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A sphere and a frustum

A sphere and a conical frustum have the same volume, the frustum has a base radius which is twice the radius of its top. The sphere has a radius which is equal to the base radius of the frustrum. What is the ratio of the diameter of sphere to the height of the frustum?

Recently the podcast “Wrong, but Useful“, (@wrongbutuseful) celebrated its 1 year anniversary. The hosts Dave (@reflectivemaths) and Colin (@icecolbeveridge) asked listeners to send in messages to be played on the anniversary show.

I thoroughly enjoy the podcast, and particularly the part where the hosts set puzzles for the listeners to have a go at. With that in mind, I decided to set a puzzle for them as part of my messages.

I know a number of puzzles, but I didn’t want this to be a problem either of them had heard before, so I set about devising one. I thought about the problems I like to discuss, and remembered that I’ve always liked the algebra based geometry problems that occur at the back of GCSE papers, you know the ones: “Three tennis balls radius r fill a cylinder… What’s proportion of the cylinder is empty space?” etc.

I thought a problem like this would be nice. I thought about it and decided I wanted to use a sphere and a frustum with the same volume. I decided that the frustums base radius should be twice the radius if the top and that I would ask them to find the ratio of diameter of the sphere to height of the frustum. I needed to think about the radius of the sphere. At first I thought it might be nice to have the sphere have the same radius as the top face of the frustum, but I reckoned that wouldn’t give a very satisfying answer as it would have hardly any height to speak of. So I tested (on a magna doodle no less) to see what it would be if I used the base radius as the same as the sphere:

image

Volume of a sphere: (4(pi)(r^3))/3

Volume of a frustum: ((pi)(h)(r^2 + rR + R^2))/3

Sub r=r, R=r/2

Equate equations

Cancel pi/3 from each side

4r^3 = (7/8)(h)(r^2)

16r = 7h

8d = 7h

Ratio d:r is 7:8

What a nice, simple, solution we get from quite a bit if algebra. I thought that as it fell out so nicely I would set it!

I realised that for Dave, Colin their listeners and the majority of people I posed this question to would not really have had too much of a problem solving it, but hopefully will have enjoyed it. I did think, however, that it would be an interesting problem to pose my yr11s (and my sixth formers). They have all the skills and knowledge required, so it will be interesting to see how they get on. I will pose it to them after the holidays!

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