Difference between revisions of "2002 AIME II Problems/Problem 11"
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Two distinct, real, infinite geometric series each have a sum of <math>1</math> and have the same second term. The third term of one of the series is <math>1/8</math>, and the second term of both series can be written in the form <math>\frac{\sqrt{m}-n}p</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers and <math>m</math> is not divisible by the square of any prime. Find <math>100m+10n+p</math>. | Two distinct, real, infinite geometric series each have a sum of <math>1</math> and have the same second term. The third term of one of the series is <math>1/8</math>, and the second term of both series can be written in the form <math>\frac{\sqrt{m}-n}p</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers and <math>m</math> is not divisible by the square of any prime. Find <math>100m+10n+p</math>. | ||
− | == Solution == | + | == Solution 1== |
Let the second term of each series be <math>x</math>. Then, the common ratio is <math>\frac{1}{8x}</math>, and the first term is <math>8x^2</math>. | Let the second term of each series be <math>x</math>. Then, the common ratio is <math>\frac{1}{8x}</math>, and the first term is <math>8x^2</math>. | ||
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The only solution in the appropriate form is <math>x = \frac{\sqrt{5}-1}{8}</math>. Therefore, <math>100m+10n+p = \boxed{518}</math>. | The only solution in the appropriate form is <math>x = \frac{\sqrt{5}-1}{8}</math>. Therefore, <math>100m+10n+p = \boxed{518}</math>. | ||
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+ | == Solution 2 == | ||
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+ | Let the two sequences be <math>a, ar, ar^2 ... \text{ }an^2</math> and <math>x, xy, xy^2 ... \text{ }xy^n</math>. We know for a fact that <math>ar = xy</math>. We also know that the sum of the first sequence = <math>\frac{a}{1-r} = 1</math>, and the sum of the second sequence = <math>\frac{x}{1-y} = 1</math>. Therefore we have <cmath>a+r = 1</cmath><cmath>x+y = 1</cmath><cmath>ar=xy</cmath> We can then replace <math>r = \frac{xy}{a}</math> and <math>y = \frac{ar}{x}</math>. We plug them into the two equations <math>a+r = 1</math> and <math>x+y = 1</math>. We then get <cmath>x^2 + ar = x</cmath><cmath>a^2 + xy = a</cmath>We subtract these equations, getting <cmath>x^2 - a^2 + ar - xy = x-a</cmath>Remember <cmath>ar=xy</cmath>, so <cmath>(x-a)(x+a-1) = 0</cmath>Then considering cases, we have either <math>x=a</math> or <math>y=a</math>. This suggests that the second sequence is in the form <math>r, ra, ra^2...</math>, while the first sequence is in the form <math>a, ar, ar^2...</math> Now we have that <math>ar^2 = \frac18</math> and we also have that <math>a+r = 1</math>. We can solve for <math>r</math> and the only appropriate value for <math>r</math> is <math>\frac{1+\sqrt{5}}{4}</math>. All we want is the second term, which is <math>ar = \frac{ar^2}{r} = \frac{\frac18}{\frac{1+\sqrt{5}}{4}} = \frac{\sqrt{5} - 1}{8}</math> | ||
+ | solution by jj_ca888 | ||
== See also == | == See also == | ||
{{AIME box|year=2002|n=II|num-b=10|num-a=12}} | {{AIME box|year=2002|n=II|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:31, 28 October 2017
Contents
Problem
Two distinct, real, infinite geometric series each have a sum of and have the same second term. The third term of one of the series is , and the second term of both series can be written in the form , where , , and are positive integers and is not divisible by the square of any prime. Find .
Solution 1
Let the second term of each series be . Then, the common ratio is , and the first term is .
So, the sum is . Thus, .
The only solution in the appropriate form is . Therefore, .
Solution 2
Let the two sequences be and . We know for a fact that . We also know that the sum of the first sequence = , and the sum of the second sequence = . Therefore we have We can then replace and . We plug them into the two equations and . We then get We subtract these equations, getting Remember , so Then considering cases, we have either or . This suggests that the second sequence is in the form , while the first sequence is in the form Now we have that and we also have that . We can solve for and the only appropriate value for is . All we want is the second term, which is solution by jj_ca888
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.