Mock Geometry AIME 2011 Problems/Problem 5
In triangle The bisector of angle meet at and the circumcircle at different from . Calculate the value of
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 .
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 .
This is the author's solution. Refer to the above diagrams.
Notice that So by AA similarity, It follows that or Now let By the previous result, we have But by the angle bisector theorem, Then, by power of a point on we have Subtracting the last two equations, we find Thus,