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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
02ft | Let $c$ be a rational. Let $f(x) = x^2 + c$. Define $f^{(0)}(x) = x$, $f^{(n+1)}(x) = f(f^{(n)}(x))$. Show that there are only finitely many rationals $x$ such that the sequence $f^{(0)}(x), f^{(1)}(x), f^{(2)}(x), \dots$ takes only finitely many values. | [
"For sake of simplicity, call $x$ a *periodic number* if $f^{(0)}(x)$, $f^{(1)}(x)$, $f^{(2)}(x)$, $\\dots$ takes finitely many values.\n\nIf $|x| > |c| + 1$ then $x^2 - |x| = |x|(|x| - 1) > (|c| + 1)|c| \\ge |c| \\implies x^2 - |c| > |x|$, so $|f(x)| = |x^2 + c| \\ge x^2 - |c| > |x|$, that is, $|f^{(n+1)}(x)| > |f... | Brazil | XIX OBM | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0kp6 | Problem:
Let $ABC$ be an acute triangle with $A$-excircle $\Gamma$. Let the line through $A$ perpendicular to $BC$ intersect $BC$ at $D$ and intersect $\Gamma$ at $E$ and $F$. Suppose that $AD = DE = EF$. If the maximum value of $\sin B$ can be expressed as $\frac{\sqrt{a}+\sqrt{b}}{c}$ for positive integers $a, b$, an... | [
"Solution:\nFirst note that we can assume $AB < AC$. Suppose $\\Gamma$ is tangent to $BC$ at $T$. Let $AD = DE = EF = x$. Then, by Power of a Point, we have $DT^2 = DE \\cdot DF = x \\cdot 2x = 2x^2 \\Longrightarrow DT = x\\sqrt{2}$. Note that $CT = s-b$, and since the length of the tangent from $A$ to $\\Gamma$ is... | United States | HMMT February | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 705 | |
0j1h | Problem:
Let $T$ be the set of numbers of the form $2^{a} 3^{b}$ where $a$ and $b$ are integers satisfying $0 \leq a, b \leq 5$. How many subsets $S$ of $T$ have the property that if $n$ is in $S$ then all positive integer divisors of $n$ are in $S$? | [
"Solution:\nAnswer: 924\n\nConsider the correspondence $(a, b) \\leftrightarrow 2^{a} 3^{b}$ for non-negative integers $a$ and $b$. So we can view $T$ as the square of lattice points $(a, b)$ where $0 \\leq a, b \\leq 5$, and subsets of $T$ as subsets of this square.\n\nNotice then that the integer corresponding to... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 924 | |
0800 | Problem:
Un cilindro retto $X$ ed un cono retto $Y$ hanno lo stesso raggio di base e la stessa altezza. Allora il rapporto fra le superfici laterali di $X$ e $Y$ :
(A) è sempre uguale al rapporto dei loro volumi
(B) può essere uguale al rapporto dei loro volumi (dipende dalle altezze)
(C) è sempre il $2 / 3$ del rappo... | [] | Italy | Italian Mathematical Olympiad - Febbraio Round | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area"
] | null | MCQ | E | |
09sw | Problem:
Bepaal het aantal verzamelingen $A=\{a_{1}, a_{2}, \ldots, a_{1000}\}$ van positieve gehele getallen met $a_{1}<a_{2}<\ldots<a_{1000} \leq 2014$, waarvoor geldt dat de verzameling
$$
S=\{a_{i}+a_{j} \mid 1 \leq i, j \leq 1000 \text{ en } i+j \in A\}
$$
een deelverzameling is van $A$. | [
"Solution:\nWe bewijzen dat er $2^{14}$ zulke verzamelingen zijn. We bewijzen in het bijzonder dat de verzamelingen $A$ die voldoen van de vorm $B \\cup C$ zijn, met $C$ een deelverzameling van $\\{2001, \\ldots, 2014\\}$ en $B=\\{1,2, \\ldots, 1000-|C|\\}$. Noem verzamelingen van die vorm \"leuk\". Omdat er $2^{14... | Netherlands | IMO-selectietoets II | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2^14 | |
0f4y | Problem:
A library is open every day except Wednesday. One day three boys, $A$, $B$, $C$ visit the library together for the first time. Thereafter they visit the library many times. $A$ always makes his next visit two days after the previous visit, unless the library is closed on that day, in which case he goes the fol... | [] | Soviet Union | 16th ASU | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | Saturday | |
01zu | Problem:
A finite set of integers is called bad if its elements add up to $2010$. A finite set of integers is a Benelux-set if none of its subsets is bad. Determine the smallest integer $n$ such that the set $\{502,503,504, \ldots, 2009\}$ can be partitioned into $n$ Benelux-sets.
(A partition of a set $S$ into $n$ sub... | [
"Solution:\nAs $502+1508=2010$, the set $S=\\{502,503, \\ldots, 2009\\}$ is not a Benelux-set, so $n=1$ does not work. We will prove that $n=2$ does work, i.e. that $S$ can be partitioned into 2 Benelux-sets.\nDefine the following subsets of $S$ :\n$$\n\\begin{aligned}\n& A=\\{502,503, \\ldots, 670\\}, \\\\\n& B=\\... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2 | |
06w0 | Let $a$, $b$, $c$, $d$ be four real numbers such that $a \geqslant b \geqslant c \geqslant d > 0$ and $a + b + c + d = 1$. Prove that
$$
(a + 2b + 3c + 4d) a^{a} b^{b} c^{c} d^{d} < 1
$$ | [
"The weighted AM-GM inequality with weights $a$, $b$, $c$, $d$ gives\n$$\na^{a} b^{b} c^{c} d^{d} \\leqslant a \\cdot a + b \\cdot b + c \\cdot c + d \\cdot d = a^{2} + b^{2} + c^{2} + d^{2}\n$$\nso it suffices to prove that $(a + 2b + 3c + 4d)(a^{2} + b^{2} + c^{2} + d^{2}) < 1 = (a + b + c + d)^{3}$. This can be ... | IMO | IMO 2020 Shortlisted Problems | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
08y3 | Reduce the following expression into the form $a + b\sqrt{2}$, where both $a$ and $b$ are rational numbers:
$$
\frac{(1 \times 4 + \sqrt{2})(2 \times 5 + \sqrt{2})\cdots(10 \times 13 + \sqrt{2})}{(2 \times 2 - 2)(3 \times 3 - 2)\cdots(11 \times 11 - 2)}
$$ | [
"$\\boxed{11+5\\sqrt{2}}$\n$$\n\\text{For } k = 1, 2, \\dots, 10, \\text{ we have}\n$$\n\\frac{k(k+3)+\\sqrt{2}}{(k+1)^2-2} = \\frac{(k+1+\\sqrt{2})(k+2-\\sqrt{2})}{(k+1+\\sqrt{2})(k+1-\\sqrt{2})} = \\frac{k+2-\\sqrt{2}}{k+1-\\sqrt{2}}\n$$\n\\text{Therefore, we obtain}\n$$\n\\begin{aligned}\n\\frac{(1 \\times 4 + \... | Japan | Japan 2015 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 11 + 5√2 | |
02gw | Let $x_1, x_2, \dots, x_{2004}$ be a sequence of integer numbers such that $x_{k+3} = x_{k+2} + x_k x_{k+1}$, $1 \le k \le 2001$. Is it possible that more than half of the elements are negative? | [
"The answer is yes. For instance, consider $x_0 = -1$, $x_1 = -n$ and $x_2 = -n^2$, $n$ sufficiently large. All $x_i$'s are polynomials in $n$ and, for all large $n$ its sign is equal to the sign of the coefficient of the term of greatest degree. Let $a(x_n)$ such term. Then\n$$\n\\begin{aligned}\na(x_0) &= -1, & a... | Brazil | XXVI OBM | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | Yes | |
0l40 | Problem:
Three points, $A$, $B$, and $C$, are selected independently and uniformly at random from the interior of a unit square. Compute the expected value of $\angle A B C$. | [
"Solution:\n\nSince $\\angle A B C + \\angle B C A + \\angle C A B = 180^{\\circ}$ for all choices of $A$, $B$, and $C$, the expected value is $60^{\\circ}$."
] | United States | HMMT February 2024 Guts Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 60° | |
0edt | Problem:
Enakostranični trikotnik na sliki je s črtami, vzporednimi eni od stranic, razdeljen na 5 enako širokih pasov, od katerih so trije pobarvani. Kolikšen delež trikotnika je pobarvan?
(A) $52 \%$
(B) $58 \%$
(C) $60 \%$
(D) $68 \%$
(E) $72 \%$
| [
"Solution:\n\nEnakokrak trikotnik na sliki razdelimo na manjše skladne trikotnike, tako da dorišemo še črte vzporedne preostalima dvema stranicama. Potem je od 25 majhnih trikotnikov pobarvanih 15. Torej je pobarvanih $\\frac{15}{25}=\\frac{60}{100}$ trikotnika, kar je $60 \\%$."
] | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | C | |
050d | Consider an acute-angled triangle $ABC$ and its circumcircle. Let $D$ be a point on the arc $AB$ which does not include point $C$ and let $A_1$ and $B_1$ be points on the lines $DA$ and $DB$, respectively, such that $CA_1 \perp DA$ and $CB_1 \perp DB$. Prove that $|AB| \ge |A_1B_1|$. | [
"If $CD$ is the diameter of the circumcircle of triangle $ABC$, then $A_1 = A$ and $B_1 = B$ and the statement holds. Assume that $CD$ is not the diameter (Fig. 3). Then $A_1 \\ne A$ and $B_1 \\ne B$. The point $A_1$ lies on the ray $AD$ if and only if the point $B_1$ does not lie on the ray $BD$ (depending on whic... | Estonia | Selected Problems from Open Contests | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
0fw4 | Problem:
Für eine natürliche Zahl $n$ sei
$$
f(n)=\frac{1}{n} \sum_{k=1}^{n}\left\lfloor\frac{n}{k}\right\rfloor
$$
Beweise, dass es unendlich viele natürliche Zahlen $m$ gibt, für die die Ungleichung $f(m)<f(m+1)$ gilt, und dass es unendlich viele natürlichen Zahlen $m$ gibt, für die die Ungleichung $f(m)>f(m+1)$ gil... | [
"Solution:\n\nWir geben einen kombinatorischen Beweis. Dazu müssen wir zuerst eine geeignete Interpretation der Funktion $f$ finden. Für natürliche Zahlen $n, k$ ist $\\left\\lfloor\\frac{n}{k}\\right\\rfloor$ gleich der Anzahl natürlicher Zahlen $\\leq n$, die durch $k$ teilbar sind. Oder anders formuliert, gleich... | Switzerland | IMO Selektion | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0jj6 | Problem:
Let $ABC$ be a triangle with $AB = AC = \frac{25}{14} BC$. Let $M$ denote the midpoint of $\overline{BC}$ and let $X$ and $Y$ denote the projections of $M$ onto $\overline{AB}$ and $\overline{AC}$, respectively. If the areas of triangle $ABC$ and quadrilateral $AXMY$ are both positive integers, find the minim... | [
"Solution:\n\nBy similar triangles, one can show that $[AXMY] = 2 \\cdot [AMX] = \\left(\\frac{24}{25}\\right)^2 \\cdot 2[ABM] = \\left(\\frac{24}{25}\\right)^2 \\cdot [ABC]$. Thus the answer is $25^2 + 24^2 = 1201$."
] | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 1201 | |
00vx | A positive integer $n > 1$ is called good if there exists some permutation of the numbers $1, 2, 3, \dots, n$, denoted by $(a_1, a_2, a_3, \dots, a_n)$ such that $a_i$ and $a_{i+1}$ have different parities for every $1 \le i \le n-1$; and for every $1 \le k \le n$, the sum $a_1 + a_2 + \dots + a_k$ is a quadratic resid... | [
"First, we will show that all numbers $n = 4^m$ with $m \\in \\mathbb{Z}^+$ are not good. Indeed, consider the last sum in the given condition\n$$\na_1 + a_2 + \\dots + a_n = 1 + 2 + \\dots + n = \\frac{4^m (4^m + 1)}{2}\n$$\nSuppose that there exists $x \\in \\mathbb{Z}$ such that\n$$\n\\frac{4^m (4^m + 1)}{2} \\e... | Balkan Mathematical Olympiad | 42nd Balkan Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0ix0 | Problem:
Suppose $a$, $b$ and $c$ are integers such that the greatest common divisor of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x+1$ (in the ring of polynomials in $x$ with integer coefficients), and the least common multiple of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x^{3}-4 x^{2}+x+6$. Find $a+b+c$. | [
"Solution:\n\nSince $x+1$ divides $x^{2}+a x+b$ and the constant term is $b$, we have $x^{2}+a x+b=(x+1)(x+b)$, and similarly $x^{2}+b x+c=(x+1)(x+c)$. Therefore, $a=b+1=c+2$.\n\nFurthermore, the least common multiple of the two polynomials is $(x+1)(x+b)(x+b-1)=x^{3}-4 x^{2}+x+6$, so $b=-2$. Thus $a=-1$ and $c=-3$... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | -6 | |
0bzw | Show that there are infinitely many natural numbers $a$ and $b$ so that:
$$
a \cdot \gcd(a, b) = b + \text{lcm}(a, b),
$$
where $\gcd(a, b)$ is the greatest common divisor and $\text{lcm}(a, b)$ is the lowest common multiple of $a$ and $b$. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
059v | Find the least possible sum of 2021 terms of the sequence $a_1, a_2, a_3, \dots$, where $a_1 = 0$, $a_2 = a_3 = 1$ and $a_{i+j} > a_i + a_j$ for every $i, j \ge 2$. | [
"**Answer:** $2 \\cdot 1010^2$.\n\nWe show that the least sum arises in the case of the sequence $0, 1, 1, 3, 3, 5, 5, \\dots$ ($a_{2i} = a_{2i+1} = 2i - 1$ for every $i \\ge 1$). Firstly, we show that this sequence meets the conditions of the problem. Indeed, if $j$ and $k$ are of the same parity then $a_{j+k} = j... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 2040200 | |
059s | Let $m, n > 2$ be positive integers. In each cell of an $m \times n$ grid there is a lamp, which can either be turned on or off. In one switch, we can change the state of five lamps, which are placed in cells forming a cross (see the diagram). Initially all the lamps are turned off. How many possible arrangements of la... | [
"Answer: $2^{(m-2)(n-2)}$.\n\nFor convenience, assume that the switch for any cross is located at the central cell of that cross. Then switches are located in all cells not on the edge of the grid. There are $(m-2)(n-2)$ such cells.\n\nThe final state of each lamp is determined by whether there have been an even or... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2^{(m-2)(n-2)} | |
0dk8 | For every $n = 1, 2, 3, \ldots$, define $a_n = 3^{3^n-1} + 2$. Prove that there are infinitely many prime numbers $p$ for which there exists a natural number $n$ such that $p$ is a divisor of $a_n$. | [
"Since $a_n$ is odd and is congruent to $2$ modulo $3$, it always has a prime divisor of the form $3h + 2$ for $h \\in \\mathbb{Z}^+$. Suppose by contradiction that the sequence $(a_n)$ has finitely many prime divisors, then the number of prime divisors of the form $3h + 2$ of the sequence is clearly also finite, l... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0jpg | Problem:
A wealthy king has his blacksmith fashion him a large cup, whose inside is a cone of height $9$ inches and base diameter $6$ inches (that is, the opening at the top of the cup is $6$ inches in diameter). At one of his many feasts, he orders the mug to be filled to the brim with cranberry juice.
For each positi... | [
"Solution:\nAnswer: $\\frac{216 \\pi^{3}-2187 \\sqrt{3}}{8 \\pi^{2}}$\n\nFirst, we find the total amount of juice consumed. We can simply subtract the amount of juice remaining at infinity from the initial amount of juice in the cup, which of course is simply the volume of the cup; we'll denote this value by $V$.\n... | United States | HMMT February | [
"Geometry > Solid Geometry > Volume",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (216 pi^3 - 2187 sqrt(3)) / (8 pi^2) | |
072o | Problem:
Prove that for every positive integer $n$ there exists a unique ordered pair $(a, b)$ of positive integers such that
$$
n = \frac{1}{2}(a + b - 1)(a + b - 2) + a
$$ | [
"Solution:\nWe have to prove that $f: \\mathbb{N} \\times \\mathbb{N} \\rightarrow \\mathbb{N}$ defined by\n$$\nf(a, b) = \\frac{1}{2}(a + b - 1)(a + b - 2) + a, \\quad \\forall a, b \\in \\mathbb{N}\n$$\nis a bijection. (Note that the right side is a natural number.) To this end define\n$$\nT(n) = \\frac{n(n+1)}{2... | India | INMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0c97 | Let $n$ be a positive integer and let $M = \{1, 2, 3, \dots, n^2 + n + 2\}$. We consider the subsets $A_1, A_2, \dots, A_n$ of $M$, such that for each $k \in \{1, 2, \dots, n\}$ the set $A_k$ has $n^2 + k + 1$ elements. Prove that the intersection of the $n$ subsets contains at least two consecutive integers. | [
"Observe that $A_n$ has $n^2 + n + 1$ elements, hence contains all the elements of $M$ except one. Similarly, $A_{n-1}$ contains all the elements of $M$ except $2$, $\\dots$, $A_1$ contains all the elements of $M$, except $n$ of them. We deduce that the intersection $A := \\bigcap_{k=1}^n A_k$ contains all the elem... | Romania | Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0cou | In a boarding school, $512$ students learn $9$ disciplines. These students live in $256$ double rooms; two students are called *neighbors* if they share a room. It is known that for every two students the sets of disciplines in which they are interested are distinct (in particular, exactly one student is interested in ... | [
"Мы докажем утверждение задачи в более общем виде, для $n \\ge 2$ предметов и $2^n$ детей, произвольно разбитых на $2^{n-1}$ пар соседей. Заметим, что существует ровно $2^n$ наборов из $n$ предметов; значит, каждый набор предметов интересен ровно одному ученику.\n\nИндукция по $n$. При $n = 2$ легко проверить утвер... | Russia | Final round | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English; Russian | proof only | null | |
0549 | Find all real-valued functions $f$ defined on real numbers which satisfy $f(f(x) + f(y)) = f(x) + y$ for all real $x, y$. | [
"Let $z_1, z_2$ be real numbers for which $f(z_1) = f(z_2)$. Substituting $y = z_1$ and $y = z_2$ into the given equation we get $f(f(x) + f(z_1)) = f(x) + z_1$, and $f(f(x) + f(z_2)) = f(x) + z_2$. Since the left hand sides are equal, we have $f(x) + z_1 = f(x) + z_2$, whence $z_1 = z_2$. Hence $f$ is one-to-one. ... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x | |
0gyn | Integers $a, b, c$ satisfy the condition $ab + bc + ca = 1$. Prove that the number $(1 + a^2)(1 + b^2)(1 + c^2)$ is a perfect square of some natural number. | [
"Let us add $a^2$ to both sides of our equality:\n$$a^2 + ab + bc + ca = 1 + a^2 \\quad \\text{or} \\quad (a+b)(a+c) = 1 + a^2.$$ \nAnalogously,\n$$(a+b)(b+c) = 1 + b^2 \\quad \\text{and} \\quad (c+b)(a+c) = 1 + c^2.$$ \nIf we now multiply together all these equalities, we will obtain that $(1 + a^2)(1 + b^2)(1 + c... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
05sy | Problem:
2019 élèves participent à un concours et répondent chacun à 6 questions. À la fin du concours, on remarque que, parmi les bonnes réponses données par un quelconque groupe de 3 élèves, il y a au moins une réponse correcte à au moins 5 des 6 questions du concours. Quelle est la valeur minimale du nombre total d... | [
"Solution:\n\nOn peut commencer par ranger les données dans un tableau $2019 \\times 6$, où les 2019 lignes représentent les 2019 élèves et les 6 colonnes les 6 problèmes. Dans la case $(i, j)$, on inscrit un 1 si l'élève $i$ a répondu correctement à la question $j$ et on inscrit un 0 sinon. Le problème nous invite... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 10065 | |
02li | Problem:
Maria encomendou certo número de televisores a $R\$ 1994,00$ cada um. Ela reparou que no total a pagar, não tem nem 0, nem 7, nem 8 e nem 9. Qual foi o menor número de televisores que ela encomendou? | [
"Solution:\n\nSe Maria comprou $n$ televisores, então ela gastou $1994n$, que é um múltiplo de $1994$ onde não aparecem os algarismos $0, 7, 8$ e $9$. Vamos tentar limitar o valor de $n$. Primeiro observe que\n$$\n1994 n = 2000 n - 6 n\n$$\ne também que se\n$$\n6 n < 300\n$$\nentão o número $2000 n - 6 n$ tem $7$ o... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 56 | |
00ki | For any integer $n$, let $M(n) = \{n, n+1, n+2, n+3, n+4\}$. Let $S(n)$ denote the sum of the squares of all elements of $M(n)$ and let $P(n)$ denote the product of these squares. For which integers $n$ is $S(n)$ a divisor of $P(n)$? | [
"We substitute $k = n + 2$ such that\n$$\n\\begin{aligned}\nS(n) &= (k-2)^2 + (k-1)^2 + k^2 + (k+1)^2 + (k+2)^2 = 5k^2 + 10 = 5(k^2 + 2). \\\\\nP(n) &= (k-2)^2(k-1)^2 k^2 (k+1)^2 (k+2)^2 = k^2(k^2-1)^2(k^2-4)^2.\n\\end{aligned}\n$$\nAs $P(n)$ is the square of the product of 5 consecutive integers, it is divisible b... | Austria | Austria 2014 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | n ∈ {-7, -6, -4, -3, -2, -1, 0, 2, 3} | |
0hn6 | Problem:
Do there exist four consecutive positive integers whose product is a perfect square? | [
"Solution:\n\nThe answer is no. If $x \\geq 1$ is an integer,\n$$\n\\begin{aligned}\n& x(x+1)(x+2)(x+3) \\\\\n& = [x(x+3)] \\cdot [(x+1)(x+2)] \\\\\n& = \\left[x^2 + 3x\\right] \\cdot \\left[x^2 + 3x + 2\\right] \\\\\n& = \\left[\\left(x^2 + 3x + 1\\right) - 1\\right] \\cdot \\left[\\left(x^2 + 3x + 1\\right) + 1\\... | United States | Berkeley Math Circle | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other"
] | null | proof and answer | No | |
0gac | 平面上有一個正三角形網格，相鄰兩格點的距離為 $1$。有一個邊長為 $n$ 的正三角形，其三個頂點都在格點上，三邊都落在格線上。現在，將此正三角形分割成 $n^2$ 個面積相等的小三角形（不需為正三角形），使得每個小三角形的三個頂點都在格點上。
證明：其中至少有 $n$ 個小三角形是正三角形。
There is a grid of equilateral triangles with a distance $1$ between any two neighboring grid points. An equilateral triangle with side length $n$ lies on the grid so that... | [
"不失一般性，假設大正三角形朝上。以下證明：在所有小三角形中，朝上的正三角形比朝下的正三角形至少多 $n$ 個。此自然可證明原命題。\n\n若所有小三角形皆為正三角形顯然成立（因每一橫排向上者都比向下者多一個）。若否，則進行以下操作：找所有小三角形中最長的邊 $AB$，並考慮以它為邊的兩個小三角形 $ABC$ 和 $ABD$。可以證明：\n\n引理一：$ABCD$ 組成一個平行四邊形。\n證明：由皮克公式 (Pick's formula) 知線段 $AB$ 上沒有其他格點，且所有可能的 $C$、$D$ 位置落在平行 $AB$ 的兩條直線上（分別在 $AB$ 兩側），並以長度為 $AB$ 的間隔分布。又，由於 $AB$ 為最長... | Taiwan | 二〇一六數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
00t0 | Let $MAZN$ be an isosceles trapezium inscribed in a circle $(c)$ with centre $O$. Assume that $MN$ is a diameter of $(c)$ and let $B$ be the midpoint of $AZ$. Let $(\varepsilon)$ be the perpendicular line on $AZ$ passing through $A$. Let $C$ be a point on $(\varepsilon)$, let $E$ be the point of intersection of $CB$ wi... | [
"$$\n\\angle EPD = 90^\\circ - \\angle EDC = 90^\\circ - \\angle ACB = \\angle EAC\n$$\nSo the points $E$, $A$, $C$, $P$ are concyclic. It follows that $\\angle CPA = 90^\\circ$, therefore the triangle $APZ$ is right-angled. Since also $B$ is the midpoint of $AZ$, then $PB = AB = BZ$.\n\nWe have\n$$\n\\angle BPE = ... | Balkan Mathematical Olympiad | BMO Short List | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
057r | There are 2020 inhabitants in a town. Before Christmas, they are all happy; but if an inhabitant does not receive any Christmas card from any other inhabitant, he or she will become sad. Unfortunately, there is only one post company which offers only one kind of service: before Christmas, each inhabitant may appoint tw... | [
"Partition 2019 inhabitants into 673 groups, each containing 3 inhabitants. Suppose that each inhabitant appoints two other members of the same group. As no group member is appointed thrice, the company cannot send three cards to one group member. Hence in every group, two different members get a card and at most o... | Estonia | IMO Team Selection Contest | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 674 | |
0afx | На турнир во туркање на раце учествуваат $n$ деца. Пред почетокот на турнирот, секое од децата добило реден број (прв, втор, ..., $n$-ти натпреварувач). Турнирот ќе се одвива во два натпреварувачки дена по следниов систем на натпреварување:
првиот ден најпрво се натпреваруваат првиот и вториот натпреварувач; победникот... | [
"Првиот ден, да го разгледаме последното туркање: нека тоа е помеѓу $k$-тиот и $n$-тиот натпреварувач. Тоа значи дека $k$-тиот натпреварувач ги победил $(k+1)$-от, $(k+2)$-от, ..., $(n-1)$-натпреварувач, па затоа првиот ден ги имало следниве дуели: $\\{k, k+1\\}$, $\\{k, k+2\\}$, $\\dots$, $\\{k, n\\}$. Еден од ови... | North Macedonia | Републички натпревар по математика за основно образование | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | Macedonian, English | proof only | null | |
0kz6 | Positive integers $x$ and $y$ satisfy the equation $\sqrt{x} + \sqrt{y} = \sqrt{1183}$. What is the minimum possible value of $x + y$?
(A) 585 (B) 595 (C) 623 (D) 700 (E) 791 | [
"Observe that $1183 = 13^2 \\cdot 7$, so $\\sqrt{1183} = 13\\sqrt{7}$. Because $\\sqrt{x} + \\sqrt{y} = 13\\sqrt{7}$, it follows that $\\sqrt{x}$ and $\\sqrt{y}$ must be of the form $a\\sqrt{7}$ and $b\\sqrt{7}$, respectively, where $a$ and $b$ are positive integers and $a + b = 13$. Then $\\sqrt{x} = \\sqrt{7a^2}$... | United States | AMC 10 B | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | B | |
01lo | $AB$ and $CD$ are two parallel chords of a parabola. Circle $S_1$ passing through points $A$, $B$ intersects circle $S_2$ passing through $C$, $D$ at points $E$, $F$.
Prove that if $E$ belongs to the parabola, then $F$ also belongs to the parabola. | [
"First note that all parabolas are similar, thus we may consider the parabola $y = x^2$. Let $a$, $b$, $c$, $d$, $e$ be the abscissae of the points $A$, $B$, $C$, $D$, $E$ respectively. We use the following easy lemmas.\n\n**Lemma 1.** The chords $AB$ and $CD$ of the parabola are parallel if and only if $a + b = c ... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
01z7 | Three non-intersecting circles of radius $1$ are placed inside the triangle $ABC$. (Circles can touch each other and the sides of a triangle, but cannot share interior points.)
Find the largest value of $r$ for which we can be sure that inside the triangle it is possible to draw a fourth circle of radius $r$ that doesn... | [
"Answer: $r = 1/3$.\n\nConsider three circles $\\omega_1$, $\\omega_2$ and $\\omega_3$ of radius $1$ with the centers $O_1$, $O_2$ and $O_3$ respectively and the equilateral triangle $ABC$ such that the side $AB$ touches the circles $\\omega_1$ and $\\omega_2$, the side $BC$ touches the circles $\\omega_2$ and $\\o... | Belarus | Belarus2022 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 1/3 | |
0drq | Let $a, b, c$ be real numbers such that $0 < a, b, c < 1/2$ and $a + b + c = 1$. Prove that for all real numbers $x, y, z$,
$$
abc(x + y + z)^2 \geq ayz(1 - 2a) + bxz(1 - 2b) + cxy(1 - 2c).
$$
When does equality hold? | [
"By symmetry, we may assume that $\\frac{x}{a} \\ge \\frac{y}{b} \\ge \\frac{z}{c}$. Let $\\frac{y}{b} = q$, $\\frac{x}{a} = q + \\alpha$ and $\\frac{z}{c} = q - \\beta$ where $\\alpha, \\beta \\ge 0$. Thus $x = a(q + \\alpha)$, $y = bq$, $z = c(q - \\beta)$. We have\n$$\n\\begin{align*}\n& abc(x + y + z)^2 \\ge ay... | Singapore | Singapur | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | Equality holds if and only if x/a = y/b = z/c. | |
07k3 | Let $ABCD$ be a parallelogram. Perpendiculars $AX$ and $AY$ are drawn from $A$ to $BC$ and $CD$, respectively (so $X$ lies on $BC$ and $Y$ lies on $CD$). Parallel lines $l$ and $d$ are drawn such that $l$ is perpendicular to $XY$. Line $l$ intersects $AB$ and $BC$ at $K$ and $L$ respectively, and line $d$ intersects $A... | [
"Let $M$ be the intersection of diagonals $AC$ and $BD$. $H$ is the foot of the perpendicular from $A$ to $BD$, and $R$ is the intersection of $KL$ and $XY$. $E$ and $F$ are the midpoints of $AX$ and $AY$ respectively. Since $ME$ is a line segment in $\\triangle AXC$ (connecting $M$ on $AC$ to $E$ midpoint of $AX$)... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0j3x | Problem:
Three unit circles $\omega_{1}$, $\omega_{2}$, and $\omega_{3}$ in the plane have the property that each circle passes through the centers of the other two. A square $S$ surrounds the three circles in such a way that each of its four sides is tangent to at least one of $\omega_{1}$, $\omega_{2}$ and $\omega_{3... | [
"Solution:\nAnswer: $\\frac{\\sqrt{6}+\\sqrt{2}+8}{4}$\n\n\n\nBy the Pigeonhole Principle, two of the sides must be tangent to the same circle, say $\\omega_{1}$. Since $S$ surrounds the circles, these two sides must be adjacent, so we can let $A$ denote the common vertex of the two sides t... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principl... | null | proof and answer | (\sqrt{6}+\sqrt{2}+8)/4 | |
0k2m | Problem:
Arnold and Kevin are playing a game in which Kevin picks an integer $1 \leq m \leq 1001$, and Arnold is trying to guess it. On each turn, Arnold first pays Kevin 1 dollar in order to guess a number $k$ of Arnold's choice. If $m \geq k$, the game ends and he pays Kevin an additional $m-k$ dollars (possibly zer... | [
"Solution:\n\nWe let $f(n)$ denote the smallest amount we can guarantee to pay at most if Arnold's first choice is $n$. For each $k < n$, if Arnold's first choice is $k+1$, in both worst case scenarios, he could end up paying either $n-k$ or $11+f(k)$. It is then clear that\n$$\nf(n) = \\min_{k+1 < n} \\max \\{ n-k... | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 859 | |
04lf | Twenty points with integer coordinates are given in the coordinate plane so that no three of them are collinear. Prove that there exists a triangle with vertices chosen among the given points whose centroid is also a point with integer coordinates. | [
"Let the given points be $A_1, A_2, \\ldots, A_{20}$, with $A_i = (x_i, y_i)$, where $x_i, y_i$ are integers.\n\nThe centroid of a triangle with vertices $A_i = (x_i, y_i)$, $A_j = (x_j, y_j)$, $A_k = (x_k, y_k)$ is\n$$\n\\left(\\frac{x_i + x_j + x_k}{3}, \\frac{y_i + y_j + y_k}{3}\\right).\n$$\n\nWe want to find t... | Croatia | Mathematical competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
0hxc | Problem:
Let $A B C D E$ be a convex pentagon circumscribed around a circle $\omega$ such that $A B \| C D$ and $B C \| D E$. Locate points $X$ and $Y$ on rays $A B$ and $E D$, respectively, such that $B X = A B$ and $D Y = D E$. Prove that $X Y$ is tangent to $\omega$. | [
"Solution:\n\nFor convenience, let $Z$ be the intersection of sides $A B$ and $D E$ (so we have rhombus $Z B C D$), and let $P$ and $Q$ be the tangency points of $\\omega$ with $A B$ and $D E$, respectively. Then it is evident that a necessary condition for $X Y$ to touch $\\omega$ is\n$$\nP X + Q Y = X Y\n$$\nTo s... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0jqk | Problem:
Let $a, b, c, d, e$ be nonnegative integers such that $625 a + 250 b + 100 c + 40 d + 16 e = 15^{3}$. What is the maximum possible value of $a + b + c + d + e$? | [
"Solution:\n\nAnswer: $153$\n\nThe intuition is that as much should be in $e$ as possible. But divisibility obstructions like $16 \\nmid 15^{3}$ are in our way. However, the way the coefficients $5^{4} > 5^{3} \\cdot 2 > \\cdots$ are set up, we can at least easily avoid having $a, b, c, d$ too large (specifically, ... | United States | HMMT February 2015 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 153 | |
0j5s | Problem:
Let $ABCD$ be a rectangle with $AB = 3$ and $BC = 7$. Let $W$ be a point on segment $AB$ such that $AW = 1$. Let $X, Y, Z$ be points on segments $BC, CD, DA$, respectively, so that quadrilateral $WXYZ$ is a rectangle, and $BX < XC$. Determine the length of segment $BX$. | [
"Solution:\n\nAnswer: $\\frac{7-\\sqrt{41}}{2}$\n\nWe note that\n$$\n\\angle YXC = 90^\\circ - \\angle WXB = \\angle XWB = 90^\\circ - \\angle AWZ = \\angle AZW\n$$\ngives us that $XYC \\cong ZWA$ and $XYZ \\sim WXB$. Consequently, we get that $YC = AW = 1$. From $XYZ \\sim WXB$, we get that\n$$\n\\frac{BX}{BW} = \... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (7-\sqrt{41})/2 | |
0bwz | Let $f: [0, \infty) \to (0, \infty)$ be a continuous function. Prove that:
a) if $n$ is a large enough integer, say, $n > n_0$, then $n \int_0^{x_n} f(t) \, dt = 1$ for a unique positive real number $x_n$;
b) the sequence $(nx_n)_{n>n_0}$ is convergent and evaluate its limit. | [
"Let $F: [0, \\infty) \\to \\mathbb{R}$ be the antiderivative of $f$ vanishing at $0$. Since $f$ takes on positive values, $F$ is strictly increasing, hence injective, and $\\alpha = \\sup \\text{im } F > 0$, the supremum being considered on the extended line.\n\na) Fix a positive integer $n_0$ such that $n_0\\alph... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series"
] | English | proof and answer | 1/f(0) | |
0iqg | Problem:
Let $n>2$ be a positive integer. Prove that there are $\frac{1}{2}(n-2)(n+1)$ ways to walk from $(0,0)$ to $(n, 2)$ using only up and right unit steps such that the walk never visits the line $y=x$ after it leaves the origin. | [
"Solution:\n\nThe first two steps can only go to the right. Then we need to compute the number of ways of walking from $(2,0)$ to $(n, 2)$ which does not pass through the point $(2,2)$. There are $\\binom{n}{2}$ ways to walk from $(2,0)$ to $(n, 2)$, and exactly one of those paths passes through the point $(2,2)$. ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | (1/2)(n-2)(n+1) | |
0jq5 | Problem:
Yan and Jacob play the following game. Yan shows Jacob a weighted 4-sided die labelled $1$, $2$, $3$, $4$, with weights $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{7}$, $\frac{1}{42}$, respectively. Then, Jacob specifies $4$ positive real numbers $x_{1}, x_{2}, x_{3}, x_{4}$ such that $x_{1}+x_{2}+x_{3}+x_{4}=1$.... | [
"Solution:\n\nJacob should pick $\\left(x_{1}, x_{2}, x_{3}, x_{4}\\right)=\\left(\\frac{1}{2}, \\frac{1}{3}, \\frac{1}{7}, \\frac{1}{42}\\right)$. More generally, suppose the weights are $p_{1}, \\ldots, p_{4}$. Then Jacob's expected payoff is\n\n$$\n10+\\sum_{i=1}^{4} p_{i} \\log \\left(x_{i}\\right)=10+\\sum_{i=... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | (1/2, 1/3, 1/7, 1/42) | |
02g4 | Planet *Zork* is spherical and has many towns. For each town there is a corresponding antipodal town (i.e. symmetric in relation to the centre of the planet).
There are roads connecting pairs of towns in Zork. If there is a road connecting towns $P$ and $Q$ then there is also a road connecting towns $P'$ and $Q'$, wher... | [
"Let $a_i \\ge b_i$ be the prices of Kriptonita in the antipodal towns $A_i$ and $B_i$ respectively. Suppose that the prices differ by more than 100 Urghs at all antipodal towns. Thus $a_i - b_i > 100$ for each $i$.\nIn a sequence of roads connecting $A_0$ to $B_0$, we can find a road connecting $A_j$ and $B_k$ for... | Brazil | XXI OBM | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0khi | Problem:
Find the number of ways in which the nine numbers
$$
1, 12, 123, 1234, \ldots, 123456789
$$
can be arranged in a row so that adjacent numbers are relatively prime. | [
"Solution:\nThe six numbers $12, 123, 12345, 123456, 12345678$, and $123456789$ are divisible by $3$, so they cannot be adjacent. However, arranging six numbers in a row with no two adjacent requires at least $11$ numbers, which is impossible."
] | United States | HMMT November 2021 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 0 | |
017g | Let $x$ be a positive acute angle. Prove that
$$
\cos^2(x) \cot(x) + \sin^2(x) \tan(x) \ge 1
$$ | [
"The geometric-arithmetic inequality gives\n$$\n\\cos x \\sin x \\le \\frac{\\cos^2 x + \\sin^2 x}{2} = \\frac{1}{2}.\n$$\nIt follows that\n$$\n1 = (\\cos^2 x + \\sin^2 x)^2 = \\cos^4 x + \\sin^4 x + 2 \\cos^2 x \\sin^2 x \\le \\cos^4 x + \\sin^4 x + \\frac{1}{2}\n$$\nso\n$$\n\\cos^4 x + \\sin^4 x \\ge \\frac{1}{2}... | Baltic Way | BALTIC WAY | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0fnp | Determinar cuántas soluciones reales tiene la ecuación
$$
\sqrt{2 - x^2} = \sqrt[3]{3 - x^3}
$$ | [
"Para que existan soluciones reales tiene que ser $x \\in [-\\sqrt{2}, \\sqrt{2}]$. Ahora bien, si $x \\in [-\\sqrt{2}, 0]$ se tiene que\n$$\n2 - x^2 \\leq 2, \\quad 3 - x^3 \\geq 3,\n$$\npero $\\sqrt{2} < \\sqrt[3]{3}$, por lo que no hay soluciones cuando $x \\in [-\\sqrt{2}, 0]$.\n\nPor otra parte, cuando $x \\in... | Spain | L Olimpiada Matemática Española | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | Spanish | proof and answer | 0 | |
02j1 | Problem:
Correndo com velocidade de $10~\mathrm{km}/\mathrm{h}$, João completa uma certa distância em 6 minutos. A qual velocidade ele pode completar a mesma distância em 8 minutos?
(A) $7{,}5~\mathrm{km}/\mathrm{h}$
(B) $7{,}75~\mathrm{km}/\mathrm{h}$
(C) $8~\mathrm{km}/\mathrm{h}$
(D) $8{,}25~\mathrm{km}/\mathrm{h}$... | [
"Solution:\n\n6 minutos é $1/10$ da hora, logo a distância corrida em 6 minutos é $10:10 = 1~\\mathrm{km}$. Como, espaço $=$ velocidade $\\times$ tempo, temos $1~\\mathrm{km} = v \\times 8~\\mathrm{min} \\Rightarrow v = 1~\\mathrm{km}/8~\\mathrm{min}$ (onde $v$ é a velocidade). Logo, João corre $1~\\mathrm{km}$ em ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | A | |
0epb | If the average of four different positive integers is $8$, what is the largest possible value of any one of these integers? | [
"If the average of four numbers is $8$, their sum must be $8 \\times 4 = 32$. To maximize the largest one of these numbers, we choose the smallest possible values for the other three, which are $1$, $2$ and $3$ (since the integers must be positive and different). The remaining number is then $32 - (1 + 2 + 3) = 26$... | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 26 | |
0cuw | Initially, we put 100 cards on a table, each containing a positive integer. Exactly 28 of these cards contain odd numbers. Then, on each minute the following operation has been performed. We compute the product of numbers on every set of 12 cards on the table, add up all these products, write this number onto a new car... | [
"Answer. No.\nIf the table contains $k$ odd numbers, then the parity of a new one coincides with the parity of $\\binom{k}{12}$; so, on the first four minutes $k$ increases by 1, and then it is always equal to 32. Consider any $n \\ge 4$. Let $E_n$ ($T_n$) be the sum of products of all 11-tuples (12-tuples) of the ... | Russia | XLIII Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English; Russian | proof and answer | No | |
06mv | Given that $22! = 1124000727777 \boxed{\phantom{000000}}680000$, where the box contains three missing digits, write down the missing digits from left to right. | [
"Answer: 607\nLet the digits in the box form the number $\\overline{abc}$. Since $22!$ is divisible by $7 \\times 11 \\times 13 = 1001$ and $1000 \\equiv -1 \\pmod{1001}$, we have\n$$\n\\begin{aligned}\n0 &\\equiv \\overline{1124000727777abc680000} \\\\\n &= 1(1000)^7 + 124(1000)^6 + 727(1000)^4 + 777(1000)^3 + \\... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | final answer only | 607 | |
04ua | There are $2018$ players sitting around a round table. At the beginning of the game we arbitrarily deal all the cards from a deck of $K$ cards to the players (some players may receive no cards). In each turn we choose a player who draws one card from each of the two neighbours. It is only allowed to choose a player who... | [
"The answer is $K = 2017$.\nFor $K = 2018$, we deal $2$ cards to one player, $0$ cards to one of his neighbours and $1$ card to everyone else. Then in each turn we choose the player with $0$ cards:\n$$\n\\dots 11 \\overbrace{\\underbrace{20}_{\\sim}}^{\\sim} 11 \\dots \\rightarrow \\dots 11 \\overbrace{\\underbrace... | Czech Republic | Czech-Polish-Slovak Match | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 2017 | |
0fgk | Problem:
Sea $p_{n}(k)$ el número de permutaciones del conjunto $\{1,2, \ldots, n\}$, $n \geq 1$, que tienen exactamente $k$ puntos fijos. Demostrar que
$$
\sum_{k=0}^{n} k p_{n}(k) = n!
$$
(Nota: Una permutación $f$ de un conjunto $S$ es una aplicación biyectiva de $S$ sobre sí mismo. Un elemento $i$ de $S$ se llama... | [] | Spain | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | n! | |
04wo | Let $ABCD$ be a given convex quadrilateral. Determine the locus of the points $P$ lying inside the quadrilateral $ABCD$ and satisfying
$$
[PAB] \cdot [PCD] = [PBC] \cdot [PDA],
$$
where $[XYZ]$ denotes the area of triangle $XYZ$. | [
"If $P$ lies on one of the diagonals $AC$ or $BD$, let's say on $AC$, then\n$$\n\\frac{[PAB]}{[PBC]} = \\frac{AP}{PC} = \\frac{[PDA]}{[PCD]},\n$$\nwhich is the desired equality. We prove that no other point lying inside $ABCD$ satisfies the conditions of the problem.\n\nDenote by $O$ the point of the intersection o... | Czech-Polish-Slovak Mathematical Match | Czech-Polish-Slovak Match | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Quadrilaterals"
] | English | proof and answer | The locus consists of the two diagonals AC and BD. | |
015m | Two boys $A$ and $B$ have a bag with $2009$ coloured balls: $2007$ balls are green and two are blue. They play a game with the following rules: When a boy gets the bag he draws two balls from it. If the two balls have the same colour he continues to draw one ball at a time until he draws a ball with the other colour th... | [
"If we imagine all the balls are drawn one at a time, and placed in a long row in the same order as drawn, then there are $\\binom{2009}{2} = 2008 \\times 1004 = 2017036$ different ways to place the blue balls.\n\nWe divide in four cases depending on the first two drawn balls ($b$ for blue and $g$ for green) and co... | Baltic Way | Baltic Way SHL | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1005/4018 | |
0d4h | Let $n > 3$ be an odd positive integer not divisible by $3$. Determine if it is possible to form an $n \times n$ array of numbers such that
a. The set of the numbers in each row is a permutation of $0, 1, \ldots, n-1$; the set of the numbers in each column is a permutation of $0, 1, \ldots, n-1$;
b. The board is tota... | [
"For $1 \\leq i, j \\leq n$, choose $a_{i, j}$ equal to the remainder of $3i + j$ when divided by $n$. Clearly, $0 \\leq a_{i, j} \\leq n-1$, for $1 \\leq i, j \\leq n$.\n\nIf we fix a row $1 \\leq i \\leq n$, $a_{i, j_1} = a_{i, j_2}$, for $1 \\leq j_1, j_2 \\leq n$, implies $3i + j_1 \\equiv 3i + j_2 \\pmod{n}$, ... | Saudi Arabia | SAMC | [
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English, Arabic | proof only | null | |
09x9 | For the integers $a$, $b$, $c$, and $d$ the difference between $a$ and $b$ equals $2$, the difference between $b$ and $c$ equals $3$, and the difference between $c$ and $d$ equals $4$. Which of the following values **cannot** be the difference between $a$ and $d$?
A) $1$ B) $3$ C) $5$ D) $7$ E) $9$ | [
"D) $7$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | D | |
0jb0 | Problem:
Maria is hopping up a flight of stairs with 100 steps. At every hop, she advances some integer number of steps. Each hop she makes has fewer steps. However, the positive difference between the length of consecutive hops decreases. Let $P$ be the number of distinct ways she can hop up the stairs. Find lower an... | [
"Solution:\n\nAnswer: 6922\n\nConsider the sequence of hops backwards. It is an increasing sequence where the first finite differences are increasing, so all the second finite differences are all positive integers. Furthermore, given positive integers $a, e_{0}$ (representing the initial value and initial first fin... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Expected values",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | final answer only | 6922 | |
06oz | Determine all pairs $(x, y)$ of integers satisfying the equation
$$
1 + 2^{x} + 2^{2x+1} = y^{2}
$$ | [
"If $(x, y)$ is a solution then obviously $x \\geq 0$ and $(x, -y)$ is a solution too. For $x = 0$ we get the two solutions $(0, 2)$ and $(0, -2)$.\n\nNow let $(x, y)$ be a solution with $x > 0$; without loss of generality confine attention to $y > 0$. The equation rewritten as\n$$\n2^{x}\\left(1 + 2^{x+1}\\right) ... | IMO | IMO 2006 Shortlisted Problems | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | [(0, 2), (0, -2), (4, 23), (4, -23)] | |
09ns | The numbers $1, 2, \dots, 2025$ are placed in the cells of a $45 \times 45$ grid so that each pair of consecutive integers occupies adjacent cells (i.e., sharing a common edge). Is it possible that all perfect squares $1^2, 2^2, \dots, 45^2$ lie in the same row?
(Nursoltan Khavalbolot) | [] | Mongolia | MMO2025 Round 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No, it is impossible. | |
06d3 | For a graph $G$,
$$
\begin{align*}
\chi_G &= \min \{k : G \text{ has a } k \text{ colouring}\}, \\
\Delta_G &= \max \{\text{degrees of vertices of } G\}.
\end{align*}
$$
a. Prove that for any simple connected graph $G$, $\chi_G \le \Delta_G + 1$.
b. A regular graph is a graph that every vertex has the same degree. Pr... | [
"a.\n(1999 Test 1 Problem 7(b)) See the solution above.\n\nb.\n(1999 Test 1 Problem 7(d) simplified) We give an alternative proof for this simpler version. We prove that there is a $\\Delta_G$-colouring by induction on the number $n$ of vertices.\n\nThe base case is $n = 3$, and the graph is $A - B - C$. Clearly, w... | Hong Kong | HKG TST | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
06lg | Given a list of integers $2^1 + 1, 2^2 + 1, \dots, 2^{2019} + 1$, Adam chooses two different integers from the list and computes their greatest common divisor. Find the sum of all possible values of this greatest common divisor. | [
"The answer is $2^{674} + 672$.\n\nWe claim that both $(2^r + 1, 2^s + 1)$ and $(2^r + 1, 2^s - 1)$ are of the form $2^a + 1$ or $1$ for any $r, s \\in \\mathbb{Z}^+$. We prove this by induction on $\\min\\{r, s\\}$. The base case $\\min\\{r, s\\} = 1$ is trivial since the greatest common divisor can only be $1$ or... | Hong Kong | IMO HK TST | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2^{674} + 672 | |
0hbs | Find all tuples of positive integers $(m, n, k)$ that satisfy the equation
$$
(m! + m)(n! + n) = (k! + k).
$$ | [
"From the problem statement, it is obvious that $k > m$ and $k > n$. We rewrite the given equation as\n$$\nm n ((m-1)! + 1)((n-1)! + 1) = k ((k-1)! + 1).\n$$\n\nSince $(k-1)!$ is divisible by $m$ and $n$, $(k-1)! + 1$ is not divisible by any factor of these numbers. Hence, $k \\nmid m n$. Suppose $n \\ge m$. If $m ... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (1, 1, 2) | |
0alu | Problem:
If $2 \sin (3x) = a \cos (3x + c)$, find all values of $ac$. In the choices below, $k$ runs through all integers.
(a) $-\frac{\pi}{2}$
(b) $2k\pi$
(c) $-\pi$
(d) $(4k-1)\pi$ | [] | Philippines | QUALIFYING STAGE | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | MCQ | (d) | |
087c | Problem:
Quattro interi positivi $a_{1}<a_{2}<a_{3}<a_{4}$ sono tali che, dati due qualunque di essi, il loro massimo comun divisore è maggiore di $1$, ma $\operatorname{mcd}(a_{1}, a_{2}, a_{3}, a_{4})=1$. Qual è il minimo valore che può assumere $a_{4}$?
(A) 10
(B) 12
(C) 15
(D) 30
(E) 105. | [
"Solution:\n\nLa risposta è $(\\mathbf{C})$. Innanzitutto, c'è almeno un numero $a_{k}$ che non è divisibile per $2$, altrimenti $2$ divide $\\operatorname{mcd}(a_{1}, a_{2}, a_{3}, a_{4})$. Vogliamo provare che $a_{k}$ deve avere almeno due fattori primi distinti: difatti, se così non fosse, dovremmo avere $a_{k}=... | Italy | Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | MCQ | C | |
0fyd | Problem:
Seien $x, y, z > 0$ reelle Zahlen mit $x y z = 1$. Beweise die Ungleichung
$$
\frac{(x+y-1)^2}{z} + \frac{(y+z-1)^2}{x} + \frac{(z+x-1)^2}{y} \geq x + y + z
$$ | [
"Solution:\n\nDie Nebenbedingung liefert zusammen mit AM-GM die Abschätzung $x + y + z \\geq 3 \\sqrt[3]{x y z} = 3$. Damit und mit CS erhält man nun für die linke Seite $A$ der Ungleichung\n$$\n\\begin{aligned}\n(x + y + z) \\cdot A &\\geq (|x + y - 1| + |y + z - 1| + |z + x - 1|)^2 \\\\\n&\\geq (2(x + y + z) - 3)... | Switzerland | SMO Finalrunde | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
068w | Let $\xi$ be the positive root of the equation $x^2 + x - 4 = 0$. The polynomial $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, where $n$ is a positive integer, has nonnegative integer coefficients and $P(\xi) = 2017$.
(i) Prove that: $a_0 + a_1 + \dots + a_n \equiv 1 \pmod{2}$
(ii) Find the least possible v... | [
"**(i)** Since $\\xi = \\frac{-1 + \\sqrt{17}}{2}$ is irrational and the polynomial $F x = P x - 2017$ has rational coefficients and $\\xi$ as a root, then it will have also the conjugate $\\frac{-1 - \\sqrt{17}}{2}$ as a root, and therefore it is divided by the polynomial $\\varphi x = x^2 + x - 4$. It comes easil... | Greece | 34th Hellenic Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | English | proof and answer | The sum of coefficients is odd; the least possible sum is 23. | |
0185 | Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
x f(f(y)) + y f(y - x) = f(f(x + y) - x) f(y)
$$
for all $x, y \in \mathbb{R}$. | [
"Let us denote $f(0) = c$. Assume that $c \\neq 0$. Taking $x = y = 0$ in the initial equation we get $c f(c) = 0$. Hence, $f(c) = 0$. Taking $y = c$ and $c - x$ instead of $x$ in the initial equation and dividing it by $c$ gives us the equality $f(x) = x - c$. Direct verification shows that no such function satisf... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 for all real x; f(x) = x for all real x | |
0ax5 | Problem:
Find all positive real numbers $a, b, c \leq 1$ such that
$$
\min \left\{\sqrt{\frac{a b+1}{a b c}}, \sqrt{\frac{b c+1}{a b c}}, \sqrt{\frac{a c+1}{a b c}}\right\}=\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}
$$ | [
"Solution:\nLet $r, s, t \\geq 0$ such that\n$$\na=\\frac{1}{1+r^{2}},\\ b=\\frac{1}{1+s^{2}},\\ c=\\frac{1}{1+t^{2}}\n$$\nAlso, WLOG, suppose $t=\\min \\{r, s, t\\}$. The required equation can then be rewritten as\n$$\n\\sqrt{\\left(1+t^{2}\\right)\\left\\{1+\\left(1+r^{2}\\right)\\left(1+s^{2}\\right)\\right\\}}=... | Philippines | 19th Philippine Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | All triples are given by a = 1/(1 + r^2), b = 1/(1 + 1/r^2), c = (r + 1/r)^2 / (1 + (r + 1/r)^2) for any positive r, together with all permutations of these three expressions. | |
04zd | Circle $c$ passes through vertices $A$ and $B$ of an isosceles triangle $ABC$, whereby line $AC$ is tangent to it. Prove that circle $c$ passes through the circumcenter or the incenter or the orthocenter of triangle $ABC$. (Seniors.) | [
"Consider three cases $|AB| = |AC|$, $|BC| = |BA|$, and $|CA| = |CB|$.\n\n1. We show that if $|AB| = |AC|$ (Fig. 4), then circle $c$ passes through the circumcenter of $ABC$. Let $O$ be the point at the same side from $AB$ as $C$ that is the intersection of the perpendicular bisector of side $AB$ and circle $c$. Th... | Estonia | Estonija 2010 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03kt | Problem:
The point $O$ is situated inside the parallelogram $ABCD$ so that
$$
\angle AOB + \angle COD = 180^\circ
$$
Prove that $\angle OBC = \angle ODC$. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ded | Let $(a_n)_{n \ge 1}$ be a sequence given by $a_1 = 45$ and
$$
a_n = a_{n-1}^2 + 15a_{n-1}
$$
for $n > 1$. Prove that the sequence contains no perfect squares. | [
"By induction we can show that $a_n > 0$ for all positive integer $n$. Suppose that there exists a positive integer $n$ such that $a_n$ is a perfect square.\n$$\na_n = a_{n-1}^2 + 15a_{n-1} = a_{n-1}(a_{n-1} + 15)\n$$\nso $a_{n-1} \\mid a_n$ for all positive integers $n$. Thus, it follows that\n$$\na_1 = 45 \\mid a... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
04s9 | A triangle $ABC$ with obtuse angle at $C$ is given. Axis $o_1$ of side $AC$ intersects side $AB$ in point $K$, axis $o_2$ of side $BC$ intersects side $AB$ in point $L$. Denote $O$ intersection of the axes $o_1$ and $o_2$. Prove that centre of the incircle of triangle $KLC$ lies on the circumcircle of triangle $OKL$. | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous >... | English | proof only | null | |
0irl | Problem:
Let $\mathcal{C}$ be the hyperbola $y^{2}-x^{2}=1$. Given a point $P_{0}$ on the $x$-axis, we construct a sequence of points $\left(P_{n}\right)$ on the $x$-axis in the following manner: let $\ell_{n}$ be the line with slope 1 passing through $P_{n}$, then $P_{n+1}$ is the orthogonal projection of the point of... | [
"Solution:\nLet $P_{n}=(x_{n}, 0)$. Then the $\\ell_{n}$ meet $\\mathcal{C}$ at $(x_{n+1}, x_{n+1}-x_{n})$. Since this point lies on the hyperbola, we have $(x_{n+1}-x_{n})^{2}-x_{n+1}^{2}=1$. Rearranging this equation gives\n$$\nx_{n+1}=\\frac{x_{n}^{2}-1}{2 x_{n}}\n$$\nChoose a $\\theta_{0} \\in(0, \\pi)$ with $\... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theore... | null | proof and answer | 254 | |
0hwt | Problem:
Aerith and Bob are playing tag at Lake Round, a perfectly circular lake. Aerith tags Bob right next to the lake and dives in. Aerith can swim at a speed of $2 \mathrm{mph}$, while Bob can't swim but runs at a speed of $9 \mathrm{mph}$. Can Aerith leave the lake without getting tagged? | [
"Solution:\n\nWe claim that Aerith can escape, even when Bob plays optimally and always runs towards Aerith's current location.\nLet a dash (d) be a unit of distance and a tick ($t$) be a unit of time such that the radius of the lake is $9$ dashes and such that one mph is equal to one dash per tick $(\\mathrm{d} / ... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | Yes, Aerith can escape. | |
0hsc | Problem:
Prove that each nonnegative integer can be represented in the form $a^{2}+b^{2}-c^{2}$, where $a, b, c$ are positive integers with $a<b<c$. | [
"Solution:\nWe note that\n$$\n0=3^{2}+4^{2}-5^{2}, \\quad 2=5^{2}+11^{2}-12^{2},\n$$\nand for $n>1$, we can write $2 n$ as\n$$\n2 n=(3 n)^{2}+(4 n-1)^{2}-(5 n-1)^{2},\n$$\nand $3 n<4 n-1<5 n-1$. For odd numbers greater than $7$, we can use\n$$\n2 n+3=(3 n+2)^{2}+(4 n)^{2}-(5 n+1)^{2},\n$$\nand for the first four od... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Number Theory > Diophantine Equations > Pythagorean triples"
] | null | proof only | null | |
0j8o | Problem:
Find the number of integers $x$ such that the following three conditions all hold:
- $x$ is a multiple of $5$
- $121 < x < 1331$
- When $x$ is written as an integer in base $11$ with no leading $0$s (i.e. no $0$s at the very left), its rightmost digit is strictly greater than its leftmost digit. | [
"Solution:\nWe will work in base $11$, so let $x = \\overline{def}_{11}$ such that $d > 0$. Then, based on the first two conditions, we aim to find multiples of $5$ between $100_{11}$ and $1000_{11}$. We note that\n$$\n\\overline{def}_{11} = 11^2 \\cdot d + 11 \\cdot e + f \\equiv d + e + f \\pmod{5}\n$$\nHence, $x... | United States | Harvard-MIT November Tournament | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics"
] | null | final answer only | 99 | |
03nb | Problem:
Find all polynomials $p(x)$ with real coefficients that have the following property: There exists a polynomial $q(x)$ with real coefficients such that
$$
p(1)+p(2)+p(3)+\cdots+p(n)=p(n) q(n)
$$
for all positive integers $n$. | [
"Solution:\nThe property clearly holds whenever $p(x)$ is a constant polynomial, since we can take $q(x)=x$. Assume henceforth that $p(x)$ is nonconstant and has the stated property. Let $d$ be the degree of $p(x)$, so $p(x)$ is of the form\n$$\np(x)=c x^{d}+\\cdots .\n$$\nBy a Lemma (which we will prove at the end... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All such polynomials are p(x)=c(x+r-1)(x+r-2)\cdots(x+r-d) with real c, integer r in {0,1,2,...,d}, and d the degree (including the constant case d=0). | |
09b0 | Let $p$ be a prime number. Prove that
$$
\sum_{k=0}^{p} (-1)^k \binom{p}{k} \binom{p+k}{k} \equiv -1 \pmod{p^3}
$$ | [
"The sum $\\sum_{k=0}^{p} (-1)^k \\binom{p}{k} \\binom{p+k}{k}$ is the coefficient of $x^p$ in the expansion of\n$$\n\\sum_k \\binom{p}{k} (x-1)^{p+k}.\n$$\nThis can be rewritten as\n$$\n\\left\\{ \\sum_{k=0}^{p} \\binom{p}{k} (x-1)^j \\right\\} (x-1)^p = x^p (x-1)^p\n$$\nHence the sum is actually $(-1)^p = -1$. He... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
0in2 | Problem:
Eric and Greg are watching their new favorite TV show, The Price is Right. Bob Barker recently raised the intellectual level of his program, and he begins the latest installment with bidding on the following question: How many Carmichael numbers are there less than 100,000?
Each team is to list one nonnegati... | [
"Solution:\n\n16? There are 16 such numbers: 561, 1105, 1729, 2465, 2821, 6601, 8911, 10585, 15841, 29341, 41041, 46657, 52633, 62745, 63973, and 75361. The next, 101101, is too large to be counted. Their distribution is considerably more subtle than that of the primes, and it was only recently proven that there ar... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | final answer only | 16 | |
04im | A grasshopper is sitting in the origin of the number line, at number $0$, and then it jumps, always in the same direction. For a positive integer $k$, in the first jump the grasshopper jumps to number $1$, and every following jump is exactly $k$ times longer than the previous jump. There is a hole in place of all multi... | [
"$$\na_1 = 1, \\quad a_n = 1 + k + \\dots + k^{n-1}, \\quad n \\ge 2.\n$$\nWe are looking for all numbers $k$ such that $2015 \\nmid a_n$ for all $n = 1, \\dots, 2015$.\nSuppose that $M(k, 2015) = d > 1$. Then every $a_n$ divided by $d$ gives the remainder $1$, and since $2015$ is divisible by $d$ we have that $201... | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | All positive integers k with gcd(k, 2015) > 1 | |
0b63 | If $a$, $b$, $c$ are complex numbers of modulus $1$, prove that
$$
|a - b|^2 + |a - c|^2 - |b - c|^2 \ge -1.
$$ | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
04pd | Solve
$$
\cos(2x) + \cos(2y) + 2 \sin x \sin y + 2 \cos x \cos y = 4.
$$
(Petar Bakić) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Precalculus > Trigonometric functions"
] | English | proof and answer | All solutions are x = mπ and y = nπ with integers m, n of the same parity (equivalently, x − y and x + y are both integer multiples of 2π). | |
0l0z | Integers $a$, $b$, and $c$ satisfy $ab + c = 100$, $bc + a = 87$, and $ca + b = 60$. What is $ab + bc + ca$?
(A) 212 (B) 247 (C) 258 (D) 276 (E) 284 | [
"**Answer (D):** Notice that the difference between $100$ and $87$ is $13$, a prime number. This fact will help to simplify the problem. Subtract the second equation from the first to get\n$$\n\\begin{aligned}\n13 &= (ab + c) - (bc + a) \\\\\n&= ab - bc - a + c \\\\\n&= b(a - c) - (a - c) \\\\\n&= (b - 1)(a - c).\n... | United States | AMC 10 A | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | MCQ | D | |
07s6 | Let $a$, $b$, $c$ be the side lengths of a triangle. Prove that
$$
2 (a^3 + b^3 + c^3) < (a + b + c) (a^2 + b^2 + c^2) \le 3 (a^3 + b^3 + c^3).
$$ | [
"Consider the inequality on the left. We prove this using the triangle inequality according to which $a < b + c$, $b < c + a$, and $c < a + b$. Applying this,\n$$\n\\begin{aligned}\n& (a + b + c)(a^2 + b^2 + c^2) - 2(a^3 + b^3 + c^3) \\\\\n&= a^2(b + c) + b^2(c + a) + c^2(a + b) - a^3 - b^3 - c^3 \\\\\n&= a^2(b + c... | Ireland | Irish | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0hav | Find all pairs of positive integers $(a, b)$, which satisfy the equation:
$$
ab^3 + a^3 + b + 1 = 2019.
$$ | [
"**Answer:** $a = 2, b = 10$.\n\nIt can be seen from the problem statement that both unknowns are even, because otherwise we would have an equality of an even and an odd number. Let us denote $a = 2n$ and $b = 2k \\Rightarrow$\n$$\n2n \\cdot 8k^3 + 8n^3 + 2k = 2018 \\text{ or } 8nk^3 + 4n^3 + k = 1009.\n$$\nHere, i... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | a = 2, b = 10 | |
0hcq | Let's consider the sequence of positive integers $(x_n)$, that is given by the formula:
$x_n = 5 \cdot 2^n - 1$, $n \in N$. Prove, that there is an infinite number of pairs $(x_i, x_j)$ of elements that are mutually-prime, and at the same time none of the elements $x_k$ in this infinite sequence of pairs is included at... | [
"Everything follows from this equation: $\\forall n \\in N$\n$$\nx_{n+1} - 2x_n = 5 \\cdot 2^{n+1} - 1 - 2 \\cdot (5 \\cdot 2^n - 1) = 10 \\cdot 2^n - 1 - 10 \\cdot 2^n + 2 = 1.\n$$\nSo, GCF of $(x_{n+1}, x_n)$ is a divisor of $1$, what means that they're coprime. What had to be shown."
] | Ukraine | Ukrainian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series"
] | English | proof only | null | |
092k | Problem:
Let $n \geqslant 2$ be an integer and $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers satisfying
(a) $x_{j}>-1$ for $j=1,2, \ldots, n$ and
(b) $x_{1}+x_{2}+\cdots+x_{n}=n$.
Prove the inequality
$$
\sum_{j=1}^{n} \frac{1}{1+x_{j}} \geqslant \sum_{j=1}^{n} \frac{x_{j}}{1+x_{j}^{2}}
$$
and determine when equality h... | [
"Solution:\nWe have to prove\n$$\n\\sum_{j=1}^{n} \\frac{1}{1+x_{j}}-\\sum_{j=1}^{n} \\frac{x_{j}}{1+x_{j}^{2}}=\\sum_{j=1}^{n} \\frac{1-x_{j}}{\\left(1+x_{j}\\right)\\left(1+x_{j}^{2}\\right)} \\geqslant 0\n$$\nWe use the supporting line method and consider the function $f$ defined by\n$$\nf(x)=\\frac{1-x}{(1+x)\\... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | Equality holds if and only if all the numbers are equal to 1. | |
03ec | Given the functions $f(x) = |x-2| - |x-4|$ and $g(x) = |x-8| - 2$. Find the area of the figure with vertices, the intersection points of the graphs of the functions $f(x)$ and $g(x)$ and the intersection points of the graph of $g(x)$ with the x-axis.
(Nedyalka Dimitrova) | [
"The graph of $g(x)$ consists of two rays with a common vertex at $x = 8$. We remove the module and easily calculate its intersection points with the x-axis through the equations $6 - x = 0$ and $x - 10 = 0$ — $A(6, 0)$ and $B(10, 0)$.\n\nAfter removing the modules in $f(x)$ we see that in the interval $(-\\infty, ... | Bulgaria | Autumn tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 12 | |
0j7y | Problem:
For any finite sequence of positive integers $\pi$, let $S(\pi)$ be the number of strictly increasing subsequences in $\pi$ with length 2 or more. For example, in the sequence $\pi=\{3,1,2,4\}$, there are five increasing sub-sequences: $\{3,4\},\{1,2\},\{1,4\},\{2,4\}$, and $\{1,2,4\}$, so $S(\pi)=5$. In an ei... | [
"Solution:\nAnswer: 8287\n\nFor each subset of Joy's set of cards, we compute the number of orders of cards in which the cards in the subset are arranged in increasing order. When we sum over all subsets of Joy's cards, we will obtain the desired sum.\n\nConsider any subset of $k$ cards. The probability that they a... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 8287 | |
08ec | Problem:
I numeri reali $x_{1}, x_{2}, x_{3}, \ldots, x_{30}$ verificano le seguenti condizioni:
$$
\left\{\begin{array}{l}
20^{3} x_{1}+21^{3} x_{2}+22^{3} x_{3}+\cdots+49^{3} x_{30}=13 \\
21^{3} x_{1}+22^{3} x_{2}+23^{3} x_{3}+\cdots+50^{3} x_{30}=1 \\
22^{3} x_{1}+23^{3} x_{2}+24^{3} x_{3}+\cdots+51^{3} x_{30}=19
\... | [
"Solution:\n\nLa risposta è (E). Sommando la prima equazione alla terza, si ottiene un'equazione della forma $a_{1} x_{1}+a_{2} x_{2}+\\ldots+a_{30} x_{30}=32$, dove $a_{1}=(21+1)^{3}-(21-1)^{3}=21^{3}+3 \\cdot 21^{2}+3 \\cdot 21+1+21^{3}-3 \\cdot 21^{2}+3 \\cdot 21-1=2 \\cdot 21^{3}+6 \\cdot 21$, $a_{2}=(22+1)^{3}... | Italy | Olimpiadi della Matematica | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | E | |
00rf | The positive real numbers $a$, $b$, $c$ satisfy the equality $a + b + c = 1$. For every natural number $n$ find the minimal possible value of the expression
$$
E = \frac{a^{-n} + b}{1 - a} + \frac{b^{-n} + c}{1 - b} + \frac{c^{-n} + a}{1 - c}
$$ | [
"We transform the first term of the expression $E$ in the following way:\n$$\n\\frac{a^{-n} + b}{1 - a} = \\frac{1 + a^n b}{a^n (b + c)} = \\frac{a^{n+1} + a^n b + 1 - a^{n+1}}{a^n (b + c)} = \\frac{a^n (a + b) + (1 - a)(1 + a + a^2 + \\dots + a^n)}{a^n (b + c)} \\\\\n\\frac{a^n (a + b)}{a^n (b + c)} + \\frac{(b + ... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (3^{n+2} + 3)/2 | |
0gyf | Prove that there exist infinitely many natural numbers $n$ with the following properties: we can represent number $n$ as a sum $n = a^2 + b^2$, and a sum $n = c^3 + d^3$, and can't represent it as a sum $n = x^6 + y^6$, where $a,b,c,d,x,y$ natural numbers. | [
"Consider the equality: $n = 8(s^6 + t^6)$. We have a representation of number $n$ as a sum of cubes. For the representation of number $n$ as a sum of squares we rewrite last equality as $n = 8(s^6 + t^6) = 2(2s^3)^2 + 2(2t^3)^2 = (2s^3 - 2t^3)^2 + (2s^3 + 2t^3)^2$.\n\nLet us show that it is impossible to represent... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0dug | Problem:
Naj bo $n \geq 4$. Oštevilčimo $n$ točk na krožnici s številkami od 1 do $n$. Rečemo, da sta nesosednji točki, oštevilčeni z $a$ in $b$, pravilen par, če so vsaj na enem izmed lokov, ki ju določata $a$ in $b$, vse točke ostevilčene s števili, manjšimi od $a$ in $b$. Dokaži, da je število pravilnih parov enako... | [
"Solution:\n\nNalogo bomo dokazali z indukcijo. Dokaz začnemo pri $n=4$. Sosedi od točke 1 sta pravilen par. Edini drug kandidat vsebuje točko 1, ki pa nikoli ni v pravilnem paru. Torej imamo le en pravilen par.\n\nRecimo, da formula velja za $n$. Oglejmo si $n+1$ točk na krožnici. Označimo sosedi točke 1 z $a$ in ... | Slovenia | 45. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null |
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