AMC 12 Mock Exam 3 Problem 23

Place the right angle at $O=(0,0)$ with the legs along the axes to $P=(3,0)$ and $Q=(0,4)$. The center of an equilateral triangle on a segment of length $s$ lies on the perpendicular bisector of the segment, at distance $\frac{s}{2\sqrt3}$ from its midpoint, on the outer side.

• On $\overline{OP}$ ($s=3$, outward is below): $G_1=\left(\tfrac32,-\tfrac{\sqrt3}{2}\right)$.
• On $\overline{OQ}$ ($s=4$, outward is to the left): $G_2=\left(-\tfrac{2}{\sqrt3},2\right)$.
• On $\overline{PQ}$ ($s=5$, midpoint $\left(\frac32,2\right)$, outward unit normal $\left(\frac45,\frac35\right)$): $G_3=\left(\tfrac32+\tfrac{2}{\sqrt3},\,2+\tfrac{\sqrt3}{2}\right)$.

By the shoelace formula, $$[G_1G_2G_3]=\tfrac12\left|(x_2-x_1)(y_3-y_1)-(x_3-x_1)(y_2-y_1)\right| ,$$ with $x_2-x_1=-\frac32-\frac2{\sqrt3}$, $y_2-y_1=2+\frac{\sqrt3}2$, $x_3-x_1=\frac2{\sqrt3}$, $y_3-y_1=2+\sqrt3$. The product terms are $\left(-\frac32-\frac2{\sqrt3}\right)(2+\sqrt3)=-3-\frac{3\sqrt3}{2}-\frac4{\sqrt3}-2$ and $\frac2{\sqrt3}\left(2+\frac{\sqrt3}{2}\right)=\frac4{\sqrt3}+1$, so the bracket is $-6-\frac{3\sqrt3}2-\frac{8}{\sqrt3}=-6-\frac{25\sqrt3}{6}$ and the area is $3+\frac{25\sqrt3}{12}$, i.e., $\boxed{3+\tfrac{25\sqrt3}{12}}$.

(Alternative. By Napoleon's theorem the centers form an equilateral triangle with side $t$, where $t^2=\frac{a^2+b^2+c^2}{6}+\frac{2}{\sqrt3}K=\frac{50}6+\frac{12}{\sqrt3}=\frac{25}{3}+4\sqrt3$ for this triangle of area $K=6$; then $\frac{\sqrt3}4t^2=\frac{25\sqrt3}{12}+3$. Choice (A) is the inward construction.)