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

Revision as of 18:25, 2 February 2007

Problem

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

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

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 $x$ ?

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

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 $x=\sqrt{2}-1$ .

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AMC 12 Problems and Solutions

@@ Line 18: / Line 18: @@
 == See also ==
 * [[2006 AMC 12A Problems]]
-*[[2006 AMC 12A Problems/Problem 22|Previous Problem]]
-*[[2006 AMC 12A Problems/Problem 24|Next Problem]]
+{{AMC12 box|year=2006|ab=A|num-b=22|num-a=24}}

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

Revision as of 18:25, 2 February 2007

Problem

Solution

See also