Difference between revisions of "2016 AIME II Problems/Problem 5"
m (→Solution) |
Mathgeek2006 (talk | contribs) m |
||
Line 5: | Line 5: | ||
Solution modified/fixed from Shaddoll's solution. | Solution modified/fixed from Shaddoll's solution. | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2016|n=II|num-b=4|num-a=6}} |
Revision as of 21:29, 16 May 2016
Triangle has a right angle at . Its side lengths are pariwise relatively prime positive integers, and its perimeter is . Let be the foot of the altitude to , and for , let be the foot of the altitude to in . The sum . Find .
Solution
Note that by counting the area in 2 ways, the first altitude is . By similar triangles, the common ratio is for reach height, so by the geometric series formula, we have . Multiplying by the denominator and expanding, the equation becomes . Cancelling and multiplying by yields , so and . Checking for Pythagorean triples gives and , so
Solution modified/fixed from Shaddoll's solution.
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |