This paper describes computations of the circumradius of cyclic polygons given by the lengths of the sides. For cyclic pentagons, the defining polynomial of $r^2$ has degree $7$, so there are at most $7$ pentagons of given side lengths.
The convex inscribed polygon of given side lengths (if exists) is unique (see this question).
My question is about star pentagons, there are at least three cases:
- Given side lengths $(2.6,2.7,2.8,2.9,3)$ there are $7$ pentagons, $2$ of them are star pentagons.
| r=2.38343 | r=1.69187 | r=1.66074 | r=1.62688 | r=1.58755 | r=1.52302 | r= 1.50091 |
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In the figure the order of vertices on the circle for the convex pentagon is ABCDEF, for the star pentagons is ADBEC. (code)
- Changing $e=3$ to $3.1$, for $(2.6,2.7,2.8,2.9,3.1)$ there exist only $5$ pentagons, no star pentagons.
- Between $3$ and $3.1$ there is a value $e\approx3.0065$ such that the star pentagon of given side lengths is unique.
I want to prove:
Given side lengths $(a,b,c,d,e)$, $e$ largest, and there is a star pentagon. If the pentagon formed by tangents at $A,B,C,D,E$ has signed area zero, then it is unique star pentagon.
[In the below picture, let the self-intersection point be $P$, then the signed area of $T_{AB}T_{BC}T_{CD}T_{DE}T_{EA}$ is $\text{area}(PT_{BC}T_{CD})-\text{area}(PT_{AB}T_{EA}T_{DE})$.]
My partial solution: For star pentagons inscribed in a circle of radius $r$, $$\tag1\label1\arcsin\left(\frac{a}{2r}\right)+\arcsin\left(\frac{b}{2r}\right)+\arcsin\left(\frac{c}{2r}\right)+\arcsin\left(\frac{d}{2r}\right)-\arcsin\left(\frac{e}{2r}\right)=\pi$$ If the pentagon $T_{AB}T_{BC}T_{CD}T_{DE}T_{EA}$ has signed area zero, then (in unsigned area): $$\text{area}(OT_{AB}T_{BC})+\text{area}(OT_{BC}T_{CD})+\text{area}(OT_{CD}T_{DE})\\-\text{area}(OT_{DE}T_{EA})-\text{area}(OT_{EA}T_{AB})=0$$ so $$\frac12r|T_{AB}T_{BC}|+\frac12r|T_{BC}T_{CD}|+\frac12r|T_{CD}T_{DE}|-\frac12r|T_{DE}T_{EA}|-\frac12r|T_{EA}T_{AB}|=0$$ Dividing by $\frac12r$ and rewriting distances, $$|T_{AB}B|+|BT_{BC}|+|T_{BC}C|+|CT_{CD}|+|T_{CD}D|+|DT_{DE}|\\-|T_{DE}E|+|ET_{EA}|-|T_{EA}A|+|AT_{AB}|=0$$ Since $|T_{AB}B|=|AT_{AB}|=r\tan\frac{\angle AOB}2,\ldots$ we get $$\tan\frac{\angle AOB}2+\tan\frac{\angle BOC}2+\tan\frac{\angle COD}2+\tan\frac{\angle DOE}2-\tan\frac{\angle EOA}2=0$$ Since $\frac{d}{dr}\arcsin\left(\frac{a}{2r}\right)=-\frac{\frac a2}{r \sqrt{r^2-(\frac a2)^2}}=-\frac1r\tan\frac{\angle AOB}2$, the LHS of \eqref{1} has a stationary point at the solution $r$. From this can we conclude the star pentagons of given side lengths is unique?
Also, in \eqref{1} the last term has negative sign where I assumed the arc $\overparen{ADE}$ is smaller than half a circle. For $(2.7, 2.75, 2.8, 2.9, 3)$, and \eqref{1} has unique solution but there are $2$ star pentagons (for one of them $\overparen{ADE}$ is larger than half a circle). How can I exclude this?
Some plots: ($r$ increases from its minimum $e/2$.)
(Let $A'$ be a point on the circle such that $EA'=e$. If $A'=A$ we get a solution.)
For $(a,b,c,d,e)=(2.6,2.7,2.8,2.9,3)$
(code)
$A'$ passes through $A$, reverses direction, and passes through $A$, so there are $2$ solutions.
For $(a,b,c,d,e)=(2.6,2.7,2.8,2.9,3.0065)$
(code)
when $A'$ reaches $A$ it reverses direction, so there is a unique solution.