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Eight identical spheres fit inside a cube with side length of 1 unit. What is the largest possible radius for a sphere which would fit between those 8 spheres (in the center)? There are no gaps between spheres as well as between each sphere and the face of the cube.

Sketch of the problem

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  • $\begingroup$ Pretty sure its (sqrt3-1)/2 $\endgroup$
    – Beastly Gerbil
    Commented Mar 4, 2017 at 18:50
  • $\begingroup$ @Beastly You changed your comment after seeing my answer :-P $\endgroup$
    – Rand al'Thor
    Commented Mar 4, 2017 at 18:51
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    $\begingroup$ This is not so much a puzzle as a mathematical question. Vote to close. $\endgroup$
    – wbogacz
    Commented Mar 4, 2017 at 19:23
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    $\begingroup$ @wbogacz I disagree. It's not trivial to solve, and some geometric insight is needed, plus the "aha" which is in the first spoilertag in my answer. $\endgroup$
    – Rand al'Thor
    Commented Mar 4, 2017 at 19:41
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    $\begingroup$ @rand, needing geometric insight doesn't take it out of the realm of standard geometry textbook problem and into the realm of puzzles. $\endgroup$
    – Peter Taylor
    Commented Mar 4, 2017 at 23:08

1 Answer 1

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  • The large cube has side-length $1$.
  • Two touching spheres stretch exactly across the length of the cube, so each of the spheres has diameter $\frac{1}{2}$ and radius $\frac{1}{4}$.

  • The largest possible small cube which can fit between the spheres is such that each of its faces is tangential to one of the spheres.

  • Thus the side-length of the small cube must fit exactly between two diagonally opposite spheres. By symmetry of the spheres, the distance between the centres of these two spheres is exactly half the diagonal of the large cube. So the side-length of the small cube is $\frac{\sqrt{3}}{2}-\frac{1}{4}-\frac{1}{4}=\frac{\sqrt{3}-1}{2}$.

To visualise this, it might help to consider the 3D analogue of the following not-very-good 2D picture:

square within cubes within square

EDIT: after the OP changed from a small cube to a small sphere, my answer remains the same except that instead of the side-length of the small cube, we should consider the diameter of the small sphere; this length should still be $\frac{\sqrt{3}-1}{2}$.

EDIT 2: actually my answer for a small cube was invalid (touching all eight spheres would make it an octahedron, not a cube), but the version with a small sphere is still fine.

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  • $\begingroup$ He's edited it to make it a sphere not a cube in the middle $\endgroup$
    – Beastly Gerbil
    Commented Mar 4, 2017 at 19:06
  • $\begingroup$ @Beastly Yep, just saw, and fixed my answer accordingly. (Fortunately not much had to change.) $\endgroup$
    – Rand al'Thor
    Commented Mar 4, 2017 at 19:06
  • $\begingroup$ The funny thing was I didn't read the question properly and thought it was a sphere from the start, even when I was arguing my point with you. Lucky for me it works for both! :P $\endgroup$
    – Beastly Gerbil
    Commented Mar 4, 2017 at 19:07
  • $\begingroup$ I'd hate to see the cube solution be forgotten. It is more subtle. Please leave it on display, rand al'thor, or create a new puzzle for it. $\endgroup$
    – humn
    Commented Mar 4, 2017 at 19:08
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    $\begingroup$ Ah yes, the famous eight-faced cube. Less flippantly: unless I'm missing something, the polyhedron in the middle with faces tangent to all the spheres is an octahedron, not a cube. This doesn't invalidate the solution to the modified puzzle with a sphere in the middle, of course. $\endgroup$
    – Gareth McCaughan ♦
    Commented Mar 4, 2017 at 21:20

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