A very simple concept from geometry: Given a number of points with no three collinear, that are the midpoints of the sides of a polygon in rotational order, a unique solution generally exists for the polygon if the number of points is odd. With an even number of points, however, either the polygon fails to exist or is nonunique.

The claim is proven via a descent argument, which is demonstrated in the picture below for the pentagon having midpoints $A,B,C,D,E$.

The pentagon is divided into a quadrilateral with three midpoints $C,D,E$ and a triangle itself midpoints $A,B$ which share a diagonal. The quadrilateral must have its fourth midpoint at $F$ which is then the third midpoint of the triangle. Assuming $F$ is not collinear with $A$ and $E$, The triangle is then found by drawing a line through $A$ parallel to $E\overline{EF}$ and cyclic permutations. Thus gives vertices $G,H,K$ of the pentagon. Then the collinearity and distance-doubling criteria give the rremainingpentagonal vertices $I,J$.

Thus with the parallelogram construction to determine $F$ the pentagonal problem is reduced to the simpler triangular one. In a similar way any set of $n$ points is reduced to $n-2$ points. Thus we can reduce all odd cases to a triangle (if the vertices of this triangle are not collinear), thus a unique solution; butcall even cases reduce to a quadrilateral for which the solution exists only if the final quadrilateral is a parallelogram and then is nonunique.

A very simple concept from geometry: Given a number of points with no three collinear, that are the midpoints of the sides of a polygon in rotational order, a unique solution generally exists for the polygon if the number of points is odd. With an even number of points, however, either the polygon fails to exist or is nonunique.

The claim is proven via a descent argument, which is demonstrated in the picture below for the pentagon having midpoints $A,B,C,D,E$.

The pentagon is divided into a quadrilateral with three midpoints $C,D,E$ and a triangle itself midpoints $A,B$ which share a diagonal. The quadrilateral must have its fourth midpoint at $F$ which is then the third midpoint of the triangle. Assuming $F$ is not collinear with $A$ and $E$, The triangle is then found by drawing a line through $A$ parallel to $E\overline{EF}$ and cyclic permutations. Thus gives vertices $G,H,K$ of the pentagon. Then the collinearity and distance-doubling criteria give the rremainingpentagonal vertices $I,J$.

Thus with the parallelogram construction to determine $F$ the pentagonal problem is reduced to the simpler triangular one. In a similar way any set of $n$ points is reduced to $n-2$ points. Thus we can reduce all odd cases to a triangle (if the vertices of this triangle are not collinear), thus a unique solution; butcall even cases reduce to four points for which the solution exists only if the final four points are vertices of a parallelogram and then is nonunique.