Two Mating Parabolas

By symmetry, the graph is reflective in $y = x$ , and the slopes of the three lines will be $\tan^{-1} 135° = -1$ , $\tan^{-1} 15° = 2 - \sqrt{3}$ , and $\tan^{-1} 75° = 2 + \sqrt{3}$ .

Let the line with a slope of $2 + \sqrt{3}$ have an equation of $y = (2 + \sqrt{3})x + b$ . Since it is tangent to both parabolas, the discriminants of both $(2 + \sqrt{3})x + b = x^2 + a$ and $x = ((2 + \sqrt{3})x + b)^2 + a$ will be zero, so $-4a + 4b + 4\sqrt{3} + 7 = 0$ and $-4(7 + 4\sqrt{3})a - 4(2 + \sqrt{3})b + 1 = 0$ .

These two equations solve to $a = \cfrac{3}{4}$ and $b = -1 - \sqrt{3}$ , so the equations of the parabolas are $y = x^2 + \cfrac{3}{4}$ and $x = y^2 + \cfrac{3}{4}$ , and the equation of this line is $y = (2 + \sqrt{3})x -1 - \sqrt{3}$ .

The equilateral triangle has a vertex on the lines $y = (2 + \sqrt{3})x -1 - \sqrt{3}$ and $y = x$ , which solves to $(1, 1)$ .

Now let the line with a slope of $-1$ have an equation of $y = -x + c$ . Since it is tangent to the parabola $y = x^2 + \cfrac{3}{4}$ , the discriminant of $-x + c = x^2 + \cfrac{3}{4}$ will be zero, so $c - \cfrac{1}{2} = 0$ , which solves to $c = \cfrac{1}{2}$ , making the equation of this line $y = -x + \cfrac{1}{2}$ .

The midpoint of the base of the equilateral triangle is on the lines $y = -x + \cfrac{1}{2}$ and $y = x$ , which solves to $\bigg(\cfrac{1}{4}, \cfrac{1}{4}\bigg)$ .

By the distance formula, the height of the equilateral triangle between $\bigg(\cfrac{1}{4}, \cfrac{1}{4}\bigg)$ and $(1, 1)$ is $\cfrac{3\sqrt{2}}{4}$ , which makes one of its sides $\cfrac{2}{\sqrt{3}} \cdot \cfrac{3\sqrt{2}}{4} = \cfrac{\sqrt{6}}{2}$ , and its area $\cfrac{\sqrt{3}}{4}\bigg(\cfrac{\sqrt{6}}{2}\bigg)^2 = \cfrac{3\sqrt{3}}{8}$ .

Therefore, $A = 3$ , $B = 8$ , $C = 3$ , and $ABC = \boxed{72}$ .

Note that the graph is symmetric in the line $y=x$ . One nice approach here is to rotate everything by $45^\circ$ . We can do this with the variable change $x=\frac{u+v}{\sqrt2},\;\;\;\;y=\frac{v-u}{\sqrt2}$

(the $\sqrt2$ s in the substitution ensure there's no scaling). Our equations become $\frac{v-u}{\sqrt2} = \frac{(u+v)^2}{2}+a$ and $\frac{u+v}{\sqrt2} = \frac{(v-u)^2}{2}+a$

These look worse, but the graph (in $uv$ coordinates) is now

There is a shared horizontal tangent. To form an equilateral triangle, all we need to do is make sure one of the other shared tangents has gradient $\tan 60^\circ=\sqrt3$ .

There are several ways to go from here; implicit differentiation works well but we can try without calculus.

For a line $v=mu+c$ to be tangent to a parabola, it has to meet it in exactly one point. This means the equation we get by substituting for $v$ has to have exactly one root. Since with a parabola we expect a quadratic equation, this is the same as saying the discriminant of that quadratic has to be zero.

In this case, substituting $v=mu+c$ into the equations for the first parabola, we get $\frac{mu+c-u}{\sqrt2} = \frac{(u+mu+c)^2}{2}+a$

(which is indeed a quadratic in $u$ ). Setting the discriminant to zero gives $1 - 4 a + 4 \sqrt2 c - 2 m - 8 a m + 4 \sqrt2 c m + m^2 - 4 a m^2 = 0$

Now, when $m=0$ , this is just $1 - 4 a + 4 \sqrt2 c = 0$

so the horizontal tangent has equation $v=\frac{4a-1}{4\sqrt2}$

Substituting $m=\sqrt3$ , we get $4 - 2 \sqrt3 - 16 a - 8 \sqrt3 a + 4 \sqrt2 c + 4 \sqrt6 c=0$

Doing the same for the second parabola, we get $4 + 2 \sqrt3 - 16 a + 8 \sqrt3 a + 4 \sqrt2 c - 4 \sqrt6 c=0$

These are just linear simultaneous equations in $a$ and $c$ ; solving, we find $a=\frac34$ and $c=\sqrt2$ .

Finally, the three tangents have equations $v=\frac{1}{2\sqrt2},\;\;\;\;v=\sqrt2 \pm u \sqrt3$

The easiest way to work out the area of the equilateral triangle is just to note its height is $\sqrt2-\frac{1}{2\sqrt2}$ , from which the area is $\frac{3\sqrt3}{8}$ and the answer to the problem is $\boxed{72}$ .