Constructing bicentric pentagon

Question

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?

Answer 1

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

$\tan(45°-\theta/4)=(1-x)/(1+x)$

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

leading to a quintic equation:

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

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 $0<x<\tan(22.5°)=\sqrt2-1$. 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.