2019 AIME I Problems/Problem 13
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Problem 13
Triangle has side lengths , , and . Points and are on ray with . The point is a point of intersection of the circumcircles of and satisfying and . Then can be expressed as , where , , , and are positive integers such that and are relatively prime, and is not divisible by the square of any prime. Find .
Solution 1
Notice that By the Law of Cosines, Then, Let , , and . Then, However, since , , but since , and the requested sum is .
(Solution by TheUltimate123)
Solution 2
Define to be the circumcircle of and to be the circumcircle of .
Because of exterior angles,
But because is cyclic. In addition, because is cyclic. Therefore, . But , so . Using Law of Cosines on , we can figure out that . Since , . We are given that and , so we can use Law of Cosines on to find that .
Let be the intersection of segment and . Using Power of a Point with respect to within , we find that . We can also apply Power of a Point with respect to within to find that . Therefore, .
Note that is similar to . . Also note that is similar to , which gives us . Solving this system of linear equations, we get . Now, we can solve for , which is equal to . This simplifies to , which means our answer is .
Solution 3
Construct and let . Let . Using , Using , it can be found that This also means that . It suffices to find . It is easy to see the following: Using reverse Law of Cosines on , . Using Law of Cosines on gives , so . -franchester
Solution 4 (No <C = <DFE, no LoC)
Let . Let and ; from we have and . From we have giving . So and . These similar triangles also gives us so . Now, Stewart's Theorem on and cevian tells us that so . Then so the answer is as desired. (Solution by Trumpeter, but not added to the Wiki by Trumpeter)
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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