Difference between revisions of "2002 AIME II Problems/Problem 11"

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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

Solution 3

Let's ignore the "two distinct, real, infinite geometric series" part and focus on what it means to be a geometric series.

Let the first term of the series with the third term equal to $\frac18$ be $a,$ and the common ratio be $r.$ Then, we get that $\frac{a}{1-r} = 1 \implies a = 1-r,$ and $ar^2 = \frac18 \implies (1-r)(r^2) = \frac18.$

We see that this cubic is equivalent to $r^3 - r^2 + \frac18 = 0.$ Through experimenting, we find that one of the solutions is $r = \frac12.$ Using synthetic division leads to the quadratic $4x^2 - 2x - 1 = 0.$ This has roots $\dfrac{2 \pm \sqrt{4 - 4(4)(-1)}}{8},$ or, when reduced, $\dfrac{1 \pm \sqrt{5}}{4}.$

It becomes clear that the two geometric series have common ratio $\frac{1 + \sqrt{5}}{4}$ and $\frac{1 - \sqrt{5}}{4}.$ Let $\frac{1 + \sqrt{5}}{4}$ be the ratio that we are inspecting. We see that in this case, $a = \dfrac{3 - \sqrt{5}}{4}.$

Since the second term in the series is $ar,$ we compute this and have that $ar = \left(\dfrac{3 - \sqrt{5}}{4} \right)\left(\dfrac{1+\sqrt{5}}{4}\right) = \dfrac{\sqrt{5} - 1}{8},$ for our answer of $100 \cdot 5 + 1 \cdot 10 + 8 = \boxed{518}.$

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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
+== Solution 3 ==
+Let's ignore the "two distinct, real, infinite geometric series" part and focus on what it means to be a geometric series.
+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>
+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>
+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>
+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>
 == See also ==
 {{AIME box|year=2002|n=II|num-b=10|num-a=12}}
 {{MAA Notice}}

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