Hyperbolic Cliffs

Let the variable equilateral triangle be $ABC$ , $M$ be the midpoint of the horizontal side $AB$ , and $r$ be the radius of the incircle. Since $\triangle ABC$ is equilateral, its incenter is also its centroid. Then $MC = 3r$ , vertex $C = (0, -2r$ , and the equation of $BC$ is given by:

$\frac {y+2r}x = \sqrt 3 \implies y = \sqrt 3 x - 2r$

At the point $P$ on the hyperbola, where $BC$ is tangent to, has a gradient same as the line $BC$ , which is $\sqrt 3$ . Then we have:

$\begin{aligned} 2x - 2y \frac {dy}{dx} & = 0 \\ \implies \frac {dy}{dx} & = \frac xy & \small \blue{\text{When gradient }\frac {dy}{dx} = \sqrt 3} \\ \frac xy & = \sqrt 3 \\ \implies x & = \sqrt 3 y \end{aligned}$

Substitute $x = \sqrt 3 y$ in the $BC$ line equation, $y = 3y - 2r \implies y = r$ . This means that the tangent point $P$ coincides with $B$ . Then $x=\sqrt 3 r$ . Substitute in the hyperbola equation, $3r^2 - r^2 = 1 \implies r = \dfrac 1{\sqrt 2}$ . And the area of $\triangle ABC$ is $\dfrac 12 \cdot 2x \cdot 3r = 3 \sqrt 3 r^2 = \dfrac {3\sqrt 3}2$ . The required answer is $A+B+C = 3+2+3 = \boxed 8$ .

If we divide the equilateral triangle into 6 parts by its heights, it's clear that x = √3y leading to y = 1/√2 and x = √(3/2) when plugged into x² – y² = 1.

Max area
= 6 × little right triangle's area
= 6 × (1/2) × √(3/2) × √(1/2)
= 3√3/2

Answer = 3 + 2 + 3 = 8

By symmetry, the right side of the equilateral triangle will follow the equation $y = \sqrt{3}x + b$ , where $b$ is the $y$ -intercept.

Substituting this into $x^2 - y^2 = 1$ gives $x^2 - (\sqrt{3}x + b)^2 = 1$ , which rearranges to $x = \frac{1}{2}(-\sqrt{3}b \pm \sqrt{b^2 - 2})$ .

Since the equilateral triangle touches the hyperbola at exactly one point, there is only one answer for $x$ , so $\sqrt{b^2 - 2} = 0$ , which means $b = -\sqrt{2}$ for $b < 0$ .

The circumradius of the equilateral triangle is $R = |b| = \sqrt{2}$ , and the area of the equilateral triangle is $A = 3 \cdot \frac{1}{2}R^2 \sin 120° = \frac{3}{2}\sqrt{3}$ .

Therefore, $A = 3$ , $B = 2$ , $C = 3$ , and $A + B + C = \boxed{8}$ .