I am unknown to 3-D Geometry. Can I use " a triangle a vertex at the centre of a circle and the opposite side (with a given fixed length) being tangential to the circle" as a 2-D analogue of this problem and try to minimize the area of the triangle external to the circle(in which case, the point of contact must be the centre of the base of the triangle...I think).

This analogue if retraced into a 3-D problem would involve a triangular prism (and not a pyramid) and a cylinder(and not a sphere) with the cylinder touching the base surface of the prism in a line and one ege of the prism coinciding the axis of the cylinder(that axis which does not intersect the curved surface of the cylinder). Any analytical solution(not involving rigors of solving a differential equation)???

I hope i am clear in conveying my ideas. This wild fragment of imagination has just popped up in my head out of nowhere. But nevertheless coaxes me to think. plz help.

>I am unknown to 3-D Geometry. Can I use " a triangle a
>vertex at the centre of a circle and the opposite side (with
>a given fixed length) being tangential to the circle" as a
>2-D analogue of this problem and try to minimize the area of
>the triangle external to the circle(in which case, the point
>of contact must be the centre of the base of the
>triangle...I think).
>
This idea should also work in 3-D, but the calculation is a little more involved. First, I must assume that you are not varying the radius of the sphere, only the position of the point where it is tangent to your square. In that case, the volume of the oblique pyramid is constant, since its base is the area of the square and its altitude is the radius of the sphere. (Volume = 1/3*BaseArea*Height for a pyramid.)

Therefore, to minimize the volume outside the sphere, you should maximize the volume inside the sphere. This is an easier shape to deal with. For any cone with its vertex at the center of the sphere, the volume of the part of the cone that is inside the sphere is (the solid angle) * (the radius)^2, i.e. WR^2. W is the solid angle (usually called omega, but I can't get omega with this keyboard).

There are forumlas for the solid angle, but they are messy, and there is a better way. Think about placing your eye at the center of the sphere, and looking at the square. Since light travels in straight lines that cross at the pupil of your eye, the image of the square on your retina has the same solid angle as the pyramid. This angle will be largest exactly when the square appears largest to you (since the apparent size of an object is essentially the same thing as the amout of room it takes up on your retina). For a flat object, it will appear largest when two things are true: you are looking at it along a line perpendicular to its plane, and you are as close as possible to its center. These both happen simultaneously when the point of tangency is at the cente of the square.

Therefore, this is the point where the square will look largest, so it will have the biggest solid angle, so the volume inside the sphere will be maximal, so the volume outside the sphere will be minimal.

Hope this helps.

--Stuart Anderson