Hex bound

Lemma: We first show that in an equilateral triangle of side length $20$ , given any 2 points $P, Q$ in the triangle, then $|PQ| \leq 20$ .

Proof: WLOG, we may assume that the points $P, Q$ lie on the perimeter of the triangle. If they lie on the same side, then clearly $|PQ| \leq 20$ . If they lie on different sides, let $A$ be the common vertex of these sides. Consider triangle $APQ$ . Since $\angle PAQ = 60^\circ$ , one of the other angles (WLOG, $APQ$ ) must be greater than or equal to $\frac {180^\circ - 60^\circ} {2} = 60 ^ \circ$ . As such, we have $|PQ| \leq |AQ| \leq 20$ . $_\square$

Now, joining opposite vertices of the hexagon breaks it into $6$ equilateral triangles of side length $20$ . We apply the pigeonhole principle by letting the number of points (7) be pigeons and the number of triangles (6) be pigeonholes. Thus, we have that one of the triangles must contain at least $2$ points. By the above lemma, they can be at most $20$ apart. Hence the maximum possible value of $m$ is $20$ . This can be achieved by placing 1 point in the center, and 1 point on each vertex.