque que\n$$\n\\widehat{DPC} = 180^{\\circ} - \\widehat{APD} = 180^{\\circ} - \\widehat{CQB} = \\widehat{DQC},\n$$\nce qui signifie que les points $C$, $D$, $P$ et $Q$ sont cocycliques.\n\nOn en déduit que\n$$\n\\wid... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09ey | The graph $G$ with $2014$ vertices does not contain a triangle as subgraph. If set of degrees of vertices of $G$ is $\{1, 2, \ldots, k\}$ then find maximal value of $k$. | [
"Suppose that $k \\geq 1343$ and denote $A$ a vertex degree of which equals to $1343$.\n\nLet's consider a vertex degree of which is no greater than $672$ and denote it by $B$. Suppose that vertices $A, B$ are connected by an edge. Then from remaining $2012$ vertices $1342$ are connected with $A$ and $671$ are conn... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1342 | |
0asf | Problem:
Let $\llbracket x \rrbracket$ denote the greatest integer less than or equal to the real number $x$. What is the largest two-digit integral value of $x \llbracket x \rrbracket$? | [
"Solution:\n\n99"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 99 | |
0gyj | Find all pairs of natural numbers $n$, $k$, for which the following equality holds:
$$
n^3 - 2 = k!
$$ | [
"If $k \\ge 2$ then the sum $(k!+2)$ is divisible by $2$, and so $n=2m$. Then for $k \\ge 4$ in the equality $(2m)^3 - k! = 2$ the left part is divisible by $4$, but the right one is not. Thus, all that we need is to consider the cases $k=1,2,3$. An easy check shows that there is an only solution $n=2, k=3$."
] | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | n=2, k=3 | |
05si | Problem:
De combien de façons peut-on placer 7 tours sur un échiquier $7 \times 7$ telle qu'aucune tour ne puisse en attaquer une autre?
Une tour peut attaquer une autre tour si elle se situe sur la même ligne ou la même colonne. | [
"Solution:\n\nSi on place 7 tours de sorte que deux tours ne soient pas sur la même colonne, sachant qu'il y a 7 colonnes, alors il y aura exactement une tour sur chaque colonne (et de même sur chaque ligne).\n\nIl y a 7 positions possibles pour placer une tour sur la première colonne. Il y a ensuite 6 positions po... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 5040 | |
06jg | $A$, $B$ and $C$ are three persons among a set $P$ of $n$ ($n \ge 3$) persons. It is known that $A$, $B$ and $C$ are friends of one another, and that every one of the three persons has already made friends with more than half the total number of people in $P$. Given that every three persons who are friends of one anoth... | [
"The answer is $\\frac{n-1}{2}$ if $n$ is odd, $\\frac{n+2}{2}$ if $n \\ge 6$ is even, and $4$ if $n = 4$.\n\nLet $A$, $B$, $C$ be the sets of friends of $A$, $B$, $C$ respectively excluding $A$, $B$, $C$ themselves.\n\n**Case 1.** $n$ is odd\n\nSince each of $A$, $B$, $C$ has at least $\\frac{n+1}{2}$ friends, we ... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | If n is odd: (n−1)/2. If n is even and at least 6: (n+2)/2. If n = 4: 4. | |
022f | Problem:
Um tabuleiro quadrado de 3 linhas por 3 colunas contém nove casas. De quantos modos diferentes podemos escrever as três letras $A$, $B$ e $C$ em três casas diferentes, de modo que em cada linha esteja escrita exatamente uma letra? | [
"Solution:\n\nComeçando com a letra $A$, ela pode ser escrita em qualquer uma das 9 casas do tabuleiro. Uma vez escrita a letra $A$, sobram 6 casas onde a letra $B$ pode ser escrita. Uma vez escritas as letras $A$ e $B$ no tabuleiro, sobram 3 casas para a letra $C$ ser escrita.\nAssim, pelo princípio multiplicativo... | Brazil | Nível 3 | [
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 162 | |
0aoi | Problem:
Placed on a really long table are 2011 boxes each containing a number of balls. The 1st and the 2nd box together contain 15 balls, the 2nd and the 3rd box together contain 16 balls, the 3rd and the 4th box together contain 17 balls, and so on. If the first and the last box together contain a total of 1023 bal... | [] | Philippines | Area Stage | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1014 | |
0gg2 | 令正方形 $ABCD$ 的內切圓為 $\Gamma$。設點 $M$ 為線段 $CD$ 的中點, 點 $P$ 為線段 $AB$ 上異於 $B$ 的一點。設點 $E$ 為圓 $\Gamma$ 上異於 $M$ 的一點, 且直線 $DP$ 平行於直線 $EM$。設直線 $CP$ 與 $AD$ 交於點 $F$。證明直線 $EF$ 與圓 $\Gamma$ 相切。 | [] | Taiwan | 2022 數學奧林匹亞競賽第二階段培訓營, 國際競賽實作(二) | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate ... | Chinese; English | proof only | null | |
0h51 | Numbers $x_1, x_2, \ldots, x_{2015}$ fulfill both equalities simultaneously:
$x_1^{2014} + x_2^{2014} + \ldots + x_{2015}^{2014} = 1$ and $x_1^{2015} + x_2^{2015} + \ldots + x_{2015}^{2015} = -1$. | [
"From the first equation $-1 \\le x_i \\le 1$, $i = 1, \\ldots, 2015$, so for all $i$: $0 \\le 1 + x_i \\le 2$. Add both equations and get\n$$\nx_1^{2014}(1+x_1) + x_2^{2014}(1+x_2) + \\ldots + x_{2015}^{2014}(1+x_{2015}) = 0.\n$$\nSince every item $x_i^{2014}(1+x_i) \\ge 0$, then their sum equals zero if and only ... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour) | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | proof and answer | Exactly one number equals negative one and all the others equal zero. | |
0cny | Given triangles $ABC$ and $A_1B_1C_1$ with equal areas. Determine if it is always possible to construct (using a ruler and a compass) a triangle $A_2B_2C_2$ such that the triangles $A_1B_1C_1$ and $A_2B_2C_2$ are congruent, and the lines $AA_2$, $BB_2$ and $CC_2$ are parallel. (D. Tereshin) | [
"**Answer.** Yes, always.\nIf $\\Delta ABC = \\Delta A_1B_1C_1$, then the construction method is obvious. Otherwise, we may assume that $AB \\neq A_1B_1$ (let's suppose $AB < A_1B_1$ for definiteness).\n\nConstruct triangle $A'B'C$ such that $AB \\parallel A'B'$, $A'B' = A_1B_1$, $B'C = B_1C_1$, $CA' = C_1A_1$ (see... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof and answer | Yes, always. | |
04xn | Let $a$ be a given integer. Prove that there exist infinitely many prime numbers $p$ such that
$$
p \mid n^2 + 3, \quad p \mid m^3 - a
$$
for some integers $n$ and $m$. | [
"Let $k$ be an arbitrary integer. Note that\n$$\n(9a^2k^3)^2 + 3 = 3(27a^4k^6 + 1)\n$$\nand\n$$\n(9a^3k^4)^3 - a = a(3^6a^8k^{12} - 1) = a(27a^4k^6 - 1)(27a^4k^6 + 1)\n$$\nIt follows that for every $k \\in \\mathbb{Z}$ the number $27a^4k^6 + 1$ is a common divisor of the numbers $n^2 + 3$ and $m^3 - a$ with $n = 9a... | Czech-Polish-Slovak Mathematical Match | 11-th Czech-Slovak-Polish Match, 2011 | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
00jz | Let $a_1, a_2, \dots, a_n$ be non-negative integers such that for all real numbers $x_1 > x_2 > x_3 > \dots > x_n > 0$ with $x_1 + x_2 + \dots + x_n < 1$ it holds that $\sum_{k=1}^n a_k x_k^3 < 1$.
Show that
$$
na_1 + (n-1)a_2 + \dots + (n-j+1)a_j + \dots + a_n \le \frac{n^2(n+1)^2}{4}.
$$
G. Baron, Vienna | [
"The assertion can be rewritten as\n$$\n\\sum_{k=1}^{n} \\sum_{j=1}^{k} a_j \\le \\sum_{k=1}^{n} k^3.\n$$\nIt therefore suffices to prove\n$$\n\\sum_{j=1}^{k} a_j \\le k^3 \\quad (6)\n$$\nfor every $k = 1, \\dots, n$.\nIn order to prove (6) we fix $k$ and set\n$$\nx_i = \\frac{1}{k} - \\frac{i}{N} \\quad \\text{for... | Austria | AustriaMO2013 | [
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof only | null | |
01er | Let's say that a digit is *eternal* for a positive integer $n$, if it is contained in every multiple of $n$. Find all digits which are eternal for at least one positive integer. | [
"The only such a digit is $0$, it is contained in every multiple of $10$. Let's show that no other digit is eternal for any positive integer.\n\nAssume that some digit is eternal for integer $n$. Consider remainders of numbers\n$$\n1, 11, 111, \\dots, \\underbrace{11\\dots11}_{n+1}\n$$\nmodulo $n$. By the pigeonhol... | Baltic Way | Baltic Way shortlist | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 0 | |
07td | A school has $2020$ pupils, no two of whom are the same height. It also has a long bench on which the pupils can all sit in some order. An *arrangement* means some ordering of pupils on the bench. A *swap* means two pupils exchanging places while the other $2018$ stay fixed. An arrangement is said to be *unimodal* if t... | [
"We will prove the following stronger statement: For any positive integer $n$ and any two unimodal arrangements of $n$ pupils, there is a sequence of swaps leading from one of the arrangements to the other, via unimodal arrangements.\n\nWe show this by induction on $n$. The result plainly holds when $n = 1$, as the... | Ireland | IRL_ABooklet_2020 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Yes | |
0gib | 令 $n$ 與 $k$ 為正整數。寶寶用 $n^2$ 個數字積木拼成一個 $n \times n$ 的方陣，每塊積木都是一個不超過 $k$ 的正整數。路過的奶爸一看，發現：
1. 方陣上每一橫列的數字都可以視為以最左方數字為首項的等差數列，且其公差都不同；
2. 方陣上每一直排的數字都可以視為以最上方數字為首項的等差數列，且其公差都不同，
試求 $k$ 的最小可能值 (以 $n$ 的函數表示。)
註：公差可能非正。
Let $n$ and $k$ be positive integers. A baby uses $n^2$ blocks to form a $n \times n$ grid, with each of the ... | [
"對於一個 $n \\times n$ 的方陣，當 $n = 2m + 1$ 時，$k$ 的最小可能值是 $2m^2 + 1$；當 $n = 2m$ 時，$k$ 的最小可能值是 $2m^2 - m + 1$。\n\n估計：首先考慮令 $[x]$ 為不大於 $x$ 的最大整數，並令 $m = [n/2]$。由於每一橫列的公差都不同，因此需要有 $n \\ge 2m$ 個不同的公差；又因為從 $-m + 1$ 到 $m - 1$ 全部也只有 $2(m-1) + 1 = 2m - 1 < n$ 個數，因此必然有一個橫列的公差 $\\ge m$ 或 $\\le -m$。不失一般性，假設該橫列的公差 $\\ge m$，則由於首項 $\... | Taiwan | Taiwan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | Chinese; English | proof and answer | If n = 2m + 1, the minimum k is 2m^2 + 1. If n = 2m, the minimum k is 2m^2 − m + 1. | |
0hwl | Problem:
Decide whether there exist positive integers $a, b, c$ such that $3(a b + b c + c a)$ divides $a^{2} + b^{2} + c^{2}$. | [
"Solution:\n\nThe answer is no: such integers do not exist. In what follows, $\\nu_{p}(n)$ will denote the exponent of $p$ in the prime factorization of $n$.\n\nAssume without loss of generality that $a, b, c$ do not have some common divisor, and $a^{2} + b^{2} + c^{2} = 3k(a b + b c + c a)$. Write\n$$\n(3k + 2)\\l... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Mu... | null | proof and answer | No | |
0ah0 | Find all integers $m$ for which $m^3 + m^2 + 7$ is divisible by $m^2 - m + 1$. | [
"From $m^3 + m^2 + 7 = (m^2 - m + 1)(m+2) + (m+5)$ it follows that $m^2 - m + 1$ is a divisor of $m+5$. Obviously, one solution is $m = -5$.\n\nLet $m \\neq -5$. Then $|m^2 - m + 1| \\le |m+5|$. From $m^2 - m + 1 = (m - \\frac{1}{2})^2 + \\frac{3}{4} > 0$ it follows that $m^2 - m + 1 \\le m + 5$ or $m^2 - m + 1 \\l... | North Macedonia | XV Junior Macedonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | {-5, 0, 1} | |
00qw | A chessboard of size $1000 \times 1000$ is tiled with tiles of size $1 \times 10$. You do not know the tiling but wish to uncover it. In order to do so, you can choose some $N$ cells on the board, following which you will learn what the positions of the tiles that cover those cells are. What is the least $N$ such that ... | [
"We will show that $N = k^2$ is the desired least number for any $kn \\times kn$ board tiled with $1 \\times n$ tiles, $n \\ge 2$, thus the answer to our problem is $N = 100^2 = 10000$. From now on consider a $kn \\times kn$ board tiled with $1 \\times n$ tiles ($n \\ge 2$), with $k > 1$ fixed.\n\nFor a lower bound... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 10000 | |
06a4 | Find all values of the positive integer $\kappa$ satisfying the following property:
There do not exist positive integers $\alpha, \beta$ such that the number
$$
A(\kappa, \alpha, \beta) = \frac{\alpha + \beta}{\alpha^2 + \kappa^2 \beta^2 - \kappa^2 \alpha \beta}
$$
is a composite positive integer. | [
"We fix $\\kappa > 1$. The idea for the solution of the problem is to find $\\alpha, \\beta$ with respect to $\\kappa$ such that the denominator of the fraction is equal to $1$. In other words, we seek polynomials $\\alpha = P(\\kappa)$, $\\beta = Q(\\kappa)$ such that\n$$\nP(\\kappa)^2 + \\kappa^2 Q(\\kappa)^2 - \... | Greece | 37th Hellenic Mathematical Olympiad 2020 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 1 | |
0dwj | Problem:
V telenoveli o dogodkih $v$ zarotniškem mestecu nastopa $n$ meščanov, $n \geq 3$. Vsaka 2 meščana skupaj kujeta zaroto proti enemu izmed ostalih meščanov. Dokaži, da obstaja tak meščan, da je vsaj $\sqrt{n}$ meščanov vpletenih v zaroto proti njemu. | [
"Solution:\n\nVseh (neurejenih) parov meščanov je $\\frac{n(n-1)}{2}$. Vsak par označimo s tistim izmed $n-2$ meščanov, proti kateremu par kuje zaroto. Potem obstaja $m$ parov, ki so vsi označeni z istim meščanom, in je $m \\geq \\frac{n-1}{2}$. Denimo, da v teh $m$ parih sodeluje $k$ meščanov. Tedaj velja $\\frac{... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
05yu | Problem:
Trouver tous les nombres premiers $p, q$ vérifiant $p^{5}+p^{3}+2=q^{2}-q$. | [
"Solution:\n\nCe qu'on connaît pour des nombres premiers, c'est leurs propriétés de divisibilité. Une expression développée n'est donc pas très pratique pour étudier cela : on va d'abord chercher à factoriser.\n\nRemarquons que $p^{5}+p^{3}+2=p^{3}\\left(p^{2}+1\\right)+2$. On veut résoudre $p^{3}\\left(p^{2}+1\\ri... | France | Préparation Olympique Française de Mathématiques - Envoi 3: Arithmétique | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis... | null | proof and answer | [(2, 7), (3, 17)] | |
0ddi | Let $ABC$ be a triangle with $AB < AC$ and incircle $(I)$ tangent to $BC$ at $D$. Take $K$ on $AD$ such that $CD = CK$. Suppose that $AD$ cuts $(I)$ at $G$ and $BG$ cuts $CK$ at $L$. Prove that $K$ is the midpoint of $CL$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"G... | null | proof only | null | |
0c2x | Let $a, b \in \mathbb{R}$, $a < b$, and let $f : [a, b] \to [0, \infty)$ be a continuous nondecreasing function. Prove that
$$
\lim_{n \to \infty} \int_a^b \sqrt[n]{f^n(x) + f^n(a + b - x)} \, dx = 2 \int_{\frac{a+b}{2}}^b f(x) \, dx.
$$ | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Precalculus > Limits",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof and answer | 2 ∫_{(a+b)/2}^b f(x) dx | |
065u | We consider a $11 \times 10$ cm rectangular table. The table is divided by parallel lines to $110$ squares of side $1$ cm. We have tiles of cross shape, consisting of $6$ squares of side $1$ cm, as in the figure. Determine the maximal possible number of non-overlapping tiles we can pack in the table such that each tile... | [
"We observe that for the covering of the squares having a side on the border of the table and especially for the four corner squares we can put on each corner one tile covering only two squares, while four squares is not possible to be covered. Therefore in the four corners will be uncovered $16$ squares.\n\nFor th... | Greece | SELECTION EXAMINATION | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 14 | |
0gdp | 令 $\Omega$ 為三角形 $ABC$ 的 $A$-旁切圓, 並設其分別切直線 $BC$, $CA$, $AB$ 於點 $D$, $E$, $F$。
設 $M$ 為線段 $EF$ 的中點。在 $\Omega$ 上再取兩點 $P$, $Q$ 使得 $EP$ 與 $FQ$ 皆平行於 $DM$。
令 $BP$ 與 $CQ$ 交於點 $X$。證明：直線 $AM$ 是 $\angle XAD$ 的內角平分線。
註：$A$-旁切圓指的是在 $\angle A$ 內部的旁切圓。
Let $\Omega$ be the $A$-excircle of triangle $ABC$, and suppose that $\Omega$ is ... | [
"延長 $DM$ 交 $\\Omega$ 另一點於 $R$。設 $AR$ 分別交 $EF$ 與 $\\Omega$ 於 $U$, $V \\ne R$, 那麼由 $FREV$ 為調和四邊形知\n$$\n\\frac{\\sin \\angle BAR}{\\sin \\angle RAC} = \\frac{\\sin \\angle FAU}{\\sin \\angle UAE} = \\frac{\\overline{FU}}{\\overline{UE}} = \\frac{\\overline{FR} \\cdot \\overline{FV}}{\\overline{RE} \\cdot \\overline{VE... | Taiwan | 2020 Taiwan IMO 3J | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcent... | null | proof only | null | |
081k | Problem:
Un atleta ha appena affrontato una gara di triathlon. Questa competizione si divide in tre fasi: la prima è una gara di nuoto, la seconda di ciclismo e la terza di corsa. Sapendo che la sua velocità media nei tre tratti è stata rispettivamente di $3\ \mathrm{Km}/\mathrm{h}$, $30\ \mathrm{Km}/\mathrm{h}$ e $17... | [
"Solution:\n\nLa risposta è $\\mathbf{( E )}$. Un modo per rendersi conto di ciò consiste nel mostrare due diverse suddivisioni del tracciato in tre parti per cui siano soddisfatte le condizioni del problema. Ad esempio: l'atleta potrebbe aver nuotato per $\\frac{7}{27}$ di ora, corso per un'ora e corso in biciclet... | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | E | |
0731 | Find all integer solutions $(x, y)$ of the equation $y^2 = x^3 - p^2x$, where $p$ is a prime such that $p \equiv 3 \pmod{4}$. | [
"$$\ny^2 = (x-p)(x)(x+p).\n$$\nWe split the analysis into the following cases.\n\nCase $p \\nmid y$. In this case, we have $(x-p,x) = (x,x+p) = 1$. If $x$ is even, then $(x-p,x+p) = 1$, so it follows that $x-p,x$ and $x+p$ are all squares. This is not possible since $x+p \\equiv 3 \\pmod{8}$. Thus, $x$ has to be od... | India | Indija TS 2007 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization >... | null | proof and answer | (x, y) = (0, 0), (p, 0), (-p, 0) | |
0bpl | Problem:
Matricele $A \in \mathcal{M}_{m, n}(\mathbb{C}), B \in \mathcal{M}_{n, m}(\mathbb{C})$, unde $m \geq n \geq 2$, au proprietatea că există $k \in \mathbb{N}^{*}$ și $a_{0}, a_{1}, \ldots, a_{k} \in \mathbb{C}$ astfel încât
$$
a_{k}(A B)^{k}+a_{k-1}(A B)^{k-1}+\cdots+a_{1}(A B)+a_{0} I_{m}=O_{m}
$$
iar
$$
a_{k}... | [
"Solution:\n\nPresupunem că $a_{0} \\neq 0$. Scriind egalitatea din ipoteză sub forma\n$$\nA B\\left(-\\frac{a_{k}}{a_{0}}(A B)^{k-1}-\\frac{a_{k-1}}{a_{0}}(A B)^{k-2}-\\cdots-\\frac{a_{1}}{a_{0}} I_{m}\\right)=I_{m}\n$$\ndeducem că $A B$ este inversabilă, deci $\\operatorname{rang}(A B)=m \\geq n$.\nPe de altă par... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
07o1 | a. Let $f(x) = x^2 + 8$. Prove that $f(x\sqrt{2})$ is the product $P(x)Q(x)$ of two polynomials of degree two with integer coefficients.
b. Let $g(x) = x^4 + x^3 + x^2 + x + 1$. Prove that $g(5x^2)$ is the product $R(x)S(x)$ of two polynomials of degree four with integer coefficients.
c. Let $h(x) = (x+1)^{13} - x^{1... | [
"a.\n$$(x^2\\sqrt{2})^2 + 8 = 2(x^4+4) = 2((x^2+2)^2 - 4x^2) = 2(x^2-2x+2)(x^2+2x+2).$$\n\nb.\n$g(x) = x^4 + x^3 + x^2 + x + 1 = (x - \\zeta)(x - \\zeta^2)(x - \\zeta^3)(x - \\zeta^4)$, where $\\zeta$ is a primitive fifth root of unity. Using $\\zeta = \\zeta^6$ and $\\zeta^3 = \\zeta^8$ we have\n$$\ng(x) = (x - \\... | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0dnq | Problem:
Нека је $O$ центар кружнице описане око $\triangle A B C$. Права $t$ додирује кружницу описану око $\triangle B O C$ и сече странице $A B$ и $A C$ у тачкама $D$ и $E$, редом $(D, E \not \equiv A)$. Тачка $A'$ је симетрична тачки $A$ у односу на праву $t$. Доказати да се кружнице описане око $\triangle A' D E$... | [
"Solution:\n\nОзначимо са $K$ додирну тачку праве $t$ и круга $B O C$. Нека се кругови описани око троуглова $B D K$ и $C E K$ секу у тачки $X \\neq K$. Како је $\\varangle B X C = \\varangle B X K + \\varangle K X C = \\varangle A D K + \\varangle K E A = 180^\\circ - \\varangle C A B$, тачка $X$ лежи на описаном ... | Serbia | 10. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid,... | null | proof only | null | |
009i | Given a positive integer $N$, we subtract from it its greatest proper divisor (different from $N$), then do the same with the new number and repeat the operation until $1$ is obtained. Find how many subtractions are there if the process starts with $N = 19^{19}$. | [
"Since $19$ is a prime, the greatest proper divisor of $N = 19^{19}$ is $19^{18}$ and the first number obtained is $N_1 = 19^{19} - 19^{18} = 18 \\cdot 19^{18}$. Note that if a number $m$ is even then the operation gives $m/2$. So the next number is $N_2 = N_1/2 = 9 \\cdot 19^{18}$. The greatest proper divisor of $... | Argentina | NATIONAL XXX OMA | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 114 | |
0011 | En una circunferencia $\Gamma$ se considera una cuerda $PQ$ tal que el segmento que une el punto medio del menor arco $\overarc{PQ}$ y el punto medio del segmento $PQ$ mide $1$. Sean $\Gamma_1$, $\Gamma_2$ y $\Gamma_3$ tres circunferencias tangentes a la cuerda $PQ$ que están en el mismo semiplano que el centro de $\Ga... | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | español | proof and answer | 1/2 | |
0f9w | Problem:
Can the squares of a $1990 \times 1990$ chessboard be colored black or white so that half the squares in each row and column are black and cells symmetric with respect to the center are of opposite color? | [
"Solution:\n\nAnswer no\n\n\nSuppose it can be done. Divide the board into 4 quadrants. Suppose there are $b$ black and $995^2 - b$ white squares in the top left quadrant. Then there are $995^2 - b$ black and $b$ white squares in the bottom right quadrant (by the symmetry property).\n\nIf h... | Soviet Union | 24th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No | |
0k2q | Problem:
A $4 \times 4$ window is made out of 16 square windowpanes. How many ways are there to stain each of the windowpanes, red, pink, or magenta, such that each windowpane is the same color as exactly two of its neighbors? Two different windowpanes are neighbors if they share a side. | [
"Solution:\n\nFor the purpose of explaining this solution, let's label the squares as\n11121314\n21222324\n31323334\n41424344\n\nNote that since the corner squares $11,14,41,44$ each only have two neighbors, each corner square is the same color as both of its neighbors (for example, $11,12$, and $21$ are the same c... | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 24 | |
05om | Problem:
Soit $ABC$ un triangle acutangle. On note $O$ le centre de son cercle circonscrit. On considère le cercle $\Gamma_{B}$ passant par $A$ et $B$ et qui est tangent à $(AC)$, ainsi que le cercle $\Gamma_{C}$ passant par $A$ et $C$ qui est tangent à $(AB)$. Une droite $\Delta$ passant par $A$ recoupe $\Gamma_{B}$ ... | [
"Solution:\n\nSoit $Z$ le point d'intersection du cercle circonscrit de $ABC$ avec $\\Delta$. En faisant une chasse aux angles, on trouve que $\\widehat{XBA} = \\widehat{ZAC} = \\widehat{ZBC}$. On en déduit que $\\widehat{XBZ} = \\widehat{ABC}$. Or, comme $\\widehat{BCA} = \\widehat{BZA}$, on en déduit que les tria... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle... | null | proof only | null | |
06ms | In a chess tournament there are $100$ players. On each day of the tournament, each player is designated to be 'white', 'black' or 'idle', and each 'white' player will play a game against every 'black' player. (You may assume that all games fixed for the day can be finished within that day.) At the end of the tournament... | [
"Clearly the tournament can be run in $99$ days if we assign, on day $k$ (where $1 \\le k \\le 99$),\n\n* player $k$ to be 'white';\n* players $k + 1$, $k + 2$, $\\ldots$, $100$ to be 'black'; and\n* all other players to be 'idle'.\n\nIn this way player $i$ and player $j$ (where $1 \\le i < j \\le 100$) would have ... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | 99 | |
09v8 | On each of the twelve edges of a cube we write the number $1$ or $-1$. For each face of the cube, we multiply the four numbers on the edges of this face and write the outcome on this face. Finally, we add the eighteen numbers that we wrote down.
What is the smallest (most negative) result we... | [
"$-12$"
] | Netherlands | Second Round, March 2019 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | -12 | |
04vf | Find all triples of positive integers $a$, $b$, $c$ for which the product
$$(a+b)(b+c)(c+a)(a+b+c+2036)$$
is equal to the power of a prime number with an integer exponent. | [
"First, let us note that at least one of the numbers $a+b$, $a+c$ and $b+c$ must be even. Indeed, two of the three numbers $a$, $b$, $c$ have the same parity, so their sum is even.\nIf the product we investigate is a power of a prime $p$, then each of the four factors must be a power of $p$. As we already know, one... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | (4, 4, 4) | |
0eta | Find the smallest and largest integers with decimal representation of the form $ababa$ ($a \neq 0$) that are divisible by 11. | [
"The number $N = \\underline{ababa}$ is divisible by 11 if and only if $a-b+a-b+a = 3a-2b$ is divisible by 11. To find the smallest possible $N$ with this property, we observe that $a=1$ (the smallest possible value for $a$) and $b=7$ offer a solution: $3 \\cdot 1 - 2 \\cdot 7 = -11$ is divisible by 11 (and no smal... | South Africa | The South African Mathematical Olympiad, Third Round | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English, Afrikaans | proof and answer | 17171 and 98989 | |
0dbt | Find all primes $p$ such that there exist integers $m$ and $n$ satisfying $p = m^{2} + n^{2}$ and $p \mid m^{3} + n^{3} + 8 m n$. | [
"If $m = 0$ then $p = n^{2}$ is a prime, which is impossible. Similar case happens when $n = 0$. So we may assume $m n \\neq 0$. Then $p > |m|$ and $p > |n|$. Note that\n$$\nm^{3} + n^{3} = (m + n)\\left(m^{2} - m n + n^{2}\\right) \\equiv -m n(m + n) \\pmod{p}\n$$\nSo the problem statement is equivalent to\n$$\nm ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 2, 5, 13 | |
07mx | Suppose the function $f$ is defined and real valued on the real numbers, and its graph is symmetric about the lines $x = 1/4$ and $x = 3/4$. Prove that $f(x) = f(x+1)$ for all real numbers $x$. Exhibit a non-constant, nonnegative function with the given properties. | [
"To say that the graph of $f$ is symmetric about the line $x = a$ is to mean that $f(x_1) = f(x_2)$ whenever $x_1, x_2$ are real numbers whose sum is $2a$. Hence, by hypothesis,\n$$\nf\\left(\\frac{1}{4} - x\\right) = f\\left(\\frac{1}{4} + x\\right) \\quad \\text{and} \\quad f\\left(\\frac{3}{4} - x\\right) = f\\l... | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = sin^2(2πx) |
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