2001 AIME II Problems/Problem 13
In quadrilateral , and , , , and . The length may be written in the form , where and are relatively prime positive integers. Find .
Extend and to meet at . Then, since and , we know that . Hence , and is isosceles. Then .
Using the similarity, we have:
The answer is .
Extension: To Find , use Law of Cosines on to get Then since use Law of Cosines on to find
Draw a line from , parallel to , and let it meet at . Note that is similar to by AA similarity, since and since is parallel to then . Now since is an isosceles trapezoid, . By the similarity, we have , hence .
Since , if we extend AB and DC, they must meet at one point to form a isosceles triangle .Now, since the problem told that , we can imply that Since , so . Assume the length of ;Since we can get , we get that .So similarly, we use the same pair of similar triangle we get , we get that . Finally, ~bluesoul
Denote , and . Note that , and . This motivates us to draw the angle bisector of because , so we do so and consider the intersection with as . By the angle bisector theorem, we have , so we write and . We also know that and , so . Hence, , so we have . As , it must be that , so the final answer is .
Video Solution by OmegaLearn
|2001 AIME II (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|