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

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==Solution==
 
==Solution==
 
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{ac}{b}}{1-\dfrac{a}{c}}</math>. Testing triangles of the form <math>2p+1, 2p^{2}+2p, 2p^{2}+2p+1</math>, we have <math>13, 84, 85</math> satisfy this equation, so <math>p=13+84+85=\boxed{182}</math>.
 
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{ac}{b}}{1-\dfrac{a}{c}}</math>. Testing triangles of the form <math>2p+1, 2p^{2}+2p, 2p^{2}+2p+1</math>, we have <math>13, 84, 85</math> satisfy this equation, so <math>p=13+84+85=\boxed{182}</math>.
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Solution by Shaddoll

Revision as of 19:17, 17 March 2016

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 $p$.

Solution

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 $6p=\dfrac{\dfrac{ac}{b}}{1-\dfrac{a}{c}}$. Testing triangles of the form $2p+1, 2p^{2}+2p, 2p^{2}+2p+1$, we have $13, 84, 85$ satisfy this equation, so $p=13+84+85=\boxed{182}$.

Solution by Shaddoll

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