AMC12 2020 12A, Problem 24

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 $s$?

Solution: 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 $\angle APC = 120^{\circ}$. [asy] size(200); pen p = fontsize(10pt)+gray+0.5; pen q = fontsize(13pt); pair A,B,C,D,P,Q; real s=sqrt(7); B=origin; A=s*dir(60); C=s*right; P=IP(CR(A,1),CR(C,2)); Q=rotate(60,A)*P; draw(A--B--C--A, black+0.8); draw(A--P--B^^P--C^^A--Q--C, p); draw(P--Q, p+dashed); //draw(A--A+C--C, p); label("$A$", A, up, q);  label("$B$", B, 0.5*(B-P), q); label("$C$", C, 0.5*(C-P), q);   label("$P$", P, dir(180), q);   label("$Q$", Q, 0.25*(Q-B), q);  label("$\sqrt{3}$",B--P, right, p);  label("$2$",C--P, 2*left, p);  label("$1$",A--P, 1.5**dir(-10), p);   label("$1$", A--Q, dir(250), p);  label("$1$",P--Q, down, p);  label("$\sqrt{3}$",C--Q, right, p);  [/asy] 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 $s = \fbox{\sqrt{7}}$.

Answer: \sqrt{7}