# Difference between revisions of "2016 AIME II Problems/Problem 5"

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− | Note that by counting the area in 2 ways, the first altitude is <math>\dfrac{ac}{b}</math>. By similar triangles, the common ratio is <math>\dfrac{a}{c}</math> for reach height, so by the geometric series formula, we have <math>6p=\dfrac{\dfrac{ | + | Note that by counting the area in 2 ways, the first altitude is <math>\dfrac{ac}{b}</math>. By similar triangles, the common ratio is <math>\dfrac{a}{c}</math> for reach height, so by the geometric series formula, we have <math>6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}</math>. Multiplying by the denominator and expanding, the equation becomes <math>\dfrac{ab}{c}=6a+6b+6c-\dfrac{6a^2}{c}-\dfrac{6ab}{c}-6a</math>. Cancelling <math>6a</math> and multiplying by <math>c</math> yields <math>ab=6bc+6c^2-6a^2-6ab</math>, so <math>7ab = 6bc+6b^2</math> and <math>7a=6b+6c</math>. Checking for Pythagorean triples gives <math>13,84,</math> and <math>85</math>, so <math>p=13+84+84=\boxed{182}</math> |

− | Solution | + | Solution modified/fixed from Shaddoll's solution. |

## Revision as of 15:10, 25 March 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.