Difference between revisions of "2002 AIME II Problems/Problem 11"
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== Problem == | == Problem == | ||
+ | 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 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>. | ||
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+ | So, the sum is <math>\frac{8x^2}{1-\frac{1}{8x}}=1</math>. Thus, <math>64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}</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>. | ||
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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 | ||
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+ | == Solution 3 == | ||
+ | |||
+ | Let's ignore the "two distinct, real, infinite geometric series" part for now and focus on what it means to be a geometric series. | ||
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+ | Let the first term of the series with the third term equal to <math>\frac18</math> be <math>a,</math> and the common ratio be <math>r.</math> Then, we get that <math>\frac{a}{1-r} = 1 \implies a = 1-r,</math> and <math>ar^2 = \frac18 \implies (1-r)(r^2) = \frac18.</math> | ||
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+ | We see that this cubic is equivalent to <math>r^3 - r^2 + \frac18 = 0.</math> Through experimenting, we find that one of the solutions is <math>r = \frac12.</math> Using synthetic division leads to the quadratic <math>4x^2 - 2x - 1 = 0.</math> This has roots <math>\dfrac{2 \pm \sqrt{4 - 4(4)(-1)}}{8},</math> or, when reduced, <math>\dfrac{1 \pm \sqrt{5}}{4}.</math> | ||
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+ | It becomes clear that the two geometric series have common ratio <math>\frac{1 + \sqrt{5}}{4}</math> and <math>\frac{1 - \sqrt{5}}{4}.</math> Let <math>\frac{1 + \sqrt{5}}{4}</math> be the ratio that we are inspecting. We see that in this case, <math>a = \dfrac{3 - \sqrt{5}}{4}.</math> | ||
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+ | Since the second term in the series is <math>ar,</math> we compute this and have that <cmath>ar = \left(\dfrac{3 - \sqrt{5}}{4} \right)\left(\dfrac{1+\sqrt{5}}{4}\right) = \dfrac{\sqrt{5} - 1}{8},</cmath>for our answer of <math>100 \cdot 5 + 1 \cdot 10 + 8 = \boxed{518}.</math> | ||
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+ | Solution by Ilikeapos | ||
+ | |||
+ | == Sidenote == | ||
+ | One of the geometric series has first term <math>\frac{3 - \sqrt{5}}{4}</math> and common ratio <math>\frac{1 + \sqrt{5}}{4}</math>, and its third term is <math>\frac{1}{8}</math>. The other series has first term <math>\frac{1 + \sqrt{5}}{4}</math> and common ratio <math>\frac{3 - \sqrt{5}}{4}</math>. | ||
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== See also == | == See also == | ||
− | + | {{AIME box|year=2002|n=II|num-b=10|num-a=12}} | |
− | + | ||
− | + | [[Category: Intermediate Algebra Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 16:27, 22 March 2022
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
Solution 3
Let's ignore the "two distinct, real, infinite geometric series" part for now and focus on what it means to be a geometric series.
Let the first term of the series with the third term equal to be and the common ratio be Then, we get that and
We see that this cubic is equivalent to Through experimenting, we find that one of the solutions is Using synthetic division leads to the quadratic This has roots or, when reduced,
It becomes clear that the two geometric series have common ratio and Let be the ratio that we are inspecting. We see that in this case,
Since the second term in the series is we compute this and have that for our answer of
Solution by Ilikeapos
Sidenote
One of the geometric series has first term and common ratio , and its third term is . The other series has first term and common ratio .
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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