2017 AMC 10B Problem 24

The hyperbola $xy = 1$ has two vertices (the points where its two branches come closest to each other): $(1, 1)$ and $(-1, -1)$. The picture for $(-1, -1)$ is just the picture for $(1, 1)$ turned halfway around the origin, so it is enough to take the centroid at $G = (1, 1)$.

1. Use the circle. In an equilateral triangle the centroid is also the center of the circle through the three corners. So all three corners are the same distance, call it $R$, from $(1, 1)$.
2. Distance from $(1, 1)$ to a point on the hyperbola. For a point $(x, y)$ with $xy = 1$:
$$(x-1)^2 + (y-1)^2 = x^2 + y^2 - 2(x+y) + 2.$$
Since $(x+y)^2 = x^2 + 2xy + y^2 = x^2 + y^2 + 2$, we have $x^2 + y^2 = (x+y)^2 - 2$. So the squared distance is
$$(x+y)^2 - 2(x+y).$$
It depends only on the sum $x + y$.
3. Only two possible sums. All three corners have squared distance $R^2$, so their sums $x + y$ all solve $t^2 - 2t = R^2$. That is a quadratic equation, so there are at most two possible sums.
4. Points with the same sum. If $x + y$ and $xy = 1$ are both fixed, the pair $x, y$ is fixed too, just possibly swapped. So the only points of the hyperbola with a given sum are some point $(p, q)$ and its mirror image $(q, p)$ across the line $y = x$.
5. Three corners, two sums. So two of the corners must be a mirror pair $(p, q)$ and $(q, p)$. The third corner is the same distance from both of them (the triangle is equilateral), so it lies on the line of points equally far from $(p, q)$ and $(q, p)$. That line is $y = x$.
6. On the hyperbola, $y = x$ gives $x^2 = 1$: the points $(1, 1)$ and $(-1, -1)$. The corner cannot be the center $(1, 1)$ itself, so it is $(-1, -1)$.
7. The radius. $R$ is the distance from $(1, 1)$ to $(-1, -1)$: $R = \sqrt{2^2 + 2^2} = 2\sqrt2$.
8. The triangle. The centroid sits $\frac23$ of the way down each height, so the height is $h = \frac32 R = 3\sqrt2$. In an equilateral triangle, height $= \frac{\sqrt3}{2} \times$ side, so the side is $\frac{2h}{\sqrt3} = \frac{6\sqrt2}{\sqrt3} = 2\sqrt6$.
9. Area: $\frac12 \cdot 2\sqrt6 \cdot 3\sqrt2 = 3\sqrt{12} = 6\sqrt3$. Its square is $36 \cdot 3 = 108$.

The answer is $\boxed{\textbf{(C)}\ 108}$.

Two symmetries do the work. An equilateral triangle's centroid is the center of its circle, and the hyperbola $xy = 1$ looks the same after flipping over the line $y = x$, which passes through that center. Together they force one corner onto the line $y = x$, where the only choice is the other vertex of the hyperbola. After that, standard equilateral-triangle facts finish the job.

Guess and check. Guess that $(-1, -1)$ is a corner. The other two corners come from turning it $120^\circ$ each way around $(1, 1)$, which gives $\left(2 + \sqrt3,\ 2 - \sqrt3\right)$ and $\left(2 - \sqrt3,\ 2 + \sqrt3\right)$. Check them: $(2 + \sqrt3)(2 - \sqrt3) = 4 - 3 = 1$, so both are on the hyperbola. The distance between them is $\sqrt{(2\sqrt3)^2 + (2\sqrt3)^2} = \sqrt{24} = 2\sqrt6$, so the area is $\frac{\sqrt3}{4} \cdot 24 = 6\sqrt3$ and its square is $108$.