Maths prize 29 June 2017: a train on a circular track

A train goes round a circular track at a varying speed, but in a constant pattern (i.e. its speed at each point on the track is the same every time it goes round). Prove there must be at least one pair of diametrically-opposite points on the track at both of which the train has exactly the same speed.

The prize was won by Mohaned al-Bassam, who pointed this out:

For every point x on the track, define

v(x) = speed at x

x* = point diametrically opposite x

f(x) = v(x) − v(x*)

f(x*) = − f(x) by definition, so f(x0) > 0 for some x0 (unless f(x) = 0 for all x)

The function is continuous. Therefore

at some point x1 between x0 and x0*, f(x1) = 0

(same reason as: if you travel in London from South London to North London, at some time you must cross the river). ▇

This is the simplest case of the Borsuk-Ulam theorem, proved by the Polish mathematician Karol Borsuk in the 1930s. The more general case says that on a sphere (e.g. the Earth) there must be at least one pair of diametrically-opposite points which have identical temperature and barometric pressure, that for any three continuous functions defined on a hypersphere (4-dimensional “sphere”) there must be at least one pair of diametrically-opposite points which have identical values of those three functions, and so on.

The general case requires ideas from algebraic topology, a branch of maths often taught as a third-year option in uni maths courses.

Borsuk developed many of his ideas at the Scottish Café in Lviv (a city then in Poland, now in Ukraine), a meeting place for mathematicians where they kept a list of mathematical problems (“the Scottish book”: bit.ly/sc-bk) and prizes were given for solving them. The café is shown below.