Given semiperimeter and radii of inscribed and circumscribed circles, find the side lengths of triangle

Question

Consider $\triangle ABC$ with side lengths $a,b,c$, semiperimeter $\rho=\tfrac12\,(a+b+c)$, inradius $r$ and circumradius $R$. Let $u=\rho/R,\ v=r/R$, $a'=a/R,\ b'=b/R,\ c'=c/R$.

Given $u,\,v$, we can find $a',\,b',\,c'$ as roots of cubic equation

\begin{align} x^3-2\,u\,x^2+(u^2+v^2+4\,v)\,x-4\,u\,v&=0 \tag{1}\label{1} , \end{align}

and the side lengths in increasing order $a\le b\le c$ can be expressed as

\begin{align} a&= \tfrac23\,R\,\left(u+\sqrt{u^2-3\,v\,(v+4)}\, \cos(\phi+\tfrac{2\pi}3) \right) \tag{2a}\label{2a} ,\\ b&= \tfrac23\,R\,\left(u+\sqrt{u^2-3\,v\,(v+4)}\, \cos(\phi-\tfrac{2\pi}3) \right) \tag{2b}\label{2b} ,\\ c&= \tfrac23\,R\,\left(u+\sqrt{u^2-3\,v\,(v+4)}\, \cos(\phi) \right) \tag{2c}\label{2c} ,\\ \phi&= \tfrac13\,\arctan\left( {3\,\sqrt3\,v\,\sqrt{u^2\,(2\,(27-(5-v)^2)-u^2)-v\,(v+4)^3}} , {u\,(9\,v\,(2-v)-u^2)} \right) \tag{2phi}\label{2phi} . \end{align}

Numerical tests demonstrate that in general expressions \eqref{2a}-\eqref{2c} work fine for the valid range of $u,\,v$, except for an annoying special case, which corresponds to equilateral triangle, when $v=\tfrac12$, $u=\tfrac{3\sqrt3}2$, $a=b=c=\tfrac23\,u\,R$. In this case \eqref{1} collapses to

\begin{align} (x-\sqrt3)^3&=0 \tag{3}\label{3} \end{align}

and direct numeric calculation of $\phi$ fails.

Of course, it's not a big deal and this case can be algorithmically isolated and handled differently, but,

the question is: Is it possible to come up with some modified version of expression for $\phi$, which would work gracefully for all valid pairs of $u,\,v$?

This probably could be useful: the valid range is found as follows: for $v\in(0,\tfrac12]$,

\begin{align} u&\in[u_{\min},u_{\max}] ,\\ u_{\min}(v)&= \sqrt{27-(5-v)^2-2\,\sqrt{(1-2\,v)^3}} ,\\ u_{\max}(v)&= \sqrt{27-(5-v)^2+2\,\sqrt{(1-2\,v)^3}} . \end{align}

Both boundary curves $u_{\max}(v)$ and $u_{\min}(v)$, correspond to isosceles triangles, $u_{\max}(v)$ represents all isosceles triangles with the base as the smallest side, while $u_{\min}(v)$ represents all isosceles triangles with the base as the largest side. The area, bounded by $u_{\min}(v)$, $u_{\max}(v)$ and the $u$-axis covers all possible types of valid triangles with $R=1$ and can be used as a "map of the Trianglia" to make a quick crude tests of some general properties of triangles, for example, to estimate $\max(\sin\alpha+\sin\beta+\sin\gamma)$, etc.

Answer 1

(Converting comment to answer, as requested.)

Triangles aside, you are effectively asking for a universal solution to the (depressed) cubic $t^3+pt+q=0$ that works even when $p=q=0$. I'm personally not aware of one. Solutions such as those in Wikipedia separate-out such special cases. (Of course, when $p=0$, the solution is clear enough not to need a "formula".)

For completeness: Your cubic's corresponding "depressed" coefficients are $$\begin{align} p &=\frac13 (3v^2+12 v-u^2) = \frac1{3R^2}\left(3r^2+12\rho R−\rho^2\right) \\[6pt] &=−\frac1{3R^2}\left(a^2+b^2+c^2−ab−bc−ca\right) \\[10pt] q &=\frac2{27} u \left(u^2 - 18 v + 9 v^2\right) = \frac2{27R^3}\rho\left(9r^2−18rR+ρ^2\right) \\[6pt] &=−\frac1{27R^3}(2a−b−c)(−a+2b−c)(−a−b+2c) \end{align}$$ which (obviously) vanish when $a=b=c$.

Note that (a multiple of) $\sqrt{-p}$ is the coefficient of the $\cos(\phi+\cdots)$ terms in OP's expressions for the roots. So, maybe the fact that the $\phi$ formula fails when $p=0$ doesn't really matter, since those terms simply vanish.

Answer 2

EDIT1:

Not an answer really. As an aside like to how geometrically how $ (R,r,\rho?,a,b,c) $ can be represented by a polar diagram from a cubic.

Mathematica outputs a big analytical solution.. may be relatable to irreducible case of Cardano's solution of the cubic equation.. that yields three real roots.

$ (R,r,s) = (4.04145, 0.866025, 7.5)$ numerical values belong to an obtuse angled triangle of sides and semi-perimeter $(a= 7, b= 5,c= 3, \rho=7.5)$

Answer 3

One alternative is to test for the possibility that we will get an undefined or ill-conditioned expression for $\phi$.

When er have an equilateral triangle, the multiplier $\sqrt{u^2-3v(v+4)}$ vanishes. So we test whether the radicand $u^2-3v(v+4)$ is close to zero ("close" to be defined by the user, based on for instance computer precision), and if so we will default to the equilateral solution.