# Constructing bicentric pentagon

I'm trying to construct a bicentric pentagon in geogebra. I read on Wikipedia that a Pentagon is bicentric if and only if it satisfies this formula $$r(R-x)=(R+x)\left(\sqrt{(R-r)^2-x^2}+\sqrt{2R(R-r-x)}\right)$$ where $$R,r$$ and $$x$$ are respectively the radius of the circumcircle, the radius of the incircle and the distance between the centers. So one way to do this is by finding three (exact) values $$R,r$$ and $$x$$ and plugging them in geogebra. Is there any known values that work?

• FYI, eliminating the square roots gives a polynomial equation that factors as $$y\left(y^3 - 8 y^2 r R + 16 y r^2 R^2 + 8 y r^3 R - 16 r^4 R^2\right)=0, \quad y:=x^2-R(R-2r)$$ The first factor corresponds to Euler's Theorem for triangles, so is extraneous. The second factor is cubic in $y$, hence is explicitly solvable in terms of $r$ and $R$; from $y$, one can obtain $x$.
– Blue
Commented Dec 2, 2023 at 11:56
• Well $y$ is not $0$ @Blue
– PNT
Commented Dec 2, 2023 at 12:25
• "Well $y$ is not $0$" ... That's why I said that the first factor is extraneous. :)
– Blue
Commented Dec 2, 2023 at 12:26
• @Blue Your remark is very interesting. But I am faced with it with the following problem : I have attempted to browse many cases, with a program covering a large "spectrum" of values of $R$ and $r$. The third degree equation has always a single real root for $y$ (two complex roots ; I haven't computed the discriminant, but surely it is negative). But, the values of $x$ one gets from the $y$ values do not lead to a good between-centers distance for having a bicentric polygon. Why that ? Is it that the relationship is a necessary only (non sufficient) condition for having a bicentric pentagon ? Commented Dec 3, 2023 at 10:12
• @JeanMarie: Hmmm ... Well, I don't vouch for the Wikipedia relation, which appears to be taken from MathWorld's "Poncelet's Porism" entry. (Of course, I don't even vouch for my own work! :) I've just tried a quick GeoGebra sketch using the cubic solution to calculate $y$ (and $x$), but it seems I'm not getting any bicentric pentagons, either. Curious.
– Blue
Commented Dec 3, 2023 at 11:01

We pursue an alternate construction in which the parameters are angles of the pentagon.

Let $$ABCDE$$ be a mirror-symmetric pentagon inscribed in the unit circle, with vertices

$$A=(0,1), B=(\sin\theta,\cos\theta), C=(\sin\theta,-\cos\theta), D=(-\sin\theta,-\cos\theta), E=(-\sin\theta,\cos\theta)$$

Then the $$y$$ axis is an angle bisector and the other angle bisectors form two pairs concurrent with this axis at the points indicated:

$$B,E\to(0,\cos\theta-\tan(45°-\theta/4)\sin\theta)$$

$$C,D\to(0,\sin\theta-\cos\theta)$$

To make the pentagon bicentric these concurrency points are to be matched. This leads to

$$2\cos\theta-([1+\tan(45°-\theta/4)]\sin\theta)=0$$

$$[1+\tan(45°-\theta/4)]\tan\theta=2$$

We convert this to a polynomial equation by defining $$x=\tan(\theta/4)$$. From trigonometric identities we have

$$\tan(45°-\theta/4)=(1-x)/(1+x)$$

$$\tan\theta=4(x-x^3)/(1-6x^2+x^4)$$

$$x^5+x^4-2x^3-6x^2-3x+1=0.$$

This quintic turns out to be reducible, with $$x+1$$ as a factor. Dividing this out leaves

$$x^4-2x^2-4x+1=0,$$

in which the resolvent cubic has no rational roots and therefore this pentagon will not be constructible via unmarked straightedge and compasses (and algebraically, the roots are unwieldy). To get a convex polygon we must find a root corresponding to $$0°<\theta<90°$$, from which $$0. Numerical calculation reveals that to five decimal places the appropriate root is $$x\approx0.22527$$ from which $$\theta\approx50°47'$$.

We may compare this with what we would have gotten with nicer values of $$\theta$$. On the left below, we have what would be obtained with $$\theta=45°$$ by matching the vertices of the pentagon to those of a regular octagon; the proposed incircle does not reach the bottom side of the pentagon because we made $$\theta$$, the arc from $$A$$ to $$B$$, too small. If we try $$\theta=60°$$ by matching the vertices of the pentagon to those of a regular hexagon, we see the opposite problem. The calculated root falls in the expected range between these two misses.

The implication is that if (except for the regular case) we want "nice" values for any angles, we have to get not-so-nice ones for the other geometric properties.

UPDATE: I have found an irregular bicyclic pentagon with integer-degree angles. See this answer from MO.

• Do you mean that non-regular bicentric pentagons are not constructible ?
– PNT
Commented Dec 2, 2023 at 20:42
• That would not surprise me. Given a general set of side lengths, pentagons inscribed in a circle turn out not to be constructibel — even with neusis. Commented Dec 2, 2023 at 21:05
• Is this the case with any non-regular polygon? (n>4)
– PNT
Commented Dec 2, 2023 at 22:17
• It only gets worse as $n$ increases. Commented Dec 2, 2023 at 23:27