Competition · AMC preparation · step 4 of 4

AMC 10 · 2019A · #14

Grade 8 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

combinations-basiccaseworkcontradiction-elementarysystematic-enumeration caseworksystematic-enumerationcontradiction-elementary ↑ Prerequisites: combinations-basic

📏 Long solution 💡 4 insights

Problem

Take four distinct straight lines in the plane. Let N be the number of points that lie on at least two of them (the intersection points, each counted once regardless of how many lines pass through it). Find every value of N that can occur for some choice of four lines, then add those values up.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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.

1STEP 1

Bound the maximum

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.

N ≤ C(4, 2) = 6

2STEP 2

Build 0 intersections

Make four lines all parallel (say four horizontals) — no two ever meet, so N = 0.

4 parallel lines → N = 0

3STEP 3

Build 1 intersection

Send all four lines through one point (a pencil at the origin) — every pair meets only there, so N = 1.

4 lines through one point → N = 1

4STEP 4

Rule out 2 intersections

Suppose only two crossing points existed — chasing the parallels in every case forces a third point, so N = 2 is impossible.

N = 2 → contradiction in both cases

5STEP 5

Build 3 intersections

Three parallel lines plus one transversal cutting all three give three distinct crossings, so N = 3.

3 parallel lines + 1 transversal → N = 3

6STEP 6

Build 4 intersections

Three concurrent lines through P (one point) plus a fourth in general position add three new crossings, so N = 4.

3 concurrent + 1 in general position → N = 1 + 3 = 4

7STEP 7

Build 5 intersections

One parallel pair with no three concurrent erases just one of the six crossings, so N = 5.

exactly one parallel pair, no concurrence → N = 6 - 1 = 5

8STEP 8

Build 6 intersections

Four lines in general position (no parallels, no concurrence) let every pair meet separately, so N = 6.

4 lines in general position → N = C(4, 2) = 6

9STEP 9

Add the possible values

Add the attainable values: 0 + 1 + 3 + 4 + 5 + 6 = 19.

0 + 1 + 3 + 4 + 5 + 6 = 19

10STEP 10

Match 19 to the choices

The sum 19 matches choice (D).

19 → (D)

Answer

N ranges from 0 (all parallel) to 6 (general position), so the sum sits between 0 + 1 + … + 6 = 21 (if every value worked) and something smaller (we have to drop unreachable values). Only N = 2 is impossible — that subtracts 2, leaving 21 - 2 = 19. Matches.

💡Key takeaway

This 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).

Bound the maximum
Build 0 intersections
Build 1 intersection
Rule out 2 intersections
Build 3 intersections
Build 4 intersections
Build 5 intersections
Build 6 intersections
Add the possible values
Match 19 to the choices

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