Suppose that $\triangle ABC$ is an equilateral triangle of side length $s$, with the property that there is a unique point $P$ inside the triangle such that $AP = 1$, $BP = \sqrt{3}$, and $CP = 2$. What is $s?$
This is a typical problem that can be solved by the rotation method. The rotation method is discussed in the book Geometry Techniques . Such a solution can be further confirmed by noting that $1$, $\sqrt{3}$ and $2$ can form a Pythagorean triangle.
Rotate $\triangle{CAP}$ for $60^{\circ}$ about point $C$. Denote the new position of $P$ as $P'$ and then the new position of $A$ will be $B$. Because $CP=CP'$ and $\angle{PCP'}=60^{\circ}$, therefore $\triangle{CPP'}$ is equilateral. It follows that $\angle{CP'P}=60^{\circ}$. Furthermore, $BP=\sqrt{3}$, $BP'=AP=1$, and $CP'=CP=2$, therefore $\triangle{PBP'}$ is a $30^{\circ} - 60^{\circ} - 90^{\circ}$ triangle, or $\angle{PP'B}=60^{\circ}$. This means that $\angle{CP'B}=120^{\circ}$. Now applying law of cosines leads to $$BC=\sqrt{2^2 + 1^2 - 2\times 2\times 1\times\cos 120^{\circ}}=\boxed{\sqrt{7}}$$
