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.
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 $2a = \boxed{16 \sqrt{3}}.$