Mock Geometry AIME 2011 Problems/Problem 5
Contents
Problem
In triangle The bisector of angle meet at and the circumcircle at different from . Calculate the value of
Solution
Solution 1
because they are both subscribed by arc . because they are both subscribed by arc . Hence , because . Then is isosceles.
Let be the foot of the perpendicular from to . As is isosceles, it follows that is the midpoint of , and so . From the angle bisector theorem, . We have . Solving this system of equations yields . Thus, .
because they are vertical angles. It was shown , and so by similarity. Then and so .
Then by the Pythagorean Theorem on , . Also from , . Subtracting these equations yields , and so .
Solution 2
Let , so that . From the Angle Bisector Theorem, . Cross-multilplying and solving for , we find that . Thus, and .
Now, from Stewart's Theorem, . Plugging in values and letting , we find that .
Dividing both sides by gives .Factoring a out of the numerator of the fraction and continuing to simplify, we find that .
Now, from Power of a Point on , we have . Now, let , so we have . From here, we can find that .