AMC 10 · 2019 · #14
Grade 8 arithmeticPick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Each pair of lines meets at most once, and four lines make C(4, 2) = 6 pairs, so N ≤ 6; the candidates are 0–6.
Grade 7 organized counting: count pairs first to set the ceiling.
7.SP.C.8Make A Systematic ListMake four lines all parallel (say four horizontals) — no two ever meet, so N = 0.
Grade 4 parallel lines never meet — zero intersections is the cleanest case.
4.G.A.2Draw A DiagramSend all four lines through one point (a pencil at the origin) — every pair meets only there, so N = 1.
Grade 4 lines through a point — every pair meets there and nowhere else.
4.G.A.1Draw A DiagramSuppose only two crossing points existed — chasing the parallels in every case forces a third point, so N = 2 is impossible.
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 lines plus one transversal cutting all three give three distinct crossings, so N = 3.
Grade 4 parallel and transversal: one extra line crossing three parallels makes exactly three intersection points.
4.G.A.2Draw A DiagramThree concurrent lines through P (one point) plus a fourth in general position add three new crossings, so N = 4.
Grade 4: a fourth line slicing three concurrent lines adds three new crossings.
4.G.A.1Draw A DiagramOne parallel pair with no three concurrent erases just one of the six crossings, so N = 5.
Grade 4: each parallel pair erases one would-be intersection.
4.G.A.2Draw A DiagramFour lines in general position (no parallels, no concurrence) let every pair meet separately, so N = 6.
Grade 4: when no shortcuts apply, every pair gives its own intersection.
4.G.A.1Draw A DiagramAdd the attainable values: 0 + 1 + 3 + 4 + 5 + 6 = 19.
Grade 4 multi-digit addition: add the six surviving values.
4.NBT.B.4Make A Systematic ListThe sum 19 matches choice (D).
Grade 4: read the choice list, pick 19.
4.NBT.A.2Eliminate PossibilitiesThis AMC 10 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).