AMC 10 · 2019 · #8
Grade 8 countingPick an answer.
Tool #1 (Draw): sketch sample configurations. Tool #2 (Systematic List): list candidate values N = 0, 1, 2, 3, 4, 5, 6 (the maximum is C(4, 2) = 6 pair-intersections) and decide which can be built. Tool #9 (Easier): for each candidate, build a small concrete picture — all-parallel, pencil through one point, two parallel pairs, etc. Tool #3 (Eliminate): rule out N = 2 by a short impossibility argument, then sum the survivors.
Cap the count
The number of pairs is the ceiling.
Grade 7 organized counting: count pairs first to set the ceiling.
Counting the pairs of lines first sets the ceiling on how many crossings there can be.
▸ Why?
Each pair is counted once from either line, so naming both and halving counts every pair once.
▸ Why?
Every crossing needs its own pair, so the crossings can never outnumber the pairs.
All four parallel
There are no crossings at all.
Grade 4 parallel lines never meet — zero intersections is the cleanest case.
4.G.A.2Draw A DiagramAll four through one point
There is exactly one crossing.
Grade 4 lines through a point — every pair meets there and nowhere else.
4.G.A.1Draw A DiagramThe impossible value
No arrangement gives exactly two.
Grade 8 informal argument: chase parallels and see that every configuration with exactly two intersection points forces a third.
8.G.A.5Solve An Easier Related ProblemThree parallel
The crossing line makes three points.
Grade 4 parallel and transversal: one extra line crossing three parallels makes exactly three intersection points.
4.G.A.2Draw A DiagramThree concurrent
The fourth line adds three more.
Grade 4: a fourth line slicing three concurrent lines adds three new crossings.
4.G.A.1Draw A DiagramExactly one parallel pair
That removes one crossing.
Grade 4: each parallel pair erases one would-be intersection.
4.G.A.2Draw A DiagramGeneral position
Every pair meets at a distinct point.
Grade 4: when no shortcuts apply, every pair gives its own intersection.
4.G.A.1Draw A DiagramAdd the possible values
Adding them gives 19.
Grade 4 multi-digit addition: add the six surviving values.
4.NBT.B.4Make A Systematic ListMatch the choice
It matches a choice exactly.
Grade 4: read the choice list, pick 19.
4.NBT.A.2Eliminate PossibilitiesThis AMC 12 problem only needs Grade 8 line-and-parallel reasoning you already know! With four lines you can hit 0, 1, 3, 4, 5, 6 intersection points — all parallel for 0, all through one point for 1, three parallels plus a transversal for 3, three concurrent plus a general line for 4, one parallel pair for 5, and general position for 6. Only N = 2 is unreachable (every attempt forces a third intersection). Sum: 0 + 1 + 3 + 4 + 5 + 6 = 19, answer (D).