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.
I know that the vertex of the parabola is at $(0,0),$ so that is one of the vertices of the triangle. To find the other two vertices, I need to find the points where the parabola intersects a line that is at a 60-degree angle to the x-axis, passing through the origin. I can use the slope-intercept form of a line to write the equation of this line as $y = \sqrt{3} x,$ where $\sqrt{3}$ is the tangent of 60 degrees. To find the intersection points, I need to solve the system of equations $x^2 = 8y$ and $y = \sqrt{3} x.$ I can substitute $y = \sqrt{3} x$ into the first equation to get $x^2 = 8 \sqrt{3} x.$ I can divide both sides by $x$ to get $x = 8 \sqrt{3},$ as long as $x \neq 0.$ I can plug this value of $x$ into the second equation to get $y = \sqrt{3} (8 \sqrt{3}) = 24.$ So the intersection points are $(8 \sqrt{3}, 24)$ and $(-8 \sqrt{3}, 24).$ The distance between these points is the side length of the equilateral triangle, which I can find using the distance formula. The distance is $\sqrt{(8 \sqrt{3} - (-8 \sqrt{3}))^2 + (24 - 24)^2} = \sqrt{(16 \sqrt{3})^2} = 16 \sqrt{3}.$ 