The vertex of the parabola is the origin.  Let $A = \left( a, \frac{a^2}{8} \right)$ be one vertex of the triangle.  Then by symmetry, $B = \left( -a, \frac{a^2}{8} \right)$ is another vertex of the triangle.

[asy]
unitsize(0.2 cm);

real parab (real x) {
  return(x^2/8);
}

pair A, B, O;

A = (8*sqrt(3),24);
B = (-8*sqrt(3),24);
O = (0,0);

draw(O--A--B--cycle);
draw(graph(parab,-15,15));

dot("$A = (a,\frac{a^2}{8})$", A, E);
dot("$B = (-a,\frac{a^2}{8})$", B, W);
dot("$O = (0,0)$", O, S);
[/asy]

Then $AB^2 = (a + a)^2 = 4a^2,$ and
\[OA^2 = a^2 + \frac{a^4}{64}.\]Hence, $4a^2 = a^2 + \frac{a^4}{64}.$  Then $3a^2 = \frac{a^4}{64},$ so $a^2 = 3 \cdot 64,$ which means $a = 8 \sqrt{3}.$

Therefore, the side length of the triangle is $2a = \boxed{16 \sqrt{3}}.$
Question:
An equilateral triangle is inscribed in the parabola $x^2 = 8y,$ such that one of the vertices of the triangle coincides with the vertex of the parabola.  Find the side length of this equilateral triangle.
Answer:
Let's think step by step.

user
16 \sqrt{3}
An equilateral triangle is inscribed in the parabola $x^2 = 8y,$ such that one of the vertices of the triangle coincides with the vertex of the parabola.  Find the side length of this equilateral triangle.
test