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# CPhill's Challenge - Sphere Packing Problem

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I can't find a picture of this but maybe I can explain it........

We have a cube that is 1 inch  on  each side......inside the cube are nine identical spheres......one of the spheres occupies the exact center of the cube and the other eight spheres are positioned at the eight corner "vertexes" of the cube such that  each one touches three sides of the cube and the central sphere but none  touches any of the other seven "corner" spheres...

The question is......what is the radius of any one of these identical spheres......????

P.S. - I get an answer of ≈  .232 in  .....does anyone know if this is correct, or not ???

Thanks for playing  !!!!!!   Nov 30, 2016

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Trippy!! Dec 2, 2016

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So the diagonal length is from an upper vertex to one lower and on the opposite side.

Diagonal Length is sqrt(3).

3 spheres touch across this length. Each sphere has a diameter of 2r. Making a total of 6r to get from that vertex to the other.

Diagonal length = 6r

sqrt(3) = 6r

sqrt(3) /6 =r

sqrt(3)/6 = 0.2886751345948129

Dec 1, 2016
edited by JonathanB  Dec 1, 2016
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The radius can't be even as great as 0.25 inches......if it were, the total width of any two spheres next to each other would be greater than 1 inch   [ they could not fit in the cube next to each other ]   Dec 1, 2016
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Trippy!! JonathanB  Dec 2, 2016
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Thanks Chris, I always like it when these sorts of problems are uploaded.

There is usually at least one person, like Jonatho,n that takes an obvious interest in these puzzles.

Thanks Jonathon But for every person who displays interest there are bound to be a few more (like me) that think about it more quietly. :))

Dec 2, 2016
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I just now actually found this problem  online :

http://artofproblemsolving.com/wiki/index.php/1990_AHSME_Problems/Problem_25

The evaluation of the correct answer, "B,"    agrees  with mine  :

https://www.wolframalpha.com/input/?i=%5B2sqrt(3)+-+3%5D+%2F+2   Dec 2, 2016