2019 AIME I Problems/Problem 13
Contents
Problem
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)
Solution 5
Connect meeting at . We can observe that Getting that . We can also observe that , getting that
Assume that , since , we can get that , getting that
Using Power of Point, we can get that Assume that , getting that
Now applying Law of Cosine on two triangles, separately, we can get two equations
Since , we can use to eliminate the term
Then we can get that , getting
, so the desired answer is , which leads to the answer
~bluesoul
Solution 6
Nice problem!
First, let and intersect at . Our motivation here is to introduce cyclic quadrilaterals and find useful relationships in terms of angles. Observe that By the so-called "Reverse Law of Cosines" on we have Applying on gives So , now by our cyclic quadrilaterals again, we are motivated by the multiple appearances of similar triangles throughout the figure. We want some that are related to and , which are crucial lengths in the problem. Suppose for simplicity. We have:
So So . The requested sum is .
~CoolJupiter
Video Solution by MOP 2024
https://youtube.com/watch?v=B7rFw05AYQ0
~r00tsOfUnity
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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