Triangle Inside Parabola

The equilateral triangle will be inverted and symmetric about the $y$ -axis. The side in the first quadrant will thus make an angle of $60$ degrees with the positive $x$ -axis, and thus have the equation $y = \sqrt{3}x$ , (as it also has one end at the origin).

We then need to find where this side and the parabola intersect. This will occur when

$\sqrt{3}x = \dfrac{x^{2}}{8} \Longrightarrow x = 8\sqrt{3}$ ,

where we have disregarded the trivial solution $x = 0$ .

Now for this value of $x$ we have that the point of intersection is $(8\sqrt{3}, 24)$ , so the side length of the equilateral triangle is the distance between this point and the origin. This distance is

$\sqrt{(8\sqrt{3})^{2} + 24^{2}} = 16\sqrt{3}$ .

Thus $q = 16, p = 3$ and $p + q = \boxed{19}$ .

Just a small variation of @brian charlesworth 's good solution:

In the first quadrant, the squared length of the inscribed triangle's side can be expressed as

$x^2 + (\frac{1}{8} x)^2$

The squared length of the inscribed triangle's side crossing the $y$ -axis can be expressed as

$(2x)^2$

The triangle is equilateral, hence

$x^2 + (\frac{1}{8} x^2)^2 = (2x)^2 \Rightarrow x^2((\frac{1}{8}x)^2 - 3) = 0 \Rightarrow x = 8\sqrt{3}$

So the length of any side in the triangle is $2 x = 16\sqrt{3}$ . This means $p = 3$ and $q = 16$ , and the solution to the problem is

$p + q = 19$