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Spherplex (Posted on 2019-11-25) Difficulty: 3 of 5
A sphere is reconstituted into n identical smaller spheres, keeping the total volume constant.

Let a be the ratio of the total surface area of the smaller spheres to the surface area of the original sphere.

Let b be the ratio of the radius of the original sphere to the radius of each of the smaller spheres.

Let c be the ratio of the volume of the original sphere to the volume of each of the smaller spheres.

What is the minimum integer value of a+b+c?

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution solution | Comment 2 of 3 |
The volume of one of the smaller spheres will be 1/n that of the larger so the surface area will be 1/n^(2/3). There will be n of them so the total surface area will be n^(1/3), which is then called a.

The ratio of the radius of the larger to the smaller will be n^(1/3), now called b.

The volume ratio is of course n, now referred to as c.

a+b+c = n + 2 * n^(1/3)

The smallest value will be when n is the smallest possible, which is 2, in which case n = 2 + 2*cuberoot(2) ~=  4.519842099789747.

But wait, the smallest integer value is wanted. We need a perfect cube for n. The smallest cube is 8:

When n = 8,

a+b+c = 8 + 2 * 2 = 12.

The answer is 12.

Note that in a, the ratio is the area of the smaller ones to the larger, rather than the other way around, and the ratio is for all of them summed, rather than for just one.

  Posted by Charlie on 2019-11-25 10:37:42
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