Prove that no more than 8 points can fit in a rectangle with sides d and 2d if any 2 points have to be at least d units away from each other.
I have proved that no more than 6 points can fit in such a rectangle using the fact that between 3 points (not on a straight line) there is a triangle with area at least sqrt(3)/4 d^2 (this is the area of an equilateral triangle with side d).
Does anybody know if that's correct and if there is a more formal way of proving that?
Many thanks
Using triangles does not work quite that simply, because an isosceles triangle with two sides of length $d$, the angle between them $\theta$, has the third side $\ge d$, if $\theta\in[\pi/3,\pi)$ (or between $60$ and $180$ degrees). But if $\theta$ is close to $180$ degrees its area $A=\frac12d^2\sin\theta$ will be close to zero.
Instead you can divide the rectangle into 8 squares with side length $d/2$. If there were more than 8 points by the pigeonhole principle at least two of them will fall into the same square and consequently have distance $\le d/\sqrt2<d$ violating the given rules.