In the cartesian plane consider rectangles with sides parallel to the coordinate axes. We say that one rectangle is [i]below[/i] another rectangle if there is a line $g$ parallel to the $x$-axis such that the first rectangle is below $g$, the second one above $g$ and both rectangles do not touch $g$. Similarly, we say that one rectangle is [i]to the right of[/i] another rectangle if there is a line $h$ parallel to the $y$-axis such that the first rectangle is to the right of $h$, the second one to the left of $h$ and both rectangles do not touch $h$. Show that any finite set of $n$ pairwise disjoint rectangles with sides parallel to the coordinate axes can be enumerated as a sequence $(R_1,\dots,R_n)$ so that for all indices $i,j$ with $1 \le i<j \le n$ the rectangle $R_i$ is to the right of or below the rectangle $R_j$

Let $D$ and $E$ be points on $[BC]$ and $[AC]$ of acute $\triangle ABC$, respectively. $AD$ and $BE$ meet at $F$. If $|AF|=|CD|=2|BF|=2|CE|$, and $Area(\triangle ABF) = Area(\triangle DEC)$, then $Area(\triangle AFC)/Area(\triangle BFC) = ?$ $ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 2\sqrt2 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \sqrt2 \qquad \textbf{(E)}\ 1$

On a line, there are $n$ closed intervals (none of which is a single point) whose union we denote by $S$. It's known that for every real number $d$, $0<d\le 1$, there are two points in $S$ that are a distance $d$ from each other. [list](a) Show that the sum of the lengths of the $n$ closed intervals is larger than $\frac{1}{n}$. (b) Prove that, for each positive integer $n$, the $\frac{1}{n}$ in the statement of part (a) cannot be replaced with a larger number.[/list]

Find all values ​​of the real parameter $a$, for which the equation $(x -6\sqrt{x} + 8)\cdot \sqrt{x- a} = 0$ has exactly two distinct real solutions.

Suppose $P(x)$ is a polynomial with real coefficients, satisfying the condition $P(\cos \theta+\sin \theta)=P(\cos \theta-\sin \theta)$, for every real $\theta$. Prove that $P(x)$ can be expressed in the form$$P(x)=a_0+a_1(1-x^2)^2+a_2(1-x^2)^4+\dots+a_n(1-x^2)^{2n}$$for some real numbers $a_0, a_1, \dots, a_n$ and non-negative integer $n$. [i]Proposed by C.R. Pranesacher[/i]

Let a regular polygon with $p$ vertices be given, where $p$ is an odd prime number. At every vertex there is one monkey. An owner of monkeys takes $p$ peanuts, goes along the perimeter of polygon clockwise and delivers to the monkeys by the following rule: Gives the first peanut for the leader, skips the two next vertices and gives the second peanut to the monkey at the next vertex; skip four next vertices gives the second peanut for the monkey at the next vertex ... after giving the $k$-th peanut, he skips the $2 \cdot k$ next vertices and gives $k+1$-th for the monkey at the next vertex. He does so until all $p$ peanuts are delivered. [b]I.[/b] How many monkeys are there which does not receive peanuts? [b]II.[/b] How many edges of polygon are there which satisfying condition: both two monkey at its vertex received peanut(s)?

A ticket to a school play costs $ x$ dollars, where $ x$ is a whole number. A group of 9th graders buys tickets costing a total of $ \$48$, and a group of 10th graders buys tickets costing a total of $ \$64$. How many values of $ x$ are possible? $ \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

In the game of [i]Ring Mafia[/i], there are $2019$ counters arranged in a circle. $673$ of these counters are mafia, and the remaining $1346$ counters are town. Two players, Tony and Madeline, take turns with Tony going first. Tony does not know which counters are mafia but Madeline does. On Tony’s turn, he selects any subset of the counters (possibly the empty set) and removes all counters in that set. On Madeline’s turn, she selects a town counter which is adjacent to a mafia counter and removes it. Whenever counters are removed, the remaining counters are brought closer together without changing their order so that they still form a circle. The game ends when either all mafia counters have been removed, or all town counters have been removed. Is there a strategy for Tony that guarantees, no matter where the mafia counters are placed and what Madeline does, that at least one town counter remains at the end of the game? [i]Proposed by Andrew Gu[/i]

Let $n$ be the smallest positive integer that is a multiple of $75$ and has exactly $75$ positive integral divisors, including $1$ and itself. Find $n/75$.

Let $H$ be the orthocenter of an acute-angled triangle $ABC$. Point $X_A$ lying on the tangent at $H$ to the circumcircle of triangle $BHC$ is such that $AH=AX_A$ and $X_A \not= H$. Points $X_B,X_C$ are defined similarly. Prove that the triangle $X_AX_BX_C$ and the orthotriangle of $ABC$ are similar.

[url=https://artofproblemsolving.com/community/c677808][b]Cyprus IMO TST 2018[/b][/url] [url=https://artofproblemsolving.com/community/c6h1666662p10591751][b]Problem 1.[/b][/url] Determine all integers $n \geq 2$ for which the number $11111$ in base $n$ is a perfect square. [url=https://artofproblemsolving.com/community/c6h1666663p10591753][b]Problem 2.[/b][/url] Consider a trapezium $AB \Gamma \Delta$, where $A\Delta \parallel B\Gamma$ and $\measuredangle A = 120^{\circ}$. Let $E$ be the midpoint of $AB$ and let $O_1$ and $O_2$ be the circumcenters of triangles $AE \Delta$ and $BE\Gamma$, respectively. Prove that the area of the trapezium is equal to six time the area of the triangle $O_1 E O_2$. [url=https://artofproblemsolving.com/community/c6h1666660p10591747][b]Problem 3.[/b][/url] Find all triples $(\alpha, \beta, \gamma)$ of positive real numbers for which the expression $$K = \frac{\alpha+3 \gamma}{\alpha + 2\beta + \gamma} + \frac{4\beta}{\alpha+\beta+2\gamma} - \frac{8 \gamma}{\alpha+ \beta + 3\gamma}$$obtains its minimum value. [url=https://artofproblemsolving.com/community/c6h1666661p10591749][b]Problem 4.[/b][/url] Let $\Lambda= \{1, 2, \ldots, 2v-1,2v\}$ and $P=\{\alpha_1, \alpha_2, \ldots, \alpha_{2v-1}, \alpha_{2v}\}$ be a permutation of the elements of $\Lambda$. (a) Prove that $$\sum_{i=1}^v \alpha_{2i-1}\alpha_{2i} \leq \sum_{i=1}^v (2i-1)2i.$$(b) Determine the largest positive integer $m$ such that we can partition the $m\times m$ square into $7$ rectangles for which every pair of them has no common interior points and their lengths and widths form the following sequence: $$1,2,3,4,5,6,7,8,9,10,11,12,13,14.$$

For a natural number $n$, let $$C_n=\frac{1}{n+1}\binom{2n}{n}=\frac{(2n)!}{n!(n+1)!}$$ be the $n$-th Catalan number. Prove that for any natural number $m$, $$\sum_{i+j+k=m} C_{i+j}C_{j+k}C_{k+i}=\frac{3}{2m+3}C_{2m+1}.$$ [i]Proposed by Bin Wang[/i]

Let $f(x)=(x-a)^3$. If the sum of all $x$ satisfying $f(x)=f(x-a)$ is $42$, find $a$.

Find the number of rationals $\frac{m}{n}$ such that (i) $0 < \frac{m}{n} < 1$; (ii) $m$ and $n$ are relatively prime; (iii) $mn = 25!$.

The sum of squares of two consecutive integers can be a square, as in $3^2+4^2 =5^2$. Prove that the sum of squares of $m$ consecutive integers cannot be a square for $m = 3$ or $6$ and find an example of $11$ consecutive integers the sum of whose squares is a square.

A point $E$ is located inside a parallelogram $ABCD$ such that $\angle BAE = \angle BCE$. The centers of the circumcircles of the triangles $ABE,ECB, CDE$ and $DAE$ are concyclic.

Let $S$ be the value of \[\sum_{n=1}^\infty \frac{d(n) + \sum_{m=1}^{\nu_2(n)}(m-3)d\left(\frac{n}{2^m}\right)}{n},\] where $d(n)$ is the number of divisors of $n$ and $\nu_2(n)$ is the exponent of $2$ in the prime factorization of $n$. If $S$ can be expressed as $(\ln m)^n$ for positive integers $m$ and $n$, find $1000n + m$. [i]Proposed by Robin Park[/i]

Let $S$ be the set of 81 points $(x, y)$ such that $x$ and $y$ are integers from $-4$ through $4$. Let $A$, $B$, and $C$ be random points chosen independently from $S$, with each of the 81 points being equally likely. (The points $A$, $B$, and $C$ do not have to be different.) Let $K$ be the area of the (possibly degenerate) triangle $ABC$. What is the expected value (average value) of $K^2$ ?

Find all positive real numbers $\lambda$ such that for all integers $n\geq 2$ and all positive real numbers $a_1,a_2,\cdots,a_n$ with $a_1+a_2+\cdots+a_n=n$, the following inequality holds: $\sum_{i=1}^n\frac{1}{a_i}-\lambda\prod_{i=1}^{n}\frac{1}{a_i}\leq n-\lambda$.

In trapezoid $ABCD$ shown, $AD$ is parallel to $BC$, and $AB = 6$, $BC = 7$, $CD = 8$; $AD = 17$. If sides $AB$ and $CD$ are extended to meet at $E$, find the resulting angle at $E$ (in degrees). [img]https://cdn.artofproblemsolving.com/attachments/c/c/987dde1d841a7b976b01f597511e5c1e828e5a.png[/img]

Define $S(N)$ to be the sum of the digits of $N$ when it is written in base $10$, and take $S^k(N) = S(S(\dots(N)\dots))$ with $k$ applications of $S$. The \textit{stability} of a number $N$ is defined to be the smallest positive integer $K$ where $S^K(N) = S^{K+1}(N) = S^{K+2}(N) = \dots$. Let $T_3$ be the set of all natural numbers with stability $3$. Compute the sum of the two least entries of $T_3$. Proposed by Albert Wang (awang11)

A square of $3 \times 3$ is subdivided into 9 small squares of $1 \times 1$. It is desired to distribute the nine digits $1, 2, . . . , 9$ in each small square of $1 \times 1$, a number in each small square. Find the number of different distributions that can be formed in such a way that the difference of the digits in cells that share a side in common is less than or equal to three. Two distributions are distinct even if they differ by rotation and/or reflection.

Let $ P(n)$ and $ S(n)$ denote the product and the sum, respectively, of the digits of the integer $ n$. For example, $ P(23) \equal{} 6$ and $ S(23) \equal{} 5$. Suppose $ N$ is a two-digit number such that $ N \equal{} P(N) \plus{} S(N)$. What is the units digit of $ N$? $ \textbf{(A)} \ 2 \qquad \textbf{(B)} \ 3 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 8 \qquad \textbf{(E)} \ 9$

Let $N$ be the number of (positive) divisors of $2010^{2010}$ ending in the digit $2$. What is the remainder when $N$ is divided by 2010?

Prove that there are only two real numbers $x$ such that \[(x-1)(x-2)(x-3)(x-4)(x-5)(x-6) = 720\]