Difference between revisions of "2016 AIME II Problems/Problem 5"
(→Solution) |
(→Solution 2) |
||
Line 34: | Line 34: | ||
<math>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</math>. | <math>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</math>. | ||
− | Setting <math>a=C_{0}B</math>, <math>b=C_{0}A</math>, and <math>c=AB</math>, we can now proceed as in Shaddoll's solution. | + | Setting <math>a=C_{0}B</math>, <math>b=C_{0}A</math>, and <math>c=AB</math>, we can now proceed as in Shaddoll's solution, and our answer is <math>p=13+84+85=\boxed{182}</math>. |
Solution by brightaz | Solution by brightaz |
Revision as of 23:56, 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 are 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 |