Difference between revisions of "2009 AMC 10B Problems/Problem 13"
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And finally, the polygon rotates around <math>C</math> until point <math>D</math> hits the <math>x</math> axis at <math>(2014,0)</math>. | And finally, the polygon rotates around <math>C</math> until point <math>D</math> hits the <math>x</math> axis at <math>(2014,0)</math>. | ||
− | At this point the side <math>\boxed{\overline{CD}}</math> touches the point <math>(2009,0)</math>. | + | At this point the side <math>\boxed{\overline{CD}}</math> touches the point <math>(2009,0)</math>. So the answer is <math>\boxed{C}</math> |
== See Also == | == See Also == | ||
{{AMC10 box|year=2009|ab=B|num-b=12|num-a=14}} | {{AMC10 box|year=2009|ab=B|num-b=12|num-a=14}} |
Revision as of 18:39, 22 February 2011
Problem
As shown below, convex pentagon has sides , , , , and . The pentagon is originally positioned in the plane with vertex at the origin and vertex on the positive -axis. The pentagon is then rolled clockwise to the right along the -axis. Which side will touch the point on the -axis?
Solution
The perimeter of the polygon is . Hence as we roll the polygon to the right, every units the side will be the bottom side.
We have . Thus at some point in time we will get the situation when and is the bottom side. Obviously, at this moment .
After that, the polygon rotates around until point hits the axis at .
And finally, the polygon rotates around until point hits the axis at . At this point the side touches the point . So the answer is
See Also
2009 AMC 10B (Problems • Answer Key • 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 | ||
All AMC 10 Problems and Solutions |