2009 AMC 10B Problem 13

As the pentagon rolls, each side in turn lies flat on the $x$-axis, and it covers a stretch of the axis as long as the side. So first work out which stretch each side covers.

1. At the start, $AB$ is on the axis from $0$ to $3$.
2. Rolling clockwise to the right, the pentagon tips over the corner $B$. The next side to land is $BC$, then $CD$, then $DE$, then $EA$, and then $AB$ again.
3. Add up the lengths to see where each side lands:

1. After $23$ units (the perimeter) the pentagon is back to the start position, just shifted, and the pattern repeats.
2. Divide: $2009 = 23 \times 87 + 8$. So $x = 2009$ is in the same spot of the pattern as $x = 8$.
3. $8$ is between $7$ and $13$, so side $CD$ touches it.

The answer is $\boxed{\textbf{(C)}\ \overline{CD}}$.

A rolling shape "unrolls" its edge onto the line: the sides land one after another, each covering its own length, and the whole pattern repeats every perimeter. That turns a geometry question into a division with remainder. The angles of the pentagon never matter, only the order and lengths of the sides.

As the pentagon rolls clockwise to the right, it pivots about $B$ and the next side to lie flat is $BC$, then $CD$, then $DE$, then $EA$, then $AB$ again. Each side covers a stretch of the axis equal to its length, so the sides tile the positive $x$-axis in this order:

• $AB$: $[0, 3]$
• $BC$: $[3, 7]$
• $CD$: $[7, 13]$
• $DE$: $[13, 16]$
• $EA$: $[16, 23]$

After one full turn ($23$ units, the perimeter) the pattern repeats. Since $2009 = 23 \cdot 87 + 8$, the point $x = 2009$ corresponds to $x = 8$ in the first cycle, which lies strictly inside $[7, 13]$.

The answer is $\boxed{\textbf{(C)}\ \overline{CD}}$.