Rainbows in the plane

As usual in Lisbon, I have been working with João Araújo on semigroups. But sometimes research has unusual spin-offs, such as the following curious fact:

Fact If the points of a projective plane are coloured with four colours (all of which are used), there is a set of four points, three on a line and the fourth off it, which get all four colours.

Spoiler alert: The proof is below. You might want to try it yourself first.

This is the only configuration which has this property. Sets of four collinear points are defeated by a colouring which uses three colours on a line and the fourth everywhere else, while sets of four points with no three collinear are defeated by a colouring which has three collinear singleton colour classes and everything else of the fourth colour.

The statement has a kind of anti-Ramsey feel to it, and I thought it might be difficult; but two simple steps get us there.

First, there is a line which sees at least three colours. For choose any line L. If it sees at least three colours, we are done, so suppose it sees only red and blue. Take a green point and a yellow point. The line through these points meets L; the intersection is either red or blue, so use this line instead.

Now let L be a line seeing at least three colours. If L sees exactly three, choose one point of each colour on L and any point of the remaining colour; if L sees all four, choose a point off L, and three points on L of the other three colours.

Is there a more general theory lurking here, or is this just an isolated curiosity?

About Peter Cameron

I count all the things that need to be counted.
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