Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.9 AIME, harder Prove it Bulgaria

A regular heptagon $ABCDEFG$ is given. The sides $AB$ , $BC$ , $CD$ , $DE$ , $EF$ , $FG$ and $GA$ are called opposite to the vertices $E$ , $F$ , $G$ , $A$ , $B$ , $C$ and $D$ , respectively. If $M$ is an interior point of $ABCDEFG$ , we say that a line through $M$ and a vertex of $ABCDEFG$ intersects the boundary of $ABCDEFG$ at a good point if this point is interior for the side which is opposite to the vertex. Prove that for every point $M$ the number of the good points is odd.

Solution

If $AM$ intersects the segment $DE$ in an interior point (i.e. we get a good point) then $M$ is interior for the triangle $ADE$ . The number of the good points which can be assigned to a fixed point $M$ is therefore equal to the number of the triangles amongst $ADE$ , $BEF$ , $CFG$ , $DGA$ , $EAB$ , $FBC$ and $GCD$ which contain $M$ as interior point.

These triangles determine 22 parts in the interior of $ABCDEFG$ (Fig. 1) as follows:

Fig. 1
* 7 triangles with a side which is a side of $ABCDEFG$ ,
* 7 quadrilaterals with one vertex which is a vertex of $ABCDEFG$ ,
* 7 triangles with two sides which are sides of the quadrilaterals,
* 1 heptagon.
Each one of these parts is common for exactly 1, 3, 5 or 7 of the triangles. Hence the number of the good points is 1, 3, 5 or 7.

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