2016 AIME II Problems/Problem 5

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 $n-2$ are 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 $C_{n-2}C_{n-1}C_{n}$ and $C_{n-1}C_{n}C_{n+1}$ . 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.

Solution by brightaz

2016 AIME II (Problems • Answer Key • Resources)
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2016 AIME II Problems/Problem 5

Solution 1

Solution 2

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