# Answer to Riddle #86: Frosty The Snowman

86. Frosty the Snowman wants to create a small snowman friend for himself. The new snowman needs a base, torso, and a head, all three of which should be spheres. The torso should be no larger than the base and the head should be no larger than the torso.

For building material, Frosty has a spherical snowball with a 6 inch radius. Since Frosty likes to keep things simple, he also wants the radius of each of the three pieces to be a positive integer. Can Frosty accomplish this?

For building material, Frosty has a spherical snowball with a 6 inch radius. Since Frosty likes to keep things simple, he also wants the radius of each of the three pieces to be a positive integer. Can Frosty accomplish this?

Fairly easy this puzzle, a useful exercise notwithstanding: (this puzzle comes from the NSA)

Before reading the answer can I interest you in a clue?

So the formula for the volume of a sphere is

We can cancel the

Which, being 1 equation with 3 unknowns is not solvable. But there are other things we know, we know they are all integers. We also know

Let's try subtracting 5

Now

Fortunately for frosty

So the formula for the volume of a sphere is

*, lets label the base '*^{4}/_{3}π r^{3}*', the torso '***a***' and the head '***b***.' So we can equate the volume of our 6 inch ball, with the volumes of the other 3.***c**^{4}/_{3}π 6^{3}=^{4}/_{3}πa^{3}+^{4}/_{3}π b^{3}+^{4}/_{3}π c^{3}^{4}/_{3}π 6^{3}=^{4}/_{3}π (a^{3}+ b^{3}+ c^{3})We can cancel the

*which gives us:*^{4}/_{3}π**216 = a**^{3}+ b^{3}+ c^{3}Which, being 1 equation with 3 unknowns is not solvable. But there are other things we know, we know they are all integers. We also know

*, it also stands to reason that***a ≥ b ≥ c***. At this point the puzzle is almost solved, but for the sake of completeness we will continue... It seems worth trying***6 ≥ a ≥ b ≥ c***there are a finite number of permutations that satisfy our criteria.***a = 5***could be 5, 4 & 3 (and, in fact are,) or 4, 3 & 2, or 5, 4 & 2 but not much else.***a, b & c**Let's try subtracting 5

^{3}from each side:**91 = b**^{3}+ c^{3}Now

**b = 4****27 = c**^{3}Fortunately for frosty

**216 = 5**^{3}+ 4^{3}+ 3^{3}© Nigel Coldwell 2004 - – The

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