AMC 10 · 2009 · #13

Grade 5 geometry-2d

Archetype Modular Cycle Frequency Count └ Sub-archetype Position-in-Cycle Lookup

perimetermodular-arithmeticpattern-recognition pattern-recognition ↑ Prerequisites: perimeter

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A convex pentagon ABCDE with side lengths AB=3, BC=4, CD=6, DE=3, EA=7 starts with side AB flat on the x-axis, A at the origin and B on the positive x-axis. It is rolled clockwise to the right along the x-axis. Determine which side is lying flat on the ground at the moment the pentagon covers the point x=2009.

Pick an answer.

(A)

$\overline{AB}$

(B)

$\overline{BC}$

(C)

$\overline{CD}$

(D)

$\overline{DE}$

(E)

$\overline{EA}$

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

Rolling the pentagon prints the sides onto the axis one after another, and after one full turn the whole print repeats. Tool #5 (Look for a Pattern) captures that the ground is tiled by copies of the same 23-unit block, so we only need to know where 2009 falls inside one block. Tool #1 (Draw a Diagram) records the exact stretch each side covers by adding up the side lengths. Tool #9 (Solve an Easier Related Problem) shrinks the giant 2009 to its remainder after one perimeter, turning a huge number into a spot inside a single small block.

1STEP 1

Add the sides for one full turn

One full turn lays every side flat exactly once, so the repeat length is the perimeter 3+4+6+3+7=23.

3+4+6+3+7=23

2STEP 2

Mark where each side ends

Stack the lengths from x=0: AB covers 0 to 3, BC 3 to 7, CD 7 to 13, DE 13 to 16, EA 16 to 23 - then the blocks repeat.

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

3STEP 3

Reduce 2009 to one block

Whole turns cancel, so only the leftover counts: 2009=23×87+8, meaning x=2009 acts exactly like x=8 in one block.

2009=23 × 87+8, 2009≡ 8 (mod 23)

4STEP 4

Locate the leftover 8

Which stretch holds 8? CD runs from 7 to 13, so the side flat at x=2009 is CD, choice (C).

7 ＜ 8 ＜ 13 → side CD → (C)

Answer

Check the block boundaries: the running ends are 3,7,13,16,23, and their gaps 3,4,6,3,7 match the given side lengths and sum to the perimeter 23, so nothing is double-counted or missed. The leftover 8 sits between 7 and 13, strictly inside CD's stretch (not on a vertex), so the choice is clean. Rebuilding the check as 23 × 87=2001 and 2001+8=2009 confirms the remainder is 8, landing on CD, choice (C).

💡Key takeaway

When a shape rolls and repeats, add up one full turn, divide out the whole turns, and just see where the leftover lands.

Add the sides for one full turn
Mark where each side ends
Reduce 2009 to one block
Locate the leftover 8

Add the sides for one full turn

Mark where each side ends

Reduce 2009 to one block

Locate the leftover 8

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