[ Extreme properties (continued).]

If five points are given on a plane, then by considering all possible triples of these points, one can form 30 angles. Let the smallest of these angles be $\alpha$. Find the maximum value of $\alpha$.

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First, assume that the points are the vertices of a convex pentagon. The sum of the angles of a pentagon is $540^{\circ}$, so one of its angles does not exceed $540^{\circ} / 5=108^{\circ}$. The diagonals divide this angle into three angles, so one of them does not exceed $108^{\circ} / 3=36^{\circ}$. In this case, $\alpha \leq 36^{\circ}$.

If the points are not the vertices of a convex pentagon, then one of them lies inside the triangle formed by the other three. One of the angles of this triangle does not exceed $60^{\circ}$. The segment connecting the corresponding vertex to the internal point divides this angle into two angles, so one of them does not exceed $30^{\circ}$. In this case, $\alpha \leq 30^{\circ}$. In all cases, $\alpha \leq 36^{\circ}$. It is clear that for a regular pentagon, $\alpha$ $=36^{\circ}$.