A $4*4*4$ cubical box contains 64 identical small cubes that exactly fill the box. How many of these small cubes touch a side or the bottom of the box? $ \text{(A)}\ 48\qquad\text{(B)}\ 52\qquad\text{(C)}\ 60\qquad\text{(D)}\ 64\qquad\text{(E)}\ 80 $

Ike and Mike go into a sandwich shop with a total of $\$30.00$ to spend. Sandwiches cost $\$4.50$ each and soft drinks cost $\$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can and use any remaining money to buy soft drinks. Counting both soft drinks and sandwiches, how many items will they buy? $\textbf{(A) } 6 \qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$

In a right triangle with sides $ a$ and $ b$, and hypotenuse $ c$, the altitude drawn on the hypotenuse is $ x$. Then: $ \textbf{(A)}\ ab \equal{} x^2 \qquad\textbf{(B)}\ \frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {1}{x} \qquad\textbf{(C)}\ a^2 \plus{} b^2 \equal{} 2x^2$ $ \textbf{(D)}\ \frac {1}{x^2} \equal{} \frac {1}{a^2} \plus{} \frac {1}{b^2} \qquad\textbf{(E)}\ \frac {1}{x} \equal{} \frac {b}{a}$

The circle is divided into $49$ areas so that no three areas touch at one point. The resulting “map” is colored in three colors so that no two adjacent areas have the same color. The border of two areas is considered to be colored in both colors. Prove that on the circle there are two diametrically opposite points, colored in one color.

Let $r$ be a root of $P(x)=x^3+ax^2+bx-1=0$ and $r+1$ be a root of $y^3+cy^2+dy+1=0,$ where $a,b,c$ and $d$ are integers. Also let $P(x)$ be irreducible over the rational numbers. Express another root $s$ of $P(x)=0$ as a function of $r$ which does not explicitly involve $a,b,c$ or $d.$

Let $p(x) = x^2 + x + 1$. Find the fourth smallest prime $q$ such that $p(n)$ is divisble by $q$ for some integer $n$.

Find all functions $f : R\rightarrow R$ such that $f ( f (x)+y) = f (x^2 -y)+4 f (x)y$ for all $x,y \in R$ .

Evaluate \[S=\sum_{k=1}^n k(k+1)\ldots (k+p), \] where $n$ and $p$ are positive integers.

Let $\overline{AD}$ be a diameter of a circle. Let point $B$ be on the circle, point $C$ on $\overline{AD}$ such that $A, B, C$ form a right triangle at $C$. The value of the hypotenuse of the triangle is $4$ times the square root of its area. If $\overline{BC}$ has length $30$, what is the length of the radius of the circle?

All faces of a convex polyhedron are parallelograms. Can the polyhedron have exactly 1992 faces?

Given is a natural number $n>4$. There are $n$ points marked on the plane, no three of which lie on the same line. Vasily draws one by one all the segments connecting pairs of marked points. At each step, drawing the next segment $S$, Vasily marks it with the smallest natural number, which hasn't appeared on a drawn segment that has a common end with $S$. Find the maximal value of $k$, for which Vasily can act in such a way that he can mark some segment with the number $k$?

Ingrid and Erik are playing a game. For a given odd prime $p$, the numbers $1, 2, 3, ..., p-1$ are written on a blackboard. The players take turns making moves with Ingrid starting. A move consists of one of the players crossing out a number on the board that has not yet been crossed out. If the product of all currently crossed out numbers is $1 \pmod p$ after the move, the player whose move it was receives one point, otherwise, zero points are awarded. The game ends after all numbers have been crossed out. The player who has received the most points by the end of the game wins. If both players have the same score, the game ends in a draw. For each $p$, determine which player (if any) has a winning strategy

Let there be a quadrilateral $ABCD$ such that $\angle CAD=45, \angle ACD=30, \angle BAC=\angle BCA=15$. Find $\angle DBC$.

Do there exist continuous functions $f, g: \mathbb{R} \to \mathbb{R}$, both nowhere differentiable, such that $f \circ g$ is differentiable?

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?