4 points inside a triangle Using pigeonhole principle, it's possible to show that for any $n^2+1$ points in an equilateral triangle of side length $1$, there exists two points with distance at most $\frac{1}{n}$. How do I select $n^2$ points such that the distance between any two points have a distance of more than $\frac{1}{n}$?
 A: Bit of a trick question. Got it wrong the first time. Then drew some pictures.
This is impossible for large $n,$ although easy enough for $n=2.$ For each point, draw a circle of radius $1/(2n)$ around it. No other circle can overlap this area, which is $\pi/(4 n^2).$ As we have $n^2$ points, we have a total area of at least $\pi / 4 \approx 0.785.$ The majority of this area is completely contained inside the triangle. This poses a problem, as the triangle of edge length $1$ has area $\sqrt 3 / 4 \approx 0.433$
A careful proof would take into account the amount of area that can be outside the big triangle. This is, essentially, the area of $3n$ semicircles, or roughly $3 \pi / (8 n) \approx 1.178/n.$ We should also add nearly three entire circles, for the three vertices.
The task is impossible once $$  \frac{\pi}{4} - \frac{3 \pi}{8n} - \frac{3 \pi}{4 n^2} > \frac{\sqrt 3}{4} $$
This can be made more accurate using the area of the regular hexagon that circumscribes each circle. 
Added: this problem can be described as an ordinary circle packing problem if we take the (slightly larger) equilateral triangle with edges moved a distance of $1/(2n)$ outwards.  The way it works out, a reasonable problem is to try to place $n^2$ points in a regular hexagon of exactly double the are of the triangle. this means the edges are $1/\sqrt 3.$ See diagrams


Alright, gave it a try. With $n=10,$ I can get more than $100$ points at pairwise distance exactly $1/10.$ However, in order to have a little remaining room so as to expand slightly and get the points more than $1/10$ apart, i can fit only $97$ points. 

