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Sphere-Cone (Posted on 2016-12-25) Difficulty: 3 of 5

Given a sphere of fixed radius a. A right circular cone is is to be
found which will enclose the sphere such that the sphere is tangent
to the cone's lateral surface, the sphere is tangent to the cone's
base at its center, and the ratio Ac/As is minimised where Ac is the
cone's surface area (both lateral and base) and As the sphere's
surface area.

To keep solutions uniform let's denote the cone's altitude by h, base
radius by r and slant height by L.

Also, give the minimised ratio.

See The Solution Submitted by Bractals    
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Solution Solution Comment 2 of 2 |
The lateral area of the cone is pi*r*L
The base area of the cone is pi*r^2
The surface area of the insphere is 4*pi*a^2

Then the ratio Ac/As = (pi*r*L+pi*r^2)/(4*pi*a^2) = r(L+r)/(4*a^2)

Take a cross section of the cone-sphere containing the altitude.  That creates an isosceles triangle and its incircle.  The triangle has sides L, L, and 2r and the circle has radius a.

The semiperimeter of the triangle is s=r+L.  Then the inradius a = sqrt[(s-L)*(s-L)*(s-r)/s] = r*sqrt[(L-r)/(L+r)]

Then Ac/As = r(L+r)/(4*r^2*(L-r)/(L+r)) = (L+r)^2/(4*r*(L-r))

Let x=L/r.  Then Ac/As = r^2*(x+1)^2/(4*r^2*(x-1)) = (x+1)^2/(4*(x-1))

The derivative d(Ac/As)/dx = (x+1)*(x-3)/(4*(x-1)^2). It has a positive root of x=3, which is the local minimum.

With x=3 then Ac/As=2.  Then a : r : L : h = 1 : sqrt(2) : 3*sqrt(2) : 4

Edited on December 25, 2016, 1:37 pm
  Posted by Brian Smith on 2016-12-25 12:33:20

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