David recently visited Pop's Ice Cream Shoppe and ordered the $1 sugar cone with one scoop of vanilla ice cream placed firmly atop the cone.

Pop told David that he would receive a percentage discount in the cost of the treat equal to the closest integer value to the answer of the following question if he answered it correctly:

"If the sugar cone is a right circular cone with a height of 10 inches, and the scoop of vanilla ice cream is a perfect sphere with a diameter of 4 inches, and both the cone and sphere are equal in spatial volume, what percentage of ice cream is above the base of the cone?"

David, a bright student, gave a correct answer. How much did David pay for the ice cream cone?

Let H be the height of the cone and r
the radius of its base. Let R be the
radius of the scoop of ice cream and
p the height of its center above the
cone. 

If you cut a sphere, of radius R, with
a plane, that is a distance b from the
center of the sphere, then the volume 
of the largest portion is

  (pi/3)*[2*a^3 + b*(3*a^2 - b^2)]  (1)

The volume of the sphere and the cone
are equal, so

  4*(pi/3)*R^3 = (pi/3)*r^2*H

               or

           r^2 = 4*R^3/H            (2)

The relation of R, p, and r is   

     p^2 + r^2 = R^2                (3)

Combining (2) and (3) we get

  p = R*sqrt(1 - 4*R/H)

Plugging R and p into (1) for a and b
we get the amount of ice cream above the
cone

  (pi/3)*R^3*[2 + sqrt(1 - 4*R/H)*(2 + 4*R/H)]

Multiplying by 100 and dividing by the
volume of the sphere we get the percent

  25*[2 + sqrt(1 - 4*R/H)*(2 + 4*R/H)]

For our problem with H = 10 and R = 2,

  50 + 14*sqrt(5) =~ 81.30495.

Posted by Bractals on 2010-03-01 16:14:57