# The concurrency property of the diagonals (semidiagonals) of bicentric polygons.

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.

• 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. – brainjam Mar 26 at 2:58
• This would be very surprising. – user Mar 26 at 5:17
• 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. – brainjam Mar 26 at 12:48
• This means that in a triangle the semi-diagonals are still concurrent, is not it? For which $n$ they cancel to be concurrent? – user Mar 26 at 13:10
• Yes, they are concurrent for $n=3$. They are not concurrent for $n=5$. – 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.

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.

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.

• 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. – user Apr 11 '20 at 7:04
• @user, my impression as well. And it doesn't get you any closer to your goal of a mapping from the regular polygon. – brainjam Apr 11 '20 at 17:40
• You are absolutely correct. I would appreciate any hint how to prove or disprove the mapping conjecture. – user Apr 12 '20 at 8:15
• 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. – brainjam Apr 12 '20 at 16:44
• 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. – user Apr 12 '20 at 19:29