Difference between revisions of "2016 AIME II Problems/Problem 5"
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− | + | First consider the sum of the lengths of the segments for which <math>n-2</math> is odd for each <math>n\geq2</math>. The perimeters of these triangles can be expressed using <math>p</math> and ratios that result because of similar triangles. Considering triangles where <math>n-2</math> is odd, we find that the perimeter for each such <math>n</math> 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>. | ||
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Simplifying, | Simplifying, | ||
− | <math>\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})</math>. (1) | + | <math>\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})</math>. (1) |
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Simplifying, | Simplifying, | ||
− | <math>\sum_{n=1}^{\infty}C_{2n-2}C_{2n-1}=C_{0}B(7-\frac{C_{0}B}{p})</math>. (2) | + | <math>\sum_{n=1}^{\infty}C_{2n-2}C_{2n-1}=C_{0}B(7-\frac{C_{0}B}{p})</math>. (2) |
Revision as of 00:05, 19 March 2017
Triangle has a right angle at . Its side lengths are pariwise relatively prime positive integers, and its perimeter is . Let be the foot of the altitude to , and for , let be the foot of the altitude to in . The sum . Find .
Solution 1
Note that by counting the area in 2 ways, the first altitude is . By similar triangles, the common ratio is for reach height, so by the geometric series formula, we have . Multiplying by the denominator and expanding, the equation becomes . Cancelling and multiplying by yields , so and . Checking for Pythagorean triples gives and , so
Solution modified/fixed from Shaddoll's solution.
Solution 2
We start by splitting the sum of all into two parts: those where is odd and those where is even.
First consider the sum of the lengths of the segments for which is odd for each . The perimeters of these triangles can be expressed using and ratios that result because of similar triangles. Considering triangles where is odd, we find that the perimeter for each such is . Thus,
.
Simplifying,
. (1)
Continuing with a similar process for the sum of the lengths of the segments for which is even, the following results:
.
Simplifying,
. (2)
Adding (1) and (2) together, we find that
.
Setting , , and , we can now proceed as in Shaddoll's solution, and our answer is .
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 |