# Let P1, P2, P3, P4, P5, and P6 be six parabolas in the plane, each congruent to the parabola y=(x^2) /16 .The vertices of the six parabolas are evenly spaced around a circle. The parabolas open outward with their axes being extensions of six of the circle's radii. Parabola P1 is tangent to P2, which is tangent to P3, which is tangent to P4, which is tangent to P5, which is tangent to P6, which is tangent to P1. What is the diameter of the circle? Let one parabola be vertical; in other words, its equation be `y = R + x^2 / 16 , ` where R is the radius of the circle in question. It is clear that each next parabola is rotated by `pi / 3 ` from the previous. Also, it is clear from the symmetry that the vertical parabola and its right neighbor are symmetric over the line `y = sqrt ( 3 ) x ,` which goes through the origin and makes the angle `pi / 6 ` with the positive half of the y-axis.

Because of this, all intersection points of these two parabolas lie on this line. It is given that the parabolas touch each other, so they have exactly one point of intersection.

This way, this point is also the intersection of `y = R + x^2 / 16 ` and `y = sqrt ( 3 ) x ,` so the equation `x^2 / 16 - sqrt ( 3 ) x + R = 0 ` has only one root. This means that its discriminant is zero, or `3 - R / 4 = 0 , ` so `R = 12` and the diameter is 24.