Solving optimum Bézier approximation of a circle

[Adlibber]

I am trying to determine the optimum length of the control vectors for a cubic Bézier approximation of a $90^\circ$ circular arc. However to my limited mathematical mind the equation cannot be solved algebraically. Wolfram Alpha even finds it too complicated. However I am told that it IS solvable! I remain unconvinced.

Can anyone spot a way of simplifying this monster?

\begin{align*}
&\left(u_\mu^6+(\textstyle\frac{1}{2})^6\right)(18\kappa^2-24\kappa+8)+\left(u_\mu^5+(\textstyle\frac{1}{2})^5\right)(-54\kappa^2+72\kappa-24)+\\
&\left(u_\mu^4+(\textstyle\frac{1}{2})^4\right)(63\kappa^2-66\kappa+18)+\left(u_\mu^3+(\textstyle\frac{1}{2})^3\right)(-36\kappa^2+12\kappa-4)+\\
&\left(u_\mu^2+(\textstyle\frac{1}{2})^2\right)(9\kappa^2+6\kappa-6)=0\\
&\text{where }\;\;u_\mu=\frac{1}{2}-\frac{\sqrt{12-20\kappa-3\kappa^2}}{4-6\kappa}
\end{align*}

 

[Ackbach]

If you look at these equations, you will see that it is at least a quintic in $\kappa$. Quintics are not, in general, solvable exactly. You can solve the quadratic exactly, and the cubic exactly, and even the quartic exactly. But not the quintic. Abel and Galois proved that. However, there are specific quintics that you can solve. I'm going to give this problem to Mathematica, to see if it will solve it. Stay tuned...

...Mathematica chickens out, and gives the answers using the Root function - basically telling you that it couldn't find the roots exactly. Mathematica does give one interesting piece of information: there are at least 7 solutions. Now, if I ask for approximations to these roots, Mathematica gives me the following:
\begin{align*}
\kappa_1&= -3.45296\\
\kappa_2&= 0.366686\\
\kappa_3&= 0.518404\\
\kappa_4&= 0.347233-1.1646i\\
\kappa_5&= 0.347233+1.1646i\\
\kappa_6&= 0.556569-0.00384123i\\
\kappa_7&= 0.556569+0.00384123i.
\end{align*}
The complex solutions come in conjugate pairs, as you'd expect if all the coefficients are real.

If Mathematica can't solve it exactly, I don't think you're going to be able to solve it exactly.

 

[Adlibber]

Thanks very much for your response and effort, Ackbach. Good to know that simple intuition still has value!

I must assume that whoever found the correct value (0.551915024494) did so using a more clever route than I chose. Although I did find a method that gave a 96% result (but not a 100%!).