Can you show how to fit nine points, whose pairwise distances apart are all greater than 1, in an equilateral triangle with side length 3 or can you prove that it isn't possible to do so?
It is a well-known problem to show, using the pigeonhole principle, that there cannot be five points in an equilateral triangle of side length $2$ whose pairwise distances apart are greater than $1$, and that there cannnot be ten points in an equilateral triangle of side length $3$ whose pairwise distances apart are greater than $1$.
In the case of side length $2$, dividing the triangle into four equilateral triangles of side length $1$ and placing points at every vertex of these triangles allows six points to be placed whose pairwise distances apart are no more than $1$, and it is easy, by placing points at the three vertices and centre of the original triangle, to fit four points whose pairwise distances apart are greater than $1$.
However, in the case of side length $3$, while dividing the triangle into nine equilateral triangles of side length $1$ and placing points at every vertex of these triangles allows ten points to be placed whose pairwise distances apart are no more than $1$, I cannot see how to fit nine points whose pairwise distances apart are greater than $1$ (eight seems quite straightforward). Is this possible, or can you prove that it isn't?
I tried taking the ten-point pattern, removing a point and then shifting the remaining nine slightly to increase the distances between them, but there appears to be a residual rigidity in the structure with one point removed that prevents you doing this. Of course, there might be a way of starting from scratch and achieving the desired outcome with a pattern which is not a perturbation of the ten-point one.
Best Answer
It is impossible. Split your side-length-three triangle into 9 side-length-one triangles. We know each triangle can have at most one point. Consider then the center hexagon. We have to fit six points in it. Try instead to fit these 6 points in the unit disc containing this hexagon. This was proven impossible in the reference for 6 on the Wikipedia page for circle packing in a circle. (Note that the circle-packing-in-a-circle problem is the same as this problem, by considering packing circles of radius 1/2 inside a disc of radius less than 3/2, rather than packing points with distance at least 1 inside a disc of radius less than 1, so this Wikipedia page alone contains enough information to prove that the construction is impossible).