2016 AIME II Problems/Problem 5

Revision as of 21:29, 16 May 2016 by Mathgeek2006 (talk | contribs)

Triangle $ABC_0$ has a right angle at $C_0$. Its side lengths are pariwise relatively prime positive integers, and its perimeter is $p$. Let $C_1$ be the foot of the altitude to $\overline{AB}$, and for $n \geq 2$, let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$. The sum $\sum_{i=1}^\infty C_{n-2}C_{n-1} = 6p$. Find $p$.


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

Solution modified/fixed from Shaddoll's solution.

See also

2016 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AIME Problems and Solutions
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