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Posers and Puzzles

Posers and Puzzles

  1. Standard member wolfgang59
    Infidel
    12 Mar '15 05:16
    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"?
  2. 12 Mar '15 18:07
    Originally posted by wolfgang59
    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"?
    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 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.
  3. Subscriber joe shmo On Vacation
    Strange Egg
    13 Mar '15 01:34
    Originally posted by Shallow Blue
    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.
    Yep... I get 3Y

    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
  4. Standard member wolfgang59
    Infidel
    13 Mar '15 08:56
    Yep.
    Easy answer is 3.

    But convoluted proof.