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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
07sg | Let $x_1, x_2, \dots, x_{2020}$ be non-negative real numbers such that
$$
x_i + x_{i+1} + x_{i+2} \le 2 \quad \text{for } i = 1, 2, \dots, 2018.
$$
Show that
$$
\sum_{i=1}^{2018} x_i x_{i+2} \le 1009.
$$ | [
"Let us consider the products in pairs, starting with $x_1x_3 + x_2x_4$. This relates to four consecutive terms. Setting $c = \\max\\{x_1, x_4\\}$ gives $x_1x_3 \\le c x_3$ and $x_2x_4 \\le c x_2$. Moreover, the assumptions imply $c + x_2 + x_3 \\le 2$ and so\n$$\nx_1x_3 + x_2x_4 \\le c(x_2 + x_3) \\le \\left(\\fra... | Ireland | IRL_ABooklet_2020 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0j2z | Problem:
Suppose that a polynomial of the form $p(x) = x^{2010} \pm x^{2009} \pm \cdots \pm x \pm 1$ has no real roots. What is the maximum possible number of coefficients of $-1$ in $p$? | [
"Solution:\nLet $p(x)$ be a polynomial with the maximum number of minus signs.\n\n$p(x)$ cannot have more than $1005$ minus signs, otherwise $p(1) < 0$ and $p(2) \\geq 2^{2010} - 2^{2009} - \\ldots - 2 - 1 = 1$, which implies, by the Intermediate Value Theorem, that $p$ must have a root greater than $1$.\n\nLet $p(... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1005 | |
080q | Problem:
Si scelgano i punti $H, K, M$ sui lati di un triangolo $A B C$ in modo tale che $A H$ sia un'altezza, $B K$ sia una bisettrice e $C M$ sia una mediana. Si indichi con $D$ l'intersezione tra $A H$ e $B K$, e con $E$ l'intersezione tra $H M$ e $B K$. Sapendo che $K D=2, D E=1, E B=3$ :
(i) si dimostri che $H M$... | [
"Solution:\n\ni) I triangoli $E M B$ e $K A B$ sono simili perché\n$$\nM B : A B = E B : K B.\n$$\ne l'angolo in $B$ è in comune. Quindi $K \\hat{A} B = E \\hat{M} B$ e $C A \\parallel M H$.\n\nii) Per il teorema di Talete si ha $C B = 2 H B$, da cui deduciamo che $C H = H B$ e che $A H$ è la mediana relativa a $C ... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"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 | |
0a2y | At Matthijs's table tennis club, one keeps the ping-pong balls on a table with cylindrical ball holders. Here is the side view of a ping-pong ball on top of a ball holder. The underside of the ball exactly touches the table. It is known that the ball holder is $4$ centimetres wide and $1$ centimetre high.
How many cent... | [] | Netherlands | Dutch Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | B | |
07vh | Aisling and Brendan take alternate moves in the following game. Before the game starts, the number $x = 2023$ is written on a piece of paper. Aisling makes the first move. A move from a positive integer $x$ consists of replacing $x$ either with $x + 1$ or with $x/p$ where $p$ is a prime factor of $x$.
The winner is th... | [
"The game is a win for Aisling. Aisling wins by forcing Brendan to write down a prime number, which allows Aisling to claim the prize by writing 1 at the next step.\n\nWe say an integer is a *2g-position* if it is of the form $2g$ where both $g$ and $2g+1$ are primes (such $g$ are known as Sophie Germain primes). I... | Ireland | IRL_ABooklet_2023 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | Aisling | |
08m5 | Problem:
In a right trapezoid $ABCD$ ($AB \parallel CD$) the angle at vertex $B$ measures $75^{\circ}$. Point $H$ is the foot of the perpendicular from point $A$ to the line $BC$. If $BH = DC$ and $AD + AH = 8$, find the area of $ABCD$. | [
"Solution:\n\nProduce the legs of the trapezoid until they intersect at point $E$. The triangles $ABH$ and $ECD$ are congruent (ASA). The area of $ABCD$ is equal to area of triangle $EAH$ of hypotenuse\n$$\nAE = AD + DE = AD + AH = 8\n$$\nLet $M$ be the midpoint of $AE$. Then\n$$\nME = MA = MH = 4\n$$\nand $\\angle... | JBMO | 2009 Shortlist JBMO | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 8 | |
0brn | Let $ABCD$ be a cyclic quadrilateral whose diagonals are not perpendicular and intersect at $X$. Let $A', C'$ be the projections of $A$ and $C$ onto the line $BD$ and let $B', D'$ be the projections of $B$ and $D$ onto $AC$. Prove that:
a) the perpendicular lines drawn from the midpoints of the sides onto the opposite... | [
"a) Let $O$ be the circumcenter of $ABCD$. It is well known that the midpoints of the sides of a quadrilateral $ABCD$ are the vertices of a parallelogram, hence the line segments joining the midpoints of two opposite sides have the same midpoint, $G$. The perpendicular lines from $O$ to $AB$ and $CD$ pass through t... | Romania | 67th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Pla... | English | proof only | null | |
05r3 | Problem:
Trouver tous les entiers strictement positifs $p, q$ tels que
$$
p 2^{q} = q 2^{p}.
$$ | [
"Solution:\nPremier cas : si $p = q$, alors l'égalité est vraie.\n\nSecond cas : si $p \\neq q$, on peut supposer sans perte de généralité que $p > q$, le cas $q > p$ se traitant de même. On remarque que tout diviseur impair de $p$ est un diviseur impair de $q$, et réciproquement. Ainsi, si on écrit $p = a 2^{b}$ e... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 5 : Pot-POURRI | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All pairs with equal entries (p, p) for any positive integer p, together with (2, 1) and (1, 2). | |
0ixv | Problem:
There are five guys named Alan, Bob, Casey, Dan, and Eric. Each one either always tells the truth or always lies. You overhear the following discussion between them:
```
Alan: "All of us are truth-tellers."
Bob: "No, only Alan and I are truth-tellers."
Casey: "You are both liars."
Dan: "If Casey is a truth-t... | [
"Solution:\n\nAlan, Bob, Dan, and Eric are liars.\n\nAlan and Bob each claim that both of them are telling the truth, but they disagree on the others. Therefore, they must both be liars, and Casey must be a truth-teller. If Dan is a truth-teller, then so is Eric, but then there would only be two truth-tellers, cont... | United States | 2nd Annual Harvard-MIT November Tournament | [
"Discrete Mathematics > Logic"
] | null | proof and answer | Alan, Bob, Dan, and Eric are liars; Casey is a truth-teller. | |
0400 | Given a prime number $p$, let $A$ be a $p \times p$ matrix such that its entries are exactly $1, 2, \dots, p^2$ in some order. The following operation is allowed for a matrix: add one to each number in a row or a column, or subtract one from each number in a row or a column. The matrix $A$ is called “good” if one can t... | [
"We may combine the operations on the same row or column, thus the final result of a series of operations can be realized as subtracting integer $x_i$ from each number of $i$-th row and subtracting integer $y_j$ from each number of $j$-th column. Thus, the matrix $A$ is good if and only if there exist integers $x_i... | China | China Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 2(p!)^2 | |
03c5 | The sequence $a_1, a_2, \dots$ is defined by the equalities $a_1 = 2$, $a_2 = 12$ and $a_{n+1} = 6a_n - a_{n-1}$ for every positive integer $n \ge 2$. Prove that no member of this sequence is equal to a perfect power (greater than one) of a positive integer. | [
"We shall use the following assertion.\n\n**Lemma.** Let $k \\ge 2$ be a positive integer. Then the equation $2x^{2k} + 1 = y^2$ does not have solutions in positive integers.\n\n**Proof.** Assume that $x, y$ and $k \\ge 2$ are positive integers such that $2x^{2k} + 1 = y^2$ and $x$ is minimum possible. It is obviou... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Infinite descent / root flipping"
] | English | proof only | null | |
06mp | Let $S$ be the set of all integers of the form $x^2 + 3xy + 8y^2$ where $x$ and $y$ are integers.
a. Show that if $u$ and $v$ are in $S$, then so is $uv$.
b. Can an integer of the form $23k + 7$, with $k$ an integer, belong to $S$? | [
"a.\nThe roots of $z^2 + 3z + 8 = 0$ are $\\frac{-3 \\pm \\sqrt{23}i}{2}$. Let $\\alpha = \\frac{-3 + \\sqrt{23}i}{2}$. Then $\\bar{\\alpha} = \\frac{-3 - \\sqrt{23}i}{2}$, and hence $x^2 + 3xy + 8y^2 = (x - \\alpha y)(x - \\bar{\\alpha}y)$. Note that\n$$\n\\begin{aligned}\n(x_1 - \\alpha y_1)(x_2 - \\alpha y_2) &=... | Hong Kong | IMO HK TST | [
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity"
] | null | proof and answer | a: yes; b: no | |
0hb1 | Consider the collection of lines of the form $y = (k+n)x + (k-n)$ on a plane, where $k, n$ are any integers. Is there a point with integer coordinates that doesn't belong to any of such lines?
**Answer:** yes, there is. | [
"Let $x=1$, then the second coordinate of all the points that belong to the lines is $y = (k+n) + (k-n) = 2k$ which is even. Therefore, the point $(1, 1)$ doesn't belong to any of the lines."
] | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | yes, there is | |
0jbf | Problem:
Let $x_{1}, x_{2}, \ldots, x_{k}$ be a sequence of integers. A rearrangement of this sequence (the numbers in the sequence listed in some other order) is called a scramble if no number in the new sequence is equal to the number originally in its location. For example, if the original sequence is $1,3,3,5$ then... | [
"Solution:\nFor the scrambles, we need to choose two locations from the $n-1$ numbers $2,3, \\ldots, n$ to be occupied by the two $1$s. Once this has been done, we are left with $n-1$ numbers, exactly two of which (the numbers whose locations were occupied by the $1$s) can be placed freely while all the rest have e... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0hl4 | Problem:
Find the exact value of $1+\frac{1}{1+\frac{2}{1+\frac{1}{1+\frac{2}{1+\ldots}}}}$. | [
"Solution:\nLet $x$ be what we are trying to find.\n\n$x - 1 = \\frac{1}{1 + \\frac{2}{1 + \\frac{1}{1 + \\frac{2}{1 + \\ldots}}}}$\n\n$\\Rightarrow \\frac{1}{x - 1} - 1 = \\frac{2}{1 + \\frac{1}{1 + \\frac{2}{1 + \\cdots}}}$\n\n$\\Rightarrow \\frac{2}{\\frac{1}{x - 1} - 1} = x$\n\n$\\Rightarrow x^2 - 2 = 0$\n\nso ... | United States | null | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | sqrt(2) | |
04w0 | Let $ABC$ be a triangle with side lengths $AB = 13$, $BC = 14$, $CA = 15$. Let $A'B'C'$ be a translate of $ABC$ by some unit vector. Find the largest possible area of the intersection of triangles $ABC$ and $A'B'C'$. | [] | Czech Republic | First Round (take-home) | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 5488/75 | |
0k95 | Problem:
The pairwise products $ab$, $bc$, $cd$, and $da$ of positive integers $a$, $b$, $c$, and $d$ are $64$, $88$, $120$, and $165$ in some order. Find $a+b+c+d$. | [
"Solution:\n\nThe sum $ab + bc + cd + da = (a + c)(b + d) = 437 = 19 \\cdot 23$, so $\\{a + c, b + d\\} = \\{19, 23\\}$ as having either pair sum to $1$ is impossible. Then the sum of all $4$ is $19 + 23 = 42$. (In fact, it is not difficult to see that the only possible solutions are $(a, b, c, d) = (8, 8, 11, 15)$... | United States | HMMT February 2019 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 42 | |
0bjp | Find the smallest integer $n$ for which the set $A = \{n, n+1, n+2, \dots, 2n\}$ contains five elements $a < b < c < d < e$ so that
$$
\frac{a}{c} = \frac{b}{d} = \frac{c}{e}.
$$ | [
"Let $p, q \\in \\mathbb{N}^*$, $(p, q) = 1$ so that $\\frac{a}{c} = \\frac{b}{d} = \\frac{c}{e} = \\frac{p}{q}$. Obviously, $p < q$. Since $a, b$ and $c$ are divisible by $p$ and $c, d, e$ are divisible by $q$, there exists $m \\in \\mathbb{N}^*$ so that $c = mpq$.\nLet us find the minimal value of the difference ... | Romania | 65th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 16 | |
0gk0 | Show that $\sum_{k=0}^{n} (-1)^k \binom{2n+1}{2k+1} 2008^k$ is not divisible by $19$ for every positive integer $n$. | [
"Observe that $-2008 \\equiv 6 \\equiv 5^2 \\pmod{19}$. Thus,\n$$\n\\begin{aligned}\n2 \\sum_{k=0}^{n} \\binom{2n+1}{2k+1} (-2008)^k &\\equiv 2 \\sum_{k=0}^{n} \\binom{2n+1}{2k+1} 5^{2k} \\pmod{19} \\\\\n&\\equiv (1+5)^{2n+1} - (1-5)^{2n+1} \\pmod{19} \\\\\n&\\equiv 6^{2n+1} + 4^{2n+1} \\\\\n&\\equiv 2^{2n+1} (3^{2... | Thailand | Thai Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
05dl | Problem:
Snow White and the Seven Dwarves are living in their house in the forest. On each of 16 consecutive days, some of the dwarves worked in the diamond mine while the remaining dwarves collected berries in the forest. No dwarf performed both types of work on the same day. On any two different (not necessarily con... | [
"Solution:\n\nWe define $V$ as the set of all 128 vectors of length 7 with entries in $\\{0,1\\}$. Every such vector encodes the work schedule of a single day: if the $i$-th entry is 0 then the $i$-th dwarf works in the mine, and if this entry is 1 then the $i$-th dwarf collects berries. The 16 working days corresp... | European Girls' Mathematical Olympiad (EGMO) | EGMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
07e9 | The Euler circle of the acute-angled triangle $ABC$ is reflected with respect to the altitude from $A$ to $BC$ and intersected the circumcircle of the triangle $ABC$ at distinct points $X$ and $Y$ ($X \neq Y$). Let $H$ be the orthocenter of triangle $ABC$. Prove that $AH$ is the external angle bisector of $\angle XHY$. | [
"Let $X'$ be the reflection of $X$ with respect to $AH$ and $X''$ be the reflection of $H$ with respect to $X'$. Notice that $X'$ lies on the nine point circle (Euler circle) and $X''$ lies on the circumcircle.\n\nLines $X''H$ and $XH$ intersect the circumcircle of triangle $ABC$ for the se... | Iran | Iranian 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 > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0glm | Let $a \neq 0$ be a real number. Determine all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying
$$
f(x)f(y) + f(x+y) = axy
$$
for all real numbers $x$ and $y$. | [
"Substituting $(x, y) = (0, 0)$ in (8) yields $f(0)^2 + f(0) = 0$; that is, $f(0) = 0$ or $f(0) = -1$. If $f(0) = 0$, then the substitution $y = 0$ in (8) yields $f(x) = 0$ for all $x \\in \\mathbb{R}$. However, the zero function does not satisfy (8), so we must have $f(0) = -1$.\n\nWe consider two cases regarding ... | Thailand | The 13th Thailand Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | If a > 0: f(x) = sqrt(a) x - 1 or f(x) = -sqrt(a) x - 1. If a < 0: no solution. | |
05m9 | Problem:
a) Prouver que, pour tous réels strictement positifs $a, b, k$ tels que $a < b$, on a
$$
\frac{a}{b} < \frac{a + k}{b + k}
$$
b) Prouver que
$$
\frac{1}{100} + \frac{4}{101} + \frac{7}{102} + \frac{10}{103} + \cdots + \frac{148}{149} > 25
$$ | [
"Solution:\na) On a $a k < b k$, donc $a b + a k < a b + b k$. Ceci s'écrit $a(b + k) < b(a + k)$, ou encore $\\frac{a}{b} < \\frac{a + k}{b + k}$.\n\nb) Soit $A = \\frac{1}{100} + \\frac{4}{101} + \\frac{7}{102} + \\frac{10}{103} + \\cdots + \\frac{148}{149}$. En appliquant ce qui précède, on en déduit\n$$\nA > \\... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
00ll | A necklace contains $2016$ pearls, each of which has one of the colours black, green or blue. In each step we replace simultaneously each pearl with a new pearl, where the colour of the new pearl is determined as follows: If the two original neighbours were of the same colour, the new pearl has their colour. If the nei... | [
"a. Since $2016$ is divisible by $4$, we can alternatingly take two black and two green pearls. In the first step, all pearls are already replaced by blue pearls.\n\nb. If we assign to each blue pearl the number $0$, to each green pearl the number $1$ and to each black pearl the number $2$, then it holds in each st... | Austria | 48th Austrian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a: yes; b: no; c: no | |
05td | Problem:
Au pays des merveilles se trouvent $n$ villes. Chaque paire de villes est reliée par une route à sens unique, qui part d'une des deux villes et arrive à l'autre. Afin de s'y retrouver, Alice interroge le roi de cœur : à chaque question, Alice choisit une paire de villes, et le roi de cœur lui dit quelle est l... | [
"Solution:\n\nNous allons décrire une stratégie qu'Alice peut mettre en place pour aboutir à ses fins en $5 n$ questions ou moins. À tout moment, on dira qu'une ville $v$ est mauvaise si Alice a déjà trouvé deux routes qui partent de $v$, et que $v$ est bonne sinon. De même, on dira qu'une paire de villes $\\{v, w\... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
00sf | Let $a_{ij}$, $i = 1, 2, \dots, m$ and $j = 1, 2, \dots, n$, be positive real numbers. Prove that
$$
\sum_{i=1}^{m} \left( \sum_{j=1}^{n} \frac{1}{a_{ij}} \right)^{-1} \le \left( \sum_{j=1}^{n} \left( \sum_{i=1}^{m} a_{ij} \right)^{-1} \right)^{-1}.
$$
When does the equality hold? | [
"We will use the following\n**Lemma.** If $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ are positive real numbers then\n$$\n\\frac{1}{\\sum_{j=1}^{n} \\frac{1}{a_j}} + \\frac{1}{\\sum_{j=1}^{n} \\frac{1}{b_j}} \\le \\frac{1}{\\sum_{j=1}^{n} \\frac{1}{a_j + b_j}}.\n$$\nThe equality holds when $\\frac{a_1}{b_1} = \\... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | Equality holds if and only if, for each row, the ratios of its entries to the corresponding entries of a fixed reference row are all equal; equivalently, all rows are proportional: a_{i1}/a_{11} = a_{i2}/a_{12} = ... = a_{in}/a_{1n} for every i. | |
0gkt | Each point of the plane is colored either red or blue. Show that there exists a triangle with side lengths $1$, $2$, $\sqrt{3}$, and its three vertices are of the same color. | [
"Assume on the contrary that there is a coloring for which any triangle with side lengths $1$, $2$, $\\sqrt{3}$ has at least one red vertex and one blue vertex. Consider an equilateral triangle $ABC$ with side length $2$.\n\n\n\nThere are at least two vertices among $A$, $B$, $C$ with the s... | Thailand | The 10th Thailand Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
08fp | Problem:
Siano $a$ e $b$ interi positivi tali che
$$
54^{a} = a^{b}.
$$
Dimostrare che $a$ è una potenza di $54$, cioè esiste un intero positivo $c$ tale che $a = 54^{c}$.
Problem:
Let $a$ and $b$ be positive integers such that
$$
54^{a} = a^{b}.
$$
Show that $a$ is a power of $54$, that is, that there exists a posi... | [
"Solution:\n\nOsserviamo che $54 = 2 \\cdot 3^{3}$, e pertanto $a$ è divisibile sia per $2$ sia per $3$, e non ha altri fattori primi oltre a $2$ e $3$. In altri termini, $a$ si scrive nella forma $a = 2^{x} \\cdot 3^{y}$ per opportuni interi positivi $x$ e $y$. Ne segue che\n$$\n54^{a} = \\left(2 \\cdot 3^{3}\\rig... | Italy | XXXIX Olimpiade Italiana di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0djr | Find all positive integer $n > 1$ such that: there exist some real coefficient polynomial $P(x)$ of degree $n$, having the leading coefficient as $1$ and there exist distinct real numbers $r, s, t$ with the sum is $-2023$ and $P(k) \in \{r, s, t\}$ for all $k = 1, 2, 3, \dots, 3n - 1, 3n$. | [
"For $n = 2$, we have $P(k) \\in \\{r, s, t\\}$ with $k = 1, 2, 3, 4, 5, 6$. Since $\\deg P = 2$, there are no more than $2$ value of $k$ such that $P(k) = r$. Similarly for $P(k) = s$, $P(k) = t$. From this it follows that each of the above equations must have exactly $2$ solutions. Notice that all three equations... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 2 | |
0d9g | Find all pairs of integers $(x, y)$ such that
$$
y^{3} = 8x^{6} + 2x^{3}y - y^{2}.
$$ | [
"We rewrite the equation:\n$$\ny^{3} + y^{2} - 2x^{3}y - 8x^{6} = 0.\n$$\n\nConsider this as a cubic in $y$:\n$$\ny^{3} + y^{2} - 2x^{3}y - 8x^{6} = 0.\n$$\n\nLet us try small integer values for $x$.\n\nIf $x = 0$:\n$$\ny^{3} + y^{2} = 0 \\implies y^{2}(y + 1) = 0 \\implies y = 0 \\text{ or } y = -1.\n$$\nSo $(0, 0... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | [(0, 0), (0, -1), (1, 2)] | |
0arn | Problem:
The system $x^{2}-y^{2}=0$, $(x-a)^{2}+y^{2}=1$ has generally at most four solutions. Find the values of $a$ so that the system has two or three solutions. | [
"Solution:\n\n(ans. $a= \\pm 1$ for two solutions, $a= \\pm \\sqrt{2}$ for three solutions.\nThe solutions are given by $x=\\frac{a \\pm \\sqrt{2-a^{2}}}{2},\\ y= \\pm x$. There are two solutions if the quadratic equation involving $x$ has a single solution $\\Rightarrow$ $2-a^{2}=0 \\Rightarrow a= \\pm \\sqrt{2}$.... | Philippines | 13th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | Two solutions when a = ±√2; three solutions when a = ±1 | |
0cqo | Let $a$, $b$ be real numbers such that $a^3 - b^3 = 2$ and $a^5 - b^5 \ge 4$. Prove that $a^2 + b^2 \ge 2$. (I. Bogdanov)
Числа $a$ и $b$ таковы, что $a^3 - b^3 = 2$, $a^5 - b^5 \ge 4$. Докажите, что $a^2 + b^2 \ge 2$. (И. Богданов) | [
"Заметим, что $2(a^2 + b^2) = (a^2 + b^2)(a^3 - b^3) = (a^5 - b^5) + a^2b^2(a - b) \\ge 4 + a^2b^2(a - b)$. Поскольку $a^3 > b^3$, мы имеем $a > b$, а значит, $a^2b^2(a - b) \\ge 0$. Итак, $2(a^2 + b^2) \\ge 4$, откуда и следует утверждение задачи."
] | Russia | Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English; Russian | proof only | null | |
0ghp | 以三角形 $ABC$ 的三條邊為邊，分別向 $ABC$ 的外面作正三角形 $ABC_1, BCA_1, CAB_1$。設點 $P$ 為 $ABC_1$ 的外接圓與 $CAB_1$ 的外接圓的另一個交點 ($P \neq A$)。在 $CAB_1$ 的外接圓上找一點 $Q$ 使得 $PQ$ 平行於 $BA_1$。在 $ABC_1$ 的外接圓上找一點 $R$ 使得 $PR$ 平行於 $CA_1$。
證明：三角形 $ABC$ 的重心，與三角形 $PQR$ 的重心連線，會平行於直線 $BC$。
Let $ABC$ be a triangle. Let $ABC_1, BCA_1, CAB_1$ be three equilateral t... | [
"暴力三角解析法. 給直線 $PA$, $PB$, $PC$ 定向, 使得由 $PA$ 到 $PB$、由 $PB$ 到 $PC$、由 $PC$ 到 $PA$ 的角度皆為 $\\frac{2\\pi}{3}$。另外也給直線 $PQ$, $PR$ 定向, 使得 $\\overrightarrow{BC}$ 與 $\\overrightarrow{PQ}$、$\\overrightarrow{PR}$ 與 $\\overrightarrow{BC}$ 的夾角也是 $\\frac{2\\pi}{3}$。令 $\\angle(\\overrightarrow{BC}, PA) = \\alpha$, $\\angle(\\overri... | Taiwan | 2023 數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geome... | Chinese (Traditional) | proof only | null | |
064h | Problem:
Es sei $ABCD$ ein konvexes Parallelogramm, das bei $A$ spitzwinklig ist. Die Spiegelpunkte von $A$ an den Geraden $BC$ und $CD$ seien mit $P$ bzw. $Q$ bezeichnet. Außerdem schneidet die Gerade $BD$ die Strecken $\overline{AP}$ und $\overline{AQ}$ im Inneren in den Punkten $R$ bzw. $S$.
Beweisen Sie, dass sich... | [
"Solution:\n\nWir betrachten den Spiegelpunkt $Z$ von $A$ bei Spiegelung an $BD$. Wir werden beweisen, dass sich die beiden in der Aufgabenstellung erwähnten Kreise in $Z$ berühren.\n\n\n\nZunächst ist wegen $\\measuredangle BZR = \\measuredangle RAB = \\measuredangle BPR$ und der Umkehrung... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0km8 | Problem:
Let $ABCD$ be a parallelogram with $AB = 480$, $AD = 200$, and $BD = 625$. The angle bisector of $\angle BAD$ meets side $CD$ at point $E$. Find $CE$. | [
"Solution:\n\n\n\nFirst, it is known that $\\angle BAD + \\angle CDA = 180^\\circ$. Further, $\\angle DAE = \\frac{\\angle BAD}{2}$. Thus, as the angles in triangle $ADE$ sum to $180^\\circ$, this means $\\angle DEA = \\frac{\\angle BAD}{2} = \\angle DAE$. Therefore, $DAE$ is isosceles, mak... | United States | HMMT November 2021 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof and answer | 280 | |
0g3z | Problem:
Given a (simple) graph $G$ with $n \geq 2$ vertices $v_{1}, v_{2}, \ldots, v_{n}$ and $m \geq 1$ edges, Joël and Robert play the following game with $m$ coins:
i) Joël first assigns to each vertex $v_{i}$ a non-negative integer $w_{i}$ such that $w_{1}+\cdots+w_{n}=m$.
ii) Robert then chooses a (possibly em... | [
"Solution:\n\nWe refer to Joël as $A$ and Robert as $B$; furthermore, the condition of coin placement can just be paraphrased as directing the edges, with the in-degree remaining inferior to $w(i)$, which we will henceforth refer to as a function, for simplicity's sake.\n\nThe solution has two key parts: proving $A... | Switzerland | IMO Selection | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Other"
] | null | proof and answer | Exactly the bipartite graphs | |
02qv | Problem:
A Princesa Telassim cortou uma folha de papel retangular em 9 quadrados de lados $1, 4, 7, 8, 9, 10, 14, 15$ e $18$ centímetros.
a) Qual era a área da folha antes de ser cortada?
b) Quais eram as medidas da folha antes de ser cortada?
c) A Princesa Telassim precisa montar a folha de novo. Ajude-a mostrando... | [
"Solution:\n\na) A área da folha era igual à soma das áreas dos nove quadrados, que é (em centímetros quadrados):\n$$\n1^2 + 4^2 + 7^2 + 8^2 + 9^2 + 10^2 + 14^2 + 15^2 + 18^2 = 1056\n$$\n\nb) Sejam $a$ e $b$ as dimensões da folha, onde supomos $a \\leq b$. Como a área de um retângulo é o produto de suas dimensões, ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Area: 1056 square centimeters. Dimensions: 32 cm by 33 cm. The tiling is unique up to rotations and reflections (one explicit arrangement exists). | |
05hi | Problem:
Si $n>0$ est un entier, on désigne par $d(n)$ le nombre de diviseurs strictement positifs de $n$.
a) Existe-t-il une suite $\left(a_{i}\right)_{i \geqslant 1}$ strictement croissante d'entiers strictement positifs tels que, pour tout $i$ suffisamment grand, le nombre $a_{i}$ soit divisible par exactement $d(... | [
"Solution:\n\na) La réponse est oui.\nIl suffit de trouver une suite $\\left(a_{i}\\right)$ qui vérifie les deux conditions suivantes :\n- le nombre $a_{1}$ ne divise aucun autre terme de la suite,\n- le nombre $i$ divise $j$ si et seulement si $a_{i}$ divise $a_{j}$, pour tous $i, j \\geqslant 2$.\nOr, il est bien... | France | Olympiades Françaises de Mathématiques, Envoi No. 6 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0esw | Determine all sequences $a_1, a_2, a_3, \dots$ of nonnegative integers such that $a_1 < a_2 < a_3 < \dots$ and $a_n$ divides $a_{n-1} + n$ for all $n \ge 2$. | [
"We claim that the only possible sequences are the following:\n* $a_n = n - 1$ for all $n$, or\n* $a_n = \\frac{n^2+n}{2} + k$ for all $n$, where $k$ is a fixed nonnegative integer, or\n* $a_n = \\begin{cases} n-1 & n \\le N, \\\\ \\frac{n^2+n}{2} - \\frac{N^2-N+2}{2} & n > N, \\end{cases}$ where $N$ is a fixed non... | South Africa | The South African Mathematical Olympiad Third Round | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | All such sequences are exactly the following three families:
1) a_n = n − 1 for all n.
2) a_n = (n^2 + n)/2 + k for all n, where k is a fixed nonnegative integer.
3) For a fixed nonnegative integer N,
a_n = n − 1 for n ≤ N, and a_n = (n^2 + n)/2 − (N^2 − N + 2)/2 for n > N. | |
04wd | A team consists of $7$ players. In each round of the tournament, five of them play and two sit on the bench. Prove that, regardless of the (positive) number of rounds and the choice of who plays in what round, at the end of the tournament there are two players who have been together (either on the field or on the bench... | [] | Czech Republic | District Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
00c1 | By writing the digits $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ and $9$ in the cells of a $3 \times 3$ board, without repetitions, $6$ numbers of $3$ digits each are formed: one in each row and one in each column. For instance, if the board is filled in like in this picture
| | Column 1 | Column 2 | Column 3 |
|-... | [
"In order for the number in the first column to be a multiple of $5$, we must write $5$ in the cell corresponding to its units, that is, in row $3$ and column $1$. The digits in the cells corresponding to the units of the numbers that are multiple of $2$, $4$ and $6$ must be even. Then, the number in the third row ... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All valid boards (rows listed top to bottom):
1) 3 1 2 / 7 8 9 / 5 6 4
2) 3 7 2 / 1 8 9 / 5 6 4
3) 1 7 2 / 3 6 9 / 5 8 4
4) 7 1 2 / 3 6 9 / 5 8 4
5) 1 7 2 / 6 3 9 / 5 8 4
6) 7 1 2 / 6 3 9 / 5 8 4 | |
03ii | Problem:
Among all triangles having (i) a fixed angle $A$ and (ii) an inscribed circle of fixed radius $r$, determine which triangle has the least perimeter. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0b2m | Problem:
In the figure below, $BC$ is the diameter of a semicircle centered at $O$, which intersects $AB$ and $AC$ at $D$ and $E$ respectively. Suppose that $AD=9$, $DB=4$, and $\angle ACD=\angle DOB$. Find the length of $AE$.
(a) $\frac{117}{16}$
(b) $\frac{39}{5}$
(c) $2 \sqrt{13}$
(d) $3 ... | [
"Solution:\n\nLet $\\angle DOB=\\angle EOC=\\alpha$. Note that $\\angle DCB=\\frac{\\alpha}{2}$. Also, note that $\\tan \\frac{\\alpha}{2}=\\frac{4}{DC}$ and $\\tan \\alpha=\\frac{9}{DC}=\\frac{9}{4} \\tan \\frac{\\alpha}{2}$. Let $x=\\tan \\frac{\\alpha}{2}$. By the double-angle formula,\n$$\n\\begin{aligned}\n\\f... | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | (b) | |
0jtj | Problem:
Find the smallest possible area of an ellipse passing through $(2,0)$, $(0,3)$, $(0,7)$, and $(6,0)$. | [
"Solution:\nLet $\\Gamma$ be an ellipse passing through $A=(2,0)$, $B=(0,3)$, $C=(0,7)$, $D=(6,0)$, and let $P=(0,0)$ be the intersection of $AD$ and $BC$. $\\frac{\\text{Area of } \\Gamma}{\\text{Area of } ABCD}$ is unchanged under an affine transformation, so we just have to minimize this quantity over situations... | United States | HMMT February | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 56π√3/9 | |
0bzt | Prove that there are infinitely many sets of four positive integers so that the sum of the squares of any three elements is a perfect square. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | null | proof only | null | |
02dl | Three numbered tiles are arranged in a tray as shown:
| 1 | 2 |
|---|---|
| 3 | |
Show that we cannot interchange the $1$ and the $3$ by a sequence of moves where we slide a tile to the adjacent vacant space. | [
"Write down the order of the tiles reading clockwise around the perimeter, starting at $1$. We get $123$ and no move changes that, so we will always get $123$ after any sequence of moves. But the desired arrangement would give $132$, so it is not possible."
] | Brazil | IV OBM | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0ftg | Problem:
Bestimme alle Funktionen $f: \mathbb{R} \rightarrow \mathbb{R}$, für die gilt:
a. $f(x-1-f(x))=f(x)-1-x$ für alle $x \in \mathbb{R}$,
b. Die Menge $\{f(x) / x \mid x \in \mathbb{R}, x \neq 0\}$ ist endlich. | [
"Solution:\n\nWir zeigen, dass $f(x)=x$ die einzige Lösung ist. Setze $g(x)=f(x)-x$. Substituiert man $f(x)=g(x)+x$ in (a), folgt für $g$ die einfachere Gleichung\n$$\ng(-1-g(x))=2 g(x)\n$$\nWegen $f(x) / x=(g(x)+x) / x=g(x) / x+1$ und (b) folgt, dass auch die Menge $\\{g(x) / x \\mid x \\in \\mathbb{R}, x \\neq 0\... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x)=x | |
05vn | Problem:
Soient $\omega_{1}$ et $\omega_{2}$ deux cercles de centres respectifs $O_{1}$ et $O_{2}$. On suppose que $\omega_{1}$ et $\omega_{2}$ se coupent en les points $A$ et $B$. La droite $(O_{1}A)$ recoupe le cercle $\omega_{2}$ en $C$ tandis que la droite $(O_{2}A)$ recoupe le cercle $\omega_{1}$ en $D$. Montrer ... | [
"Solution:\n\n\n\nNotons que puisque les angles $\\widehat{DAO_{1}}$ et $\\widehat{CAO_{2}}$ sont opposés par le sommet, ils sont égaux. D'autre part, puisque les points $A$ et $D$ appartiennent au cercle $\\omega_{1}$, le triangle $A O_{1} D$ est isocèle en $O_{1}$. De même, le triangle $C... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bwn | Let $ABC$ be a triangle which is not isosceles, with $G$ its centroid and $I$ its incenter. Prove that $GI \perp BC$ if and only if $AB + AC = 3BC$. | [
"Using the usual notations for a triangle, we have:\n$$\n\\begin{aligned} \\overline{GI} &= \\overline{AI} - \\overline{AG} = \\left( \\frac{b}{a+b+c} - \\frac{1}{3} \\right) \\overline{AB} - \\left( \\frac{c}{a+b+c} - \\frac{1}{3} \\right) \\overline{AC} \\\\ &= \\frac{1}{3(a+b+c)} \\left( (2b-a-c)\\overline{AB} +... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
0l3n | Problem:
Let $\mathcal{P}$ be a regular 10-gon in the coordinate plane. Mark computes the number of distinct $x$ coordinates that vertices of $\mathcal{P}$ take. Across all possible placements of $\mathcal{P}$ in the plane, compute the sum of all possible answers Mark could get. | [
"Solution:\n\n\n\n10 distinct coordinates\n\n\n\n5 distinct coordinates\n\n\n\n6 distinct coordinates\n\nLet $\\mathcal{P}$ have vertices $P_{1} P_{2} \\ldots P_{10}$. If no two vertices have the same $x$-coordinate, then Mark gets $10$.\n\n... | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | 21 | |
0l3t | What value of $x$ satisfies
$$
\frac{\log_2 x \cdot \log_3 x}{\log_2 x + \log_3 x} = 2?
$$
(A) 25 (B) 32 (C) 36 (D) 42 (E) 48 | [
"**Answer (C):** Observe that\n$$\n\\begin{aligned}\n2 &= \\frac{\\log_2 x \\cdot \\log_3 x}{\\log_2 x + \\log_3 x} \\\\\n&= \\frac{1}{\\frac{1}{\\log_3 x} + \\frac{1}{\\log_2 x}} \\\\\n&= \\frac{1}{\\log_4 3 + \\log_4 2} \\\\\n&= \\frac{1}{\\log_4 6} = \\log_6 x.\n\\end{aligned}\n$$\nIt follows that $x = 6^2 = 36$... | United States | 2024 AMC 12 B | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | MCQ | C | |
03vd | Given an integer $m \ge 2$, and two real numbers $a, b$ with $a > 0$ and $b \ne 0$, the sequence $\{x_n\}$ is such that $x_1 = b$ and $x_{n+1} = a x_n^m + b$, $n = 1, 2, \dots$. Prove that:
(1) When $b < 0$ and $m$ is even, the sequence $\{x_n\}$ is bounded if and only if $a b^{m-1} \ge -2$;
(2) When $b < 0$ and $m$ ... | [
"(1) When $b < 0$ and $m$ is even, in order that $a b^{m-1} < -2$, we should first have $a b^m + b > -b > 0$, and therefore $a(a b^m + b)^m + b > a b^m + b > 0$, i.e. $x_3 > x_2 > 0$. Using the fact that $a x^m + b$ is monotonically increasing on $(0, +\\infty)$, it can be established that each succeeding term of t... | China | China Western Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0ewd | Problem:
Given the lengths $AB$ and $BC$ and the fact that the medians to those two sides are perpendicular, construct the triangle $ABC$. | [
"Solution:\nLet $M$ be the midpoint of $AB$ and $X$ the midpoint of $MB$. Construct the circle center $B$, radius $BC/2$ and the circle diameter $AX$. If they do not intersect (so $BC < AB/2$ or $BC > AB$) then the construction is not possible. If they intersect at $N$, then take $C$ so that $N$ is the midpoint of ... | Soviet Union | 2nd ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler lin... | null | proof only | null | |
01o6 | Given a trapezoid $ABCD$ ($BC \parallel AD$) with $CD = AO$ and $BC = OD$, where $O$ is a point of intersection of the diagonals of the trapezoid. $CA$ is a bisectrix of $\angle BCD$.
Find the angles of $ABCD$. | [
"Answer: $\\angle A = 54^\\circ$, $\\angle B = 126^\\circ$, $\\angle C = 72^\\circ$, $\\angle D = 108^\\circ$.\nLet $CD = AO = a$, $BC = OD = b$ and $\\angle BCD = 2\\alpha$. By condition,\n\n\n\n$BD = y$, $\\angle DCA = \\angle ACB = \\alpha$. Since $BC \\parallel AD$, we have $\\angle CAD... | Belarus | Belorusija 2012 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | ∠A = 54°, ∠B = 126°, ∠C = 72°, ∠D = 108° | |
02ef | Let $N$ be the natural numbers and $N' = N \setminus \{0\}$. Find all functions $f: N' \to N$ such that $f(xy) = f(x) + f(y)$, $f(30) = 0$ and $f(x) = 0$ for all $x \equiv 7 \pmod{10}$. | [
"$f(30) = f(2) + f(3) + f(5) = 0$ and $f(n)$ is non-negative, so $f(2) = f(3) = f(5) = 0$. For any positive integer $n$ not divisible by $2$ or $5$ we can find a positive integer $m$ such that $mn \\equiv 7 \\pmod{10}$. But then $f(mn) = 0$, so $f(n) = 0$.\nIt is a trivial induction that $f(2^a 5^b n) = f(5^b n) = ... | Brazil | X OBM | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | f(x) = 0 for all positive integers x | |
0c0c | Given two positive integers $m$ and $n$, show that there exist a positive integer $k$ and a set $S$ of at least $m$ multiples of $n$ such that the numbers $2^k \sigma(s)/s$, $s \in S$, are all odd; $\sigma(s)$ is the sum of all positive divisors of $s$ (1 and $s$ inclusive). | [
"Let $n = 2^a n'$, where $a$ is a non-negative integer and $n'$ is odd, and let $2^b$ be the highest power of $2$ dividing $\\sigma(n')$. Let $p_1, \\dots, p_\\ell$ be odd primes not dividing $n'$ (e.g., let each $p_i > n'$), and let $N$ be an integer such that $r = N\\varphi(p_1^2 \\cdots p_\\ell^2 n') - 1 > \\max... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0kf1 | Problem:
Ana and Banana are rolling a standard six-sided die. Ana rolls the die twice, obtaining $a_{1}$ and $a_{2}$, then Banana rolls the die twice, obtaining $b_{1}$ and $b_{2}$. After Ana's two rolls but before Banana's two rolls, they compute the probability $p$ that $a_{1} b_{1} + a_{2} b_{2}$ will be a multiple... | [
"Solution:\n\nIf either $a_{1}$ or $a_{2}$ is relatively prime to $6$, then $p = \\frac{1}{6}$. If one of them is a multiple of $2$ but not $6$, while the other is a multiple of $3$ but not $6$, we also have $p = \\frac{1}{6}$. In other words, $p = \\frac{1}{6}$ if $\\operatorname{gcd}(a_{1}, a_{2})$ is coprime to ... | United States | HMMT February 2020 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 2/3 | |
07x1 | Let $A$, $B$, $C$ be three points on a circle $\Gamma$, and let $L$ denote the midpoint of segment $BC$. The perpendicular bisector of $BC$ intersects the circle $\Gamma$ at two points $M$ and $N$, such that $A$ and $M$ are on different sides of line $BC$. Let $S$ denote the point where the segments $BC$ and $AM$ inter... | [
"(a) By construction, $M$, $N$ are on $\\Gamma = (ABC)$ and $D$, $E$ are on $(ALM)$. The circles $\\Gamma = (ABC)$ and $(BCD)$ have radical axis $BC$, and the circles $(ABC)$ and $(ALM)$ have radical axis $AM$.\nSince $S$ is the intersection of these two radical axes, this point is the radical centre of the three c... | Ireland | IRL_ABooklet_2024 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point... | null | proof only | null | |
05ty | Problem:
Soit $ABC$ un triangle. On note $H_{A}$ le pied de la hauteur de $ABC$ issue de $A$, et $A'$ le milieu du segment $[BC]$. On note ensuite $Q_{A}$ le symétrique de $H_{A}$ par rapport à $A'$. On définit de même les points $Q_{B}$ et $Q_{C}$. Enfin, on note $R$ le point d'intersection, autre que $Q_{A}$, entre ... | [
"Solution:\n\nSoit $H$ l'orthocentre de $ABC$, $O$ le centre du cercle circonscrit à $ABC$, et $R'$ le symétrique de $H$ par rapport à $O$. Les projetés orthogonaux de $H$ et $O$ sur $(BC)$ sont $H_{A}$ et $A'$, donc le projeté orthogonal de $R'$ sur $(BC)$ est $Q_{A}$. De même, les projetés orthogonaux de $R'$ sur... | France | Préparation Olympique Française de Mathématiques - Test du 14 et du 21 Février 2021 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
00f8 | Let $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+ \cdots+b_{n}$. Show that
$$
\frac{a_{1}^{2}}{a_{1}+b_{1}}+\frac{a_{2}^{2}}{a_{2}+b_{2}}+\cdots+\frac{a_{n}^{2}}{a_{n}+b_{n}} \geq \frac{a_{1}+a_{2}+\cdots+a_{n}}{2} .
$$ | [
"By the Cauchy-Schwartz inequality,\n\n$$\n\\left(\\frac{a_{1}^{2}}{a_{1}+b_{1}}+\\frac{a_{2}^{2}}{a_{2}+b_{2}}+\\cdots+\\frac{a_{n}^{2}}{a_{n}+b_{n}}\\right)\\left(\\left(a_{1}+b_{1}\\right)+\\left(a_{2}+b_{2}\\right)+\\cdots+\\left(a_{n}+b_{n}\\right)\\right) \\geq \\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{2} .\n... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1991 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
07i7 | We call a natural number $m$ "interesting", if for all natural numbers $1 \le n \le m$, we can write $n$ as the sum of distinct divisors of $m$. Prove that there are infinitely many interesting numbers of the form $k^2 + k + 2022$. | [
"**First solution**\nWe shall firstly prove the following lemmas,\n\n**Lemma 1.** If $x$ is interesting and $y < x$ then $xy$ is also interesting.\n*Proof.* If $n < xy$, by division algorithm we have $n = yq + r$ where $q < x$, $r < y$. Now, we can write $q = \\sum d_i$ and $r = \\sum d'_i$ where $\\{d_i\\}$ and $\... | Iran | 40th Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Co... | null | proof only | null | |
09sm | Problem:
Voor een positief geheel getal $n$ dat geen tweemacht is, definiëren we $t(n)$ als de grootste oneven deler van $n$ en $r(n)$ als de kleinste positieve oneven deler van $n$ die ongelijk aan 1 is. Bepaal alle positieve gehele getallen $n$ die geen tweemacht zijn en waarvoor geldt
$$
n=3 t(n)+5 r(n)
$$ | [
"Solution:\nAls $n$ oneven is, geldt $t(n)=n$ dus is $3 t(n)$ groter dan $n$, tegenspraak. Als $n$ deelbaar door 2 is maar niet deelbaar door 4, dan geldt $t(n)=\\frac{1}{2} n$ en is $3 t(n)$ weer groter dan $n$, opnieuw tegenspraak. We kunnen concluderen dat $n$ in elk geval deelbaar door 4 moet zijn. Als $n$ deel... | Netherlands | Selectietoets | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 60, 100, and all numbers of the form 8p where p is an odd prime | |
0ar1 | Problem:
A circle with center $C$ and radius $r$ intersects the square $EFGH$ at $H$ and at $M$, the midpoint of $EF$. If $C$, $E$ and $F$ are collinear and $E$ lies between $C$ and $F$, what is the area of the region outside the circle and inside the square in terms of $r$? | [] | Philippines | 13th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | r^2(22/25 - (1/2) arctan(4/3)) | |
03ww | Given points $P$, $Q$ on an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a > b > 0$), satisfying $OP \perp OQ$, the minimum of $|OP| \times |OQ|$ is ____. | [
"Define\n$$\nP(|OP| \\cos \\theta, |OP| \\sin \\theta), \\\\\nQ(|OQ| \\cos(\\theta \\pm \\frac{\\pi}{2}), |OQ| \\sin(\\theta \\pm \\frac{\\pi}{2})).\n$$\nWe have\n$$\n\\frac{1}{|OP|^2} = \\frac{\\cos^2\\theta}{a^2} + \\frac{\\sin^2\\theta}{b^2}, \\qquad \\textcircled{1}\n$$\n$$\n\\frac{1}{|OQ|^2} = \\frac{\\sin^2\\... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 2a^2b^2/(a^2+b^2) | |
01b2 | Let $a_0, a_1, \dots, a_N$ be real numbers where $a_0 = a_N = 0$. Prove the inequality
$$ a_1^2 + a_2^2 + \dots + a_{N-1}^2 \le C \left((a_1 - a_0)^2 + (a_2 - a_1)^2 + \dots + (a_N - a_{N-1})^2\right), $$
where $C = \frac{N^2}{4}$. | [
"Let $b_i = a_i - a_{i-1}$, and note that $a_0 = a_N = 0$ implies $a_i = \\sum_{k=1}^i b_k = -\\sum_{k=i+1}^N b_k$.\n\nCase 1: $C = \\frac{N^2}{4}$.\nSplit the left hand side of\n$$\na_1^2 + a_2^2 + \\dots + a_{N-1}^2 \\le \\frac{N^2}{4} \\left( (a_1 - a_0)^2 + (a_2 - a_1)^2 + \\dots + (a_N - a_{N-1})^2 \\right)\n$... | Baltic Way | Baltic Way | [
"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",
"Algebra > Linear Algebra > Vectors",
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
0e8r | We have $10$ balls in a bowl, some of them are blue, some of them are yellow and the others are green. They can be put in a line in $360$ different ways. At most how many blue balls are there in the bowl?
(A) $4$
(B) $5$
(C) $6$
(D) $7$
(E) $8$ | [
"Denote the numbers of blue, yellow and green balls by $b$, $y$ and $g$, where $b + y + g = 10$. The balls can be put in a line in $\\frac{10!}{b!y!g!}$ different ways. So, $\\frac{10!}{b!y!g!} = 360$. This equality can be rewritten as $10 \\cdot 9 \\cdots (b+1) = 360 \\cdot y! \\cdot g!$. This implies $10 \\cdot 9... | Slovenia | National Math Olympiad 2013 - First Round | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | D | |
0fdo | Problem:
Sea $O$ el circuncentro de un triángulo $ABC$. La bisectriz que parte de $A$ corta al lado opuesto en $P$.
Probar que se cumple:
$$
AP^{2} + OA^{2} - OP^{2} = bc
$$ | [
"Solution:\nProlongamos $AP$ hasta que corte en $M$ al circuncírculo.\n\n\n\nLos triángulos $ABM$ y $APC$ son semejantes al tener dos ángulos iguales. ($\\angle ACB = \\angle AMB$ por inscritos en el mismo arco y $\\angle BAN = \\angle CAN$ por bisectriz).\n\nEntonces:\n$$\n\\frac{c}{AM} = ... | Spain | null | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04sd | Positive real numbers $a$, $b$, $c$, $d$ satisfy equalities
$$
a = c + \frac{1}{d} \quad \text{and} \quad b = d + \frac{1}{c}.
$$
Prove an inequality $ab \ge 4$ and find a minimum of $ab + cd$. | [
"To prove the inequality $ab \\ge 4$ we substitute from the equalities. We so obtain an estimate\n$$\nab = \\left(c + \\frac{1}{d}\\right)\\left(d + \\frac{1}{c}\\right) = cd + 1 + 1 + \\frac{1}{cd} \\ge 4,\n$$\nwhere we use in the last inequality well-known fact that $x + 1/x \\ge 2$ holds for all positive reals $... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | ab ≥ 4; minimum of ab + cd is 2(1 + sqrt(2)). | |
03l4 | Problem:
Let $A = (a_{1}, a_{2}, \ldots, a_{2000})$ be a sequence of integers each lying in the interval $[-1000, 1000]$. Suppose that the entries in $A$ sum to $1$. Show that some nonempty subsequence of $A$ sums to zero. | [
"Solution:\n\nWe may assume no entry of $A$ is zero, for otherwise we are done. We sort $A$ into a new list $B = (b_{1}, \\ldots, b_{2000})$ by selecting elements from $A$ one at a time in such a way that $b_{1} > 0$, $b_{2} < 0$ and, for each $i = 2, 3, \\ldots, 2000$, the sign of $b_{i}$ is opposite to that of th... | Canada | Canadian Mathematics Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0chy | Let $n \ge 2$ be an integer. A Welsh darts board is a disc divided into $2n$ equal sectors, half of them being red and the other half being white. Two Welsh darts boards are matched if they have the same radius and they are superimposed so that each sector of the first board comes exactly over a sector of the second bo... | [
"For any two Welsh darts boards that can be matched, if two of their superimposed sectors have the same color, we call it a *concordance*, and if two of their superimposed sectors have different colors, we call it a *non-concordance*.\n\nOn each of the two Welsh darts boards that can be matched, we write $+1$ on ev... | Romania | 74th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0cr3 | На стороне $AB$ треугольника $ABC$ выбраны точки $C_1$ и $C_2$. Аналогично, на стороне $BC$ выбраны точки $A_1$ и $A_2$, а на стороне $AC$ — точки $B_1$ и $B_2$. Оказалось, что отрезки $A_1B_2$, $B_1C_2$ и $C_1A_2$ имеют равные длины, пересекаются в одной точке, и угол между любыми двумя из них равен $60^\circ$. Докажи... | [
"Заметим, что\n$$\n\\overrightarrow{A_1B_2} + \\overrightarrow{B_2A_1} + \\overrightarrow{B_1C_2} + \\overrightarrow{C_2C_1} + \\overrightarrow{C_1A_2} + \\overrightarrow{A_2A_1} = \\overrightarrow{0}. \\quad (*)\n$$\nПо условию имеем $A_1B_2 = B_1C_2 = C_1A_2$, и угол между любыми двумя из трех прямых $A_1B_2$, $B... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0h3n | Let $O$ be the circumcenter of an acute non-isosceles triangle $ABC$. Lines $BO$ and $CO$ meet sides $AC$ and $AB$ respectively at points $K$ and $N$. Points $P$ and $T$, different from $K$ and $N$, are chosen respectively on $AC$ and $AB$ so that $OK = OP$ and $ON = OT$. A line through $P$ parallel to $BK$ and a line ... | [
"Нехай $H$ — ортоцентр трикутника $ABC$, точка $D$ симетрична $H$ відносно прямої $AC$. Як відомо, точка $D$ лежить на описаному кілі трикутника $ABC$. Позначимо через $Q$ точку перетину відрізків $OD$ і $AC$. Тоді маємо: $\\angle QDH = \\angle QHD = \\angle OBD$. Звідси випливає, що $HQ \\parallel BK$, і $\\angle ... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08ct | Problem:
Sia $ABC$ un triangolo isoscele con base $BC = 10$ e $AB = AC$. Si costruiscano esternamente sui suoi due lati obliqui altri due triangoli isosceli $DAB$ e $EAC$, entrambi simili ad $ABC$, con $DA = DB$ e $EA = EC$. Sapendo che $DE = 45$, trovare la lunghezza di $AB$.
(A) 15
(B) 20
(C) $9\sqrt{5}$
(D) 22.5
(... | [
"Solution:\n\nLa risposta è (A). Poiché i triangoli $ABC$, $DAB$, $EAC$ sono simili, si ha che i rispettivi angoli alla base sono congruenti; questo significa che $\\widehat{DAB} = \\widehat{ABC} = \\widehat{ACB} = \\widehat{CAE}$. Adesso si può calcolare l'ampiezza dell'angolo $\\widehat{DAE} = \\widehat{DAB} + \\... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | MCQ | A | |
0ho7 | Problem:
Let $A_{1}, B_{1}, C_{1}$ be points in the interior of sides $BC, CA, AB$, respectively, of equilateral triangle $ABC$. Prove that if the radii of the inscribed circles of $\triangle C_{1}AB_{1}$, $\triangle B_{1}CA_{1}$, $\triangle A_{1}BC_{1}$, $\triangle A_{1}B_{1}C_{1}$ are equal, then $A_{1}, B_{1}, C_{1}... | [
"Solution:\nFirst, suppose that $BA_{1} > CB_{1}$. We claim this forces $CB_{1} > AC_{1}$. If we rotate triangle $A_{1}CB_{1}$ by $2\\pi/3$ radians about the center of $\\triangle ABC$, we get $\\triangle A_{2}AB_{2}$ ($A_{2} \\in CA$, $B_{2} \\in AB$) whose incircle coincides with that of $\\triangle B_{1}AC_{1}$ ... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"... | null | proof only | null | |
0gz7 | Find all solutions of the equation $\sqrt[3]{x} + \sqrt[3]{y} = \sqrt[3]{z}$, where $x$, $y$, $z$ are integer numbers. | [
"Any group of three $(da^3, db^3, dc^3)$ if $a + b = c$ satisfies the condition of the problem.\n\nLet's find solutions of the equation $\\sqrt[3]{x} + \\sqrt[3]{y} + \\sqrt[3]{z} = 0$. From the known equation\n$$\na^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\n$$\nit follows that $x + y + z ... | Ukraine | The Problems of Ukrainian Authors | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All integer solutions are x = d a^3, y = d b^3, z = d (a + b)^3 for integers a, b, d. | |
01c1 | Show that there exist two distinct positive integers $a, b$, each having exactly 2014 digits (in base ten; initial zeroes disallowed), with the following properties.
* The digits of $b$ are those of $a$ in reverse order.
* When a digit in each of $a$ and $b$ is deleted at random, and the resulting numbers are denoted $... | [
"Note that, for any natural number $m$, we have\n$$\n1 \\overbrace{33 \\dots 33}^{m} 2 = 1 \\overbrace{11 \\dots 11}^{m+1} \\cdot 12 \\quad \\text{and} \\quad 2 \\overbrace{33 \\dots 33}^{m} 1 = 1 \\overbrace{11 \\dots 11}^{m+1} \\cdot 21,\n$$\nso that\n$$\n\\frac{1 \\overbrace{33 \\dots 33}^{m} 2}{2 \\overbrace{33... | Baltic Way | Baltic Way | [
"Number Theory > Other"
] | null | proof only | null | |
0fuw | Problem:
Sei $n \geq 1$ eine natürliche Zahl. Bestimme alle positiven ganzzahligen Lösungen der Gleichung
$$
7 \cdot 4^{n}=a^{2}+b^{2}+c^{2}+d^{2}
$$ | [
"Solution:\n\nSei zuerst $n \\geq 2$. Dann ist die linke Seite durch $8$ teilbar, also auch die rechte. Insbesondere sind $a, b, c, d$ alle gerade, alle ungerade oder genau zwei davon gerade und zwei ungerade. Betrachte die Gleichung modulo $8$. Die einzigen quadratischen Reste $(\\bmod\\ 8)$ sind $0,1,4$. Wären $a... | Switzerland | SMO Finalrunde | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All solutions, for n ≥ 1, are the permutations of:
(5·2^{n−1}, 2^{n−1}, 2^{n−1}, 2^{n−1}),
(2^{n+1}, 2^{n}, 2^{n}, 2^{n}),
(3·2^{n−1}, 3·2^{n−1}, 3·2^{n−1}, 2^{n−1}). | |
01m5 | Let $x_1, \ldots, x_{100}$ be nonnegative real numbers such that $x_i + x_{i+1} + x_{i+2} \le 1$ for all $i = 1, \ldots, 100$ (we put $x_{101} = x_1, x_{102} = x_2$).
Find the maximal possible value of the sum $S = \sum_{i=1}^{100} x_i x_{i+2}$. | [
"3. See IMO-2010 Shortlist, Problem A3."
] | Belarus | Selection and Training Session | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 25/2 | |
02bu | Problem:
Um anel simétrico com $m$ polígonos regulares de $n$ lados cada é formado de acordo com as regras:
i) cada polígono no anel encontra dois outros;
ii) dois polígonos adjacentes têm apenas um lado em comum;
iii) o perímetro da região interna delimitada pelos polígonos consiste em exatamente dois lados de cada p... | [
"Solution:\n\nSeja $\\alpha=\\frac{360^{\\circ}}{n}$ a medida de cada ângulo externo de um polígono regular $A B C D E \\ldots$. Suponha que o caminho $B C D$ faz parte do perímetro da região interna de algum anel. Se $O$ é o encontro dos prolongamentos de $A B$ e $D E$, por simetria, ele é o centro da região inter... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 6 | |
0ku0 | Problem:
Tyler has an infinite geometric series with sum $10$. He increases the first term of his sequence by $4$ and swiftly changes the subsequent terms so that the common ratio remains the same, creating a new geometric series with sum $15$. Compute the common ratio of Tyler's series. | [
"Solution:\n\nLet $a$ and $r$ be the first term and common ratio of the original series, respectively. Then $\\frac{a}{1-r} = 10$ and $\\frac{a+4}{1-r} = 15$. Dividing these equations, we get that\n\n$$\n\\frac{a+4}{a} = \\frac{15}{10} \\Longrightarrow a = 8\n$$\n\nSolving for $r$ with $\\frac{a}{1-r} = \\frac{8}{1... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1/5 | |
044o | In a plane rectangular coordinate system $xOy$, the focus of parabola $\Gamma: y^2 = 2px$ ($p > 0$) is $F$. Make a tangent line to $\Gamma$ passing through point $P$ (different from $O$) on $\Gamma$ and it intersects the $y$-axis at point $Q$. If $|FP| = 2$, $|FQ| = 1$, then the dot product of vectors $\overrightarrow{... | [
"Let $P(\\frac{t^2}{2p}, t)$ ($t \\neq 0$), and then the equation of the tangent line of $\\Gamma$ is $yt = p(x + \\frac{t^2}{2p})$.\nLet $x = 0$, and we get $yt = \\frac{t}{2}$. The coordinates of $F$ are $(\\frac{p}{2}, 0)$, and thus\n$$\n|FP| = \\sqrt{\\left(\\frac{p}{2} - \\frac{t^2}{2p}\\right)^2 + t^2} = \\fr... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 3/2 | |
06pn | Point $P$ lies on side $AB$ of a convex quadrilateral $ABCD$. Let $\omega$ be the incircle of triangle $CPD$, and let $I$ be its incenter. Suppose that $\omega$ is tangent to the incircles of triangles $APD$ and $BPC$ at points $K$ and $L$, respectively. Let lines $AC$ and $BD$ meet at $E$, and let lines $AK$ and $BL$ ... | [
"Let $\\Omega$ be the circle tangent to segment $AB$ and to rays $AD$ and $BC$; let $J$ be its center. We prove that points $E$ and $F$ lie on line $IJ$.\n\n\n\nDenote the incircles of triangles $ADP$ and $BCP$ by $\\omega_{A}$ and $\\omega_{B}$. Let $h_{1}$ be the homothety with a negative... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0252 | Problem:
Um número de quatro algarismos $a b c d$ é chamado balanceado se
$$
a+b=c+d
$$
Calcule as seguintes quantidades:
a) Quantos números $a b c d$ são tais que $a+b=c+d=8$ ?
b) Quantos números $a b c d$ são tais que $a+b=c+d=16$ ?
c) Quantos números balanceados existem? | [
"Solution:\n\na) Vamos contar primeiro os valores possíveis para o par $(a, b)$. Observe que $a$ não pode ser igual a zero, por ser o primeiro algarismo em $a b c d$. Mas $a$ pode tomar qualquer valor em\n$$\n\\{1,2,3,4,5,6,7,8\\}\n$$\nporque por cada um desses valores, o número $8 - a$ dá como resultado um valor a... | Brazil | null | [
"Discrete Mathematics > Combinatorics",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | a) 72; b) 9; c) 615 | |
01eg | Given two circles on the plane do not intersect. We choose diameters $A_1B_1$ and $A_2B_2$ of these circles such that the segments $A_1A_2$ and $B_1B_2$ intersect. Let $A$ and $B$ be the midpoints of segments $A_1A_2$ and $B_1B_2$, $C$ be its intersection point. Prove that the orthocenter of the triangle $ABC$ belongs ... | [
"\n\nProve that the orthocenter $H$ of $\\triangle ABC$ belongs to their radical axis.\nDenote the circles by $s_1$ and $s_2$. Let the line $A_1A_2$ intersect circles $s_1$ and $s_2$ second time in points $X_1$ and $X_2$ respectively, and the line $B_1B_2$ intersect the circles second time ... | Baltic Way | Baltic Way shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08gs | Problem:
Sono dati tre numeri reali positivi $a$, $b$, $c$ con $a c = 9$. Si sa che per tutti i numeri reali $x$, $y$ con $x y \neq 0$ vale
$$
\frac{a}{x^{2}} + \frac{b}{x y} + \frac{c}{y^{2}} \geq 0.
$$
Qual è il massimo valore possibile per $b$?
(A) 1
(B) 3
(C) 6
(D) 9
(E) Non esiste nessun tale $b$. | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Per ipotesi, per $x$, $y$ tali che $x y \\neq 0$, si ha\n$$\n\\frac{a y^{2} + b x y + c x^{2}}{x^{2} y^{2}} = \\frac{a}{x^{2}} + \\frac{b}{x y} + \\frac{c}{y^{2}} \\geq 0.\n$$\nDimostriamo che tale disuguaglianza è sempre soddisfatta per $b \\leq 6$: infatti, se $x y > 0... | Italy | Olimpiadi di Matematica | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | C | |
06kn | Find the number of nonnegative integers $k$, $0 \le k \le 2188$, such that $\binom{2188}{k}$ is divisible by $2188$.
(Note that the binomial coefficient is defined by $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.) | [
"The answer is $2146$.\n\nNote that $2188 = 4 \\times 547$, where $547$ is a prime. So $2188 = 40_{(547)}$ (base $547$ representation). Let $k = \\overline{ab}_{(547)}$. If $k$ is not divisible by $547$, then $b > 0$. By Lucas' theorem,\n$$\n\\binom{2188}{k} = \\binom{40_{(547)}}{\\overline{ab}_{(547)}} \\equiv \\b... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Other"
] | null | proof and answer | 2146 | |
0gz2 | Parabola $y = ax^2 + bx + c$ passes through the points $A(-2, 1)$ and $B(2, 9)$, and does not intersect $x$-axis. Find all possible values of $x$ coordinate of the vertex of the parabola. | [
"We first write analytically the conditions that our parabola passes through the given points:\n$$\n\\begin{cases} 4a - 2b + c = 1, \\\\ 4a + 2b + c = 9, \\end{cases}\n$$\nWe can now find $b = 2$ and $4a + c = 5$. From the condition, that our parabola does not have real zeros we get $D = b^2 - 4ac = 4 - 4a(5 - 4a) ... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010) | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (-4, -1) | |
0am3 | Problem:
What is the constant term in the expansion of $\left(2 x^{2}+\frac{1}{4 x}\right)^{6}$?
(a) $\frac{15}{32}$
(b) $\frac{12}{25}$
(c) $\frac{25}{42}$
(d) $\frac{15}{64}$ | [] | Philippines | Qualifying Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | d | |
07f4 | Let $\triangle ABC$ be an isosceles triangle ($AB = AC$) with incenter $I$. Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$. The circle $\omega$ intersects $AC$ and circumcircle of $\triangle ABC$ at $Q$ and $D$, respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$. Prove that... | [
"Let $P$ be the midpoint of segment $BC$ and $J$ be the midpoint of arc $\\widearc{BC}$ ($J \\neq A$).\n\nWe call the circumcircle of triangle $\\triangle CID$, $\\omega$ and the intersection point of $\\omega$ and segment $BC$, $R$. We have\n$$\n\\angle QIC = 180^{\\circ} - (\\angle IQC + ... | Iran | 37th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > ... | English | proof only | null | |
0ifz | Problem:
How many regions of the plane are bounded by the graph of
$$
x^{6}-x^{5}+3 x^{4} y^{2}+10 x^{3} y^{2}+3 x^{2} y^{4}-5 x y^{4}+y^{6}=0 ?
$$ | [
"Solution: 5\nThe left-hand side decomposes as\n$$\n\\left(x^{6}+3 x^{4} y^{2}+3 x^{2} y^{4}+y^{6}\\right)-\\left(x^{5}-10 x^{3} y^{2}+5 x y^{4}\\right)=\\left(x^{2}+y^{2}\\right)^{3}-\\left(x^{5}-10 x^{3} y^{2}+5 x y^{4}\\right) .\n$$\nNow, note that\n$$\n(x+i y)^{5}=x^{5}+5 i x^{4} y-10 x^{3} y^{2}-10 i x^{2} y^{... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 5 | |
08vc | Determine the smallest positive integer with the last 4 digits 9999, which is divisible by 2011. | [
"Let $n$ be a positive integer for which the last 4 digits of $2011n$ is $9999$. Since the one's digit of $2011n$ is $9$, the one's digit of $n$ has to be $9$. Therefore, we can represent $n$ in the form $n = 10k + 9$ where $k$ is a non-negative integer. Then we must have $2011n = 10 \\cdot 2011k + 18099$ and since... | Japan | Japan Junior Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | 5849999 | |
0d9w | Let $a$, $b$, $c$ be real numbers such that $a+b+c+ab+bc+ca+abc \geq 7$. Prove that
$$
\sqrt{a^{2}+b^{2}+2}+\sqrt{b^{2}+c^{2}+2}+\sqrt{c^{2}+a^{2}+2} \geq 6
$$ | [
"First, by AM-GM we can show that\n$$\nx^{2}+y^{2}+1 \\geq xy+x+y \\text{ for all } x, y, z \\in \\mathbb{R}.\n$$\nHence, $\\sqrt{a^{2}+b^{2}+2} \\geq \\sqrt{|ab|+|a|+|b|+1} = \\sqrt{(|a|+1)(|b|+1)}$. Construct similar inequalities and take the sum, we get\n$$\n\\begin{aligned}\n& \\sqrt{a^{2}+b^{2}+2}+\\sqrt{b^{2}... | Saudi Arabia | Team selection tests for JBMO 2018 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0kdj | Problem:
Let $a_{1}, a_{2}, a_{3}, \ldots$ be a sequence of positive real numbers that satisfies
$$
\sum_{n=k}^{\infty}\binom{n}{k} a_{n}=\frac{1}{5^{k}}
$$
for all positive integers $k$. The value of $a_{1}-a_{2}+a_{3}-a_{4}+\cdots$ can be expressed as $\frac{a}{b}$, where $a, b$ are relatively prime positive integers... | [
"Solution:\nLet $S_{k}=\\frac{1}{5^{k}}$. In order to get the coefficient of $a_{2}$ to be $-1$, we need to have $S_{1}-3 S_{3}$. This subtraction makes the coefficient of $a_{3}$ become $-6$. Therefore, we need to add $7 S_{3}$ to make the coefficient of $a_{4}$ equal to $1$. The coefficient of $a_{4}$ in $S_{1}-3... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 542 | |
0kwz | Problem:
On an $8 \times 8$ chessboard, 6 black rooks and $k$ white rooks are placed on different cells so that each rook only attacks rooks of the opposite color. Compute the maximum possible value of $k$.
(Two rooks attack each other if they are in the same row or column and no rooks are between them.) | [
"Solution:\n\nThe answer is $k=14$. For a valid construction, place the black rooks on cells $(a, a)$ for $2 \\leq a \\leq 7$ and the white rooks on cells $(a, a+1)$ and $(a+1, a)$ for $1 \\leq a \\leq 7$.\n\n\n\nNow, we prove the optimality. As rooks can only attack opposite color rooks, t... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 14 | |
0bxy | Determine the positive integers $n > 1$ such that, for any divisor $d$ of $n$, the numbers $d^2 - d + 1$ and $d^2 + d + 1$ are prime. | [
"First, we prove that $n$ is square-free. If $d^2$ divides $n$ for a positive integer $d > 1$, then $(d^2)^2+d^2+1$ would be a prime number. But $d^4+d^2+1 = (d^2-d+1)(d^2+d+1)$, with both factors larger than $1$, which is a contradiction.\n\nThus, $n = p_1 \\cdot p_2 \\cdot \\dots \\cdot p_s$, where $s \\in \\math... | Romania | THE DANUBE MATHEMATICAL COMPETITION | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | English | proof and answer | 2, 3, 6 | |
03s2 | In an isosceles right angled triangle $\triangle ABC$, $CA = CB = 1$, and $P$ is an arbitrary point on the perimeter of $\triangle ABC$. Find the maximum value of $PA \cdot PB \cdot PC$. (posed by Li Weigu) | [
"(1) In the first diagram, if $P \\in AC$, we have $PA \\cdot PC \\le \\frac{1}{4}$ and $PB \\le \\sqrt{2}$. Thus $PA \\cdot PB \\cdot PC \\le \\frac{\\sqrt{2}}{4}$. The equality is not valid, since the two equality signs cannot be valid at the same time. Therefore $PA \\cdot PB \\cdot PC < \\frac{\\sqrt{2}}{4}$.\n... | China | China Western Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | √2/4 | |
051j | Let $x$, $y$, $z$ be positive real numbers whose sum is $2012$. Find the maximum value of
$$
\frac{(x^2 + y^2 + z^2)(x^3 + y^3 + z^3)}{(x^4 + y^4 + z^4)}
$$ | [
"If $x = y = z = \\frac{2012}{3}$, then\n$$\n\\frac{(x^2 + y^2 + z^2)(x^3 + y^3 + z^3)}{(x^4 + y^4 + z^4)} = 2012.\n$$\nNow we prove that for all $x$, $y$, $z$ satisfying the premises we have\n$$\n\\frac{(x^2 + y^2 + z^2)(x^3 + y^3 + z^3)}{(x^4 + y^4 + z^4)} \\le 2012\n$$\nIt suffices to show that $(x^2 + y^2 + z^2... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 2012 | |
0297 | Problem:
Defina $f(n, k)$ como o número de maneiras de distribuir $k$ chocolates para $n$ crianças em que cada criança recebe 0, 1 ou 2 chocolates. Por exemplo, $f(3,4)=6$, $f(3,6)=1$ e $f(3,7)=0$.
a) Exiba todas as 6 maneiras de distribuir 4 chocolates para 3 crianças com cada uma ganhando no máximo dois chocolates.... | [
"Solution:\n\n(a) Vamos representar cada distribuição por uma tripla ordenada de números $(a, b, c)$ em que cada número representa a quantia de chocolates que cada criança receberá. As seis possibilidades são:\n$$\n(2,2,0);\\ (2,0,2);\\ (0,2,2);\\ (2,1,1);\\ (1,2,1);\\ (1,1,2)\n$$\n\n(b) Se $k \\geq 2 \\cdot 2015 +... | Brazil | NÍVEL 3 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 3^2015 | |
023p | Problem:
(a) Mostre que não existem dois pontos com coordenadas inteiras no plano cartesiano que estão igualmente distanciados do ponto $\left(\sqrt{2}, 1/3\right)$.
(b) Mostre que existe um círculo no plano cartesiano que contém exatamente 2011 pontos com coordenadas inteiras em seu interior. | [
"Solution:\n\n(a) Suponhamos que os $(a, b)$ e $(c, d)$ são pontos com coordenadas inteiras que estão igualmente distanciados do ponto $\\left(\\sqrt{2}, 1/3\\right)$. Assim,\n$$\n\\sqrt{(a-\\sqrt{2})^{2}+\\left(b-\\frac{1}{3}\\right)^{2}}=\\sqrt{(c-\\sqrt{2})^{2}+\\left(d-\\frac{1}{3}\\right)^{2}}\n$$\nDeste modo,... | Brazil | Desafios | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0l99 | In plane, let be given two fixed circles $ (O_1) $ and $ (O_2) $ touching each other at $ M $, the radius of $ (O_2) $ is greater than that of $ (O_1) $. $ A $ is the point on $ (O_2) $ such that the points $ O_1 $, $ O_2 $, $ A $ are not collinear. Let $ AB $ and $ AC $ be the tangents to $ (O_1) $ with touching point... | [
"We consider two cases:\n\n* **First case:** The circles $ (O_1), (O_2) $ touch each other externally at $ M $.\nLet $ xy $ be the common tangent at $ M $ of $ (O_1), (O_2) $. Since $ CA $ and $ My $ are tangents to $ (O_1) $ at $ C $ and $ M $, we have $ \\angle FCA = \\angle CMy $. But $ \\angle CMy = \\angle FMx... | Vietnam | 2003 Vietnamese Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null |
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