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> | + | <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> \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 == |
Revision as of 19:23, 5 January 2008
Problem
Given a finite sequence of
real numbers, let
be the sequence
of real numbers. Define
and, for each integer
,
, define
. Suppose
, and let
. If
, then what is
?
Solution
In general, such that
has
terms. Specifically,
To find x, we need only solve the equation
. Algebra yields
.
See also
2006 AMC 12A (Problems • Answer Key • 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 | |
All AMC 12 Problems and Solutions |