Maximize length of side of triangle from points on a circle Given a circle with $n$ points, among all triangles we can make using these points, we want to find a triangle with maximum length of its shortest side in $o(n^2)$.
We try to make a relation between this problem and convex hull but we can't. Any help be appreciated.
You could transform the question into finding the biggest minimum angle of the triangle vertices relative to the centre.
Preparatory work:

Find centre $(x_C, y_C)$, if not already known (intersect bisectors of randomly-chosen pairs of points). $o(1)$

a) Make list of angle position for each point, $a_i = \text{atan2}(y_i-y_C, x_i-x_C)$. $o(n)$
b) Sort angle list. $o(n \log n)$


Main determination:

In succession for each point:
a)  find points closest to $\pm 120^\circ$ points  from the list, $o(\log n)$ per point
b) record max minimum angle with this point, update overall maximum if needed. $o(1)$ per point

Post process

Convert angle back into distance (either directly or by retrieving the indices of the appropriate points). $o(1)$

Explaining #3 a little more: you should find four points from the two binary searches, one either side of the $+120^\circ$ &amp; $-120^\circ$ points. That gives you four triangles to examine for angles (differences), composed of the base point plus choices from left/right of each of the $\pm 120^\circ$ lines.  (If any of the points are the same (few/crowded points) , there will be fewer triangles to check).

For convenience you might add extra copies of the points either side of the wraparound $0^\circ=360^\circ$ to avoid special cases in the search - eg. duplicate the point at (say) $+358^\circ$ to also be seen at $-2^\circ$.