, 2011

Denote by $A_1, A_2, \dots, A_{66}$ the points from $T$ and by $a_i$ the number of lines from $P$ containing $A_i$. Then the number of pairs of lines intersecting at $A_i$ equals $\binom{a_i}{2}$, and the number of incident pairs $I = \sum_{i=1}^{66} a_i$. Since any two lines meet in at most one point, we have $\sum_{i=1}^{66} \binom{a_i}{2} \le \binom{16}{2} = 120$. Let $b_k$ be the number of points from $T$ which lie on exactly $k$ lines from $P$. Then $\sum b_k = 66$, $\sum \binom{k}{2} b_k \le 120$ and $I = \sum k b_k \le \sum \frac{1}{2} \left(3 + \binom{k}{2}\right) = \frac{1}{2}(3 \cdot 66 + 120) = 159$, because $3 + \binom{k}{2} \ge 2k$. Equality is attained when $b_k = 0$ for $k \notin \{2, 3\}$, $b_2 = 39$ and $b_3 = 27$ - i.e. when the lines from $P$ determine exactly 39 double and 27 triple intersection points.

An example of a configuration with 159 incident pairs can be constructed using Pappus' Theorem. Take points $A_1, A_2, A_3$ on a line $a$ and $B_1, B_2, B_3$ on line $b \parallel a$, then draw 9 lines $A_jB_j$, $i, j \in \{1, 2, 3\}$. For instance, in the diagram we set $A_1A_2 : A_2A_3 : B_1B_2 : B_2B_3 = 2 : 2 : 3 : 6$, so among these lines no two are parallel and no three concurrent. By Pappus' Theorem, the 98 lines determine 18 intersection points which are collinear in triples - so these determine another 6 lines. Together with these 6 lines, we have 15 lines and 24 triple intersections. Moreover, three lines obtained by Pappus' Theorem meet in a point (denoted by $K$), which gives us 25 triple intersections. Draw one more line through two double intersection points only. The set $P$ of the 16 drawn lines and set $T$ consisting of the 27 triple intersection points and the 39 remaining double intersection points determine 159 incident pairs.