Find the side of an equilateral triangle given only the distance of an arbitrary point to its vertices Triangle $ABC$ is an equilateral triangle and $P$ is an arbitrary point inside it. The distance from $P$ to $A$ is $4$ and the distance from $P$ to $B$ is $6$ and the distance from $P$ to $C$ is $5$. How to find the side of an equilateral triangle from this information?

Let $P=0\in{\mathbb C}$, and let $M={d\over\sqrt{3}}>0$ be the center of the triangle. When $s$ is the side length of the triangle then there is an argument $\phi$ such that the three vertices are given by
$$A={1\over\sqrt{3}}(d+se^{i\phi}),\quad B={1\over\sqrt{3}}(d+s\omega e^{i\phi}),\quad C={1\over\sqrt{3}}(d+s\bar\omega e^{i\phi})\ ,$$
where $\omega=e^{2\pi i/3}$ is a third root of unity. When $a$, $b$, $c$ are the given distances we therefore have
$$\eqalign{
a^2&={1\over3}(d^2 + ds(e^{i\phi}+e^{-i\phi})+s^2) \cr
b^2&={1\over3}(d^2 + ds(\omega e^{i\phi}+\bar\omega e^{-i\phi})+s^2) \cr
c^2&={1\over3}(d^2 + ds(\bar\omega e^{i\phi}+\omega e^{-i\phi})+s^2) \cr}\tag{1}$$
Adding these three equations gives
$$a^2+b^2+c^2=d^2+s^2\ .\tag{2}$$
Multiplying the equations $(1)$  with $1$, $\omega$, $\bar\omega$ (resp. with $1$, $\bar\omega$, $\omega$), and adding we obtain
$$\eqalign{
a^2+\omega b^2+\bar\omega c^2&=dse^{-i\phi}\cr 
a^2+\bar\omega b^2+\omega c^2&=dse^{i\phi}\ .\cr}\tag{3}$$ 
In order to eliminate $\phi$ we now multiply the two equations $(3)$. Using $\omega+\bar\omega=-1$ we get
$$a^4+b^4+c^4-(a^2b^2+b^2c^2+c^2a^2)=d^2s^2\ .\tag{4}$$
Writing 
$$a^2+b^2+c^2=:\sigma_1,\qquad a^2b^2+b^2c^2+c^2a^2=:\sigma_2$$
we can condense $(2)$ and $(4)$ to
$$d^2+s^2=\sigma_1,\qquad d^2s^2=\sigma_1^2-3\sigma_2\ .$$
This leads to
$$s^4-\sigma_1s^2+\sigma_1^2-3\sigma_2=0\ ,$$
from which $s$ can be determined. 
For the given data we obtain $\sigma_1=77$, $\sigma_2=1876$, and this leads to
$s^4-77s^2+301=0$, or $s_1=8.536$, $s_2=2.0324$. (It seems that $s_2$ corresponds to a solution where $P$ is outside the triangle.)
