After answering a recent question where the collinearity of the points $X,O,I$ was proven in an elementary way, I rather soon came to the conclusion that the intersection point $X$ of the diagonals is completely fixed by the positions ($O,I$) and the radii ($R,r$) of the circumscribed and inscribed circles (i.e. the product $\sin\alpha\sin\beta$ in the cited answer is a constant for given $R,r$). After some algebra I obtained the following simple formula: $$ XI=\frac rR\sqrt{\frac{R^2+d^2-2r^2}2}, $$ where $d$ is the distance between the centers of the circumscribed and inscribed circles $OI$ (given $R$ and $r$, it is fixed by the Poncelet porism).

To my surprise I did not find any mention of the above formula in the online sources. But during the search I realized that the formula follows from a much more general statement cited in the above reference:

For an even-sided [bicentric] polygon, the diagonals are concurrent at the limiting point of the two circles, whereas for an odd-sided polygon, the lines connecting the vertices to the opposite points of tangency are concurrent at the limiting point.

I am looking for a (possibly simple) proof of the above statement and/or reference to the original publication on this result.

  • $\begingroup$ I revisited this and found (through Geogebra experiments) that the statement is not true for odd-sided bicentric polygons. In general the diagonals don't go through the limiting point and are not concurrent. I've contacted Mathworld about this. $\endgroup$ – brainjam Mar 26 at 2:58
  • $\begingroup$ This would be very surprising. $\endgroup$ – user Mar 26 at 5:17
  • $\begingroup$ For example, for the case $n=3$, the semi-diagonals intersect at the Gergonne point, which is different from (but quite close to) the limiting point. $\endgroup$ – brainjam Mar 26 at 12:48
  • $\begingroup$ This means that in a triangle the semi-diagonals are still concurrent, is not it? For which $n$ they cancel to be concurrent? $\endgroup$ – user Mar 26 at 13:10
  • 1
    $\begingroup$ Yes, they are concurrent for $n=3$. They are not concurrent for $n=5$. $\endgroup$ – brainjam Mar 26 at 13:27

Note: After I gave my partial answer, which included some handwaving, I found a reference to the requested proof, which I've added at the end.

The 4 vertex case isn't too difficult, and it suggests ways to attack the general case.

Bicentric quadrilateral

In the above diagram, $ABCD$ is a bicentric quadrilateral. Points $K,L,M,N$ are the touch points with the incircle. The diagonals of $ABCD$ and $KLMN$ meet at $X$. (see Yiu, Euclidean Geometry Notes, pg 157).

We want to show that $X$ is a limit point of the two circles.

Polar of X

To do this we construct the polars of $X$ for the two circles. (see Weisstein, Polar ) Build the complete quadrilateral $ABCDEF$. Regarding the sides of $ABCD$ as tangents to the incircle, we get the construction of the polar $EF$ of $X$ with respect to the incircle. Regarding $A,B,C,D$ as the points where two lines through $X$ cut the circumcircle, we get the polar $EF$ of $X$ with respect to the circumcircle. Evidently the two polars are the same, which implies that both circles invert $X$ to the same point $X'$ (which lies on the polar). So $X$ is a limit point of the two circles.

Assume $n$ is even. For the case of the general $n$-sided bicentric polygon, if we assume that all principle diagonals are concentric at a point $X$, we can use a similar argument to show that $X$ is a limit point. This of course is just a partial result, because it remains to prove that the diagonals are concurrent.

Some further empirical observations and speculations. The setup for the quadrilateral case suggests a construction for bicentric-adjacent polygons (they are tangential, but not necessarily cyclic) that may be a useful avenue to a proof. Start with a circle $C$ (the incircle) and a line $p$(the polar). For even $n$ place $\frac{n}{2}$ points $P_i$ on $p$ and draw the $n$ tangents from these points to $C$. Then the intersections of adjacent tangents form a tangential polygon $P$ with the property that the principal diagonals are concurrent. But $P$ will generally not be cyclic. For example, when $n=4$ the polygon $P$ will be cyclic only if $\angle{P_1IP_2}$, where $I$ is the center of $C$, is a right angle. For general $n$ it remains to show that certain configurations of $P_i$ result in cyclic $P$, and that for a given combination of incircle and circumcircle if one bicentric polygon has concurrent diagonals they all do.

I've ignored the case $n$ odd. With any luck it follows from $n$ even.

Update: There's a proof at Halbeisen and Norbert, A Simple Proof of Poncelet’s Theorem (on the occasion of its bicentennial), Theorems 4.1,4.2

  • $\begingroup$ Thank you very much for your solution and for the reference! My first impression that the proof from the reference is not quite simple. Besides the case of odd-sided polygons seems to be not considered. $\endgroup$ – user Apr 11 '20 at 7:04
  • $\begingroup$ @user, my impression as well. And it doesn't get you any closer to your goal of a mapping from the regular polygon. $\endgroup$ – brainjam Apr 11 '20 at 17:40
  • $\begingroup$ You are absolutely correct. I would appreciate any hint how to prove or disprove the mapping conjecture. $\endgroup$ – user Apr 12 '20 at 8:15
  • $\begingroup$ Exactly what is your mapping conjecture? If it's "[there exists] a mapping from a regular polygon onto arbitrary bicentric polygon with the same number of sides" I would say it is obviously true. $\endgroup$ – brainjam Apr 12 '20 at 16:44
  • $\begingroup$ You are right. A real conjecture of the mapping does not exist. It would therefore suffice to prove or disprove that if three (semi-)diagonals are (non-)concurrent in the case of regular polygon their "images" in an equal-sized bicentric polygon are also (non-)concurrent. $\endgroup$ – user Apr 12 '20 at 19:29

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