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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
01j4 | In an acute triangle $\triangle ABC$ with $|AB| \neq |AC|$, the angle bisector of $\angle BAC$ intersects side $BC$ and $\odot(ABC)$ at the points $D$ and $M_A$, respectively. Let points $X$ and $Y$ be the feet of perpendiculars from $M_A$ to sides $AB$ and $AC$, respectively. The tangent of $\odot(BXM_A)$ at the point... | [
"Therefore, quadrilateral $M_AXTY$ is cyclic as well. Combining this with $AXM_AY$ being cyclic, gives us that $M_AXTAY$ is cyclic. Moreover, note that\n$$\n\\angle TAM_A = \\angle TYM_A = \\angle YCM_A = \\angle ACM_A.\n$$\nThis means that $AT$ is tangent to $\\odot(ABC)$. Also note that $\\angle STM_A = 180^\\cir... | Baltic Way | Baltic Way 2023 Shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0a4d | Problem:
Zij $\triangle ABC$ een scherphoekige driehoek zo dat $|AB| + |BC| = 4|AC|$ en $|AB| < |BC|$. Zij $D$ het snijpunt van de bissectrice van $\angle ABC$ met de zijde $AC$. Punten $P$ en $Q$ liggen op het lijnstuk $BD$ zo dat $|BP| = 2|DQ|$. Zij $\ell$ de lijn door $P$ parallel aan $AC$. De lijn door $Q$ loodrec... | [
"Solution:\n\nLaat $R$, $S$ en $T$ respectievelijk de punten zijn waar de mier de eerste keer op $AC$ is, op $\\ell$ is en de tweede keer op $AC$ is. Laat $\\ell'$ en $S'$ de spiegelingen van $\\ell$ en $S$ in $AC$ zijn, en zij $Y'$ de spiegeling van $Y$ in $\\ell'$. Dan geldt wegens de driehoekongelijkheid voor de... | Netherlands | IMO-selectietoets I | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hyl | Problem:
Let $A$ and $B$ be two different hospitals that treat exactly the same number of patients during a year. Each patient suffers from one of two diseases, $X$ or $Y$. Hospital $A$ cures a greater percentage of its patients than hospital $B$. Is it possible that hospital $B$ cures both a greater percentage of $X$... | [
"Solution:\n\nThis is the well-known Simpson's Paradox: just make $B$ specialize in a riskier disease. For example, let $B$ treat 90 cancer patients and 10 acne patients, with respective cure rates of $50\\%$ and $100\\%$. Let $A$ treat 10 cancer and 90 acne patients, with cure rates of $0\\%$ and $70\\%$, respecti... | United States | BAMO | [
"Statistics > Mathematical Statistics"
] | null | proof and answer | Yes | |
0gjr | 令 $a_1 < a_2 < a_3 < \dots$ 為正整數數列, 其中每個 $k \ge 1$, $a_{k+1}$ 都整除 $2(a_1 + a_2 + \dots + a_k)$。假設對於無窮多個質數 $p$, 存在某個 $k$ 使得 $p$ 整除 $a_k$。證明對於每一個正整數 $n$, 都存在某個 $k$ 使得 $n$ 整除 $a_k$。
Let $a_1 < a_2 < a_3 < \dots$ be positive integers such that $a_{k+1}$ divides $2(a_1 + a_2 + \dots + a_k)$ for every $k \ge 1$. Suppose tha... | [
"For every $k \\ge 2$ define the quotient $b_k = 2(a_1+\\cdots+a_{k-1})/a_k$, which must be a positive integer. We first prove the following properties of the sequence $(b_k)$:\n\n*Claim 1.* We have $b_{k+1} \\le b_k + 1$.\n\n*Proof.* By subtracting $b_k a_k = 2(a_1 + \\cdots + a_{k-1})$ from $b_{k+1} a_{k+1} = 2(a... | Taiwan | IMO 1J, Independent Study 2 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Chinese; English | proof only | null | |
000a | Un punto $P$ es interior al triángulo equilátero $ABC$ y cumple que $\angle APC = 120^\circ$. Sean $M$ la intersección de $CP$ con $AB$ y $N$ la intersección de $AP$ con $BC$. Hallar el lugar geométrico del circuncentro del triángulo $MBN$ al variar $P$. | [] | Argentina | XVII Olimpíada Iberoamericana de Matemática | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | español | proof only | null | |
0gwd | Mariyka has drawn a square grid $2006 \times 2006$ on a blackboard. During one step, it is allowed to choose any unit segment of that grid and erase it together with all adjacent segments, which were not erased before (so, at most $7$ segments can be erased in one step). Is it possible to erase the whole drawing in no ... | [
"Відповідь: ні, не можна. Задача полягає в тому, чи можемо ми покрити всю \"сітку\" таблиці $1300000$ фігурками, зображеними на рисунку. Якщо така фігурка не лежить повністю в таблиці, то вона покриває не більше $6$ відрізків. Якщо фігурка повністю знаходиться в таблиці, тоді інші фігурки повинні покрити відрізки $... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | No | |
007h | Consider the following sequence of tables:
<table><tr><td></td></tr></table>
1-table
<table><tr><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td></tr></table>
2-table
<table><tr><td></td><td></... | [
"A $k$-table has $2k^2 + 2k$ cells. Imagine them colored black and white in chessboard pattern, with the top right cell black. The white part can be regarded as the union of $k$ white diagonals with $k+1$ cells in each, running from top left to bottom right. Likewise the black part is the union of $k$ black diagona... | Argentina | Mathematical Olympiad Rioplatense | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Maximum = 2k^2 + 2k if k is odd; Maximum = 2k^2 if k is even. | |
00r1 | A quadrilateral $ABCD$ is given with $AD \parallel BC$. The midpoints of $AD$ and $BC$ are denoted by $M$ and $N$, respectively. The line $MN$ intersects the diagonals $AC$ and $BD$ in points $K$ and $L$, respectively. Prove that the circumcircles of the triangles $AKM$ and $BNL$ have a common point on the line $AB$. | [
"\nLet these two circles intersect also at $Q$. Then $\\angle AQP = \\angle PMD = \\angle PNB$ (by the concyclicity of $A$, $B$, $P$, $M$ and the similarity of $\\triangle ADP$, $\\triangle CBP$) and $\\angle BQP = \\angle PNB$ (by the concyclicity of $B$, $Q$, $N$, $P$), thus $\\angle AQP ... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point"
] | null | proof only | null | |
02yl | Problem:
Uma competição de matemática consiste de três problemas, cada um dos quais recebe uma nota inteira de 0 a 7. Para quaisquer dois competidores, sabemos que existe no máximo um problema em que eles obtiveram a mesma pontuação. Encontre o maior número possível de competidores nessa competição. | [
"Solution:\n\nExistem 8 pontuações possíveis para cada problema e, consequentemente, $8 \\cdot 8 = 64$ pontuações distintas possíveis para os dois primeiros problemas. Como não podem existir dois competidores com exatamente as mesmas pontuações nos dois primeiros problemas, o total de competidores não pode ser maio... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 64 | |
003o | Alrededor de una circunferencia están escritos los números $1, 2, \ldots, 2006$. Una operación permitida es intercambiar dos números adyacentes. Al cabo de una secuencia de tales intercambios, cada número quedó ubicado $13$ posiciones hacia la derecha de su posición inicial. Partimos los números $1, 2, \ldots, 2006$ en... | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Español | proof only | null | |
0ja2 | Problem:
Let $ABC$ be a triangle with $\angle A = 90^{\circ}$, $AB = 1$, and $AC = 2$. Let $\ell$ be a line through $A$ perpendicular to $BC$, and let the perpendicular bisectors of $AB$ and $AC$ meet $\ell$ at $E$ and $F$, respectively. Find the length of segment $EF$. | [
"Solution:\nAnswer: $\\frac{3 \\sqrt{5}}{4}$\nLet $M, N$ be the midpoints of $AB$ and $AC$, respectively. Then we have $\\angle EAB = \\angle ACB$ and $\\angle EAC = \\angle ABC$, so $AEM \\sim CBA \\Rightarrow AE = \\frac{\\sqrt{5}}{4}$ and $FAN \\sim CBA \\Rightarrow AF = \\sqrt{5}$. Consequently, $EF = AF - AE =... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3*sqrt(5)/4 | |
08uz | Let $H$ be the orthocenter of an acute triangle $ABC$, and $M$ be the midpoint of the side $BC$. Let $P$ be the point of intersection of the line $AM$ and the line through $H$ and perpendicular to the line $AM$. Prove that $AM \cdot PM = BM^2$ holds. Here for a line segment $XY$ its length is also denoted by $XY$. | [
"Let $X$ be the point of intersection of the lines $BH$ and $AC$, and let $N$ be the midpoint of the line segment $AH$. Since $\\angle AXH = \\angle APH = 90^\\circ$, the points $P$, $X$ lie on the circle having $AH$ as its diameter (if $AB = AC$, then $P$ coincides with $H$ and it is clear that $X$ lies on the cir... | Japan | Japan 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 > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kcs | Problem:
In triangle $ABC$, $AB = 32$, $AC = 35$, and $BC = x$. What is the smallest positive integer $x$ such that $1 + \cos^2 A$, $\cos^2 B$, and $\cos^2 C$ form the sides of a non-degenerate triangle? | [
"Solution:\n\nBy the triangle inequality, we wish $\\cos^2 B + \\cos^2 C > 1 + \\cos^2 A$. The other two inequalities are always satisfied, since $1 + \\cos^2 A \\geq 1 \\geq \\cos^2 B, \\cos^2 C$. Rewrite the above as\n$$\n2 - \\sin^2 B - \\sin^2 C > 2 - \\sin^2 A\n$$\nso it is equivalent to $\\sin^2 B + \\sin^2 C... | United States | HMMO 2020 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof and answer | 48 | |
0le0 | Assume $a$ is a real number in $[\frac{1}{2}, \frac{2}{3}]$. Consider two sequences $(u_n), (v_n), (n = 0, 1, \dots)$, defined by:
$$
u_n = \frac{3}{2^{n+1}} \cdot (-1)^{\lfloor 2^{n+1}a \rfloor}, \quad v_n = \frac{3}{2^{n+1}} \cdot (-1)^{n+\lfloor 2^{n+1}a \rfloor}.$$
a. Prove that
$$
\left(\sum_{i=0}^{2018} u_i\righ... | [
"(a) By assumption, we have $v_i = u_i$ for even $i$ and $v_i = -u_i$ for odd $i$. Thus, the inequality can be rewritten as\n$$\n\\left(\\sum_{i=0}^{1009} u_{2i} + \\sum_{i=0}^{1008} u_{2i+1}\\right)^2 + \\left(\\sum_{i=0}^{1009} u_{2i} - \\sum_{i=0}^{1008} u_{2i+1}\\right)^2 \\le 72a^2 - 48a + 10 + \\frac{2}{4^{20... | Vietnam | VN IMO Booklet | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a = (2/3) * (1 - 1/4^1010) | |
0e68 | Problem:
V trikotniku $ABC$ je kot $\alpha$ velik $30^{\circ}$, stranica $a$ je dolga $4~\mathrm{cm}$, stranica $c$ je dolga dvakrat toliko kot težiščnica na stranico $c$. Natančno izračunaj dolžine stranic trikotnika $ABC$. Nariši skico. | [
"Solution:\n\nTočka $D$ naj bo razpolovišče stranice $c$. Trikotnik $CAD$ je enakokrak, zato velja $\\angle ACD = 30^{\\circ}$ in $\\angle CDA = 120^{\\circ}$. Torej je $\\angle BDC = 60^{\\circ}$. Ker je tudi trikotnik $BCD$ enakokrak, velja $\\angle CBD = \\angle DCB = 60^{\\circ}$. To pomeni, da je trikotnik $AB... | Slovenia | Državno tekmovanje | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | a = 4 cm, b = 4√3 cm, c = 8 cm | |
0jwh | Problem:
Let $ABCD$ be a quadrilateral with an inscribed circle $\omega$. Let $I$ be the center of $\omega$ and let $IA = 12$, $IB = 16$, $IC = 14$, and $ID = 11$. Let $M$ be the midpoint of segment $AC$. Compute $\frac{IM}{IN}$, where $N$ is the midpoint of segment $BD$. | [
"Solution:\n\nLet points $W, X, Y, Z$ be the tangency points between $\\omega$ and lines $AB, BC, CD, DA$ respectively. Now invert about $\\omega$. Then $A'$, $B'$, $C'$, $D'$ are the midpoints of segments $ZW, WX, XY, YZ$ respectively. Thus by Varignon's Theorem $A'B'C'D'$ is a parallelogram. Then the midpoints of... | United States | February 2017 | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Transformations > Inversion"
] | null | proof and answer | 21/22 | |
0is3 | Problem:
Kermit the frog enjoys hopping around the infinite square grid in his backyard. It takes him 1 Joule of energy to hop one step north or one step south, and 1 Joule of energy to hop one step east or one step west. He wakes up one morning on the grid with 100 Joules of energy, and hops till he falls asleep with... | [
"Solution:\n\nIt is easy to see that the coordinates of the frog's final position must have the same parity. Suppose that the frog went to sleep at $(x, y)$. Then, we have that $-100 \\leq y \\leq 100$ and $|x| \\leq 100 - |y|$, so $x$ can take on the values $-100 + |y|, -98 + |y|, \\ldots, 100 - |y|$. There are $1... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 10201 | |
0e03 | Find all positive integers $m$ and $n$ such that $m^2 + n^5 = 252$. | [
"Since $m^2 = 252 - n^5$ is non-negative, we have $n^5 \\le 252$, so $n < 4$.\n\nIf $n=1$ we get $m^2 = 251$,\n\nif $n=2$ we have $m^2 = 220$,\n\nand if $n=3$ we have $m^2 = 9$.\n\nThe only possible solution is $m = n = 3$."
] | Slovenia | National Math Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | m=3, n=3 | |
0gk6 | Let $f: \mathbb{R} \to \mathbb{R}$ be a function satisfying
$$
|f(x+y) - f(x) - f(y)| < 1 \text{ for all } x, y \in \mathbb{R}.
$$
Prove that $\left| f\left(\frac{x}{2008}\right) - \frac{f(x)}{2008} \right| < 1$ for all $x \in \mathbb{R}$. | [
"$$\n\\begin{aligned}\n\\left| f(2008x) - 2008f(x) \\right| &= \\left| \\sum_{k=1}^{2007} \\left( f((k+1)x) - f(x) - f(kx) \\right) \\right| \\\\\n&\\le \\sum_{k=1}^{2007} \\left| f((k+1)x) - f(x) - f(kx) \\right| < 2007.\n\\end{aligned}\n$$\nReplacing $x$ with $\\frac{x}{2008}$ and simplifying, one gets\n$$\n\\lef... | Thailand | Thai Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null | |
0agv | Given real numbers $x, y, z$ such that $x + y + z = 0$, show that
$$
\frac{x(x+2)}{2x^2+1} + \frac{y(y+2)}{2y^2+1} + \frac{z(z+2)}{2z^2+1} \ge 0.
$$
When does equality hold? | [
"The inequality is clear if $xyz = 0$, in which case equality holds if and only if $x = y = z = 0$.\n\nHenceforth assume $xyz \\neq 0$ and rewrite the inequality as\n$$\n\\frac{(2x+1)^2}{2x^2+1} + \\frac{(2y+1)^2}{2y^2+1} + \\frac{(2z+1)^2}{2z^2+1} \\ge 3.\n$$\nNotice that (exactly) one of the products $xy, yz, zx$... | North Macedonia | XXVIII-th Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | Equality holds if and only if either all three are zero, or one of them is one and the other two are negative one half (in any order). | |
0hwz | Problem:
Are there positive integers $a$ and $b$ satisfying $a^{2}-23=b^{11}$? | [
"Solution:\n\nThe answer is no. We may write the given equation as\n$$\na^{2}+45^{2}=b^{11}+2^{11}.\n$$\nTaking modulo $4$, we have $b^{11} \\equiv a^{2}+45^{2} \\equiv a^{2}+1 \\pmod{4}$ which forces $b \\equiv 1 \\pmod{4}$ and $a \\equiv 0 \\pmod{2}$. Thus $b+2 \\equiv 3 \\pmod{4}$.\n\nLet $\\nu_{p}$ be the usual... | United States | Berkeley Math Circle | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Other"
] | null | proof and answer | No | |
0h92 | Find all triples pairwise distinct positive integer numbers $(a, b, c)$ that satisfy the condition: number $2a-1$ is divisible by $b$, number $2b-1$ is divisible by $c$ and number $2c-1$ is divisible by $a$. | [
"Let's rewrite the conditions in form of a system: there exist natural numbers $k, m, n$, for which equalities are true:\n$$\n2a-1=kb, \\quad 2b-1=nc \\text{ and } 2c-1=ma.\n$$\n\nIt is obvious, that all the numbers $k, m, n$ and $a, b, c$ are odd.\nThen we have, that\n$$\nb = \\frac{2a-1}{k} \\Rightarrow 2b-1 = \\... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (25, 7, 13) and its cyclic permutations: (7, 13, 25) and (13, 25, 7) | |
09js | Let $F$ be a point outside the square $ABCD$ and $E$ be a point inside the square $ABCD$ such that triangle $BCF$ and $ABE$ are equilateral triangles. If $M$ is the midpoint of $EF$, determine $\angle AMD$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 45° | |
01zc | Prove the inequality
$$
\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{2022!} > \frac{1^2}{2!} + \frac{2^2}{3!} + \frac{3^2}{4!} + \dots + \frac{2022^2}{2023!}.
$$ | [
"Let us prove that $\\sum_{n=1}^{2022} \\frac{n^2}{(n+1)!} - \\sum_{n=1}^{2022} \\frac{1}{n!} < 0$, which is equivalent to the required. Note that\n$$\n\\frac{n^2}{(n+1)!} - \\frac{1}{n!} = \\frac{n(n+1) - 2(n+1) + 1}{(n+1)!} = \\frac{1}{(n-1)!} - \\frac{2}{n!} + \\frac{1}{(n+1)!}\n$$\nWith this identity in mind, w... | Belarus | Belarus2022 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null | |
0k4a | Problem:
Farmer James invents a new currency, such that for every positive integer $n \leq 6$, there exists an $n$-coin worth $n!$ cents. Furthermore, he has exactly $n$ copies of each $n$-coin. An integer $k$ is said to be nice if Farmer James can make $k$ cents using at least one copy of each type of coin. How many ... | [
"Solution:\n\nWe use the factorial base, where we denote\n$$\n\\left(d_{n} \\ldots d_{1}\\right)_{*}=d_{n} \\times n!+\\cdots+d_{1} \\times 1!\n$$\nThe representation of $2018_{10}$ is $244002_{*}$ and the representation of $720_{10}$ is $100000_{*}$. The largest nice number less than $244002_{*}$ is $243321_{*}$. ... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Other"
] | null | proof and answer | 210 | |
08rf | Find three distinct positive integers which minimize their sum under the condition that any two of them add up to a perfect square. | [
"Let $a$, $b$ and $c$ be distinct positive integers with sum of any two of them being squares. We may assume that $a < b < c$. Write $a + b = x^2$, $b + c = y^2$, $c + a = z^2$. Then we shall minimize $x^2 + y^2 + z^2$ under the conditions $x < y < z$, $z^2 < x^2 + y^2$, and $x^2 + y^2 + z^2$ even. $z > 5$, since i... | Japan | The 16th Japanese Mathematical Olympiad - The First Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 6, 19, 30 | |
00ec | Let $\mathbb{Z}$ be the set of integer numbers. Determine all functions $f : \mathbb{Z} \to \mathbb{Z}$
such that
$$
f(x + f(y + 1)) + f(xy) = f(x + 1)(f(y) + 1)
$$
for any integers $x, y$. | [
"Let $P(x, y)$ denote the assertion\n$$\nf(x + f(y + 1)) + f(xy) = f(x + 1)(f(y) + 1).\n$$\nIf $f$ is a constant $c$, we have $2c = c(c+1)$. This implies $c = 0$ or $c = 1$. Hence, there are two constant solutions, $f \\equiv 0$ and $f \\equiv 1$.\n\n$$\nP(0, y) : \\quad f(f(y + 1)) + f(0) = f(1)(f(y) + 1), \\qquad... | Argentina | Rioplatense Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(n) = 0 for all integers n; f(n) = 1 for all integers n; f(n) = n for all integers n | |
0362 | Problem:
A triangle $ABC$ with centroid $G$ and incenter $I$ is given. If $AB = 42$, $GI = 2$ and $AB \parallel GI$, find $AC$ and $BC$. | [
"Solution:\nLet $CM$ be the median and $CL$ be the bisector of $\\triangle ABC$ ($I \\in CL$). Using the standard notation for $\\triangle ABC$ we have $\\frac{AL}{BL} = \\frac{AC}{BC} = \\frac{b}{a}$, whence $AL = \\frac{bc}{a+b}$.\n\nSince $AI$ is the bisector of $\\triangle ALC$ through $A$ we get $\\frac{CI}{IL... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | AC = 48, BC = 36 or AC = 36, BC = 48 | |
0iom | Problem:
I have four distinct rings that I want to wear on my right hand (five distinct fingers). One of these rings is a Canadian ring that must be worn on a finger by itself, the rest I can arrange however I want. If I have two or more rings on the same finger, then I consider different orders of rings along the sam... | [
"Solution:\n\nAnswer: $600$. First we pick the finger for the Canadian ring. This gives a multiplicative factor of $5$. For distributing the remaining $3$ rings among $4$ fingers, they can either be all on the same finger ($4 \\cdot 3!$ ways), all on different fingers ($\\binom{4}{3} \\cdot 3!$ ways), or two on one... | United States | $10^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics"
] | null | final answer only | 600 | |
042s | In ellipse $\Gamma$, $A$ is an endpoint of the major axis, $B$ is an endpoint of the minor axis, and $F_1, F_2$ are the foci. If $\overrightarrow{AF_1} \cdot \overrightarrow{AF_2} + \overrightarrow{BF_1} \cdot \overrightarrow{BF_2} = 0$, then the value of $\frac{|AB|}{|F_1F_2|}$ is ______. | [
"Without loss of generality, suppose the equation of $\\Gamma$ is $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ ($a > b > 0$), and $A(a, 0)$, $B(0, b)$, $F_1(-c, 0)$, $F_2(c, 0)$. By the given conditions, we get\n$$ \\overrightarrow{AF_1} \\cdot \\overrightarrow{AF_2} + \\overrightarrow{BF_1} \\cdot \\overrightarrow{BF... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | final answer only | sqrt(2)/2 | |
03yj | Given eight points $A_1, A_2, \dots, A_8$ on a circle, determine the smallest positive integer $n$ such that among any $n$ triangles with vertices in these eight points, there are two which have a common side. | [
"First, we consider the maximal number of triangles with no common side.\nConsider the maximal number of triangles with no common side pairwise. There are $C_8^2 = 28$ chords by connecting eight points. If each chord only belongs to one triangle, then these chords can only form $r \\le \\lfloor \\frac{28}{3} \\rflo... | China | China Southeastern Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 9 | |
0kbx | Problem:
Let $\triangle ABC$ be a triangle with $AB = 7$, $BC = 1$, and $CA = 4\sqrt{3}$. The angle trisectors of $C$ intersect $\overline{AB}$ at $D$ and $E$, and lines $\overline{AC}$ and $\overline{BC}$ intersect the circumcircle of $\triangle CDE$ again at $X$ and $Y$, respectively. Find the length of $XY$. | [
"Solution:\nLet $O$ be the circumcenter of $\\triangle CDE$. Observe that $\\triangle ABC \\sim \\triangle XYC$. Moreover, $\\triangle ABC$ is a right triangle because $1^{2} + (4\\sqrt{3})^{2} = 7^{2}$, so the length $XY$ is just equal to $2r$, where $r$ is the radius of the circumcircle of $\\triangle CDE$. Since... | United States | HMMT February 2020 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 112/65 | |
0h4y | Let $AD$ be a bisector in an isosceles triangle $ABC$ ($AB = BC$), and let $DE$ be another bisector in the triangle $ABD$. Find out the measures of all angles in $ABC$ if the bisectors of $ABD$ and $AED$ intersect on the straight line $AD$. | [
"Let $K$ be the intersection point for the bisectors of the angles $ABD$ and $AED$ (Fig. 3). Then this point lies on the segment $AD$ and is equidistant from rays $BA$ and $BC$, as well as from $EA$ and $ED$. Hence, it's equidistant from rays $DE$ and $DC$. Then $DA$ is the bisector of $\\angle CED$ (in other words... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"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 | ∠BAC = 80°, ∠BCA = 80°, ∠ABC = 20° | |
020u | Problem:
Let $N$ be a positive integer. A collection of $4 N^{2}$ unit tiles with two segments drawn on them as shown is assembled into a $2 N \times 2 N$ board. Tiles can be rotated.
The segments on the tiles define paths on the board. Determine the least possible number and the largest possi... | [
"Solution:\nLet $p$ denote the number of paths. Notice that there are two types of paths: (1) those that start and end at a point on the boundary of the board and (2) closed paths in the interior of the board. Let $p_{1}, p_{2}$ denote the respective numbers of paths of either type. There are $8 N$ points on the bo... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | minimum 4N, maximum N^2 + (N+1)^2 | |
01lw | Find the least positive integer $n$ for which there exists a set $\{s_1, \dots, s_n\}$ consisting of $n$ distinct positive integers such that
$$
\left(1 - \frac{1}{s_1}\right) \left(1 - \frac{1}{s_2}\right) \cdots \left(1 - \frac{1}{s_n}\right) = \frac{51}{2010}
$$
(IMO-2010 Shortlist, Problem N1) | [
"2. See IMO-2010 Shortlist, Problem N1."
] | Belarus | Selection and Training Session | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | 59 | |
082n | Problem:
Un dodecaedro è un solido regolare con 12 facce pentagonali. Una diagonale di un solido è un segmento che ha per estremi due vertici del solido che non appartengono ad una stessa faccia. Quante sono le diagonali del dodecaedro? | [
"Solution:\n\nLa risposta è 100. È necessario contare le coppie (non ordinate) di vertici non appartenenti ad una stessa faccia. In ogni vertice si incontrano 3 facce e in ciascuna di esse ci sono 2 vertici che non sono su una faccia che contiene anche il vertice iniziale (totale 6) più 2 in comune con un'altra fac... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 100 | |
0l04 | The first three terms of a geometric sequence are the integers $a$, $720$, and $b$, where $a < 720 < b$. What is the sum of the digits of the least possible value of $b$?
(A) 9 (B) 12 (C) 16 (D) 18 (E) 21 | [
"The prime factorization of $720$ is $2^4 \\cdot 3^2 \\cdot 5$. Let $r = \\frac{m}{n}$ be the common ratio of the geometric sequence, where $m$ and $n$ are relatively prime positive integers. If $n$ had any prime factor greater than $5$, then $b = 720r$ would not be an integer. Analogously, if $m$ had any prime fac... | United States | AMC 10 A | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | E | |
0knr | Let $n \ge 4$ be an integer. Find all positive real solutions to the following system of $2n$ equations:
$$
\begin{aligned}
a_1 &= \frac{1}{a_{2n}} + \frac{1}{a_2}, & a_2 &= a_1 + a_3, \\
a_3 &= \frac{1}{a_2} + \frac{1}{a_4}, & a_4 &= a_3 + a_5, \\
a_5 &= \frac{1}{a_4} + \frac{1}{a_6}, & a_6 &= a_5 + a_7, \\
\vdots & &... | [] | United States | USAMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | For all k = 1, 2, ..., n: a_{2k-1} = 1 and a_{2k} = 2. | |
00u6 | Let $n$ be a positive integer. What is the smallest sum of digits of $5^n + 6^n + 2022^n$? | [
"We will prove that the smallest sum is equal to $8$. One case when it is achieved is for $n = 1$.\n\nSuppose that for some $n > 1$ it is possible to obtain a smaller sum than $8$. Observing the last digit of the number $5^n + 6^n + 2022^n$, we can easily conclude that\n$$\n5^n + 6^n + 2022^n \\equiv \\begin{cases}... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | 8 | |
0jbl | Problem:
Dizzy Daisy is standing on the point $(0,0)$ on the $xy$-plane and is trying to get to the point $(6,6)$. She starts facing rightward and takes a step 1 unit forward. On each subsequent second, she either takes a step 1 unit forward or turns 90 degrees counterclockwise then takes a step 1 unit forward. She ma... | [
"Solution:\n\nAnswer: $131922$\n\nBecause Daisy can only turn in one direction and never goes to the same square twice, we see that she must travel in an increasing spiral about the origin. Clearly, she must arrive at $(6,6)$ coming from below. To count her paths, it therefore suffices to consider the horizontal an... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | final answer only | 131922 | |
0bho | Find the minimum value of the expression
$$
E = \sqrt{x^2 + \frac{1}{y^2}} + \sqrt{y^2 + \frac{1}{z^2}} + \sqrt{z^2 + \frac{1}{x^2}},
$$
taken for all $x, y, z \in \mathbb{R}^*$. | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 3*sqrt(2) | |
036n | Problem:
Find all pairs $(a, b)$ of non-negative real numbers such that the equations $x^{2} + a^{2} x + b^{3} = 0$ and $x^{2} + b^{2} x + a^{3} = 0$ have a common real root. | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | All pairs with a = b and either a = b = 0 or a = b ≥ 4. | |
0jmr | Problem:
Given that $a$, $b$, and $c$ are complex numbers satisfying
$$
\begin{aligned}
a^{2}+a b+b^{2} & =1+i \\
b^{2}+b c+c^{2} & =-2 \\
c^{2}+c a+a^{2} & =1,
\end{aligned}
$$
compute $(a b+b c+c a)^{2}$. (Here, $i=\sqrt{-1}$.) | [
"Solution:\nAnswer: $\\quad \\frac{-11-4 i}{3}$ OR $-\\frac{11+4 i}{3}$\n\nMore generally, suppose $a^{2}+a b+b^{2}=z$, $b^{2}+b c+c^{2}=x$, $c^{2}+c a+a^{2}=y$ for some complex numbers $a, b, c, x, y, z$.\nWe show that\n$$\nf(a, b, c, x, y, z)=\\left(\\frac{1}{2}(a b+b c+c a) \\sin 120^{\\circ}\\right)^{2}-\\left(... | United States | HMMT 2014 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | (-11 - 4 i)/3 | |
0j1f | Problem:
In the game of projective set, each card contains some nonempty subset of six distinguishable dots. A projective set deck consists of one card for each of the 63 possible nonempty subsets of dots. How many collections of five cards have an even number of each dot? The order in which the cards appear does not ... | [
"Solution:\n\nAnswer: 109368\n\nWe'll first count sets of cards where the order does matter. Suppose we choose the first four cards. Then there is exactly one card that can make each dot appear twice. However, this card could be empty or it could be one of the cards we've already chosen, so we have to subtract for ... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 109368 | |
07ph | Let $n > 1$ be an integer and $\Omega := \{1, 2, \dots, 2n-1, 2n\}$ the set of all positive integers that are not larger than $2n$.
A non-empty subset $S$ of $\Omega$ is called *sum-free* if, for all elements $x, y$ belonging to $S$, $x+y$ does not belong to $S$. We allow $x=y$ in this condition.
Prove that $\Omega$ ha... | [
"Any non-empty subset of $\\Psi = \\{n+1, n+2, \\dots, 2n\\}$ is obviously a sum-free subset of $\\Omega$, and there are $2^n - 1$ of these. Also, every non-empty subset of the set $\\Phi$ of odd numbers in $\\Omega$ is also a sum-free subset and there are $2^n - 1$ of these. The number of common subsets in the uni... | Ireland | Ireland | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof only | null | |
0edv | The value of the expression $10^{2016} - 10^{15}$ is a positive integer. Determine the sum of its digits.
(A) 1
(B) 17
(C) 2001
(D) 18\,000
(E) 18\,009 | [
"The positive integer which represents the value of the expression $10^{2016} - 10^{15}$ has 2016 digits. The last 15 of them are 0 and the rest are equal to 9. The sum of the digits is therefore $(2016 - 15) \\cdot 9 = 18\\,009$."
] | Slovenia | Slovenija 2016 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | E | |
0jym | Problem:
Tetrahedron $A B C D$ with volume $1$ is inscribed in circumsphere $\omega$ such that $A B = A C = A D = 2$ and $B C \cdot C D \cdot D B = 16$. Find the radius of $\omega$. | [
"Solution:\n\nLet $X$ be the foot of the perpendicular from $A$ to $\\triangle B C D$. Since $A B = A C = A D$, it follows that $X$ is the circumcenter of $\\triangle B C D$. Denote $X B = X C = X D = r$. By the Pythagorean Theorem on $\\triangle A B X$, we have $A X = \\sqrt{4 - r^{2}}$. Now, from the extended law... | United States | HMMT November | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 5/3 | |
0dgp | Consider non-negative real numbers $a, b, c$ satisfying the condition $a^2 + b^2 + c^2 = 2$. Find the maximum value of the following expression
$$
P = \frac{\sqrt{b^2 + c^2}}{3 - a} + \frac{\sqrt{c^2 + a^2}}{3 - b} + a + b - 2022c.
$$ | [
"First, we will show that $4\\sqrt{b^2+c^2} \\le (3-a)^2$. Notice that $b^2+c^2 = 2-a^2 \\ge 0$, so we need to prove $4\\sqrt{2-a^2} \\le (3-a)^2$. According to the AM-GM inequality, we have\n$$\n4\\sqrt{2-a^2} = 4\\sqrt{1 \\cdot (2-a^2)} \\le 4 \\cdot \\frac{1+2-a^2}{2} = 2(3-a^2).\n$$\nWe need to prove\n$$\n2(3-a... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 3 | |
0elq | Given $n$ positive real numbers satisfying $x_1 \ge x_2 \ge \dots \ge x_n \ge 0$ and $x_1^2 + x_2^2 + \dots + x_n^2 = 1$, prove that
$$
\frac{x_1}{\sqrt{1}} + \frac{x_2}{\sqrt{2}} + \dots + \frac{x_n}{\sqrt{n}} \ge 1.
$$ | [
"Note that for any $k \\le n$, we have\n$$\nkx_k^2 \\le x_1^2 + x_2^2 + \\dots + x_k^2 \\le 1\n$$\nby the given conditions. This implies that $x_k \\le \\frac{1}{\\sqrt{k}}$ and thus $x_k^2 \\le \\frac{x_k}{\\sqrt{k}}$. We conclude that\n$$\n\\frac{x_1}{\\sqrt{1}} + \\frac{x_2}{\\sqrt{2}} + \\cdots + \\frac{x_n}{\\... | South Africa | The South African Mathematical Olympiad Third Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0f8g | Problem:
Show that there are infinitely many triples of distinct positive integers $a$, $b$, $c$ such that each divides the product of the other two and $a + b = c + 1$. | [
"Solution:\n\n$\\{ n(n + 1),\\ n(n^2 + n - 1),\\ (n + 1)(n^2 + n - 1) \\}$."
] | Soviet Union | 22nd ASU | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
09yk | Joah has a very long liquorice lace. He keeps taking bites out of the lace (but not from the very beginning or end of the lace), each time eating $2$ cm of the liquorice, creating two smaller pieces in the process. He repeats this several times. At the end, he is left with pieces of liquorice lace of $1$, $2$, $3$, $4$... | [] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | MCQ | C | |
0ec4 | Let $D$ and $E$ be the midpoints of sides $BC$ and $CA$ of the triangle $ABC$ respectively. The lines $AD$ and $BE$ intersect the circumcircle of the triangle $ABC$ additionally in points $P$ and $Q$ respectively. Suppose that $|DP| = |EQ|$. Prove that the triangle $ABC$ is isosceles with apex $C$. | [
"\n\nSolution:\n\nSince $D$ and $E$ are the midpoints of the segments $BC$ and $AC$ the lines $DE$ and $AB$ are parallel. It follows that $\\angle EDA = \\angle BAD$, and by the Angles Subtended by Same Arc Theorem we have $\\angle BAD = \\angle BAP = \\angle BQP$. Therefore\n\n$\\angle EQP... | Slovenia | National Math Olympiad 2015 – Final Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Pla... | null | proof only | null | |
0dq4 | Let $p > 200$ be a prime number. We call a positive integer $n$ *good* if $p$ divides the numerator of the irreducible fraction $\frac{a_n}{b_n} = 1 + \frac{1}{2} + \cdots + \frac{1}{n}$. Prove that for all large enough $N$ the number of good numbers not exceeding $N$ is not greater than $C N^{\frac{3}{4}}$, where $C$ ... | [
"We will use congruences modulo $p$ for fractions, writing $\\frac{a}{b} \\equiv x \\pmod{p}$ for $b \\not\\equiv 0 \\pmod{p}$ if $b x \\equiv a \\pmod{p}$. A sum of such fractions is congruent to $0 \\pmod{p}$ if and only if $p$ divides the numerator of the (reduced) sum of respective usual fractions. We shall pro... | Silk Road Mathematics Competition | SILK ROAD MATHEMATICAL COMPETITION | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof only | null | |
0anz | Problem:
Let $R A L P$ be a trapezoid with $R A \parallel L P$. Let $H$ be the intersection of its diagonals. If the area of $\triangle R A H$ is $9$ and the area of $\triangle L P H$ is $16$, find the area of the trapezoid. | [] | Philippines | Area Stage | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 49 | |
05w7 | Problem:
Soit $n \geqslant 1$ un entier strictement positif. Sur un mur, $n$ clous sont plantés. Chaque paire de clous est reliée par une corde coloriée à l'aide d'une des $n$ couleurs. On dit que le mur est coloré si pour tout triplet de couleurs deux à deux distinctes $a, b, c$, il existe trois clous tels que les tr... | [
"Solution:\n\nC'est en fait la parité de $n$ qui est cruciale.\n\nCas $n$ pair : Il y a $\\frac{n(n-1)}{2}$ cordes, donc en moyenne il y a $\\frac{n-1}{2}$ cordes de chaque couleur. Ce nombre n'étant pas entier, il y a des couleurs avec plus de cordes que la moyenne et d'autres avec moins. On pourrait tout à fait c... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Number Theory > Modul... | null | proof and answer | No for n = 6; Yes for n = 7. | |
05ut | Problem:
Déterminer tous les couples d'entiers $(x, y)$ tels que $x^{2}+73=y^{2}$. | [
"Solution:\n\nComme souvent pour une équation diophantienne, on cherche à réarranger l'équation de sorte à avoir des produits de facteurs des deux côtés de l'égalité. Lorsque l'on est en présence de carrés parfaits, on peut utiliser l'identité remarquable $y^{2}-x^{2}=(y-x)(y+x)$. Ceci permet de réécrire l'équation... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (-36, -37), (-36, 37), (36, -37), (36, 37) | |
04g8 | Show that there are no positive integers $m$ and $n$ such that $3^m + 3^n + 1$ is a perfect square. | [
"Assume that there is $k$ such that $3^m + 3^n + 1 = k^2$. Obviously, $k$ is odd. The last equation is equivalent to $3^m + 3^n = k^2 - 1$.\n\nIt is easy to see that for odd number $k$, $8$ divides $k^2 - 1$. Since powers of $3$ are congruent to $1$ or $3$ modulo $8$, the number $3^m + 3^n$ is congruent to $2$, $4$... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
014j | Problem:
For a positive integer $n$ let $a_{n}$ denote the last digit of $n^{\left(n^{n}
ight)}$. Prove that the sequence $\left(a_{n}\right)$ is periodic and determine the length of the minimal period. | [
"Solution:\n\nLet $b_{n}$ and $c_{n}$ denote the last digit of $n$ and $n^{n}$, respectively. Obviously, if $b_{n}=0,1,5,6$, then $c_{n}=0,1,5,6$ and $a_{n}=0,1,5,6$, respectively.\nIf $b_{n}=9$, then $n^{n} \\equiv 1(\\bmod 2)$ and consequently $a_{n}=9$. If $b_{n}=4$, then $n^{n} \\equiv 0$ $(\\bmod 2)$ and conse... | Baltic Way | Baltic Way | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | 20 | |
05ph | Problem:
Prouver qu'il existe un entier $n>0$ tel que parmi les 2016 chiffres de droite dans l'écriture décimale de $2^{n}$, il y a au moins 1008 chiffres 9. | [
"Solution:\n\nOn peut légitimement se demander quand trouver des 9 à la droite de l'écriture décimale de $2^{n}$. On peut penser que c'est quand la puissance de 2 est légèrement inférieure à une puissance de 10. On va donc chercher des nombres de la forme $2^{n}+1$ qui sont divisibles par 5 selon une puissance élev... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
02dc | For which $k$ does the system $x^2 - y^2 = 0$, $(x - k)^2 + y^2 = 1$ have exactly
(1) two, (2) three real solutions? | [
"We have $(x - k)^2 + x^2 = 1$, so $2x^2 - 2k x + k^2 - 1 = 0$. This has 0, 1 or 2 real solutions according as $k^2 > 2$, $k^2 = 2$ or $k^2 < 2$.\n\n$k = \\sqrt{2}$ gives $x = \\frac{1}{\\sqrt{2}}$, $y = \\frac{1}{\\sqrt{2}}$ or $y = -\\frac{1}{\\sqrt{2}}$, so there are two solutions to the original set. Similarly ... | Brazil | III OBM | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (1) Two solutions: k = ±√2. (2) Three solutions: k = ±1. | |
0d9o | Find the smallest positive integer $n$ which can not be expressed as $n=\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$ for some positive integers $a, b, c, d$. | [
"Let $S$ be the set of positive integers which can be written as $s=\\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$ for some positive integers $a, b, c, d$.\nSince $s>0$, we can assume that $a>b, c>d$ and write $s=2^{b-d} \\frac{2^{a-b}-1}{2^{c-d}-1}$. It's now clear that $b-d=v_{2}(x)$.\nSo if we set $x=2^{v_{2}(x)} y$ then $y=... | Saudi Arabia | Team selection tests for BMO 2018 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 11 | |
0676 | Let $AB\Gamma\Delta$ be a quadrilateral inscribed into the circle. With centers $A, B, \Gamma, \Delta$ we draw circles $C_A, C_B, C_\Gamma, C_\Delta$ respectively, not having common points. The circle $C_A$ intersects the sides of the quadrilateral at the points $A_1, A_2$, the circle $C_B$ at the points $B_1, B_2$, th... | [
"Since the triangles $AA_1A_2, BB_1B_2, \\Gamma\\Gamma_1\\Gamma_2$ and $\\Delta\\Delta_1\\Delta_2$ are isosceles, using small letters for their equal angles we have the equalities:\n\nFigure 4\n$$\n\\hat{A} + \\hat{x} + \\hat{x} = 180^\\circ \\Leftrightarrow \\hat{x} = 90^\\circ - \\frac{\\... | Greece | SELECTION EXAMINATION | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
057t | Let $p, q$ be prime numbers and $a$ be an integer such that $p > 2$ and $a \neq 1 \pmod{q}$ but $a^p \equiv 1 \pmod{q}$. Prove that
$$
(1+a^1)(1+a^2)\dots(1+a^{p-1}) \equiv 1 \pmod{q}.
$$ | [
"As $a^p \\equiv 1 \\pmod{q}$ while $a \\neq 1 \\pmod{q}$, the case $q = 2$ is impossible. Thus, the desired equation is equivalent to\n$$\n(1 + a^0)(1 + a^1)(1 + a^2)\\dots(1 + a^{p-1}) \\equiv 2 \\pmod{q}. \\quad (13)\n$$\nRemoving parentheses in the l.h.s. of (13) gives all monomials of the form $a^{i_1+\\dots+i... | Estonia | IMO Team Selection Contest | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
03kf | Problem:
Let $ABC$ be an acute angled triangle. Let $AD$ be the altitude on $BC$, and let $H$ be any interior point on $AD$. Lines $BH$ and $CH$, when extended, intersect $AC$ and $AB$ at $E$ and $F$, respectively. Prove that $\angle EDH = \angle FDH$. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0j0n | Problem:
Define a sequence of polynomials as follows: let $a_{1}=3 x^{2}-x$, let $a_{2}=3 x^{2}-7 x+3$, and for $n \geq 1$, let $a_{n+2}=\frac{5}{2} a_{n+1}-a_{n}$. As $n$ tends to infinity, what is the limit of the sum of the roots of $a_{n}$? | [
"Solution:\n\nAnswer: $\\frac{13}{3}$ By using standard methods for solving linear recurrences $\\{ \\}^2$, we see that this recurrence has a characteristic polynomial of $x^{2}-\\frac{5}{2} x+1=\\left(x-\\frac{1}{2}\\right)(x-2)$, hence $a_{n}(x)=c(x) \\cdot 2^{n}+d(x) \\cdot 2^{-n}$ for some polynomials $c$ and $... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 13/3 | |
0bbo | Let $k$ and $n$ be integer numbers with $2 \le k \le n - 1$. Consider a set $A$ of $n$ real numbers such that the sum of any $k$ distinct elements of $A$ is a rational number. Prove that all elements of the set $A$ are rational numbers. | [
"The difference of any two elements from $A$ is a rational number. To show this, let $x \\ne y \\in A$ and choose other $k-1$ elements of $A$ – the choice can be made, for $k-1 \\le n-2$. Denote $s$ the sum of the $k-1$ elements and apply the hypothesis to infer that $x+s$ and $y+s$ are both rational numbers. Subtr... | Romania | 62nd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Other"
] | null | proof only | null | |
04mi | Determine all pairs $(x, y)$ of real numbers such that $x + y = x^2 + y^2 = x^3 + y^3$. | [
"Let $S = x + y$ and $P = x y$.\n\nWe are given:\n\n$$\nS = x + y = x^2 + y^2 = x^3 + y^3\n$$\n\nRecall:\n$$\nx^2 + y^2 = (x + y)^2 - 2 x y = S^2 - 2P\n$$\nx^3 + y^3 = (x + y)^3 - 3 x y (x + y) = S^3 - 3 P S\n$$\n\nSo, the system becomes:\n\n1. $S = S^2 - 2P$\n2. $S = S^3 - 3 P S$\n\nFrom (1):\n$$\nS = S^2 - 2P \\i... | Croatia | Croatia_2018 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | (0, 0), (0, 1), (1, 0), (1, 1) | |
094n | Problem:
We are given a convex quadrilateral $A B C D$ whose angles are not right. Assume there are points $P, Q, R, S$ on its sides $A B, B C, C D, D A$, respectively, such that $P S \| B D$, $S Q \perp B C$, $P R \perp C D$. Furthermore, assume that the lines $P R, S Q$, and $A C$ are concurrent. Prove that the poin... | [
"Solution:\n\nLet the intersection point of $P R$, $Q S$, $A C$ be $T$ and let $H$ be the orthocenter of $B C D$. Since $\\angle B C D$ is not right, $H \\neq C$. Notice that triangles $H B D$ and $T P S$ are homothetic due to their corresponding sides being parallel. This means that $H T$, $B P$, $D S$ are concurr... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"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 ... | null | proof only | null | |
086t | Problem:
Eleonora gioca con un dado e un orologio (fermo) che all'inizio segna le 12. Per 2008 volte tira il dado e porta le lancette avanti di tante ore quanto è il risultato. Qual è alla fine la probabilità che la lancetta delle ore sia orizzontale?
(A) 0
(B) $\frac{1}{2008}$
(C) $\frac{1}{1004}$
(D) $\frac{1}{12}$
... | [
"Solution:\n\nLa risposta è $\\mathbf{( E )}$. La lancetta delle ore è orizzontale se l'orologio segna le 3 o le 9. Sia $K$ l'ora segnata dall'orologio subito prima dell'ultimo lancio di dado. Se $K$ è uno dei 6 numeri compresi tra 3 e 8, vi è esattamente 1 risultato del dado su 6 che permetterebbe di raggiungere l... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | E | |
0b02 | Problem:
Consider all the subsets of $\{1,2,3, \ldots, 2018,2019\}$ having exactly 100 elements. For each subset, take the greatest element. Find the average of all these greatest elements. | [
"Solution:\n\nLet $M$ be the average that we are computing. First, there are $\\binom{2019}{100}$ ways to choose a 100-element subset. Next, if $x$ is the largest element, then $x \\geq 100$, and there are $\\binom{x-1}{99}$ subsets having $x$ as the largest element. Hence\n$$\nM=\\frac{\\sum_{x=100}^{2019} x\\bino... | Philippines | Philippines Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 2000 | |
0klh | Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$st pole along this road is exactly one mile ($5280$ feet) from the first pole. How much longer, in fee... | [] | United States | AMC 12 A | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | B | |
0a0x | In a room there are $2023$ vases numbered from $1$ to $2023$. In each vase we want to put a note with a positive integer from $1, 2, \ldots, 2023$ on it. The numbers on the notes do *not* necessarily have to be distinct. The following should now apply to each vase. Look at the note inside the vase, find the (not necess... | [
"A possible way to provide each vase with a note is to put in vase $1$ a note with $1$, in vase $2$ a note with $2$, in vase $3$ a note with $3$, $\\ldots$, and in vase $2023$ a note with $2023$. We will use induction to show that this is the only distribution. Note that for a valid distribution it does not matter ... | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Each vase contains the note equal to its own label; that is, for every k from 1 to 2023, vase k contains k. | |
0i7r | Problem:
Evaluate
$$
\int_{-\infty}^{\infty} \frac{1-x^{2}}{1+x^{4}} d x
$$ | [
"Solution:\n\n$0$\n\nLet $S=\\int_{0}^{\\infty} \\frac{1}{x^{4}+1} d x$; note that the integral converges absolutely. Substituting $x=1/u$, so that $d x=-1/u^{2} d u$, we have\n\n$$\n\\begin{gathered}\nS=\\int_{0}^{\\infty} \\frac{1}{1+x^{4}} d x=\\int_{\\infty}^{0} \\frac{1}{1+u^{-4}} \\frac{d u}{-u^{2}}=\\int_{\\... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Functions"
] | null | proof and answer | 0 | |
04si | A parallelogram $ABCD$ with $|AB| = 2|BC|$ is given. Determine all the lines that divide the parallelogram into two tangential quadrilaterals. (Jaroslav Švrček) | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | The unique line parallel to the shorter side of the parallelogram that passes through its center, i.e., the line through the midpoints of the longer sides (parallel to BC). | |
0j5t | Problem:
Find the least positive integer $N$ with the following property: If all lattice points in $[1,3] \times [1,7] \times [1, N]$ are colored either black or white, then there exists a rectangular prism, whose faces are parallel to the $xy$, $xz$, and $yz$ planes, and whose eight vertices are all colored in the sam... | [
"Solution:\nAnswer: $127$\n\nFirst we claim that if the lattice points in $[1,3] \\times [1,7]$ are colored either black or white, then there exists a rectangle whose faces are parallel to the $x$ and $y$ axes, whose vertices are all the same color (a.k.a. monochromatic). Indeed, in every row $y = i$, $1 \\leq i \\... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 127 | |
0ilg | Problem:
Find the sum of the positive integer divisors of $2^{2007}$. | [
"Solution:\n\n$2^{2007}$ has divisors $1, 2, 2^2, \\ldots, 2^{2007}$. The sum is\n$$\n1 + 2 + 2^2 + \\cdots + 2^{2007} = 2^{2008} - 1.\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2^{2008} - 1 | |
0k2h | Problem:
Fran writes the numbers $1,2,3, \ldots, 20$ on a chalkboard. Then she erases all the numbers by making a series of moves; in each move, she chooses a number $n$ uniformly at random from the set of all numbers still on the chalkboard, and then erases all of the divisors of $n$ that are still on the chalkboard ... | [
"Solution:\n\nFor each $n, 1 \\leq n \\leq 20$, consider the first time that Fran chooses one of the multiples of $n$. It is in this move that $n$ is erased, and all the multiples of $n$ at most $20$ are equally likely to be chosen for this move. Hence this is the only move in which Fran could possibly choose $n$; ... | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Number Theory > Divisibility / Factorization"
] | null | final answer only | 131/10 | |
0acc | Let $n$ be a positive integer. The rectangle $ABCD$ with side lengths $AB = 90n + 1$ and $BC = 90n + 5$ is partitioned into unit squares with sides parallel to the sides of $ABCD$. Let $S$ be the set of all points which are vertices of these unit squares. Prove that the number of lines which pass through at least two p... | [
"Denote $90n+1 = m$. We investigate the number of the lines modulo $4$ consecutively reducing different types of lines. The vertical and horizontal lines are $(m+5)+(m+1)=2(m+3)$ which is divisible by $4$. Moreover, every line which makes an acute angle to the axis $Ox$ (i.e. that line has a positive angular coeffi... | North Macedonia | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Rotation",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorizati... | null | proof only | null | |
0bb3 | Let $n$ be a positive integer and let $x_1, x_2, \dots, x_n$ and $y_1, y_2, \dots, y_n$ be real numbers. Prove that there exists a number $i$, $i = 1, 2, \dots, n$, such that
$$
\sum_{j=1}^{n} |x_i - x_j| \le \sum_{j=1}^{n} |x_i - y_j|.
$$ | [
"Without the loss of generality, suppose $x_1 \\le x_2 \\le \\dots \\le x_n$. For each $k = 1, 2, \\dots, n$ we have $|x_1 - x_k| + |x_n - x_k| = |x_1 - x_n| \\le |x_1 - y_k| + |x_n - y_k|$, hence\n$$\n\\sum_{k=1}^{n} |x_1 - x_k| + \\sum_{k=1}^{n} |x_n - x_k| \\le \\sum_{k=1}^{n} |x_1 - y_k| + \\sum_{k=1}^{n} |x_n ... | Romania | 62nd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0fag | Problem:
$ABCD$ is a rectangle. Points $K$, $L$, $M$, $N$ are chosen on $AB$, $BC$, $CD$, $DA$ respectively so that $KL$ is parallel to $MN$, and $KM$ is perpendicular to $LN$. Show that the intersection of $KM$ and $LN$ lies on $BD$. | [
"Solution:\n\n\n\nLet $LN$ and $KM$ meet at $O$. $\\angle NOM = \\angle NDM = 90^{\\circ}$, so $OMDN$ is cyclic. Hence $\\angle NOD = \\angle NMD$. Similarly, $BLOK$ is cyclic and $\\angle LOB = \\angle LKB$. But $NM$ is parallel to $LK$ and $AB$ is parallel to $CD$, so $\\angle LKB = \\ang... | Soviet Union | 25th ASU | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | null | proof only | null | |
02hn | Problem:
Quantos são os pares diferentes de inteiros positivos $(a, b)$ tais que $a+b \leq 100$ e $\frac{a+\frac{1}{b}}{\frac{1}{a}+b}=13$? | [
"Solution:\n\nTemos: $13=\\frac{a+\\frac{1}{b}}{\\frac{1}{a}+b}=\\frac{\\frac{a b+1}{b}}{\\frac{1+a b}{a}}=\\frac{a}{b}$. Logo, $a=13 b$ e como $a+b \\leq 100$ segue que $14 b \\leq 100 \\Rightarrow b \\leq 7,14$. Como $b$ é inteiro devemos ter $b \\leq 7$. Logo os pares são em número de 7, a saber:\n$$\n(13,1),\\ ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 7 | |
0ghg | 令 $a > 1$ 為正整數, 而 $d > 1$ 為與 $a$ 互質的正整數。令 $x_1 = 1$ 並對於所有 $k \ge 1$ 以遞迴方式定義
$$
x_{k+1} = \begin{cases} x_k + d & \text{若 } a \text{ 不整除 } x_k, \\ x_k/a & \text{若 } a \text{ 整除 } x_k. \end{cases}
$$
求最大的正整數 $n$ (以 $a$ 和 $d$ 的函數表示), 使得存在足標 $k$, 滿足 $x_k$ 被 $a^n$ 整除。
Let $a > 1$ be a positive integer, and let $d > 1$ be ... | [
"$n = \\max\\{m: am < ad\\}$; 注意到這表示 $a^{n+1} > ad \\Rightarrow a^n > d$.\n\n**解法一、由數歸知 $x_k$ 與 $d$ 互質。** 此外, 注意到 $x_k$ 至多只有連續 $a-1$ 個遞增, 故由數歸知\n$$\n\\begin{cases} x_k < da & \\text{若 } x_k = x_{k-1} + d, \\\\ x_k < d & \\text{若 } x_k = x_{k-1}/a \\text{或 } k = 1. \\end{cases} \\quad (1)\n$$\n這意味著 $a^n < da$。此給出了 $... | Taiwan | 2023 數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | Chinese (Traditional) | proof and answer | The largest n is the unique integer with d < a^n < a d, equivalently n = floor(log_a(ad)). | |
0j62 | Problem:
Let $a \star b = \sin a \cos b$ for all real numbers $a$ and $b$. If $x$ and $y$ are real numbers such that $x \star y - y \star x = 1$, what is the maximum value of $x \star y + y \star x$? | [
"Solution:\nWe have $x \\star y + y \\star x = \\sin x \\cos y + \\cos x \\sin y = \\sin(x + y) \\leq 1$.\n\nEquality is achieved when $x = \\frac{\\pi}{2}$ and $y = 0$. Indeed, for these values of $x$ and $y$, we have $x \\star y - y \\star x = \\sin x \\cos y - \\cos x \\sin y = \\sin(x - y) = \\sin \\frac{\\pi}{... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 1 | |
01mb | 2500 chess kings have to be placed on a $100 \times 100$ chessboard so that
1) no king can capture any other one (i.e. no two kings are placed in two squares sharing a common vertex);
2) each row and each column contains exactly $25$ kings;
Find the number of such arrangements. (Two arrangements differing by rotation o... | [
"3. See IMO-2010 Shortlist, Problem C3."
] | Belarus | Selection and Training Session | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | binomial(50, 25)^100 | |
0hlh | Problem:
For which positive integers $n$ is $n^{4}+4$ equal to a prime number? | [
"Solution:\nFor $n=1$ we get $1^{4}+4=5$, which works.\n\nFor all other values of $n$, the key idea is that\n$$\nn^{4}+4=n^{4}+4 n^{2}+4-4 n^{2}=(n^{2}+2)^{2}-(2 n)^{2}=(n^{2}+2 n+2)(n^{2}-2 n+2)\n$$\nwhich is the product of two integers greater than 1, and hence cannot be prime."
] | United States | Berkeley Math Circle: Monthly Contest 8 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1 | |
0jg1 | Problem:
Pick a subset of at least four of the following seven numbers, order them from least to greatest, and write down their labels (corresponding letters from $A$ through $G$) in that order:
$(A)$ $\pi$;
$(B)$ $\sqrt{2}+\sqrt{3}$;
$(C)$ $\sqrt{10}$;
$(D)$ $\frac{355}{113}$;
$(E)$ $16 \tan^{-1} \frac{1}{5} - 4 ... | [
"Solution:\nAnswer: $F, G, A, D, E, B, C$ OR $F<G<A<D<E<B<C$ OR $C>B>E>D>A>G>F$\n\nWe have $\\ln(23) < 2^{\\sqrt{e}} < \\pi < \\frac{355}{113} < 16 \\tan^{-1} \\frac{1}{5} - 4 \\tan^{-1} \\frac{1}{240} < \\sqrt{2} + \\sqrt{3} < \\sqrt{10}$."
] | United States | HMMT November 2013 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | F, G, A, D, E, B, C | |
0b1u | Problem:
What is the remainder when $3^{2020}$ is divided by $73$? | [
"Solution:\nBy Fermat's Little Theorem, $3^{2016} = \\left(3^{72}\\right)^{28} \\equiv 1 \\pmod{73}$. Therefore, $3^{2020} \\equiv 3^{4} \\equiv 8 \\pmod{73}$."
] | Philippines | 22nd Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | final answer only | 8 | |
0503 | Let $a$, $b$, $c$ be fixed real numbers, where $0 \le a, b, c \le 4$. Prove that the system of equations
$$
\begin{cases}
p^2 - a q = -3 \\
q^2 - b r = -4 \\
r^2 - c p = -5
\end{cases}
$$
has no real solutions ($p$, $q$, $r$). | [
"Adding up all equations gives $p^2 - c p + q^2 - a q + r^2 - b r = -12$. From the inequality $(p - \\frac{c}{2})^2 \\ge 0$ we have $p^2 - c p \\ge -\\frac{c^2}{4} \\ge -4$ and similarly, $q^2 - a q \\ge -4$ and $r^2 - b r \\ge -4$. Adding up these inequalities, we see that to avoid a contradiction with the equalit... | Estonia | Selected Problems from Open Contests | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
0h01 | We are given irrational number $\alpha$ for which there exist real $x, y$, such that $x + y = \alpha$ and $x^k + y^k$ is rational for all $k$ from $2$ to $n$. Find maximal $n$ for which it is possible?
**Answer:** $n = 3$. | [
"We show that for $n = 4$ it cannot hold.\n\nSuppose that $xy = 0$, or $y = 0$. For $n = 2$ it is possible, as an example we can take $x = \\sqrt{2} \\in \\mathbb{R} \\setminus \\mathbb{Q}$, $x^2 = 2 \\in \\mathbb{Q}$. But, if $x^2$ and $x^3$ are rational, then $\\frac{x^3}{x^2} = x$ is also rational.\n\nLet us con... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010) | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | 3 | |
07sx | The lengths of the sides of a triangle are consecutive integers and its inradius is $4$. Find the lengths of the sides and the circumradius. | [
"Let $a$, $b$, $c$ be the sides such that $a = b - 1$ and $c = b + 1$. Recall Heron's Formula and two other well known formulae for the area of a triangle:\n$$\n|ABC| = \\sqrt{s(s-a)(s-b)(s-c)} = \\frac{abc}{4R} = rs,\n$$\nwhere $r = 4$ is the inradius and $R$ the circumradius.\nWe obtain $s = ((b - 1) + b + (b + 1... | Ireland | IRL_ABooklet_2020 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | Side lengths: 13, 14, 15. Circumradius: 65/8. | |
0agg | Given the positive numbers $a_1, a_2, \dots, a_n$, such that $n > 2$ and $a_1 + a_2 + \dots + a_n = 1$, prove that the inequality
$$
\frac{a_2 a_3 \dots a_n}{a_1 + n - 2} + \frac{a_1 a_3 \dots a_n}{a_2 + n - 2} + \frac{a_1 a_2 a_4 \dots a_n}{a_3 + n - 2} + \dots + \frac{a_1 a_2 \dots a_{n-1}}{a_n + n - 2} \le \frac{1}{... | [
"Suppose first $n \\ge 4$. Then we have\n\n$$\n\\frac{a_1 a_2 \\dots a_{k-1} a_{k+1} \\dots a_n}{a_k + n - 2} \\le \\frac{\\left( \\frac{a_1 + a_2 + \\dots + a_{k-1} + a_{k+1} + \\dots + a_n}{n-1} \\right)^{n-1}}{a_k + n - 2} < \\\\\n< \\frac{\\left( \\frac{a_1 + a_2 + \\dots + a_n}{n-1} \\right)^{n-1}}{n-2} = \\fr... | North Macedonia | Mediterranean Mathematics Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0keu | Problem:
Let $n > 1$ be a positive integer and $S$ be a collection of $\frac{1}{2}\binom{2n}{n}$ distinct $n$-element subsets of $\{1,2, \ldots, 2n\}$. Show that there exists $A, B \in S$ such that $|A \cap B| \leq 1$. | [
"Solution:\nAssume for the sake of contradiction that there exist no such $A, B$. Pair up each subset with its complement, like so:\n$$\n\\begin{aligned}\n\\{1,2,3, \\ldots, n\\} & \\leftrightarrow \\{n+1, n+2, \\ldots, 2n\\} \\\\\n\\{1,2,3, \\ldots, n-1, n+1\\} & \\leftrightarrow \\{n, n+2, \\ldots, 2n\\} \\\\\n\\... | United States | HMMT February 2020 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0e2w | Problem:
Reši enačbo $\log_{4}\left(1+\log_{4}\left(3^{x}-\sqrt{\left(5^{0}+4^{2}\right)^{2}}\right)\right)=e^{0}$ | [
"Solution:\n\nUgotovimo, da je $e^{0}=1$. Poenostavimo tudi korenjenec $\\sqrt{\\left(5^{0}+4^{2}\\right)^{2}}=\\sqrt{(17)^{2}}=17$. Uporabimo zvezo $1=\\log_{4} 4$ in dobimo $1+\\log_{4}\\left(3^{x}-17\\right)=4$.\n\nUredimo $\\log_{4}\\left(3^{x}-17\\right)=3$.\n\nUporabimo definicijo logaritma $64=3^{x}-17$.\n\n... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 4 | |
0cls | Let $ABCD$ be a convex quadrilateral with the property that the circles having the segments $AB$ and $CD$ as diameters are tangent externally at a point $M$, different from the intersection point of the diagonals of the quadrilateral.
Let $K$ be the second point of intersection of the circumcircle of triangle $AMC$ wit... | [
"Let $F$ and $E$ be the midpoints of chords $ML$ and $MK$, respectively. Then $O_3F \\perp ML$ and $O_4E \\perp MK$. Since $O_4O_1O_2O_3$ is cyclic with $\\angle O_4O_2O_3 = \\angle O_4O_1O_3 = 90^\\circ$, the midpoint $X$ of segment $O_3O_4$ is the center of the circumscribed circle of $O_4O_1O_2O_3$. Moreover, if... | Romania | 75th NMO Selection Tests | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0le7 | There are several identical caro papers of size $5 \times 5$. Someone uses $n$ colors to fill in each paper such that two cells at the same position on two sides share the same color. Two papers are considered congruent if they can be stacked together in such a way that the pairs of squares at the same position have th... | [
"We will prove the following lemma\n\n**Lemma.** Consider positive integer $m = 2k+1$ with $k \\ge 2$, and the square table of size $m \\times m$ in which each cell is filled by one of $n$ colors. Then the number of different ways to color (not duplicated by the rotation) is equal to\n$$\n\\frac{n(a^4 + a^2 + 2a)}{... | Vietnam | VMO | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Abstract Algebra > Group Theory"
] | English | proof only | null | |
02is | Problem:
Se dois lados de um triângulo medem $5~\mathrm{cm}$ e $7~\mathrm{cm}$, então o terceiro lado não pode medir:
(A) $11~\mathrm{cm}$
(B) $10~\mathrm{cm}$
(C) $6~\mathrm{cm}$
(D) $3~\mathrm{cm}$
(E) $1~\mathrm{cm}$ | [
"Solution:\n\nLembre que num triângulo a soma de dois lados quaisquer tem que ser maior que o terceiro lado. Como $1+5$ não é maior do que $7$, o terceiro lado não pode ser $1$."
] | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | MCQ | E | |
09v3 | Five smart students are sitting in a circle. The teacher gives one or more marbles to each of them. He explains that he has handed out a total of 18 marbles, and that everyone got a different number of marbles. Each student is allowed to see his own number of marbles, as well as the number of marbles of his neighbour o... | [
"D) 3"
] | Netherlands | Junior Mathematical Olympiad, September 2019 | [
"Discrete Mathematics > Logic"
] | English | MCQ | D) 3 | |
0fuo | Problem:
Seien $m$ und $n$ teilerfremde natürliche Zahlen. Zeige, dass dann auch die beiden Zahlen
$$
m^{3}+m n+n^{3} \quad \text{ und } \quad m n(m+n)
$$
teilerfremd sind. | [
"Solution:\n\nWir zeigen zuerst, dass $m^{3}+m n+n^{3}$ teilerfremd zu $m$ ist. Nehme an nicht, dann gibt es eine Primzahl $p$ die beide Zahlen teilt. Dann teilt $p$ aber auch $\\left(m^{3}-m n+n^{3}\\right)-m\\left(m^{2}+n\\right)=n^{3}$, also auch $n$, im Widerspruch dazu, dass $m$ und $n$ teilerfremd sind. Analo... | Switzerland | Vorrundenprüfung | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0913 | Problem:
Consider a chessboard $n \times n$ where $n>1$ is a positive integer. We select the centers of $2 n-2$ squares. How many selections are there such that no two selected centers lie on a line parallel to one of the diagonals of the chessboard? | [
"Solution:\n\nBy a $k$-diagonal we mean any chessboard diagonal formed by $k$ squares, where $1 \\leqslant k \\leqslant n$. Since the number of stones is $2 n-2$, while the number of chessboard diagonals in one direction is $2 n-1$ and two of them, which are 1-diagonals, must not be occupied by stones simultaneousl... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 2^n | |
0dkl | Let $n$ be a positive integer. In a family of finite sets, let a splitting element be an element that belongs to at least two of the sets and is omitted by at least two of the sets. Determine the maximum size of a family of subsets of $\{1, \dots, n\}$ for which there is no splitting element. | [] | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n + 1 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.