A bored pirate...

Geometry Level pending

"All hands on deck ye scurvy dogs!!!" yells the captain of a sea faring pirate ship.

One of the crew, Niccolò Fontana Tartaglia, thinks, "Why, oh why, did I sign up to be a deck hand on this wretched pirate ship?"

He looks down at a pile of four cannonballs forming a pyramid and his mind starts to wander...

He soon decides it would be much more fun to calculate the radius of the largest sphere he can squeeze in the small space in the middle of these cannonballs.

Without the aid of a calculator, he comes up with the right answer...

Assuming the cannonballs have unit radius, what answer did he arrive at?

Image credit: http://serioussam.wikia.com


The answer is 0.225.

Geoff Pilling by Geoff Pilling , 53, USA

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1 solution

Geoff Pilling
Apr 7, 2017

The centers of the balls form the vertices of a regular tetrahedron with side length of two.

In a suitably chosen coordinate system, these coordinates could be defined as follows:

  • 1 2 ( 1 , 1 , 1 ) \frac{1}{\sqrt2} (1,1,1)
  • 1 2 ( − 1 , − 1 , 1 ) \frac{1}{\sqrt2} (-1,-1,1)
  • 1 2 ( − 1 , 1 , − 1 ) \frac{1}{\sqrt2} (-1,1,-1)
  • 1 2 ( 1 , − 1 , − 1 ) \frac{1}{\sqrt2} (1,-1,-1)

And the tetrahedron will be centered at the origin, so the distance from any vertex to the center of the tetrahedron will be given by:

d = 1 2 + 1 2 + 1 2 = 3 2 d = \sqrt{\frac{1}{2} + \frac{1}{2} + \frac{1}{2}} = \sqrt{\frac{3}{2}}

And, since the radius of each ball is one, then the distance from the center to any of the four balls (the radius of the small ball one would put in the middle) is given by:

r = 3 2 − 1 = 0.225 r = \sqrt{\frac{3}{2}} - 1 = \boxed{0.225}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Funny how your problem and this one showed up at almost the same time. Great minds .... Anyway, I solved Shubham's problem first, which made this one much easier. :)

Brian Charlesworth - 4 years, 2 months ago

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Hahaha.... Tetrahedron overload! :-)

Geoff Pilling - 4 years, 2 months ago

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