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Difference between revisions of "2012 AIME II Problems/Problem 11"

(Created page with "== Problem 11 == Let <math>f_1(x) = \frac23 - \frac3{3x+1}</math>, and for <math>n \ge 2</math>, define <math>f_n(x) = f_1(f_{n-1}(x))</math>. The value of <math>x</math> that sa...")
 
(→‎Video Solution)
 
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== Problem 11 ==
 
== Problem 11 ==
 
Let <math>f_1(x) = \frac23 - \frac3{3x+1}</math>, and for <math>n \ge 2</math>, define <math>f_n(x) = f_1(f_{n-1}(x))</math>. The value of <math>x</math> that satisfies <math>f_{1001}(x) = x-3</math> can be expressed in the form <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
Let <math>f_1(x) = \frac23 - \frac3{3x+1}</math>, and for <math>n \ge 2</math>, define <math>f_n(x) = f_1(f_{n-1}(x))</math>. The value of <math>x</math> that satisfies <math>f_{1001}(x) = x-3</math> can be expressed in the form <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
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== Solution ==
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After evaluating the first few values of <math>f_k (x)</math>, we obtain <math>f_4(x) = f_1(x) = \frac{2}{3} - \frac{3}{3x+1} = \frac{6x-7}{9x+3}</math>. Since <math>1001 \equiv 2 \mod 3</math>, <math>f_{1001}(x) = f_2(x) = \frac{3x+7}{6-9x}</math>. We set this equal to <math>x-3</math>, i.e.
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<math>\frac{3x+7}{6-9x} = x-3 \Rightarrow x = \frac{5}{3}</math>. The answer is thus <math>5+3 = \boxed{008}</math>.
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==Video Solution==
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https://www.youtube.com/watch?v=zBKm3M71K4c&t=47s
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This video is now private.
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== See Also ==
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{{AIME box|year=2012|n=II|num-b=10|num-a=12}}
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{{MAA Notice}}

Latest revision as of 16:03, 29 December 2023

Problem 11

Let $f_1(x) = \frac23 - \frac3{3x+1}$, and for $n \ge 2$, define $f_n(x) = f_1(f_{n-1}(x))$. The value of $x$ that satisfies $f_{1001}(x) = x-3$ can be expressed in the form $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.


Solution

After evaluating the first few values of $f_k (x)$, we obtain $f_4(x) = f_1(x) = \frac{2}{3} - \frac{3}{3x+1} = \frac{6x-7}{9x+3}$. Since $1001 \equiv 2 \mod 3$, $f_{1001}(x) = f_2(x) = \frac{3x+7}{6-9x}$. We set this equal to $x-3$, i.e.


$\frac{3x+7}{6-9x} = x-3 \Rightarrow x = \frac{5}{3}$. The answer is thus $5+3 = \boxed{008}$.

Video Solution

https://www.youtube.com/watch?v=zBKm3M71K4c&t=47s

This video is now private.

See Also

2012 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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