2019 AMC 10A Problem 14

Four lines form $\binom42 = 6$ pairs, and each pair contributes at most one point, so $N \le 6$. Points are lost when a pair is parallel (no point) or when three or more lines pass through the same point (several pairs share one point). Organize by how many lines are mutually parallel.

• $N = 6$: four lines in general position.
• $N = 5$: two parallel lines plus two crossing lines, meeting away from the parallels ($2 + 2 + 1$).
• $N = 4$: two pairs of parallel lines (a parallelogram), or three concurrent lines plus a fourth line crossing all of them.
• $N = 3$: three parallel lines and one transversal; or three concurrent lines plus a line parallel to one of them.
• $N = 1$: all four lines through one point.
• $N = 0$: all four parallel.

Why not $N = 2$? With two points, every pair of lines is parallel or meets at one of those two points. If some three lines are concurrent at $P$, the fourth line meets at least two of them (it is parallel to at most one), producing at least two more points unless it also passes through $P$, which gives $N = 1$. If no three lines are concurrent, each point accounts for exactly one pair, so only two of the six pairs meet and the other four are parallel; but with four lines, four parallel pairs force all four lines parallel (or three parallel plus one, which gives three points), a contradiction.

Possible values: $0, 1, 3, 4, 5, 6$, with sum $19$.

The answer is $\boxed{\textbf{(D)}\ 19}$.

Start from the maximum $\binom{n}{2}$ and understand the only two ways to lose intersections: parallelism (a pair contributes nothing) and concurrency (several pairs collapse to one point). Sketching each configuration is fast; the real content is proving the one missing value impossible rather than assuming the list is complete.

• Assuming every value from 0 through 6 occurs (sum 21) without checking that 2 intersection points is impossible.
• Missing configurations like three concurrent lines plus a parallel line ($N = 3$) and concluding $3$ is impossible, which gives $16$.