id stringlengths 4 4 | problem_markdown stringlengths 36 3.59k | solutions_markdown listlengths 0 10 | images images listlengths 0 15 | country stringclasses 58
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|---|---|---|---|---|---|---|---|---|---|
0b69 | Let $ABCD$ be a square, and the points $M \in [BC]$, $N \in [CD]$, $P \in [DA]$, such that
$$
\angle(\overrightarrow{AB}, \overrightarrow{AM}) = x, \quad \angle(\overrightarrow{BC}, \overrightarrow{MN}) = 2x, \quad \angle(\overrightarrow{CD}, \overrightarrow{NP}) = 3x.
$$
i) Show that, for any $x \in [0, \pi/8]$, such... | [
"i) Assume $AB = 1$, and denote $t = \\tan x$. Since\n$$\n1 = \\tan \\frac{\\pi}{4} = \\frac{2 \\tan \\frac{\\pi}{8}}{1 - \\tan^2 \\frac{\\pi}{8}}\n$$\nit follows $\\tan \\frac{\\pi}{8}$ is the positive root of $t^2 + 2t - 1 = 0$, and so $\\tan \\frac{\\pi}{8} = \\sqrt{2} - 1$.\n\nNow $BM = t$, hence $M$ takes once... | Romania | 2010 Fourth STARS OF MATHEMATICS COMPETITION | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 1 | |
0an3 | Problem:
Define the following operation for real numbers: $a \star b = ab + a + b$. If $x \star y = 11$, $y \star z = -4$, and $x \star z = -5$. What is the difference between the maximum and minimum elements of the solution set $\{x, y, z\}$? | [] | Philippines | AREA STAGE | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 5 | |
06dt | Show that there exists a positive integer $k$ such that $k \times 22^n + 1$ is not prime (composite) for every positive integer $n$. | [
"We can take $k = 20$. Then for any $n \\in \\mathbb{Z}^+$, we have\n$$\n20 \\cdot 22^n + 1 \\equiv (-1)(1) + 1 = 0 \\pmod{21},\n$$\nand hence $20 \\cdot 22^n + 1$ is a composite number."
] | Hong Kong | IMO HK TST | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | k = 20 | |
01y4 | Each cell of an $n \times n$ table is filled with one of the two signs: «+» and «-». For each $k$ from $1$ to $n$ the amount of pluses in the first $k$ rows is greater than the amount of minuses in the first $k$ columns.
Find the maximal possible number of minuses in the table. | [
"Answer: $\\left[ \\frac{n^2 - 1}{2} \\right]$."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | floor((n^2 - 1)/2) | |
0a7g | Problem:
Determine all real numbers $x > 1$, $y > 1$, and $z > 1$, satisfying the equation
$$
x + y + z + \frac{3}{x-1} + \frac{3}{y-1} + \frac{3}{z-1} = 2(\sqrt{x+2} + \sqrt{y+2} + \sqrt{z+2})
$$ | [
"Solution:\nConsider the function $f$,\n$$\nf(t) = t + \\frac{3}{t-1} - 2 \\sqrt{t+2}\n$$\ndefined for $t > 1$. The equation of the problem can be written as\n$$\nf(x) + f(y) + f(z) = 0\n$$\nWe reformulate the formula for $f$:\n$$\n\\begin{aligned}\nf(t) & = \\frac{1}{t-1} \\left( t^2 - t + 3 - 2(t-1) \\sqrt{t+2} \... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 6 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | x = y = z = (3 + sqrt(13)) / 2 | |
06wx | For a positive integer $n$, an $n$-sequence is a sequence $(a_{0}, \ldots, a_{n})$ of non-negative integers satisfying the following condition: if $i$ and $j$ are non-negative integers with $i+j \leqslant n$, then $a_{i}+a_{j} \leqslant n$ and $a_{a_{i}+a_{j}}=a_{i+j}$.
Let $f(n)$ be the number of $n$-sequences. Prove ... | [
"In order to solve this, we will give a complete classification of $n$-sequences.\nLet $k=\\lfloor n / 2\\rfloor$. We will say that an $n$-sequence is large if $a_{i}>k$ for some $i$, and small if no such $i$ exists. For now we will assume that $(a_{i})$ is not the identity sequence (in other words, that $a_{i} \\n... | IMO | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
07h9 | We call a natural number $n$ *nice* if it has no zero digit in its decimal representation and it has a digit $b$ in this representation such that the resulting number would be a divisor of $n$, after its removal. Prove that there are only finitely many *nice* numbers. | [
"We prove that a nice number has at most 20 digits which proves that there are finitely many nice numbers. Assume that $n = \\overline{abc}$ where $a, c$ can have more than one digit and $b$ is a digit such that $\\overline{ac} = nk$ for some natural number $k$. If $c$ has $t$ digits, then $n = \\overline{ac}$ (mod... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
06wb | Let $n \geqslant 1$ be an integer, and let $x_{0}, x_{1}, \ldots, x_{n+1}$ be $n+2$ non-negative real numbers that satisfy $x_{i} x_{i+1}-x_{i-1}^{2} \geqslant 1$ for all $i=1,2, \ldots, n$. Show that
$$
x_{0}+x_{1}+\cdots+x_{n}+x_{n+1}>\left(\frac{2 n}{3}\right)^{3 / 2}
$$ | [
"Solution 1.\nLemma 1.1. If $a, b, c$ are non-negative numbers such that $a b-c^{2} \\geqslant 1$, then\n$$\n(a+2 b)^{2} \\geqslant(b+2 c)^{2}+6\n$$\nProof. $(a+2 b)^{2}-(b+2 c)^{2}=(a-b)^{2}+2(b-c)^{2}+6\\left(a b-c^{2}\\right) \\geqslant 6$.\n\nLemma 1.2. $\\sqrt{1}+\\cdots+\\sqrt{n}>\\frac{2}{3} n^{3 / 2}$.\nPro... | IMO | IMO 2021 Shortlisted Problems | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
075u | In a triangle $ABC$ with $B = 90^\circ$, $D$ is a point on the segment $BC$ such that the inradii of triangles $ABD$ and $ADC$ are equal. If $\angle ADB = \varphi$ then prove that $\tan^2(\varphi/2) = \tan(C/2)$. $\square$ | [
"This follows easily from formulae for inradius of a triangle and simple trigonometric identities. $\\square$"
] | India | Indija TS 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
06kl | Find all positive integer(s) $n$ such that $n^2 + 32n + 8$ is a perfect square. | [
"$n$ can be $47$ or $17$.\n\nSuppose $n^2 + 32n + 8 = m^2$ where $m$ is a positive integer. Note that this can be rewritten as $(n+16)^2 - 248 = m^2$, which implies\n$$\n(n + m + 16)(n - m + 16) = 248 = 2^3 \\times 31.\n$$\nHence, $(n+m+16, n-m+16) = (248, 1), (124, 2), (62, 4), (31, 8)$. For each pair, we can solv... | Hong Kong | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 17, 47 | |
04cu | Let $n$ be a positive integer. $2^n$ players participated in a tennis tournament. Every player played one match against each of the remaining players. Prove that we can choose $n+1$ players and arrange them in a line in such a way that each of them defeated all players in line behind him. | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
014p | Problem:
An occasionally unreliable professor has devoted his last book to a certain binary operation $*$. When this operation is applied to any two integers, the result is again an integer. The operation is known to satisfy the following axioms:
a. $x *(x * y)=y$ for all $x, y \in \mathbb{Z}$;
b. $(x * y) * y=x$ fo... | [
"Solution:\n\nWrite $(x, y, z)$ for $x * y = z$. So the axioms can be formulated as\n$$\n\\begin{aligned}\n& (x, y, z) \\Longrightarrow (x, z, y) \\\\\n& (x, y, z) \\Longrightarrow (z, y, x) .\n\\end{aligned}\n$$\n(C1) is proved by the sequence $(x, y, z) \\xrightarrow{(2)} (z, y, x) \\xrightarrow{(1)} (z, x, y) \\... | Baltic Way | Baltic Way | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof and answer | Commutativity follows; associativity does not (for example, define the operation by sending a pair to the negative of their sum). | |
04fz | Let $ABC$ be a triangle whose longest side is $BC$, and the angle $\angle BCA$ is three times greater than $\angle ABC$. The exterior angle bisector at vertex $A$ meets the line $BC$ at point $A_0$, and the exterior angle bisector at vertex $B$ meets the line $AC$ at point $B_0$. If $|AA_0| = |BB_0|$, determine the ang... | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | Angle A = 132°, Angle B = 12°, Angle C = 36° | |
03zc | Given real number $\alpha$, please find the minimum real number $\lambda = \lambda(\alpha)$, such that for any complex numbers $z_1, z_2$ and real number $x \in [0, 1]$, if $|z_1| \le \alpha |z_1 - z_2|$ then $|z_1 - xz_2| \le \lambda |z_1 - z_2|$. (posed by Li Shenghong)
 | [
"As shown in the figure, in the complex plane, points $A$, $B$ and $C$ denote the complex numbers $z_1$, $z_2$ and $xz_2$, respectively. $C$ is obviously on the segment $OB$. The vectors $\\vec{BA}$ and $\\vec{CA}$ are represented by complex numbers $z_1 - z_2$ and $z_1 - xz_2$, respectively. As $|z_1| \\le \\alpha... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | max{α, 1} | |
08nw | Problem:
All possible pairs of $n$ apples are weighed and the results are given to us in an arbitrary order. Can we determine the weights of the apples if
a. $n=4$
b. $n=5$
c. $n=6$? | [
"Solution:\n\na. No. Four apples with weights $1, 5, 7, 9$ and with weights $2, 4, 6, 10$ both give the results $6, 8, 10, 12, 14, 16$ when weighed in pairs.\n\nb. Yes. Let $a \\leq b \\leq c \\leq d \\leq e$ be the weights of the apples. As each apple is weighed 4 times, by adding all 10 pairwise weights and divid... | JBMO | 17th Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a: No; b: Yes; c: Yes | |
034e | Problem:
Given a quadrilateral $ABCD$ set $AB = a$, $BC = b$, $CD = c$, $DA = d$, $AC = e$ and $BD = f$. Prove that:
a) $a^{2} + b^{2} + c^{2} + d^{2} \geq e^{2} + f^{2}$;
b) if the quadrilateral $ABCD$ is cyclic then $|a - c| \geq |e - f|$. | [
"Solution:\n\na. Denote the midpoints of $AC$ and $BD$ by $M$ and $N$, respectively. Using the median formula for $\\triangle BMD$ we have\n$$\nMN^{2} = \\frac{2 MB^{2} + 2 MD^{2} - f^{2}}{4}\n$$\nWe also have\n$$\nMB^{2} = \\frac{2 a^{2} + 2 b^{2} - e^{2}}{4}, \\quad MD^{2} = \\frac{2 c^{2} + 2 d^{2} - e^{2}}{4}\n... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0i8m | Problem:
A point in three-space has distances $2, 6, 7, 8, 9$ from five of the vertices of a regular octahedron. What is its distance from the sixth vertex? | [
"Solution:\nBy a simple variant of the British Flag Theorem, if $ABCD$ is a square and $P$ any point in space, $AP^{2} + CP^{2} = BP^{2} + DP^{2}$. Four of the five given vertices must form a square $ABCD$, and by experimentation we find their distances to the given point $P$ must be $AP = 2$, $BP = 6$, $CP = 9$, $... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | sqrt(21) | |
0gey | 對於任一質數 $p$ 以及任一整數 $n$, 將 $n$ 除以 $p$ 的餘數記為 $d_p(n) \in \{0, 1, ..., p-1\}$。若正整數數列 $(a_0, a_1, a_2, ...)$ 符合 $a_0$ 與 $p$ 互質且當 $n \ge 0$ 時有 $a_{n+1} = a_n + d_p(a_n)$ 這樣的條件, 則我們稱這個正整數數列為一個 $p$-數列。
(a) 是否存在無窮多個質數 $p$, 可以找到 $p$-數列 $(a_0, a_1, a_2, ...)$ 和 $(b_0, b_1, b_2, ...)$, 使得對於無窮多個 $n$ 有 $a_n > b_n$ 且對於無窮多個 $n$ 有 $b_n... | [
"Answer: Yes, for both parts.\n\nFix some odd prime $p$, and let $T$ be the smallest positive integer such that $p \\mid 2^T - 1$; in other words $T$ is the multiplicative order of $2$ modulo $p$.\n\nConsider any $p$-sequence $(x_n) = (x_0, x_1, x_2, \\dots)$. Obviously, $x_{n+1} \\equiv 2x_n \\pmod p$ and therefor... | Taiwan | 2021 數學奧林匹亞競賽第二階段選訓營, 獨立研究(二) | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | Yes, for both parts. | |
0b0b | Problem:
How many positive rational numbers less than $1$ can be written in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime integers and $p+q=2020$? | [
"Solution:\n\nLet $\\frac{p}{q}$ be a positive rational number less than $1$ with $p$ and $q$ relatively prime and $p+q=2020$.\n\nSince $\\frac{p}{q} < 1$, we have $p < q$. Also, $p$ and $q$ are positive integers, so $p \\geq 1$ and $q \\geq 1$.\n\nFrom $p+q=2020$, we have $q=2020-p$.\n\nSo $\\frac{p}{q} = \\frac{p... | Philippines | Philippines Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | 400 | |
0e7k | Natural numbers $m$ and $n$ satisfy $n > m \ge 1$. Let $S$ be the set of all pairs of natural numbers $(x, y)$ where $1 \le x, y \le n$. Determine the least natural number $k$ such that for each subset $P \subseteq S$ with cardinality $k$ there exist $m+1$ pairs $(x_1, y_1), (x_2, y_2), \dots, (x_{m+1}, y_{m+1}) \in P$... | [
"We shall prove that the solution of the problem is $k = nm + 1$.\n\nSuppose that the problem were solvable for a $k \\le nm$. Let $A$ be a subset of $S$ that contains all pairs of natural numbers $(x, y)$ such that $1 \\le x \\le m$ and $1 \\le y \\le n$. Then $|A| = mn$. Consider $m+1$ arbitrary elements of this ... | Slovenia | Selection Examinations for the IMO 2013 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | nm + 1 | |
06hu | Let $a$, $b$, $c$ be distinct nonzero real numbers. If the equations $ax^3 + bx + c = 0$, $bx^3 + cx + a = 0$ and $cx^3 + ax + b = 0$ have a common root, prove that at least one of these equations has three real roots (not necessarily distinct). | [
"Let the common root be $t$. Adding the three equations, we get\n$$\n(a + b + c)(t^3 + t + 1) = 0.\n$$\nIf $t^3 + t + 1 = 0$, then $at^3 + bt + c = 0$ becomes $(b-a)t + (c-a) = 0$. Similarly, $bt^3 + ct + a = 0$ becomes $(c-b)t + (a-b) = 0$. As $a$, $b$, $c$ are distinct, we obtain $\\frac{c-a}{a-b} = t = \\frac{a-... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
02d8 | Box $A$ contains black balls and box $B$ contains white balls. Take a certain number of balls from $A$ and place them in $B$. Then take the same number of balls from $B$ and place them in $A$. Is the number of white balls in $A$ then greater, equal to, or less than the number of black balls in $B$? | [
"Suppose we move $n$ balls from $A$ to $B$, then $x$ white balls from $B$ and $n-x$ black balls from $B$. That leaves $x$ black balls in $B$. So both numbers equal $x$."
] | Brazil | II OBM | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | equal | |
0be3 | Find all real numbers $x$ for which the number $a = \frac{2x+1}{x^2+2x+3}$ is an integer. | [
"Suppose the integer $a$ is not equal to $0$. Then $|2x+1| \\ge |x^2+2x+3|$. Notice that $x^2+2x+3 = (x+1)^2 + 2 > 0$ to get $2x+1 \\ge x^2+2x+3$ or $2x+1 \\le -(x^2+2x+3)$, hence $x^2+2 \\le 0$ or $x^2+4x+4 \\le 0$. The former leads to no solutions and the latter gives $x = -2$ and then $a = -1 \\in \\mathbb{Z}$.\... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | {-2, -1/2} | |
099s | Find all integer solutions of the equation.
$$
x^3 = y^{16} + y^{15} + \ldots + y + 9.
$$
(proposed by Ts. Dashdorj) | [
"This equation is same as $x^3 - 8 = \\Phi_{17}(y)$, $\\Phi_{17}(y)$ is 17th cyclotomic polynomial.\n$p$ is prime number, if $p|\\Phi_{17}(x)$ then $p \\equiv 1 \\pmod{17}$ or $p|17$. So if for arbitrary $d|\\Phi_{17}(y)$ then $d \\equiv 1 \\pmod{17}$ or $d|17$.\n$(x-2)(x^2+2x+4) = \\Phi_{17}(y)$. If we have for $d... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operat... | English | proof and answer | No integer solutions | |
0h2d | Inside an inscribed quadrilateral $ABCD$ a point $P$ is chosen so that $\angle PBC = \angle PDA$, $\angle PCB = \angle PAD$. Prove that there exists a circle that touches the lines $AB$, $CD$ and also touches the circumscribed circles of the triangles $ABP$, $CDP$. | [
"Let $E$, $F$ be the points of intersection of the lines $AB$, $CD$ and $AD$, $BC$ respectively, and $O$ be the center of the circumscribed circle of the quadrilateral $ABCD$ (Fig. 50). On the half-line $OE$ we choose the point $P'$, such that $OE \\cdot OP' = OA^2$. Let $\\angle AOB = 2\\alpha$, $\\angle BOC = 2\\... | Ukraine | Problems of Ukrainian Authors | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
014n | Problem:
Prove that for every polynomial $P(x)$ with real coefficients there exist a positive integer $m$ and polynomials $P_{1}(x), P_{2}(x), \ldots, P_{m}(x)$ with real coefficients such that
$$
P(x)=\left(P_{1}(x)\right)^{3}+\left(P_{2}(x)\right)^{3}+\cdots+\left(P_{m}(x)\right)^{3} .
$$ | [
"Solution:\nWe will prove by induction on the degree of $P(x)$ that all polynomials can be represented as a sum of cubes. This is clear for constant polynomials.\n\nNow we proceed to the inductive step. It is sufficient to show that if $P(x)$ is a polynomial of degree $n$, then there exist polynomials $Q_{1}(x), Q_... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0cf9 | Let $k \ge 2$ be a given positive integer. Determine the number of pairs $(m, n)$ of positive integers, with $m, n < 2^k$, such that the equation $z^{2^k} - z^m - z^n = 1$ has at least one solution with modulus $1$.
Bogdan Blaga | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Intermediate Algebra > Complex numbers",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | (2^k - 1)(2^k - 2) | |
082k | Problem:
Per ogni intero $n$, sia $S(n)$ la somma delle cifre di $n$ (in base decimale). Qual è il più piccolo intero $N$ per cui $S(S(N)) \geq 10$ ? | [
"Solution:\n\nLa risposta è $199$. Infatti se $n$ avesse $1$ o $2$ cifre, $n \\leq 99$, e allora $S(n) \\leq 18$ e $S(S(n)) \\leq 9$, dunque $n$ ha almeno tre cifre. Osserviamo che per $n \\leq 198$, ancora $S(n) \\leq 18$. Invece se $n=199$, $S(n)=19$ e $S(S(n))=10$, come cercato."
] | Italy | Progetto Olimpiadi di Matematica 2003 GARA di SECONDO LIVELLO TRIENNIO | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 199 | |
02mo | Problem:
Os algarismos $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ e $9$ foram escritos (numa ordem desconhecida) ao redor de uma circunferência. Lendo esses algarismos de três em três no sentido horário, formam-se nove números de três algarismos. Determine a soma desses nove números. | [
"Solution:\n\nConsidere os algarismos $a_1, a_2, \\ldots, a_9$ dispostos ao redor da circunferência, em ordem. Os nove números formados são:\n\n$N_1 = 100a_1 + 10a_2 + a_3$\n\n$N_2 = 100a_2 + 10a_3 + a_4$\n\n$N_3 = 100a_3 + 10a_4 + a_5$\n\n$\\ldots$\n\n$N_9 = 100a_9 + 10a_1 + a_2$\n\nA soma total é:\n\n$S = N_1 + N... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 4995 | |
0f8q | Problem:
Do there exist two reals whose sum is rational, but the sum of their $n$th powers is irrational for all $n > 1$? Do there exist two reals whose sum is irrational, but the sum of whose $n$th powers is rational for all $n > 1$? | [] | Soviet Union | 23rd ASU | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | Yes; No | |
0bxu | Find all pairs of integers $(x, y)$, for which
$$
2^x + \log_3 x = y^2 \text{ and } 2^y + \log_3 y = x^2.
$$ | [
"The equality $2^x + \\log_3 x + x^2 = 2^y + \\log_3 y + y^2$ is readily obtained, and since the function $f: (0, +\\infty) \\to \\mathbb{R}, f(t) = 2^t + \\log_3 t + t^2$ is increasing, hence one-to-one, we deduce that $x = y$.\n\nObserve that $x = 3$ is a solution for the equation $2^x + \\log_3 x = x^2$, while $... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | (3, 3) | |
0jfv | Problem:
For an even positive integer $n$, Kevin has a tape of length $4n$ with marks at $-2n, -2n+1, \ldots, 2n-1, 2n$. He then randomly picks $n$ points in the set $-n, -n+1, -n+2, \ldots, n-1, n$, and places a stone on each of these points. We call a stone 'stuck' if it is on $2n$ or $-2n$, or either all the points... | [
"Solution:\n\nAnswer: $\\frac{1}{n-1}$\n\nAfter we have selected the positions of the initial $n$ stones, we number their positions: $a_{1} < a_{2} < \\ldots < a_{n}$. The conditions on how we move the stones imply that the expected value of $\\left(a_{i}-a_{j}\\right)$ after $t$ minutes is still equal to $a_{i}-a_... | United States | HMMT 2013 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 1/(n-1) | |
04nl | A positive integer is called *Babylonic* if it is greater than $9$, and if its base-$60$ representation is the same as its decimal representation without the first digit. For example, the number $123$ is Babylonic since $123 = (23)_{60}$. How many Babylonic numbers less than $10\,000$ are there? (Mea Bombardelli) | [] | Croatia | Croatia_2018 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 50 | |
08k3 | Problem:
If the positive integers $x$ and $y$ are such that both $3x + 4y$ and $4x + 3y$ are perfect squares, prove that both $x$ and $y$ are multiples of $7$. | [
"Solution:\n\nLet\n$$\n3x + 4y = m^2, \\quad 4x + 3y = n^2\n$$\nThen\n$$\n7(x + y) = m^2 + n^2 \\Rightarrow 7 \\mid m^2 + n^2\n$$\nConsidering $m = 7k + r$, $r \\in \\{0, 1, 2, 3, 4, 5, 6\\}$, we find that $m^2 \\equiv u \\pmod{7}$, $u \\in \\{0, 1, 2, 4\\}$ and similarly $n^2 \\equiv v \\pmod{7}$, $v \\in \\{0, 1,... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0ars | Problem:
Find the sum $\sum_{k=1}^{19} k\binom{19}{k}$. | [
"Solution:\n$\\binom{n}{k} = \\frac{n}{k} \\binom{n-1}{k-1} \\Rightarrow k\\binom{n}{k} = n\\binom{n-1}{k-1}$. Then, $n=19$:\n\n$$\n\\sum_{k=1}^{19} k\\binom{19}{k} = \\sum_{k=1}^{19} 19\\binom{18}{k-1} = 19 \\sum_{k=1}^{19} \\binom{18}{k-1}\n$$\n\nNow, $\\sum_{k=1}^{19} \\binom{18}{k-1} = \\sum_{j=0}^{18} \\binom{... | Philippines | 13th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 19·2^{18} | |
0gzs | Let $P$ and $Q$ be two points inside parallelogram $ABCD$, that are symmetric with respect to the point of intersection of the diagonals. Prove that the circumcircles of triangles $ABP$, $CDP$, $BCQ$ and $ADQ$ have a common point. | [
"Denote by $X$ the second point of intersection of the circumcircles of $\\triangle ADQ$ and $BCQ$, $QE$ the ray which has the same direction as ray $CB$ (Fig.12).\n\nThen $\\angle EQB = \\angle QBC$, $\\angle EQA = \\angle QAD$\n$\\Rightarrow \\angle AQB = \\angle QAD + \\angle QBC = 180^\\circ - \\angle QXD + 180... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010) | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0cv1 | 100 dwarves whose weights are $1$, $2$, $\ldots$, $100$ lb came to the left bank of a river. They cannot swim, but they have a boat which can take up to $100$ lb. When a boat crosses the river, one of the dwarves in it is an oarsman; while performing one crossing, the oarsman remains the same. Due to the stream, it is ... | [
"Answer. No.\nCall the dwarves weighing at least $50$ lb heavy; assume they oared backwards $d$ times. Then there were at least $51 + d$ their forward passages, and hence at least $50$ backward passages with no heavy dwarves oaring. But who could oar during these $50$ passages?\n\nПервое решение. Предположим, что п... | Russia | XLIII Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English; Russian | proof and answer | No | |
0az8 | Problem:
Let $n$ be a positive integer. An $n \times n$ matrix (a rectangular array of numbers with $n$ rows and $n$ columns) is said to be a platinum matrix if
(i) the $n^2$ entries are integers from $1$ to $n$;
(ii) each row, each column, and the main diagonal (from the upper left corner to the lower right corner) co... | [
"Solution:\nThere is no platinum matrix for $n=1$ and for $n=2$. We claim that a platinum matrix exists for all integers $n \\geq 3$.\n\nDefine a transversal as a collection of $n$ matrix entries which are taken from distinct rows and columns, and which contains each of the numbers $1$ to $n$.\n\nWhen $n \\geq 3$ i... | Philippines | 20th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Other"
] | null | proof and answer | all integers n ≥ 3 | |
0dxi | Problem:
Poišči vsa praštevila $p$, za katera je število $7^{p-2}+9 p^{4}$ popoln kvadrat. | [
"Solution:\n\nNaj bo $7^{p-2}+9 p^{4}=m^{2}$. Potem je $7^{p-2}=m^{2}-9 p^{4}=(m-3 p^{2})(m+3 p^{2})$. Torej je $m-3 p^{2}=7^{r}$ in $m+3 p^{2}=7^{p-r-2}$ za nek $r \\geq 0$ in velja $p-r-2>r$. Enakosti odštejemo in dobimo $6 p^{2}=7^{r}(7^{p-2 r-2}-1)$. Torej je $p=7$ ali $r=0$. Prvi primer je res rešitev, saj dob... | Slovenia | 51. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | p = 7 | |
01ps | Let $I$ be the center of inscribed circle of the non-isosceles triangle $ABC$. The ray $AI$ meets circumscribed circle of the triangle $ABC$ at point $D$. The circle passing through $C$, $D$, and $I$ meets again the ray $BI$ at point $K$.
Prove that $BK = CK$. | [
"Let $O$ be the circumcenter of the triangle $\\triangle ABC$. Let $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$, $\\angle ACB = \\gamma$. We construct the line passing through $D$ and $O$. Let $L$ be the point of intersection of this line and the ray $BI$. Since $AI$ is a bisector of the angle $BAC$, we have $BD... | Belarus | BelarusMO 2013_s | [
"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"
] | null | proof only | null | |
08o5 | Problem:
Find all ordered triples $(x, y, z)$ of integers satisfying the following system of equations:
$$
\begin{aligned}
x^{2}-y^{2} & =z \\
3 x y+(x-y) z & =z^{2}
\end{aligned}
$$ | [
"Solution:\nIf $z=0$, then $x=0$ and $y=0$, and $(x, y, z)=(0,0,0)$.\n\nLet us assume that $z \\neq 0$, and $x+y=a$ and $x-y=b$ where $a$ and $b$ are nonzero integers such that $z=a b$. Then $x=(a+b)/2$ and $y=(a-b)/2$, and the second equation gives $3 a^{2}-3 b^{2}+4 a b^{2}=4 a^{2} b^{2}$.\n\nHence\n$$\nb^{2}=\\f... | JBMO | 17th Junior Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (0,0,0), (1,0,1), (0,1,-1), (1,2,-3), (2,1,3) | |
03fp | Let $N$ be a positive integer. The sequence $x_1, x_2, \dots$ of non-negative reals is defined by
$$
x_n^2 = \sum_{i=1}^{n-1} \sqrt{x_i x_{n-i}}
$$
for all positive integers $n > N$. Show that there exists a constant $c > 0$, such that $x_n \le \frac{n}{2} + c$ for all positive integers $n$. | [
"Applying AM-GM inequality to each term of the right-hand side, we get,\n$$\nx_n^2 \\le \\sum_{i=1}^{n-1} x_i, \\quad n > N.\n$$\nLet us set $x_n = a \\cdot n + \\Delta(n)$, where $a$ is a constant that will be determined later. Vaguely speaking, we want to pick a value for $a$ such that $\\Delta(n)$ has a smaller ... | Bulgaria | 4 Bulgarian National Olympiad - Regional Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0jrw | Problem:
For positive integers $n$, let $S_{n}$ be the set of integers $x$ such that $n$ distinct lines, no three concurrent, can divide a plane into $x$ regions (for example, $S_{2} = \{3, 4\}$, because the plane is divided into 3 regions if the two lines are parallel, and 4 regions otherwise). What is the minimum $i... | [
"Solution:\n\nFor $S_{3}$, either all three lines are parallel (4 regions), exactly two are parallel (6 regions), or none are parallel (6 or 7 regions, depending on whether they all meet at one point), so $|S_{3}| = 3$.\n\nThen, for $S_{4}$, either all lines are parallel (5 regions), exactly three are parallel (8 r... | United States | HMMT February 2016 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 4 | |
017o | In an acute triangle $ABC$, the segment $CD$ is an altitude and $H$ is the orthocenter. Given that the circumcenter of the triangle lies on the line containing the bisector of the angle $DHB$, determine all possible values of $\angle CAB$. | [
"The value is $\\angle CAB = 60^\\circ$.\nDenote by $\\ell$ the line containing the angle bisector of $DHB$, and let $E$ be the point where the ray $CD \\to$ intersects the circumcircle of the triangle $ABC$ again. The rays $HD \\to$ and $HB \\to$ are symmetric with respect to $\\ell$ by the definition of $\\ell$. ... | Baltic Way | BALTIC WAY | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 60° | |
07sa | A round table has $2N$ chairs around it. Due to social distancing guidelines, no two people are allowed to sit next to each other. How many different ways are there to choose seats around the table on which $N-1$ guests can be seated? | [
"**Solution 1.** Either there are 3 consecutive empty seats or two separate pairs of empty seats. In the first case, there are $2N$ possibilities as the middle of the three empty seats can be any of the $2N$ seats. In the second case, there are two pairs of empty seats. If the seats are numbered 1 to $2N$ then one ... | Ireland | IRL_ABooklet_2020 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | N^2 | |
055i | The leader of an IMO team chooses positive integers $n$ and $k$ with $n > k$, and announces them to the deputy leader and a contestant. The leader then secretly tells the deputy leader an $n$-digit binary string, and the deputy leader writes down all $n$-digit binary strings which differ from the leader's in exactly $k... | [
"See IMO 2016 shortlist, problem C1."
] | Estonia | IMO Team Selection Contest I | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 2 if n = 2k; otherwise 1 | |
0jq4 | Problem:
James writes down three integers. Alex picks some two of those integers, takes the average of them, and adds the result to the third integer. If the possible final results Alex could get are $42$, $13$, and $37$, what are the three integers James originally chose? | [
"Solution:\nLet $x$, $y$, $z$ be the integers. We have\n$$\n\\begin{aligned}\n& \\frac{x+y}{2} + z = 42 \\\\\n& \\frac{y+z}{2} + x = 13 \\\\\n& \\frac{x+z}{2} + y = 37\n\\end{aligned}\n$$\nAdding these three equations yields $2(x+y+z) = 92$, so $x+y+z = 46$.\n\nNow, from $\\frac{x+y}{2} + z = 42$:\n$$\n\\frac{x+y}{... | United States | HMMT November 2015 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | -20, 28, 38 | |
0iye | Problem:
Which is greater, $\log_{2008}(2009)$ or $\log_{2009}(2010)$? | [
"Solution:\n\nLet $f(x) = \\log_{x}(x+1)$. Then\n$$\nf'(x) = \\frac{x \\ln x - (x+1) \\ln (x+1)}{x(x+1) (\\ln x)^2} < 0\n$$\nfor any $x > 1$, so $f$ is decreasing. Thus $\\log_{2008}(2009)$ is greater."
] | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | log_{2008}(2009) | |
02xi | Problem:
Dois candidatos participaram de uma eleição com $p+q$ eleitores. O candidato $A$ recebeu $p$ votos e o candidato $B$ recebeu $q$ votos, com $p>q$. Durante a apuração, é registrado apenas um voto de cada vez em um quadro. Seja $r$ a probabilidade de que o número associado ao candidato $A$ no quadro seja sempre... | [
"Solution:\n\na) Podemos fazer listas com as letras $A$ e $B$ representando as possíveis ordens de votos apurados. Por exemplo, a lista $A A B A B$ indica que os dois primeiros e o quarto voto apurados foram para o candidato $A$, o terceiro e o quinto para o candidato $B$. Existem exatamente 10 listas com 3 letras ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | a) 1/5; b) 1/2019 | |
032u | Problem:
Let $k$ be an integer, $1 < k < 100$. For every permutation $a_{1}, a_{2}, \ldots, a_{100}$ of the integers $1, 2, \ldots, 100$, set $a_{101} = 0$ and choose the least integer $m > k$ such that $a_{m}$ is less than at least $k-1$ of the numbers $a_{1}, a_{2}, \ldots, a_{k}$. Find all $k$ for which the number o... | [
"Solution:\nConsider the more general problem for $n+1$ numbers instead of $100$ and denote by $p_{k, n+1}$ the probability that $a_{m}$ be equal to $1$. If $1$ is not the last number of a permutation of the positive integers $1, 2, \\ldots, n+1$, then $a_{m} = 1$ with probability $p_{k, n}$ (we assume that $p_{n, ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 45 and 55 | |
0321 | Problem:
Let $A_{1}$, $B_{1}$ and $C_{1}$ be respectively the midpoints of the sides $BC$, $CA$ and $AB$ of $\triangle ABC$ with centroid $M$. The line through $A_{1}$ and parallel to $BB_{1}$ meets the line $B_{1}C_{1}$ at a point $D$. Prove that if the points $A$, $B_{1}$, $M$ and $C_{1}$ are concyclic, then $\Varang... | [
"Solution:\nSince $A_{1}D \\parallel MB_{1}$ and $A$, $B_{1}$, $M$, $C_{1} \\in k$, it follows that\n$$\n\\Varangle AA_{1}D = \\Varangle AMB_{1} = \\Varangle AC_{1}B_{1} = \\Varangle ABC.\n$$\nUsing that $B_{1}C_{1} \\parallel BA_{1}$ and $A_{1}D \\parallel BB_{1}$ we conclude that $BA_{1}DB_{1}$ is a parallelogram... | Bulgaria | Bulgarian Mathematical Competitions | [
"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"
] | null | proof only | null | |
0g27 | Problem:
Sei $k$ der Inkreis des Dreiecks $A B C$ mit Inkreismittelpunkt $I$. Der Kreis $k$ berühre die Seiten $B C$, $C A$ und $A B$ in den Punkten $D$, $E$, respektive $F$. Sei $G$ der Schnittpunkt der Geraden $A I$ und des Kreises $k$, der zwischen $A$ und $I$ liegt. Nehme an, $B E$ und $F G$ seien parallel. Zeige,... | [
"Solution:\n\nDa $A I$ die Winkelhalbierende von $\\angle C A B$ ist und $A E=A F$ gilt, sind die Dreiecke $A G E$ und $A G F$ kongruent. Somit gilt $\\angle A E G=\\angle A F G$. Da $A E=A F$, hat man auch $\\angle A E F=\\angle A F E$ und somit\n$$\n\\angle F E G=\\angle F E A-\\angle G E A=\\angle F E G=\\angle ... | Switzerland | SMO - Finalrunde | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jre | Problem:
Let $ABCD$ be a regular tetrahedron with side length $1$. Let $X$ be the point in triangle $BCD$ such that $[XBC] = 2[XBD] = 4[XCD]$, where $[\varpi]$ denotes the area of figure $\varpi$. Let $Y$ lie on segment $AX$ such that $2AY = YX$. Let $M$ be the midpoint of $BD$. Let $Z$ be a point on segment $AM$ such... | [
"Solution:\n\nWe apply three-dimensional barycentric coordinates with reference tetrahedron $ABCD$. The given conditions imply that\n\n$$\n\\begin{aligned}\nX & =(0: 1: 2: 4) \\\\\nY & =(14: 1: 2: 4) \\\\\nM & =(0: 1: 0: 1) \\\\\nZ & =(t: 1: 0: 1)\n\\end{aligned}\n$$\n\nfor some real number $t$. Normalizing, we obt... | United States | HMMT February 2015 | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | 4/7 | |
04ot | 42 pebbles are placed on a table. Two players take turns making moves. When making a move, a player has to take at least one pebble, but no more than half of the remaining pebbles. The player after whose move there is exactly one pebble left on the table, wins.
Which player can win regardless of how his opponent plays?... | [] | Croatia | Croatian Mathematical Society Competitions | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | First player | |
0jnq | Problem:
Let $ABC$ be an acute triangle with orthocenter $H$. Let $D, E$ be the feet of the $A$-, $B$-altitudes respectively. Given that $AH = 20$ and $HD = 15$ and $BE = 56$, find the length of $BH$. | [
"Solution:\nLet $x$ be the length of $BH$. Note that quadrilateral $ABDE$ is cyclic, so by Power of a Point, $x(56 - x) = 20 \\cdot 15 = 300$. Solving for $x$, we get $x = 50$ or $6$. We must have $BH > HD$ so $x = 50$ is the correct length."
] | United States | HMMT November 2015 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof and answer | 50 | |
0637 | Problem:
Es seien $m, n \geqslant 4$ zwei ganze Zahlen. Wir betrachten ein $m \times n$-Gitterrechtteck, das von $m+1$ horizontalen und $n+1$ vertikalen Strecken gebildet wird. Die Schnittpunkte dieser Strecken heißen Ecken. Es sei $P$ ein von Selbstüberschneidungen freier, geschlossener Weg, der durch jede der $(m-1)... | [
"Solution:\n\nEs sei $D$ die Anzahl der Gitterquadrate, von denen genau eine Seite zu $P$ gehört, $E$ die Anzahl der Gitterquadrate, von denen genau zwei Seiten zu $P$ gehören, die zudem benachbart sind, und schließlich $F$ die Anzahl der Gitterquadrate von denen genau drei Seiten zu $P$ gehören. Wegen $m, n \\geqs... | Germany | 1. Auswahlklausur | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
09ze | On a long street, there are four houses, numbered from $1$ to $4$, where the distances between the houses are all distinct. The houses have their front door directly on the street. There are eight people living in the first house, two people each in the second and the third house, and three people in the fourth house. ... | [] | Netherlands | Junior Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | MCQ | A | |
023j | Problem:
a) Leandro desenha uma Lúnula de Hipócrates como mostrado na figura a seguir:
Nesta figura, o triângulo $A B C$ é retângulo e isósceles. A lúnula é a região em forma de lua crescente interna a uma semicircunferência e externa à outra semicircunferência, como mostra a figura. A primeir... | [
"Solution:\na) Denotemos por $r$ o comprimento dos catetos $A B$ e $A C$. Aplicando o Teorema de Pitágoras no triângulo retângulo $A B C$, temos que o comprimento da hipotenusa $B C$ é igual a $\\sqrt{2} r$. Observe a figura abaixo:\n\nComo mostrado acima, o comprimento do raio da semicircu... | Brazil | null | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
02gz | Let $a$ and $b$ be real numbers. Define $f_{a,b}: \mathbb{R}^2 \to \mathbb{R}^2$ by $f_{a,b}(x, y) = (a - b y - x^2, x)$. If $P = (x, y) \in \mathbb{R}^2$, define $f_{a,b}^0(P) = P$ and $f_{a,b}^{k+1}(P) = f_{a,b}(f_{a,b}^k(P))$ for all nonnegative integers $k$.
The set $\operatorname{per}(a, b)$ of the *periodic poin... | [
"Let the *orbit* $\\operatorname{orb}(P)$ of $P = (x_0, y_0)$ be the least $n > 0$ such that $f_{a,b}^n(P) = P$.\n\nAlso, let $f_{a,b}^k(P) = (x_k, y_k)$. We have $(x_{k+1}, y_{k+1}) = (a - b y_k - x_k^2, x_k)$, that is,\n$y_{k+1} = x_k$ and, consequently, $x_{k+1} = a - b x_{k-1} - x_k^2$.\n\nSum this relation ove... | Brazil | XXVI OBM | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | -((b-1)^2)/4 | |
08t5 | In the diagram given below, the triangles $\triangle OAB$, $\triangle OBC$, $\triangle OCD$ are isosceles right triangles with $\angle OAB$, $\angle OBC$, $\angle OCD$ being the right angle, respectively. Find the area of the triangle $\triangle OAB$ if the area of the triangle $\triangle OCD$ is 12.
![](attached_imag... | [
"Since the length of the hypotenuse of an isosceles right triangle is $\\sqrt{2}$ times the length of each of the other sides, by letting $OA = x$ we have $OB = \\sqrt{2}x$, $OC = \\sqrt{2} \\cdot \\sqrt{2}x = 2x$. Then the ratio of the area of the triangle $\\triangle OAB$ to that of the triangle $\\triangle OCD$ ... | Japan | Japan Junior Mathematical Olympiad First Round | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3 | |
0967 | Problem:
În triunghiul $A B C$ mediana $A M$ şi bisectoarea $B N$ se intersectează în punctul $P$. Determinaţi măsurile unghiurilor triunghiului $A B C$, dacă se ştie că dreptele $M N$ şi $B C$ sunt reciproc perpendiculare, iar $B P: A N=3: 2$. | [
"Solution:\n\nFie $N Q \\| A M, Q \\in B C$. Din ipoteză avem $B P=3 x, A N=2 x$. Fie $P N=y$. În triunghiul $B C N$ segmentul $M N$ este mediană şi înălțime, atunci triunghiul este isoscel, iar $B N=C N=3 x+y$. Întrucât $P M \\| N Q$, conform teoremei lui Thales în\n\n\n\ntriunghiul $B N Q... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | angle BAC = 90°, angle ACB = 30°, angle ABC = 60° | |
07ft | Let $a$, $b$, $c$ be non-zero distinct real numbers so that there exist functions $f$, $g : \mathbb{R}^+ \to \mathbb{R}$ so that
$$
a f(xy) + b f\left(\frac{x}{y}\right) = c f(x) + g(y)
$$
for all positive real $x$ and large enough $y$.
Prove that there exists a function $h : \mathbb{R}^+ \to \mathbb{R}$ so that
$$
f(x... | [
"Putting $xy$, $\\frac{x}{y}$ instead of $x$, we find that\n$$\na f(xy^2) + b f(y) = c f(xy) + g(y)\n$$\nand,\n$$\na f(x) + b f\\left(\\frac{x}{y^2}\\right) = c f\\left(\\frac{x}{y}\\right) + g(y).\n$$\nMultiplying the first equation by $a$ and the second one by $b$, then, adding and using the original equation, we... | Iran | 37th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof only | null | |
00w6 | Problem:
Prove that the equation $x^{2}-7 y^{2}=1$ has infinitely many solutions in natural numbers. | [
"Solution:\nFor any solution $(m, n)$ of the equation we have $m^{2}-7 n^{2}=1$ and\n$$\n1=\\left(m^{2}-7 n^{2}\\right)^{2}=\\left(m^{2}+7 n^{2}\\right)^{2}-7 \\cdot(2 m n)^{2} .\n$$\nThus $\\left(m^{2}+7 n^{2}, 2 m n\\right)$ is also a solution. Therefore it is sufficient to note that the equation $x^{2}-7 y^{2}=1... | Baltic Way | Baltic Way | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Algebraic Number Theory > Quadratic fields"
] | null | proof only | null | |
0d61 | Let $S$ be a positive integer divisible by all the integers $1, 2, \ldots, 2015$ and $a_{1}, a_{2}, \ldots, a_{k}$ numbers in $\{1, 2, \ldots, 2015\}$ such that
$$
2S \leq a_{1} + a_{2} + \ldots + a_{k},
$$
Prove that we can select from $a_{1}, \ldots, a_{k}$ some numbers so that the sum of these selected numbers is eq... | [
"Notice that because $S$ is divisible by $2013$, $2014$ and $2015$, we have $S \\geq 2013 \\times 2014 \\times 2015$.\n\nAssume that each integer $1, 2, \\ldots, 2015$ appears at most $2013$ times in the list of numbers $a_{1}, a_{2}, \\ldots, a_{k}$. We have\n$$\n2S \\leq a_{1} + \\cdots + a_{k} \\leq 2013(1 + 2 +... | Saudi Arabia | SAMC 2015 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Other"
] | English, Arabic | proof only | null | |
0hkj | Problem:
A building has seven rooms numbered $1$ through $7$, all on one floor, and any number of doors connecting these rooms. These doors may be one-way, admitting motion in only one of the two directions, or two-way. In addition, there is a two-way door between room $1$ and the outside, and a treasure in room $7$. Y... | [
"Solution:\nOn the way to the treasure, no room need be entered twice; otherwise the path could be shortened by skipping the loop. Thus, the minimal path to the treasure, if it exists, is at most $7$ steps long. Similarly, the minimal path from the treasure to the outside is at most $7$ steps long, so the total num... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 14 | |
0jc3 | Problem:
$ABCD$ is a parallelogram satisfying $AB = 7$, $BC = 2$, and $\angle DAB = 120^\circ$. Parallelogram $ECFA$ is contained in $ABCD$ and is similar to it. Find the ratio of the area of $ECFA$ to the area of $ABCD$. | [
"Solution:\n\nAnswer: $\\boxed{\\dfrac{39}{67}}$ First, note that $BD$ is the long diagonal of $ABCD$, and $AC$ is the long diagonal of $ECFA$. Because the ratio of the areas of similar figures is equal to the square of the ratio of their side lengths, we know that the ratio of the area of $ECFA$ to the area of $AB... | United States | HMMT November 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | 39/67 | |
06eh | For every positive integer $n$, find (with proof) the number of ways of sitting $2n$ people $A_1, B_1, A_2, B_2, \dots, A_n, B_n$ around a round table so that $A_i$ and $B_i$ are not next to each other for $i = 1, 2, \dots, n$. Two sitting arrangements are considered the same if one can be obtained from another by a ro... | [
"The answer is $0$ if $n = 1$, and is $$\\sum_{k=0}^{n} (-1)^k \\binom{n}{k} 2^k (2n-k-1)!$$ if $n \\ge 2$, or any equivalent expression.\n\nThe result is obvious for $n = 1$. Consider $n \\ge 2$. For each $k = 0, 1, \\dots, n$, let $N_k$ be the number of ways to arrange the $2n$ people such that $A_i, B_i$ are nex... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 0 for n = 1; for n ≥ 2: ∑_{k=0}^{n} (-1)^k * C(n, k) * 2^k * (2n - k - 1)! | |
00qa | The International Mathematical Olympiad is being organized in Japan, where a folklore belief is that the number 4 brings bad luck. The opening ceremony takes place at the Grand Theatre where each row has the capacity of 55 seats. What is the maximum number of contestants that can be seated in a single row with the rest... | [
"Denote the desired number by $t$. Consider the set $B$ consisting of all integers of the form $10a + b$, for $a = 0, 1, 2, 3, 4, 5$ and $b = 1, 2, 3, 4, 5$, i.e.\n$$\nB = \\{1, 2, 3, 4, 5, 11, 12, 13, 14, 15, 21, 22, 23, 24, 25, \\dots, 51, 52, 53, 54, 55\\}.\n$$\nIf $(10a_1 + b_1) - (10a_2 + b_2) = 5$ then $10(a_... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 30 | |
0g9i | 給定一圓 $\Gamma$ 以及 $\Gamma$ 上的三個定點 $A, B, C$, 同時給定一實數 $\lambda$, $0 < \lambda < 1$. 設 $P$ 為 $\Gamma$ 上不等於 $A, B, C$ 的一個動點, 並讓 $M$ 是 $CP$ 線段上滿足 $CM = \lambda \cdot CP$ 的點。令 $Q$ 為三角形 $AMP$ 與三角形 $BMC$ 的兩外接圓的第二個交點。證明: 當 $P$ 變動時, $Q$ 會落在一定圓上。 | [
"在證明中，我們將用 $\\angle(a, b)$ 代表直線 $a$ 與直線 $b$ 所夾的有向角。\n令 $D$ 是在 $AB$ 線段上滿足 $BD = \\lambda \\cdot BA$ 的點。我們將證明：要不\n$Q = D$，不然 $\\angle(DQ, QB) = \\angle(AB, BC)$；這兩個情形都會保證 $Q$ 點在某\n個通過 $D$ 點、並且與直線 $BC$ 在 $B$ 點相切的圓上變動。這就我們想證明\n的敘述。\n\n將三角形 $AMP$ 與三角形 $BMC$ 的外接圓分別記為 $\\omega_A, \\omega_B$。三條直線 $AP$, $BC$, $MQ$ 兩兩成為三個圓 $... | Taiwan | 二〇一五數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03tm | Suppose line $l$ through point $(0, 1)$ and curve $C: y = x + \frac{1}{x}$ ($x > 0$) intersect at two different points $M$ and $N$. Find the locus of the intersection points of two tangent lines of curve $C$ at $M$ and $N$ respectively. | [
"Denote the coordinates of $M$ and $N$ as $(x_1, y_1)$ and $(x_2, y_2)$ respectively. Denote the tangent lines of $C$ at $M$ and $N$ by $l_1$ and $l_2$ respectively, with their intersection point being $P(x_p, y_p)$. Suppose the slope ratio of line $l$ is $k$. Then we can write the equation of $l$ as $y = kx + 1$.\... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | All points with x = 2 and 2 < y < 2.5; equivalently, the open segment from (2, 2) to (2, 2.5). | |
0igf | Problem:
You and I play the following game on an $8 \times 8$ square grid of boxes: Initially, every box is empty. On your turn, you choose an empty box and draw an $X$ in it; if any of the four adjacent boxes are empty, you mark them with an $X$ as well. (Two boxes are adjacent if they share an edge.) We alternate tu... | [
"Solution:\n\nI can follow a symmetry strategy: whenever you play in the box $S$, I play in the image of $S$ under the $180^{\\circ}$ rotation about the center of the board. This ensures that the board will always be centrally symmetric at the beginning of your turn. Thus, if you play in an empty box $S$, its symme... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | 0 | |
0jo2 | Problem:
For positive integers $n$, prove that $\operatorname{gcd}(6 n+1,15 n+2)=1$. | [
"Solution:\n\nLet $a=6 n+1$ and $b=15 n+2$, and notice that $5 a-2 b=5(6 n+1)-2(15 n+2)=1$. Thus, if a positive integer $d$ divides both $a$ and $b$, we must have $d$ dividing $5 a-2 b=1$ as well. This implies $d=1$, as desired."
] | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
03wh | For a convex pentagon $ABCDE$, $AB = DE = EA$, $AB \neq EA$, and $B$, $C$, $D$, $E$ are concyclic. Prove that $A$, $B$, $C$, $D$ are concyclic if and only if $AC = AD$. (Posed by Xiong Bin) | [
"First, if $A$, $B$, $C$, $D$ are concyclic, by $AB = DE$ and $BC = EA$ we have $\\angle BAC = \\angle EDA$, $\\angle ACB = \\angle DAE$, so $\\angle ABC = \\angle DEA$, which means that $AC = AD$.\n\n\n\nSecond, if $AC = AD$, let $O$ be the center of the circle ($B$, $C$, $D$, $E$ are on t... | China | China Southeastern Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0awp | Problem:
Given $a_{n} \in \mathbb{Z}$ with $a_{10}=11$ and $a_{9}=-143$, determine the number of polynomials of the form
$$
P(x)=\sum_{n=0}^{10} a_{n} x^{n}
$$
such that the zeros of $P(x)$ are all positive integers. | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | final answer only | 3 | |
0bet | Find all increasing functions $f: \mathbb{N}^* \to \mathbb{N}^*$ with the property: for every $n \in \mathbb{N}^*$, the number $\frac{1}{n} \sum_{k=1}^{n} k f(k)$ is a perfect square not larger than $(n+1)^2$. | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | f(n) = 3n + 1 for all positive integers n | |
01ki | Is it possible to mark 8 distinct points (1 red, 3 blue, and 4 green points) on a plane so that the sum of the distances between the red point and the blue points is 6, the sum of the distances between the red point and the green points is 16, and the sum of the distances between the blue points and the green points is... | [
"For example, the points can be marked on the straight line in the following way: the blue points have the coordinates $-3$, $-1$, $2$, the red point has the coordinate $0$, and the green points have the coordinates $3$, $3.5$, $4.5$, $5$."
] | Belarus | 60th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | Yes; for example, place the blue points at -3, -1, 2; the red point at 0; and the green points at 3, 3.5, 4.5, 5. | |
00vk | Consider a table with $m$ rows and 22 columns. Each cell is filled with a number from the set $A = \{1, 2, 3, \dots, 2025\}$ (numbers may be repeated), such that for every pair of distinct numbers in $A$, there exists a row that contains exactly one of these two numbers. Find the minimum value of $m$. | [
"Let $x$ be the number of elements of $A$ that appear exactly once, and $y$ be the number of elements of $A$ that appear at least twice in the table. There is at most one number that does not appear in the table, otherwise, there would be two numbers that do not appear in any row, a contradiction. From there, we de... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | 176 | |
058r | Let $ABC$ be a triangle such that $AB = AC$. Point $K$ lies on the altitude drawn from vertex $A$ and point $L$ is chosen on the line $BK$ in such a way that $AL \parallel BC$. Prove that if $KC \perp CL$ then point $L$ lies on the bisector of the external angle on vertex $C$ of triangle $ABC$. | [
"As $AL \\parallel BC$ and $AK \\perp BC$ (Fig. 11), we have $\\angle KAL = 90^\\circ$. As $KAL$ and $KCL$ are both right angles, points $A$ and $C$ lie on circle with diameter $KL$. By inscribed angles, $\\angle KCA = \\angle KLA$. On the other hand, $\\angle KLA = \\angle KBC$ and, by symmetry of isosceles triang... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0jvx | Problem:
The vertices of a regular nonagon are colored such that 1) adjacent vertices are different colors and 2) if 3 vertices form an equilateral triangle, they are all different colors.
Let $m$ be the minimum number of colors needed for a valid coloring, and $n$ be the total number of colorings using $m$ colors. D... | [
"Solution:\n\nAnswer: $54$\n\nIt's clear that $m$ is more than $2$ since it's impossible to alternate the color of the vertices without having two of the same color adjacent (since the graph is not bipartite). However, it's possible to use $3$ colors. Number the vertices $1$ through $9$ in order and let the colors ... | United States | HMMT November 2016 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 54 | |
0acs | Mile imagined a number and said: "If I multiply the number by two and add this product to the half of the number I will get a number that is for 26 bigger than the third of the imagined number." What number did Mile imagine? | [
"Let $x$ be the number that Mile imagined. Then we solve the equation $2x + \\frac{1}{2}x = \\frac{1}{3}x + 26$, from where we get $12x + 3x = 2x + 156$ or $13x = 156$, $x = 12$. Mile imagined the number 12."
] | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 12 | |
0bki | Let $n$ be an integer greater than $1$ and let $S$ be a finite set containing more than $n+1$ elements. Consider the collection of all sets $\mathcal{A}$ of subsets of $S$ satisfying the following two conditions:
(a) Each member of $\mathcal{A}$ contains at least $n$ elements of $S$; and
(b) Each element of $S$ is cont... | [
"The required number is $m = |S| - n$. We begin by showing that any set $\\mathcal{A}$ of subsets of $S$ satisfying the two conditions in the statement has a subcover of cardinality at most $m$.\n\nThis is clear if $S$ is a member of $\\mathcal{A}$.\n\nAssume henceforth that $\\mathcal{A}$ does not contain $S$. If ... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | |S| - n | |
0kis | Problem:
Let $n$ be the answer to this problem. An urn contains white and black balls. There are $n$ white balls and at least two balls of each color in the urn. Two balls are randomly drawn from the urn without replacement. Find the probability, in percent, that the first ball drawn is white and the second is black. | [
"Solution:\n\nLet the number of black balls in the urn be $k \\geq 2$. Then the probability of drawing a white ball first is $\\frac{n}{n+k}$, and the probability of drawing a black ball second is $\\frac{k}{n+k-1}$. This gives us the equation\n$$\n\\frac{n k}{(n+k)(n+k-1)} = \\frac{n}{100}\n$$\nfrom which we get\n... | United States | HMMT November 2021 | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 19% | |
0e9o | Problem:
Za realni števili $a$ in $b$, kjer je $|a| \neq |b|$ in $a \neq 0$, velja
$$
\frac{a-b}{a^{2}+a b}+\frac{a+b}{a^{2}-a b}=\frac{3 a-b}{a^{2}-b^{2}}
$$
Določi vrednost izraza $\frac{b}{a}$. | [
"Solution:\n\nEnačbo pomnožimo z $a(a+b)(a-b)$ in dobimo\n$$\n(a-b)^{2}+(a+b)^{2}=a(3 a-b)\n$$\nKo odpravimo oklepaje in vse člene nesemo na desno stran, dobimo\n$$\n0=a^{2}-a b-2 b^{2}=(a-2 b)(a+b)\n$$\nKer je $a \\neq -b$, mora biti $a-2 b=0$ oziroma $a=2 b$. Ker $a$ ni enak 0 je torej $\\frac{b}{a}=\\frac{1}{2}$... | Slovenia | 58. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1/2 | |
06m7 | In $\triangle ABC$, $D$, $E$, $F$ are points on $BC$, $CA$ and $AB$ respectively such that the line segments $AD$, $BE$ and $CF$ meet at $G$. If the lengths of $DG$, $EG$, $BG$, $AG$ and $CF$ are $1$, $2$, $3$, $4$, $5$ respectively, find the area of $\triangle ABC$. | [
"Answer: $10\\sqrt{2}$\n\nWe use $[XYZ]$ to denote the area of $XYZ$. Suppose $[GDB] = b$ and $[GDC] = c$. We now make use of the side length ratios to get the following:\n\n* Using $AG:GD = 4:1$, we have $[AGB] = 4b$ and $[AGC] = 4c$.\n* Using $BG:GE = 3:2$, we have $[AGE] = \\frac{2}{3}(4b)$ and $[CGE] = \\frac{2... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem"
] | English | proof and answer | 10√2 | |
0eni | A poker set contains three chips of each of $2n$ different colours for a complete set of $6n$ chips. Let $p_n$ be the number of ways in which the chips can be partitioned into two piles of $3n$ chips each in such a way that no pile contains three chips of the same colour. Prove that $p_n$ is odd if and only if $n$ is a... | [
"No three chips of the same colour should be in the same pile, so each pile must contain a chip of each colour. So essentially we need to partition $2n$ chips into two piles of $n$ each. This can be done in $p_n = \\binom{2n}{n}/2$ ways.\n\nNow,\n$$\n\\begin{align*}\np_n &= \\frac{1}{2} \\binom{2n}{n} = \\frac{(2n)... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0525 | Each unit square in a $5 \times 5$ table is coloured either blue or yellow. Prove that there exists a rectangle with sides parallel to the edges of the table, such that the four unit squares in its corners have the same colour. | [
"Each row contains at least 3 squares with the same colour. Similarly, the dominating colour must be the same in at least 3 rows. W.l.o.g., suppose that the first 3 rows contain at least 3 blue squares each. If the first two rows contain two blue squares in the same columns then the desired rectangle exists. Otherw... | Estonia | Final Round of National Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0hfd | At the beginning of work, the memory of computer contained a single polynomial $x^2 - 1$. Each minute, the computer can do one of two things: choose any polynomial $f(x)$ from its memory and memorize also polynomials $f^2(x) - 1$ and $f(x^2 - 1)$, or chooses any two different polynomials $g(x)$ and $h(x)$ from its memo... | [] | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No | |
07dt | Let $\mathbb{Z}[x]$ be the set of polynomials with integer coefficients. Let $q$ be a prime number. A polynomial $P \in \mathbb{Z}[x]$ is called **q-reducible** if there exists $f_q, g_q \in \mathbb{Z}[x]$ such that $f_q, g_q$ have at least one non-constant coefficient which is not divisible by $q$, and all the coeffic... | [
"a.\n$$\nx^4 - 2x^3 + 3x^2 - 2x - 5 = (x^2 - x + 1)^2 - 6\n$$\nTherefore, it suffices to prove that there exists infinitely many prime numbers $q$ such that $q \\mid r^2 - 6$ for some $r \\in \\mathbb{Z}$, as in that case we have\n$$\nr^2 - 6 = (x^2 - x + 1)^2 - 6 - ((x^2 - x + 1)^2 - r^2) \\\\ = P(x) - (x^2 - x + ... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciproci... | null | proof and answer | a: There are infinitely many primes q for which the given polynomial is q-reducible. b: Yes; for example, x^6 + 3 is irreducible over the integers and is q-reducible for every prime q. | |
0acg | Let $ABC$ be a given acute angled triangle. $D$ is the foot for the altitude from $A$ to $BC$. Let $E$ be a point on $AD$ such that $\frac{AE}{ED} = \frac{CD}{DB}$ and $F$ is the foot for the altitude from $D$ to $BE$. Prove that $\angle AFC = 90^\circ$. | [
"Let $P$ be a point such that $ADCP$ is a rectangle. Then\n$$\n\\frac{AE}{ED} = \\frac{CD}{DB} = \\frac{AP}{DB}.\n$$\nSo we obtain that $B$, $E$ and $P$ are colinear, hence $\\angle DFP = 90^\\circ$. Because $\\angle DCP = 90^\\circ$, we obtain that $F$, $D$, $C$, $P$ lie on the circumference of the circumscribe ci... | North Macedonia | Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0cp5 | A positive integer $n$ is called *unfortunate* if it cannot be represented in a form $n = \frac{x^2 - 1}{y^2 - 1}$, where $x > 1$, $y > 1$ are integers. Determine whether the set of all unfortunate positive integers is finite or infinite.
Назовём натуральное число $n$ неудачным, если его нельзя представить в виде $n =... | [
"Докажем, что неудачным является любое число вида $n = p^2$, где $p$ — нечётное простое число. Предположим противное, т. е.\n$$\n(y^2 - 1)p^2 = x^2 - 1 \\quad (1)\n$$\nпри некоторых натуральных $x, y \\neq 1$. Тогда либо $x + 1$, либо $x - 1$ делится на $p$.\nПусть $x + 1 \\neq p$. Тогда $x - 1 = (x + 1) - 2$ не де... | Russia | Final round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English; Russian | proof and answer | infinite | |
0fbj | Problem:
Sea $(G, \cdot)$ un grupo y $e$ un elemento neutro. Probar que si todos los elementos $x$ de $G$ cumplen
$$
x \cdot x = e
$$
entonces $(G, \cdot)$ es abeliano (o sea, conmutativo). | [
"Solution:\n\nSean $a, b \\in G$. Entonces $a \\cdot b \\in G$ y $(a \\cdot b)(a \\cdot b) = e$. Esta igualdad se puede escribir, por la propiedad asociativa, $a(b \\cdot a) b = e$, y sale, multiplicando a la derecha por $b$, $a(b \\cdot a)(b \\cdot b) = e \\cdot b = b$. Volviendo a aplicar la propiedad asociativa,... | Spain | OME 11 | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
05ah | Kadi has a square pasture with a side length of $2$, where a goat lives. Since there is no fence around the pasture, Kadi fears that the goat might run away at any moment. Instead of getting a fence, Kadi decided that it would be cheaper to place scarecrows around the pasture, which the goat is afraid of. It is known t... | [
"Let $P$ be the center of the square pasture and let some scarecrow be at point $O$. Since the area bounded by the scarecrows must entirely contain the interior of the square and the points of the square are at least distance $1$ from the center of the square, we have $PO \\ge 1+1=2$ with the equality holding only ... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 7 | |
0636 | Problem:
Es sei $n$ eine ungerade natürliche Zahl und $x$ und $y$ seien zwei rationale Zahlen mit
$$
x^{n}+2 y=y^{n}+2 x
$$
Man zeige, dass $x=y$. | [
"Solution:\n\nWir überlegen uns zuerst\n($\\boxtimes$) Es sei $n \\geqslant 3$ eine ungerade natürliche Zahl. Weiterhin seien $a, b$ und $c$ drei ganze Zahlen mit\n$$\na^{n-1}+a^{n-2} b+\\ldots+b^{n-1}=2 c^{n-1}\n$$\nSodann ist $a=b=c=0$.\nWenn dies bei einem gewissen $n$ falsch wäre, könnten wir zu diesem ein Gege... | Germany | 1. Auswahlklausur | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
09q3 | Problem:
Vind alle functies $f: \mathbb{R} \rightarrow \mathbb{R}$ die voldoen aan
$$
f(x) f(y)=f(x+y)+x y
$$
voor alle $x, y \in \mathbb{R}$. | [
"Solution:\nMerk eerst op dat de functie $f(x)=0$ voor alle $x \\in \\mathbb{R}$ niet voldoet. Er is dus zeker een $x_{0} \\in \\mathbb{R}$ waarvoor $f\\left(x_{0}\\right) \\neq 0$. Vul nu $x=x_{0}$ en $y=0$ in: $f\\left(x_{0}\\right) f(0)=f\\left(x_{0}\\right)$. We mogen delen door $f\\left(x_{0}\\right)$, waardoo... | Netherlands | Dutch TST | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = x + 1 or f(x) = 1 − x | |
0k5u | Problem:
Let $ABCD$ be a parallelogram. Points $X$ and $Y$ lie on segments $AB$ and $AD$ respectively, and $AC$ intersects $XY$ at point $Z$. Prove that
$$
\frac{AB}{AX} + \frac{AD}{AY} = \frac{AC}{AZ}
$$ | [
"Solution:\nLet $X'$ and $Y'$ lie on segments $AB$ and $AD$ respectively such that $ZX' \\parallel AD$ and $ZY' \\parallel AB$. We note that triangles $AXY$ and $Y'YZ$ are similar, and that triangles $AY'Z$ and $ADC$ are similar. Thus, we have\n$$\n\\frac{AC}{AZ} = \\frac{AD}{AY'} \\text{ and } \\frac{AY'}{AY} = \\... | United States | HMMT February 2019 Team Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jyv | Problem:
Find the number of ordered triples of nonnegative integers $(a, b, c)$ that satisfy
$$
(a b+1)(b c+1)(c a+1)=84.
$$ | [
"Solution:\nThe solutions are $(0,1,83)$ and $(1,2,3)$ up to permutation.\n\nFirst, we do the case where at least one of $a, b, c$ is $0$. WLOG, say $a=0$. Then we have $1+b c=84 \\Longrightarrow b c=83$. As $83$ is prime, the only solution is $(0,1,83)$ up to permutation.\n\nOtherwise, we claim that at least one o... | United States | February 2017 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 12 | |
0cxj | Find all $n$ for which there are $n$ consecutive integers whose sum of squares is a prime. | [] | Saudi Arabia | SAMC | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | n = 2, 3, 6 | |
0hmz | Problem:
A finite number of points are drawn in the plane. Prove that one can select two of them, $A$ and $B$, such that:
a. $A$ and $B$ are not the same point.
b. No drawn point, other than $A$ itself, is closer to $A$ than $B$ is.
c. No drawn point, other than $B$ itself, is closer to $B$ than $A$ is. | [
"Solution:\n\nConsider all the distances $X Y$ between two different drawn points. Since a finite number of points are drawn, there are only finitely many distances, and one of them, say $d$, is minimal. Let $A$ and $B$ be two of the drawn points such that $A B = d$. Now, by definition, $A$ and $B$ are not the same... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
05ll | Problem:
Trouver tous les nombres premiers $p$ et $q$ tels que $p$ divise $5^{q}+1$ et $q$ divise $5^{p}+1$. | [
"Solution:\n\nNotons $\\alpha$ l'ordre de $5$ modulo $q$ et $\\beta$ l'ordre de $5$ modulo $p$. Comme $5^{2p} \\equiv 1 \\bmod q$, on en déduit que $\\alpha$ divise $2p$. Donc $\\alpha = 1, 2, p$ ou $2p$.\n\nSi $\\alpha = 1$, alors $5 \\equiv 1 \\bmod q$, donc $q = 2$. Si $\\alpha = 2$, alors $25 \\equiv 1 \\bmod q... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (2,2), (3,3), (13,2), (2,13), (3,7), (7,3) | |
0izd | Problem:
Let $f(n) = \sum_{k=1}^{n} \frac{1}{k}$. Then there exist constants $\gamma$, $c$, and $d$ such that
$$
f(n) = \ln(n) + \gamma + \frac{c}{n} + \frac{d}{n^{2}} + O\left(\frac{1}{n^{3}}\right)
$$
where the $O\left(\frac{1}{n^{3}}\right)$ means terms of order $\frac{1}{n^{3}}$ or lower. Compute the ordered pair ... | [
"Solution:\n\nAnswer: $\\left(\\frac{1}{2}, -\\frac{1}{12}\\right)$\n\nFrom the given formula, we pull out the term $\\frac{k}{n^{3}}$ from $O\\left(\\frac{1}{n^{4}}\\right)$, making $f(n) = \\log(n) + \\gamma + \\frac{c}{n} + \\frac{d}{n^{2}} + \\frac{k}{n^{3}} + O\\left(\\frac{1}{n^{4}}\\right)$. Therefore,\n$$\n... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | (1/2, -1/12) |
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