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A cube is inscribed in a sphere if all its vertices lie on the sphere's surface (Figure 1).

Figure 1. A cube inscribed in a sphere

If we colour a sphere's surface so 90% of it is red, then amongst the population of cubes it is possible to inscribe in different orientations, there must be at least one cube with 8 red vertices.

We can prove this rather surprising fact by a clever statistical argument.

From the given information we know that a randomly chosen point on the sphere is coloured red with probability.

Now let's introduce a so-called indicator variable which is 1 if a point is red and 0 otherwise. The mean average of this across all points must be .

There is a very useful principle called Linearity of Expectation, which says that given the average value of a number of variables, we can find the average of their combined value simply by adding together the individual averages and this will be true even if they are not independent of one another. 

So we can add together the expected averages for the 8 vertices to get as the expected value for the whole cube, and this is the average number of red vertices across the whole population of inscribed cubes.

But if 7.2 is the average, some cubes must have a higher value and because the number of red vertices for any particular cube is a whole number, this means some cubes must have 8 red vertices.

This just proves such cubes must exist. It doesn't say where they are or how we might find them. In particular cases it may be possible but it's conceivable no general procedure exists.

See the Comments for notes on the origins of this inscribed cube problem.

We can apply the same ideas to show that if more than 75% of a circle is coloured red, there must be a square with 4 red cornes. What about 4D and higher dimensions? Leave your answer in the comments.