Question about the Solution to the Moving Sofa Problem

Question

I have a question about the solution to this question posed by @newzad: Modelling the "Moving Sofa"

In the answer, @Intelligenti pauca solves the equation

$$ (x_{env}/r)^2+(y_{env}/t)^2=1 $$

where

$$ x_{env}=(r−t)\cosα+\frac12(t−r)\cos(2α)+\frac12(r+t) $$ $$ y_{env}=4(t−r)\sin\fracα2\cos^3\fracα2 $$

And obtains the solution

$$ \begin{cases} \displaystyle\bar\alpha= 2\arccos\sqrt{t\over{t+r}}, &\text{for $t\le3r$;}\\ \displaystyle\bar\alpha= \arccos\sqrt{t\over{2(t-r)}}, &\text{for $t\ge3r$.}\\ \end{cases} $$

I can't seem to solve the equation analytically, and was hoping someone could help me with it.

Thank you very much for your time!

Answer 1

This works.

Use the definition of $y_{\text{env}}$, plug in the first equation, make $\alpha=2 \cos ^{-1}(z)$. Now use the trigonometric identities, factor the result (since $z=\pm 1$ are trivial solutions) and what remains is $$-4 t^3-4 z^2 \left(t^2 (3 r-5 t)\right)-32 z^4 \left(t^2 (t-r)\right)+z^6 (r-4 t) (r-t) (r+4 t)=0$$ which is a cubic equation in $z^2$.

Using the method for solving cubic equations (pretty tedious here), I suppose that we shall get the (nice and simple) result.