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

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(Solution)
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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>.
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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{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 by Shaddoll
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Solution modified/fixed from Shaddoll's solution.

Revision as of 15:10, 25 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 $6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}$. 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 $p=13+84+84=\boxed{182}$

Solution modified/fixed from Shaddoll's solution.

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