Fall Mathematical Competition

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.