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 Sphere-Cone (Posted on 2016-12-25)

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.

 Submitted by Bractals No Rating Solution: (Hide) The surface area of a right circular cone is generated by rotating a right triangle about one of its legs. The lateral surface area is generated by the hypotenuse and the base by the other side. Pappus' rule: If a plane curve is rotated about a line in its plane such that the curve lies on one side of the line, then the area of the resulting surface of rotation is equal to the product of the length of the generating curve and the length of the path of the center of gravity of the curve. For our problem the center of gravity of both the hypotenuse L and the leg r with respect to leg h is r/2. Therefore,      Ac = (L + r)*2π*(r/2) = πr(L+r)                                         (1) The leg h is called the axis of the cone. The intersection of the cone and any plane containing the axis is an isosceles triangle. Where the length of the two equal sides is L and the base is 2r. The intersection of the sphere with the plane is the incircle of of this triangle. The axis of the triangle bisects the apex angle and the bisector of either of the other two angles intersects the axis at the center of the incircle. Therefore,      (h-a)/a = L/r                                                                      (2) Solving (2) for L gives      L = r(h-a)/a                                                                       (3)  Using this L and the equation L2 = h2 + r2 gives      h = 2ar2/(r2-a2)                                                                 (4) Using this h and equation (3) gives      L = r(r2+a2)/(r2-a2)                                                           (5) Using this L and equation (1) gives      Ac = 2πr4/(r2-a2)                                                              (6) Since the sphere is of fixed radius a, As is also fixed. Therefore,      d(Ac)/dr = [(r2-a2)(8πr3)-(2πr4)(2r)] / [r2-a2]2                   (7) Therefore, the minimum of Ac occurs at r = a√2                      (8) Thus, using equations (4), (5), (6) we get      h = 4a, L = 3a√2, Ac = 8πa2, and Ac/As = 2. QED

 Subject Author Date Solution Brian Smith 2016-12-25 12:33:20 Possible solution broll 2016-12-25 08:56:05

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