Fermat-Torricelli minimum distance

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.

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.