AoPSWiki

Do you have what it takes to be the next brilliant trader, researcher, or developer at Jane Street Capital? Find out in the Careers in Mathematics Forum.

Personal tools

Log in

Search

Toolbox

2006 AMC 12A Problems/Problem 23

From AoPSWiki

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 $x$ ?

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

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 Problems

2006 AMC 12A (Problems • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2006_AMC_12A_Problems/Problem_23"

Visit the AoPS Book Store.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us