Two Spheres in a Cube - Answer
The largest space for two spheres to fit inside of a cube will be what is called the “space
diagonal” of the cube. This is the line that starts from one corner of the cube, goes through the
center, and to the far corner of the cube.
Since any point on the space diagonal is equidistant from the three edges that run from that
corner of the cube, any sphere which lies in that corner will have its center on the space
diagonal.
After a little thought about the situation, we know that the centers of both spheres and the point
of tangency between the spheres will all three lie directly on the space diagonal. Therefore, to
solve the puzzle, we need to calculate the length of the space diagonal, the diameters of the
spheres, and the two lengths of those portions of the space diagonal that will not be inside a
sphere.
To find the length of the space diagonal we first break one of the sides of the cube into two right
triangles. By applying the Pythagorean theorem, we find the hypotenuse (or the face diagonal)
to be √2. Next we find the hypotenuse of a right triangle with a leg of √2 and a leg of 1 (see the
upright red triangle in the image below). We find that the space diagonal of our cube is:
√3
We let D equal the diameter of the large sphere, d equal the diameter of the small sphere, x
equal that portion of the space diagonal between the largest sphere and the closest corner, and
we let y equal that portion of the space diagonal between the smallest sphere and its closest
corner. So:
0 = √3 − 𝐷 − 𝑑 − 𝑥 − 𝑦
© 2011, Leon Hostetler, www.leonhostetler.com
Two Spheres in a Cube - Answer
In order to find x we pretend that we have a two-dimensional problem. Since the center of a
sphere in the corner of a cube is equidistant from all three edges going away from that corner,
the whole thing can be viewed as a circle in the corner of a square.
We do this by simply imagining a circle of radius ½ meter which fills the square. If x is that
portion of the diagonal which lies outside of the circle, then we know it equals ½ the hypotenuse
minus the radius.
Next we find that x is:
We need to put this in terms of D:
𝑥=
√8𝑟 2
−𝑟
2
�8(𝐷 )^2 𝐷
2
𝑥=
−
2
2
and:
�8(𝑑)^2 𝑑
2
𝑦=
−
2
2
Now we can put these into our original equation:
2
𝑑 2
�8 �𝐷 �
�
8
�
𝐷
𝑑
2
2�
0 = √3 − 𝐷 − 𝑑 − ⎛
− ⎞−⎛
− ⎞
2
2
2
2
⎝
⎠ ⎝
⎠
We know that 𝐷 = 2𝑑, so:
© 2011, Leon Hostetler, www.leonhostetler.com
Two Spheres in a Cube - Answer
Which simplifies to:
2
2
�8 �2𝑑�
�8 �𝑑�
2𝑑
𝑑
2
2
⎞−⎛
0 = √3 − 2𝑑 − 𝑑 − ⎛
−
− ⎞
2
2
2
2
⎝
⎠ ⎝
⎠
Which means:
So our answer is 0.95659 meter
𝑑=
𝐷=
© 2011, Leon Hostetler, www.leonhostetler.com
2√3
3 + 3√2
4√3
3 + 3√2