Other than triangle, H. Eves' tricky quickie can be used for n gon. I recommend one better geometric inlustration for following page:
We denote X(Y) as a circle centered at X passing Y.
A1A2... An is any n-gon
We select on line AnA1 any point P1
A circle A1(P1) cuts line A1A2 at P2
A circle A2(P2) cuts line A2A3 at P3
A circle A3(P3) cuts line A3A4 at P4
A circle An(Pn) cuts line AnA1 at Q1
We finish construction of the first chain of circles with:
n circles Ai(Pi), n points Pi
Point Q1 is generally not P1 and use for starting point of second chain of circles:
A circle A1(Q1) cuts line A1A2 at Q2
A circle A2(Q2) cuts line A2A3 at Q3
A circle A3(Q3) cuts line A3A4 at Q4
A circle An(Qn) cuts line AnA1 at R
Case A: If n odd
a). R=P1 and we can not make more another chain of circles.
b). P1Q1 = P2Q2 = P3Q3 = ... = PnQn but changed when we change starting point P1.
c). Midpoints Mi of segments PiQi are fixed with any starting point P1
d). If we select P1 = M1 then Pi=Qi, Ai(Pi) = Ai(Qi) and they are sequence of circles as required in the problem. It is also geometric procedure to construct these circles.
Case B: If n even
a). R<>P1 and we can make chains of circles infinitely.
b). P1Q1 = P2Q2 = P3Q3 = ... = PnQn = t = constant with any starting point P1
c). Hence, we can see one pair of circles Ai(Pi)Ai(Qi) at vertex Ai as one circle with thickness t.
d). If we change n-gon until thickness t=0 then we get n-gon which has sequence of circles as required in the problem. In this case, we can select any starting point P1.
Please note that all results are already proved in explaining text of the applet. Here is only geometric inlustration by using H. Eves' tricky quickie.
By the way: please check to see and correct page:
Some items are empty, for example: item No. 860, item No. 780. I think it is because auto indexing.
Bui Quang Tuan