a.) For all positive integer $k$ find the smallest positive integer $f(k)$ such that $5$ sets $s_1,s_2, \ldots , s_5$ exist satisfying: i. each has $k$ elements; ii. $s_i$ and $s_{i+1}$ are disjoint for $i=1,2,...,5$ ($s_6=s_1$) iii. the union of the $5$ sets has exactly $f(k)$ elements. b.) Generalisation: Consider $n \geq 3$ sets instead of $5$.

Click for solution Just a simple observation: if $n$ is even, we can take $s_1=s_3=...s_{n-1}$ and $s_2=s_4=...s_n$ so we need at most $2k$ different elements! It is obvius we cant do it with less than that number of elements! Now for odd $n$, each element appears at most in $\frac{n-1}{2}$ sets, so we get $f(k)\geq \frac{2n}{n-1}k$ I am not awake like to give a construction right now...

A closed recticular polygon with 100 sides (may be concave) is given such that it's vertices have integer coordinates, it's sides are parallel to the axis and all it's sides have odd length. Prove that it's area is odd.

Let $r_1=2$ and $r_n = \prod^{n-1}_{k=1} r_i + 1$, $n \geq 2.$ Prove that among all sets of positive integers such that $\sum^{n}_{k=1} \frac{1}{a_i} < 1,$ the partial sequences $r_1,r_2, ... , r_n$ are the one that gets nearer to 1.

Given a convex figure in the Cartesian plane that is symmetric with respect of both axis, we construct a rectangle $A$ inside it with maximum area (over all posible rectangles). Then we enlarge it with center in the center of the rectangle and ratio lamda such that is covers the convex figure. Find the smallest lamda such that it works for all convex figures.

Find all positive integer $n$ such that the equation $x^3+y^3+z^3=n \cdot x^2 \cdot y^2 \cdot z^2$ has positive integer solutions.

Click for solution let's assume that $x \geq y \geq z$, thus $3x^3 \geq x^3+y^3+z^3=nx^2y^2z^2$, so $x \geq \frac{ny^2z^2}{3}$. Because $x^3+y^3+z^3=nx^2y^2z^2$ thus $x^2|y^3+z^3$, so $2y^3 \geq y^3+z^3 \geq x^2 \geq \frac{n^2y^4z^4}{9}$. If $z>1$ then $y \leq \frac{18}{16n^2}$ so $y=1$ but $y \geq z$ - contradiction. If $z=1$ then $y^3+1 \geq \frac{n^2y^4}{9}$, so $9+\frac{9}{y^3} \geq n^2y$. -If $y=1$ then $x^3+2=nx^2$ so the only solution is $(x,y,z,n)=(1,1,1,3)$ $(x^2|2=>x=1)$ so $n=3$ satisfies. -If $y>1$ then $10 \geq n^2y$ so a) $n=1$ then for instance $(x,y,z,n)=(3,2,1,1)$ is a solution so $n=1$ also satisfies b) $n=2$ then $y=2$ or $y=1$ then there are no solutions-easy to check. c) $n=3$ and $y=1$ that gives $(1,1,1,3)$ what we had before. (q.e.d.) Answer : $n=1,3$

Let $ G$ be a simple graph with $ 2 \cdot n$ vertices and $ n^{2}+1$ edges. Show that this graph $ G$ contains a $ K_{4}-\text{one edge}$, that is, two triangles with a common edge.

Click for solution Induction works. Assume we have shown the result to be true for small $n$. Suppose we are given $2(n+1)$ points with $(n+1)^2+1$ edges between them. By Turan's Theorem, there is a triangle. This means that we can choose two vertices $x,y$ of the triangle with degrees of the same parity. They are, of course, connected, since they form a side of the triangle. If the other $2n$ points have $n^2+1$ sides between them, then we're done by the induction hypothesis. If, on the other hand, they do not, there are at least $2n+2$ edges adjacent to either $x$ or $y$. One of these edges is between the two, and there can't be only $2n+1$ more because the degrees of $x,y$ have the same parity. This means that there are at least $2n+2$ edges adjacent to either $x$ or $y$ but not both. From here we easily conclude that $x,y$ have at least two common neighbours $u,v$, and $x,y,u,v$ is the "$K_4-$ one edge" .