\section*{Exercise 5 - 291045}

Determine a distribution of five different points on the area of an equilateral triangle (including its boundary) such that the smallest distance between any two of these points is maximized!
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(I) The connecting lines of the midpoints of the 3 sides of the given equilateral triangle with side length \(a\) divide it into four congruent sub-triangles with side length \(\frac{a}{2}\).

Therefore, in any distribution of 5 points, according to the Dirichlet drawer principle, there is a sub-triangle in which - including the boundary - at least two of the 5 points lie.

These two have a distance that is not greater than the side length, i.e., not greater than \(\frac{a}{2}\). Consequently, in no distribution of 5 points can the smallest distance between 2 of them be greater than \(\frac{a}{2}\).

(II) We place 2 of the 5 points at the vertices and the remaining 3 at the midpoints of the sides of the given triangle. Then the smallest distance between 2 of them is equal to \(\frac{a}{2}\).

Due to (I), this is the desired distribution.