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

Revision as of 21:29, 16 May 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 $p=13+84+85=\boxed{182}$

Solution modified/fixed from Shaddoll's solution.

@@ Line 5: / Line 5: @@
 Solution modified/fixed from Shaddoll's solution.
+== See also ==
+{{AIME box|year=2016|n=II|num-b=4|num-a=6}}

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

Revision as of 21:29, 16 May 2016

Solution

See also