solution: 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 $2a = \boxed{16 \sqrt{3}}.$
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  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.
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 16 \sqrt{3}
extra_info.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.
extra_info.split: test