## Symmetries on vertices (even)

ABCDEFGH is a polygon (octagon) symmetric w.r. to X (center of symmetry).
Select a point I and get successive symmetric points w.r. to the vertices of the polygon, producing points I, J, K, L, ... etc.
Prove that the last point (P) has symmetric the original point I. Thus the polygon of I and its symmetrics on the vertices A, B, C, ...etc. is closed (non-symmetric in general).
Show that the segment IM has constant length and direction (look at SymmetriesOnVerticesEven.html for the idea). Find the condition that the resulting polygon IJKLMNOP is again symmetric (Move IM, so that its middle becomes X).

Polygon IJKL... is a solution of the so-called "Carnot's problem" : to find a polygon whose sides have middle-points some arbitrary given points A, B, C, ... (in our case A, B, C, ... are vertices of a symmetric polygon). We see here that the problem has many solutions. However odd-sided polygons accept exactly one solution. An odd-sided case is handled in the file: SymmetriesOnVerticesOdd.html .

Non-symmetric even-sided polygons have, in general, no solution to Carnot's problem. An example is given in Carnot.html .
Free movable are :
Point I , A, B, C, D and X.
Switch to the selection-tool (Ctrl+1), catch and modify these points.

For examples of non-symmetric even-sided polygons that do have solutions to Carnot's problem look at the file: NonSymmetricClosed.html .