2009 AMC 10B Problems/Problem 13

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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?

$\text{(A) } \overline{AB}\qquad\text{(B) } \overline{BC}\qquad\text{(C) } \overline{CD}\qquad\text{(D) } \overline{DE}\qquad\...$

Solution

The perimeter of the polygon is $3+4+6+3+7 = 23$ . Hence as we roll the polygon to the right, every $23$ units the side $\overline{AB}$ will be the bottom side.

We have $2009 = 23 \times 87 + 8$ . Thus at some point in time we will get the situation when $A=(2001,0)$ and $\overline{AB}$ is the bottom side. Obviously, at this moment $B=(2004,0)$ .

After that, the polygon rotates around $B$ until point $C$ hits the $x$ axis at $(2008,0)$ .

And finally, the polygon rotates around $C$ until point $D$ hits the $x$ axis at $(2014,0)$ . At this point the side $\boxed{\overline{CD}}$ touches the point $(2009,0)$ .

2009 AMC 10B (Problems • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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2009 AMC 10B Problems/Problem 13

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Problem

Solution

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