2017 AMC 10B Problem 24

By symmetry, assume that the centroid is the hyperbola vertex $G=(1,1).$ At least two triangle vertices lie on the same branch of the hyperbola. They cannot both lie on the negative branch: if two of their $x$-coordinates were negative, the third would exceed $3,$ while the sum of the three $y$-coordinates would be less than $\frac{1}{3},$ contradicting that their centroid is $(1,1).$ Thus two vertices lie on the positive branch. Write these vertices as $P=(a,\frac{1}{a})$ and $Q=(b,\frac{1}{b}),$ where $a,b>0.$ The centroid of an equilateral triangle is also its circumcenter, so $P$ and $Q$ are equidistant from $G.$ For $t>0,$ the squared distance from $(t,\frac{1}{t})$ to $G$ is $$\left(t+\dfrac1t\right)\left(t+\dfrac1t-2\right).$$ This is strictly increasing as $t+\frac{1}{t}$ increases from $2,$ so the distinct points must satisfy $b=\frac{1}{a}.$ Hence $P$ and $Q$ are reflections across $y=x.$ The third vertex lies on the perpendicular bisector $y=x.$ Its coordinates also satisfy $xy=1,$ so it is either $(1,1)$ or $(-1,-1).$ It cannot equal the centroid, so it is $(-1,-1).$ Therefore, the circumradius is the distance from $(1,1)$ to $(-1,-1),$ namely $2\sqrt2.$ Dividing the equilateral triangle into three triangles at its center gives its area as $$3\cdot\dfrac12(2\sqrt2)^2\sin120^\circ=6\sqrt3.$$ The square of the area is $(6\sqrt3)^2=108.$ Thus, the correct answer is C .