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

Revision as of 15:27, 22 March 2022

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 $\frac{3 - \sqrt{5}}{4}$ .

@@ Line 32: / Line 32: @@
 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>.
-== See also =
+== See also ==
 {{AIME box|year=2002|n=II|num-b=10|num-a=12}}
 [[Category: Intermediate Algebra Problems]]
 {{MAA Notice}}

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

Page

Toolbox

Search