17. (MON 1) Given seven points in the plane, some of them are connected by segments so that:
(i) among any three of the given points, two are connected by a segment;
(i) the number of segments is minimal.
How many segments does a figure satisfying (i) and (ii) contain? Give an example of such a figure.
17. A figure consisting of 9 lines is shown below.

Now we show that 8 lines are not sufficient. Assume the opposite. By the pigeonhole principle, there is a vertex, say $A$, that is joined to at most 2 other vertices. Let $B, C, D, E$ denote the vertices to which $A$ is not joined, and $F, G$ the other two vertices. Then any two vertices of $B, C, D, E$ must be mutually joined for an edge to exist within the triangle these two points form with A. This accounts for 6 segments. Since only two segments remain, among $A, F$, and $G$ at least two are not joined. Taking these two and one of $B, C, D, E$ that is not joined to any of them (it obviously exists), we get a triple of points, no two of which are joined; a contradiction.
Second solution. Since (a) is equivalent to the fact that no three points make a "blank triangle," by Turan's theorem the number of "blank edges" cannot exceed $\left[7^{2} / 4\right]=12$, leaving at least $7 \cdot 6 / 2-12=9$ segments. For general $n$, the answer is $[(n-1) / 2]^{2}$.