We consider a cube with sides of length 1. Prove that a tetrahedron with vertices in the set of the vertices of the cube has the volume $\dfrac 16$ if and only if 3 of the vertices of the tetrahedron are vertices on the same face of the cube. [i]Dinu Serbanescu[/i]

Hesam chose $10$ distinct positive integers and he gave all pairwise $\gcd$'s and pairwise ${\text lcm}$'s (a total of $90$ numbers) to Masoud. Can Masoud always find the first $10$ numbers, just by knowing these $90$ numbers? [i]Proposed by Morteza Saghafian [/i]

In how many ways can you divide the set of eight numbers $\{2,3,\cdots,9\}$ into $4$ pairs such that no pair of numbers has $\text{gcd}$ equal to $2$?

All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$, with $A$ at the origin and $\overline{BC}$ parallel to the $x$-axis. The area of the triangle is $64$. What is the length of $BC$? $\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$

Let ABC be a triangle with $\angle A = 90^o$, $\angle B = 75^o$, and $AB = 2$. Points $P$ and $Q$ of the sides $AC$ and $BC$ respectively, are such that $\angle APB = \angle CPQ$ and $\angle BQA = \angle CQP$. Calculate the lenght of $QA$.

[b](9.7)[/b] On the segment $[0, 2002]$ its ends and the point with coordinate $d$ are marked, where $d$ is a coprime number to $1001$. It is allowed to mark the midpoint of any segment with ends at the marked points, if its coordinate is integer. Is it possible, by repeating this operation several times, to mark all the integer points on a segment? [b](10.7)[/b] On the segment $[0, 2002]$ its ends and $n-1 > 0$ integer points are marked so that the lengths of the segments into which the segment $ [0, 2002]$ is divided are corpime in the total (i.e., have no common divisor greater than $1$). It is allowed to divide any segment with marked ends into $n$ equal parts and mark the division points if they are all integers. (The point can be marked a second time, but it remains marked.) Is it possible, by repeating this operation several times, mark all the integer points on the segment? [b](11.8)[/b] On the segment $ [0,N]$ its ends and $2 $ more points are marked so that the lengths segments into which the segment $[0,N]$ is divided are integer and coprime in total. If there are two marked points $A$ and $B$ such that the distance between them is a multiple of $3$, then we can divide from cutting $AB$ by $3$ equal parts, mark one of the division points and erase one of the points $A, B$. Is it true that for several such actions you can mark any predetermined integer point of the segment $[0,N]$?

Show that there exist two consecutive squares such that there are at least $1000$ primes between them.

Let $ABC$ be an acute-angled triangle in which no two sides have the same length. The reflections of the centroid $G$ and the circumcentre $O$ of $ABC$ in its sides $BC,CA,AB$ are denoted by $G_1,G_2,G_3$ and $O_1,O_2,O_3$, respectively. Show that the circumcircles of triangles $G_1G_2C$, $G_1G_3B$, $G_2G_3A$, $O_1O_2C$, $O_1O_3B$, $O_2O_3A$ and $ABC$ have a common point. [i]The centroid of a triangle is the intersection point of the three medians. A median is a line connecting a vertex of the triangle to the midpoint of the opposite side.[/i]

Adam places down cards one at a time from a standard 52 card deck (without replacement) in a pile. Each time he places a card, he gets points equal to the number of cards in a row immediately before his current card that are all the same suit as the current card. For instance, if there are currently two hearts on the top of the pile (and the third card in the pile is not hearts), then placing a heart would be worth 2 points, and placing a card of any other suit would be worth 0 points. What is the expected number of points Adam will have after placing all 52 cards? [i]Proposed by Adam Bertelli[/i]

Determine which of the following numbers is greater $$10^{10^{10^{10}}}, (10^{10})!$$

Let $n!=n\cdot (n-1)\cdot (n-2)\cdot \ldots \cdot 2\cdot 1$.For example, $5! = 5\cdot 4\cdot 3 \cdot 2\cdot 1 = 120.$ Compute $\frac{(6!)^2}{5!\cdot 7!}$.

A sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50}=m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Given $ n$ countries with three representatives each, $ m$ committees $ A(1),A(2), \ldots, A(m)$ are called a cycle if [i](i)[/i] each committee has $ n$ members, one from each country; [i](ii)[/i] no two committees have the same membership; [i](iii)[/i] for $ i \equal{} 1, 2, \ldots,m$, committee $ A(i)$ and committee $ A(i \plus{} 1)$ have no member in common, where $ A(m \plus{} 1)$ denotes $ A(1);$ [i](iv)[/i] if $ 1 < |i \minus{} j| < m \minus{} 1,$ then committees $ A(i)$ and $ A(j)$ have at least one member in common. Is it possible to have a cycle of 1990 committees with 11 countries?

For any positive integer $k$ define $f_1(k)$ as the square of the digital sum of $k$ in the decimal system, and $f_{n}(k)=f_1(f_{n-1}(k))$ $\forall n>1$. Compute $f_{1992}(2^{1991})$.

Let $a, b, c$ be reals such that some two of them have difference greater than $\frac{1}{2 \sqrt{2}}$. Prove that there exists an integer $x$, such that $$x^2-4(a+b+c)x+12(ab+bc+ca)<0.$$

In $\triangle ABC$, let $X$ and $Y$ be points on segment $BC$ such that $AX=XB=20$ and $AY=YC=21$. Let $J$ be the $A$-excenter of triangle $\triangle AXY$. Given that $J$ lies on the circumcircle of $\triangle ABC$, the length of $BC$ can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. Compute $m+n$. [i]Proposed by Andrew Wen[/i]

Find the smallest prime number that does not divide \[9+9^2+9^3+\cdots+9^{2010}.\]

Let $I$ be the incenter of triangle $ABC$, and $\ell$ be the perpendicular bisector of $AI$. Suppose that $P$ is on the circumcircle of triangle $ABC$, and line $AP$ and $\ell$ intersect at point $Q$. Point $R$ is on $\ell$ such that $\angle IPR = 90^{\circ}$.Suppose that line $IQ$ and the midsegment of $ABC$ that is parallel to $BC$ intersect at $M$. Show that $\angle AMR = 90^{\circ}$ (Note: In a triangle, a line connecting two midpoints is called a midsegment.)

Find a polynomial $Q(x)$ such that $(2x^2 - 6x + 5)Q(x)$ is a polynomial with all positive coefficients.

A square $ (n \minus{} 1) \times (n \minus{} 1)$ is divided into $ (n \minus{} 1)^2$ unit squares in the usual manner. Each of the $ n^2$ vertices of these squares is to be coloured red or blue. Find the number of different colourings such that each unit square has exactly two red vertices. (Two colouring schemse are regarded as different if at least one vertex is coloured differently in the two schemes.)

Let $(x,y,z)$ be an ordered triplet of real numbers that satisfies the following system of equations: \begin{align*}x+y^2+z^4&=0,\\y+z^2+x^4&=0,\\z+x^2+y^4&=0.\end{align*} If $m$ is the minimum possible value of $\lfloor x^3+y^3+z^3\rfloor$, find the modulo $2007$ residue of $m$.

A polyhedron $W$ has the following properties: (i) It possesses a center of symmetry. (ii) The section of $W$ by a plane passing through the center of symmetry and one of its edges is always a parallelogram. (iii) There is a vertex of $W$ at which exactly three edges meet. Prove that $W$ is a parallelepiped.

In a mathematical challenge, positive real numbers $a_{1}\geq a_{2} \geq ... \geq a_{n}$ and an initial sequence of positive real numbers $(b_{1}, b_{2},...,b_{n+1})$ are given to Secco. Let $C$ a non-negative real number. In a sequence $(x_{1},x_{2},...,x_{n+1})$, consider the following operation: Subtract $1$ of some $x_{j}$, $j \in \{1,2,...,n+1\}$, add $C$ to $x_{n+1}$ and replace $(x_{1},x_{2},...,x_{j-1})$ for $(x_{1}+a_{\sigma (1)}, x_{2}+a_{\sigma (2)}, ..., x_{j-1}+a_{\sigma (j-1)})$, where $\sigma$ is a permutation of $(1,2,...,j-1)$. Secco's goal is to make all terms of sequence $(b_{k})$ negative after a finite number of operations. Find all values of $C$, depending of $a_{1}, a_{2},..., a_{n}, b_{1}, b_{2}, ..., b_{n+1}$, for which Secco can attain his goal.

Quadrilateral $ABCD$ is cyclic. Line through point $D$ parallel with line $BC$ intersects $CA$ in point $P$, line $AB$ in point $Q$ and circumcircle of $ABCD$ in point $R$. Line through point $D$ parallel with line $AB$ intersects $AC$ in point $S$, line $BC$ in point $T$ and circumcircle of $ABCD$ in point $U$. If $PQ=QR$, prove that $ST=TU$

Find the limit \[\lim_{n\to\infty} \frac{1}{n\ln n}\int_{\pi}^{(n+1)\pi} (\sin ^ 2 t)(\ln t)\ dt.\]