Point $ E$ is selected on side $ AB$ of triangle $ ABC$ in such a way that $ AE: EB \equal{} 1: 3$ and point $ D$ is selected on side $ BC$ such that $ CD: DB \equal{} 1: 2$. The point of intersection of $ AD$ and $ CE$ is $ F$. Then $ \frac {EF}{FC} \plus{} \frac {AF}{FD}$ is: $ \textbf{(A)}\ \frac {4}{5} \qquad \textbf{(B)}\ \frac {5}{4} \qquad \textbf{(C)}\ \frac {3}{2} \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \frac {5}{2}$

Let $A$ be a square matrix of odd order (at least $3$) whose entries are odd integers. Prove that if $A$ is invertible, then it is not possible for all the minors of the entries of a row to have equal absolute values.

[b]Q3.[/b] For any possitive integer $a$, let $\left[ a\right]$ denote the smallest prime factor of $a.$ Which of the following numbers is equal to $\left[ 35 \right]$ ? \[(A) \; \left[10 \right]; \qquad (B) \; \left[ 15 \right]; \qquad (C ) \; \left[45 \right]; \qquad (D) \; \left[ 55 \right]; \qquad (E) \; \left[75 \right].\]

Let $AD$ and $BE$ be angle bisectors in triangle $ABC$. Let $x$, $y$ and $z$ be distances from point $M$, which lies on segment $DE$, from sides $BC$, $CA$ and $AB$, respectively. Prove that $z=x+y$

Given polynomial $P(x) = a_{0}x^{n}+a_{1}x^{n-1}+\dots+a_{n-1}x+a_{n}$. Put $m=\min \{ a_{0}, a_{0}+a_{1}, \dots, a_{0}+a_{1}+\dots+a_{n}\}$. Prove that $P(x) \ge mx^{n}$ for $x \ge 1$. [i]A. Khrabrov [/i]

Each of six fruit baskets contains pears, plums and apples. The number of plums in each basket equals the total number of apples in all other baskets combined while the number of apples in each basket equals the total number of pears in all other baskets combined. Prove that the total number of fruits is a multiple of $31$.

In all tetrahedron $ABCD$ holds [list=1] [*] $\displaystyle{\sum \limits_{cyc}\frac{h_a-r}{h_a+r}\geq \sum \limits_{cyc}\frac{h_a^t-r^t}{(h_a+r)^t}}$ [*] $\displaystyle{\sum \limits_{cyc}\frac{2r_a-r}{2r_a+r}\geq \sum \limits_{cyc}\frac{2r_a^t-r^t}{(2r_a+r)^t}}$ [/list] for all $t\in [0,1]$

Let $x,y,z>0$ such that $\frac1x+\frac1y+\frac1z<\frac{1}{xyz}$. Show that \[\frac{2x}{\sqrt{1+x^2}}+\frac{2y}{\sqrt{1+y^2}}+\frac{2z}{\sqrt{1+z^2}}<3.\]

Given distinct real numbers $a_1,a_2,...,a_n$, find the minimum value of the function $$y = |x-a_1|+|x-a_2|+...+|x-a_n|, \,\,\, x \in R.$$

In the quadrilateral $ABCD $, $AB = BC $, the point $K $ is the midpoint of the side $CD $, the rays $BK $ and $AD $ intersect at the point $M $ , the circumscribed circle $ \Delta ABM $ intersects the line $AC $ for the second time at the point $P $. Prove that $\angle BKP = 90 {} ^ \circ $. (Anton Trygub)

A grandpa has a finite number of boxes in his attic. Each box is a straight rectangular prism with integer edge lengths. For every box its width is greater or equal to its height and its length is greater or equal to its width. A box can be put inside another box if and only if all of its dimensions are respectively smaller than the other one's. You can put two or more boxes in a bigger box only if the smaller boxes are all already inside one of the boxes. The grandpa decided to put the boxes in each other so that there would be a minimal number of visible boxes in the attic (boxes that have not been put inside another). He decided to use the following algorithm: at each step he finds the longest sequence of boxes so that the first can be put in the second, the second can be put in the third, etc., and then he puts them inside each other in the aforementioned order. The grandpa used the algorithm until no box could be put inside another. It is known that at each step the longest sequence of boxes was unique, e.g., at no moment were there two different sequences with the same length. The grandpa now claims that he has the minimal possible number of visible boxes in his attic. Is the claim necessarily true?

Find all three digit numbers, for which the sum of squares of each digit is $90$ .

Let $ABC$ be a triangle with $\angle B > 90^o$ such that there is a point $H$ on side $AC$ with $AH = BH$ and BH perpendicular to $BC$. Let $D$ and $E$ be the midpoints of $AB$ and $BC$ respectively. A line through $H$ parallel to $AB$ cuts $DE$ at $F$. Prove that $\angle BCF = \angle ACD$.

There are 5 monkeys and 5 ladders and at the top of each ladder there is a banana. A number of ropes connect the ladders, each rope connects two ladders. No two ropes are attached to the same rung of the same ladder. Each monkey starts at the bottom of a different ladder. The monkeys climb up the ladders but each time they encounter a rope they climb along it to the other ladder at the end of the rope and then continue climbing upwards. Show that, no matter how many ropes there are, each monkey gets a banana.

Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen? $ \textbf{(A)}\ 60 \qquad\textbf{(B)}\ 170 \qquad\textbf{(C)}\ 290 \qquad\textbf{(D)}\ 320 \qquad\textbf{(E)}\ 660 $

Let $a,b,c,d \in R^+$ with $abcd=1$. Prove that $$\left(\frac{1+ab}{1+a}\right)^{2008}+\left(\frac{1+bc}{1+b}\right)^{2008}+\left(\frac{1+cd }{1+c}\right)^{2008}+\left(\frac{1+da}{1+d}\right)^{2008} \geq 4$$ [i](dektep)[/i]