Difference between revisions of "2016 AIME II Problems/Problem 5"
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− | We start by splitting the sum of all <math>C_{n-2}C_{n-1}</math> into two parts: those where <math>n-2</math> | + | We start by splitting the sum of all <math>C_{n-2}C_{n-1}</math> into two parts: those where <math>n-2</math> is odd and those where <math>n-2</math> is even. |
Revision as of 22:57, 18 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 and . 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 |