8. For a square field with side length 1 (including all four sides), if 3 points are placed arbitrarily, the minimum value of $a$ such that the distance between at least two points does not exceed $a$ is


Solve: According to the problem, the minimum value of $a$ is actually the maximum of the shortest sides of triangles within a unit square. For this, we only need to examine the maximum triangle with the shortest side in the square $ABCD$.

Assume $\triangle AEF$ is the triangle that meets the requirement, with $A$ being one of the vertices of the square. It is evident that $E$ and $F$ should be on the sides $BC$ and $CD$ of the square. To maximize the shortest side, we only need to examine the equilateral triangle.

Let $AB=1, \angle BAE=15^{\circ}$, it is easy to calculate $AE=\sqrt{6}-\sqrt{2}$.