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 $s$?
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$
Solution 1(a)
We begin by rotating $\triangle{ APB}$ counterclockwise by $60^{\circ}$ about $A$, such that $P\mapsto Q$ and $B\mapsto C$. We see that $\triangle{ APQ}$ is equilateral with side length $1$, meaning that $\angle APQ = 60^{\circ}$. We also see that $\triangle{CPQ}$ is a $30$-$60$-$90$ right triangle, meaning that $\angle CPQ= 60^{\circ}$. Thus, by adding the two together, we see that $\angle APC = 120^{\circ}$.

We can now use the law of cosines as following:
\begin{align*} s^2 &= (AP)^2 + (CP)^2 - 2\cdot AP\cdot CP\cdot \cos{\angle{APC}} \\ &= 1 + 4 - 2\cdot 1\cdot 2\cdot \cos{120^{\circ}} \\ &= 5 - 4\left(-\frac{1}{2}\right) \\ &= 7, \end{align*}
giving us that $s = \boxed{ \sqrt{7} \ \textbf{(B)}}$.
~ciceronii

Solution 1(b)
Rotate $\triangle CPA$ counterclockwise $60^\circ$ around point $C$ to $\triangle CQB$. Then $CP=CQ, \angle PCQ=60^\circ$, so $\triangle CPQ$ is an equilateral triangle.

Note that $\triangle PQB$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, hence $\angle BPQ=30^\circ$, and $\angle BPC=90^\circ$, so \[BC^2=PC^2+PB^2=2^2+3=7,\] and the answer is $\boxed{\textbf{(B) } \sqrt{7}}$.
~szhangmath

Solution 1(c)
Rotate $\triangle BPC$ counterclockwise by $60^\circ$ around point $B$ to $\triangle BQA$. Then $BP=BQ$, and $\angle PBQ=60^\circ$, so $\triangle BPQ$ is an equilateral triangle.

Note that $\triangle QAP$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, hence $\angle PQA=30^\circ$, and $\angle BQA=90^\circ$, so \[AB^2=PQ^2+AQ^2=2^2+3=7,\] and the answer is $\boxed{\textbf{(B) } \sqrt{7}}$.