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Tuesday, September 29, 2009

Black and white painted sphere

The surface of a sphere is painted with black and white paints such that the area of black paint is less than 1/8 of the area of the sphere (the rest of the sphere is painted with white). Prove that one can inscribe a rectangular box in the sphere such that all eight corners land on white points.

More formally, if each point on the sphere is assigned a color black or white such that there is an injective but not surjective one-to-eight unordered mapping from the black points to the white points, prove that there are eight white points on the sphere that form a rectangular box.

Solution


Let $P_1, P_2, P_3$ be three fixed planes such that they are perpendicular to each other and each divides the sphere in half. For each point $p$, construct an unordered 8-tuple $(p, p_1, p_2, \cdots, p_7)$ consisting of the reflections of $p$ with respect of the three planes, and their respective reflections. This 8-tuple is unordered, meaning that if we take $p_1$ and perform the reflections, we get the 8-tuple $(p_1,p,p_2, \cdots, p_7)$ and we consider these two tuples to be the same.

This construction partitions the surface of the sphere into infinitely many 8-tuples where the points in the same tuple form a rectangular box. Because the area of white paint is greater than 7/8, then there exists at least one tuple where all the points are white. This is our white rectangular box.

Remark: The proof here is very "loose / lax." We chose the three planes arbitrarily, so there was a lot of degrees of freedom. The problem could probably be strengthened by asserting that there is a cube with all white points, but I was unable to find the suitable tuple construction to make sure that they're all disjoint.

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