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2009 AMC 10B Problem 12

2009 AMC 10B Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Triangles: Area & Pythagorean · difficulty 2 of 5 · about 3 min · position P11-15

Distinct points $A$ , $B$ , $C$ , and $D$ lie on a line, with $AB=BC=CD=1$ . Points $E$ and $F$ lie on a second line, parallel to the first, with $EF=1$ . A triangle with positive area has three of the six points as its vertices. How many possible values are there for the area of the triangle?

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

Answer

Key insight

Every triangle has its base on one line and apex on the other, so its height is the fixed gap and only the base length, 1, 2 or 3, matters.

Solution

Let the two parallel lines be a distance $h$ apart. Three collinear points give no triangle, so every valid triangle has two vertices on one line (the base) and the third on the other line (the apex). The height is always $h$ , regardless of where the apex is, so the area is $\frac12 \cdot (\text{base}) \cdot h$ and depends only on the base length.

Base on the line through $A, B, C, D$ : the base is one of $AB, BC, CD$ (length $1$ ), $AC, BD$ (length $2$ ), or $AD$ (length $3$ ). Areas $\frac h2, h, \frac{3h}{2}$ .
Base $EF$ on the other line: length $1$ , area $\frac h2$ , already counted.

The possible areas are $\frac h2$ , $h$ , $\frac{3h}{2}$ : three values.

The answer is $\boxed{\textbf{(A)}\ 3}$ .

Why this works

With parallel lines, the apex position is irrelevant to the area, which collapses a large family of triangles onto a few base lengths. Ask "what actually varies?" before enumerating; here only the base length does, and the segment lengths among equally spaced points are just $1, 2, 3$ .

The trap

Counting triangles (or where the apex sits) instead of distinct areas, or overlooking that a base EF of length 1 duplicates the area from base AB.

Common mistakes

Counting triangles (or where the apex sits) instead of distinct areas, or overlooking that a base EF of length 1 duplicates the area from base AB.
Including "triangles" with all three vertices on the first line, which have zero area.

Techniques

Split into exhaustive cases and handle each