Let $n > 3$ be a positive integer. Find all integers $k$ such that $1 \le k \le n$ and for which the following property holds: If $x_1, . . . , x_n$ are $n$ real numbers such that $x_i + x_{i + 1} + ... + x_{i + k - 1} = 0$ for all integers $i > 1$ (indexes are taken modulo $n$), then $x_1 = . . . = x_n = 0$. Proposed by [i]Vincent Jugé and Théo Lenoir, France[/i]

Let $k$ and $s$ be positive integers such that $s<(2k + 1)^2$. Initially, one cell out of an $n \times n$ grid is coloured green. On each turn, we pick some green cell $c$ and colour green some $s$ out of the $(2k + 1)^2$ cells in the $(2k + 1) \times (2k + 1)$ square centred at $c$. No cell may be coloured green twice. We say that $s$ is $k-sparse$ if there exists some positive number $C$ such that, for every positive integer $n$, the total number of green cells after any number of turns is always going to be at most $Cn$. Find, in terms of $k$, the least $k$-sparse integer $s$. [I]Proposed by Nikolai Beluhov.[/i]

Given the integer $n\geq 2$ and a integer ${a}$, which is coprime with ${n}$. A country has ${n}$ islands $D_1$, $D_2$, $\cdots$, $D_n$. For any $1\leq i\neq j\leq n$, there is a one-way ferry $D_i$ to $D_j$ if and only if $ij\equiv ia\pmod n$. A tourist can initially fly to any of the islands, and then he can only take a one-way ferry. What is the maximum number of islands he can visit? [i]Created by Zhenhua Qu[/i]

Let $ABC$ be a triangle and let $D$ be a point on $BC$ so that $AD$ bisects the angle $\angle BAC$. The common tangents of the circles $(BAD)$, $(CAD)$ meet at the point $A'$. The points $B'$, $C'$ are defined similarly. Show that $A'$, $B'$, $C'$ are collinear.

Let $n$ be a positive integer. Prove that the numbers $n^2(n^2 + 2)^2$ and $n^4(n^2 + 2)^2$ are written in base $n^2 +1$ with the same digits but in opposite order.

Consider the shape shown below, formed by gluing together the sides of seven congruent regular hexagons. The area of this shape is partitioned into $21$ quadrilaterals, all of whose side lengths are equal to the side length of the hexagon and each of which contains a $60^{\circ}$ angle. In how many ways can this partitioning be done? (The quadrilaterals may contain an internal boundary of the seven hexagons.) [asy] draw(origin--origin+dir(0)--origin+dir(0)+dir(60)--origin+dir(0)+dir(60)+dir(0)--origin+dir(0)+dir(60)+dir(0)+dir(60)--origin+dir(0)+dir(60)+dir(0)+dir(60)+dir(120)--origin+dir(0)+dir(60)+dir(0)+dir(60)+dir(120)+dir(60)--origin+dir(0)+dir(60)+dir(0)+dir(60)+dir(120)+dir(60)+dir(120)--origin+dir(0)+dir(60)+dir(0)+dir(60)+dir(120)+dir(60)+dir(120)+dir(180)--origin+dir(0)+dir(60)+dir(0)+dir(60)+dir(120)+dir(60)+dir(120)+dir(180)+dir(120)--origin+dir(0)+dir(60)+dir(0)+dir(60)+dir(120)+dir(60)+dir(120)+dir(180)+dir(120)+dir(180)--origin+dir(0)+dir(60)+dir(0)+dir(60)+dir(120)+dir(60)+dir(120)+dir(180)+dir(120)+dir(180)+dir(240)--origin+dir(0)+dir(60)+dir(0)+dir(60)+dir(120)+dir(60)+dir(120)+dir(180)+dir(120)+dir(180)+dir(240)+dir(180)--origin+dir(0)+dir(60)+dir(0)+dir(60)+dir(120)+dir(60)+dir(120)+dir(180)+dir(120)+dir(180)+dir(240)+dir(180)+dir(240)--origin+dir(0)+dir(60)+dir(0)+dir(60)+dir(120)+dir(60)+dir(120)+dir(180)+dir(120)+dir(180)+dir(240)+dir(180)+dir(240)+dir(300)--origin+dir(0)+dir(60)+dir(0)+dir(60)+dir(120)+dir(60)+dir(120)+dir(180)+dir(120)+dir(180)+dir(240)+dir(180)+dir(240)+dir(300)+dir(240)--origin+dir(0)+dir(60)+dir(0)+dir(60)+dir(120)+dir(60)+dir(120)+dir(180)+dir(120)+dir(180)+dir(240)+dir(180)+dir(240)+dir(300)+dir(240)+dir(300)--origin+dir(0)+dir(60)+dir(0)+dir(60)+dir(120)+dir(60)+dir(120)+dir(180)+dir(120)+dir(180)+dir(240)+dir(180)+dir(240)+dir(300)+dir(240)+dir(300)+dir(0)--origin+dir(0)+dir(60)+dir(0)+dir(60)+dir(120)+dir(60)+dir(120)+dir(180)+dir(120)+dir(180)+dir(240)+dir(180)+dir(240)+dir(300)+dir(240)+dir(300)+dir(0)+dir(300)--cycle); draw(2*dir(60)+dir(120)+dir(0)--2*dir(60)+dir(120)+2*dir(0),dashed); draw(2*dir(60)+dir(120)+dir(60)--2*dir(60)+dir(120)+2*dir(60),dashed); draw(2*dir(60)+dir(120)+dir(120)--2*dir(60)+dir(120)+2*dir(120),dashed); draw(2*dir(60)+dir(120)+dir(180)--2*dir(60)+dir(120)+2*dir(180),dashed); draw(2*dir(60)+dir(120)+dir(240)--2*dir(60)+dir(120)+2*dir(240),dashed); draw(2*dir(60)+dir(120)+dir(300)--2*dir(60)+dir(120)+2*dir(300),dashed); draw(dir(60)+dir(120)--dir(60)+dir(120)+dir(0)--dir(60)+dir(120)+dir(0)+dir(60)--dir(60)+dir(120)+dir(0)+dir(60)+dir(120)--dir(60)+dir(120)+dir(0)+dir(60)+dir(120)+dir(180)--dir(60)+dir(120)+dir(0)+dir(60)+dir(120)+dir(180)+dir(240)--dir(60)+dir(120)+dir(0)+dir(60)+dir(120)+dir(180)+dir(240)+dir(300),dashed); [/asy]

Let $C$ be the circle of radius $12$ centered at $(0, 0)$. What is the length of the shortest path in the plane between $(8\sqrt3, 0)$ and $(0, 12 \sqrt2)$ that does not pass through the interior of $C$?

Fix $a_1, . . . , a_n \in (0, 1)$ and define $$f(I) = \prod_{i \in I} a_i \cdot \prod_{j \notin I} (1 - a_j)$$ for each $I \subseteq \{1, . . . , n\}$. Assuming that $$\sum_{I\subseteq \{1,...,n\}, |I| odd} {f(I)} = \frac12,$$ show that at least one $a_i$ has to be equal to $\frac12$. (Paolo Leonetti)

Let three circles be externally tangent in pairs, with parallel diameters $A_1A_2, B_1B_2, C_1C_2$ (i.e. each of the quadrilaterals $A_1B_1B_2A_2$ and $A_1C_1C_2A_2$ is a parallelogram or trapezoid, which segment $A_1A_2$ is the base). Prove that $A_1B_2, B_1C_2, C_1A_2$ intersect at one point. (Yuri Biletsky )

A pyramid of non-negative integers is constructed as follows (a) The first row consists of only $0$, (b) The second row consists of $1$ and $1$, (c) The $n^{th}$ (for $n > 2$) is an array of $n$ integers among which the left most and right most elements are equal to $n - 1$ and the interior numbers are equal to the sum of two adjacent numbers from the $(n - 1)^{th}$ row (see Figure). Let $S_n$ be the sum of numbers in row $n^{th}$. Determine the remainder when dividing $S_{2018}$ by $2018$: A. $2$ B. $4$ C. $6$ D. $11$ E. $17$

Compute $\left\lceil\displaystyle\sum_{k=2018}^{\infty}\frac{2019!-2018!}{k!}\right\rceil$. (The notation $\left\lceil x\right\rceil$ denotes the least integer $n$ such that $n\geq x$.) [i]Proposed by Tristan Shin[/i]

Each point of the 3-dimensional space is coloured with exactly one of the colours red, green and blue. Let $R$, $G$ and $B$, respectively, be the sets of the lengths of those segments in space whose both endpoints have the same colour (which means that both are red, both are green and both are blue, respectively). Prove that at least one of these three sets includes all non-negative reals.

Given a,b c are lenth of a triangle (If ABC is a triangle then AC=b, BC=a, AC=b) and $a+b+c=2$. Prove that $1+abc<ab+bc+ca\leq \frac{28}{27}+abc$

Let $ T$ denote the set of all ordered triples $ (p,q,r)$ of nonnegative integers. Find all functions $ f: T \rightarrow \mathbb{R}$ satisfying \[ f(p,q,r) = \begin{cases} 0 & \text{if} \; pqr = 0, \\ 1 + \frac{1}{6}(f(p + 1,q - 1,r) + f(p - 1,q + 1,r) & \\ + f(p - 1,q,r + 1) + f(p + 1,q,r - 1) & \\ + f(p,q + 1,r - 1) + f(p,q - 1,r + 1)) & \text{otherwise} \end{cases} \] for all nonnegative integers $ p$, $ q$, $ r$.

Let $P(x)$ be a quadratic polynomial with real coefficients such that $P(3) = 7$ and \[P(x) = P(0) + P(1)x + P(2)x^2\] for all real $x$. What is $P(-1)$?

Suppose $E$, $I$, $L$, $V$ are (not necessarily distinct) nonzero digits in base ten for which [list] [*] the four-digit number $\underline{E}\ \underline{V}\ \underline{I}\ \underline{L}$ is divisible by $73$, and [*] the four-digit number $\underline{V}\ \underline{I}\ \underline{L}\ \underline{E}$ is divisible by $74$. [/list] Compute the four-digit number $\underline{L}\ \underline{I}\ \underline{V}\ \underline{E}$.

How many ordered triples $(a,b,c)$ of integers satisfy the inequality \[a^2+b^2+c^2 \leq a+b+c+2?\] Let $T = TNYWR$. David rolls a standard $T$-sided die repeatedly until he first rolls $T$, writing his rolls in order on a chalkboard. What is the probability that he is able to erase some of the numbers he's written such that all that's left on the board are the numbers $1, 2, \dots, T$ in order?

Let be a nonnegative integer $ n $ and three real numbers $ a,b,c $ satisfying $$ a^n+c=b^n+a=c^n+b=a+b+c. $$ Show that $ a=b=c. $ [i]Gheorghe Iurea[/i]

A semicircle with diameter length $16$ contains a circle radius $3$ tangent both to the inside of the semicircle and its diameter as shown. A second larger circle is tangent to the inside of the semicircle, the outside of the circle, and the diameter of the semicircle. The diameter of the second circle can be written as $\frac{n + k\sqrt{2}}{m}$ where $m$, $n$, and $k$ are positive integers and $m$ and $n$ have no factors in common. Find $m + n + k$. [asy] size(200); pair O=(0,0); real R=10, r=4.7; draw(arc(O,R,0,180)--cycle); pair P=(sqrt((R-r)^2-r^2),r),Q; draw(circle(P,r)); real a=0,b=r,c; for(int k=0;k<20;++k) { c=(a+b)/2; Q=(-sqrt((R-c)^2-c^2),c); if(abs(P-Q)>c+r) a=c; else b=c; } draw(circle(Q,c));[/asy]

Calculate the sum $$ 2^n+2^{n-1}\cos\alpha +2^{n-2} \cos2\alpha +\cdots +2\cos (n-1)\alpha +\cos n\alpha , $$ where $ \alpha $ is a real number and $ n $ a natural one. [i]Dan Negulescu[/i]

Square $A$ is adjacent to square $B$ which is adjacent to square $C$. The three squares all have their bottom sides along a common horizontal line. The upper left vertices of the three squares are collinear. If square $A$ has area $24$, and square $B$ has area $36$, find the area of square $C$. [asy] import graph; size(8cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real xmin = -4.89, xmax = 13.61, ymin = -1.39, ymax = 9; draw((0,0)--(2,0)--(2,2)--(0,2)--cycle, linewidth(1.2)); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle, linewidth(1.2)); draw((5,4.5)--(5,0)--(9.5,0)--(9.5,4.5)--cycle, linewidth(1.2)); draw((2,0)--(2,2), linewidth(1.2)); draw((2,2)--(0,2), linewidth(1.2)); draw((0,2)--(0,0), linewidth(1.2)); draw((2,0)--(5,0), linewidth(1.2)); draw((5,0)--(5,3), linewidth(1.2)); draw((5,3)--(2,3), linewidth(1.2)); draw((2,3)--(2,0), linewidth(1.2)); draw((5,4.5)--(5,0), linewidth(1.2)); draw((5,0)--(9.5,0), linewidth(1.2)); draw((9.5,0)--(9.5,4.5), linewidth(1.2)); draw((9.5,4.5)--(5,4.5), linewidth(1.2)); label("A",(0.6,1.4),SE*labelscalefactor); label("B",(3.1,1.76),SE*labelscalefactor); label("C",(6.9,2.5),SE*labelscalefactor); draw((13.13,8.56)--(-3.98,0), linewidth(1.2)); draw((-3.98,0)--(15.97,0), linewidth(1.2));[/asy]

Prove that does not exist positive integers $a$, $b$ and $k$ such that $4abk-a-b$ is a perfect square.

Consider and odd prime $p$. For each $i$ at $\{1, 2,..., p-1\}$, let $r_i$ be the rest of $i^p$ when it is divided by $p^2$. Find the sum: $r_1 + r_2 + ... + r_{p-1}$

Solve the equation in the set of real numbers: $$\left[ x+\frac{1}{x} \right] = \left[ x^2+\frac{1}{x^2} \right]$$ where $[a]$, represents the integer part of the real number $a$.

$(a)$ Prove that $\sqrt{2}(\sin t + \cos t) \ge 2\sqrt[4]{\sin 2t}$ for $0 \le t \le\frac{\pi}{2}.$ $(b)$ Find all $y, 0 < y < \pi$, such that $1 +\frac{2 \cot 2y}{\cot y} \ge \frac{\tan 2y}{\tan y}$. .