2016 AIME I Problems/Problem 6
In let be the center of the inscribed circle, and let the bisector of intersect at . The line through and intersects the circumscribed circle of at the two points and . If and , then , where and are relatively prime positive integers. Find .
Suppose we label the angles as shown below. As and intercept the same arc, we know that . Similarly, . Also, using , we find . Therefore, . Therefore, , so must be isosceles with . Similarly, . Then , hence . Also, bisects , so by the Angle Bisector Theorem . Thus , and the answer is .
This is a cheap solution.
WLOG assume is isosceles. Then, is the midpoint of , and . Draw the perpendicular from to , and let it meet at . Since , is also (they are both inradii). Set as . Then, triangles and are similar, and . Thus, . , so . Thus . Solving for , we have: , or . is positive, so . As a result, and the answer is
WLOG assume is isosceles (with vertex ). Let be the center of the circumcircle, the circumradius, and the inradius. A simple sketch will reveal that must be obtuse (as an acute triangle will result in being greater than ) and that and are collinear. Next, if , and . Euler gives us that , and in this case, . Thus, . Solving for , we have , then , yielding . Next, so . Finally, gives us , and . Our answer is then .
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