## Symmetries on vertices

Draw an arbitrary pentagon ABCDE. Then an arbitrary point F. Then find the points G, H, I, J, K as follows:
G is the symmetric of F w.r. to A, H is the symmetric of G w.r. to B, etc.
The final point K and the original point F define a segment whose middle M is a fixed point, independent of the position of A (Prove it).
Of course M depends on the shape of the pentagon but not on the position of A.

This generalizes to arbitrary odd-sided polygons. For a hint to the even-sided polygons look at the reference below.

The key-fact is that an odd number of point-symmetries gives a composition that is a point symmetry. Thus for odd-sided polygons there is exactly one point (F) for which the resulting polygon FGH... is closed. This gives the way to solve Carnot's problem: to determine the odd-sided polygon given the middles A,B,C ... of its sides. Simply find the center of symmetry M, which is the composition of symmetries on A, B, C, ... . There is exactly one solution (F must coincide with M).
For even sided polygons, there can be more solutions.
A discussion for the even case starts at: SymmetriesOnVerticesEven2.html .

The general case for polygons with even number of sides is considered in: Carnot.html .