The Fermat – Torricelli point minimizes sum of distances $S$ taken from vertices of a triangle of sides $a,b,c. $ Find $S$ in terms of $a,b,c$.
Am trying to set up problem with a Lagrange multiplier or partial derivatives for extremization but it seems tedious even with a CAS.
Although it is supposed to be known from the earliest Greek times, it is not seen (by me) in these modern times.
Best Answer
The above picture depicts the Fermat-Torricelli point. From there we see that $$\begin{aligned}S^2 &= b^2+c^2-2bc\cos(A+\frac\pi3)\\ &=b^2+c^2-2bc(\cos A\cos\frac\pi3 - \sin A\sin\frac\pi3)\\ &= b^2+c^2-bc\cos A +\sqrt{3} bc\sin A\\ &= \frac{a^2+b^2+c^2}2 +\frac{\sqrt 3abc}{2R}, \end{aligned}$$ where $R$ is the radius of the circumcircle of $\triangle ABC$.
Then we can use, for example Heron's formula, to obtain a formula in $a$, $b$, and $c$ only.