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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>.
  
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== Solution ==
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== 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>
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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 21:31, 28 October 2017

Problem

Two distinct, real, infinite geometric series each have a sum of $1$ and have the same second term. The third term of one of the series is $1/8$, and the second term of both series can be written in the form $\frac{\sqrt{m}-n}p$, where $m$, $n$, and $p$ are positive integers and $m$ is not divisible by the square of any prime. Find $100m+10n+p$.

Solution 1

Let the second term of each series be $x$. Then, the common ratio is $\frac{1}{8x}$, and the first term is $8x^2$.

So, the sum is $\frac{8x^2}{1-\frac{1}{8x}}=1$. Thus, $64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}$.

The only solution in the appropriate form is $x = \frac{\sqrt{5}-1}{8}$. Therefore, $100m+10n+p = \boxed{518}$.


Solution 2

Let the two sequences be $a, ar, ar^2 ... \text{ }an^2$ and $x, xy, xy^2 ... \text{ }xy^n$. We know for a fact that $ar = xy$. We also know that the sum of the first sequence = $\frac{a}{1-r} = 1$, and the sum of the second sequence = $\frac{x}{1-y} = 1$. Therefore we have \[a+r = 1\]\[x+y = 1\]\[ar=xy\] We can then replace $r = \frac{xy}{a}$ and $y = \frac{ar}{x}$. We plug them into the two equations $a+r = 1$ and $x+y = 1$. We then get \[x^2 + ar = x\]\[a^2 + xy = a\]We subtract these equations, getting \[x^2 - a^2 + ar - xy = x-a\]Remember \[ar=xy\], so \[(x-a)(x+a-1) = 0\]Then considering cases, we have either $x=a$ or $y=a$. This suggests that the second sequence is in the form $r, ra, ra^2...$, while the first sequence is in the form $a, ar, ar^2...$ Now we have that $ar^2 = \frac18$ and we also have that $a+r = 1$. We can solve for $r$ and the only appropriate value for $r$ is $\frac{1+\sqrt{5}}{4}$. All we want is the second term, which is $ar = \frac{ar^2}{r} = \frac{\frac18}{\frac{1+\sqrt{5}}{4}} = \frac{\sqrt{5} - 1}{8}$ 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

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