Difference between revisions of "2006 AMC 12A Problems/Problem 23"

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Given a finite sequence <math>S=(a_1,a_2,\ldots ,a_n)</math> of <math>n</math> real numbers, let <math>A(S)</math> be the sequence  
 
Given a finite sequence <math>S=(a_1,a_2,\ldots ,a_n)</math> of <math>n</math> real numbers, let <math>A(S)</math> be the sequence  
  
<math>(\frac{a_1+a_2}{2},\frac{a_2+a_3}{2},\ldots ,\frac{a_{n-1}+a_n}{2})</math>
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<math>\left(\frac{a_1+a_2}{2},\frac{a_2+a_3}{2},\ldots ,\frac{a_{n-1}+a_n}{2}\right)</math>
  
 
of <math>n-1</math> real numbers. Define <math>A^1(S)=A(S)</math> and, for each integer <math>m</math>, <math>2\le m\le n-1</math>, define <math>A^m(S)=A(A^{m-1}(S))</math>. Suppose <math>x>0</math>, and let <math>S=(1,x,x^2,\ldots ,x^{100})</math>. If <math>A^{100}(S)=(1/2^{50})</math>, then what is <math>x</math>?
 
of <math>n-1</math> real numbers. Define <math>A^1(S)=A(S)</math> and, for each integer <math>m</math>, <math>2\le m\le n-1</math>, define <math>A^m(S)=A(A^{m-1}(S))</math>. Suppose <math>x>0</math>, and let <math>S=(1,x,x^2,\ldots ,x^{100})</math>. If <math>A^{100}(S)=(1/2^{50})</math>, then what is <math>x</math>?
  
<math> \mathrm{(A) \ } 1-\frac{\sqrt{2}}{2}\qquad \mathrm{(B) \ } \sqrt{2}-1\qquad \mathrm{(C) \ } \frac{1}{2}</math><math>\mathrm{(D) \ } 2-\sqrt{2}\qquad \mathrm{(E) \ }  \frac{\sqrt{2}}{2}</math>
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<math> \mathrm{(A) \ } 1-\frac{\sqrt{2}}{2}\qquad \mathrm{(B) \ } \sqrt{2}-1\qquad \mathrm{(C) \ } \frac{1}{2}\qquad \mathrm{(D) \ } 2-\sqrt{2}\qquad \mathrm{(E) \ }  \frac{\sqrt{2}}{2}</math>
  
 
== Solution ==
 
== Solution ==
{{solution}}
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<cmath>A^1(S)=\left(\frac{1+x}{2},\frac{x+x^2}{2},...,\frac{x^{99}+x^{100}}{2}\right)</cmath>
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<cmath>A^2(S)=\left(\frac{1+2x+x^2}{2^2},\frac{x+2x^2+x^3}{2^2},...,\frac{x^{98}+2x^{99}+x^{100}}{2^2}\right)</cmath>
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<cmath>\Rightarrow A^2(S)=\left(\frac{(x+1)^2}{2^2},\frac{x(x+1)^2}{2^2},...,\frac{x^{98}(x+1)^2}{2^2}\right)</cmath>
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In general, <math>A^n(S)=\left(\frac{(x+1)^n}{2^n},\frac{x(x+1)^n}{2^n},...,\frac{x^{100-n}(x+1)^n}{2^n}\right)</math> such that <math>A^n(s)</math> has <math>101-n</math> terms. Specifically, <math>A^{100}(S)=\frac{(x+1)^{100}}{2^{100}}</math> To find x, we need only solve the equation <math>\frac{(x+1)^{100}}{2^{100}}=\frac{1}{2^{50}}</math>. Algebra yields <math>x=\sqrt{2}-1</math>.
  
 
== See also ==
 
== See also ==
* [[2006 AMC 12A Problems]]
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*[[2006 AMC 12A Problems/Problem 22|Previous Problem]]
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{{AMC12 box|year=2006|ab=A|num-b=22|num-a=24}}
*[[2006 AMC 12A Problems/Problem 24|Next Problem]]
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{{MAA Notice}}

Latest revision as of 20:53, 18 December 2020

Problem

Given a finite sequence $S=(a_1,a_2,\ldots ,a_n)$ of $n$ real numbers, let $A(S)$ be the sequence

$\left(\frac{a_1+a_2}{2},\frac{a_2+a_3}{2},\ldots ,\frac{a_{n-1}+a_n}{2}\right)$

of $n-1$ real numbers. Define $A^1(S)=A(S)$ and, for each integer $m$, $2\le m\le n-1$, define $A^m(S)=A(A^{m-1}(S))$. Suppose $x>0$, and let $S=(1,x,x^2,\ldots ,x^{100})$. If $A^{100}(S)=(1/2^{50})$, then what is $x$?

$\mathrm{(A) \ } 1-\frac{\sqrt{2}}{2}\qquad \mathrm{(B) \ } \sqrt{2}-1\qquad \mathrm{(C) \ } \frac{1}{2}\qquad \mathrm{(D) \ } 2-\sqrt{2}\qquad \mathrm{(E) \ }  \frac{\sqrt{2}}{2}$

Solution

\[A^1(S)=\left(\frac{1+x}{2},\frac{x+x^2}{2},...,\frac{x^{99}+x^{100}}{2}\right)\] \[A^2(S)=\left(\frac{1+2x+x^2}{2^2},\frac{x+2x^2+x^3}{2^2},...,\frac{x^{98}+2x^{99}+x^{100}}{2^2}\right)\] \[\Rightarrow A^2(S)=\left(\frac{(x+1)^2}{2^2},\frac{x(x+1)^2}{2^2},...,\frac{x^{98}(x+1)^2}{2^2}\right)\]

In general, $A^n(S)=\left(\frac{(x+1)^n}{2^n},\frac{x(x+1)^n}{2^n},...,\frac{x^{100-n}(x+1)^n}{2^n}\right)$ such that $A^n(s)$ has $101-n$ terms. Specifically, $A^{100}(S)=\frac{(x+1)^{100}}{2^{100}}$ To find x, we need only solve the equation $\frac{(x+1)^{100}}{2^{100}}=\frac{1}{2^{50}}$. Algebra yields $x=\sqrt{2}-1$.

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 12 Problems and Solutions

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