12 Mar 15
Consider a cube "C" with surface area X.
What is the surface area of the smallest cube which can
contain the smallest sphere which contains "C"?
Similarly, consider the sphere "S" with surface area Y.
What is the surface are of the smallest sphere which can
contain the smallest smallest cube which contains "S"?
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12 Mar 15
Originally posted by wolfgang59Surface area X, therefore surface of one face is X/6, edge is sqrt(X/6), diagonal is sqrt(3)*sqrt(X/6) or sqrt(3*X/6), or sqrt(X/2). The smallest sphere which contains that cube has a diameter as long as that diagonal, therefore a surface area of 4pi r^2 i.e. 4pi (sqrt(X/2)/2)^2, or 4 pi X/2/4, or pi/2 X, which is just over 1½X.
Consider a cube "C" with surface area X.
What is the surface area of the smallest cube which can
contain the smallest sphere which contains "C"?
Similarly, consider the sphere "S" with surface area Y.
What is the surface are of the smallest sphere which can
contain the smallest smallest cube which contains "S"?
The smallest cube which contains that sphere which contains the original cube has the sphere's diameter, i.e. the original cube/s diagonal, as its edge length, therefore a surface area of 6*X/2 or 3X.
I may have dropped a figure somewhere, but I don't think so. Someone else can do the sphere-in-cube-in-sphere one, but my intuition says it has to be 3Y as well.
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13 Mar 15
Originally posted by Shallow BlueYep... I get 3Y
Surface area X, therefore surface of one face is X/6, edge is sqrt(X/6), diagonal is sqrt(3)*sqrt(X/6) or sqrt(3*X/6), or sqrt(X/2). The smallest sphere which contains that cube has a diameter as long as that diagonal, therefore a surface area of 4pi r^2 i.e. 4pi (sqrt(X/2)/2)^2, or 4 pi X/2/4, or pi/2 X, which is just over 1½X.
The smallest [i]cube[/i ...[text shortened]... one else can do the sphere-in-cube-in-sphere one, but my intuition says it has to be 3Y as well.
Let d = diameter of "S"
pi*d² = Y
The cube of minimum diameter enclosing "S" will have side length "d"
The sphere enclosing the cube that encloses sphere "S" will have diameter "Z" from:
Z² = 3*d²
Z² = 3*Y/pi
The surface area will be:
= pi*Z²
= 3*Y