**From New Scientist #1362, 16th June 1983** [link]

As a challenging geometry puzzle, I asked my son to mark a prescribed number of points on a piece of paper, no three of them being in a straight line, and then to join each of the points to each of the others by straight lines. I knew by my choice of the number of points that he would not be able to do this without at least two of these lines crossing.

But I asked him to do it with some of his lines drawn in red, and the rest drawn in blue, and in such a way that it would be impossible to find a red triangle or a blue triangle in the whole configuration. This he managed.

How many points had I asked him to draw?

**Note:** I am waiting for a phone line to be connected at my new house, so I only have sporadic access to the internet at the moment.

[enigma216]

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Any three non-collinear points can be joined with non-crossing lines, which form the edges of a triangle. A fourth point can be added to give a lattice in the shape of a triangle with an additional interior point, so it is possible to have 4 points that are joined with non-crossing lines. But it is not possible to add a fifth point without requiring that the lines joining them must cross. So there must be at least 5 points.

This Python program considers increasing sets of points, starting from 5, to find a colouring of some of the edges such that no triangles formed from the points have all sides the same colour. There is a solution for 5 points, and it finds the first solution in 32ms.

Solution:He was asked to draw 5 points.There are several colourings of the edges that work, but they basically boil down to the following two arrangements:

Here’s a recursive version of my program that runs under Python 3.

I guess I misread the enigma. I got the same answer but looking at your illustrations in both cases there are red triangles present and in the first illustration there is a blue and red triangle. I interpreted the enigma to mean you could not find red or blue triangles in the configuration.

I agree that the wording on this problem could be clearer. My program only looks for solutions using triangles with vertices at the selected points (the black blobs in the diagram). But if you want to look for solutions where the vertices of triangles may be formed by the intersections of the lines, then obviously there are more triangles to consider, but each line between selected points can only be of one colour, so any solution to the second problem must also be a solution to the first.

By considering the following equivalence of graphs, the solution I found also gives a solution to this second problem:

There is a single additional crossing point X (the intersection of AC and BD). This adds 4 new triangles: XAB, XAD, XBC, XCD to be considered.

Either way both puzzles can be solved by selecting 5 points.

Ramsey’s Theoremprovides a proof that 6 or more points must contain a triangle of one colour.