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2009 AMC 10B Problem 13

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2009 AMC 10B Problem 13AMC 10 Step by Step
Name:                             Date: September 13, 2026
Modular Arithmetic · difficulty 2 of 5 · about 3 min · position P11-15

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

figure

  1. A)

    AB‾\overline{AB}

  2. B)

    BC‾\overline{BC}

  3. C)

    CD‾\overline{CD}

  4. D)

    DE‾\overline{DE}

  5. E)

    EA‾\overline{EA}

Answer

C

Key insight

Rolling lays the sides along the axis in order AB, BC, CD, DE, EA with period 23; 2009 = 23*87 + 8 falls in CD's span 7 to 13.

Solution

As the pentagon rolls clockwise to the right, it pivots about BB and the next side to lie flat is BCBC, then CDCD, then DEDE, then EAEA, then ABAB again. Each side covers a stretch of the axis equal to its length, so the sides tile the positive xx-axis in this order:

  • ABAB: [0,3][0, 3]
  • BCBC: [3,7][3, 7]
  • CDCD: [7,13][7, 13]
  • DEDE: [13,16][13, 16]
  • EAEA: [16,23][16, 23]

After one full turn (2323 units, the perimeter) the pattern repeats. Since 2009=23⋅87+82009 = 23 \cdot 87 + 8, the point x=2009x = 2009 corresponds to x=8x = 8 in the first cycle, which lies strictly inside [7,13][7, 13].

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

Why this works

A rolling polygon unrolls its perimeter onto the line: the sides appear consecutively, each spanning its own length, and the whole pattern has period equal to the perimeter. That turns a geometry question into a remainder computation. The shape of the pentagon (and its angles) never matters, only the side order and lengths.

The trap

Rolling the sides in the wrong order (AB, EA, DE, ...) or miscounting the cumulative endpoints 3, 7, 13, 16, 23.

Common mistakes

  • Rolling the sides in the wrong order (AB, EA, DE, ...) or miscounting the cumulative endpoints 3, 7, 13, 16, 23.
  • Computing 2009 mod 232009 \bmod 23 incorrectly; 23⋅87=200123 \cdot 87 = 2001, so the remainder is 88, not 99 or 77.

Techniques

Set up the equation/formula and compute; no special trick needed