Puzzlet 120 was originally about a sphere in a cube. Colin Hughes has pointed that my solution had an error in it, and it could be solved by hand anyway, so I've replaced it with the interesting problem you can see in the archives about primes. The old Puzzlet 120 will re-appear at some future date redesigned to need a computer solution. This was the original:colin@mathschallenge.net

"A hollow sphere's material thickness is t mm. Within the hollow space inside the sphere is a cube of side 1 metre. The cube is an exact fit so that each of its eight corners just touches the sphere's inside surface. Both objects are made from the same material. If the ratio of the sphere's material volume to the cube's material volume is exactly 4, what is the sphere's thickness t to within one-tenth of a millimeter? Take pi to be 3.14159."

Contribution for Puzzlet #120

Hi Dave,

In looking at your solution to Puzzlet 120, and maybe I am misunderstanding the problem, but I wonder if you have made a mistake …

Using the Pythagorean Theorem, the radius of the internal sphere, r=sqrt(3(1/2)^3)=sqrt(3)/2, so the volume, v1=(4/3)pi*3sqrt(3)/8.

Let R be radius of external sphere, so v2=(4/3)pi*R^3.

Volume of material, V=v2-v1=(4/3)pi(R^3-3sqrt(3)/8), and as volume of cube is 1 and ratio of volumes is 4, V=4. Solving this we get:

(4/3)pi(R^3-3sqrt(3)/8)=4

R^3-3sqrt(3)/8=3/pi

R^3=3(sqrt(3)/8+1/pi)

Which leads to R, and as thickness, t=R-r, we get t=cbrt(3(sqrt(3)/8+1/pi))-sqrt(3)/2~=0.305m=305mm.

Colin

Colin's solution is correct. I originally published this Puzzlet in 1997, and the solution was correct there. But I still missed the fact that it could be solved by hand. Thanks, Colin, for pointing this out.

Dave.

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