Suppose that the vertices of a polygon with four or more sides lie on a circle, and that another (possibly non-concentric) circle touches each of its sides. Intuitively, it seems to me that the polygon has to be regular, and the circles concentric, but I don't see a way to go about proving it. Does anyone have an idea how to do this?
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Have you know about Poncelet's Porism? – Somos Oct 20 '22 at 20:05
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@Somos : Thanks for the steer. I looked at the Wikipedia article on it, but couldn't see how to get quite the result I mentioned out of the article. – John Bentin Oct 20 '22 at 20:14
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Did you look at the MathWorld article also? The polygon need not be regular. – Somos Oct 20 '22 at 20:17
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No. The four-sided case is given below, followed by a mathematical-induction approach that can be used to generate bicentric polygons with more sides.
To see what happens with four sides, you first observe that a convex quadrilateral with sides $a,b,c,d$ in rotational order can always be made cyclic by adjusting either diagonal until one vertex hits the same circle as the other three.
Now suppose that $a+c=b+d$, not necessarily equal. For instance, you might select $a=b\not=c=d$, a kite. Then the quadrilateral always has an incircle, but we have just seen it can be made cyclic too by adjusting the diagonals (in the case of a kite, you end up with one pair of opposing angles being right angles, but the others need not be so). So you have a quadrilateral with both an incircle and a circumcircle. Such quadrilateral is called bicentric.
Now suppose that, having constructed a bicentric quadrilateral $ABCD$, you hold the circumcircle fixed and move the incircle, without changing its radius, towards vertex $A$. Vertices $B$ and $D$ are adjusted along the circumcircle so that $\overline{AB}$ and $\overline{AD}$ remain tangent to the incircle. Similarly $C$ is adjusted so that $\overline{BC}$ is tangent to the incircle, but the point on the circumcircle that makes "$\overline{CD}$" tangent to the incircle is no longer $C$ but a fifth point $C'$ such that a pentagon $ABCC'D$ is formed. Initially $\overline{CC'}$ will lie outside the incircle, but ultimately it must pass through as the incircle approaches tangency at $A$ and the pentagon collapses to that point. So there must be an intermediate position where $\overline{CC'}$ just touches the incircle and you now have a bicentric pentagon with the same incircle/circumcircle radius ratio as the original quadrilateral.
A similar procedure can be used to add another side to the pentagon, then the hexagon, and so on.
Oscar Lanzi
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Thank you. I can see now that any cyclic kite has an incircle. But I don't see how the condition can hold for a nonregular pentagon. – John Bentin Oct 20 '22 at 20:53
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An example of an irregular bicentric pentagon is given in this post. – John Bentin Oct 21 '22 at 06:37