2013 AMC 12A Problems/Problem 9
Problem
In , and . Points and are on sides , , and , respectively, such that and are parallel to and , respectively. What is the perimeter of parallelogram ?
Solution
Note that because and are parallel to the sides of , the internal triangles and are similar to , and are therefore also isosceles triangles.
It follows that . Thus, .
Since opposite sides of parallelograms are equal, the perimeter is .
Solution(Cheese)
We can set point to be on point , and point to be on point .
This makes a degenerate parallelogram with sides of length and , so it has a perimeter of .
Video Solution
~sugar_rush
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 |
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.