A glass Christmas tree decoration is made from three different types of sphere. Seven small spheres, eight middle-sized spheres and one large sphere.

Image credit: Phil Chaffé, MEI

Spheres of each type are identical. The eight middle sized spheres are arranged around one of the small spheres. The small sphere touches each of the spheres around it and each middle-sized sphere touches three other middle-sized spheres. The remaining six small spheres are arranged so that they each touch four middle sized spheres. The large sphere surrounds all of the other spheres touching every other sphere except for the small one in the centre. A cross-section of the decoration on this plane baubes-additional

looks like this:

Assuming the glass to be of negligible thickness, find the ratio of the radii of the three sizes of sphere. What fraction of the volume of the large sphere is occupied by the small (not middle-sized) spheres?

Reveal Solution
Drawing a cube connecting the centres of the middle-sized squares provides a good starting point:baubles cube

Using the cross-section shown in the problem:

baubles solution cross section

The radius of the large outside sphere will be the sum of the distance from the centre to a vertex of the cube and the radius of a middle-sized sphere.

The radius of the central small sphere (and therefore all of the other small spheres) will be the difference between the same two distances.

Let the radius of the middle-sized spheres be r. The cube would then have side lengths of 2r.

The distance from the centre of the cube to one of the vertices is therefore \frac{1}{2} \sqrt{(2r)^2+(2r)^2+(2r)^2} which simplifies to \sqrt3r.

The radius of the large outside sphere is therefore \sqrt3r + r = r (\sqrt3 +1).

The radius of a small sphere is \sqrt3r - r = r (\sqrt3 -1).

The ratio of the radii is r(\sqrt3 -1):r:r(\sqrt3 +1) which simplifies to (\sqrt3-1):1:(\sqrt3+1).

The scale factor of enlargement from the small sphere’s radius to the large sphere’s radius is \frac{\sqrt3 +1}{\sqrt3-1} which simplifies by rationalizing the denominator to 2+\sqrt3.

The scale factor for the volume is therefore {(2+\sqrt3)}^3.

There are 7 small spheres so the fraction of the volume they occupy is \frac{7}{{(2+\sqrt3)}^3}.

If you wish to rationalize the denominator here, expanding the denominator gives \frac{7}{26+15\sqrt3} which simplifies to 7(26-15\sqrt3).

Problem Page Coordinator: Claire Baldwin – Mathematics in Education and Industry
Acknowledgement: The IMA are indebted to MEI for sourcing and supplying Mathematics Today with these well-known puzzles.
First published in Mathematics Today (December 2017)
Image credit: baubles by Judy van der Velden / Flickr / CC BY-NC-ND 2.0

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