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The ice cream cone (Posted on 2010-03-01) Difficulty: 3 of 5
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?

See The Solution Submitted by Dej Mar    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 2

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
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
           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
  (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
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