I am looking into a question of how many points can be put on a plane that pairwise distances between them are as close to a constant as possible. As the first step, it was rather easy to figure out that only 3 points can be put on a plane so that all pairwise distances between them are equal. As the next step, I think about 4 points.


Formally, if there are four distinct points $A\_1(x\_1, y\_1)$, $A\_2(x\_2, y\_2)$, $A\_3(x\_3, y\_3)$, $A\_4(x\_4, y\_4)$, and we define pairwise distances $d\_{ij} = d(A\_i, A\_j) = \sqrt{(x\_i - x\_j)^2 + (y\_i - y\_j)^2}$ and $d\_\text{min} = \min\_{i \neq j} d\_{ij}$, $d\_\text{max} = \max\_{i \neq j} d\_{ij}$, what is the minimum ratio between the largest and smallest distances:
$$
\min\_{\text{distinct } A\_1, \dotsc, A\_4} \frac{d\_\text{max}}{d\_\text{min}} = ?
$$


At the moment, I cannot even figure out if it is in fact a minimum or infimum (i.e., this ratio can get as close to 1 as you want).


Geometry is not really my area, so perhaps it is even some known result...



4 points on a plane with (almost) equal pairwise distances
It is $\sqrt{2}$. For 4 vertices of a square, you get this value. For proving that it is always not less than $\sqrt{2}$, note that one of angles $\angle A\_iA\_jA\_k$ is not less than $\pi/2$ (if $A\_1A\_2A\_3A\_4$ is a convex quadrilateral, the sum of angles equals $2\pi$, thus one of them is at least $\pi/2$; if $A\_4$ lies in a a triangle $A\_1A\_2A\_3$, the sum of angles $\angle A\_iA\_4A\_j$, $1\leqslant i<j\leqslant 3$, equals $2\pi$, thus one of them is at least $2\pi/3>\pi/2$). Now $A\_iA\_k^2\geqslant A\_iA\_j^2+A\_jA\_k^2$, hence $\max (d\_{ik}/d\_{ij},d\_{ik}/d\_{jk})\geqslant \sqrt{2}$.


