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 | + | 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 00:00, 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.
Considering the sum of the lengths of the segments for which is odd, for each , consider the perimeters of the triangles . The perimeters of these triangles can be expressed using and ratios that result because of similar triangles. Considering triangles of the form , we find that the perimeter 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 |