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?

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## About Peter Cameron

I count all the things that need to be counted.