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2002 AIME II Problems/Problem 11: Difference between revisions

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== Solution 3 ==  
== 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'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 <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>
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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Solution by Ilikeapos
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>.
== Video Solution 1 by SpreadTheMathLove==
https://www.youtube.com/watch?v=gIPLYMe6pqQ&t=15s
== 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}}
[[Category: Intermediate Algebra Problems]]
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 15:06, 2 June 2024

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 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 $\frac18$ 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 $100 \cdot 5 + 1 \cdot 10 + 8 = \boxed{518}.$

Solution by Ilikeapos

Sidenote

One of the geometric series has first term $\frac{3 - \sqrt{5}}{4}$ and common ratio $\frac{1 + \sqrt{5}}{4}$, and its third term is $\frac{1}{8}$. The other series has first term $\frac{1 + \sqrt{5}}{4}$ and common ratio $\frac{3 - \sqrt{5}}{4}$.

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=gIPLYMe6pqQ&t=15s

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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