We want to draw a number of straight lines such that for each square of a chessboard, at least one of the lines passes through an interior point of the square. What is the smallest number of lines needed for a (a) $3\times 3$; (b) $4\times 4$ chessboard? Use a picture to show that this many lines are enough, and prove that no smaller number would do. (M Vyalyi)

At one of George Washington's parties, each man shook hands with everyone except his spouse, and no handshakes took place between women. If $13$ married couples attended, how many handshakes were there among these $26$ people? $\text{(A)} \ 78 \qquad \text{(B)} \ 185 \qquad \text{(C)} \ 234 \qquad \text{(D)} \ 312 \qquad \text{(E)} \ 325$

In the simple and connected graph $G$ let $x_i$ be the number of vertices with degree $i$. Let $d>3$ be the biggest degree in the graph $G$. Prove that if : $$x_d \ge x_{d-1} + 2x_{d-2}+... +(d-1)x_1$$ Then there exists a vertex with degree $d$ such that after removing that vertex the graph $G$ is still connected. Proposed by [i]Ali Mirzaie[/i]

Let $m$ be the smallest integer whose cube root is of the form $n+r$, where $n$ is a positive integer and $r$ is a positive real number less than $1/1000$. Find $n$.

Let $ABCD$ be a parallelogram and $M$ is the intersection of $AC$ and $BD$. The point $N$ is inside of the $\triangle AMB$ such that $\angle AND=\angle BNC$. Prove that $\angle MNC=\angle NDA$ and $\angle MND=\angle NCB$.

In a convex $n$-gon, several diagonals are drawn. Among these diagonals, a diagonal is called [i]good[/i] if it intersects exactly one other diagonal drawn (in the interior of the $n$-gon). Find the maximum number of good diagonals.

In equilateral triangle $ABC$, $AB=2$ and $M$ is the midpoint of $AB$. A laser is shot from $M$ in a certain direction, and whenever it collides with a side of $ABC$ it will reflect off the side such that the acute angle formed by the incident ray and the side is equal to the acute angle formed by the reflected ray and the side. Once the laser coincides with a vertex, it stops. Find the sum of the smallest three possible integer distances that the laser could have traveled. [i]Proposed by Jerry Xu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{21}$ Whenever the laser hits a side of the triangle, reflect the laser's path over that side so that the path of the laser forms a straight line. We want the path of the laser to coincide with a vertex of one of the reflected triangles. Thus, we can restate the problem as follows: Tessellate the plane with equilateral triangles of side length $3$. Consider one of these equilateral triangles $ABC$ with $M$ being the midpoint of $AB=2$. Find the sum of the three minimum integer distances from $M$ to any vertex in the plane. [asy] import geometry; size(8cm); pair A = (0,sqrt(3)); pair B = (-1,0); pair C = (1,0); pair M = (0,0); for (int i = -1; i <= 2; ++i) { draw((i-3,i*sqrt(3))--(-i+3,i*sqrt(3))); draw(((i-1)*2,-sqrt(3))--(i+1,(2-i)*sqrt(3))); draw((-i-1,(2-i)*sqrt(3))--((1-i)*2,-sqrt(3))); } draw(A--B--C--A, red); dot(M); label("$A$",A+(0,0.25),N); label("$B$",B-(0.25,0),SW); label("$C$",C+(0.25,0),SE); label("$M$",M,S); [/asy] It is trivial to see that the vertical distance between $M$ and a given vertex is $n\sqrt{3}$ for $n \in \mathbb{N}^{0}$. If $n$ is even, the horizontal distance between $O$ and a given vertex is $1+2m$ for $m \in \mathbb{N}^{0}$. If $n$ is odd, the horizontal distance is $2m$ for $m \in \mathbb{N}^{0}$. We consider two separate cases: $1.$ $n$ is even. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(1+2m)^2=l^2.$$Make the substitution $1+2m=k$ to get that $$3n^2+k^2=l^2.$$Notice that these equations form a family of generalized Pell equations $y^2-3x^2=N$ with $N=k^2$. We can find some set of roots to these equations using the multiplicative principle: we will use this idea to find three small $l$ values, and that gives us an upper bound on what the three $l$ values can be. From there, a simple bash of lower $l$ values to see if solutions to each generalized Pell equation not given by the multiplicative principle exist finishes this case. By the multiplicative principle some set of solutions $(x_n,y_n)$ to the above equation with sufficiently small $x_n$ follow the formula$$x_n\sqrt{3}+y_n=\left(x_0\sqrt{3}+y_0\right)\left(u_n\sqrt{3}+v_n\right),$$where $\left(x_0,y_0\right)$ is a solution to the generalized Pell equation and $\left(u_n,v_n\right)$ are solutions to the Pell equation $y^2-3x^2=1$. Remember that the solutions to this last Pell equation satisfy$$u_n\sqrt{3}+v_n=\left(u_0\sqrt{3}+v_0\right)^k$$where the trivial positive integer solution $$\left(u_0, v_0\right)=(1,2)$$(this can easily be found by inspection or by taking the convergents of the continued fraction expansion of $\sqrt{3}$). We thus get that$$\left(u_1,v_1\right)=(4,7),\left(u_2,v_2\right)=(15,26),\left(u_2,v_2\right)=(56,97)\dots$$(also don't forget that $(u,v)=(0,1)$ is another solution). From here, note that $k$ must be odd since $k=1+2m$ for $m \in \mathbb{N}^{0}$. For $k=1$, the smallest three solutions to the Pell equation with $n$ even are \begin{align*} (x,y)&=(0,1),(4,7),(56,97) \\ \longrightarrow (n,m,l)&=(0,0,1),(4,0,7),(56,0,97) \end{align*}Our current smallest three values of $l$ are thus $1,7,97$. A quick check confirms that all of these solutions are not extraneous (extraneous solutions appear when the path taken by the laser prematurely hits a vertex). For $k=3$, using the multiplicative principle we get two new smaller solutions \begin{align*} (x,y)&=(0,3),(12,21) \\ \longrightarrow (n,m,l)&=(0,1,3),(12,1,21) \end{align*}However, note that $(n,m,l)=(0,1,3)$ is extraneous since is equivalent to the path that is traced out by the solution $(n,m,l)=(0,0,1)$ found previously and will thus hit a vertex prematurely. Thus, our new three smallest values of $l$ are $1,7,21$. For $k \ge 5$, it is evident that there are no more smaller integral values of $l$ that can be found using the multiplicative principle: the solution set $(n,m,l)=\left(0,\dfrac{k-1}{2},k\right)$ is always extraneous for $k > 1$ since it is equivalent to the path traced out by $(0,0,1)$ as described above, and any other solutions will give larger values of $l$. Thus, we now only need to consider solutions to each generalized Pell equation not found by the multiplicative principle. A quick bash shows that $l=3,5,9,11$ gives no solutions for any odd $k$ and even $n$, however $n=13$ gives $k=11$ and $n=4$, a non-extraneous solution smaller than one of the three we currently have. Thus, our new three smallest $l$ values are $1,7,13$. $2$. $n$ is odd. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(2m)^2=l^2.$$Make the substitution $2m=k$ to get that $$3n^2+k^2=l^2.$$This is once again a family of generalized Pell equations with $N=k^2$, however this time we must have $k$ even instead of $k$ odd. However, note that there are no solutions to this family of Pell equation with $n$ odd: $k^2 \equiv 0 \text{ (mod }4)$ since $k$ is even, and $3n^2 \equiv 3 \text{ (mod }4)$ since $n$ is odd, however $0+3 \equiv 3 \text{ (mod }4)$ is not a possible quadratic residue mod $4$. Thus, this case gives no solutions. Our final answer is thus $1+7+13=\boxed{21}$. [/hide]

Let $Y_1 , ..., Y_n$ be exchangeable random variables, ie for all permutations $\pi$ , the distribution of $(Y_{\pi (1)}, \dots, Y_{\pi (n)} )$ is equal to the distribution of $(Y_1 , ..., Y_n)$. Let $S_0 = 0$ and $$S_j = \sum_{i = 1}^j Y_i \qquad j = 1,\dots,n$$ Denote $S_{(0)} , ..., S_{(n)}$ by the ordered statistics formed by the random variables $S_0 , ..., S_n$. Show that the distribution of $S_{(j)}$ is equal to the distribution of $\max_{0 \le i \le j} S_i + \min_ {0 \le i \le n-j} (S_{j + i} -S_j)$.

Let $x$ and $y$ be positive real numbers such that $x + y = 1$. Show that $$\left( \frac{x+1}{x} \right)^2 + \left( \frac{y+1}{y} \right)^2 \geq 18.$$

Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying \[ f(f(x \plus{} y)) \equal{} f(x \plus{} y) \plus{} f(x)f(y) \minus{} xy\] for all real numbers $x$ and $y$.

We call a rectangle of the size $1 \times 2$ a domino. Rectangle of the $2 \times 3$ removing two opposite (under center of rectangle) corners we call tetramino. These figures can be rotated. It requires to tile rectangle of size $2008 \times 2010$ by using dominoes and tetraminoes. What is the minimal number of dominoes should be used?

Given is a convex pentagon $ABCDE$ in which $\angle A = 60^o$, $\angle B = 100^o$, $\angle C = 140^o$. Show that this pentagon can be placed in a circle with a radius of $\frac23 AD$.

For the numbers $ a$, $ b$, $ c$, $ d$, $ e$ define $ m$ to be the arithmetic mean of all five numbers; $ k$ to be the arithmetic mean of $ a$ and $ b$; $ l$ to be the arithmetic mean of $ c$, $ d$, and $ e$; and $ p$ to be the arithmetic mean of $ k$ and $ l$. Then, no matter how $ a$, $ b$, $ c$, $ d$, and $ e$ are chosen, we shall always have: $ \textbf{(A)}\ m \equal{} p \qquad \textbf{(B)}\ m \ge p \qquad \textbf{(C)}\ m > p \qquad \textbf{(D)}\ m < p \qquad \textbf{(E)}\ \text{none of these}$

The base of the pyramid $ABCD$ is an equilateral triangle $ABC$ with side length $1$. Additionally, \[\angle ADB=\angle BDC=\angle CDA=90^\circ\,.\] Calculate the volume of pyramid $ABCD$.

The figure below shows a dotted grid $8$ cells wide and $3$ cells tall consisting of $1''\times1''$ squares. Carl places $1$-inch toothpicks along some of the sides of the squares to create a closed loop that does not intersect itself. The numbers in the cells indicate the number of sides of that square that are to be covered by toothpicks, and any number of toothpicks are allowed if no number is written. In how many ways can Carl place the toothpicks? [asy] size(6cm); for (int i=0; i<9; ++i) { draw((i,0)--(i,3),dotted); } for (int i=0; i<4; ++i){ draw((0,i)--(8,i),dotted); } for (int i=0; i<8; ++i) { for (int j=0; j<3; ++j) { if (j==1) { label("1",(i+0.5,1.5)); }}} [/asy] $\textbf{(A) }130\qquad\textbf{(B) }144\qquad\textbf{(C) }146\qquad\textbf{(D) }162\qquad\textbf{(E) }196$

Bhairav runs a 15-mile race at 28 miles per hour, while Daniel runs at 15 miles per hour and Tristan runs at 10 miles per hour. What is the greatest length of time, in [i]minutes[/i], between consecutive runners' finishes? [i]2016 CCA Math Bonanza Lightning #2.1[/i]

A sequence $P_1, \dots, P_n$ of points in the plane (not necessarily diferent) is [i]carioca[/i] if there exists a permutation $a_1, \dots, a_n$ of the numbers $1, \dots, n$ for which the segments $$P_{a_1}P_{a_2}, P_{a_2}P_{a_3}, \dots, P_{a_n}P_{a_1}$$ are all of the same length. Determine the greatest number $k$ such that for any sequence of $k$ points in the plane, $2023-k$ points can be added so that the sequence of $2023$ points is [i]carioca[/i].

In the morning, three people, $A$, $B$ and $C$ run in a same line at a beach in Albufeira. Some day, the three people were in the same point of the beach and then they started to run at the same time, but in different velocities. For each person, the velocity was constant. When someone arrived in an extreme of the beach, he/she turned back and runned in the opposite direction. In the moment in that the three people were in the same point of the beach again, the running finished. Not counting with the beginning and the final of the running, $A$ met $B$ six times and $A$ met $C$ eight times. How many times did $B$ and $C$ meet?

In the triangle $ABC$, where $<$ $ACB$ $=$ $45$, $O$ and $H$ are the center the circumscribed circle, respectively, the orthocenter. The line that passes through $O$ and is perpendicular to $CO$ intersects $AC$ and $BC$ in $K$, respectively $L$. Show that the perimeter of $KLH$ is equal to the diameter of the circumscribed circle of triangle $ABC$.

For any real number $t$, let $\lfloor t \rfloor$ denote the largest integer $\le t$. Suppose that $N$ is the greatest integer such that $$\left \lfloor \sqrt{\left \lfloor \sqrt{\left \lfloor \sqrt{N} \right \rfloor}\right \rfloor}\right \rfloor = 4$$Find the sum of digits of $N$.

Determine the smallest natural number $a\geq 2$ for which there exists a prime number $p$ and a natural number $b\geq 2$ such that \[\frac{a^p - a}{p}=b^2.\]

Real numbers $a$, $b$, $c$ which are differ from $1$ satisfies the following conditions; (1) $abc =1$ (2) $a^2+b^2+c^2 - \left( \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} \right) = 8(a+b+c) - 8 (ab+bc+ca)$ Find all possible values of expression $\dfrac{1}{a-1} + \dfrac{1}{b-1} + \dfrac{1}{c-1}$.

Triangle $ABC$ is inscribed in circle $\omega$. Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$. Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$, then $ST=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

What is the value of $(8 \times 4 + 2) - (8 + 4 \times 2)?$ $\textbf{(A)}~0\qquad\textbf{(B)}~6\qquad\textbf{(C)}~10\qquad\textbf{(D)}~18\qquad\textbf{(E)}~24$

Ali puts $5$ points on the plane such that no three of them are collinear. Ramtin adds a sixth point that is not collinear with any two of the former points.Ali wants to eventually construct two triangles from the six points such that one can be placed inside another. Can Ali put the 5 points in such a manner so that he would always be able to construct the desired triangles? (We say that triangle $T_1$ can be placed inside triangle $T_2$ if $T_1$ is congruent to a triangle that is located completely inside $T_2$.)