2017 AMC 10B Problem 24

The vertices of the hyperbola are $(1,1)$ and $(-1,-1)$; by symmetry take the centroid to be $G = (1,1)$. In an equilateral triangle the centroid is also the circumcenter, so the three vertices lie on a circle centered at $G$ with some radius $R$.

Intersect this circle with the hyperbola. Writing $(x-1)^2 + (y-1)^2 = R^2$ and using $x^2 + y^2 = (x+y)^2 - 2xy = (x+y)^2 - 2$:
$$(x+y)^2 - 2(x+y) + 2 - 2 = R^2 \quad\Longrightarrow\quad (x+y)^2 - 2(x+y) - R^2 = 0.$$
So every intersection point lies on one of two lines $x + y = k_1$ or $x + y = k_2$. Each such line is perpendicular to $y = x$ and meets the hyperbola in at most two points, which are mirror images across $y = x$.

Three vertices on two such lines means one line carries a single vertex. The other two vertices are mirror images across $y = x$, so the third vertex lies on the perpendicular bisector of that pair, which is the line $y = x$ itself; on the hyperbola that forces it to be $(-1,-1)$ (not $(1,1)$, the center itself).

Hence $R =$ distance from $(1,1)$ to $(-1,-1) = 2\sqrt2$. An equilateral triangle with circumradius $R$ has side $R\sqrt3 = 2\sqrt6$ and area
$$\frac{\sqrt3}{4}(2\sqrt6)^2 = \frac{\sqrt3}{4} \cdot 24 = 6\sqrt3.$$
Its square is $36 \cdot 3 = 108$.

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

Two symmetries carry the problem: an equilateral triangle's centroid is its circumcenter, and the hyperbola is symmetric about $y = x$, a line through that center. Together they force one vertex onto the axis of symmetry, where the only available point is the other vertex of the hyperbola. Once $R$ is known, the standard equilateral-triangle formulas finish it.

Trust the symmetry and verify: with $(-1,-1)$ as a vertex and center $(1,1)$, the other two vertices are at $120^\circ$ rotations, namely $(2+\sqrt3,\ 2-\sqrt3)$ and $(2-\sqrt3,\ 2+\sqrt3)$. Their coordinates multiply to $4 - 3 = 1$, so they do lie on the hyperbola, and the side length $\sqrt{(2\sqrt3)^2 + (2\sqrt3)^2} = 2\sqrt6$ gives area $6\sqrt3$.

• Trying to place three unknown points (t, 1/t) on the hyperbola and solving a system, rather than using the symmetry of the hyperbola about y = x.
• Reporting the area $6\sqrt3$ or its square wrongly as $36$, or using $R = \sqrt2$ (distance from $(1,1)$ to the origin) and getting $27$.