AMC 10 · 2019 · #14
Grade 8 arithmeticProblem
For a set of four distinct lines in a plane, there are exactly N distinct points that lie on two or more of the lines. What is the sum of all possible values of N?
Pick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: Four distinct lines in a plane; $N = $ number of distinct points where two or more lines meet; Answer choices: $14, 16, 18, 19, 21$
Unknowns: The sum of all attainable values of $N$
Understand
Restated: 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.
Givens: Four distinct lines in a plane; $N = $ number of distinct points where two or more lines meet; Answer choices: $14, 16, 18, 19, 21$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #2 Make a Systematic List, #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
Tool #1 (Draw): sketch sample configurations. Tool #2 (Systematic List): list candidate values $N = 0, 1, 2, 3, 4, 5, 6$ (the maximum is $\binom{4}{2} = 6$ 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.
Execute — Answer: D
7.SP.C.8 Step 1 - Bound the maximum first.
- Each pair of lines meets in at most one point, and four lines have $\binom{4}{2} = 6$ pairs, so $N \le 6$.
- So the candidate values are $0, 1, 2, 3, 4, 5, 6$.
💡 Grade 7 organized counting: count pairs first to set the ceiling.
4.G.A.2 Step 2 - Build $N = 0$.
- Take four lines all parallel to each other (for example, four horizontal lines).
- No two ever meet, so $N = 0$.
- Achievable.
💡 Grade 4 parallel lines never meet — zero intersections is the cleanest case.
4.G.A.1 Step 3 - Build $N = 1$.
- Take a pencil of four lines all through a single point (for example, four lines through the origin at different slopes).
- Every pair meets at that one point, so $N = 1$.
💡 Grade 4 lines through a point — every pair meets there and nowhere else.
8.G.A.5 Step 4 - Show $N = 2$ is impossible.
- Suppose only two intersection points $P, Q$ exist.
- Each line contributes to at least one intersection (or it would be parallel to all three others, forcing the other three pairwise parallel, which makes them all parallel and the picture have $0$ intersections, not $2$).
- Case I: no line passes through both $P$ and $Q$.
- Then two lines meet at $P$ and two at $Q$; for no new intersection to appear, the $P$-pair must be parallel to the $Q$-pair line-by-line.
- But then the two $P$-lines are both parallel to the same $Q$-line, hence parallel to each other — they can't meet at $P$, contradiction.
- Case II: one line $\ell$ passes through both $P$ and $Q$.
- Then a second line through $P$ (call it $m$) and a second line through $Q$ (call it $n$) must miss each other, so $m \parallel n$.
- The fourth line must pass through $P$ or $Q$; whichever it is, it can't be parallel to its partner, so it intersects the partner somewhere — creating a third intersection point, contradiction.
- So $N = 2$ is unreachable.
💡 Grade 8 informal argument: chase parallels and see that every configuration with exactly two intersection points forces a third.
4.G.A.2 Step 5 - Build $N = 3$.
- Take three lines forming a triangle (three intersection points), and add a fourth line parallel to one of the triangle's sides — it misses one side entirely and crosses the other two at vertices the triangle already has?
- No: the fourth must cross the other two at new points unless...
- easier construction: three concurrent lines through a common point $P$ (one intersection), plus a fourth line parallel to none — it crosses each of the three lines at a different point, adding $3$ new intersections, for $N = 1 + 3 = 4$, not $3$.
- Use this instead: three concurrent lines through $P$, plus a fourth line that passes through $P$ as well?
- That gives $N = 1$.
- To get $N = 3$, take three parallel lines plus one transversal crossing all three — the transversal meets each parallel line at a distinct point, giving $3$ intersections; the parallels do not meet each other.
- $N = 3$.
💡 Grade 4 parallel and transversal: one extra line crossing three parallels makes exactly three intersection points.
4.G.A.1 Step 6 - Build $N = 4$.
- Take three concurrent lines through a point $P$ (that gives $1$ intersection) and add a fourth line that is parallel to none of them and avoids $P$.
- The fourth line then meets each of the other three at a distinct new point, adding $3$ intersections, for $N = 1 + 3 = 4$.
💡 Grade 4: a fourth line slicing three concurrent lines adds three new crossings.
4.G.A.2 Step 7 - Build $N = 5$.
- Take four lines in which exactly one pair is parallel and no three are concurrent.
- The maximum $6$ pair-intersections drop by $1$ (the parallel pair makes no point), giving $N = 6 - 1 = 5$.
- Concretely: two parallel horizontal lines, plus two other lines in general position.
💡 Grade 4: each parallel pair erases one would-be intersection.
4.G.A.1 Step 8 - Build $N = 6$.
- Four lines in general position (no two parallel, no three concurrent) — for example, four lines with distinct slopes that avoid any triple meeting.
- Every pair gives a distinct intersection, so $N = 6$.
💡 Grade 4: when no shortcuts apply, every pair gives its own intersection.
4.NBT.B.4 Step 9 Sum the attainable values $\{0, 1, 3, 4, 5, 6\}$: $0 + 1 + 3 + 4 + 5 + 6 = 19$.
💡 Grade 4 multi-digit addition: add the six surviving values.
4.NBT.A.2 Step 10 Match $19$ to the choices: $(D)$.
💡 Grade 4: read the choice list, pick $19$.
7.SP.C.8 Bound the maximum first. Each pair of lines meets in at most one point, and four 4.G.A.2 Build $N = 0$. Take four lines all parallel to each other (for example, four hor 4.G.A.1 Build $N = 1$. Take a pencil of four lines all through a single point (for examp 8.G.A.5 Show $N = 2$ is impossible. Suppose only two intersection points $P, Q$ exist. E 4.G.A.2 Build $N = 3$. Take three lines forming a triangle (three intersection points), 4.G.A.1 Build $N = 4$. Take three concurrent lines through a point $P$ (that gives $1$ i 4.G.A.2 Build $N = 5$. Take four lines in which exactly one pair is parallel and no thre 4.G.A.1 Build $N = 6$. Four lines in general position (no two parallel, no three concurr 4.NBT.B.4 Sum the attainable values $\{0, 1, 3, 4, 5, 6\}$: $0 + 1 + 3 + 4 + 5 + 6 = 19$. 4.NBT.A.2 Match $19$ to the choices: $(D)$. Review
Reasonableness: $N$ ranges from $0$ (all parallel) to $6$ (general position), so the sum sits between $0 + 1 + \dots + 6 = 21$ (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.
Alternative: Tool #16 (Complement / Change Focus): instead of building each $N$, start with the maximum $6$ and ask how many intersections you 'lose' to parallelism or concurrence. Each parallel pair removes $1$ intersection; making $k$ lines concurrent through one point removes $\binom{k}{2} - 1$ (because $k$ pair-intersections collapse into one). Track all achievable losses from $0$ to $6$, exclude the loss $= 4$ (which would give $N = 2$), and sum the survivors. Same result, different bookkeeping.
CCSS standards used (min grade 8)
4.G.A.1Draw points, lines, line segments, rays, angles, and identify in figures (Sketching concurrent-pencil, general-position, and other configurations.)4.G.A.2Classify two-dimensional figures based on presence of parallel or perpendicular lines (Building all-parallel, three-parallel-plus-transversal, and one-parallel-pair configurations.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Matching the final sum $19$ to the answer choices.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Computing $0 + 1 + 3 + 4 + 5 + 6 = 19$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the $\binom{4}{2} = 6$ pairs of lines as the upper bound for $N$.)8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Proving $N = 2$ is impossible by a two-case parallel-chasing argument.)
⭐ 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: $0 + 1 + 3 + 4 + 5 + 6 = \mathbf{19}$, answer $(D)$.
⭐ 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: $0 + 1 + 3 + 4 + 5 + 6 = \mathbf{19}$, answer $(D)$.
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