Difference between revisions of "2016 AIME II Problems/Problem 5"
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Triangle <math>ABC_0</math> has a right angle at <math>C_0</math>. Its side lengths are pariwise relatively prime positive integers, and its perimeter is <math>p</math>. Let <math>C_1</math> be the foot of the altitude to <math>\overline{AB}</math>, and for <math>n \geq 2</math>, let <math>C_n</math> be the foot of the altitude to <math>\overline{C_{n-2}B}</math> in <math>\triangle C_{n-2}C_{n-1}B</math>. The sum <math>\sum_{i=1}^\infty C_{n-2}C_{n-1} = 6p</math>. Find <math>p</math>. | Triangle <math>ABC_0</math> has a right angle at <math>C_0</math>. Its side lengths are pariwise relatively prime positive integers, and its perimeter is <math>p</math>. Let <math>C_1</math> be the foot of the altitude to <math>\overline{AB}</math>, and for <math>n \geq 2</math>, let <math>C_n</math> be the foot of the altitude to <math>\overline{C_{n-2}B}</math> in <math>\triangle C_{n-2}C_{n-1}B</math>. The sum <math>\sum_{i=1}^\infty C_{n-2}C_{n-1} = 6p</math>. Find <math>p</math>. | ||
− | ==Solution== | + | ==Solution 1== |
Note that by counting the area in 2 ways, the first altitude is <math>\dfrac{ac}{b}</math>. By similar triangles, the common ratio is <math>\dfrac{a}{c}</math> for reach height, so by the geometric series formula, we have <math>6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}</math>. Multiplying by the denominator and expanding, the equation becomes <math>\dfrac{ab}{c}=6a+6b+6c-\dfrac{6a^2}{c}-\dfrac{6ab}{c}-6a</math>. Cancelling <math>6a</math> and multiplying by <math>c</math> yields <math>ab=6bc+6c^2-6a^2-6ab</math>, so <math>7ab = 6bc+6b^2</math> and <math>7a=6b+6c</math>. Checking for Pythagorean triples gives <math>13,84,</math> and <math>85</math>, so <math>p=13+84+85=\boxed{182}</math> | Note that by counting the area in 2 ways, the first altitude is <math>\dfrac{ac}{b}</math>. By similar triangles, the common ratio is <math>\dfrac{a}{c}</math> for reach height, so by the geometric series formula, we have <math>6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}</math>. Multiplying by the denominator and expanding, the equation becomes <math>\dfrac{ab}{c}=6a+6b+6c-\dfrac{6a^2}{c}-\dfrac{6ab}{c}-6a</math>. Cancelling <math>6a</math> and multiplying by <math>c</math> yields <math>ab=6bc+6c^2-6a^2-6ab</math>, so <math>7ab = 6bc+6b^2</math> and <math>7a=6b+6c</math>. Checking for Pythagorean triples gives <math>13,84,</math> and <math>85</math>, so <math>p=13+84+85=\boxed{182}</math> | ||
Solution modified/fixed from Shaddoll's solution. | Solution modified/fixed from Shaddoll's solution. | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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> are odd and those where <math>n-2</math> is even. | ||
+ | |||
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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}</math> and <math>C_{n-1}C_{n}C_{n+1}</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, | ||
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+ | <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, | ||
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+ | <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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+ | Continuing with a similar process for the sum of the lengths of the segments for which <math>n-2</math> is even, the following results: | ||
+ | |||
+ | <math>p\sum_{n=1}^{\infty}\frac{C_{2n-2}C_{2n-1}}{C_{0}B}=6p+C_{0}A+AB=7p-C_{0}B</math>. | ||
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+ | |||
+ | Simplifying, | ||
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+ | <math>\sum_{n=1}^{\infty}C_{2n-2}C_{2n-1}=C_{0}B(7-\frac{C_{0}B}{p})</math>. (2) | ||
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+ | |||
+ | Adding (1) and (2) together, we find that | ||
+ | |||
+ | <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>. | ||
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+ | 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. | ||
+ | |||
+ | Solution by brightaz | ||
== See also == | == See also == | ||
{{AIME box|year=2016|n=II|num-b=4|num-a=6}} | {{AIME box|year=2016|n=II|num-b=4|num-a=6}} |
Revision as of 22:55, 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.
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 |