1996 AIME Problems/Problem 13
Contents
Problem
In triangle , , , and . There is a point for which bisects , and is a right angle. The ratio can be written in the form , where and are relatively prime positive integers. Find .
Solution
Let be the midpoint of . Since , then and share the same height and have equal bases, and thus have the same area. Similarly, and share the same height, and have bases in the ratio , so (see area ratios). Now,
By Stewart's Theorem, , and by the Pythagorean Theorem on ,
Subtracting the two equations yields . Then , and .
Solution 2
Because the problem asks for a ratio, we can divide each side length by to make things simpler. We now have a triangle with sides , , and .
We use the same graph as above.
Draw perpendicular from to . Denote this point as . We know that and and also let .
Using Pythagorean theorem, we get three equations,
Adding the first and second, we obtain , and then subtracting the third from this we find that . (Note, we could have used Stewart's Theorem to achieve this result).
Subtracting the first and second, we see that , and then we find that
Using base ratios, we then quickly find that the desired ratio is so our answer is
See also
1996 AIME (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.