Difference between revisions of "2013 AMC 12A Problems/Problem 11"
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Revision as of 09:55, 7 April 2013
Problem
Triangle is equilateral with
. Points
and
are on
and points
and
are on
such that both
and
are parallel to
. Furthermore, triangle
and trapezoids
and
all have the same perimeter. What is
?
Solution
Let , and
. We want to find
, which is nothing but
.
Based on the fact that ,
, and
have the same perimeters, we can say the following:
Simplifying, we can find that
Since ,
.
After substitution, we find that , and
=
.
Again substituting, we find =
.
Therefore, =
, which is
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 12 Problems and Solutions |