AMC 10 · 2009 · #12

Grade 6 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Axis-Aligned Figure Area from Coordinates
area-trianglessystematic-enumerationcasework physical-representation ↑ Prerequisites: area-triangles
📏 Medium solution 💡 2 insights
Problem
Four distinct points A, B, C, D sit on one line with AB=BC=CD=1. Two more points E and F sit on a second line parallel to the first, with EF=1. Three of these six points are chosen as the vertices of a triangle with positive area. How many different values can the area of that triangle take?

Pick an answer.

(A)
3
(B)
4
(C)
5
(D)
6
(E)
7

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

A picture of the two parallel lines (Tool #1) reveals the hidden structure: every triangle has one whole side sitting on a parallel line, and its third vertex on the other line. Because parallel lines stay the same distance apart, the height of every such triangle is the same. That collapses the whole problem into one question about the base. Tool #7 (Identify Subproblems) splits the base into two cases (base on the crowded line vs. base on the two-point line), and Tool #2 (Make a Systematic List) enumerates the distinct base lengths, which map one-to-one onto the distinct areas.

1STEP 1

Draw the two parallel lines

Put A,B,C,D at 0,1,2,3 on the bottom line and E,F on the top line. With only two points up there, every triangle uses both lines.

A,B,C,D=0,1,2,3; E,F on top line, EF=1; gap=h
2STEP 2

The height is always the same

Take the side on a parallel line as the base. The opposite vertex is always h away, so the area is half of base times h.

Area=1/2 · base · h, h fixed
3STEP 3

Split the base into two cases

The base lies either on the bottom line, with length 1, 2, or 3, or on the top line, where the only base available is EF=1.

bottom base∈{|1-0|,|2-0|,|3-0|}={1,2,3}; top base=EF=1
4STEP 4

List the distinct areas and count

EF=1 repeats a length already there, so the distinct bases are 1, 2, 3 and the areas number 3 — answer (A).

areas={1/2h, h, 3/2h}→ 3 values→(A)
Answer
3
The base on the top line (EF=1) matches a base already available on the bottom line (the unit gap like AB), so it adds no new area — a good sign the count stays small. The only base lengths anywhere are 1, 2, and 3, giving exactly 3 areas h/2, h, 3h/2. Choice (A) 3 fits; the larger choices would require base lengths that simply do not exist among these points.
💡Key takeaway

When all your triangles have a side on one of two parallel lines, the height never changes, so counting different areas is just counting different base lengths.

  • Draw the two parallel lines
  • The height is always the same
  • Split the base into two cases
  • List the distinct areas and count

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