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

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?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: 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 $x=2009$.

Givens: The five side lengths in order are $AB=3$, $BC=4$, $CD=6$, $DE=3$, $EA=7$.; The pentagon starts with $AB$ flat on the ground, $A=(0,0)$ and $B=(3,0)$.; It rolls clockwise without slipping to the right along the $x$-axis.; Answer choices name a side: (A) $\overline{AB}$, (B) $\overline{BC}$, (C) $\overline{CD}$, (D) $\overline{DE}$, (E) $\overline{EA}$.

Unknowns: Which side of the pentagon lies flat on the $x$-axis at the point $x=2009$.

Understand

Plan

Primary tool: #5 Look for a Pattern

Secondary: #1 Draw a Diagram, #9 Solve an Easier Related Problem

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.

Execute — Answer: C

#1 Draw a Diagram 3.MD.D.8 Step 1

Add the sides for one full turn

As the pentagon rolls, one complete turn lays every side flat exactly once, so the ground it covers in one turn equals the perimeter.
Add the five side lengths: $3+4+6+3+7=23$.
So every $23$ units along the $x$-axis the pattern of flat sides repeats exactly.

$$3+4+6+3+7=23$$

💡 One full roll uses each side once, so the repeat length is just the perimeter.

#5 Look for a Pattern 4.OA.C.5 Step 2

Mark where each side ends

Track the right-hand end of each flat side, starting from $x=0$.
Side $AB$ (length $3$) covers $0$ to $3$.
Then $BC$ (length $4$) covers $3$ to $7$.
Then $CD$ (length $6$) covers $7$ to $13$.
Then $DE$ (length $3$) covers $13$ to $16$.
Then $EA$ (length $7$) covers $16$ to $23$, finishing the first full turn.
After that the same blocks repeat.

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

💡 Stacking the side lengths end to end tells you exactly which stretch of ground each side owns.

#9 Solve an Easier Related Problem 5.NBT.B.6 Step 3

Reduce 2009 to one block

Because the pattern repeats every $23$ units, only the leftover after removing whole turns matters.
Divide $2009$ by $23$: it fits $87$ whole times since $23\times 87=2001$, leaving $2009-2001=8$.
So the point $x=2009$ behaves exactly like the point $x=8$ inside a single $23$-unit block.

$$2009=23\times 87+8,\qquad 2009\equiv 8 \pmod{23}$$

💡 Whole turns land the pentagon back where it started, so only the remainder decides the answer.

#5 Look for a Pattern 4.OA.C.5 Step 4

Locate the leftover 8

Find which side's stretch contains $x=8$.
From the marks, $CD$ covers $7$ to $13$, and $8$ lies inside that stretch.
So the side lying flat at $x=8$ - and therefore at $x=2009$ - is $\overline{CD}$.
That is choice $\textbf{(C)}$.

$$7<8<13\ \Rightarrow\ \text{side } \overline{CD}\ \Rightarrow\ \textbf{(C)}$$

💡 The remainder is just a spot on the first block, and reading off which side owns that spot gives the answer.

[1] #1 3.MD.D.8 As the pentagon rolls, one complete turn lays every side flat exactly once, so t

[2] #5 4.OA.C.5 Track the right-hand end of each flat side, starting from $x=0$. Side $AB$ (leng

[3] #9 5.NBT.B.6 Because the pattern repeats every $23$ units, only the leftover after removing w

[4] #5 4.OA.C.5 Find which side's stretch contains $x=8$. From the marks, $CD$ covers $7$ to $13

Review

Reasonableness: 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 $23\times 87=2001$ and $2001+8=2009$ confirms the remainder is $8$, landing on $\overline{CD}$, choice (C).

Alternative: Instead of dividing, subtract full perimeters until the number is small: $2009-23=1986$, and repeating (or noting $2009-2001=8$) leaves $8$. Then walk the sides from the start, subtracting lengths: $8-3=5$ (past $AB$), $5-4=1$ (past $BC$), and $1$ is less than $CD$'s length $6$, so the point lands on $\overline{CD}$ - again choice (C).

CCSS standards used (min grade 5)

3.MD.D.8 Solve real-world problems involving perimeters of polygons (Adding the five side lengths $3+4+6+3+7=23$ to get the perimeter, which is the repeat length of the rolling pattern.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Building the repeating block of ground stretches each side covers and reading off which side owns a given position.)
5.NBT.B.6 Find whole-number quotients with up to four-digit dividends and two-digit divisors (Dividing $2009$ by $23$ to get quotient $87$ and remainder $8$, collapsing the big position into one block.)

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

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Add the sides for one full turn

Mark where each side ends

Reduce 2009 to one block

Locate the leftover 8

Review

CCSS standards used (min grade 5)

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