2009 AMC 10B Problem 12

There are many triangles, but the question only asks about their areas, so figure out what the area depends on.

1. Three points on the same line do not make a triangle. So every triangle has two corners on one line and the third corner on the other line.
2. Use the two corners on the same line as the base. The height is the distance from the third corner to that line, which is always the gap between the two parallel lines. Call that gap $h$.
3. So area $= \frac12 \times \text{base} \times h$. Only the base length can change.
4. Bases on the line with $A, B, C, D$ (spaced $1$ apart):
- $AB$, $BC$, $CD$ have length $1$.
- $AC$, $BD$ have length $2$.
- $AD$ has length $3$.
5. The only base on the other line is $EF$, with length $1$. That is a length we already have.
6. So the base can be $1$, $2$ or $3$, giving areas $\frac12 h$, $h$ and $\frac32 h$. These are three different values.

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

Between two parallel lines, sliding the top corner along its line does not change the height, so it does not change the area. Before listing many cases, ask what actually varies. Here only the base length varies, and equally spaced points only make a few different lengths.

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 $\boxed{\textbf{(A)}\ 3}$.