Difference between revisions of "2009 AMC 10B Problems/Problem 13"
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==Solution 2: Mod Arithmetic== | ==Solution 2: Mod Arithmetic== | ||
− | The perimeter is 23 and 2009 mod 23 | + | The perimeter is <math>23</math> and <math>2009\equiv8(</math>mod <math>23)</math>, so it will end up on side <math>AB</math> + a total of 8 more units. <math>4<8</math>, but <math>4+6=10>8</math>, so it ends on side <math>CD</math> for an answer of <math>\boxed{C}</math>. |
== See Also == | == See Also == |
Revision as of 01:48, 7 November 2020
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
Solution 2: Mod Arithmetic
The perimeter is and
mod
, so it will end up on side
+ a total of 8 more units.
, but
, so it ends on side
for an answer of
.
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 |
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