2016 AIME II Problems/Problem 5
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 .
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.
We start by splitting the sum of all into two parts: those where is odd and those where is even.
Considering the sum of the lengths of the segments for which is odd, for each , first look at perimeters of the triangles . The perimeters of these triangles can be expressed using and ratios that result because of similar triangles. Considering triangles of the form , we find that the perimeter is . Thus,
Continuing with a similar process for the sum of the lengths of the segments for which is even, the following results:
Adding (1) and (2) together, we find that
Setting , , and , we can now proceed as in Shaddoll's solution, and our answer is .
Solution by brightaz
|2016 AIME II (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|