Fermat-Torricelli minimum distance 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.
 A: 
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.
A: Let


*

*$a, b, c$ be the sides of $\triangle ABC$ whose angles are smaller than $120^\circ$. 

*$P$ be the Fermat-Torricelli point for $\triangle ABC$. 

*$\alpha = |AP|$, $\beta = |BP|$, $\gamma = |CP|$ and $S = \alpha+\beta+\gamma$.

*$\mathcal{A}$ be the area of $\triangle ABC$.


It is known that for such a triangle, $P$ is lying in its interior and
$$\angle APB = \angle BPC = \angle CPA = 120^\circ$$
Using these, we can express the sides and area of triangle as
$$\begin{cases}
a^2 = \beta^2 + \beta\gamma + \gamma^2\\
b^2 = \gamma^2 + \gamma\alpha + \alpha^2\\
c^2 = \alpha^2 + \alpha\beta + \beta^2
\end{cases}
\quad\text{and}\quad
\mathcal{A} = \frac{\sqrt{3}}{4}(\alpha\beta + \beta\gamma+\gamma\alpha)
$$
Summing the three equations from left and apply the equation from right, we find
$$a^2+b^2+c^2 = 2(\alpha^2 + \beta^2 + \gamma^2) + \frac{4}{\sqrt{3}}\mathcal{A}$$
As a result,
$$\begin{align}
S^2 &= (\alpha+\beta+\gamma)^2 
= \alpha^2 + \beta^2 + \gamma^2 + \frac{8}{\sqrt{3}}\mathcal{A}
= \frac12\left(a^2 + b^2 + c^2 - \frac{4}{\sqrt{3}}\mathcal{A}\right) + 
\frac{8}{\sqrt{3}}\mathcal{A}\\
&= \frac12\left(a^2 + b^2 + c^2\right) + 2\sqrt{3}\mathcal{A}\\
&= \frac12\left(a^2+b^2+c^2 + \sqrt{3(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}\,\right)
\end{align}
$$
An expression equivalent to what Quang Hoang obtained in another answer.
A: It is worthy to notice that absolutely no calculation is required in order to figure out some geometric properties of the Torricelli Point (for instance, how to construct it with ruler and compass).
Well, denote $f(X)=|XA|+|XB|+|XC|$. $f$ is convex since it is continuous and it is very easy to verify (with no calculation, only triangle inequality) that $f(X+Y/2)\geq \frac{f(X)+f(Y)}2$.
As such, $f$ obtains a global minima at an interior point $T$ of the triangle.
So we have $\nabla f(T)=0$. On the other hand, $\nabla|XA|$ is a unit vector in $T_X\mathbb R^2$ in the direction $AX$, which we will denote $n_A(X)$. This fact follows from simple calculation ($\nabla|x|=x/|x|$) or from thinking about the level curves of the distance function and its linear growth.
In any case, we therefor have: $n_A(T)+n_B(T)+n_C(T)=0$. The sum of three unit vectors is zero if and only if the angles between any two of them are $120$.
Consequences:

*

*The angles $ATB, BTC, CTA$ are all equal to $120$ degrees.


*If we construct equilateral triangles $t_1, t_2 ,t_3$ having their base concurrent with the sides of our original triangle, and their circumscribed circles are $\pi_1, \pi_2, \pi_3$ respectively then $\pi_1\cap\pi_2\cap\pi_3={T}$.
These 2 are enough to calculate all else you might need as was demonstrated above, and I hope you enjoyed the geometric flavor of this argument.
