2019 AMC 10A Problem 14

Most points possible. Every two lines cross at most once. Four lines make $6$ pairs (with lines $a, b, c, d$: $ab, ac, ad, bc, bd, cd$), so there are at most $6$ crossing points.

There are only two ways to lose points:
- two lines are parallel, so that pair never meets;
- three or more lines pass through the same point, so several pairs share one point.

Which counts can happen? Sketch each one.

1. $N = 6$: four lines in general position, no two parallel and no three through one point.
2. $N = 5$: two parallel lines, plus two other lines that cross each other away from them. Each of the two other lines crosses both parallels ($4$ points), and they cross each other ($1$ more).
3. $N = 4$: two pairs of parallel lines, like a tic-tac-toe board with only $2 + 2$ lines.
4. $N = 3$: three parallel lines and one line crossing all of them.
5. $N = 1$: all four lines through one point.
6. $N = 0$: all four lines parallel.

Why $N = 2$ is impossible. Suppose there are exactly two crossing points.

1. First, suppose three of the lines pass through one point $P$. The fourth line is parallel to at most one of those three, so it crosses at least two of them. If it misses $P$, those crossings are two new points, making at least $3$ in all. If it goes through $P$, all four lines meet at $P$ and $N = 1$. Either way, not $2$.
2. Now suppose no three lines share a point. Then each crossing point comes from just one pair, so only $2$ pairs cross and the other $4$ pairs are parallel. Look at how parallel lines can group: all four parallel gives $6$ parallel pairs; three parallel and one other gives $3$; two groups of two gives $2$; one parallel pair gives $1$. None of these gives exactly $4$. So this cannot happen either.

The possible values are $0, 1, 3, 4, 5, 6$, and their sum is $0 + 1 + 3 + 4 + 5 + 6 = 19$.

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

Start from the most points possible and ask exactly how points get lost: parallel pairs or several lines through one point. Drawing examples for the values that work is quick. The real work is showing that the one missing value, $2$, can never happen, instead of assuming your list is complete.

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 $\boxed{\textbf{(D)}\ 19}$.