Bulgarian Mathematical Olympiad

Solution:

LEMMA. For any five vertices of a regular 13-gon there exists an isosceles triangle with vertices amongst these points.

Proof of the lemma. Let the five points form a convex pentagon $ABCDE$. We first consider the case when there exist two pairs of parallel lines determined by some vertices of $ABCDE$. We have the following possibilities:

1) There are two pairs of parallel sides of $ABCDE$—for example $AE \parallel CD$ and $BC \parallel DE$. Then we have $\angle AED = \angle BCD$ or $\angle AED = 180^\circ - \angle BCD$, i.e. $\sin \angle AED = \sin \angle BCD$. Hence $AD = 2R \sin \angle AED = 2R \sin \angle BCD = BD$ and therefore the triangle $\triangle ABD$ is isosceles.

2) There are two diagonals that are parallel respectively to two sides of $ABCDE$. Without loss of generality we may assume that $AB \parallel CE$. If $AC \parallel DE$ we conclude as in 1) that $BC = CD$. If $AD \parallel BC$, then $EB = BD$. The cases $BE \parallel CD$ and $BD \parallel AE$ are similar.

3) There is a diagonal of $ABCDE$ that is parallel to its side and a pair of parallel sides of $ABCDE$. Without loss of generality we may assume that $AB \parallel CE$. Since $AE$ is not parallel to $BC$ (otherwise $ABCE$ is a rectangle with vertices amongst the vertices of a regular 13-gon) we may assume that $DE \parallel BC$. Then $DC = CA$.

Let us now assume that there exists at most one pair of parallel lines determined by some vertices of $ABCDE$. Since the vertices of $ABCDE$ determine 10 lines, at least 9 of them are not parallel to each other. Consider the 9 pairs of vertices determining such lines. Every such pair is a base of an isosceles triangle whose third vertex is a vertex of the 13-gon. Note that all such third vertices are different because no two of the bases are parallel. Hence at least one of these 9 vertices is a vertex of $ABCDE$ and this proves the existence of the desired isosceles triangle. This completes the proof of the lemma.

The regular 26-gon is formed by two disjoint regular 13-gons. If we choose 9 of its vertices then at least 5 of them are vertices of one of these 13-gons. Then it follows from the lemma that there exists an isosceles triangle with vertices amongst the vertices of this 13-gon.

Finally, note that there exist 8 vertices such that no three of them form an isosceles triangle. For example, if the vertices are labelled from 1 to 26, then we choose the vertices $1, 2, 4, 5, 10, 11, 13$ and $14$.