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

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Considering the sum of the lengths of the segments for which <math>n-2</math> is odd, for each <math>n\geq2</math>, consider the perimeters of the triangles <math>C_{n-2}C_{n-1}C_{n}</math> and <math>C_{n-1}C_{n}C_{n+1}</math>. The perimeters of these triangles can be expressed using <math>p</math> and ratios that result because of similar triangles. Considering triangles of the form <math>C_{n-2}C_{n-1}C_{n}</math>, we find that the perimeter is <math>p*\frac{C_{n-1}C_{n}}{C_{0}B}</math>. Thus,  
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Considering the sum of the lengths of the segments for which <math>n-2</math> is odd, for each <math>n\geq2</math>, consider the perimeters of the triangles <math>C_{n-2}C_{n-1}C_{n}</math>. The perimeters of these triangles can be expressed using <math>p</math> and ratios that result because of similar triangles. Considering triangles of the form <math>C_{n-2}C_{n-1}C_{n}</math>, we find that the perimeter is <math>p*\frac{C_{n-1}C_{n}}{C_{0}B}</math>. Thus,  
  
 
<math>p\sum_{n=1}^{\infty}\frac{C_{2n-1}C_{2n}}{C_{0}B}=6p+C_{0}B</math>.
 
<math>p\sum_{n=1}^{\infty}\frac{C_{2n-1}C_{2n}}{C_{0}B}=6p+C_{0}B</math>.

Revision as of 23:00, 18 March 2017

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 1

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.

Solution 2

We start by splitting the sum of all $C_{n-2}C_{n-1}$ into two parts: those where $n-2$ is odd and those where $n-2$ is even.


Considering the sum of the lengths of the segments for which $n-2$ is odd, for each $n\geq2$, consider the perimeters of the triangles $C_{n-2}C_{n-1}C_{n}$. The perimeters of these triangles can be expressed using $p$ and ratios that result because of similar triangles. Considering triangles of the form $C_{n-2}C_{n-1}C_{n}$, we find that the perimeter is $p*\frac{C_{n-1}C_{n}}{C_{0}B}$. Thus,

$p\sum_{n=1}^{\infty}\frac{C_{2n-1}C_{2n}}{C_{0}B}=6p+C_{0}B$.


Simplifying,

$\sum_{n=1}^{\infty}C_{2n-1}C_{2n}=6C_{0}B + \frac{(C_{0}B)^2}{p}=C_{0}B(6+\frac{C_{0}B}{p})$. (1)


Continuing with a similar process for the sum of the lengths of the segments for which $n-2$ is even, the following results:

$p\sum_{n=1}^{\infty}\frac{C_{2n-2}C_{2n-1}}{C_{0}B}=6p+C_{0}A+AB=7p-C_{0}B$.


Simplifying,

$\sum_{n=1}^{\infty}C_{2n-2}C_{2n-1}=C_{0}B(7-\frac{C_{0}B}{p})$. (2)


Adding (1) and (2) together, we find that

$6p=13C_{0}B \Rightarrow p=\frac{13C_{0}B}{6}=C_{0}B+C_{0}A+AB \Rightarrow \frac{7C_{0}B}{6}=C_{0}A+AB \Rightarrow 7C_{0}B=6C_{0}A + 6AB$.

Setting $a=C_{0}B$, $b=C_{0}A$, and $c=AB$, we can now proceed as in Shaddoll's solution, and our answer is $p=13+84+85=\boxed{182}$.

Solution by brightaz

See also

2016 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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