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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
096t | Problem:
Să se demonstreze, că
$$
\frac{1}{3!}+\frac{5}{4!}+\frac{11}{5!}+\ldots+\frac{n^{2}+n-1}{(n+2)!}<\frac{1}{2}
$$
oricare ar fi numărul natural $n$. Cu $n$ ! (se citeşte $n$ factorial) se notează produsul primelor $n$ numere naturale nenule: $n!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot n$. | [
"Solution:\n\nFie $S_{n}$ suma din partea stângă a inegalităţii, iar $a_{k}$ termenul general al sumei. Acest termen se descompune într-o sumă algebrică de câteva fracţii mai simple:\n$$\na_{k}=\\frac{k^{2}+k-1}{(k+2)!}=\\frac{(k+2)(k+1)-2(k+2)+1}{(k+2)!}=\\frac{1}{k!}-\\frac{2}{(k+1)!}+\\frac{1}{(k+2)!}\n$$\nAtunc... | Moldova | A 63-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
072q | Problem:
Let $X$ denote the set of all triples $(a, b, c)$ of integers. Define a function $f: X \rightarrow X$ by
$$
f(a, b, c) = (a + b + c, ab + bc + ca, abc)
$$
Find all triples $(a, b, c)$ in $X$ such that $f(f(a, b, c)) = (a, b, c)$. | [
"Solution:\nWe show that the solution set consists of $\\{(t, 0, 0) ; t \\in \\mathbb{Z}\\} \\cup \\{(-1, -1, 1)\\}$. Let us put $a + b + c = d$, $ab + bc + ca = e$ and $abc = f$. The given condition $f(f(a, b, c)) = (a, b, c)$ implies that\n$$\nd + e + f = a, \\quad de + ef + fd = b, \\quad def = c\n$$\nThus $abcd... | India | INMO | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All solutions are the triples (t, 0, 0) for any integer t, and the triple (-1, -1, 1). | |
0e7s | Problem:
2. Dan je pokončni valj, za katerega velja, da ploščina osnovne ploskve, ploščina plašča in površina tvorijo aritmetično zaporedje.
a) Naj bo $v$ višina danega valja. Izrazi polmer $r$ tega valja $z$ $v$.
b) Denimo, da je dani valj visok $3~\mathrm{m}$. V kolikem času bi se $v$ tem primeru valj napolnil do ... | [] | Slovenia | Državno tekmovanje | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > Volume"
] | null | proof and answer | a) r = (2/3)·v. b) Time to half-fill = 12000·π seconds ≈ 10 hours 28 minutes 19 seconds. | |
0lel | Given a fixed circle ($O$) and two fixed points $B, C$ on that circle, let $A$ be a moving point on ($O$) such that triangle $ABC$ is acute and scalene. Let $I$ be the midpoint of $BC$ and let $AD, BE$ and $CF$ be the altitudes of triangle $ABC$. On two rays $\overrightarrow{FA}$, $\overrightarrow{EA}$, take $M, N$ suc... | [
"a)\nIt is clear that $G$ is the Miquel point of completed quadrilateral $MNEF.LA$, thus $G$ lies on the circumcircle of $\\triangle AMN$, $\\triangle AEF$. Besides,\n\n$$\nBM = BF + FM = EN + CE = CN.\n$$\n\n\n\nLet $X$ be the midpoint of arc $BAC$ of $(O)$ then $\\triangle XBM = \\triangl... | Vietnam | TST | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, E... | English | proof only | null | |
0jah | Problem:
Consider the cube whose vertices are the eight points $(x, y, z)$ for which each of $x$, $y$, and $z$ is either $0$ or $1$. How many ways are there to color its vertices black or white such that, for any vertex, if all of its neighbors are the same color then it is also that color? Two vertices are neighbors ... | [
"Solution:\n\nAnswer: $118$\n\nDivide the $8$ vertices of the cube into two sets $A$ and $B$ such that each set contains $4$ vertices, any two of which are diagonally adjacent across a face of the cube. We do casework based on the number of vertices of each color in set $A$.\n\n- Case 1: $4$ black. Then all the ver... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 118 | |
0dyt | Problem:
V produktu potenc $5^{-2} \cdot 5^{-4} \cdot 5^{-8} \cdot \ldots \cdot 5^{-x}$, kjer eksponenti tvorijo geometrijsko zaporedje, določi $x$ tako, da bo $5^{-2} \cdot 5^{-4} \cdot 5^{-8} \cdot \ldots \cdot 5^{-x}=5^{-16382}$. | [
"Solution:\n\nZapišemo vsoto členov geometrijskega zaporedja $-2+(-4)+(-8)+\\ldots+(-x)=-16382$, od koder odčitamo prvi člen, količnik in $x=a_{n}$. Uporabimo obrazec za vsoto prvih členov geometrijskega zaporedja $s_{n}=a_{1} \\cdot \\frac{k^{n}-1}{k-1}$. Vstavimo podatke in izračunamo, da je $n=13$. Uporabimo zve... | Slovenia | Državno tekmovanje | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 8192 | |
0gk4 | Let $n$ be a positive integer for which $5n + 1$ is a perfect square. Show that $n + 1$ is a sum of 5 perfect squares. | [
"Let $5n + 1 = m^2 \\equiv 1 \\pmod{5}$. Thus $m = 5k \\pm 1$ for some integer $k$. We have\n$$\nn + 1 = \\frac{(5k \\pm 1)^2 + 4}{5} = 5k^2 \\pm 2k + 1 = 4k^2 + (k \\pm 1)^2\n$$\nwhich can be written as a sum of 5 perfect squares as desired."
] | Thailand | Thai Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
07v3 | Define $f$ on the unit square $S = [0, 1] \times [0, 1] = \{(x, y) \in \mathbb{R}^2 : 0 \le x, y \le 1\}$ by
$$
f(x, y) = x\sqrt{1-y^2} + y\sqrt{1-x^2}.
$$
Prove that $f$ maps $S$ onto the unit interval $[0, 1]$. Is $f$ one-to-one on $S$? | [
"Clearly, $f$ is real valued, non-negative, and, by Cauchy-Schwarz,\n$$\nf(x, y) = x \\cdot \\sqrt{1 - y^2} + \\sqrt{1 - x^2} \\cdot y \\le \\sqrt{x^2 + (1 - x^2)} \\cdot \\sqrt{(1 - y^2) + y^2} = 1,\n$$\nso that $f$ maps $S$ into $[0, 1]$. But if $0 \\le z \\le 1$, then $(z, 0) \\in S$ and $f(z, 0) = z$, and so $f... | Ireland | IRL_ABooklet | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | f maps the unit square onto the unit interval [0, 1], and f is not one-to-one on the square. | |
0ig1 | Problem:
What is the probability that in a randomly chosen arrangement of the numbers and letters in "HMMT2005," one can read either "HMMT" or "2005" from left to right? (For example, in "5HM0M20T," one can read "HMMT.") | [
"Solution:\n$\\frac{23}{144}$\nTo read \"HMMT,\" there are $\\binom{8}{4}$ ways to place the letters, and $\\frac{4!}{2}$ ways to place the numbers. Similarly, there are $\\binom{8}{4} \\frac{4!}{2}$ arrangements where one can read \"2005.\" The number of arrangements in which one can read both is just $\\binom{8}{... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 23/144 | |
0i8r | Problem:
The sets $A$ and $B$ form a partition of positive integers if $A \cap B = \emptyset$ and $A \cup B = \mathbb{N}$. The set $S$ is called prohibited for the partition, if $k + l \neq s$ for any $k, l \in A, s \in S$ and any $k, l \in B, s \in S$.
a) Define Fibonacci numbers $f_{i}$ by letting $f_{1} = 1$, $f_{... | [
"Solution:\n\nb) We prove the following: Given a partition of the set of all powers of $2$ (i.e. two sets $Q$ and $R$ such that each $2^{k}$ is in exactly one of $Q, R$) there exists a unique partition $A, B$ of positive integers with all powers of $2$ prohibited and with $Q \\subseteq A, R \\subseteq B$. Note that... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | a) Exactly one partition. b) Infinitely many partitions; if all powers of two must lie in A, exactly one partition. | |
0c4z | Show that, if $a, b, c \in (0, \infty)$ and $a + b + c = 3$, then
$$
a^5 + b^5 + c^5 + 8(ab + bc + ca) \ge 27.
$$ | [
"Let $a, b, c \\in (0, \\infty)$ and $a + b + c = 3$.\n\nBy the Power Mean inequality, since $a, b, c > 0$ and $a + b + c = 3$:\n$$\na^5 + b^5 + c^5 \\ge 3 \\left(\\frac{a + b + c}{3}\\right)^5 = 3 \\cdot 1^5 = 3.\n$$\n\nAlso, by the Cauchy-Schwarz inequality:\n$$\nab + bc + ca \\le \\frac{(a + b + c)^2}{3} = \\fra... | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
06jz | Decide if there is a permutation $a_1, a_2, \dots, a_{6666}$ of the numbers $1, 2, \dots, 6666$ with the property that the sum $k + a_k$ is a perfect square for all $k = 1, 2, \dots, 6666$. | [
"Yes. Consider the permutation\n$3, 2, 1, 12, 11, \\dots, 4, 23, 22, \\dots, 13, 57, 56, \\dots, 24, 6666, 6665, \\dots, 58.$\nMore precisely, we define\n$$\na_k = \\begin{cases} 2^2 - k & \\text{if } 1 \\le k \\le 3, \\\\ 4^2 - k & \\text{if } 4 \\le k \\le 12, \\\\ 6^2 - k & \\text{if } 13 \\le k \\le 23, \\\\ 9^... | Hong Kong | HKG TST | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | Yes | |
0j8j | Problem:
Julia is learning how to write the letter $C$. She has $6$ differently-colored crayons, and wants to write $Cc\ Cc\ Cc\ Cc\ Cc$. In how many ways can she write the ten $C$s, in such a way that each upper case $C$ is a different color, each lower case $C$ is a different color, and in each pair the upper case $... | [
"Solution:\n\nSuppose Julia writes $Cc$ a sixth time, coloring the upper-case $C$ with the unique color different from that of the first five upper-case $C$s, and doing the same with the lower-case $C$ (note: we allow the sixth upper-case $C$ and lower-case $c$ to be the same color). Note that because the colors on... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 222480 | |
05ui | Problem:
On note $\mathbb{R}_{>0}$ l'ensemble des réels strictement positifs. Trouver toutes les fonctions $f: \mathbb{R}_{>0} \mapsto \mathbb{R}_{>0}$ telles que
$$
f(x+f(x y))+y=f(x) f(y)+1
$$
pour tous les réels strictement positifs $x$ et $y$. | [
"Solution:\n\nSoit $f$ une solution éventuelle du problème. Dans la suite, on notera $\\mathbf{E}_{x, y}$ l'égalité de l'énoncé. Soit $a$ et $b$ deux réels strictement positifs tels que $f(a)=f(b)$. Les égalités $\\mathbf{E}_{1, a}$ et $\\mathbf{E}_{1, b}$ indiquent que\n$$\na=f(1) f(a)+1-f(1+f(a))=f(1) f(b)+1-f(1+... | France | Préparation Olympique Française de Mathématiques - Test du 14 et du 21 Février 2021 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x + 1 | |
0k4m | Problem:
Let $[n]$ denote the set of integers $\{1,2, \ldots, n\}$. We randomly choose a function $f:[n] \rightarrow [n]$, out of the $n^{n}$ possible functions. We also choose an integer $a$ uniformly at random from $[n]$. Find the probability that there exist positive integers $b, c \geq 1$ such that $f^{b}(1)=a$ an... | [
"Solution:\n\nAnswer: $\\frac{1}{n}$\n\nGiven a function $f$, define $N(f)$ to be the number of numbers that are in the same cycle as $1$ (including $1$ itself), if there is one, and zero if there is no such cycle. The problem is equivalent to finding $\\mathbb{E}(N(f)) / n$. Note that\n$$\nP(N(f)=k)=\\frac{n-1}{n}... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 1/n | |
0jf8 | Problem:
Let $A_{1}, A_{2}, \ldots, A_{m}$ be finite sets of size $2012$ and let $B_{1}, B_{2}, \ldots, B_{m}$ be finite sets of size $2013$ such that $A_{i} \cap B_{j} = \emptyset$ if and only if $i = j$. Find the maximum value of $m$. | [
"Solution:\nAnswer: $\\binom{4025}{2012}$\n\nIn general, we will show that if each of the sets $A_{i}$ contain $a$ elements and if each of the sets $B_{j}$ contain $b$ elements, then the maximum value for $m$ is $\\binom{a+b}{a}$.\n\nLet $U$ denote the union of all the sets $A_{i}$ and $B_{j}$ and let $|U| = n$. Co... | United States | HMMT 2013 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | binom(4025,2012) | |
0fae | Problem:
A lottery ticket has 50 cells into which one must put a permutation of $1, 2, 3, \ldots, 50$. Any ticket with at least one cell matching the winning permutation wins a prize. How many tickets are needed to be sure of winning a prize? | [
"Solution:\nAnswer 26\n\nTake the tickets:\n$1\\ 2\\ 3\\ \\ldots\\ 25\\ 26\\ 27\\ \\ldots\\ 50$\n$2\\ 3\\ 4\\ \\ldots\\ 26\\ 1\\ 27\\ \\ldots\\ 50$\n$3\\ 4\\ 5\\ \\ldots\\ 1\\ 2\\ 27\\ \\ldots\\ 50$\n$26\\ 1\\ 2\\ \\ldots\\ 24\\ 25\\ 27\\ \\ldots\\ 50$\nEach of the numbers $1, 2, \\ldots, 26$ occurs in each of the ... | Soviet Union | 25th ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 26 | |
0gpf | Graph Air (GA) is running two way flights between some cities of a country so that it is possible to travel between any two cities using GA flights. It turned out that after adding one flight, one may travel between any two cities by using at most 17 GA flights. Determine the maximal possible number (if exists) of GA f... | [
"The answer is $34$.\nLet $A_1, A_2, \\dots, A_{35}$ be cities so that only $A_i$ and $A_{i+1}$ are connected for $i = 1, 2, \\dots, 34$. The travel between $A_1$ and $A_{35}$ uses at least $34$ flights. After adding flights between $A_1$ and $A_{35}$, it is possible to travel between any pair of cities by using at... | Turkey | Team Selection Test | [
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | 34 | |
03wf | Given integer $n \ge 2$, find the largest number $\lambda(n)$ with the following property: if a sequence of real numbers $a_0, a_1, a_2, \dots, a_n$ satisfies
$$0 = a_0 \le a_1 \le a_2 \le \dots \le a_n,$$
$$a_i \ge \frac{1}{2}(a_{i+1} + a_{i-1}), \quad i = 1, 2, \dots, n-1,$$
then
$$
\left(\sum_{i=1}^{n} ia_{i}\right... | [
"The largest possible value of $\\lambda(n)$ is $\\frac{n(n+1)^2}{4}$.\n\nLet $a_1 = a_2 = \\dots = a_n = 1$. Then we have $\\lambda(n) \\le \\frac{n(n+1)^2}{4}$.\n\nWe shall show that for any real numbers $a_0, a_1, a_2, \\dots, a_n$ satisfying the indicated property in the problem, the following inequality holds:... | China | China National Team Selection Test | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | n(n+1)^2/4 | |
0e3v | A square with the sides measuring $1$ cm is divided into three triangles. Which of the following statements is certainly true?
(A) The perimeter of one of the triangles is equal to $2 + \sqrt{2}$ cm.
(B) The area of one of the triangles is equal to $0.5 \text{ cm}^2$.
(C) Two of the triangles are right triangles.
(D) N... | [
"There are only three essentially different ways of dividing a square into three triangles:\n\n\n\n\n\n\n\nboth cuts run from a vertex to a common point on one of the sides.\nNotice that in all three cases the area of the biggest of the thre... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | B | |
0eyf | Problem:
Prove that we can find positive integers $x$, $y$ satisfying $x^{2} + x + 1 = p y$ for an infinite number of primes $p$. | [
"Solution:\n\nThis is a trivial variant on the proof that there are an infinite number of primes. Suppose that we can only find $x$, $y$ for a finite number of primes $p_1$, $p_2$, ..., $p_n$. Set $x = p_1 p_2 \\dots p_n$. Then none of the $p_i$ can divide $x(x + 1) + 1$. But it must have prime factors. Contradicti... | Soviet Union | 2nd ASU | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0j1c | Problem:
A rectangular piece of paper is folded along its diagonal (as depicted below) to form a non-convex pentagon that has an area of $\frac{7}{10}$ of the area of the original rectangle. Find the ratio of the longer side of the rectangle to the shorter side of the rectangle.
 | [
"Solution:\n\n\n\nGiven a polygon $P_{1} P_{2} \\cdots P_{k}$, let $\\left[P_{1} P_{2} \\cdots P_{k}\\right]$ denote its area. Let $A B C D$ be the rectangle. Suppose we fold $B$ across $\\overline{A C}$, and let $E$ be the intersection of $\\overline{A D}$ and $\\overline{B^{\\prime} C}$. ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | sqrt(5) | |
0dot | Let $A$, $B$, $C$ be three collinear points such that the point $B$ lies between $A$ and $C$. Let $AA'$ and $BB'$ be parallel lines such that the points $A'$ and $B'$ lie on the same side of the line $AB$, and $A'$, $B'$, $C$ are not collinear. Let $O_1$ be the center of the circle passing through the points $A$, $A'$,... | [
"If $\\angle CAA'$ is acute, then $\\angle CO_1A' = 2\\angle CAA' = 2\\angle CBB' = \\angle CO_2B'$. And if $\\angle CAA'$ is obtuse, then $\\angle CO_1A' = 2(180^\\circ - \\angle CAA') = 2(180^\\circ - \\angle CBB') = \\angle CO_2B'$. In particular, $\\angle CO_1A' = \\angle CO_2B'$. (It can be easily seen that $S... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | all values strictly between 0 and 180 degrees except 90 degrees | |
08qx | Problem:
We call an even positive integer $n$ nice if the set $\{1,2, \ldots, n\}$ can be partitioned into $\frac{n}{2}$ two-element subsets, such that the sum of the elements in each subset is a power of $3$. For example, $6$ is nice, because the set $\{1,2,3,4,5,6\}$ can be partitioned into subsets $\{1,2\},\{3,6\},... | [
"Solution:\n\nFor a nice number $n$ and a given partition of the set $\\{1,2, \\ldots, n\\}$ into two-element subsets such that the sum of the elements in each subset is a power of $3$, we say that $a, b \\in \\{1,2, \\ldots, n\\}$ are paired if both of them belong to the same subset.\n\nLet $x$ be a nice number an... | JBMO | JBMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2^{2022} - 1 | |
0arf | Problem:
Seven points on a circle are numbered $1$ to $7$ in the clockwise direction. A grasshopper jumps in the counterclockwise direction, from one point to another on the circle. If the grasshopper is on an odd-numbered point, it moves one point, and moves two points if it is on an even-numbered point. If the grass... | [
"Solution:\n\nIt will only land at the points $6, 4, 2, 7$, landing at $7$ after every fourth jump. $2011 \\equiv 3 \\pmod{4}$, so it is at $2$ by the $2011$th jump."
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2 | |
0l92 | Find all positive integer $n$ such that the equation
$$
x + y + u + v = n\sqrt{xyuv}
$$
has positive integer solution $x$, $y$, $u$, $v$. | [
"For $x$, $y$, $u$, $v \\in \\mathbb{Z}^+$, we can write the given equation in the form\n$$\n(x + y + u + v)^2 = n^2xyuv\n$$\nor in the form\n$$\nx^2 + 2(y + u + v)x + (y + u + v)^2 = n^2xyuv \\quad (1)\n$$\nLet $n$ be a positive integer such that (1) has positive integer solutions $(x, y, u, v)$. Let $(x_0, y_0, u... | Vietnam | THE 2002 VIETNAMESE MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | {1, 2, 3, 4} | |
0jg4 | Problem:
Consider triangle $ABC$ with side lengths $AB = 4$, $BC = 7$, and $AC = 8$. Let $M$ be the midpoint of segment $AB$, and let $N$ be the point on the interior of segment $AC$ that also lies on the circumcircle of triangle $MBC$. Compute $BN$. | [
"Solution:\n\nAnswer: $\\frac{\\sqrt{210}}{4}$ OR $\\frac{\\sqrt{105}}{2 \\sqrt{2}}$\n\nLet $\\angle BAC = \\theta$. Then,\n$$\n\\cos \\theta = \\frac{4^2 + 8^2 - 7^2}{2 \\cdot 4 \\cdot 8}.\n$$\nSince $AM = \\frac{4}{2} = 2$, and power of a point gives $AM \\cdot AB = AN \\cdot AC$, we have\n$$\nAN = \\frac{2 \\cdo... | United States | HMMT November | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | sqrt(210)/4 | |
02ag | Problem:
Na Feira de Ciências de uma escola, observou-se que metade dos alunos do ensino fundamental e um quarto dos alunos do ensino médio presentes nesse evento compraram um adesivo cada.
| FEIRA DE CIÊNCIAS | |
| :---: | :---: |
| Preço dos Adesivos (unidade) | |
| $R\$ 0,30$ | alunos do ensino fundamental |
| $... | [
"Solution:\n\nSejam $x$ e $y$ o número de alunos do ensino fundamental e do médio respectivamente, presentes na feira. Logo, o número daqueles que compraram um adesivo é:\n$$\n\\frac{x}{2} \\text{ do ensino fundamental e } \\quad \\frac{y}{4} \\text{ do ensino médio; }\n$$\ne os que não compraram foram\n$$\n\\frac{... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | 120 alunos do ensino fundamental e 160 alunos do ensino médio | |
0dgf | Let $a$, $b$, $c$, $d$ be real numbers such that $a^4 + b^4 + c^4 + d^4 = 16$. Prove the inequality $a^5 + b^5 + c^5 + d^5 \le 32$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
03mk | Let $n$ be a positive integer. For any positive integer $j$ and positive real number $r$, define $f_j(r)$ and $g_j(r)$ by
$$
f_j(r) = \min(jr, n) + \min\left(\frac{j}{r}, n\right), \quad \text{and} \quad g_j(r) = \min(\lceil jr \rceil, n) + \min\left(\left\lfloor \frac{j}{r} \right\rfloor, n\right),
$$
where $\lceil x ... | [
"**Solution 1:** We first prove the left hand side inequality. We begin by drawing an $n \\times n$ board, with corners at $(0,0)$, $(n,0)$, $(0,n)$ and $(n,n)$ on the Cartesian plane.\nConsider the line $\\ell$ with slope $r$ passing through $(0,0)$. For each $j \\in \\{1, \\dots, n\\}$, consider the point $(j, \\... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
05cn | Is there a positive integer $n$ such that $88$ divides $2^n + n^3$? | [
"Taking $n = 10$ gives $2^n + n^3 = 1024 + 1000 = 2024 = 88 \\cdot 23$, so $88$ divides $2^{10} + 10^3$.\n\nWe consider divisibility by $11$ and by $8$ separately. By Fermat's little theorem, we have $2^{10} \\equiv 1 \\pmod{11}$, whereas $10^3 \\equiv (-1)^3 = -1 \\pmod{11}$. In summary $2^{10} + 10^3 \\equiv 1 - ... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English | proof and answer | n = 10 | |
0ilm | Problem:
Define the sequence of positive integers $a_{n}$ recursively by $a_{1}=7$ and $a_{n}=7^{a_{n-1}}$ for all $n \geq 2$. Determine the last two digits of $a_{2007}$. | [
"Solution:\n\nAnswer: 43. Note that the last two digits of $7^{4}$ are 01. Also, $a_{2006}=7^{a_{2005}}=(-1)^{a_{2005}}=-1=3$ $(\\bmod 4)$ since $a_{2005}$ is odd. Therefore, $a_{2007}=7^{a_{2006}}=7^{3}=43\\ (\\bmod\\ 100)$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 43 | |
00h1 | Let $ABC$ be an acute triangle satisfying the condition $AB > BC$ and $AC > BC$. Denote by $O$ and $H$ the circumcenter and the orthocenter, respectively, of the triangle $ABC$. Suppose that the circumcircle of the triangle $AHC$ intersects the line $AB$ at $M$ different from $A$, and that the circumcircle of the trian... | [
"In the sequel, we denote $\\angle BAC = \\alpha$, $\\angle CBA = \\beta$, $\\angle ACB = \\gamma$. Let $O'$ be the circumcenter of the triangle $MNH$. The lengths of line segments starting from the point $H$ will be treated as signed quantities.\nLet us denote by $M', N'$ the point of intersection of $CH, BH$, res... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > An... | English | proof only | null | |
02dq | Show that $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$ is not an integer for $n > 1$. | [
"Let $2^m$ be the highest power of $2$ that does not exceed $n$. Then none of the other positive integers less than or equal to $n$ are divisible by $2^m$. Let $k = \\text{lcm}(1, 2, \\dots, n)$. Now write each of the terms $1, \\frac{1}{2}, \\dots, \\frac{1}{n}$ as fractions with denominator $k$. All will have eve... | Brazil | V OBM | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof only | null | |
03zf | Seven students are arranged to attend five sporting events. It is required that students $A$ and $B$ cannot attend the same event, every event is attended by at least one student, and each student must attend one and only one event. Then the number of the arrangement plans meeting the required condition is _______. (th... | [
"There are two possible cases that satisfy the required conditions:\n\n(1) there is an event attended by three students — this case has $C_7^3 \\cdot 5! - C_5^1 \\cdot 5! = 3600$ plans;\n\n(2) there are two events each attended by two students — this case has $\\frac{1}{2}(C_7^2 \\cdot C_5^2) \\cdot 5! - C_5^2 \\cd... | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | final answer only | 15000 | |
08da | Problem:
Quante sono le coppie ordinate $(x, y)$ di interi positivi minori o uguali a $2019$ tali che $x+y$ e $x y+1$ siano potenze di $2$? | [
"Solution:\n\nLa risposta è $37$. Supponiamo $x = y$. Allora otteniamo $x^2 + 1$ pari, dunque $x$ dispari. Ma $2x$ con $x$ dispari è potenza di due se e solo se $x = 1$.\n\nOra ci basta contare tutte le coppie $(x, y)$ che verificano le ipotesi con $x < y$. Ciò significa che esistono due interi positivi $a$ e $b$ t... | Italy | Progetto Olimpiadi della Matematica | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 37 | |
0ji0 | Problem:
Evaluate the expression
where the digit $2$ appears $2013$ times. | [
"Solution:\n\n$\\boxed{\\dfrac{2013}{2014}}$\n\nLet $f(n)$ denote the corresponding expression with the digit $2$ appearing exactly $n$ times. Then $f(1) = \\dfrac{1}{2}$ and for $n > 1$, $f(n) = \\dfrac{1}{2 - f(n-1)}$.\n\nBy induction using the identity\n$$\n\\frac{1}{2 - \\frac{N-1}{N}} = \\frac{N}{N+1},\n$$\nwe... | United States | HMMT November 2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 2013/2014 | |
0flq | Problem:
Determinar razonadamente si el número $\lambda_{n}=\sqrt{3 n^{2}+2 n+2}$ es irracional para todo entero no negativo $n$. | [
"Solution:\n\nSupongamos que $n$ es par. Entonces, $3 n^{2}+2 n$ es múltiplo de $4$ y $3 n^{2}+2 n+2$ es múltiplo de $2$ pero no de $4$, con lo que no puede ser un cuadrado perfecto.\n\nSupongamos que $n$ es impar. Cualquier cuadrado perfecto impar da resto $1$ al dividir entre $8$; este resultado se demuestra triv... | Spain | 48 aME | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | It is irrational for every nonnegative integer n. | |
0ak1 | Let be given the $\triangle ABC$. On the arc $\overarc{BC}$ of the circumcircle of $\triangle ABC$, which does not contain the point $A$, points $X$ and $Y$ are chosen, such that $\angle BAX = \angle CAY$. Let $M$ be the middle point of the chord $AX$. Prove that $\overline{BM} + \overline{CM} > \overline{AY}$.
![](at... | [
"Let $O$ be the circumcenter of the circumcircle of $\\triangle ABC$. Then $OM \\perp AX$. We draw a normal line from the point $B$ at $OM$ and let it intersect the circumcircle in the point $Z$. Since $BZ \\perp OM$ we have that $OM$ is a line of symmetry of $BZ$. According to this, $\\overline{MZ} = \\overline{MB... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Misc... | null | proof only | null | |
01fe | For a positive integer $n$, consider all nonincreasing functions $f: \{1, \dots, n\} \to \{1, \dots, n\}$. Some of them have a fixed-point ($c = f(c)$), some don't have. Which are more numerous? Evaluate the difference between the sizes of the two sets of functions. | [
"Lemma. Let $A$ be a set of $a$ consecutive integers and let $B$ be a set of $b$ consecutive integers.\nThere are exactly $\\binom{a+b-1}{a}$ nonincreasing functions $f: A \\to B$.\nProof. Let (wlog) $A = \\{1, \\dots, a\\}$, $B = \\{1, \\dots, b\\}$. A function $f: A \\to B$ is nonincreasing if and only if the fun... | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | Functions with a fixed point are more numerous; the difference equals binom(2n−2, n−1) − binom(2n−2, n) = (1/n)·binom(2n−2, n−1), i.e., the Catalan number C_{n−1}. | |
07yw | Problem:
Sia dato un quadrato $ABCD$ di lato unitario e siano $M, N$ due punti rispettivamente sui lati $AB$ e $AD$ tali che $AM = AN$. Quanto può valere, al massimo, l'area del quadrilatero $CDNM$?
(A) $\frac{1}{2}$
(B) $\frac{9}{16}$
(C) $\frac{19}{32}$
(D) $\frac{5}{8}$
(E) $\frac{2}{3}$ | [] | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | MCQ | D | |
0gmf | Two circles are tangent to each other from outside at a point $A$ and to a third circle $\Gamma$ from inside at points $B$ and $C$. Let $D$ be the midpoint of the secant of $\Gamma$ which is tangent to the smaller circles at $A$. Show that $A$ is the incenter of the triangle $BCD$ if the centers of the circles are not ... | [] | Turkey | X. NATIONAL 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 > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarit... | English | proof only | null | |
0ec6 | A square $ABCD$ and two points $E$ and $F$ outside of this square are given so that the triangles $BEC$ and $CFD$ are equilateral. Prove that the triangle $AEF$ is also equilateral. | [
"Since $ABCD$ is a square and the triangles $BEC$ and $CFD$ are equilateral we have $|EB| = |BA| = |AD| = |DF|$. Therefore the triangles $EBA$ and $ADF$ are isosceles with apexes at vertexes $B$ and $D$, and they have legs of the same length. We also have $\\angle EBA = \\angle ADF = 90^\\circ + 60^\\circ = 150^\\c... | Slovenia | National Math Olympiad 2015 – Final Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0822 | Problem:
È noto che i Marziani maschi dicono sempre la verità, mentre le Marziane mentono sempre; al contrario i Venusiani maschi mentono e le Venusiane dicono sempre il vero. Atterra un'astronave piena di Marziani e Venusiani; all'ufficio immigrazione due degli occupanti, Ark e Bark, fanno le seguenti dichiarazioni:
... | [
"Solution:\n\nLa risposta è (E). Se Ark dice il vero, allora Bark è un venusiano maschio, quindi mente, ma allora Ark deve essere un venusiano maschio, assurdo. Dunque Ark mente, quindi Bark è una femmina marziana, che mente, e dunque Ark è effettivamente un venusiano maschio (che mente). Quindi tutto è determinato... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Logic"
] | null | MCQ | E | |
00nn | Let $a$ and $b$ be positive integers and $c$ be a positive real number satisfying
$$
\frac{a+1}{b+c} = \frac{b}{a}.
$$
Prove that $c \ge 1$ holds. | [
"$$\n\\begin{aligned}\na^2 + a &= b^2 + bc \\\\\n4a^2 + 4a + 1 &= 4b^2 + 4bc + 1 \\\\\n(2a + 1)^2 &= 4b^2 + 4bc + 1.\n\\end{aligned}\n$$\nAssume to the contrary that $c < 1$ holds. This yields\n$$\n(2b)^2 = 4b^2 < (2a + 1)^2 = 4b^2 + 4bc + 1 < 4b^2 + 4b + 1 = (2b + 1)^2.\n$$\nThis is a contradiction as the square o... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
00cg | Sea $n \ge 1$ un entero. Se tienen dos sucesiones, cada una de $n$ números reales positivos $a_1, a_2, ..., a_n$ y $b_1, b_2, ..., b_n$ tales que $a_1 + a_2 + ... + a_n = 1$ y $b_1 + b_2 + ... + b_n = 1$. Hallar el menor valor posible que puede tomar la suma
$$
\frac{a_1^2}{a_1+b_1} + \frac{a_2^2}{a_2+b_2} + \dots + \f... | [] | Argentina | Nacional OMA 2019 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | Spanish | proof and answer | 1/2 | |
0bs1 | A set $S = \{s_1, \dots, s_k\}$ of positive real numbers is *polygonal* if $k \ge 3$ and there is a non-degenerate planar $k$-gon whose side lengths are exactly $s_1, \dots, s_k$; the set $S$ is *multipolygonal* if in every partition of $S$ into two subsets, each of which has at least three elements, exactly one of the... | [
"Recall that a necessary and sufficient condition for $k \\ge 3$ positive real numbers $s_1, \\dots, s_k$ to be the side lengths of a non-degenerate planar $k$-gon is that a maximal $s_i$ be less than the sum of the other $s_j$.\n\na) The answer is in the affirmative. Given pairwise distinct positive real numbers $... | Romania | 67th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Yes | |
0j92 | Find all integers $n \ge 3$ such that among any $n$ positive real numbers $a_1, a_2, \ldots, a_n$ with
$$ \max(a_1, a_2, \ldots, a_n) \le n \cdot \min(a_1, a_2, \ldots, a_n), $$
there exist three that are the side lengths of an acute triangle. | [
"**Solution.** The answer is $n \\ge 13$. First, we show that any $n \\ge 13$ satisfies the desired condition. Suppose for the sake of contradiction that $a_1 \\le a_2 \\le \\dots \\le a_n$ are integers such that $\\max(a_1, a_2, \\dots, a_n) \\le n \\cdot \\min(a_1, a_2, \\dots, a_n)$ and no three are the side len... | United States | USAMO | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | n ≥ 13 | |
0a5q | Problem:
The sequence $x_{1}, x_{2}, x_{3}, \ldots$ is defined by $x_{1} = 2022$ and $x_{n + 1} = 7x_{n} + 5$ for all positive integers $n$. Determine the maximum positive integer $m$ such that
$$
\frac{x_{n}(x_{n}-1)(x_{n}-2)\ldots(x_{n}-m+1)}{m!}
$$
is never a multiple of $7$ for any positive integer $n$. | [
"Solution:\nWe claim the answer is $404$. First, we notice that $m \\leq 2022$. Otherwise,\n$$\n\\frac{x_{1}(x_{1}-1)\\cdots(x_{1}-m+1)}{m!} = \\frac{2022(2022-1)\\cdots(2022-m+1)}{m!} = 0,\n$$\nwhich is a multiple of $7$. Then, since $x_{n} \\geq x_{1} = 2022$ for all $n$, we can write\n$$\n\\frac{x_{n}(x_{n}-1)\\... | New Zealand | New Zealand Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization... | null | proof and answer | 404 | |
0f6z | Problem:
$A_1A_2\ldots A_n$ is a regular $n$-gon and $P$ is an arbitrary point in the plane. Show that if $n$ is even we can choose signs so that the vector sum $\pm PA_1 \pm PA_2 \pm \ldots \pm PA_n = 0$, but if $n$ is odd, then this is only possible for finitely many points $P$. | [] | Soviet Union | 20th ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof only | null | |
0ifi | Problem:
Find the sum
$$
\frac{2^{1}}{4^{1}-1}+\frac{2^{2}}{4^{2}-1}+\frac{2^{4}}{4^{4}-1}+\frac{2^{8}}{4^{8}-1}+\cdots .
$$ | [
"Solution:\nNotice that\n$$\n\\frac{2^{2^{k}}}{4^{2^{k}}-1}=\\frac{2^{2^{k}}+1}{4^{2^{k}}-1}-\\frac{1}{4^{2^{k}}-1}=\\frac{1}{2^{2^{k}}-1}-\\frac{1}{4^{2^{k}}-1}=\\frac{1}{4^{2^{k-1}}-1}-\\frac{1}{4^{2^{k}}-1} .\n$$\nTherefore, the sum telescopes as\n$$\n\\left(\\frac{1}{4^{2^{-1}}-1}-\\frac{1}{4^{2^{0}}-1}\\right)... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1 | |
0j7g | Consider the assertion that for each positive integer $n \ge 2$, the remainder upon dividing $2^{2n}$ by $2^n - 1$ is a power of 4. Either prove the assertion or find (with proof) a counterexample. | [
"The assertion is false, and the smallest $n$ for which it fails is $n = 25$. Given $n \\ge 2$, let $r$ be the remainder when $2^n$ is divided by $n$. Then $2^n = kn + r$ where $k$ is a positive integer and $0 \\le r < n$. It follows that\n$$\n2^{2n} = 2^{kn+r} \\equiv 2^r \\pmod{2^n - 1},\n$$\nwhere $2^r < 2^{2n} ... | United States | USAMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
0fkn | Problem:
Los puntos de una retícula $m \times n$ pueden ser de color blanco o negro. Una retícula se dice que está equilibrada si para cualquier punto $P$ de ella, la fila y columna que pasan por este punto $P$ tienen ambas el mismo número de puntos de igual color que $P$. Determinar todos los pares de enteros positiv... | [
"Solution:\n\nDenotaremos por $BF(i)$ el número de puntos de color blanco que hay en la fila $i$ y con $BC(j)$ el número de puntos blancos en la columna $j$. Análogamente, $NF(i)$ y $NC(j)$ denotarán el número de puntos negros en la fila $i$ y en la columna $j$, respectivamente. Siendo $P_{ij}$ el punto que se encu... | Spain | XLV Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | All pairs (m, n) of positive integers with m = n, m = 2n, or n = 2m; equivalently {(t, t), (t, 2t), (2t, t)} for positive t. | |
0fx0 | Problem:
Ein quadratisches Spielbrett besteht aus $2 n \times 2 n$ Feldern. Es sollen $n$ dieser Felder markiert werden, sodass keine zwei markierten Felder in derselben oder benachbarten Zeilen liegen, und sodass auch keine zwei markierten Felder in derselben oder benachbarten Spalten liegen. Auf wieviele Arten ist d... | [
"Solution:\n\nWir nennen eine Zeile oder Spalte markiert, wenn sie ein markiertes Feld enthält. Von den $2 n$ Zeilen sind genau $n$ markiert und keine zwei davon sind benachbart. Nummeriert man die Zeilen von $1$ bis $2 n$, dann sieht man leicht, dass die markierten Zeilen die Nummern\n$$\n1, 3, \\ldots, 2k+1, 2k+4... | Switzerland | Vorrundenprüfung | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | (n+1)^2 n! | |
0281 | Problem:
A figure shows a grid formed by squares of $1~\mathrm{cm}$ side. What is the ratio between the shaded area and the non-shaded area?
(a) $\frac{1}{4}$
(b) $\frac{1}{5}$
(c) $\frac{1}{6}$
(d) $\frac{2}{5}$
(e) $\frac{2}{7}$
 | [
"Solution:\nThe correct option is (a).\nThe grid is a square with side equal to $5~\\mathrm{cm}$, so its area is $25~\\mathrm{cm}^2$. The shaded part of the grid is formed by four triangles, two of which have base $1~\\mathrm{cm}$ and height $2~\\mathrm{cm}$, and the other two have base $1~\\mathrm{cm}$ and height ... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | (a) | |
015s | Suppose that every point of the plane has been coloured white or black. We say that a line $\ell$ is an antisymmetry-axis of that colouring if every two different points symmetric with respect to $\ell$ have different colours. Decide if there exists a colouring such that for every line $k$ there is an antisymmetry-axis... | [
"We will show that such a colouring does not exist.\nSuppose on the contrary that we can colour the plane so that for every line $l$ there is an antisymmetry-axis parallel to $l$. Consider antisymmetry-axes $l_1, l_2, l_3, l_4$ parallel to the lines $y = x, y = -x, x = 0, y = 0$ respectively. Without loss of genera... | Baltic Way | Baltic Way SHL | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No, such a coloring does not exist. | |
0aj8 | Determine whether there exists an infinite sequence $a_1, a_2, a_3, \dots$ of positive integers which satisfies the equality
$$
a_{n+2} = a_{n+1} + \sqrt{a_{n+1} + a_n}
$$
for every positive integer $n$. | [
"The answer is no.\nSuppose that there exist a sequence $(a_n)$ of positive integers satisfying the given condition. We will show that this will lead to a contradiction.\nFor each $n \\ge 2$ define $b_n = a_{n+1} - a_n$. Then, by assumption, for $n \\ge 2$ we get $b_n = \\sqrt{a_n + a_{n-1}}$ so that we have\n$$\nb... | North Macedonia | Girls European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
01gp | Let $n$ be a positive integer. A restaurant offers a choice of $n$ starters, $n$ main dishes, $n$ desserts and $n$ wines. A merry company dines at the restaurant, with each guest choosing a starter, main dish, dessert and wine. No two people place exactly the same order. It so happens that there is no collection of $n$... | [
"Answer: The maximal number of guests is $n^4 - n^3$.\nThe possible menus are represented by quadruples\n$$\n(a, b, c, d), \\quad 1 \\le a, b, c, d \\le n.\n$$\nLet us count those menus satisfying\n$$\na + b + c + d \\not\\equiv 0 \\pmod{n}.\n$$\nThe numbers $a, b, c$ may be chosen arbitrarily ($n$ choices for each... | Baltic Way | Baltic Way 2020 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n^4 - n^3 | |
04fq | Do there exist real numbers $x, y, z$ such that
$$
\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = -5 \quad \text{and} \quad \frac{y^2}{x^2} + \frac{z^2}{y^2} + \frac{x^2}{z^2} = 8? \quad (\text{Ilko Brneti\'c})
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
08l4 | Problem:
Find all triples $(x, y, z)$ of real positive numbers, which satisfy the system
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{4}{y}+\frac{9}{z}=3 \\
x+y+z \leq 12
\end{array}\right.
$$ | [
"Solution:\nIf we multiply the given equation and inequality $(x>0, y>0, z>0)$, we have\n$$\n\\left(\\frac{4x}{y}+\\frac{y}{x}\\right)+\\left(\\frac{z}{x}+\\frac{9x}{z}\\right)+\\left(\\frac{4z}{y}+\\frac{9y}{z}\\right) \\leq 22\n$$\nFrom AM-GM we have\n$$\n\\frac{4x}{y}+\\frac{y}{x} \\geq 4, \\quad \\frac{z}{x}+\\... | JBMO | 2008 Shortlist JBMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | (2, 4, 6) | |
0h96 | Let $ABC$ be an acute angled triangle, $AA_1$ and $CC_1$ be its bisector, $I$ be the incenter of $ABC$, $M$ and $N$ be the midpoints of $AI$ and $CI$, respectively. Inside the triangles $AC_1I$ and $A_1CI$ we choose points $K$ and $L$, such that
$\angle AKI = \angle CLI = \angle AIC$, $\angle AKM = \angle ICA$, $\angl... | [
"(Anton Trigub)\n\n**Fig. 46**\nSuppose that lines $AI$ and $CI$ meet the circumcircle of $\\triangle ABC$ (second time) in the points $W_A$ and $W_C$, respectively (Fig. 46). Since $\\angle AKI = \\angle AIC$, then circumcircle of the triangle $AKI$ tangent to the line $W_C I$. Consider th... | Ukraine | 58th Ukrainian National 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 > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellane... | English | proof only | null | |
0fhw | Problem:
Con 21 fichas de damas, unas blancas y otras negras, se forma un rectángulo $3 \times 7$. Demostrar que siempre hay cuatro fichas del mismo color situadas en los vértices de un rectángulo. | [
"Solution:\n\nDispondremos el tablero en posición vertical, es decir, con 7 filas y 3 columnas. Asignaremos el color blanco a la cifra 0 y el negro a la cifra 1. De este modo cada fila representa un número escrito en base 2.\n\nEn primer lugar es fácil ver que si en una fila se colocan todas las fichas del mismo co... | Spain | OME 30 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0ab2 | The ratios $\frac{3a5b}{36}$ and $\frac{4c7d}{45}$ are positive integers, where $a, b, c, d$ are digits. Order all numbers of this kind by size. | [
"The ratios $\\frac{3a5b}{36}$ and $\\frac{4c7d}{45}$ are positive integers if $3a5b$ is divisible by $36$ (hence by $4$ and $9$) and $4c7d$ is divisible by $45$ (hence by $5$ and $9$).\n\n$3a5b$ is divisible by $36$ if the last digit is $2$ or $6$ and because $3a5b$ is divisible by $9$ the only two possible cases ... | North Macedonia | Macedonian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 4275/45 < 3456/36 < 4770/45 < 3852/36 | |
0jbd | Let $P$ be a point in the plane of $ABC$, and $\gamma$ a line passing through $P$. Let $A'$, $B'$, $C'$ be the points where the reflections of lines $PA$, $PB$, $PC$ with respect to $\gamma$ intersect lines $BC$, $AC$, $AB$, respectively. Prove that $A'$, $B'$, $C'$ are collinear.
 | [
"There are several possible configurations depending on the location of $P$ and the orientation of $\\gamma$. We will consider the configuration above but will use directed lengths and angles so our arguments apply to all diagram configurations. By the law of sines on triangles $A_1PB$ and $A_1CP$, we have\n$$\nBP ... | United States | USAMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gho | 666 fields line up in a row, with each field being a wheat field or a rice field. Rose the hen lays an egg on each of the $W$ wheat fields and $R$ rice fields, so that for each egg, there is at most one field without egg between it and the closest egg to its right. Find the largest positive integer $S$ so that, regardl... | [
"首先證明牠總是可以取得 167 分。不失一般性假設有至少 $666/2 = 333$ 個麥田。\n考慮以下操作：讓母雞從排頭的田開始。若牠在麥田，則立刻下蛋然後前進一塊田；\n否則，她先前進一塊田，下蛋，再前進一塊田。注意到此策略保證所有的麥田都有蛋，故 $W \\ge 333$。此外，對於每一個有蛋的稻田，都保證前面有一個沒有蛋的稻田，因此 $R \\le [333/2] = 166$，故 $W - R \\ge 333 - 166 = 167$。\n\n現在證明 $S = 167$ 是最大值。考慮如下排列 (其中 R 為稻田, W 為麥田):\n$$\n\\{R\\}, \\{W, W\\}, \\{R, R\\}, ... | Taiwan | 2023 數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Chinese (Traditional) | proof and answer | 167 | |
00ow | A grasshopper jumps on the plane from an integer point (point with all integer coordinates) to another integer point according to the following rules: His first jump is of length $\sqrt{98}$, his second jump is of length $\sqrt{149}$, his next jump is of length $\sqrt{98}$, and so on, alternatively. What is the least p... | [
"Since the only representations of $98$ and $149$ as sums of two squares are $7^2 + 7^2$ and $7^2 + 10^2$, we conclude that every odd move of the grasshopper is of the form $(x, y) \\rightarrow (x \\pm 7, y \\pm 7)$ and every even one - of the form $(x, y) \\rightarrow (x \\pm 7, y \\pm 10)$ or $(x \\pm 10, y \\pm ... | Balkan Mathematical Olympiad | BMO 2010 Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | English | proof and answer | 29 | |
08b2 | Problem:
Un servizio di streaming musicale propone canzoni classificate in 10 generi musicali, in modo che ogni brano appartenga ad uno e un solo genere. Le canzoni vengono suonate una dopo l'altra: le prime 17 sono scelte dall'utente, ma a partire dalla diciottesima il servizio determina automaticamente quale canzone... | [] | Italy | XXXI Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
086c | Problem:
Il So-poko è un nuovo gioco enigmistico che si gioca su una tabella quadrata di lato $203$ caselle. Le caselle sono colorate di bianco e di nero a cornici concentriche alternate; la cornice più esterna è nera, mentre la casella centrale è bianca (vedi a fianco un esempio $7 \times 7$). Qual è la differenza tr... | [
"Solution:\n\nLa risposta è $(\\mathbf{E})$. Consideriamo come aumenta la differenza tra caselle nere e bianche $d_{k}$ in un so-poko di lato $k$. Per $k=3$, la differenza è $7$. Ogni volta che vengono aggiunte all'esterno due cornici concentriche (una bianca più interna e una nera più esterna), $d_{k}$ aumenta di ... | Italy | Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | MCQ | E | |
05ea | Problem:
For a positive integer $N$, let $c_{1} < c_{2} < \dots < c_{m}$ be all the positive integers smaller than $N$ that are coprime to $N$. Find all $N \geq 3$ such that
$$
gcd (N, c_{i} + c_{i + 1}) \neq 1$$
for all $1 \leq i \leq m - 1$.
Here $\gcd(a, b)$ is the largest positive integer that divides both $a$ and ... | [
"Solution:\nThe answer is all even integers and all powers of $3$. First we show that these work.\n\n- When $N$ is even, all $c_{i}$ are odd, and so $2 \\mid \\gcd (N, c_{i} + c_{i + 1})$ for every $i$.\n\n- When $N$ is a power of $3$, the $c_{i}$ are exactly the numbers in the range $1, 2, \\ldots, N - 1$ that are... | European Girls' Mathematical Olympiad (EGMO) | EGMO | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All even integers and all powers of 3. | |
00pc | Given a triangle $ABC$, the line parallel to the side $BC$ and tangent to the incircle of the triangle meets the sides $AB$ and $AC$ at the points $A_1$ and $A_2$; the points $B_1$, $B_2$ and $C_1$, $C_2$ are defined similarly. Show that
$$
AA_1 \cdot AA_2 + BB_1 \cdot BB_2 + CC_1 \cdot CC_2 \geq \frac{1}{9} (AB^2 + BC... | [
"Let $D$, $E$, $F$ be the points where the incircle touches the sides $BC$, $CA$, $AB$, respectively, and let $x = AE = AF$, $y = BF = BD$, $z = CD = CE$.\nExpress all the lengths involved in the required inequality in terms of $x$, $y$ and $z$. Clearly, $AB = x+y$, $BC = y+z$, and $CA = z+x$. To express $AA_1$ and... | Balkan Mathematical Olympiad | shortlistBMO 2011 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | Equality holds if and only if the triangle is equilateral. | |
087k | Problem:
In quanti modi diversi si possono mettere in fila i numeri $\{21,31,41,51,61,71,81\}$ in modo che, comunque se ne scelgano quattro in posti consecutivi, la loro somma sia divisibile per tre? | [
"Solution:\nLa risposta è 144.\nPer brevità indicheremo con buono un modo di ordinare i numeri $a_{1}, \\ldots, a_{7}$ assegnati che soddisfi le caratteristiche richieste. Cerchiamo di stabilire alcune proprietà degli ordinamenti buoni.\n\ni) perché un ordinamento sia buono non è importante quali siano i numeri sce... | Italy | UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof and answer | 144 | |
0g9f | 令 $\mathbb{R}$ 表示實數所成的集合。給定實數 $t \neq -1$。找出所有的函數 $f: \mathbb{R} \to \mathbb{R}$ 使得
$$
(t + 1)f(1 + xy) - f(x + y) = f(x + 1)f(y + 1) \text{ 成立。}
$$ | [
"首先，易知 $f(x) \\equiv 0$ 為方程的一組解，故假設 $f(x)$ 不總是 $0$。原式代入 $(x - 1, 1)$ 得到 $f(2) = t$。\n\n設 $f(1) = a$。代入 $(x, 0)$ 得到\n$$\n(t + 1)a - f(x) = af(x + 1) \\quad (1)\n$$\n若 $a = 0$，則由 (1) 知 $f(x) \\equiv 0$，不合。\n所以\n$$\na \\neq 0, \\quad f(x + 1) = t + 1 - \\frac{f(x)}{a} \\quad (2)\n$$\n(1) 中 $x$ 代 $x - 1$ 整理得\n$$\nf(x -... | Taiwan | 2015 Math Olympiad Second Stage Training Camp | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) ≡ 0; f(x) ≡ t; f(x) = (t+1)x − (t+2), for real t ≠ −1 | |
0ini | Prove that for every nonnegative integer $n$, the number $7^{7^n} + 1$ is the product of at least $2n+3$ (not necessarily distinct) primes. | [
"The proof is by induction. The base is provided by the $n=0$ case, where $7^{7^0} + 1 = 7^1 + 1 = 2^3$.\n\nTo prove the inductive step, it suffices to show that if $x = 7^{2^{m-1}}$ for some positive integer $m$ then $(x^7+1)/(x+1)$ is composite. As a consequence, $x^7+1$ has at least two more prime factors than d... | United States | USAMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0gbs | 設 $ABCC_1B_1A_1$ 為凸六邊形, 其中 $AB = BC$, 並且線段 $\overline{AA_1}, \overline{BB_1}$ 及 $\overline{CC_1}$ 的中垂線皆相同。令對角線 $AC_1$ 與 $A_1C$ 交於點 $D$, 並將三角形 $ABC$ 的外接圓記為 $\omega$。設 $\omega$ 與三角形 $A_1BC_1$ 的外接圓再交於一點 $E \neq B$。
試證：直線 $BB_1$ 與 $DE$ 的交點落在 $\omega$ 上。 | [
"If $AA_1 = CC_1$, then the hexagon is symmetric about the line $BB_1$; in particular the circles $ABC$ and $A_1BC_1$ are tangent to each other. So $AA_1$ and $CC_1$ must be different. Since the points $A$ and $A_1$ can be interchanged with $C$ and $C_1$, respectively, we may assume $AA_1 < CC_1$.\nLet $R$ be the r... | Taiwan | 二〇一八數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dn8 | Problem:
Нека су $M$, $N$ и $P$ средишта страница $BC$, $AC$ и $AB$, редом, а $O$ центар описане кружнице оштроуглог троугла $ABC$. Кружнице описане око троуглова $BOC$ и $MNP$ секу се у различитим тачкама $X$ и $Y$ унутар троугла $ABC$. Доказати да је
$$
\varangle BAX = \varangle CAY
$$ | [
"Solution:\n\nОбележимо са $k_{1}$ и $k_{2}$ редом кругове $MNP$ и $BOC$. Круг $k_{1}$ је Ојлеров круг у $\\triangle ABC$ и пролази кроз подножја висина $D$, $E$ из $B$, $C$ и средиште $O_{1}$ дужи $AH$, где је $H$ ортоцентар $\\triangle ABC$.\n\nПокажимо да друга пресечна тачка $Z$ праве $AY$ и круга $k_{1}$ лежи ... | Serbia | Serbian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals... | null | proof only | null | |
06vq | Find all pairs $(m, n)$ of positive integers satisfying the equation
$$
\left(2^{n}-1\right)\left(2^{n}-2\right)\left(2^{n}-4\right) \cdots\left(2^{n}-2^{n-1}\right)=m!
$$ | [
"We will get an upper bound on $n$ from the speed at which $v_{2}\\left(L_{n}\\right)$ grows.\nFrom\n$$\nL_{n}=\\left(2^{n}-1\\right)\\left(2^{n}-2\\right) \\cdots\\left(2^{n}-2^{n-1}\\right)=2^{1+2+\\cdots+(n-1)}\\left(2^{n}-1\\right)\\left(2^{n-1}-1\\right) \\cdots\\left(2^{1}-1\\right)\n$$\nwe read\n$$\nv_{2}\\l... | IMO | IMO 2019 Shortlisted Problems | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | (1,1), (3,2) | |
04s6 | Let $a$, $b$ be non-negative real numbers. Prove the inequality
$$
\frac{a}{\sqrt{b^2+1}} + \frac{b}{\sqrt{a^2+1}} \ge \frac{a+b}{\sqrt{ab+1}}
$$
and find when the equality holds. | [
"It is evident that the inequality under consideration becomes an equality when $a=0$, $b=0$ or $a=b$. To prove that otherwise the strong inequality holds, it suffices to deal with the case $a > b > 0$ and (after removing the fractions) to show that\n$$\na\\sqrt{a^2+1} + b\\sqrt{b^2+1} > (a+b)\\sqrt{a^2+1}\n$$\nDis... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | Equality holds if and only if one of the numbers is zero or the two numbers are equal. | |
004l | Sean $ABC$ un triángulo con incentro $I$ y $\Gamma$ una circunferencia de centro $I$, de radio mayor al de la circunferencia inscrita y que no pasa por ninguno de los vértices. Sean $X_1$ el punto de intersección de $\Gamma$ con la recta $AB$ más cercano a $B$; $X_2$, $X_3$ los puntos de intersección de $\Gamma$ con la... | [] | Argentina | XXII Olimpiada Iberoamericana de Matemática | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane G... | Spanish | proof only | null | |
03h4 | Problem:
What is the maximum number of terms in a geometric progression with common ratio greater than $1$ whose entries all come from the set of integers between $100$ and $1000$ inclusive? | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 6 | |
08o9 | Problem:
Let $x$, $y$, $z$ be non-negative real numbers satisfying $x + y + z = x y z$. Prove that
$$
2\left(x^{2} + y^{2} + z^{2}\right) \geq 3(x + y + z)
$$
and determine when equality occurs. | [
"Solution:\nEquality holds when $x = y = z = 0$.\nApply AM-GM to $x + y + z = x y z$,\n$$\n\\begin{aligned}\n& x y z = x + y + z \\geq 3 \\sqrt[3]{x y z} \\Rightarrow (x y z)^{3} \\geq (3 \\sqrt[3]{x y z})^{3} \\\\\n& \\Rightarrow x^{3} y^{3} z^{3} \\geq 27 x y z \\\\\n& \\Rightarrow x^{2} y^{2} z^{2} \\geq 27 \\\\... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | Equality occurs only at x = y = z = 0. | |
0emi | The cells of a $10 \times 10$ square are susceptible to infection. In a unit of time, the cells with 2 or more infected neighbours (having a common side) become infected. Is it possible to start the infection in
a. nine cells,
b. ten cells,
in such a way that the infection spreads to the whole square? | [
"It is easy to observe that the infection can never spread outside of a rectangle that bounds it. A less obvious but more useful observation is that the total perimeter of the infected area will never increase.\n\nLet infected squares be black and non-infected squares be white. It is clear that the spread can be co... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) No. b) Yes, for example by infecting all ten cells on the main diagonal. | |
01vc | A checkered polygon $A$ is drawn on the checkered plane. We call a cell of $A$ *internal* if all 8 of its adjacent cells belong to $A$. All other (non-internal) cells of $A$ we call *boundary*. It is known that 1) each boundary cell has exactly two common sides with other boundary cells; and 2) the union of all boundar... | [
"It is clear that the set of boundary cells of the polygon $A$ can be divided into trapezoids so that you can go from any of them to another, going only through the sides of the trapezoids. We call a polygon satisfying this condition *good*. We will prove by induction on $k \\ge 2$ the following statement: if the a... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem"
] | English | proof only | null | |
020p | Problem:
a) Let $a, b, c, d$ be real numbers with $0 \leqslant a, b, c, d \leqslant 1$. Prove that
$$
a b(a-b)+b c(b-c)+c d(c-d)+d a(d-a) \leqslant \frac{8}{27}
$$
b) Find all quadruples $(a, b, c, d)$ of real numbers with $0 \leqslant a, b, c, d \leqslant 1$ for which equality holds in the above inequality. | [
"Solution:\n\nDenote the left-hand side by $S$. We have\n$$\n\\begin{aligned}\nS & =a b(a-b)+b c(b-c)+c d(c-d)+d a(d-a) \\\\\n& =a^{2} b-a b^{2}+b^{2} c-b c^{2}+c^{2} d-c d^{2}+d^{2} a-d a^{2} \\\\\n& =a^{2}(b-d)+b^{2}(c-a)+c^{2}(d-b)+d^{2}(a-c) \\\\\n& =(b-d)\\left(a^{2}-c^{2}\\right)+(c-a)\\left(b^{2}-d^{2}\\righ... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Maximum value is 8/27. Equality holds for the four cyclic permutations of (1, 2/3, 1/3, 0), namely (1, 2/3, 1/3, 0), (2/3, 1/3, 0, 1), (1/3, 0, 1, 2/3), and (0, 1, 2/3, 1/3). | |
018n | In a group of $n$ people everyone has at least $k$ friends. Each day every member of the group shares with all his friends all the news that he received in the previous days. Suppose that if some information is revealed to any member of the group, then after some number of days all the members will eventually know the ... | [
"By a path between members $A$ and $B$ we mean a sequence $A = A_0, A_1, A_2, \\dots, A_d = B$, where $A_i, A_{i+1}$ are friends for $i = 0, 1, 2, \\dots, d-1$. The smallest possible number $d$ in such a sequence will be called the distance between $A$ and $B$. By assumption, for any two members there exists a path... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0f9v | Problem:
An equilateral triangle of side $n$ is divided into $n^2$ equilateral triangles of side $1$. A path is drawn along the sides of the triangles which passes through each vertex just once. Prove that the path makes an acute angle at at least $n$ vertices. | [
"Solution:\n\n\n\nThe diagram has $1 + 2 + \\ldots + n = n(n + 1) / 2$ upright triangles and $1 + 2 + \\ldots + n - 1 = n(n - 1) / 2$ upside down triangles. It has $1 + 2 + \\ldots + n + 1 = (n + 1)(n + 2) / 2$ vertices. So the path must be $(n + 1)(n + 2) / 2 - 1 = (n^2 + 3n) / 2$ units lo... | Soviet Union | 24th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
03a3 | Find all values of the real parameter $a$ such that the solutions of the system
$$
\begin{cases} \frac{3x-5}{3} + \frac{3x+5}{4} \ge \frac{x}{7} - \frac{1}{15} \\ (2x-a)^3 + (2x+a)(1-4x^2) + 16x^2a - 6xa^2 + a^3 \le 2a^2 + a \end{cases}
$$
form an interval of length $\frac{32}{225}$. | [] | Bulgaria | Fall Mathematical Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | -3/5 and 3/5 | |
0l3v | How many angles $\theta$ with $0 \le \theta \le 2\pi$ satisfy $\log(\sin(3\theta)) + \log(\cos(2\theta)) = 0$?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4 | [
"Suppose $\\theta$ satisfies the equation. Then $\\log(\\sin(3\\theta) \\cdot \\cos(2\\theta)) = 0$. This implies $\\sin(3\\theta) \\cdot \\cos(2\\theta) = 1$, so applying the Product-to-Sum Formula\n$$\n\\sin a \\cdot \\cos b = \\frac{1}{2} (\\sin(a + b) + \\sin(a - b))\n$$\ngives $\\frac{1}{2}(\\sin(5\\theta) + \... | United States | AMC 12 A | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | null | MCQ | A | |
0659 | Determine the values of the positive integer $n$ for which
$$
A = \sqrt{\frac{9n-1}{n+7}}
$$
is rational. | [
"It is enough to prove that there exist $a, b \\in \\mathbb{N}^*$ with $(a, b) = 1$ such that:\n$$\n\\frac{9n-1}{n+7} = \\frac{a^2}{b^2} \\qquad (1)\n$$\nFrom this relation we get:\n$$\nn = \\frac{7a^2 + b^2}{9b^2 - a^2} = \\frac{7(a^2 - 9b^2) + 64b^2}{9b^2 - a^2} = -7 + \\frac{64b^2}{9b^2 - a^2} \\quad (2)\n$$\nSi... | Greece | 26th Hellenic Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 1, 11 | |
0j5q | Problem:
Let $z = \cos \frac{2\pi}{2011} + i \sin \frac{2\pi}{2011}$, and let
$$
P(x) = x^{2008} + 3 x^{2007} + 6 x^{2006} + \ldots + \frac{2008 \cdot 2009}{2} x + \frac{2009 \cdot 2010}{2}
$$
for all complex numbers $x$. Evaluate $P(z) P\left(z^{2}\right) P\left(z^{3}\right) \ldots P\left(z^{2010}\right)$. | [
"Solution:\n\nAnswer: $2011^{2009} \\cdot \\left(1005^{2011} - 1004^{2011}\\right)$\n\nMultiply $P(x)$ by $x-1$ to get\n$$\nP(x)(x-1) = x^{2009} + 2 x^{2008} + \\ldots + 2009 x - \\frac{2009 \\cdot 2010}{2}\n$$\nor,\n$$\nP(x)(x-1) + 2010 \\cdot 1005 = x^{2009} + 2 x^{2008} + \\ldots + 2009 x + 2010\n$$\nMultiplying... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2011^{2009} \cdot \left(1005^{2011} - 1004^{2011}\right) | |
02ja | Problem:
Um ponto $P$ está no centro de um quadrado com $10~\mathrm{cm}$ de lado. Quantos pontos da borda do quadrado estão a uma distância de $6~\mathrm{cm}$ de $P$?
A) 1
B) 2
C) 4
D) 6
E) 8 | [
"Solution:\n\nOs pontos que estão a $6~\\mathrm{cm}$ de distância do ponto $P$ formam uma circunferência de centro $P$ e raio $R=6~\\mathrm{cm}$. Se $D$ denota a diagonal do quadrado, do teorema de Pitágoras temos\n$$\nD=\\sqrt{10^{2}+10^{2}}=\\sqrt{2 \\times 10^{2}}=10 \\sqrt{2}\n$$\nA circunferência de raio $L/2=... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | E | |
0ggw | 設三角形 $ABC$ 的外心為 $O$，垂心為 $H$，並且 $OH$ 與 $BC$ 平行。設 $AH$ 與三角形 $ABC$ 的外接圓再交於點 $X$ ($X \neq A$)，並設 $XB, XC$ 分別與 $OH$ 交於點 $Y, Z$。令 $P$ 為 $Y$ 對 $AB$ 的投影點，$Q$ 為 $Z$ 對 $AC$ 的投影點。證明 $PQ$ 平分線段 $BC$。 | [
"**Lemma.** Let $ABC$ be a triangle whose circumcenter is $O$. Let $\\ell$ be an arbitrary line passing through $O$. Let $Y, Z$ be two points on $\\ell$ such that $AY, AZ$ are perpendicular to $AC, AB$, respectively. Let $P, Q$ be the projections of $Y, Z$ onto $AB, AC$, respectively. Then $PQ$ bisects $AB$.\n\n*Pr... | Taiwan | 2022 數學奧林匹亞競賽第二階段選訓營, 獨立研究 (二) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Misc... | Chinese; English | proof only | null | |
0cr8 | Треугольник $ABC$ вписан в окружность $\Omega$ с центром $O$. Окружность, построенная на $AO$ как на диаметре, пересекает описанную окружность треугольника $OBC$ в точке $S \neq O$. Касательные к $\Omega$ в точках $B$ и $C$ пересекаются в точке $P$. Докажите, что точки $A$, $S$ и $P$ лежат на одной прямой. | [
"Поскольку $CP$ и $BP$ — касательные к $\\Omega$, имеем $\\angle OBP = \\angle OCP = 90^\\circ$; значит, точка $P$ лежит на описанной окружности треугольника $OBC$, и $PO$ — диаметр этой окружности (см. рис. 4). Поэтому $\\angle OSP = 90^\\circ$.\n\nДалее, поскольку $AO$ — диаметр окружности, проходящей через $A$, ... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
076q | Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $A_1$, $B_1$ and $C_1$ be respectively the midpoints of the arcs $BAC$, $CBA$ and $ACB$ of $\Gamma$. Show that the inradius of triangle $A_1B_1C_1$ is not less than the inradius of triangle $ABC$. | [
"Rotate the triangle around the centre of $\\Gamma$ by $180^\\circ$. Then $A_1$, $B_1$, $C_1$ respectively go to $A_2$, $B_2$, $C_2$. These are respectively the midpoints of the minor arc $BC$, $CA$ and $AB$. We see that $A_1B_1C_1$ and $A_2B_2C_2$ are congruent triangles and hence have the same in-radii. But we kn... | India | Indija TS 2016 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinat... | English | proof only | null | |
0jth | Problem:
Find the number of ways to choose two nonempty subsets $X$ and $Y$ of $\{1,2, \ldots, 2001\}$, such that $|Y|=1001$ and the smallest element of $Y$ is equal to the largest element of $X$. | [
"Solution:\nAnswer: $2^{2000}$\nWe claim that there is a bijection between pairs $(X, Y)$ and sets $S$ with at least $1001$ elements. To get $S$ from $X$ and $Y$, take $S = X \\cup Y$, which contains $Y$ and thus has at least $1001$ elements. To form $(X, Y)$ from $S$, make $Y$ the largest $1001$ elements of $S$, a... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 2^2000 | |
053d | Let $D$ be the point different from $B$ on the hypotenuse $AB$ of a right triangle $ABC$ such that $|CB| = |CD|$. Let $O$ be the circumcenter of triangle $ACD$. Rays $OD$ and $CB$ intersect at point $P$, and the line through point $O$ perpendicular to side $AB$ and ray $CD$ intersect at point $Q$. Points $A$, $C$, $P$,... | [
"\nFigure 32\n\nAs $OQ$ is the perpendicular bisector of $AD$, one has $\\angle QAD = \\angle ADQ = \\angle BDC = \\angle CBD$ (Fig. 32). Therefore $AQ \\parallel BC$, whence $\\angle QAC = 180^\\circ - \\angle ACB = 90^\\circ$. From the cyclic quadrilateral $APCQ$ one also gets $\\angle CP... | Estonia | IMO Team Selection Contest | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | Yes, ACPQ is a square. | |
0bdy | Let $m$ and $n$ be integers $m, n \ge 2$. The matrices $A_1, A_2, \dots, A_m \in \mathcal{M}_n(\mathbb{R})$, are not all nilpotent. Prove that there is an integer $k > 0$ such that $A_1^k + A_2^k + \dots + A_m^k \ne O_n$.
Marius Cavachi | [
"Denote by $\\lambda_{i1}, \\lambda_{i2}, \\dots, \\lambda_{in}$ the eigenvalues of $A_i$, $i = 1, 2, \\dots, m$. Suppose $A_1^k + A_2^k + \\dots + A_m^k = O_n$, for all $k \\ge 1$. Then $\\text{tr}(A_1^k) + \\text{tr}(A_2^k) + \\dots + \\text{tr}(A_m^k) = 0$, implying $\\sum_{j=1}^m \\sum_{i=1}^n \\lambda_{ij}^k =... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0a5r | Problem:
A round-robin tournament is one where each team plays every other team exactly once. Five teams take part in such a tournament getting: 3 points for a win, 1 point for a draw and 0 points for a loss. At the end of the tournament the teams are ranked from first to last according to the number of points.
(a) I... | [
"Solution:\nWe show the answer is no for five teams, and yes for six. \nFirst, we show five teams cannot have this property. Suppose the teams had points $x, x + 2, x + 4, x + 6, x + 8$ in the final ranking. If there were $d$ draws and $10 - d$ decisive games, then the total number of points is \n$$5x + 20 = x + ... | New Zealand | New Zealand Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | For five teams: not possible. For six teams: possible. | |
017t | Let $p$ be a prime number. For each $k$, $1 \le k \le p-1$, there exists an integer $m$, $1 \le m \le p-1$, such that $mk \equiv 1 \pmod{p}$. We will denote this integer by $\frac{1}{k}$. Prove that the sequence
$$
1, 1+\frac{1}{2}, 1+\frac{1}{2}+\frac{1}{3}, \dots, 1+\frac{1}{2}+\dots+\frac{1}{p-1}
$$
(addition modulo... | [
"Calculating modulo $p$ we have that $(p-k)\\frac{1}{k} = -1$ so $\\frac{1}{p-k} = -\\frac{1}{k}$. If $p$ is odd, we set $m = \\frac{p-1}{2}$ and it follows that\n$$\n\\sum_{k=1}^{p-1} \\frac{1}{k} = \\sum_{k=1}^{m} \\left(\\frac{1}{k} + \\frac{1}{p-k}\\right) = 0.\n$$\nFor $\\ell$ such that $m < \\ell < p-1$ we ca... | Baltic Way | BALTIC WAY | [
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof only | null | |
05jt | Problem:
Soient $n$ et $m$ deux entiers strictement positifs. Montrer que $5^{m}+5^{n}$ s'écrit comme une somme de deux carrés si et seulement si $n$ et $m$ ont même parité. | [
"Solution:\nSupposons pour commencer que $n$ et $m$ n'ont pas même parité. On étudie alors la quantité $5^{n}+5^{m}$ modulo $8$. Par récurrence, on démontre facilement que $5^{k}$ est congru à $1$ (resp. $5$) modulo $8$ si $k$ est pair (resp. impair). Ainsi, d'après notre hypothèse, on a toujours $5^{n}+5^{m} \\equ... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
00if | Determine all tripels $(x, y, z)$ of positive integers $x > y > z > 0$, such that $x^2 = y \cdot 2^x + 1$ holds. | [
"We first note that the right-hand side of the equation is odd. We therefore know that $x^2$ is odd, and therefore $x$ is odd. One of the neighbors of $x$ must therefore be divisible by $4$, and we therefore have $x = 2^p a \\pm 1$ with $p > 1$ and $a$ odd.\n\nLet us first assume $x = 2^p a + 1$. If $a > 1$, we fir... | Austria | Austria 2010 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (2^p + 1, 2^{p-1} + 1, p + 1) for p > 2; (2^p - 1, 2^{p-1} - 1, p + 1) for p > 3; (2^p - 1, 2^p - 2, p) for p > 2. |
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