2020 AMC 12A Problems/Problem 24

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 $s$ ?

$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$

Solution

$[asy] draw((0,0)--(4,5.65)--(8,0)--cycle); label("A", (4,5.65), N, p = fontsize(10pt)); label("C", (8,0), SE, p = fontsize(10pt)); label("B", (0,0), SW, p = fontsize(10pt)); label("P", (3.5,3.5), NW, p = fontsize(10pt)); draw((0,0)--(3.5,3.5)); label("$\sqrt{3}$",(0,0)--(3.5,3.5), SE); draw((8,0)--(3.5,3.5)); label("2",(8,0)--(3.5,3.5), SW); draw((4,5.65)--(3.5,3.5)); label("1",(4,5.65)--(3.5,3.5), E); draw((8,0)--(4,5.65)--(11,5.65)--cycle); label("$P'$", (6,4.75), NE, p = fontsize(10pt)); draw((4,5.65)--(6,4.75)); label("1",(4,5.65)--(6,4.75), S); draw((8,0)--(6,4.75)); label("$\sqrt{3}$",(8,0)--(6,4.75), E); draw((3.5,3.5)--(6,4.75)); label("1", (3.5,3.5)--(6,4.75), SE); [/asy]$

We begin by rotating $\triangle{ABC}$ by $60^{\circ}$ about $A$ , such that in $\triangle{A'B'C'}$ , $B' = C$ . We see that $\triangle{APP'}$ is equilateral with side length $1$ , meaning that $\angle APP' = 60^{\circ}$ . We also see that $\triangle{CPP'}$ is a $30-60-90$ right triangle, meaning that $\angle CPP'= 60^{\circ}$ . Thus, by adding the two together, we see that $\angle APC = 120^{\circ}$ . We can now use the law of cosines as following: