Start from an arbitrary point G of the circle and construct a polygonal line with 2k+1 vertices.

G' --- >1' --- >2'--- >3'--- > .... --- >k'--- >1''--- >2''--- > ... --- >k''.

Where (1') is the other intersection point of c with the parallel from G to line (1). Then (2') is the other intersection point of c with the parallel from (1') to line (2), (3') is the other intersection point of c with the parallel from (2') to line (3) etc. By going two times through all lines you create a polygonal line that comes back to the starting point G.

The proof is very simple. Each point (e.g. 1') is related to the previous one (e.g. G') by a reflexion on the line through the center O of the circle, that is orthogonal to the corresponding parallel (e.g. 1). Thus the varius points of the polygonal line are the successive image points under these reflexions : r1, r2, ..., rk. i.e. for the first k vertices

1' = r1(G'), 2' = r2(1'), ..., k' = rk((k-1)') , i.e. k' = f(G'), where f = r1*...*rk is the composition of an odd number of reflexions, whose axis pass through O. But such a composition is always a reflection and consequently f(f(G')) = G'. The procedure to generate f(f(G')) creates the 2k vertices of the polygonal line.

The case of polygonal line into an ellipse is considered in the file ClosedPolygonalLines_Conics.html .

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