Let $P(x)$ be a polynomial with real coefficients. Let $\alpha_1 , \dots , \alpha_k$ be the distinct real roots of $P(x)=0$. If $P'$ is the derivative of $P$, show that for each $i=1,\dots , k$ \[\lim_{x\to \alpha_i} \frac{(x-\alpha_i)P'(x)}{P(x)} = r_i, \] for some positive integer $r_i$.

Let $O$ be the circumcentre of a triangle $ABC$ with $AB > AC > BC$. Let $D$ be a point on the minor arc $BC$ of the circumcircle and let $E$ and $F$ be points on $AD$ such that $AB \perp OE$ and $AC \perp OF$ . The lines $BE$ and $CF$ meet at $P$. Prove that if $PB=PC+PO$, then $\angle BAC = 30^{\circ}$.

Prove that the sequence $a_{n}=\lfloor n\sqrt 2 \rfloor+\lfloor n\sqrt 3 \rfloor$ contains infintely many even and infinitely many odd numbers.

A sequence of real numbers $a_1,a_2,\ldots$ satisfies the relation $$a_n=-\max_{i+j=n}(a_i+a_j)\qquad\text{for all}\quad n>2017.$$ Prove that the sequence is bounded, i.e., there is a constant $M$ such that $|a_n|\leq M$ for all positive integers $n$.

Given 10 light switches, each can be in two states: on and off. For each pair of switches there is a light bulb which is on if and only if when both switches are on (45 bulbs in total). The bulbs and the switches are unmarked so it is unclear which switches correspond to which bulb. In the beginning all switches are off. How many flips are needed to find out regarding all bulbs which switches are connected to it? On each step you can flip precisely one switch

100 people from 25 countries, four from each countries, stay on a circle. Prove that one may partition them onto 4 groups in such way that neither no two countrymans, nor two neighbours will be in the same group.

Find all polynomials $ p\in\mathbb Z[x]$ such that $ (m,n)\equal{}1\Rightarrow (p(m),p(n))\equal{}1$

For a positive integer $n$, denote by $\tau (n)$ the number of its positive divisors. For a positive integer $n$, if $\tau (m) < \tau (n)$ for all $m < n$, we call $n$ a good number. Prove that for any positive integer $k$, there are only finitely many good numbers not divisible by $k$.

Let $D$ be a point on the side $[BC]$ of $\triangle ABC$ such that $|BD|=2$ and $|DC|=6$. If $|AB|=4$ and $m(\widehat{ACB})=20^\circ$, then what is $m(\widehat {BAD})$? $ \textbf{(A)}\ 10^\circ \qquad\textbf{(B)}\ 18^\circ \qquad\textbf{(C)}\ 20^\circ \qquad\textbf{(D)}\ 22^\circ \qquad\textbf{(E)}\ 25^\circ $

Which of the following is not equal to $\dfrac{5}{4}$? $\text{(A)}\ \dfrac{10}{8} \qquad \text{(B)}\ 1\dfrac{1}{4} \qquad \text{(C)}\ 1\dfrac{3}{12} \qquad \text{(D)}\ 1\dfrac{1}{5} \qquad \text{(E)}\ 1\dfrac{10}{40}$

Is it possible to assign every integral point $(x,y)$ of the plane a positive integer $a_{x,y}$ such that for every two integers $i$ and $j$ the following equality holds $$a_{i,j}=\gcd(a_{i-1,j},a_{i+1,j})+\gcd(a_{i,j-1},a_{i,j+1})$$ [i]M. Shutro[/i]

Evaluate \[\int_{-\pi}^{\pi} (\cos2x)(\cos 2^2x)\cdots (\cos 2^{2006}x)dx\]

We will say that a convex polygon $M$ has the property $(*)$ if the straight lines on which its sides lie, being moved outward by a distance of $1$ cm, form a polygon $M'$, similar to this one. a) Prove that if a convex polygon has property $(*)$ , then a circle can be inscribed in it. b) Find the fourth side of a quadrilateral satisfying condition $(*)$ if the lengths of its three consecutive sides are $9, 7$, and $3$ cm.

Circle $\omega_1$ with center $K$ of radius $4$ and circle $\omega_2$ of radius $6$ intersect at points $W$ and $U$. If the incenter of $\triangle KWU$ lies on circle $\omega_2$, find the length of $\overline{WU}$. (Note: The incenter of a triangle is the intersection of the angle bisectors of the angles of the triangle) [i]Proposed by Bradley Guo[/i]

$ABC$ and $A'B'C'$ are equilateral triangles with parallel sides and the same center, as in the figure. The distance between side $BC$ and side $B'C'$ is $\frac{1}{6}$ the altitude of $\triangle ABC$. The ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$ is [asy] size(170); defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, B=(1,-(1/sqrt(3))), C=(-1,-(1/sqrt(3))), A=(0,(2/sqrt(3))), E=(2,-(2/sqrt(3))), F=(-2,-(2/sqrt(3))), D=(0,(4/sqrt(3))); draw(A--B--C--A^^D--E--F--D); label("$A'$", A, dir(90)); label("$B'$", B, SE); label("$C'$", C, SW); label("$A$", D, dir(90)); label("$B$", E, dir(0)); label("$C$", F, W); [/asy] $ \textbf{(A)}\ \frac{1}{36}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{\sqrt{3}}{4}\qquad\textbf{(E)}\ \frac{9+8\sqrt{3}}{36} $

Consider a polynomial \[f(x)=x^{2012}+a_{2011}x^{2011}+\dots+a_1x+a_0.\] Albert Einstein and Homer Simpson are playing the following game. In turn, they choose one of the coefficients $a_0,a_1,\dots,a_{2011}$ and assign a real value to it. Albert has the first move. Once a value is assigned to a coefficient, it cannot be changed any more. The game ends after all the coefficients have been assigned values. Homer's goal is to make $f(x)$ divisible by a fixed polynomial $m(x)$ and Albert's goal is to prevent this. (a) Which of the players has a winning strategy if $m(x)=x-2012$? (b) Which of the players has a winning strategy if $m(x)=x^2+1$? [i]Proposed by Fedor Duzhin, Nanyang Technological University.[/i]