2009 AMC 10B Problem 13

P11-15~3 minPrint

Modular Arithmetic Angles & Polygons Set up the equation/formula and compute; no special trick needed

As shown below, convex pentagon $ABCDE$ has sides $AB=3$ , $BC=4$ , $CD=6$ , $DE=3$ , and $EA=7$ . The pentagon is originally positioned in the plane with vertex $A$ at the origin and vertex $B$ on the positive $x$ -axis. The pentagon is then rolled clockwise to the right along the $x$ -axis. Which side will touch the point $x=2009$ on the $x$ -axis?

Show solution without answering

Key insight

Rolling lays the sides down in order AB, BC, CD, DE, EA, repeating every 23 units. 2009 leaves remainder 8 when divided by 23, landing on CD.

Solution

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.

At the start, $AB$ is on the axis from $0$ to $3$ .
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.
Add up the lengths to see where each side lands:

Side	Length	Covers
$AB$	$3$	$0$ to $3$
$BC$	$4$	$3$ to $7$
$CD$	$6$	$7$ to $13$
$DE$	$3$	$13$ to $16$
$EA$	$7$	$16$ to $23$

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

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic
Introduction to Geometry, Chapter 2: Angles
Introduction to Number Theory, Chapter 10: Units Digits
Introduction to Geometry, Chapter 9: Polygons (§9.1, §9.2)