1999 AIME Problems/Problem 14
Drop perpendiculars from to the three sides of and let them meet and at and respectively.
Let and . We have that We can then use the tool of calculating area in two ways On the other hand, We still need though. We have all these right triangles and we haven't even touched Pythagoras. So we give it a shot: Adding gives Recall that we found that . Plugging in , we get , giving us for an answer.
Let , , , , , and .
Now the area of a triangle, , where and are sides on either side of an angle, . So, Adding these equations yields: Dividing by , we have: Thus, .
Note: In fact, this problem is unfairly easy to those who happen to have learned about Brocard point. The Brocard Angle is given by
Let Then, using Law of Cosines on the three triangles containing vertex we have Add the three equations up and rearrange to obtain Also, using we have Divide the two equations to obtain
|1999 AIME (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|