July 19th - Day 1
Let $ A,B,C,D$ be four distinct points on a line, in that order. The circles with diameters $ AC$ and $ BD$ intersect at $ X$ and $ Y$. The line $ XY$ meets $ BC$ at $ Z$. Let $ P$ be a point on the line $ XY$ other than $ Z$. The line $ CP$ intersects the circle with diameter $ AC$ at $ C$ and $ M$, and the line $ BP$ intersects the circle with diameter $ BD$ at $ B$ and $ N$. Prove that the lines $ AM,DN,XY$ are concurrent.
Click for solution we know that ${XY}\perp{AD}$ at $Z$. let ${AM}\cap{XY} = R$. we have $\angle{RMC} = 90$ (AC is diameter) so, $RMCZ$ is cyclic, that implies $PR.PZ = PM.PC$. let ${ND}\cap{XY} = S$. analogous, we have $PS.PZ = PN.PB$. $XY$ is the radical axes of the two circles, so, $PM.PC = PN.PB \Longrightarrow PR = PS \Longrightarrow R = S$.
Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Prove that \[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}. \]
Click for solution From Cauchy we have \[ \left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\right) \Big( a(b+c)+b(c+a)+c(a+b) \Big) \geq \left( \frac 1a +\frac 1b+ \frac 1c \right)^2 \] and as $\frac 1a + \frac 1b + \frac 1c = ab+bc+ca$ we obtain that \[ \left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\right) \geq \frac{ ab+bc+ca} 2 \geq \frac{\sqrt [3]{a^2b^2c^2}}{2} = \frac 32 \] which is what we wanted to prove.
Determine all integers $ n > 3$ for which there exist $ n$ points $ A_{1},\cdots ,A_{n}$ in the plane, no three collinear, and real numbers $ r_{1},\cdots ,r_{n}$ such that for $ 1\leq i < j < k\leq n$, the area of $ \triangle A_{i}A_{j}A_{k}$ is $ r_{i} + r_{j} + r_{k}$.
Click for solution Let's prove that only $n=4$ works. Obviously, $n=4$ is Ok, since we can take $A_1A_2A_3A_4$ to be a square and all the $r_i$'s equal to one third of the area of a triangle formed by three of its vertices. Since if we have $n\ge 5$ such points we also have $5$, all we need to do is prove that $n=5$ is impossible. First notice that if $A_1A_2A_3A_4$ is a convex quadrilateral, then $r_1+r_3=r_2+r_4$ and if $A_4$ lies inside $A_1A_2A_3$, then $r_4=-\frac{r_1+r_2+r_3}3$. I'll use these without mentioning it from now on. Case 1 $A_1A_2A_3A_4A_5$ is a convex pentagon In this case $r_i+r_{i+2}$ is the same for all $i$ (the indices are modulo $5$), which, in turn, implies that all the $r_i$ are equal. This is impossible, since $[A_1A_2A_3]=[A_1A_2A_5]$ clearly implies $[A_1A_2A_4]>[A_1A_2A_3]$. Case 2 $A_5$ lies inside the convex quadrilateral $A_1A_2A_3A_4$ Suppose furthermore that $A_5$ lies inside the triangles $A_1A_2A_3,A_1A_2A_4$, and outside the other two formed by the vertices $A_1,A_2,A_3,A_4$. We then get $r_5=-\frac{r_1+r_2+r_3}3=-\frac{r_1+r_2+r_4}3\Rightarrow r_3=r_4$, which means that $A_3A_4\|A_1A_2\Rightarrow r_1=r_2$. We also have $r_5+r_3=r_2+r_4=r_2+r_3\Rightarrow r_1=r_2=r_5$. This is impossible, since it would mean that $A_5$ lies on the parallel $A_1A_2$ to $A_3A_4$. Case 3 $A_4,A_5$ lie inside $A_1A_2A_3$ We get $r_4=r_5=-\frac{r_1+r_2+r_3}3$, so the areas of the following pairs of triangles are equal: $(A_1A_2A_4,A_1A_2A_5),(A_2A_3A_4,A_2A_3A_5)$. This means that $A_4A_5$ is parallel to both $A_1A_2$ and $A_2A_3$, which is, of course, absurd.
July 20th - Day 2
Find the maximum value of $ x_{0}$ for which there exists a sequence $ x_{0},x_{1}\cdots ,x_{1995}$ of positive reals with $ x_{0} = x_{1995}$, such that \[ x_{i - 1} + \frac {2}{x_{i - 1}} = 2x_{i} + \frac {1}{x_{i}}, \] for all $ i = 1,\cdots ,1995$.
Click for solution Taking sum both side from 1 to 1995 and some simplifying job we would obtain $ \sum_{i = 0}^{1994}x_i - \frac {1}{x_i} = 0$ ..........(#) Also , simplfying the expression also gives us $ (x_i\cdot x_{i - 1} - 1)(2x_i - x_{i - 1}) = 0$ If $ x_i\cdot x_{i - 1} = 1$ then (#) would gives us $ x_{1994} = \frac {1}{x_{1994}}$ or $ x_{1994} = 1$ . So this means $ x_0 = x_{1995} = \frac {1}{x_{1994}} = 1$ If $ 2x_i = x_{i - 1}$ $ \implies x_i = x_0\left(\frac {1}{2}\right)^i$ . Hence from (#) we get $ \sum_{i = 0}^{1994}x_0\left(\frac {1}{2}\right)^i - \frac {2^i}{x_0} = 0$ simplify to get $ x_0 = 2^{997}$ . So the maximum is $ x_0 = 2^{997}$
Let $ ABCDEF$ be a convex hexagon with $ AB = BC = CD$ and $ DE = EF = FA$, such that $ \angle BCD = \angle EFA = \frac {\pi}{3}$. Suppose $ G$ and $ H$ are points in the interior of the hexagon such that $ \angle AGB = \angle DHE = \frac {2\pi}{3}$. Prove that $ AG + GB + GH + DH + HE \geq CF$.
Let $ p$ be an odd prime number. How many $ p$-element subsets $ A$ of $ \{1,2,\dots,2p\}$ are there, the sum of whose elements is divisible by $ p$?
Click for solution Consider the polynomial $f(x,y)=(1+xy)(1+x^2y)\ldots(1+x^{2p-1}y)$, the $k$-th factor refering to whether $k$ is present in the set or not. Then the monomial $x^ky^l$ will refer to the set having $l$ elements and sum of elements $k$, so we need to compute the sum of coefficients of $x^{kp}y^p$ of $f$ to find the answer. To do this, we need to compute the sum of coefficients $x^{pk}y^{pl}$ and substract 2, since there are two terms for $l\neq 1$:$1, x^{p(2p-1)}y^{2p}$. To compute the sum of coefficients of $x^{pk}y^{pl}$ we use Multisection formula that says us that it equals $\frac{1}{p^2}\sum_{i,j}f(w^i,w^j)$($w$ is $p$th root of unity). Now if $w^i$ is not $1$, $f(w^i,w^j)=\prod(1+w^{ik}w^j)=\prod (1+w^k)^2$ (every $w^m$ will be written twice as $w^{ki}w^j$). This equals $\prod (-1-w^k)^2=g(-1)^2=((-1)^p-1)^2=4$ ($g(x)=x^p-1$ is the polynomial with roots $w^i$). Since there are $p-1$ choices for $w^i$, $p$ choices for $w^j$ we get $4p(p-1)$. Finally, if $w^i=1$, $f(1,w^j)=(1+w^j)^{2p}$. To evaluate $\sum_{j=0}^{p-1} (1+w^j)^{2p}$, note that it equals $p$ times the coefficients of $x^p$ in the polynomial $(1+x)^{2p}$ by the same multisection formula, so it equals $p\binom{2p}{p}$. So our total sum is $\frac 1{p^2} (4p(p-1)+p\binom{2p}{p}) =\frac{\binom{2p}{p}-2}p+4$. Substracting 2 we get the desired answer.