Difference between revisions of "2013 AMC 12A Problems/Problem 11"
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Therefore, <math>x+y</math> = <math>\frac{21}{13}</math>, which is <math>C</math> | Therefore, <math>x+y</math> = <math>\frac{21}{13}</math>, which is <math>C</math> | ||
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+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=XQpQaomC2tA | ||
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+ | ~sugar_rush | ||
== See also == | == See also == |
Latest revision as of 19:12, 24 November 2020
Contents
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
Video Solution
https://www.youtube.com/watch?v=XQpQaomC2tA
~sugar_rush
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.