1) All the polygons P(n) of the sequence are symmetric about Y.

2) The line joining the middles of opposite sides of a polygon P(n), pass also from the middles of the after next polygon P(n+2). Here [LI] passes also through W and T.

3) Excluding P(1), polygon P(n) has sides parallel to P(n+2).

4) Excluding P(1), the ratio of two parallel sides of a polygon P(n) and its after next, P(n+2), is constant (=3/4). Consequently the sequence of perimeter-ratios s(n) = perim(P(n+2))/perim(P(n)) is constant.

5) Starting from k=1, the sequence of perimeter-ratios u(n) = perim(P(2k+1))/perim(P(2k)) is constant.

1) Is proved easily inductively.

2) First KJ is parallel and one half of AC = LI. Project K, J parallel to side MR (or OP) to Z, A' respectively. KZ, JA' are equal and consequently WR = (1/2)KZ = (1/2)JA' = PT. By the symmetry PT = WM, hence W is the middle of MR and correspondingly T is the middle of OP.

3) The sides KJ, UV are respectively parallel and half the lengths of AC and PR, proven to be parallels.

4) The distances LW = WZ = A'T = TI. The first one because R is the middle of LK and we project parallel to WR. The other equalities follow by the symmetry. Also ZY = YA' and ZA' being half the length of LI. All these imply that LY/WY = 4/3.

Notice that although the first polygon is equilateral, all other polygons of the sequence, in general, are not equilateral. The case of hexagons is a particular one of equilateral symmetric polygons of 2n sides. The results proven here are not true for n >3. Look at SymmetricNestedOctagons.html for a discussion of these cases and related open problems. Quadrangles and the middles of their sides produce sequences of polygons with similar properties. One can ask if these two cases, of quadrangles and hexagons, are the only cases of symmetric polygons that produce sequences of polygons with fixed ratios of perimeters. I don't know the answer yet.

Variable points are: A and F, using the selection tool (Ctrl+1), and B, C using the Select-on-contour tool (Ctrl+2). The two last points move on hidden circles.

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