Vijetnam 2011

To verify the claim, we will show that it is possible to cover the pentagon $ABCDE$ by 5 unit discs with center lying on the edges of the pentagon.

We have following remark: Remark: It is possible to cover a triangle $XYZ$ with edges of length not exceeding $\sqrt{3}$ by 3 unit discs with centers at the vertices of the triangle.

Proof: Assuming the contrary, there exists a point $M$ belonging to triangle $XYZ$ but not lying in the unit discs with centers at the vertices of the triangle. Then we have $MX>1$, $MY>1$ and $MZ>1$.

Clearly, among the angles $\widehat{XMY}$, $\widehat{YMZ}$ and $\widehat{ZMX}$ at least one is larger than $120^{\circ }$. Without loss of generality, assume that $\widehat{XMY}\ge 120^{\circ }$. Using the cosine theorem for triangle $XMY$, we obtain $$X{Y}^2=M{X}^2+M{Y}^2-2MX\cdot MY\cdot \cos \widehat{XMY}>1+1+2\cdot \frac{1}{2}=3(\text{since }\cos \widehat{XMY}\le -\frac{1}{2}).$$ Consequently $XY>\sqrt{3}$, contradicting the assumption. The contradiction yields the claim to be verified.

Since the triangles $ABC$, $ACD$ and $ADE$ have edges with length less than $\sqrt{3}$, according to the remark, they can be covered by the triples of unit discs $((A),(B),(C))$, $((A),(C),(D))$ and $((A),(D),(E))$. Hence the pentagon $ABCDE$ is covered by 5 unit discs with center at the vertices of the pentagon. According to Dirichlet's principle, among the 5 discs there exists one containing at least 403 of the chosen points. ■