Find the maximum value for the area of a heptagon with all vertices on a circle and two diagonals perpendicular.

How many ways are there to divide a $2\times n$ rectangle into rectangles having integral sides, where $n$ is a positive integer?

If $x,y,z$ are positive real numbers and $xy+yz+zx=1$ prove that \[ \frac{27}{4} (x+y)(y+z)(z+x) \geq ( \sqrt{x+y} +\sqrt{ y+z} + \sqrt{z+x} )^2 \geq 6 \sqrt 3. \]

Click for solution The inequality $9(x+y)(y+z)(z+x)\geq 8(xy+yz+zx)(x+y+z)$ (1) it seems to be lovely Using (1) and after Cauchy We have $\frac{27}{4}(x+y)(y+z)(z+x)\geq 6(x+y+z)\geq (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2$ Now by using AM-GM and then (1) we have $(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2\geq 9\sqrt[3]{(x+y)(y+z)(z+x)}\geq 9\sqrt[3]{\frac{8(x+y+z)}{9}}\geq 9 \sqrt[3]{\frac{8 \sqrt{3(xy+yz+zx)}}{9}}\geq 6\sqrt{3}$ and we are done

For all integers $n\geq 1$ we define $x_{n+1}=x_1^2+x_2^2+\cdots +x_n^2$, where $x_1$ is a positive integer. Find the least $x_1$ such that 2006 divides $x_{2006}$.

From a point $Q$ on a circle with diameter $AB$ different from $A$ and $B$, we draw a perpendicular to $AB$, $QH$, where $H$ lies on $AB$. The intersection points of the circle of diameter $AB$ and the circle of center $Q$ and radius $QH$ are $C$ and $D$. Prove that $CD$ bisects $QH$.

Each one of 2006 students makes a list with 12 schools among 2006. If we take any 6 students, there are two schools which at least one of them is included in each of 6 lists. A list which includes at least one school from all lists is a good list. a) Prove that we can always find a good list with 12 elements, whatever the lists are; b) Prove that students can make lists such that no shorter list is good.