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 are odd and those where is even.
Considering the sum of the lengths of the segments for which is odd, for each , consider the perimeters of the triangles and . 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
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