Vijetnam 2007

Denote the vertices of the regular 2007-polygon by $A_1,A_2,\dots ,A_{2007}$. Note that every quadrilateral has 3 edges of the given polygon if and only if its 4 vertices are consecutive vertices of the polygon.

Denote by $A$ the set of the vertices except $A_{4k}$ ($k=1,2,\dots ,501$) and $A_{2007}$, in fact, $A=\{A_1,A_2,A_3,A_5,A_6,A_7,\dots ,A_{2005},A_{2006}\}$. Let $|A|=1505$ and $A$ has no 4 consecutive vertices of the given 2007-polygon. Thus, $k\ge 1506$.

Now, we will prove that every subset $B$ of 1506 vertices contains 4 consecutive vertices of the given 2007-polygon. Let $T$ be a subset of arbitrary 1506 vertices of the given 2007-polygon. By deleting 501 vertices not belonging to $B$ we obtain a decomposition of the set of the vertices of the given 2007-polygon into the subsets $B_1,B_2,\dots ,B_m$ with $m\le 501$. By the Dirichlet principle, there is a set $B_i$ containing $\ge \frac{1506}{501}>3$ consecutive vertices.