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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0iay | Problem:
Find the value of
$$
\binom{6}{1} 2^{1}+\binom{6}{2} 2^{2}+\binom{6}{3} 2^{3}+\binom{6}{4} 2^{4}+\binom{6}{5} 2^{5}+\binom{6}{6} 2^{6} .
$$ | [
"Solution:\nThis sum is the binomial expansion of $(1+2)^{6}$, except that it is missing the first term, $\\binom{6}{0} 2^{0}=1$. So we get $3^{6}-1=728$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 728 | |
0a97 | Problem:
Determine all real numbers $A$, $B$ and $C$ such that there exists a real function $f$ that satisfies
$$
f(x+f(y))=A x+B y+C
$$
for all real $x$ and $y$. | [
"Solution:\nLet $A$, $B$ and $C$ be real numbers and $f$ a function such that $f(x+f(y))=A x+B y+C$ for all $x$ and $y$. Let $z$ be a real number and set $x=z-f(0)$ and $y=0$. Then\n$$\nf(z)=f(z-f(0)+f(0))=A(z-f(0))+B \\cdot 0+C=A z-A f(0)+C\n$$\nso there are numbers $a$ and $b$ such that $f(z)=a z+b$ for all $z$. ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 22 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | (A, B, C) = (a, a^2, c) for any real a ≠ −1 and any real c; additionally (A, B, C) = (−1, 1, 0). | |
08gy | Problem:
Given a parallelogram $ABCD$. The line perpendicular to $AC$ passing through $C$ and the line perpendicular to $BD$ passing through $A$ intersect at point $P$. The circle centered at point $P$ and radius $PC$ intersects the line $BC$ at point $X$ ($X \neq C$) and the line $DC$ at point $Y$ ($Y \neq C$). Prove... | [
"Solution:\n\nDenote the feet of the perpendiculars from $P$ to the lines $BC$ and $DC$ by $M$ and $N$ respectively and let $O = AC \\cap BD$. Since the points $O$, $M$ and $N$ are midpoints of $CA$, $CX$ and $CY$ respectively, it suffices to prove that $M$, $N$ and $O$ are collinear. According to Menelaus's theore... | JBMO | null | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0jxc | Problem:
Find the value of
$$
\sum_{1 \leq a < b < c} \frac{1}{2^{a} 3^{b} 5^{c}}
$$
(i.e. the sum of $\frac{1}{2^{a} 3^{b} 5^{c}}$ over all triples of positive integers ( $a, b, c$ ) satisfying $a < b < c$ ) | [
"Solution:\nLet $x = b - a$ and $y = c - b$ so that $b = a + x$ and $c = a + x + y$. Then\n$$\n2^{a} 3^{b} 5^{c} = 2^{a} 3^{a + x} 5^{a + x + y} = 30^{a} 15^{x} 5^{y}\n$$\nand $a, x, y$ are any positive integers. Thus\n$$\n\\begin{aligned}\n\\sum_{1 \\leq a < b < c} \\frac{1}{2^{a} 3^{b} 5^{c}} & = \\sum_{1 \\leq a... | United States | February 2017 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1/1624 | |
06px | Let $a$, $b$, $c$, $d$ be positive real numbers such that
$$
a b c d = 1 \quad \text{ and } \quad a + b + c + d > \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}
$$
Prove that
$$
a + b + c + d < \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d}
$$ | [
"We show that if $a b c d = 1$, the sum $a + b + c + d$ cannot exceed a certain weighted mean of the expressions $\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{d} + \\frac{d}{a}$ and $\\frac{b}{a} + \\frac{c}{b} + \\frac{d}{c} + \\frac{a}{d}$.\nBy applying the AM-GM inequality to the numbers $\\frac{a}{b}$, $\\frac{a}{b}... | IMO | 49th International Mathematical Olympiad Spain | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0kys | Problem:
Alice, Bob, and Charlie are playing a game with 6 cards numbered 1 through 6. Each player is dealt 2 cards uniformly at random. On each player's turn, they play one of their cards, and the winner is the person who plays the median of the three cards played. Charlie goes last, so Alice and Bob decide to tell t... | [
"Solution:\n\nIf Alice has a card that is adjacent to one of Bob's, then Alice and Bob will play those cards as one of them is guaranteed to win. If Alice and Bob do not have any adjacent cards, since Charlie goes last, Charlie can always choose a card that will win.\n\nLet $A$ denote a card that is held by Alice a... | United States | HMMT February 2024 Guts Round | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 2/15 | |
03j9 | Problem:
A Mathlon is a competition in which there are $M$ athletic events. Such a competition was held in which only $A$, $B$, and $C$ participated. In each event $p_{1}$ points were awarded for first place, $p_{2}$ for second and $p_{3}$ for third, where $p_{1} > p_{2} > p_{3} > 0$ and $p_{1}, p_{2}, p_{3}$ are integ... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | M = 5; C was second in the high jump. | |
0ag0 | Find all functions $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, for which:
$$
f(xf(x,z), yf(y,z)) = f(x,z)f(y,z)
$$
for every real numbers $x$, $y$ and $z$ and $f(0,0) \neq 0$. | [
"For $x = y = z = 0$, we get $f(0,0) = (f(0,0))^2$ and because of $f(0,0) \\neq 0$, we get that $f(0,0) = 1$. For $x = y = 0$ we get $f(0,0) = (f(0,z))^2$, from where $f(0,z) = \\pm 1$, for every real number $z$. For $x = 0$, we get $\\pm 1 = \\pm f(y,z)$, so $f(y,z) = \\pm 1$, for every real numbers $y$ and $z$.\n... | North Macedonia | null | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | All solutions are of the form f(x, y) = s(|x|) s(|y|), where s maps nonnegative real numbers (including zero) to either plus one or minus one. Equivalently, choose any subset M of the nonnegative reals and let f(x, y) = 1 if the absolute values of x and y both lie in M or both lie outside M, and f(x, y) = −1 otherwise.... | |
0dwf | Problem:
V telenoveli o dogodkih v zarotniškem mestecu nastopa $n$ meščanov, $n \geq 4$. Vsaka skupina 3 meščanov kuje zaroto proti enemu izmed ostalih meščanov. Dokaži, da obstaja tak meščan, da je vsaj $\sqrt[3]{(n-1)(n-2)}$ meščanov vpletenih v zaroto proti njemu. | [
"Solution:\n\nVseh (neurejenih) trojic meščanov je $\\frac{n(n-1)(n-2)}{6}$. Vsako trojico označimo s tistim izmed $n-3$ meščanov, proti kateremu trojica kuje zaroto. Potem obstaja $m$ trojic, ki so vse označene z istim meščanom, in je $m \\geq \\frac{(n-1)(n-2)}{6}$. Denimo, da v teh $m$ trojicah sodeluje $k$ mešč... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0hm7 | Problem:
In how many different ways one can place 3 rooks on the cells of $6 \times 2006$ chessboard such that they don't attack each other? | [
"Solution:\n\nIn order not to attack each other, the rooks cannot share a row or a column. Since there are 6 columns in total, we have 20 possibilities for choosing 3 columns in which we will place our rooks. After choosing the columns, we will place our rooks one by one in these columns. The first rook can be plac... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 20 * 2006 * 2005 * 2004 | |
0jq9 | Problem:
Let $I$ be the set of points $(x, y)$ in the Cartesian plane such that
$$
x > \left( \frac{y^{4}}{9} + 2015 \right)^{1/4}
$$
Let $f(r)$ denote the area of the intersection of $I$ and the disk $x^{2} + y^{2} \leq r^{2}$ of radius $r > 0$ centered at the origin $(0,0)$. Determine the minimum possible real numb... | [
"Solution:\n$\\quad \\frac{\\pi}{3}$\n\nLet $B(P, r)$ be the (closed) disc centered at $P$ with radius $r$. Note that for all $(x, y) \\in I$, $x > 0$, and $x > \\left( \\frac{y^{4}}{9} + 2015 \\right)^{1/4} > \\frac{|y|}{\\sqrt{3}}$. Let $I' = \\{ (x, y) : x \\sqrt{3} > |y| \\}$. Then $I \\subseteq I'$ and the int... | United States | HMMT February 2015 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | pi/3 | |
0a7m | Problem:
Find all solutions of the system of equations
$$
\left\{\begin{aligned}
s(x)+s(y) & = x \\
x+y+s(z) & = z \\
s(x)+s(y)+s(z) & = y-4
\end{aligned}\right.
$$
where $x$, $y$, and $z$ are positive integers, and $s(x)$, $s(y)$, and $s(z)$ are the numbers of digits in the decimal representations of $x$, $y$, and $z$... | [
"Solution:\nThe first equation implies $x \\geq 2$ and the first and third equation together imply\n$$\ns(z) = y - x - 4\n$$\nSo $y \\geq x + 5 \\geq 7$. From (1) and the second equation we obtain $z = 2y - 4$. Translated to the values of $s$, these equations imply $s(x) \\leq s(2y) \\leq s(y) + 1$ and $s(x) \\leq ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 7 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (2, 8, 12) | |
0hwk | Problem:
We are given a right isosceles triangle with legs of length $1$ inside which every point (including vertices, points on sides, and all points in the interior) is colored red, yellow, green, or blue. Prove that there are two points of the same color such that the distance between them is at least $2-\sqrt{2}$. | [
"Solution:\n\nDefine points $D, E, F, G$ on the sides of the triangle as shown such that $AD = AF = GB = BE = 2 - \\sqrt{2}$. Then $CD = CE = \\sqrt{2} - 1$, and since $\\triangle CDE$ is isosceles and right, $DE = \\sqrt{2}(\\sqrt{2} - 1) = 2 - \\sqrt{2}$. Now segments $DE$ and $AF$ are eq... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0idx | Problem:
A positive integer $n$ is picante if $n!$ ends in the same number of zeroes whether written in base $7$ or in base $8$. How many of the numbers $1, 2, \ldots, 2004$ are picante? | [
"Solution:\nThe number of zeroes in base $7$ is the total number of factors of $7$ in $1 \\cdot 2 \\cdots n$, which is\n$$\n\\left\\lfloor \\frac{n}{7} \\right\\rfloor + \\left\\lfloor \\frac{n}{7^2} \\right\\rfloor + \\left\\lfloor \\frac{n}{7^3} \\right\\rfloor + \\cdots.\n$$\nThe number of zeroes in base $8$ is ... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 4 | |
0l1l | Problem:
For each prime $p$, a polynomial $P(x)$ with rational coefficients is called $p$-good if and only if there exist three integers $a, b$, and $c$ such that $0 \leq a < b < c < \frac{p}{3}$ and $p$ divides all the numerators of $P(a)$, $P(b)$, and $P(c)$, when written in simplest form. Compute the number of order... | [
"Solution:\nBy Vieta, the sum of the roots is $-10 \\pmod{p}$. However, since the three roots are less than $p / 3$, it follows that the roots are $\\left(p - a'\\right) / 3$, $\\left(p - b'\\right) / 3$, and $\\left(p - c'\\right) / 3$, where there are finitely many choices $a' < b' < c'$. By pigeonhole, one choic... | United States | HMMT February 2024 Guts Round | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 12 | |
08qk | Problem:
Determine whether there is a natural number $n$ for which $8^{n}+47$ is prime. | [
"Solution:\nThe number $m = 8^{n} + 47$ is never prime.\n\nIf $n$ is even, say $n = 2k$, then $m = 64^{k} + 47 \\equiv 1 + 2 \\equiv 0 \\pmod{3}$. Since also $m > 3$, then $m$ is not prime.\n\nIf $n \\equiv 1 \\pmod{4}$, say $n = 4k + 1$, then $m = 8 \\cdot (8^{k})^{4} + 47 \\equiv 3 + 2 \\equiv 0 \\pmod{5}$. Since... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | No; there is no natural number for which the expression is prime. | |
03tb | Given a positive integer $n \ge 2$, let $B_1, B_2, \dots, B_n$ denote $n$ arbitrary subsets of set $X$, each of which contains exactly two elements. Find the minimum value of $|X|$ such that there exists a subset $Y$ of set $X$ satisfying:
(a) $|Y| = n$;
(b) $|Y \cap B_i| \le 1$ for $i = 1, 2, \dots, n$,
where $|A|$ de... | [
"We first prove that $|X| > 2n - 2$. In fact, if $|X| = 2n - 2$, let $X = \\{1, 2, \\dots, 2n - 2\\}$, $B_1 = \\{1, 2\\}$, $B_2 = \\{3, 4\\}$, $\\dots$, $B_{n-1} = \\{2n - 3, 2n - 2\\}$. Since $|Y| = n$, there exist two elements in $Y$ that belong to the same $B_i$, then $|Y \\cap B_i| > 1$, a contradiction.\n\nLet... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2n - 1 | |
08g6 | Problem:
Lucio ha due carte identiche di forma rettangolare con lati lunghi $6$ e $9$; le appoggia sul tavolo in modo che abbiano una diagonale in comune ma non risultino precisamente sovrapposte. Qual è l'area della porzione della prima carta che è coperta dalla seconda?
(A) $27$
(B) $36$
(C) $37$
(D) $38$
(E) $39$ | [
"Solution:\n\nLa risposta è $(E)$.\n\nIl problema chiede di calcolare l'area di $AGCH$. Dimostriamo che tale quadrilatero è un rombo. Detto $O$ il punto medio di $AC$, questo è centro di simmetria sia per il rettangolo $ABCD$ che per il rettangolo $AECF$, quindi è centro di simmetria anche per $AGCH$ che quindi è u... | Italy | Olimpiadi di Matematica - Febbraio | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | E | |
00ap | One of the numbers $1$, $2$, ..., $n$ is written in each cell of a $17 \times 17$ table for a certain $n \in \mathbb{N}$; all of these numbers are used. If a row contains two cells $C_1$ and $C_2$ with equal numbers $k$ and $C_1$ is to the left of $C_2$ then there are no numbers $k$ in the column of $C_1$ that are abov... | [
"The minimal $n$ in question is $n=9$.\nFirst we show that each number $k \\in \\{1, 2, ..., n\\}$ occurs in the table at most $34$ times. Let a row contain at least two $k$'s. Underline all of them except the rightmost one. The remaining numbers in the table are not underlined. By hypothesis $k$ does not occur abo... | Argentina | Argentine National Olympiad 2016 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 9 | |
00l4 | Determine all nonnegative integers $n$ having two distinct positive divisors with the same distance from $\frac{n}{2}$. | [
"Since the smallest possible divisors of an integer $n$ are $1$, $2$ and $3$, the greatest possible divisors are $n$, $\\frac{n}{2}$ and $\\frac{n}{2}$. Hence a divisor that is bigger than $\\frac{n}{2}$ can only be $n$ or $\\frac{n}{2}$. Since there is no positive divisor of $n$ having the same distance from $\\fr... | Austria | Beginners' Competition | [
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | all positive multiples of 6 | |
044i | Sequence $\{a_n\}$ satisfies $a_1 = 2$ and $a_{n+1} = (n+1)a_n - n$, $n = 1, 2, \dots$. Then the general term formula of $\{a_n\}$ is ______. | [
"By the condition, we have $a_{n+1} - 1 = (n+1)(a_n - 1)$. Therefore,\n$$\n\\begin{aligned}\na_n - 1 &= n(a_{n-1} - 1) = n(n-1)(a_{n-2} - 1) = \\dots \\\\\n&= n(n-1)\\dots2(a_1 - 1) = n!,\n\\end{aligned}\n$$\nnamely, $a_n = n! + 1$."
] | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | a_n = n! + 1 | |
0g1i | Problem:
Trouver toutes les fonctions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{\geq 0}$ telles que pour tous $x, y \in \mathbb{R}_{>0}$ :
$$
f(x)-f(x+y)=f\left(x^{2} f(y)+x\right)
$$ | [
"Solution:\n\nSoit $f$ une solution de l'équation. On commence naturellement par chercher des fonctions qui sont solution. Après une petite recherche on trouve que les fonctions $x \\mapsto 0$ et $x \\mapsto 1 / x$ marchent. De plus, comme $f\\left(x^{2} f(y)+x\\right) \\geq 0$ la fonction est décroissante. Est-ce ... | Switzerland | IMO-Selektion | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 for all x > 0; and f(x) = 1/x for all x > 0 | |
0bc6 | a) There are infinitely many positive integer numbers $n$ such that there exists a square equal to the sum of the squares of $n$ consecutive positive integer numbers. (For instance, $2$ and $11$ are such: $5^2 = 3^2 + 4^2$ and $77^2 = 18^2 + 19^2 + \dots + 28^2$.)
b) If $n$ is a positive integer number which is not a ... | [
"a.\nWe must show that there are infinitely many positive integer numbers $n$ such that the equation $y^2 = x^2 + (x+1)^2 + \\dots + (x+n-1)^2$ has positive integral solutions. To this end, rewrite the equation in the form\n$$\ny^2 = n\\left(\\left(x + \\frac{n-1}{2}\\right)^2 + \\frac{(n^2-1)}{12}\\right). \\qquad... | Romania | 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
04eg | How many ordered pairs of positive integers $(m, n)$ satisfy the equation $m^2 - n^2 = 2^{2013}$? | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1006 | |
0fp2 | El triángulo $BCF$ es rectángulo en $B$. Sea $A$ el punto de la recta $CF$ tal que $FA = FB$ y $F$ está entre $A$ y $C$. Se elige el punto $D$ de modo que $DA = DC$ y $AC$ es bisectriz del ángulo $\angle DAB$. Se elige el punto $E$ de modo que $EA = ED$ y $AD$ es bisectriz del ángulo $\angle EAC$. Sea $M$ el punto medi... | [] | Spain | LVII Olimpiada Internacional de Matemáticas | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Me... | Spanish | proof only | null | |
072l | Let $(u_{jk})$, $1 \le j \le 3$, $1 \le k \le 2$ be real numbers and let $N$ be an integer such that
$$
\max_{1 \le k \le 2} \sum_{j=1}^{3} |u_{jk}| \le N.
$$
Suppose $M$ and $l$ are positive integers such that $l^2 < (M+1)^3$. Prove that there exist integers $\xi_1, \xi_2, \xi_3$, not all zero, such that
$$
\max_{1 \l... | [
"Let $a_1, a_2, a_3$ be integers such that $0 \\le a_1, a_2, a_3 \\le M$. Let\n$$\nS_k = \\left| \\sum_{j=1}^{3} u_{jk} a_j \\right|, \\quad k = 1, 2.\n$$\nThen Cauchy-Schwartz inequality gives\n$$\nS_k \\le \\sqrt{\\sum_{j=1}^{3} a_j^2} \\sqrt{\\sum_{j=1}^{3} u_{jk}^2} \\le \\sqrt{3M^2} \\sqrt{N^2} = \\sqrt{3MN^2}... | India | Indija TS 2006 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
07pv | Suppose $a_1 = 3$, $a_{n+1} = \frac{1}{2}(a_n^2 + 1)$, $n = 1, 2, \dots$. Prove that
$$
\sum_{k=1}^{n} \frac{1}{1+a_k} < \frac{1}{2}
$$ | [
"Note that $a_n > 1$ for all $n$, and $2(a_{n+1} - 1) = a_n^2 - 1 = (a_n - 1)(a_n + 1)$, hence\n$$\n\\frac{1}{a_{n+1}-1} = \\frac{2}{a_n^2-1} > 0 \\quad \\text{for } n=1,2,\\dots\n$$\nLet $s_n = \\sum_{k=1}^{n} \\frac{1}{1+a_k}$ for $n \\ge 1$. We will prove by induction that\n$$\ns_n = \\frac{1}{2} - \\frac{2}{a_n... | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
05f8 | Problem:
Existe-t-il des entiers $a$ et $b$ pour lesquels : $a^{5} b+3$ et $a b^{5}+3$ sont tous deux des cubes parfaits?
Un cube parfait est un entier $n$ pour lequel il existe $m \in \mathbb{Z}$ tel que : $n = m^{3}$. | [
"Solution:\n\nNous allons montrer qu'il n'y a pas d'entiers $a$ et $b$ satisfaisant les conditions du problème.\n\nSoient $a$ et $b$ des entiers tels que $a^{5} b+3$ et $a b^{5}+3$ sont des cubes parfaits. Soit $m$ et $n$ des entiers tels que $a^{5} b+3=m^{3}$ et $a b^{5}+3=n^{3}$.\n\nSupposons que $3 \\mid n$. Alo... | France | ENVOi 3 : ARITHMÉTIQUE | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
02br | Problem:
Mostre que é possível construir um pentágono com todos os lados de mesma medida e cujos ângulos internos meçam $60^{\circ}$, $80^{\circ}$, $100^{\circ}$, $140^{\circ}$ e $160^{\circ}$, em alguma ordem. | [
"Solution:\n\nSuponhamos que já construímos o pentágono $A B C D E$ e que o ângulo em $A$ mede $60^{\\circ}$. Traçando a reta $B E$, concluímos que o triângulo $A B E$ é equilátero, pois $A B = A E$ e $E \\hat{A} B = 60^{\\circ}$. Logo, $B E = A B$ e, portanto, $B C D E$ tem todos os seus lados com a mesma medida, ... | Brazil | null | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0iur | Problem:
A circle passes through the points $(2,0)$ and $(4,0)$ and is tangent to the line $y = x$. Find the sum of all possible values for the $y$-coordinate of the center of the circle. | [
"Solution:\nFirst, we see that the $x$-coordinate must be $3$. Let the $y$-coordinate be $y$. Now, we see that the radius is $r = \\sqrt{1 + y^{2}}$. The line from the center of the circle to the point of tangency with the line $x = y$ is perpendicular to the line $x = y$. Hence, the distance from the center of the... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | -6 | |
0err | Let $ABCD$ be a rectangle with side lengths $AB = CD = 5$ and $BC = AD = 10$. $W, X, Y, Z$ are points on $AB, BC, CD$ and $DA$ respectively chosen in such a way that $WXYZ$ is a kite, where $\angle ZWX$ is a right angle. Given that $WX = WZ = \sqrt{13}$ and $XY = ZY$, determine the length of $XY$. | [
"\nNote that $\\angle AZW = 90^\\circ - \\angle AWZ = \\angle BWX$ and $\\angle AWZ = 90^\\circ - \\angle BWX = \\angle BXW$. Moreover, $WX = WZ$, so the two triangles $AWZ$ and $BXW$ are congruent. Let $BW = AZ = x$, so that $BX = AW = 5 - x$. Pythagoras' theorem gives us\n$$\nx^2 + (5-x)^... | South Africa | The South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | sqrt(65) | |
0az1 | Problem:
Suppose that $a, b, c$ are real numbers such that
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 4\left(\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}\right) = \frac{c}{a+b} + \frac{a}{b+c} + \frac{b}{c+a} = 4
$$
Determine the value of $abc$. | [] | Philippines | 21st PMO Area Stage | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 49/23 | |
0edl | Problem:
Katera izmed navedenih enačb premic je enačba simetrale daljice s krajiščema $A(3,-3)$ in $B(2,-2)$ ?
(A) $y=-x-3$
(B) $x+y-1=0$
(C) $x-y-5=0$
(D) $3 x-2 y=0$
(E) $\frac{x}{3}-\frac{y}{2}=1$ | [
"Solution:\nSimetrala daljice $AB$ poteka skozi razpolovišče $S\\left(\\frac{5}{2},-\\frac{5}{2}\\right)$ daljice $AB$ in je pravokotna na daljico $AB$. Smerni koeficient premice skozi točki $A$ in $B$ je $k_{1}=\\frac{-2-(-3)}{2-3}=-1$. Torej je smerni koeficient simetrale $k_{2}=-\\frac{1}{k_{1}}=1$. Upoštevamo $... | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | C | |
09tn | Problem:
Gegeven is een verzameling $A$ van functies $f: \mathbb{R} \rightarrow \mathbb{R}$. Voor alle $f_1, f_2 \in A$ bestaat er een $f_3 \in A$ zodat
$$
f_1\left(f_2(y)-x\right)+2 x=f_3(x+y)
$$
voor alle $x, y \in \mathbb{R}$. Bewijs dat voor alle $f \in A$ geldt:
$$
f(x-f(x))=0
$$ | [
"Solution:\nInvullen van $x=0$ geeft\n$$\nf_1\\left(f_2(y)\\right)=f_3(y)\n$$\ndus de $f_3$ die bij $f_1$ en $f_2$ hoort, is blijkbaar de samenstelling $f_3(x)=f_1\\left(f_2(x)\\right)$. Invullen van $x=-y$ geeft nu dat voor alle $f_1, f_2 \\in A$\n$$\nf_1\\left(f_2(y)+y\\right)-2 y=f_3(0)=f_1\\left(f_2(0)\\right)\... | Netherlands | IMO-selectietoets | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0gys | In triangle $ABC$ points $M$ and $N$ are the midpoints of $BC$ and $AC$ respectively. Inside this triangle point $P$ is chosen in such a way that $\angle BAP = \angle PCA = \angle MAC$. Prove that $\angle PNA = \angle AMB$. | [
"It is easy to see that $\\angle APC = 180^\\circ - (\\angle PCA + \\angle PAC) = 180^\\circ - \\angle BAC = \\angle ANM$, since $MN \\parallel AB$ as the centerline of the triangle (fig.14).\n\nWe know that $\\angle MAC = \\angle PCA$, and so $\\triangle MNA \\sim \\triangle APC$. $N$ and $K$ are the midpoints of ... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05a2 | Let $ABC$ be a triangle where $AB = AC$. Let $E$ be the foot of its altitude from the vertex $B$. Given that either $BEA$ or $BEC$ is an isosceles triangle, find all possibilities of the size of the angle at the vertex $A$ of the triangle $ABC$. | [
"*Answer:* $45^\\circ$, $90^\\circ$, $135^\\circ$.\n\nLet $\\alpha = \\angle BAC$. Both triangles $BEA$ and $BEC$ have right angle at vertex $E$. Hence these triangles can be isosceles only if their other angles have size $45^\\circ$.\n\nSuppose that $BEC$ is isosceles (see figure below).\nThen $\\angle BCE = 45^\\... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 45°, 90°, 135° | |
0dpk | Prove that among every 42 numbers from the interval $[1, 10^6]$ it is possible to choose four numbers in such a way that for any permutation $(a, b, c, d)$ of these numbers, the following inequality holds:
$$
25(ab + cd)(ad + bc) \geq 16(ac + bd)^2.
$$ | [
"Let $1 \\leq x_1 \\leq x_2 \\leq \\dots \\leq x_{42} \\leq 10^6$ be the given numbers. Suppose that $x_{2k+1}x_{2k+2} > 4x_{2k}x_{2k-1}$ for every $k = 1, 2, \\dots, 20$. Multiplying all these inequalities for every $k = 1, 2, \\dots, 20$ we obtain $x_{41}x_{42} > 4^{20}x_1x_2 \\geq 4^{20}$, hence $x_{42} > 2^{20}... | Silk Road Mathematics Competition | XVI Silk Road Math Competition | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0as2 | Problem:
The 2011th prime number is $17483$, and the next prime is $17489$.
Does there exist a sequence of $2011^{2011}$ consecutive positive integers that contains exactly $2011$ prime numbers? Prove your answer. | [
"Solution:\n\nLet $N = 2011^{2011}$. Since $N > 17489$, there are more than $2011$ primes in the sequence $1, 2, 3, \\ldots, N$.\n\nClaim. There exists a sequence of $N$ consecutive positive integers that are all composite.\n\nProof of the Claim. The sequence\n$$\n(N+1)! + 2,\\ (N+1)! + 3,\\ (N+1)! + 4,\\ \\ldots,\... | Philippines | 13th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Yes, such a sequence exists. | |
0cef | Find the integer solutions of the equation
$$
\left( \frac{1 + \sqrt{1 - x^2}}{2} \right)^{\sqrt{1-x}} = (\sqrt{1-x})^{\sqrt{1-x}+\sqrt{1+x}}.
$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 0 | |
01cq | In the parliament of Neverland, all legislative work is carried out in committees of three people. The constitution dictates that any four people can be in at most two committees. We call a collection of committees a *clique* if any two of them have exactly two people in common, and any manner of including another comm... | [
"It is easy to see that, given three different committees, each pair of them can have two people in common only if all three committees share the same two people, for otherwise the people in the committees would contain four people from whom three committees have been formed. As a corollary, for any clique, there a... | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0k88 | Problem:
A circle $\Gamma$ with center $O$ has radius 1. Consider pairs $(A, B)$ of points so that $A$ is inside the circle and $B$ is on its boundary. The circumcircle $\Omega$ of $O A B$ intersects $\Gamma$ again at $C \neq B$, and line $A C$ intersects $\Gamma$ again at $X \neq C$. The pair $(A, B)$ is called techy ... | [
"Solution:\nWe claim that $(A, B)$ is techy if and only if $O A = A B$.\n\nNote that $O X$ is tangent to the circle $(O B C)$ if and only if $O X$ is perpendicular to the angle bisector of $\\angle B O C$, since $O B = O C$. Thus $(A, B)$ is techy if and only if $O X$ is parallel to $B C$. Now since $O C = O X$,\n$... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 3π/4 | |
06x1 | Lucy starts by writing $s$ integer-valued 2022-tuples on a blackboard. After doing that, she can take any two (not necessarily distinct) tuples $\mathbf{v} = (v_1, \ldots, v_{2022})$ and $\mathbf{w} = (w_1, \ldots, w_{2022})$ that she has already written, and apply one of the following operations to obtain a new tuple:... | [
"We solve the problem for $n$-tuples for any $n \\geqslant 3$: we will show that the answer is $s = 3$, regardless of the value of $n$.\n\nFirst, let us briefly introduce some notation. For an $n$-tuple $\\mathbf{v}$, we will write $\\mathbf{v}_i$ for its $i$-th coordinate (where $1 \\leqslant i \\leqslant n$). For... | IMO | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Linear Algebra > Vectors"
] | English | proof and answer | 3 | |
0fzg | Problem:
Gegeben sind eine natürliche Zahl $n$ und natürliche Zahlen $a_{1}, a_{2}, \ldots, a_{n}$. Wir erweitern die Folge periodisch durch $a_{n+i}=a_{i}$ für alle $i \geq 1$. Nehme nun an, dass folgende zwei Bedingungen erfüllt sind:
(i) $a_{1} \leq a_{2} \leq \cdots \leq a_{n} \leq a_{1}+n$.
(ii) $a_{a_{i}} \leq... | [
"Solution:\n\nWir wissen, dass $a_{1} \\leq a_{a_{1}} \\leq n$ gilt. Für $a_{j}<i \\leq a_{j+1}$ haben wir $a_{i} \\leq a_{a_{j+1}} \\leq n+j$ und für $a_{a_{1}}<i \\leq n$ haben wir $a_{i} \\leq a_{n} \\leq a_{1}+n$.\n\nKombiniert erhalten wir folgende Ungleichungen:\n$$\n\\begin{aligned}\na_{1}+a_{2}+\\cdots+a_{n... | Switzerland | IMO-Selektionsprüfung | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof only | null | |
00uz | Let $n \ge 3$. Alice and Bob play the following game: Alice chooses $k \in \{3, 4, \dots, n\}$ and draws a $3 \times k$ table, then he fills the $k$ cells of the first row with different numbers from $\{1, 2, \dots, n\}$. Then, Bob fills on the second row some of the cells (eventually none) with distinct numbers from $... | [
"Let $1 \\le a_1 < a_2 < \\dots < a_k \\le n$ be the numbers Alice chose. For a sequence $x_1 < x_2 < \\dots < x_k$ of positive integers, we call its *deficit* the set $N \\cap [x_1, x_k] \\setminus \\{x_1, x_2, \\dots, x_k\\}$.\n\nBob has the following strategy: he starts with $a_1 < a_2 < \\dots < a_k$. Let $t$ b... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
052b | A convex quadrilateral $ABCD$ where $\angle DAB + \angle ABC < 180^\circ$ is given on a plane. Let $E$ be a point different from the vertices of the quadrilateral on the line interval $AB$ such that the circumcircles of triangles $AED$ and $BEC$ intersect inside the quadrilateral $ABCD$ at point $F$. Point $G$ is defin... | [
"Denote $\\angle DAB = \\alpha$ and $\\angle ABC = \\beta$ (Fig. 21). From cyclic quadrilaterals $AEFD$ and $BEFC$ one obtains\n\n$$\n\\begin{aligned}\n\\angle DFE &= 180^\\circ - \\angle DAE = 180^\\circ - \\alpha, \\\\\n\\angle CFE &= 180^\\circ - \\angle CBE = 180^\\circ - \\beta,\n\\end{aligned}\n$$\nrespective... | Estonia | Final Round of National Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fem | Problem:
Probar que hay infinitos números primos cuyo resto al dividirlos entre $3$ es $2$. | [
"Solution:\n\nEvidentemente hay al menos un primo, pues $2$ es un ejemplo. Supongamos que solo hubiese un número finito de ellos, $p_{1}, \\ldots, p_{k}$, y sea $n$ el producto de todos ellos. El producto de dos de esos primos deja resto $1$ al dividir entre $3$, por lo que, dependiendo de si $k$ es par o es impar,... | Spain | null | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0esi | A shop has a special offer on pumpkins: "Buy one, get one at half price". For this offer, the average cost per pumpkin is the same as in
(A) Two for the price of one
(B) Three for the price of one
(C) Three for the price of two
(D) Four for the price of three
(E) Five for the price of four | [
"If you buy one item and get one at half price you are getting two items for the price of one and a half items. Since $2 : 1\\frac{1}{2}$ is equivalent to $4 : 3$, the average cost per item is the same as four items for the price of three."
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | MCQ | D | |
02gs | $f(x)$ is a real-valued function defined on the positive reals such that
(1) if $x < y$, then $f(x) < f(y)$;
(2) $f\left(\frac{2xy}{x+y}\right) = \frac{f(x)+f(y)}{2}$ for all $x$.
Show that $f(x) < 0$ for some value of $x$. | [
"Put $x_n = 1/n$, $y_n = f(x_n)$. We have $\\frac{2x_{n-1}x_{n+1}}{x_{n-1} + x_{n+1}} = x_n$, so $y_n = \\frac{y_{n-1} + y_{n+1}}{2}$, or $y_n - y_{n+1} = y_{n-1} - y_n$. Now $1/2 < 1$, so $y_2 < y_1$.\n\nPut $y_1 - y_2 = d > 0$. Then $y_n - y_{n+1} = d$ for all $n$. Hence $y_{n+1} = y_1 - nd$. So $y_n$ is negative... | Brazil | XXV OBM | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0kht | Recall that the conjugate of the complex number $w = a + bi$, where $a$ and $b$ are real numbers and $i = \sqrt{-1}$, is the complex number $\bar{w} = a - bi$. For any complex number $z$, let $f(z) = 4i\bar{z}$. The polynomial $P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1$ has four complex roots: $z_1, z_2, z_3$, and $z_4$. Let $... | [
"First note that because complex roots of a polynomial with real coefficients come in conjugate pairs, the roots of $P(z)$ are $\\overline{z_1}, \\overline{z_2}, \\overline{z_3}, \\overline{z_4}$. In other words, $\\{z_1, z_2, z_3, z_4\\} = \\{\\overline{z_1}, \\overline{z_2}, \\overline{z_3}, \\overline{z_4}\\}$. ... | United States | AMC 12 A | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | MCQ | D | |
0kxq | Problem:
Suppose $a$, $b$, and $c$ are distinct positive integers such that $\sqrt{a \sqrt{b \sqrt{c}}}$ is an integer. Compute the least possible value of $a+b+c$. | [
"Solution:\nFirst, check that no permutation of $(1,2,3)$ works, so the sum must be more than $6$. Then since $(a, b, c) = (2, 4, 1)$ has $\\sqrt{2 \\sqrt{4 \\sqrt{1}}} = 2$, the answer must be $2+4+1=7$."
] | United States | HMMT February 2023 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 7 | |
051u | The midpoints of sides $C_2C_3$, $C_3C_1$ and $C_1C_2$ of a triangle $C_1C_2C_3$ are $K_1$, $K_2$ and $K_3$, respectively. The centers of circles $c_1$, $c_2$ and $c_3$ are $C_1$, $C_2$ and $C_3$, respectively, and the centers of circles $k_1$, $k_2$, $k_3$ are $K_1$, $K_2$, $K_3$, respectively. No two of the given six... | [
"Let the radii of the circles $c_1$, $c_2$, $c_3$ be $r_1$, $r_2$, $r_3$, and the radii of the circles $k_1$, $k_2$, $k_3$ be $R_1$, $R_2$, $R_3$, respectively (Fig. 28). By assumptions,\n$$\n\\begin{aligned}\nR_1 + R_2 &= |K_1K_2| = \\frac{1}{2}|C_1C_2|, \\\\\nR_2 + R_3 &= |K_2K_3| = \\frac{1}{2}|C_2C_3|, \\\\\nR_... | Estonia | Final Round of National Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof only | null | |
0c70 | Find the greatest integer $n$ for which the following condition holds: one can find $n$ distinct lines starting at the same point such that the angle made by any two of them, measured in degrees, is an integer that is not a prime. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 90 | |
0ar5 | Problem:
Let $a = 444 \cdots 444$ and $b = 999 \cdots 999$ (both have 2010 digits). What is the 2010th digit of the product $a b$? | [
"Solution:\nLet us first express $a$ and $b$ in terms of their digits.\n\n$a$ consists of $2010$ digits, all $4$'s. So,\n$$\na = \\underbrace{444\\cdots 444}_{2010\\ \\text{digits}} = 4 \\times \\underbrace{111\\cdots 111}_{2010\\ \\text{digits}}\n$$\nLet $N = \\underbrace{111\\cdots 111}_{2010\\ \\text{digits}}$. ... | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | final answer only | 3 | |
00am | Let $ABC$ be a right triangle with $\hat{C} = 90^\circ$. Points $D$ and $E$ on the hypotenuse $AB$ are such that $AD = AC$ and $BE = BC$. Points $P$ and $Q$ on $AC$ and $BC$ respectively are such that $AP = AE$ and $BQ = BD$. Let $M$ be the midpoint of segment $PQ$. Find $\hat{AM}\hat{B}$. | [
"We show that $M$ coincides with the incenter $I$ of the triangle. Since $\\hat{A} + \\hat{B} = 90^\\circ$, this implies $\\hat{AM}\\hat{B} = \\hat{AI}\\hat{B} = 180^\\circ - \\frac{1}{2}(\\hat{A} + \\hat{B}) = 135^\\circ$.\n\n\n\nBy hypothesis $AD = AC$, meaning that $D$ is the reflection ... | Argentina | Argentine National Olympiad 2016 | [
"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 | 135° | |
0bkh | Let $ABCD$ be a cyclic quadrilateral, inscribed in the circle of diameter $AC$. We suppose that there exist the points $E \in (CD)$ and $F \in (BC)$ such that the lines $AE$ and $DF$ are perpendicular to the lines $AF$ and $BE$ respectively. Prove that $AB = AD$. | [
"From the perpendicularity conditions, we have\n$$\n\\overrightarrow{AE} \\cdot \\overrightarrow{DF} = 0 \\Leftrightarrow \\overrightarrow{AE} \\cdot (\\overrightarrow{AF} - \\overrightarrow{AD}) = 0,\n$$\n$$\n\\overrightarrow{AF} \\cdot \\overrightarrow{BE} = 0 \\Leftrightarrow \\overrightarrow{AF} \\cdot (\\overr... | Romania | 65th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fgd | Problem:
Indicamos por $m(a, b)$ la media aritmética de los números reales positivos $a$ y $b$. Dada la función real positiva $g$ que tiene la primera y la segunda derivada positivas, definimos la media $\mu(a, b)$ relativa a la función $g$ mediante
$$
2 g(\mu(a, b))=g(a)+g(b)
$$
Decir, razonadamente, cuál de las dos ... | [
"Solution:\n\nSi $f$ es una función real de variable real tal que $f'>0$ y $f''>0$ entonces es una función convexa, es decir, la gráfica está por debajo de cualquier cuerda, tal como está representado en la figura.\n\n\n\nEn el caso que $\\lambda=\\frac{1}{2}$, tendremos el resultado del en... | Spain | OME 22 | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | μ(a,b) ≥ m(a,b), with equality only when a = b. | |
0cao | Problem:
Determinați funcțiile $f: \mathbb{R} \rightarrow \mathbb{R}$ pentru care există funcția $g: \mathbb{R} \rightarrow \mathbb{R}$ astfel încât
$$
f(x)+f(y)=[g(x+y)]
$$
oricare ar fi $x, y$ numere reale.
(Am notat cu $[a]$ partea întreagă a numărului real $a$.) | [
"Solution:\n\nPentru $x \\rightarrow (x+y), y \\rightarrow 0$ în relația din enunț deducem\n$$\nf(x+y)+f(0)=[g(x+y)]=f(x)+f(y)\n$$\nNotând $h(x)=f(x)-f(0)$ obținem $h(x+y)=h(x)+h(y)$ (ecuația lui Cauchy).\n\nDeducem că, pentru orice $x_{0} \\in \\mathbb{R}$ ales arbitrar și orice $n$ natural, $h\\left(n x_{0}\\righ... | Romania | Olimpiada Națională de Matematică | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All constant functions f with f(x) = k/2 for some fixed integer k; for these, one can take any g with values in the interval [k, k+1). | |
0fya | Problem:
Auf wieviele Arten kann man jeder Ecke eines Würfels eine der Zahlen $1,2,3, \ldots, 10$ zuordnen, sodass keine Zahl mehrfach verwendet wird, und so dass für jede Seitenfläche die Summe der Zahlen in den vier angrenzenden Ecken ungerade ist? | [
"Solution:\n\nWir nennen diese Summen von vier Zahlen in den Ecken einer Seitenfläche kurz Flächensummen. Ob eine Flächensumme ungerade ist oder nicht, hängt nur davon ab, wie viele der Zahlen in den Ecken ungerade sind. Für jede Fläche müssen also entweder 1 oder 3 angrenzende Eckzahlen ungerade sein.\n\nFixiere n... | Switzerland | Vorrundenprüfung | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 115200 | |
01hg | Let $a, b, c$ be real numbers, representing the side lengths of a triangle. Prove that
$$ 4(a + b)(a + c)(b + c) \ge (a + b + c)^3. $$ | [
"Given are three real numbers $a, b, c$ such that $0 < a, b, c < \\frac{a+b+c}{2}$ (the inequality $a, b, c < \\frac{a+b+c}{2}$ is equivalent to the triangle inequality). Introduce the variables\n$$\nx = \\frac{-a + b + c}{2(a + b + c)}, \\quad y = \\frac{a - b + c}{2(a + b + c)} \\quad \\text{and} \\quad z = \\fra... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
08sd | We have a square grid of $19 \times 19$. We call a rectangle **good** if its edges are all along the border of small squares. What is the smallest $n$ with the following property?
Property: If $9$ small squares are removed from the grid, we can divide remaining area into $n$ or less number of good rectangles. | [
"Let $(i, j)$ denote the square on the intersection of the $i$-th row and the $j$-th column.\n\nFirst, let us show that $n = 28$ works. Remove the squares one by one. The initial grid can be divided into $1$ good rectangle. If $1$ square is removed from a good rectangle, we can divide the remaining part into $4$ or... | Japan | Japan 2007 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 28 | |
0d2b | The quadratic equation $a x^{2} + b x + c = 0$ has its roots in the interval $[0, 1]$. Find the maximum of
$$
\frac{(a-b)(2a-b)}{a(a-b+c)}
$$ | [
"Let $u, v$ be the roots of the quadratic equation $a x^{2} + b x + c = 0$ such that $0 \\leq u \\leq v \\leq 1$. We have the relations $b = -a(u + v)$ and $c = u v$. Therefore\n$$\n\\begin{aligned}\n\\frac{(a-b)(2a-b)}{a(a-b+c)} &= \\frac{(1 + u + v)(2 + u + v)}{1 + u + v + u v} \\\\\n&= 2 + \\frac{u}{1 + v} + \\f... | Saudi Arabia | Preselection tests for the full-time training | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 3 | |
0d2x | $\triangle ABC$ is a triangle, $M$ the midpoint of $BC$, $D$ the projection of $M$ on $AC$ and $E$ the midpoint of $MD$. Prove that the lines $AE$, $BD$ are orthogonal if and only if $AB = AC$. | [
"First solution. Let $N$ be the midpoint of $CD$ and $F$ the intersection point of lines $AE$ and $MN$. Since $M$ is the midpoint of $BC$, the segment $MN$ is parallel to $BD$.\n\n\n\nTherefore, $AE$, $BD$ are perpendicular if and only if $AF$, $MN$ are perpendicular.\nEquivalently, triangl... | Saudi Arabia | Preselection tests for the full-time training | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01cr | The set $\{1, 2, \ldots, 10\}$ is split to three parts. For each part the sum of its elements, the product of its elements and the sum of the digits of all its elements are calculated.
Is it possible that the first part has the largest sum of elements, the second part has the largest product of elements, and the third ... | [
"Yes! For example\n\n| set | sum | digsum | prod |\n|-------------|------|--------|-------|\n| 1, 9, 10 | 20✓ | 11 | 90 |\n| 3, 7, 8 | 18 | 18✓ | 168 |\n| 2, 4, 5, 6 | 17 | 17 | 240✓ |"
] | Baltic Way | Baltic Way 2016 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Yes; for example, the three parts can be {1, 9, 10}, {3, 7, 8}, and {2, 4, 5, 6}. | |
0i6m | Problem:
In how many ways can the numbers $1,2, \ldots, 2002$ be placed at the vertices of a regular $2002$-gon so that no two adjacent numbers differ by more than $2$? (Rotations and reflections are considered distinct.) | [
"Solution:\n\n$4004$. There are $2002$ possible positions for the $1$. The two numbers adjacent to the $1$ must be $2$ and $3$; there are two possible ways of placing these. The positions of these numbers uniquely determine the rest: for example, if $3$ lies clockwise from $1$, then the number lying counterclockwis... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 4004 | |
073e | Given a finite string $S$ of symbols $X$ and $O$, we write $\Delta(S)$ for the number $X$'s in $S$ minus the number of $O$'s. (For example, $\Delta(XOOXOOX) = -1$.) We call a string $S$ balanced if every sub-string $T$ of (consecutive symbols) $S$ has the property $-1 \le \Delta(T) \le 2$. (Thus $XOOXOOX$ is not balanc... | [
"The balanced strings consists of $X$'s and $O$'s arranged alternately, or with as many as two consecutive letters of the same kind. Such occurrences of double letters must happen alternately with an even number (possibly zero) of single letters between any two double occurrence. If $b_n$ is the number of balanced ... | India | Indija TS 2007 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | For length 2n−1: 2^(n+1) − 2. For length 2n: 3·2^n − 2. | |
0h1l | Does there exist a quadruple with non parallel sides, that can be partitioned into 2011 isosceles triangles? | [
"**Answer:** See fig. 13."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | Yes | |
0k3e | Problem:
In a wooden block shaped like a cube, all the vertices and edge midpoints are marked. The cube is cut along all possible planes that pass through at least four marked points. Let $N$ be the number of pieces the cube is cut into. Estimate $N$.
An estimate of $E>0$ earns $\lfloor 20 \min (N / E, E / N)\rfloor$... | [
"Solution:\n\n15600"
] | United States | HMMT February 2018 | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | final answer only | 15600 | |
0iwg | Problem:
Circle $\Omega$ has radius $13$. Circle $\omega$ has radius $14$ and its center $P$ lies on the boundary of circle $\Omega$. Points $A$ and $B$ lie on $\Omega$ such that chord $AB$ has length $24$ and is tangent to $\omega$ at point $T$. Find $AT \cdot BT$. | [
"Solution:\n\nLet $M$ be the midpoint of chord $AB$; then $AM = BM = 12$ and by the Pythagorean theorem on triangle $AMO$ we have $MO = 5$.\n\nNote that $\\angle AOM = \\angle AOB / 2 = \\angle APB = \\angle APT + \\angle TPB$, or $\\tan(\\angle AOM) = \\tan(\\angle APT + \\angle TPB)$. Applying the tangent additio... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 56 | |
0fn3 | Let $p, n$ be positive integers such that $p$ is prime and $p < n$. If $p$ divides $n+1$ and $\left(\left[\frac{n}{p}\right], (p-1)!\right) = 1$, then prove that $p \cdot \left[\frac{n}{p}\right]^2$ divides $\binom{n}{p} - \left[\frac{n}{p}\right]$.
(Here $[x]$ represents the integer part of the real number $x$.) | [
"Since $p \\mid n + 1$, then $p \\mid n + 1 - p$. So, there exists $k \\in \\mathbb{N}$ such that $n = k p + p - 1$ and $\\left\\lfloor \\frac{n}{p} \\right\\rfloor = k$. Now, we have\n$$\n\\begin{aligned}\n\\binom{n}{p} - \\left\\lfloor \\frac{n}{p} \\right\\rfloor &= \\binom{k p + p - 1}{p} - k \\\\\n&= \\frac{(k... | Spain | International Mathematical Arhimede Contest | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | Spanish | proof only | null | |
0alt | Problem:
Find all ordered pairs $(m, n)$ of integers such that $4^{m} - 4^{n} = 255$. | [
"Solution:\nLet $m \\geq n$. Then $4^{m} - 4^{n} = 255$.\n\nLet $k = m - n \\geq 0$, so $4^{m} = 4^{n} \\cdot 4^{k}$, and the equation becomes:\n$$\n4^{n}(4^{k} - 1) = 255\n$$\nFactor $255 = 3 \\times 5 \\times 17$.\n\nSince $4^{n}$ is a power of $4$, i.e., $1, 4, 16, 64, \\ldots$, it must divide $255$.\n\nLet us c... | Philippines | Area Stage | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (4, 0) | |
0i5a | Problem:
How many triples $(A, B, C)$ of positive integers (positive integers are the numbers $1,2,3,4, \ldots$) are there such that $A+B+C=10$, where order does not matter (for instance the triples $(2,3,5)$ and $(3,2,5)$ are considered to be the same triple) and where two of the integers in a triple could be the sam... | [
"Solution:\n\nThe triples can merely be enumerated: $(1,1,8)$, $(1,2,7)$, $(1,3,6)$, $(1,4,5)$, $(2,2,6)$, $(2,3,5)$, $(2,4,4)$, and $(3,3,4)$. There are 8 elements."
] | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | final answer only | 8 | |
0c06 | Show that $\sqrt{n + [\sqrt{n} + \frac{1}{2}]}$ is an irrational number, for every positive integer $n$. | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0fsu | Problem:
Gegeben sind zwei Kreise $k_{1}$ und $k_{2}$, die sich in den verschiedenen Punkten $P$ und $Q$ schneiden. Konstruiere eine durch $P$ verlaufende Strecke $AB$ mit ihren Endpunkten auf $k_{1}$ und $k_{2}$, sodass das Produkt $|AP| \cdot |PB|$ maximal ist. | [
"Solution:\n\nSeien $t_{1}$ und $t_{2}$ die Tangenten an $k_{1}$ und $k_{2}$ in $P$. Diese zerlegen die Ebene in vier Gebiete, von denen eines den Punkt $Q$ enthält. Sei $G$ die Vereinigung der beiden gegenüberliegenden Gebiete, welche den Punkt $Q$ nicht enthalten. Offenbar muss die Strecke $AB$ ganz in $G$ verlau... | Switzerland | IMO - Selektion | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Draw the tangents to both circles at the common intersection and take the line through that point which bisects the angle between these tangents within the sector not containing the other intersection. Its intersections with the two circles give the segment maximizing the product. | |
07ki | Let $r$ and $n$ be nonnegative integers such that $r \le n$.
a. Prove that
$$
\frac{n+1-2r}{n+1-r} \binom{n}{r}
$$
is an integer.
b. Prove that
$$
\sum_{r=0}^{\lfloor n/2 \rfloor} \frac{n+1-2r}{n+1-r} \binom{n}{r} < 2^{n-2}
$$
for all $n \ge 9$.
(Note that $\binom{n}{r} = \frac{n!}{r!(n-r)!}$. Also, if $x$ is a real... | [
"Define $S(n,r) := \\frac{n+1-2r}{n+1-r} \\binom{n}{r}$, for all pairs $r, n$ with $-1 \\le r$ and $2r \\le n+1$. In particular $S(n,r) = 0$ if $r = -1$ or if $2r = n+1$. Note that $S(1,0) = 1$. We prove that if $n \\ge 2$ and $0 \\le r \\le n/2$ then\n$$\nS(n,r) = S(n-1, r-1) + S(n-1, r)\n$$\nThe first statement w... | Ireland | Irish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0geg | 設 $ABCD$ 為菱形, 其中心為點 $O$。設點 $P$ 落在 $AB$ 邊上。令點 $I, J, L$ 分別為三角形 $PCD, PAD, PBC$ 的內心。設點 $H$ 與 $K$ 分別是三角形 $PLB$ 與 $PJA$ 的垂心。證明直線 $OI$ 與 $HK$ 互相垂直。
Let $ABCD$ be a rhombus with center $O$. $P$ is a point lying on the side $AB$. Let $I, J$, and $L$ be the incenters of triangles $PCD, PAD$, and $PBC$, respectively. Let $H$ a... | [
"**Lemma 1.** Let $ABC$ be a triangle with incenter $I$. $IB, IC$ meet circle diameter $BC$ at $S, T$, respectively. $P$ is any point on circle diameter $BC$. $M$ is midpoint of $BC$. $MP$ meets $A$-midline at $Q$. $AQ$ meets $BC$ at $R$. $K, L$ lie on line $CS, BT$ such that $RK \\perp PC, RL \\perp PB$. Prove tha... | Taiwan | 2021 數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > M... | null | proof only | null | |
0jfn | Problem:
A 24-hour digital clock shows times $h: m: s$, where $h, m$, and $s$ are integers with $0 \leq h \leq 23$, $0 \leq m \leq 59$, and $0 \leq s \leq 59$. How many times $h: m: s$ satisfy $h+m=s$? | [
"Solution:\n\nAnswer: $1164$\n\nWe are solving $h+m=s$ in $0 \\leq s \\leq 59$, $0 \\leq m \\leq 59$, and $0 \\leq h \\leq 23$.\n\nIf $s \\geq 24$, each $h$ corresponds to exactly 1 solution, so we get $24(59-23)=24(36)$ in this case.\n\nIf $s \\leq 23$, we want the number of nonnegative integer solutions to $h+m \... | United States | HMMT November 2013 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1164 | |
08ql | Problem:
Find the largest integer $k$ ($k \geq 2$), for which there exists an integer $n$ ($n \geq k$) such that from any collection of $n$ consecutive positive integers one can always choose $k$ numbers, which verify the following conditions:
1. each chosen number is not divisible by $6$, by $7$ and by $8$;
2. the pos... | [
"Solution:\nAn integer is divisible by $6$, $7$ and $8$ if and only if it is divisible by their Least Common Multiple, which equals $6 \\times 7 \\times 4 = 168$.\nLet $n$ be a positive integer and let $A$ be an arbitrary set of $n$ consecutive positive integers. Replace each number $a_{i}$ from $A$ with its remain... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 108 | |
0gbp | 令 $a_0, a_1, a_2, \dots$ 是整數數列且 $b_0, b_1, b_2, \dots$ 是正整數數列使得 $a_0 = 0, a_1 = 1$, 以及對於所有的 $n = 1, 2, \dots$, 都有
$$
a_{n+1} = \begin{cases} a_n b_n + a_{n-1} & \text{若 } b_{n-1} = 1; \\ a_n b_n - a_{n-1} & \text{若 } b_{n-1} > 1. \end{cases}
$$
試證: 在 $a_{2017}$ 與 $a_{2018}$ 兩個數中至少有一個數大於或等於 2017. | [
"The value of $b_0$ is irrelevant since $a_0 = 0$, so we may assume that $b_0 = 1$.\n\n*Lemma*. We have $a_n \\ge 1$ for all $n \\ge 1$.\n\n*Proof*. Let us suppose otherwise in order to obtain a contradiction. Let\n$$\nn \\ge 1 \\text{ be the smallest integer with } a_n \\le 0. \\quad (1)\n$$\nNote that $n \\ge 2$.... | Taiwan | 二〇一八數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0hg7 | For any real numbers $x, y$ prove the inequality
$$
\sqrt{(x + 4)^2 + (y + 2)^2} + \sqrt{(x - 5)^2 + (y + 4)^2} \le \sqrt{(x - 2)^2 + (y - 6)^2} + \sqrt{(x - 5)^2 + (y - 6)^2} + 20.
$$ | [
"On a coordinate plane consider points $A(-4, -2)$, $B(2, 6)$, $C(5, 6)$ and $D(5, -4)$ (fig. 2). For any point $M(x, y)$ of the plane the inequality is rewritten as: $MA + MD - MB - MC \\le 20$. Let's find the largest possible value of the expression\n\n\nFig. 2\n$$\nS(M) = MA + MD - MB - ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
0gec | 設 $\triangle ABC$ 為等腰三角形, 其中 $BC = CA$, 且設 $D$ 為邊 $AB$ 內部的一點, 滿足 $AD < DB$。在邊 $BC$ 與 $CA$ 上分別取點 $P$ 與 $Q$ 滿足 $\angle DPB = \angle DQA = 90^\circ$。設線段 $PQ$ 的中垂線與直線 $CQ$ 交於點 $E$, 且設三角形 $ABC$ 的外接圓與 $CPQ$ 的外接圓再交點 $F$, 其中 $F \neq C$。
已知點 $P, E, F$ 共線。試證 $\angle ACB = 90^\circ$。
Let $ABC$ be an isosceles triangle with $BC =... | [
"Let $\\ell$ be the perpendicular bisector of $PQ$, and denote by $\\omega$ the circle $CFPQ$. By $DP \\perp BC$ and $DQ \\perp AC$, the circle $\\omega$ passes through $D$; moreover, $CD$ is a diameter of $\\omega$.\n\nThe lines $QE$ and $PE$ are symmetric about $\\ell$, and $\\ell$ is a symmetry axis of $\\omega$... | Taiwan | 2021 數學奧林匹亞競賽第二階段選訓營, 國際競賽實作 (二) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
0804 | Problem:
a. Determinare tutte le coppie $(x, k)$ di interi positivi che soddisfano l'equazione
$$
3^{k}-1=x^{3}.
$$
b. Dimostrare che se $n$ è un intero maggiore di 1 e diverso da 3 non esistono coppie $(x, k)$ di interi positivi che soddisfano l'equazione
$$
3^{k}-1=x^{n}.
$$ | [
"Solution:\n\na. Riscriviamo l'equazione nella forma\n$$\n3^{k}=x^{3}+1=(x+1)\\left(x^{2}-x+1\\right)\n$$\nPoiché gli unici divisori di una potenza di 3 sono a loro volta potenze di 3, ogni soluzione deve soddisfare il sistema\n$$\n\\left\\{\\begin{array}{l}\nx+1=3^{a} \\\\\nx^{2}-x+1=3^{b}\n\\end{array}\\right.\n$... | Italy | XV GARA NAZIONALE DI 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 | a) (x, k) = (2, 2). b) No solutions for any exponent greater than one different from three. | |
00ls | Let $ABCD$ be a cyclic quadrilateral with perpendicular diagonals and circumcenter $O$. Let $g$ be the line obtained by reflection of the diagonal $AC$ about the angle bisector of $\angle BAD$.
*Prove that the point $O$ lies on the line $g$.* | [
"Denote by $X$ the point of intersection of the diagonals $AC$ and $BD$, i.e. $AX$ is an altitude in the triangle $ABD$, see Figure 1.\n\n\n\nFigure 1: Problem 2\n\n$$\\angle ABX = \\frac{1}{2} \\angle DOA.$$ Hence\n$$\n\\angle XAB = 90^\\circ - \\angle ABX = \\frac{1}{2} \\cdot (180^\\circ... | Austria | 48th Austrian Mathematical Olympiad Regional Competition (Qualifying Round) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0jbs | Let $a$, $b$, $c$ be real numbers in the interval $[0, 1]$ with $a + b$, $b + c$, $c + a \ge 1$. Prove that
$$
1 \le (1-a)^2 + (1-b)^2 + (1-c)^2 + \frac{2\sqrt{2abc}}{\sqrt{a^2+b^2+c^2}}.
$$ | [
"We may assume without loss of generality that $a \\ge b \\ge c$. Noting that $(1-b)^2 + (1-c)^2 = (b+c-1)^2 - 2bc + 1$, we may transform the right hand side of the desired inequality into\n$$\n(1-a)^2 + (1-b)^2 + (1-c)^2 + \\frac{2\\sqrt{2abc}}{\\sqrt{a^2 + b^2 + c^2}} = \\frac{2\\sqrt{2abc}}{\\sqrt{a^2 + b^2 + c^... | United States | Team Selection Test Selection Test | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0gdd | 老趙當兵太無聊了, 因此他在地上畫了 $n + 1$ 個格子排成一列 ($n$ 為正整數), 從左至右依序編號為第 0 格到第 $n$ 格。起始的時候, 第 0 格有 $n$ 顆石頭, 其他格子則都是空的。在每一回合, 老趙先選擇一格非空的格子, 假設其中有 $k$ 顆石頭。接著, 他從選定的格子中拿取一顆石頭, 並將它向右移動至多 $k$ 格 (石頭不得超出最右邊的格子。) 老趙的目標是將所有石頭都放到第 $n$ 格。
試證明, 老趙要達成目標, 所需的回合數不少於
$$
\left\lceil \frac{n}{1} \right\rceil + \left\lceil \frac{n}{2} \right\rceil + \l... | [
"讓我們將石頭編號 1 到 $n$，並且不失一般性，假設老趙在移動石頭時，都是移動該格內編號最大的石頭。現在，注意到當老趙移動編號 $k$ 的石頭時，該格內至多只有 $k$ 顆石頭，因此編號 $k$ 的石頭每次至多只能移動 $k$ 步。換言之，編號 $k$ 的石頭要移動到第 $n$ 格，至少要 $\\left\\lceil \\frac{n}{k} \\right\\rceil$ 步，從而原命題得證。"
] | Taiwan | 二〇一九數學奧林匹亞競賽第二階段選訓營,模擬競賽(二) | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
06ed | Let $a$, $b$, $c$, $d$ be positive real numbers such that $a + b + c + d = 1$. Prove that
$$
6(a^3 + b^3 + c^3 + d^3) \ge (a^2 + b^2 + c^2 + d^2) + \frac{1}{8}.
$$ | [
"We claim that $f(x) = 6x^3 - x^2 \\ge \\frac{5x-1}{8}$ for any $x > 0$. Indeed,\n$$\n\\begin{align*}\n& 6x^3 - x^2 \\ge \\frac{5x-1}{8} \\\\\n\\Leftrightarrow \\quad & 48x^3 - 8x^2 - 5x + 1 \\ge 0 \\\\\n\\Leftrightarrow \\quad & (4x-1)^2(3x+1) \\ge 0.\n\\end{align*}\n$$\nThis clearly holds. It follows that\n$$\nf(... | Hong Kong | CHKMO | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0dni | Problem:
Дат је тетивни четвороугао $A B C D$. Тачке $M, N, P$ и $Q$ су средишта страница $D A, A B, B C$ и $C D$, редом, а тачка $E$ је пресек дијагонала $A C$ и $B D$. Кружнице описане око $\triangle E M N$ и $\triangle E P Q$ секу се у тачки $F \neq E$. Доказати да важи $E F \perp A C$. | [
"Solution:\n\nТроуглови $E A B$ и $E D C$ су слични, па су то и троуглови $E B N$ и $E C Q$. Зато у оријентисаним угловима важи $\\varangle M F E=\\varangle M N E=\\varangle B E N=\\varangle Q E C=\\varangle E Q M$. Аналогно важи $\\varangle Q F E=\\varangle E M Q$, одакле следи да је $F$ ортоцентар троугла $E M Q$... | Serbia | 9. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformat... | null | proof only | null | |
01tm | The vertices of the regular $2n+1$-gon are marked on a circle. Two players play the following game. They, in turn, delete exactly one of the vertices. The player wins if after his move all triangles with the vertices in the remained points are obtuse.
Who of the player wins if both of them play to win? | [
"4. See the 38th All-Russian Mathematical Olympiad, Final Round, Problem 11.7."
] | Belarus | 66th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | Second player | |
09mw | In a family of $35$ cats, Mau weighed $250$ grams less than the average weight of all cats in the family including Mau. After going out alone and finding food, Mau returned weighing $600$ grams more than the average weight of all cats in the family including Mau. Apart from Mau, the weight of the other cats remained un... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 875 | |
0ldo | For a positive integer $n$, consider $a_1, a_2, \dots, a_{2n}$ as a permutation of the first $2n$ positive integers. A permutation is called "nice" if $a_i + a_{n+i} = 2n+1$, $i = 1, 2, 3, \dots, n$ and $a_i - a_{i+1}$ is not congruent to $a_j - a_{j+1}$ modulo $2n+1$ for all $1 \le i < j \le 2n$ (suppose that $a_{2n+1... | [
"1) We consider the following permutation $1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7$.\nIt is easy to check that $1+12=2+11=4+9=8+5=3+10=6+7$. Furthermore, the difference between two consecutive numbers in modulo $13$ in order are $1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7$, which are all distinct. Therefore, this permutat... | Vietnam | Vietnamese Team Selection Test for IMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Modular ... | null | proof and answer | For n = 6, one nice permutation is 1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7. Moreover, for every positive integer n, a nice permutation exists. | |
0jsc | Problem:
Let $R$ be the rectangle in the Cartesian plane with vertices at $(0,0)$, $(2,0)$, $(2,1)$, and $(0,1)$. $R$ can be divided into two unit squares, as shown; the resulting figure has seven edges.
Compute the number of ways to choose one or more of the seven edges such that the resul... | [
"Solution:\n\nWe have two cases, depending on whether we choose the middle edge. If so, then either all the remaining edges are either to the left of or to the right of this edge, or there are edges on both sides, or neither; in the first two cases there are 6 ways each, in the third there are $16+1=17$ ways, and i... | United States | HMMT February | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | 61 | |
04n4 | Let $ABC$ be a right-angled triangle with the right angle at $C$, such that $|AC| = 2$ and $|BC| = 1$. A square is situated inside the triangle so that two of its vertices are on the side $AC$, the third one is on the side $AB$, and the fourth is on the circle of radius $1$ centred at $B$. Determine the length of the s... | [] | Croatia | Croatia_2018 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 2/5 | |
09gh | Find all positive integer numbers $a, b, c$ such that $a+b = (a,b)^2$, $b+c = (b,c)^2$, $c+a = (c,a)^2$. Here denoted by $(a,b)$ the greatest common divisor of $a, b$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | a = b = c = 2 | |
0jk4 | Problem:
Let $ABC$ be a triangle with $AB = 13$, $BC = 14$, and $CA = 15$. We construct isosceles right triangle $ACD$ with $\angle ADC = 90^{\circ}$, where $D$, $B$ are on the same side of line $AC$, and let lines $AD$ and $CB$ meet at $F$. Similarly, we construct isosceles right triangle $BCE$ with $\angle BEC = 90^... | [
"Solution:\n\nAnswer: $\\quad -\\frac{5}{13}$\n\nWe see that $\\angle GAF = \\angle GBF = 45^{\\circ}$, hence quadrilateral $GFBA$ is cyclic. Consequently $\\angle AGF + \\angle FBA = 180^{\\circ}$. So $\\cos \\angle AGF = -\\cos \\angle FBA$. One can check directly that $\\cos \\angle CBA = \\frac{5}{13}$ (say, by... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | -5/13 | |
06l3 | Let $\triangle ABC$ be an isosceles triangle with $AB = AC$. The incircle $\Gamma$ of $\triangle ABC$ has centre $I$, and it is tangent to the sides $AB$ and $AC$ at $F$ and $E$ respectively. Let $\Omega$ be the circumcircle of $\triangle AFE$. The two external common tangents of $\Gamma$ and $\Omega$ intersect at a po... | [
"Since $\\angle AFI = \\angle AEI = 90^\\circ$, the point $I$ lies on $\\Omega$ and the centre of $\\Omega$ is the midpoint $O$ of $AI$. Note that $P$ lies on the line joining the centres of $\\Gamma$ and $\\Omega$. In other words, $P$ lies on the line $IO$. As $AB = AC$, the line $AI$ is simply the internal angle ... | Hong Kong | CHKMO | [
"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 > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0d13 | Consider the isosceles triangle $ABC$ with $AB = AC$. A semicircle of diameter $EF$ situated on the side $BC$, is tangent to the sides $AB$ and $AC$ at $M$ and $N$, respectively. The line $AE$ intersects the semicircle at $P$. Prove that the line $PF$ passes through the midpoint of the chord $MN$. | [
"Let $O$ be the center of the semicircle and let $R$ be the midpoint of segment $MN$.\n\n\n\nIn triangle $ANO$ we have $AN^2 = AR \\cdot AO$. Using the power of the point $A$ with respect to the circle we get\n$$\nAM^2 = AP \\cdot AE = AN^2 = AR \\cdot AO. \\quad (1)\n$$\nFrom (1) it follow... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0h99 | Find pairs of positive integers $x, y$, which satisfy the system of equations:
$$
\begin{cases}
[x, y] + (x, y) = 2018, \\
x + y = 2018,
\end{cases}
$$
Where $[x, y]$ and $(x, y)$ are LCM and GCD of numbers $x, y$. | [
"Since $(x, y) \\cdot [x, y] = xy$, then the first equation of the system can be re-written as\n$$ \\frac{xy}{(x, y)} + (x, y) = 2018, $$\nwhich gives us quadratic equation with respect to $(x, y)$:\n$$\n(x, y)^2 - 2018(x, y) + xy = 0.\n$$\nIts discriminant, also taking into account the second equation of the syste... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (x, y) ∈ {(1, 2017), (2017, 1), (2, 2016), (2016, 2), (1009, 1009)} | |
08s6 | Let $[r]$ be the largest integer not exceeding real number $r$. For real positive numbers $x$, the set $A(x)$ is defined by
$$
A(x) = \{[nx] \mid n: \text{positive integer}\}.
$$
Find all irrational numbers $\alpha > 1$ satisfying the following condition.
Condition: If a positive real number $\beta$ satisfies $A(\alpha... | [
"First, we prove that any irrational number $\\alpha > 2$ satisfies the condition. Let $A(\\alpha) \\supset A(\\beta)$, $[\\beta] = [m\\alpha]$ ($m$: positive integer). We can prove this by proving that $[k\\beta] = [km\\alpha]$ for any positive integer $k$. We are going to prove this by induction on $k$. Assume th... | Japan | Japan 2007 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | All irrational real numbers greater than 2 | |
0de4 | Problem:
Let coefficients of the polynomial $P(x) = a_d x^d + \ldots + a_2 x^2 + a_0$ where $d \ge 2$, are positive integers. The sequence $(b_n)$ is defined by $b_1 = a_0$ and $b_{n+1} = P(b_n)$ for $n \ge 1$. Prove that for any $n \ge 2$, there exists a prime number $p$ such that $p \mid b_n$ but it does not divide ... | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
08d0 | Problem:
Sia $ABC$ un triangolo tale che $AB = AC$ e sia $I$ il suo incentro. Sia $\Gamma$ la circonferenza circoscritta ad $ABC$. Le rette $BI$ e $CI$ intersecano $\Gamma$ in due nuovi punti, denotati rispettivamente $M$ ed $N$. Sia $D$ un altro punto di $\Gamma$, giacente sull'arco $BC$ che non contiene $A$, e siano... | [] | Italy | Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscel... | null | proof only | null | |
0hqd | Problem:
Call a positive integer one-full if it satisfies the following criteria:
a. Every digit is either $0$, $1$, or $2$.
b. Out of every two consecutive digits, at least one of them is a $1$.
For $n \geq 2$, calculate the number of $n$-digit one-full numbers. (Numbers beginning with the digit $0$ are not allowe... | [
"Solution:\n\nThe answer is $2^{n}$. It is clear that there are four $2$-digit one-full numbers: $10$, $11$, $12$, and $21$.\n\nTo prove that the required number is $2^{n}$ for all $n \\geq 3$, it is enough to show that there are twice as many $n$-digit one-full numbers as $(n-1)$-digit one-full numbers for every $... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2^n |
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